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crescita-del-pene.info | 1,566,763,643,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330800.17/warc/CC-MAIN-20190825194252-20190825220252-00475.warc.gz | 47,509,809 | 7,922 | It can be shown that the activity of a source is. Archaeologists use the exponential, radioactive decay of carbon 14 to This half life is a relatively small number, which means that carbon 14 dating is not. Six megacuries 6.
We may use the exponential decay model when we are calculating half-lifeor the time it takes for a substance to exponentially decay to half of its original quantity. The decay rate of carbon in fresh wood today is counts per minute per gram, and the half life of carbon is years. Solution: 1) Determine decimal . Other isotopes commonly used for dating include uranium half-life of 4.
Substances must have obtained C from the atmosphere. Carbon dating is based upon the decay of 14C, a radioactive isotope of carbon with a Returning to our example of carbon, knowing that the half-life of 14C is years, we can use Problem 3- Calculate the initial amount ofC in a fossil. Express the amount of carbon remaining as a function of time, t.
Molecular clock. Explain radioactive half-life and its role in radiometric dating; Calculate radioactive half-life and solve problems associated with radiometric dating. Measuring the amount of 14 C in a sample from a dead plant or animal such as a piece of wood or a fragment of bone provides information that can be used to calculate when the animal or plant died.
While undergoing the test, I heard the technician telling somebody on the phone that "in twenty-four hours, you'll be down to background radiation levels. A radioactive half-life refers to the amount of time it takes for half of the To answer this question, there is no need to solve for the radioactive decay equation . And that's what we have here.
At the time of manufacture, such a sign contains Half-Life and Practice Problems. 1. Half-Life and Radiometric Dating; 2. Rate of Decay The time required for half the nuclei in a sample of a. This means that radiocarbon dates on wood samples can be older than the date at which the tree was felled.
As ofthe standard format required by the journal Radiocarbon is as follows. Carbon has a half-life of years and is produced in a nuclear .. Data from the appendices and the periodic table may be needed for these problems. What happens over that 5, years is that, probabilistically, some of these guys just start turning into nitrogen randomly, at random points.
R CH8. It has not been determined how the image was placed on the material. By comparing the ratio of carbon to carbon in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.
Isotopes with shorter half-lives are used to date more recent samples. Radiocarbon dating is a method for determining the age of an object containing organic .. Calculating radiocarbon ages also requires the value of the half-life for 14 Other materials can present the same problem: for example, bitumen is . Rearranging to isolate gives.
The equation for the radioactive decay of 14 C is: [17]. Half-Life and Practice Problems. 1. Half-Life and Radiometric Dating; 2. Rate of Decay The time required for half the nuclei in a sample of a. Knowledge of the U U half-life has shown, for example, that the oldest rocks on Earth solidified about 3. | 696 | 3,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-35 | longest | en | 0.920694 |
http://solanolabs.com/2020/sum-of-an-infinite-geometric-series.html | 1,591,180,395,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347432521.57/warc/CC-MAIN-20200603081823-20200603111823-00265.warc.gz | 110,532,387 | 4,586 | Sum of an infinite geometric series. Infinite Geometric Series: Sequences and Series
Infinite Geometric Series
Then what happens to r n when n gets really big? Varsity Tutors connects learners with experts. The sum of the terms between two values e. The fact that a geometric sequence has a common factor allows you to do two things. The side of this square is then the diagonal of the third square and so on, as shows the figure below. In fact, when you need the sum of a geometric series, it's usually easier add the numbers yourself when there are only a few terms.
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Sum of a Convergent Geometric Series
Since , this infinite geometric series has a sum. Simon and Lawrence Blume 1994. Geometric series are used throughout mathematics, and they have important applications in , , , , , , and. Historically, geometric series played an important role in the early development of , and they continue to be central in the study of of series. Formula 3: This form of the formula is used when the number of terms n , the first term a 1 , and the common ratio r are known. Media outlet trademarks are owned by the respective media outlets and are not affiliated with Varsity Tutors. When the ratio between each term and the next is a constant, it is called a geometric series.
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Infinite Geometric Series: Sequences and Series
Join thousands of satisfied students, teachers and parents! The idea here is similar to that of the arithmetic sequence, except each term is multiplied by an additional factor of r. Author: Page last modified: 15 April 2018. We can find the sum of all finite geometric series. Please provide the required information in the form below: More about the infinite geometric series The idea of an infinite series can be baffling at first. Chris Deziel holds a Bachelor's degree in physics and a Master's degree in Humanities, He has taught science, math and English at the university level, both in his native Canada and in Japan. These problems are appropriately applicable to analytic geometry and algebra. Basic Mathematics for Economists, 2nd ed.
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Infinite Geometric Series: Sequences and Series
} which is not geometric The Rule We can also calculate any term using the Rule: This sequence has a factor of 3 between each number. If r is greater than one or less than minus one the terms of the series become larger and larger in magnitude. The sum of an infinite converging geometric series, examples Example: Given a square with side a. With , you can get step-by-step solutions to your questions from an expert in the field. The third term has been multiplied by r twice, and so on. For example, Zeno's dichotomy paradox maintains that movement is impossible, as one can divide any finite path into an infinite number of steps wherein each step is taken to be half the remaining distance.
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Geometric Series
What happens to an exponential function when the base is between 0 and 1? A geometric sequence can have a fractional common factor, in which case each successive number is smaller than the one preceding it. Plug these values into the infinite sum formula: Keep in mind that this sum is finite only if r lies strictly between —1 and 1. His method was to dissect the area into an infinite number of triangles. You would be right, of course, but that definition doesn't mean anything unless you have some knowledge of what calculus is. A geometric series converges if the r-value i.
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Geometric series
By the way, this one was worked out by Archimedes over 2200 years ago. The 1 in the denominator is always 1 and the 3 in denominator was the ratio, r. This series would have no last term. So I have everything I need to proceed. In the study of , geometric series often arise as the , , or of a figure.
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Infinite Geometric Series
The difference between each term is 2. Introduction to Mathematics for Life Scientists, 3rd ed. Order The order of the terms can be very important! For example in an alternating series, what if we made all positive terms come first? You can write this number as 0. We will denote the n th partial sum as S n. Using the Formula Let's see the formula in action: Example: Grains of Rice on a Chess Board On the page we give an example of grains of rice on a chess board. You're probably wondering why I didn't go ahead and simplify. And you can use this method to convert any repeating decimal to its fractional form.
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Geometric series
Find the sum of all circumferences and areas of all inscribed circles. If r is equal to one, all of the terms of the series are the same. It is equivalent to the modern formula. Need help with a homework or test question? When the ratio has a magnitude greater than 1, the terms in the sequence will get larger and larger, and the if you add larger and larger numbers forever, you will get infinity for an answer. Now, if we try to figure out where the different parts of that formula come from, we can conjecture about a formula for the n th partial sum.
Next | 1,081 | 5,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-24 | latest | en | 0.950789 |
https://www.varsitytutors.com/common_core_high_school__geometry-help/transformation-and-congruence-of-rigid-motions-ccss-math-content-hsg-co-b-6 | 1,721,437,117,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514972.64/warc/CC-MAIN-20240719231533-20240720021533-00103.warc.gz | 897,440,591 | 43,806 | # Common Core: High School - Geometry : Transformation and Congruence of Rigid Motions: CCSS.Math.Content.HSG-CO.B.6
## Example Questions
### Example Question #1 : Transformation And Congruence Of Rigid Motions: Ccss.Math.Content.Hsg Co.B.6
Determine whether the statement is true or false.
For a translation to be considered rigid, the starting and ending figures must be congruent.
True
False
True
Explanation:
Recall that a rigid motion is that that preserves the distances while undergoing a motion in the plane. This is also called an isometry, rigid transformations, or congruence transformations and there are four different types. Therefore, for the translation to be considered "rigid" the two figures must be congruent by definition of a rigid motion.
Therefore, the statement, "For a translation to be considered rigid, the starting and ending figures must be congruent." is true.
### Example Question #2 : Transformation And Congruence Of Rigid Motions: Ccss.Math.Content.Hsg Co.B.6
Which of the following is NOT a rigid motion?
Translation
Expansion
Reflection
Glide Reflection
Rotation
Expansion
Explanation:
Recall that a rigid motion is that that preserves the distances while undergoing a motion in the plane. This is also called an isometry, rigid transformations, or congruence transformations and there are four different types.
These basic type of rigid motions include the following:
1. Rotation
2. Reflection
3. Translation
4. Glide Reflection
Therefore, of the answer selections, "Expansion" is the term that is NOT a rigid motion.
### Example Question #1 : Transformation And Congruence Of Rigid Motions: Ccss.Math.Content.Hsg Co.B.6
has a rigid motion applied to it that results in . What can be concluded about the relationship between these two triangles.
Explanation:
A rigid motion is that that preserves the distances while undergoing a motion in the plane. This is also called an isometry, rigid transformations, or congruence transformations and there are four different types.
Regardless of which type of rigid motion occurred, the following is true about the triangles' angles:
The following is also true about the side lengths:
Therefore, of the answer choices is the correct choice.
### Example Question #1 : Transformation And Congruence Of Rigid Motions: Ccss.Math.Content.Hsg Co.B.6
has a rigid motion applied to it that results in . What can be concluded about the relationship between these two triangles.
Explanation:
A rigid motion is that that preserves the distances while undergoing a motion in the plane. This is also called an isometry, rigid transformations, or congruence transformations and there are four different types.
Regardless of which type of rigid motion occurred, the following is true about the triangles' angles:
The following is also true about the side lengths:
Therefore, of the answer choices is the correct choice.
### Example Question #51 : Congruence
Determine whether the statement is true or false:
A glide reflection is a synonym for a rigid motion translation.
True
False
False
Explanation:
A glide reflection is a rigid motion that occurs when a figure is translated a certain distance and then reflected or reflected and then translated.
A translation is a rigid motion describing when a object is moved a certain distance.
A glide reflection contains a translation but it is not a synonym for translation therefore, the statement is false.
### Example Question #6 : Transformation And Congruence Of Rigid Motions: Ccss.Math.Content.Hsg Co.B.6
Determine whether the statement is true or false:
All glide reflections are reflections.
True
False
True
Explanation:
A glide reflection is a rigid motion that occurs when a figure is translated a certain distance and then reflected or reflected and then translated. In either case, a glide reflection aways contains a reflection.
Therefore, the statement, "All glide reflections are reflections." is true.
### Example Question #3 : Transformation And Congruence Of Rigid Motions: Ccss.Math.Content.Hsg Co.B.6
Sally has a quarter that is face up on the desk. If she slides it to Bob on the other side of the desk and he flips it over so that the tails side is facing up, is it a rigid motion?
Yes
No
Yes
Explanation:
In the situation where Sally has a quarter that is face up on the desk. The coin is the object. Then she slides it to Bob on the other side of the desk and he flips it over so that the tails side is facing up. Therefore, since the coin maintains it shape it is undergoing a translation to reach the other side of the desk and a reflection to flip the coin. This describes a glide reflection which is in fact, a rigid motion.
Therefore, the statement is describing a rigid motion. The answer is "Yes"
### Example Question #4 : Transformation And Congruence Of Rigid Motions: Ccss.Math.Content.Hsg Co.B.6
Sally has a quarter that is face up on the desk. Then she slides the coin to Bob on the other side of the desk and he flips it over so that the tails side is facing up. What type of rigid motion does this situation describe?
Translation
This is not a rigid motion.
Reflection
Glide Reflection
Rotation
Glide Reflection
Explanation:
In the situation where Sally has a quarter that is face up on the desk. The coin is the object. Then she slides it to Bob on the other side of the desk and he flips it over so that the tails side is facing up. Therefore, since the coin maintains it shape it is undergoing a translation to reach the other side of the desk and a reflection to flip the coin. This describes a glide reflection which is in fact, a rigid motion.
### Example Question #5 : Transformation And Congruence Of Rigid Motions: Ccss.Math.Content.Hsg Co.B.6
Jane and Bob are filling up water balloons for a party they are throwing. Jane thinks the balloons should have more water in them so she fills them fuller. Each water balloon's circumference is one inch greater than before. Does this describe a rigid motion?
Yes
No
No
Explanation:
Recall that a rigid motion is that that preserves the distances between points within the object while undergoing a motion in the plane. This is also called an isometry, rigid transformations, or congruence transformations and there are four different types. The water balloons are the objects in this situation. Since the they are filled with more water to increase their circumferences, it does not preserve the shape of the object and thus, is not a rigid motion.
### Example Question #6 : Transformation And Congruence Of Rigid Motions: Ccss.Math.Content.Hsg Co.B.6
Select the answer that best completes the following definition:
A motion that preserves distance in the plane is called a __________
Reflection
Transformation
Non-Rigid Motion
Rotation
Rigid Motion
Rigid Motion
Explanation:
A rigid motion is that that preserves the distances while undergoing a motion in the plane. This is also called an isometry, rigid transformations, or congruence transformations and there are four different types.
These basic type of rigid motions include the following:
1. Rotation
2. Reflection
3. Translation
4. Glide Reflection
Therefore, "A motion that preserves distance in the plane is called a rigid motion". | 1,549 | 7,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-30 | latest | en | 0.863587 |
https://estebantorreshighschool.com/interesting-about-equations/equivalence-point-equation.html | 1,685,337,785,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644683.18/warc/CC-MAIN-20230529042138-20230529072138-00373.warc.gz | 283,555,538 | 10,556 | How do you calculate the equivalence point?
Equivalence point: point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution. At the equivalence point in an acid-base titration, moles of base = moles of acid and the solution only contains salt and water.
What is equal at the equivalence point?
The equivalence point or stoichiometric point is the point in a chemical reaction when there is exactly enough acid and base to neutralize the solution. In a titration, it is where the moles of titrant equal the moles of solution of unknown concentration.
How do you calculate moles at the equivalence point?
Therefore, 100 mL ÷ 1000 mL/L = 0.1 L. Next, multiply the molarity by the volume, as follows: (0.1 L) x (0.1 M) = 0.01 moles. This provides the amount of titrant chemical added to reach the first equivalence point.
Is equivalence point always 7?
At the Equivalence point of an acid-base titration the pH is not always 7. But at the Equivalence point the reaction mixture just crosses the pH mark of 7. It will not stay still at pH=7. Consider the titration of HCl solution against NaOH.
What is the pH equivalence point?
(In an acid-base titration, there is a 1:1 acid:base stoichiometry, so the equivalence point is the point where the moles of titrant added equals the moles of substance initially in the solution being titrated.) At the equivalence point, the pH = 7.00 for strong acid-strong base titrations.
What is end point and equivalence point?
The main difference between equivalence and endpoint is that the equivalence point is a point where the chemical reaction comes to an end while the endpoint is the point where the colour change occurs in a system.
What is the half equivalence point?
The half equivalence point represents the point at which exactly half of the acid in the buffer solution has reacted with the titrant. The half equivalence point is relatively easy to determine because at the half equivalence point, the pKa of the acid is equal to the pH of the solution.
Why is the equivalence point higher than 7?
At the equivalence point, all of the weak acid is neutralized and converted to its conjugate base (the number of moles of H+ = added number of moles of OH). However, the pH at the equivalence point does not equal 7. This is due to the production of conjugate base during the titration.
What volume of NaOH is needed to reach the equivalence point?
A mole is equal to 6.022 x 1023 molecules.) By doing the titration and making a plot of the volume of NaOH added versus the resulting pH of the solution, we find that the equivalence point occurs at 0.04398 L of NaOH. (This is the point where the plot appears to increase most rapidly.)
What is the equivalence point of HCl and NaOH titration?
Beyond the equivalence point (when the sodium hydroxide is in excess) the curve is just the same as that end of the HCl – NaOH graph. The common example of this would be ethanoic acid and ammonia. It so happens that these two are both about equally weak – in that case, the equivalence point is approximately pH 7.
How do you calculate the concentration of an acid in a titration?
Use the titration formula. If the titrant and analyte have a 1:1 mole ratio, the formula is molarity (M) of the acid x volume (V) of the acid = molarity (M) of the base x volume (V) of the base. (Molarity is the concentration of a solution expressed as the number of moles of solute per litre of solution.)
How do you find pH after equivalence point?
pH after equivalence point After the equivalence point, the stoichiometric reaction has neutralized all the sample, and the pH depends on how much excess titrant has been added. After equivalence point, any excess strong base KOH determines the pH. If total KOH added was 0.150 moles, then excess OH- = 0.050 moles.
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What does Bernoulli’s equation State? Bernoulli’s principle states the following, Bernoulli’s principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed. Why is Bernoulli’s equation used? The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one […] | 1,040 | 4,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-23 | latest | en | 0.897815 |
http://marybourassa.blogspot.com/2015/05/mfm2p-day-57-pyramids.html | 1,531,736,768,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589251.7/warc/CC-MAIN-20180716095945-20180716115945-00500.warc.gz | 231,666,093 | 13,905 | Monday, 4 May 2015
MFM2P - Day 57: Pyramids
Instead of starting with a counting circle today, I had them multiply two binomials. They haven't practiced this lately (and were really doing well with counting circles) so I thought it would be more beneficial. I have 8 of these ready to go on little slips of paper (found here) so we can do one each day this week and again later in the semester.
Next up, pyramids! Alex Overwijk and I had talked about creating a new activity involving tents in the shape of pyramids, but that hasn't happened. This week is crazy for me so I went with some not-very-exciting worksheets. We did start by building pyramids. I brought in the bins of G-O-Frames and asked them to build pyramids. That's all I said. Here is some of the collection:
There were a lot of tetrahedrons and some square-based pyramids. There is one hexagonal-based pyramid and one right-angled triangular-based pyramid. I asked what they noticed about the shapes and they said that the sides are always triangles. The base could be a variety of shapes. I asked how they could find the surface area of a pyramid and took a squared-based one apart. I asked the same question while holding up the one with a hexagon as its base. They seemed to understand the idea of adding up all the areas. I had them work on the first example from this handout and I circulated helping some of them get started.
I showed them the right-angled triangular-based prism and said that if we had three of them we could make a cube. I also showed them this video. I could have done the demo myself, but I would just make a mess and I wasn't up for that first thing on a Monday morning. I tried to really emphasize that to find the volume of a prism you take the area of the base and multiply it by the height because the height represents the number of "layers". To find the volume of a pyramid, you perform that same process and divide the result by 3. Again, I let them work through some of the examples before going over this one together:
Along the way we talked about the difference between vertical height and slant height. They figured out that the slant height of the pyramid is the height of the triangle. They also were able to determine that if you were missing one of the heights in a square-based prism, you could use the sum of squares (Pythagorean theorem) to calculate it.
We will do a little more with pyramids tomorrow and work on some trig again. | 547 | 2,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-30 | latest | en | 0.986093 |
https://formulasearchengine.com/wiki/Integral_curve | 1,679,720,968,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945315.31/warc/CC-MAIN-20230325033306-20230325063306-00063.warc.gz | 312,940,048 | 10,403 | # Integral curve
Three integral curves for the slope field corresponding to the differential equation dy / dx = x2 − x − 1.
In mathematics, an integral curve is a parametric curve that represents a specific solution to an ordinary differential equation or system of equations. If the differential equation is represented as a vector field or slope field, then the corresponding integral curves are tangent to the field at each point.
Integral curves are known by various other names, depending on the nature and interpretation of the differential equation or vector field. In physics, integral curves for an electric field or magnetic field are known as field lines, and integral curves for the velocity field of a fluid are known as streamlines. In dynamical systems, the integral curves for a differential equation that governs a system are referred to as trajectories or orbits.
## Definition
Suppose that F is a vector field: that is, a vector-valued function with Cartesian coordinates (F1,F2,...,Fn); and x(t) a parametric curve with Cartesian coordinates (x1(t),x2(t),...,xn(t)). Then x(t) is an integral curve of F if it is a solution of the following autonomous system of ordinary differential equations:
{\displaystyle {\begin{aligned}{\frac {dx_{1}}{dt}}&=F_{1}(x_{1},\ldots ,x_{n})\\&\vdots \\{\frac {dx_{n}}{dt}}&=F_{n}(x_{1},\ldots ,x_{n}).\end{aligned}}}
Such a system may be written as a single vector equation
${\displaystyle \mathbf {x} '(t)=\mathbf {F} (\mathbf {x} (t)).\!\,}$
This equation says precisely that the tangent vector to the curve at any point x(t) along the curve is precisely the vector F(x(t)), and so the curve x(t) is tangent at each point to the vector field F.
If a given vector field is Lipschitz continuous, then the Picard–Lindelöf theorem implies that there exists a unique flow for small time.
## Generalization to differentiable manifolds
### Definition
Let M be a Banach manifold of class Cr with r ≥ 2. As usual, TM denotes the tangent bundle of M with its natural projection πM : TMM given by
${\displaystyle \pi _{M}:(x,v)\mapsto x.}$
A vector field on M is a cross-section of the tangent bundle TM, i.e. an assignment to every point of the manifold M of a tangent vector to M at that point. Let X be a vector field on M of class Cr−1 and let pM. An integral curve for X passing through p at time t0 is a curve α : JM of class Cr−1, defined on an open interval J of the real line R containing t0, such that
${\displaystyle \alpha (t_{0})=p;\,}$
${\displaystyle \alpha '(t)=X(\alpha (t)){\mbox{ for all }}t\in J.}$
### Relationship to ordinary differential equations
The above definition of an integral curve α for a vector field X, passing through p at time t0, is the same as saying that α is a local solution to the ordinary differential equation/initial value problem
${\displaystyle \alpha (t_{0})=p;\,}$
${\displaystyle \alpha '(t)=X(\alpha (t)).\,}$
It is local in the sense that it is defined only for times in J, and not necessarily for all tt0 (let alone tt0). Thus, the problem of proving the existence and uniqueness of integral curves is the same as that of finding solutions to ordinary differential equations/initial value problems and showing that they are unique.
### Remarks on the time derivative
In the above, α′(t) denotes the derivative of α at time t, the "direction α is pointing" at time t. From a more abstract viewpoint, this is the Fréchet derivative:
${\displaystyle (\mathrm {d} _{t}f)(+1)\in \mathrm {T} _{\alpha (t)}M.}$
In the special case that M is some open subset of Rn, this is the familiar derivative
${\displaystyle \left({\frac {\mathrm {d} \alpha _{1}}{\mathrm {d} t}},\dots ,{\frac {\mathrm {d} \alpha _{n}}{\mathrm {d} t}}\right),}$
where α1, ..., αn are the coordinates for α with respect to the usual coordinate directions.
The same thing may be phrased even more abstractly in terms of induced maps. Note that the tangent bundle TJ of J is the trivial bundle J × R and there is a canonical cross-section ι of this bundle such that ι(t) = 1 (or, more precisely, (t, 1)) for all tJ. The curve α induces a bundle map α : TJ → TM so that the following diagram commutes:
Then the time derivative α′ is the composition α′ = α o ι, and α′(t) is its value at some point t ∈ J.
## References
• {{#invoke:citation/CS1|citation
|CitationClass=book }} | 1,144 | 4,364 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 9, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-14 | latest | en | 0.896812 |
https://www.nottingham.ac.uk/xpert/scoreresults.php?keywords=Statistics%20-%20an%20intuitive%20introduction%20:%20standard%20&start=22780&end=22800 | 1,568,854,761,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573385.29/warc/CC-MAIN-20190918234431-20190919020431-00193.warc.gz | 994,518,013 | 30,186 | The same rules about the order of calculations apply to decimals as apply to whole numbers.
## Calculations are performed in the following order:
Brackets;
Powers (e.g. squaring or cubing a number);
Division and Multiplication (performed in the order written, left to
Author(s): The Open University
## Activity 31
Insert brackets in the following calculations to emphasise the order in which a scientific calculator would perform them, then do the calculations by hand and on your calculator, with and without the bracke
Author(s): The Open University
You may have noticed that sometimes the order in which calculations are carried out seems to matter and sometimes it does not. When using a calculator, it is very important to know the order in which it will do calculations. It is not always the order in which you enter them.
Although written English is read from left to right, this is not the case for all written languages (Chinese is read top to bottom, right to left). With mathematics, the order of the written operations does not alw
Author(s): The Open University
To multiply and divide by 10, 100, 1000, etc., write the digits in their place value columns. To multiply, move the digits to the left (replacing the numbers on the right with zeros) and to divide move them to the right (putting in a decimal point, and any zeros necessary for the place value).
Multiplication and division by whole numbers in general can be carried out by combining this technique with a knowledge of the multiplication tables up to 10.
Author(s): The Open University
## Activity 22
Carry out the following calculations, without using a calculator.
• (a) A million pound lottery prize minus a three hundred pound administrative charge.
• Author(s): The Open University
The basic metric unit for capacity is the litre, usually denoted by the symbol l (though sometimes an uppercase L is used to avoid confusion with the number 1).
In the SI system, units such as cubic metres (m3), cubic centimetres (cm3) and cubic millimetres (mm3) are used. These two systems are linked because:
1 ml = 1 cm3
The animation below i
Author(s): The Open University
The basic SI unit for mass is the kilogram, symbol kg
The tonne (t) which is equivalent to 1000 kg and is a metric unit is often used alongside the SI units.
The animation below illustrates how to convert between the most commonly used units of mass, the metric tonne (t); the kilogram (kg); the gram (g); the milligram (mg) and the microgram (μg).
Author(s): The Open University
A great advantage of the metric system of units is that conversion between units within the system is particularly easy. For example, ‘£1 is worth 100p’ is converting one pound into pence. To convert pounds to pence, you multiply by 100. So £2 is 200p, and £2.63 is 263p. (Remember that to multiply by 100, you move the digits two places to the left in the place value table.)
To convert from pence to pounds, you need to reverse this process, i.e. to divide by 100 (moving the
Author(s): The Open University
## Activity 15
Suggest appropriate units for each of the following:
• (a) the age of the kitten when it is weaned;
• (b) the distance between one train station and the
Author(s): The Open University
## Activity 14
Contour lines on a map show all the points at a given height above sea level. The lines are drawn for each height at 50-metre intervals, and points below sea level are shown by negative heights. The diagra
Author(s): The Open University
Numbers can be positive or negative, i.e. greater than or less than zero. Negative numbers have several uses; for example, to measure temperatures below zero, such as −3°C (‘minus 3 degrees Celsius’). They are also used to represent debts and overdrawn accounts: a bank balance of −£84.33 means ‘overdrawn by £84.33’.
Negative numbers are shown on the number line to the left of 0. The animation below shows −8, −7, −
Author(s): The Open University
A fraction is written as one number over another (such as ) and means the top number divided by the bottom number. The top number, 3, is called the numerator and the bottom number, 10, is call
Author(s): The Open University
## Activity 6
What is 370.76 grams in kilograms? There are 1000 grams in a kilogram.
370.76 ÷ 1000 = 0.370 76.
So 370.76 g = 0.370
Author(s): The Open University
## Activity 5
When dividing by 1000, move the digits 3 places to the right (past the decimal point). Divide 202.15 by 1000. What is 202.15 metres in kilometres? | 1,107 | 4,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-39 | latest | en | 0.931117 |
https://socratic.org/questions/a-charge-of-3-c-is-at-4-7-and-a-charge-of-1-c-is-at-2-3-if-both-coordinates-are- | 1,575,953,621,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540525821.56/warc/CC-MAIN-20191210041836-20191210065836-00366.warc.gz | 533,248,353 | 6,317 | # A charge of 3 C is at (4 ,-7 ) and a charge of -1 C is at (-2 , 3 ) . If both coordinates are in meters, what is the force between the charges?
net displacement between the two charges is $r = \sqrt{{\left(4 - \left(- 2\right)\right)}^{2} + {\left(- 7 - 3\right)}^{2}} = \sqrt{136} m$
So, the electrostatic force acting between them will be $9 \cdot {10}^{9} \cdot 3 \cdot \frac{- 1}{136} = - 1.97 \cdot {10}^{8} N$ (using, $F = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$) | 178 | 465 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-51 | latest | en | 0.792495 |
https://gmatclub.com/forum/two-boys-joe-and-seth-can-mow-a-lawn-in-2-hours-thirty-minutes-if-396260.html | 1,721,802,151,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518157.20/warc/CC-MAIN-20240724045402-20240724075402-00515.warc.gz | 238,776,272 | 128,551 | Last visit was: 23 Jul 2024, 23:22 It is currently 23 Jul 2024, 23:22
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# Two boys, Joe and Seth, can mow a lawn in 2 hours thirty minutes. If
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Re: Two boys, Joe and Seth, can mow a lawn in 2 hours thirty minutes. If [#permalink]
Bunuel wrote:
Two boys, Joe and Seth, can mow a lawn in 2 hours thirty minutes. If 3 other boys, Mike, Tommy and Teddy, helped, how long would it take to mow the same lawn with all of them starting at the same time?
(1) Tommy and Teddy each individually work at a rate that is twice Mike's rate.
(2) Mike would take 7 hours to mow the entire yard alone.
pratiksha1998 wrote:
How do we know the individual rates of Joe and Seth? Since we do not know that how can we find the rate of all working together?
we don't need to know the individual rates we can assume that both are one entity act as one. so now we have only 3 entity with the help of option(as (1) gives combined information of rates.
My thought process in solving this question.
there are 2 person another 3 join and we need to find time when all work together.
1- gives rate of another 2 person but what about 3rd person if the rate of 3rd person changes then time changes also. so INSUFFICIENT
2- gives rate of only 3rd person so same as 1. so INSUFFICIENT
now lets combine
now we have everyone's rate (assuming joe and Seth as one and tommy and teddy as one) and they do the same work so we can find accumulate time. | 697 | 2,541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-30 | latest | en | 0.896415 |
https://www.physicsforums.com/threads/calculating-tension-in-each-of-3-strings.263313/ | 1,544,562,299,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823702.46/warc/CC-MAIN-20181211194359-20181211215859-00550.warc.gz | 1,025,324,893 | 14,561 | # Homework Help: Calculating tension in each of 3 strings
1. Oct 10, 2008
### mathsgeek
I have got a question i am having a lot of trouble with. The question is that three strings on unequal lengths are used to support a mass of 50g. All of these strings are used to support the weight. Two of these strings cannot be in the same plane as the other string. Determine the tension in each String.
Well, i have 2 strings in the i and j plane and was thinking to have a third string in the k or k and j plane. However i do not know if this is correct. Also, could someone show me the general steps on how to calculate the tension in each string as i am having trouble with calculating this.
2. Oct 10, 2008
### Rake-MC
It's going to depend exactly on how you define your planes and the angles that follow.
3. Oct 10, 2008
### mathsgeek
well the planes are goin to be defined in terms of i, j and k.
4. Oct 10, 2008
### Rake-MC
So you're putting the j plane parallel to the acceleration due to gravity?
There are potentially infinite solutions to this question, I must be interpreting it incorrectly. Is that the exact wording of the question? Could you not have 1 string hanging from a ceiling straight down parallel to gravity, then two strings of equal tensions opposing each other perpendicular to the acceleration due to gravity?
5. Oct 10, 2008
### mathsgeek
The strings have to be of unequal length though. And yes the j component is the same as y. (vertical)
6. Oct 10, 2008
### Rake-MC
It gives no definition of where they have to be fixed though? The end that is not fixed to the mass I mean.
7. Oct 10, 2008
### mathsgeek
its basically fixed to a wall or sumthing, so it must be fixed at both ends.
8. Oct 11, 2008
### mathsgeek
any1? Its urgent. Thanks
9. Oct 11, 2008
### Redbelly98
Staff Emeritus
Draw a freebody diagram.
Fnet = 0 for x, y, and z directions.
It depends on what angles the strings are oriented, which you know (hopefully) but we don't know.
10. Oct 11, 2008
### mathsgeek
Heres a pic of 2 of the strings: | 546 | 2,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-51 | latest | en | 0.937241 |
https://www.intellectualmath.com/division-of-polynomials-by-binomials-worksheet.html | 1,722,785,975,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640404969.12/warc/CC-MAIN-20240804133418-20240804163418-00285.warc.gz | 669,857,616 | 11,879 | # DIVISION OF POLYNOMIAL BY BINOMIALS WORKSHEET
Divide. Write your answer in fraction form.
Problem 1 :
(2x5– 15x3– 9x2+ 11x + 12) ÷ (x + 2)
Solution
Problem 2 :
(x4– x3– 19x2- 3x - 19) ÷ (x - 5)
Solution
Problem 3 :
(10x4– 4x3+ 14x2- 14x - 16) ÷ (2x - 2)
Solution
Problem 4 :
(9x5– 9x4– x3- 12x2+ x - 11) ÷ (3x - 5)
Solution
Problem 5 :
(16x4+ 4x3+ 2x2- 21x + 7) ÷ (4x - 1)
Solution
Problem 6 :
(6x5+ 21x4– 14x3- 8x2+ x - 6) ÷ (x + 4)
Solution
## Answer Key
1) Quotient = 2x4 – 4x3 – 7x2 + 5x + 1
Remainder = 10
Fraction form :
2) Quotient = x3 + 4x2 + x + 2
Remainder = -9
Fraction form :
3) Quotient = 5x3 + 3x2 + 10x + 3
Remainder = -5
Fraction form :
4) Quotient = 3x4 + 2x3 + 3x2 + x + 2
Remainder = -1
Fraction form :
5) Quotient = 4x3 + 2x2 + x - 5
Remainder = 2
Fraction form :
6) Quotient = 6x4 - 3x3 - 2x2 + 0x + 1
Remainder = -10
Fraction form :
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Read More | 585 | 1,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-33 | latest | en | 0.480938 |
https://best.dujuz.com/367/ | 1,652,889,884,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00755.warc.gz | 165,762,056 | 10,885 | # If A B And B C Then A C
If A B And B C Then A C. PROOF: Suppose a, b, and c are integers where both a and b do not equal to zero. If a = b then b = a.
And I'm not sure how to figure it out. If problem B is decidable, then so is problem A. Then, x ∈ A – (B ∩ C).
## If problem B is decidable, then so is problem A.
For all real numbers a and b: a + b = b + a, and ab = ba. If the positive numbers a,b,c are the pth,qth,rth terms of a G. If a and b are any real numbers, c is any nonzero real number, and a = b, then a/c = b/c.
Supplementary Exercise Q9 – If a, b, c are coplanar, then a+b
If A+B-C=180degree and sin2A +Sin2B -sin2C=D sinA sinB …
Misc 12 – If AB = BA, then prove by induction that ABn = BnA
### Apne doubts clear karein ab Whatsapp par bhi.
The hint is confusing for me. You can only apply this to coprime a and b, all variables representing numbers are natural numbers: Define a=nk and b=mk, and define c=nmk. The proof is explained step wise below : Step-by-step explanation: Let x be any element of A – (B ∩ C).
P., then the dot product of vectors i^loga+j^ logb+k^logc and (q−r)i^+(r−p)j^ +(p−q)k^ is If a. ,b. ,c. are mutually perpendicular vectors of equal magnitude, then the angle between a. +b. +c. and a. is. Equivalently, If problem A is undecidable, then problem B is undecidable. If a and b are any real numbers, c is any nonzero real number, and a = b, then a/c = b/c.
### IF (a < c) THEN ! a < c : a the smallest.
If the positive numbers a,b,c are the pth,qth,rth terms of a G. Then I can do something like. a is odd, therefore a² is odd. PROOF: Suppose a, b, and c are integers where both a and b do not equal to zero.
How do I show for all a, b, and c? And I'm not sure how to figure it out. I've been searching online and I can find this (transitive property of equality): If a = b and b = c, then a = c. but not this Ok, that's not a very good example but it might illustrate my problem with: if a=b and a=c, then b=c.
Then, x ∈ A – (B ∩ C). P., then the dot product of vectors i^loga+j^ logb+k^logc and (q−r)i^+(r−p)j^ +(p−q)k^ is If a. ,b. ,c. are mutually perpendicular vectors of equal magnitude, then the angle between a. +b. +c. and a. is. I was looking for a tool that can convert C code expressions for the form: a = (A) ? | 722 | 2,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2022-21 | latest | en | 0.832058 |
http://www.wired.com/wiredscience/2010/03/on-the-8th-day-god-made-pi/ | 1,369,306,536,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703293367/warc/CC-MAIN-20130516112133-00092-ip-10-60-113-184.ec2.internal.warc.gz | 815,891,994 | 26,860 | # On the 8th Day, God made Pi
Pi day is March 14th – get it? (3.14) I am a big fan of Pi. Here is my first post to celebrate the awesomeness of Pi (I know this is early, but I was too excited to wait).
### How can you determine Pi?
Oh sure, tons of high schools do the classic experiment. Measure the circumference and diameter of as many round things as possible. Plot diameter vs. circumference. The slope will be Pi. Really, this is a great lab to do for all sorts of ages. The key thing is that students can see what Pi really means. I am not going to talk about this lab, I am going to do some thing cooler.
What if I had a 1 meter by 1 meter square taped out in the grass near the physics building and shot ping pong balls off the roof at this square? Ping pong balls are very difficult to aim, so you could easily imagine that they would fall in a random pattern in the square. I don’t really want to set this up, so I am going to do it with a program instead. Here is a program that will generate random points in a 1 x 1 square.
This is what the output would look like:
Now, here is the key. What if I calculate the distance of each of these point from the origin? This would simply be:
Doing this, I can separate all the points into those that are more than 1 unit away from the origin and those that are less than 1 unit away. In this plot, (of 1000 points) the red dots are more than one unit away and the blue are 1 unit or less.
Note that this plot does not have exactly the same horizontal and vertical scale – no idea why it came out like that. However, this is enough dots that maybe you can see a pattern. The blue dots are filling up 1/4th of a circle. The full area of this circle would be:
Since I am a physicist, I have trouble leaving the area as unitless. What if I look at the ratio of blue dots to total dots? This should be the same as the ratio of the area of the quarter circle to the whole square, or:
Doing this for my last random number run, I get pi = 3.04. What happens as I increase the number of random dots? This a plot of the estimation of Pi for numbers of dots starting at 1000 up to 100,000 dots. I did this 5 times because each time is different.
So, you can see that as the number of random points increases, the estimated value of Pi is a little less spread out and closer to 3.14-something.
P.S. Thanks to Dave for this idea. You know who you are.
### Update
I also made a Scratch program version of this calculation. (Scratch is a graphical programming language made at MIT – you know, for kids).
Rhett Allain is an Associate Professor of Physics at Southeastern Louisiana University. He enjoys teaching and talking about physics. Sometimes he takes things apart and can't put them back together. | 656 | 2,757 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2013-20 | latest | en | 0.971721 |
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#### K.Smithy
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##### Re: QCE Physics Questions Thread
« Reply #15 on: April 28, 2020, 10:50:56 pm »
+3
Question: A charged ball of mass 0.32 g hovers 2.0 cm above a charged plate that has a charge of −4.72 × 10-7 C. Calculate the magnitude of the charge on the ball.
Hey A.Rose!
I'm gonna start off by saying that: I am not super confident in physics right now! It is just one of those subjects I have put on the back burner, and have been meaning to work on. So, I would take my reply with a grain of salt (just incase I have no clue what I'm talking about).
Anywho, I don't remember mass ever coming into play for questions like these, so I believe that is just an added (and unnecessary) piece of information trying to throw you off. So, I would ignore the 0.32g. If you are a bit rusty, as you said, and have no clue whether or not the mass is necessary, your go to is the formula sheet If you look under the "electromagnetism" section there are no formulas involving mass. So you can just rule out that bit of information.
But lets just have a look at what we think is important:
- the ball hovers 2cm above charged plate
- plate has a charge of -4.72x10-7 C
So now we can head to the formula sheet and see what we can use.
You'll see that there are two formulas that look very good:
F = 1/4πɛ0 x Qq/r2 AND E = 1/4πɛ0 x q/r2
(keeping in mind that 1/4πɛ0 = k = 9x109 - also be careful with units, I'm fairly certain that the distance should be in meters so just make sure you do any necessary conversions)
Because we aren't looking for the force we can rule out the first formula - leaving us with the second one.
Therefore, I would start off by using: E = 1/4πɛ0 x q/r2
(Once again, I could be entirely wrong. But this is what I would do anyway... I hope I'm not wrong tho and I hope this helps - good luck!)
« Last Edit: April 29, 2020, 12:07:10 am by K.Smithy »
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#### Bri MT
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##### Re: QCE Physics Questions Thread
« Reply #16 on: April 29, 2020, 03:04:39 pm »
+2
Hello!
I am revising Electrostatics from Unit 3 Topic 2 at the moment and I am a bit rusty! I would really appreciate a little bit of help with this question, just at least to start me off
Question: A charged ball of mass 0.32 g hovers 2.0 cm above a charged plate that has a charge of −4.72 × 10-7 C. Calculate the magnitude of the charge on the ball.
Thank you!!
K. Smithy you weren't quite there but you were on the right track.
The trick to this question is recognising that since the ball is hovering, it is stationary with 0 net force. Thus, force due to gravity and force to the electric field strength cancel out.
If we take the magnitude of those two forces as being equal:
$mg = \frac{1}{4 \pi \varepsilon_0} \frac{qQ}{r^2}$
all you need to do is sub in values appropriately (as K.Smithy said, be careful with units) and solve for q
Edit: for showing fractions neatly in latex it's \frac{}{} . I didn't learn it for a while on the forums but there's a great guide by Rui here
« Last Edit: April 29, 2020, 03:10:16 pm by Bri MT »
#### K.Smithy
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##### Re: QCE Physics Questions Thread
« Reply #17 on: April 29, 2020, 07:31:08 pm »
+2
K. Smithy you weren't quite there but you were on the right track.
Aha worth a shot! Hopefully i'll get there before the externals
Thanks for clearing it up!
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#### Bri MT
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##### Re: QCE Physics Questions Thread
« Reply #18 on: April 29, 2020, 07:48:26 pm »
+1
Aha worth a shot! Hopefully i'll get there before the externals
Thanks for clearing it up!
No worries!
I'm sure you will
#### A.Rose
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##### Re: QCE Physics Questions Thread
« Reply #19 on: April 29, 2020, 08:01:18 pm »
+2
The trick to this question is recognising that since the ball is hovering, it is stationary with 0 net force. Thus, force due to gravity and force to the electric field strength cancel out.
Thank you both for your help! Yes, I did work out that I needed the mass to calculate the force eventually.
I definitely need to do some thorough electrostatics revision!
Thanks!!
#### Bri MT
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##### Re: QCE Physics Questions Thread
« Reply #20 on: April 29, 2020, 08:14:18 pm »
+1
Thank you both for your help! Yes, I did work out that I needed the mass to calculate the force eventually.
I definitely need to do some thorough electrostatics revision!
Thanks!!
No worries!
This stuff is pretty standard across states so if you want more revision you should be able to find a bunch from other states.
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##### Re: QCE Physics Questions Thread
« Reply #21 on: April 30, 2020, 12:56:38 pm »
0
Hi! Me again with another question!
I am very confused about this question. (The diagram is attached). The question is to 'explain using physics concepts how the system will operate to close and open the tubing'.
A student designs a simple solenoid valve system to control the flow of a liquid through flexible tubing by closing and opening the tubing. It is comprised of an iron rod with a spring coiled around it at one end. One end of the spring is attached at one end of the rod and its other end is attached to a rectangular housing. The tubing is fixed to a solid support.
If you could at least direct me to where I can find a good explanation of how these kinds of experiments work because I don't remember learning about solenoids in this kind of context last year. I'm assuming its something to do with the establishment of a magnetic field that opens or closes the tube - but I am not confident. Just some clue or somewhere I can research this (I didn't have much luck in my textbook or searching online) that would be amazing!!
Thank you!
#### Bri MT
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##### Re: QCE Physics Questions Thread
« Reply #22 on: April 30, 2020, 02:03:12 pm »
+1
Hi! Me again with another question!
I am very confused about this question. (The diagram is attached). The question is to 'explain using physics concepts how the system will operate to close and open the tubing'.
A student designs a simple solenoid valve system to control the flow of a liquid through flexible tubing by closing and opening the tubing. It is comprised of an iron rod with a spring coiled around it at one end. One end of the spring is attached at one end of the rod and its other end is attached to a rectangular housing. The tubing is fixed to a solid support.
If you could at least direct me to where I can find a good explanation of how these kinds of experiments work because I don't remember learning about solenoids in this kind of context last year. I'm assuming its something to do with the establishment of a magnetic field that opens or closes the tube - but I am not confident. Just some clue or somewhere I can research this (I didn't have much luck in my textbook or searching online) that would be amazing!!
Thank you!
Hey!
You're on the right track! Current carrying solenoids create magnetic fields like the field around bar magnets.
This exact question type isn't common but there are often questions involving magnetic fields around solenoids in VCE physics so it might be worth looking at some of those.
I'm aware I've been pretty vague here (you seemed to be on the right track) - please feel free to follow up with anything you're unsure of
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##### Re: QCE Physics Questions Thread
« Reply #23 on: May 02, 2020, 10:33:41 pm »
+1
Hi!
I'm confused on the working that my teacher has provided on a recent practise data test we did.
I have attached the dataset and the question. I really don't have a clue on what my teacher has done. I thought I would have seen "9.81m/s/s" somewhere in the working. But... Yeah I really have no idea what's going on here. Can anyone explain to me why the acceleration due to gravity is 12.5m/s/s?
Thank you!
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#### Bri MT
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##### Re: QCE Physics Questions Thread
« Reply #24 on: May 03, 2020, 09:57:24 am »
+1
Hi!
I'm confused on the working that my teacher has provided on a recent practise data test we did.
I have attached the dataset and the question. I really don't have a clue on what my teacher has done. I thought I would have seen "9.81m/s/s" somewhere in the working. But... Yeah I really have no idea what's going on here. Can anyone explain to me why the acceleration due to gravity is 12.5m/s/s?
Thank you!
Hey!
This is a question saying if the data we had was x, then what would that mean about y? Sometimes this will end in a reasonable value and sometimes it won't. The reason you don't see the actual value for g is because the question wants you to pretend you don't know what this is & figure out what it would be for the provided fake situation.
Do you understand where g = sin(theta) comes from?
#### A.Rose
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##### Re: QCE Physics Questions Thread
« Reply #25 on: May 05, 2020, 03:22:41 pm »
+1
Hi, I have another Electromagnetism question! Please see attached.
I just need a clue to where to start. Since its asking for speed I'm assuming I'm going to need to use F=qvB at some stage...? But it also gives me a potential difference which makes me think I need to use ∆U=∆Vq? Am I at least on the right track or is there something I'm missing?
Thank you!
#### Bri MT
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##### Re: QCE Physics Questions Thread
« Reply #26 on: May 19, 2020, 01:51:29 pm »
+1
Hi, I have another Electromagnetism question! Please see attached.
I just need a clue to where to start. Since its asking for speed I'm assuming I'm going to need to use F=qvB at some stage...? But it also gives me a potential difference which makes me think I need to use ∆U=∆Vq? Am I at least on the right track or is there something I'm missing?
Thank you!
Sorry I somehow missed this earlier!
You may have resolved this by now, but my approach would be to use the change in energy to get the kinetic energy and thus speed | 3,031 | 11,394 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-24 | latest | en | 0.954693 |
https://books.google.gr/books?id=sIQAAAAAMAAJ&lr=&hl=el&source=gbs_navlinks_s | 1,632,796,606,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058589.72/warc/CC-MAIN-20210928002254-20210928032254-00133.warc.gz | 182,118,500 | 11,005 | # Trigonometry, Plane and Spherical: With the Construction and Application of Logarithms
Kimber and Conrad, 1810 - 125 σελίδες
### Τι λένε οι χρήστες -Σύνταξη κριτικής
Δεν εντοπίσαμε κριτικές στις συνήθεις τοποθεσίες.
### Περιεχόμενα
Ενότητα 1 2 Ενότητα 2 3 Ενότητα 3 5 Ενότητα 4 25 Ενότητα 5 26
Ενότητα 6 43 Ενότητα 7 49 Ενότητα 8 51 Ενότητα 9 87
### Δημοφιλή αποσπάσματα
Σελίδα 69 - TO THEIR DIFFERENCE ; So IS THE TANGENT OF HALF THE SUM OF THE OPPOSITE ANGLES', To THE TANGENT OF HALF THEIR DIFFERENCE.
Σελίδα 79 - ... projection is that of a meridian, or one parallel thereto, and the point of sight is assumed at an infinite distance on a line normal to the plane of projection and passing through the center of the sphere. A circle which is parallel to the plane of projection is projected into an equal circle, a circle perpendicular to the plane of projection is projected into a right line equal in length to the diameter of the projected circle; a circle in any other position is projected into an ellipse, whose...
Σελίδα 25 - The cotangent of half the sum of the angles at the base, Is to the tangent of half their difference...
Σελίδα 28 - The rectangle of the radius, and sine of the middle part, is equal to the rectangle of the tangents of the two EXTREMES CONJUNCT, and to that of the cosines of the two EXTREMES DISJUNCT.
Σελίδα 7 - If the sine of the mean of three equidifferent arcs' dius being unity) be multiplied into twice the cosine of the common difference, and the sine of either extreme be deducted from the product, the remainder will be the sine of the other extreme. (B.) The sine of any arc above 60°, is equal to the sine of another arc as much below 60°, together with the sine of its excess above 60°. Remark. From this latter proposition, the sines below 60° being known, those...
Σελίδα 28 - In a right angled spherical triangle, the rectangle under the radius and the sine of the middle part, is equal to the rectangle under the tangents of the adjacent parts ; or', to the rectangle under the cosines of the opposite parts. | 547 | 2,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-39 | latest | en | 0.632811 |
https://en.wikipedia.org/wiki/Hahn_series | 1,560,729,773,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998325.55/warc/CC-MAIN-20190616222856-20190617004856-00447.warc.gz | 424,469,021 | 19,119 | # Hahn series
In mathematics, Hahn series (sometimes also known as Hahn–Mal'cev–Neumann series) are a type of formal infinite series. They are a generalization of Puiseux series (themselves a generalization of formal power series) and were first introduced by Hans Hahn in 1907[1] (and then further generalized by Anatoly Maltsev and Bernhard Neumann to a non-commutative setting). They allow for arbitrary exponents of the indeterminate so long as the set supporting them forms a well-ordered subset of the value group (typically ${\displaystyle \mathbb {Q} }$ or ${\displaystyle \mathbb {R} }$). Hahn series were first introduced, as groups, in the course of the proof of the Hahn embedding theorem and then studied by him as fields in his approach to Hilbert's seventeenth problem.
## Formulation
The field of Hahn series ${\displaystyle K\left[\left[T^{\Gamma }\right]\right]}$ (in the indeterminate T) over a field K and with value group Γ (an ordered group) is the set of formal expressions of the form
${\displaystyle f=\sum _{e\in \Gamma }c_{e}T^{e}}$
with ${\displaystyle c_{e}\in K}$ such that the support ${\displaystyle \{e\in \Gamma :c_{e}\neq 0\}}$ of f is well-ordered. The sum and product of
${\displaystyle f=\sum _{e\in \Gamma }c_{e}T^{e}}$ and ${\displaystyle g=\sum _{e\in \Gamma }d_{e}T^{e}}$
are given by
${\displaystyle f+g=\sum _{e\in \Gamma }(c_{e}+d_{e})T^{e}}$
and
${\displaystyle fg=\sum _{e\in \Gamma }\sum _{e'+e''=e}c_{e'}d_{e''}T^{e}}$
(in the latter, the sum ${\displaystyle \sum _{e'+e''=e}\cdot }$ over values ${\displaystyle (e',e'')}$ such that ${\displaystyle c_{e'}\neq 0}$ and ${\displaystyle d_{e''}\neq 0}$ is finite because a well-ordered set cannot contain an infinite decreasing sequence).
For example, ${\displaystyle T^{-1/p}+T^{-1/p^{2}}+T^{-1/p^{3}}+\cdots }$ is a Hahn series (over any field) because the set of rationals
${\displaystyle \left\{-{\frac {1}{p}},-{\frac {1}{p^{2}}},-{\frac {1}{p^{3}}},\ldots \right\}}$
is well-ordered; it is not a Puiseux series because the denominators in the exponents are unbounded. (And if the base field K has characteristic p, then this Hahn series satisfies the equation ${\displaystyle X^{p}-X=T^{-1}}$ so it is algebraic over ${\displaystyle K(T)}$.)
## Properties
The valuation ${\displaystyle v(f)}$ of a non zero Hahn series
${\displaystyle f=\sum _{e\in \Gamma }c_{e}T^{e}}$
is defined as the smallest ${\displaystyle e\in \Gamma }$ such that ${\displaystyle c_{e}\neq 0}$ (in other words, the smallest element of the support of ${\displaystyle f}$): this makes ${\displaystyle K[[T^{\Gamma }]]}$ into a spherically complete valued field with value group ${\displaystyle \Gamma }$ and residue field ${\displaystyle K}$ (justifying a posteriori the terminology). In fact, if ${\displaystyle K}$ has characteristic zero, then ${\displaystyle (K[[T^{\Gamma }]],v)}$ is up to (non unique) isomorphism the only spherically complete valued field with residue field ${\displaystyle K}$ and value group ${\displaystyle \Gamma }$[2]. The valuation ${\displaystyle v}$ defines a topology on ${\displaystyle K\left[\left[T^{\Gamma }\right]\right]}$. If ${\displaystyle \Gamma \subseteq \mathbb {R} }$, then v corresponds to an ultrametric absolute value ${\displaystyle |f|=\exp(-v(f))}$, with respect to which ${\displaystyle K\left[\left[T^{\Gamma }\right]\right]}$ is a complete metric space. However, unlike in the case of formal Laurent series or Puiseux series, the formal sums used in defining the elements of the field do not converge: in the case of ${\displaystyle T^{-1/p}+T^{-1/p^{2}}+T^{-1/p^{3}}+\cdots }$ for example, the absolute values of the terms tend to 1 (because their valuations tend to 0), so the series is not convergent (such series are sometimes known as "pseudo-convergent"[3]).
If K is algebraically closed (but not necessarily of characteristic zero) and Γ is divisible, then ${\displaystyle K\left[\left[T^{\Gamma }\right]\right]}$ is algebraically closed.[4] Thus, the algebraic closure of ${\displaystyle K((T))}$ is contained in ${\displaystyle K[[T^{\mathbb {Q} }]]}$ (when K is of characteristic zero, it is exactly the field of Puiseux series): in fact, it is possible to give a somewhat analogous description of the algebraic closure of ${\displaystyle K((T))}$ in positive characteristic as a subset of ${\displaystyle K\left[\left[T^{\Gamma }\right]\right]}$.[5]
If K is an ordered field then ${\displaystyle K\left[\left[T^{\Gamma }\right]\right]}$ is totally ordered by making the indeterminate T infinitesimal (greater than 0 but less than any positive element of K) or, equivalently, by using the lexicographic order on the coefficients of the series. If K is real-closed and Γ is divisible then ${\displaystyle K\left[\left[T^{\Gamma }\right]\right]}$ is itself real closed.[6] This fact can be used to analyse (or even construct) the field of surreal numbers (which is isomorphic, as an ordered field, to the field of Hahn series with real coefficients and value group the surreal numbers themselves[7]).
If κ is an infinite regular cardinal, one can consider the subset of ${\displaystyle K\left[\left[T^{\Gamma }\right]\right]}$ consisting of series whose support set ${\displaystyle \{e\in \Gamma :c_{e}\neq 0\}}$ has cardinality (strictly) less than κ: it turns out that this is also a field, with much the same algebraic closedness properties as the full ${\displaystyle K\left[\left[T^{\Gamma }\right]\right]}$: e.g., it is algebraically closed or real closed when K is so and Γ is divisible.[8]
## Hahn–Witt series
The construction of Hahn series can be combined with Witt vectors (at least over a perfect field) to form twisted Hahn series or Hahn–Witt series:[9] for example, over a finite field K of characteristic p (or their algebraic closure), the field of Hahn–Witt series with value group Γ (containing the integers) would be the set of formal sums ${\displaystyle \sum _{e\in \Gamma }c_{e}p^{e}}$ where now ${\displaystyle c_{e}}$ are Teichmüller representatives (of the elements of K) which are multiplied and added in the same way as in the case of ordinary Witt vectors (which is obtained when Γ is the group of integers). When Γ is the group of rationals or reals and K is the algebraic closure of the finite field with p elements, this construction gives a (ultra)metrically complete algebraically closed field containing the p-adics, hence a more or less explicit description of the field ${\displaystyle \mathbb {C} _{p}}$ or its spherical completion.[10]
## Examples
• The field ${\displaystyle K((T))}$ of formal Laurent series over ${\displaystyle K}$ can be described as ${\displaystyle K[[T^{\mathbb {Z} }]]}$.
• The field of surreal numbers can be regarded as a field of Hahn series with real coefficients and value group the surreal numbers themselves.[11]
• The Levi-Civita field can be regarded as a subfield of ${\displaystyle \mathbb {R} [[T^{\mathbb {Q} }]]}$, with the additional imposition that the coefficients be a left-finite set: the set of coefficients less than a given coefficient ${\displaystyle c_{q}}$ is finite.
• The field of transseries ${\displaystyle \mathbb {T} }$ is a directed union of Hahn fields (and is an extension of the Levi-Civita field). The construction of ${\displaystyle \mathbb {T} }$ resembles (but is not literally) ${\displaystyle T_{0}=\mathbb {R} }$, ${\displaystyle T_{n+1}=\mathbb {R} \left[\left[\varepsilon ^{T_{n}}\right]\right]}$.
## Notes
1. ^ Hahn (1907)
2. ^ Kaplansky, Irving, Maximal fields with valuation, Duke Mathematical Journal, vol. 1, n°2, 1942.
3. ^ Kaplansky (1942, Duke Math. J., definition on p.303)
4. ^ MacLane (1939, Bull. Amer. Math. Soc., theorem 1 (p.889))
5. ^ Kedlaya (2001, Proc. Amer. Math. Soc.)
6. ^ Alling (1987, §6.23, (2) (p.218))
7. ^ Alling (1987, theorem of §6.55 (p. 246))
8. ^ Alling (1987, §6.23, (3) and (4) (p.218–219))
9. ^ Kedlaya (2001, J. Number Theory)
10. ^ Poonen (1993)
11. ^ Alling (1987)
## References
• Hahn, Hans (1907), "Über die nichtarchimedischen Größensysteme", Sitzungsberichte der Kaiserlichen Akademie der Wissenschaften, Wien, Mathematisch - Naturwissenschaftliche Klasse (Wien. Ber.), 116: 601–655, JFM 38.0501.01 (reprinted in: Hahn, Hans (1995), Gesammelte Abhandlungen I, Springer-Verlag)
• MacLane, Saunders (1939), "The universality of formal power series fields", Bulletin of the American Mathematical Society, 45: 888–890, doi:10.1090/s0002-9904-1939-07110-3, Zbl 0022.30401
• Kaplansky, Irving (1942), "Maximal fields with valuations I", Duke Mathematical Journal, 9: 303–321, doi:10.1215/s0012-7094-42-00922-0
• Alling, Norman L. (1987). Foundations of Analysis over Surreal Number Fields. Mathematics Studies. 141. North-Holland. ISBN 0-444-70226-1. Zbl 0621.12001.
• Poonen, Bjorn (1993), "Maximally complete fields", L'Enseignement mathématique, 39: 87–106, Zbl 0807.12006
• Kedlaya, Kiran Sridhara (2001), "The algebraic closure of the power series field in positive characteristic", Proceedings of the American Mathematical Society, 129: 3461–3470, doi:10.1090/S0002-9939-01-06001-4
• Kedlaya, Kiran Sridhara (2001), "Power series and 𝑝-adic algebraic closures", Journal of Number Theory, 89: 324–339, arXiv:math/9906030, doi:10.1006/jnth.2000.2630 | 2,752 | 9,262 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 58, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-26 | latest | en | 0.882065 |
https://math.stackexchange.com/questions/1993112/no-subsequence-of-a-sequence-x-n-such-that-lim-n-rightarrow-infty-x/1993118 | 1,726,587,726,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00258.warc.gz | 352,506,051 | 36,870 | # No subsequence of a sequence $\{x_n\}$ such that $\lim_{n \rightarrow \infty} |x_n-x_{n+1}|=1$ converges to 3/4
Let $\{x_n\}$ be a sequence of real numbers in the interval $[0, 3/2]$. Suppose that $\lim_{n \rightarrow \infty} |x_n-x_{n+1}|=1$. Prove that no subsequence of $\{x_n\}$ converges to $3/4$.
My inclination is to prove this by contradiction. Suppose, to the contrary, that there exists a subsequence $\{x_{n_k}\}$ such that $x_{n_k}\rightarrow 3/4$. Then for every $\epsilon >0$ there exists an $N$ such that $n_k\geq N$ implies $|x_{n_k}-3/4|<\epsilon$. Then for all $n_k \geq N$, $$-\epsilon +3/4 \leq x_{n_k} \leq 3/4 + \epsilon.$$
Not sure what to do from here. I also tried using that convergent implies Cauchy, but I was also not sure what to do. Any help is appreciated, thank you.
If there is such a subsequence, then there are terms very close to 3/4. But since $\lim_{n \to \infty}|x_n - x_{n+1}| = 1$, there must be terms very close to $3/4 \pm 1$. This would be a contradiction because $x_n \in [0,3/2]$ | 356 | 1,032 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-38 | latest | en | 0.883167 |
https://www.ee-diary.com/2023/05/what-is-telegraphers-equation.html | 1,719,294,937,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00629.warc.gz | 639,923,858 | 49,424 | # What is Telegrapher's Equation
The Telegrapher's equations are a set of partial differential equations that describe the behavior of electrical signals traveling along a transmission line. They are widely used in the analysis and modeling of transmission lines, including homogeneous transmission lines like coaxial cables and parallel-plate transmission lines.
The Telegrapher's equations consist of two equations:
Voltage Equation (also known as the "Kirchhoff's voltage law" equation):
∂V(x,t)/∂x = -L ∂I(x,t)/∂t - R I(x,t)
This equation relates the voltage (V) along the transmission line with the current (I) flowing through it. The left side of the equation represents the rate of change of voltage with respect to the position (x) along the line, while the right side represents the effects of inductance (L) and resistance (R) on the current.
Current Equation (also known as the "Kirchhoff's current law" equation):
∂I(x,t)/∂x = -C ∂V(x,t)/∂t - G V(x,t)
This equation relates the current (I) along the transmission line with the voltage (V) across it. The left side represents the rate of change of current with respect to the position (x), while the right side represents the effects of capacitance (C) and conductance (G) on the voltage.
In these equations, x represents the position along the transmission line, t represents time, and L, R, C, and G are the per-unit-length inductance, resistance, capacitance, and conductance of the line, respectively.
By solving the Telegrapher's equations, various properties of the transmission line can be determined, such as the propagation velocity, impedance, reflection coefficient, and attenuation of electrical signals. These equations serve as fundamental tools for the analysis and design of transmission lines in various fields, including telecommunications, RF engineering, and signal integrity analysis.
References | 406 | 1,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-26 | latest | en | 0.935705 |
https://brainly.com/question/121507 | 1,485,183,006,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282932.75/warc/CC-MAIN-20170116095122-00568-ip-10-171-10-70.ec2.internal.warc.gz | 797,500,896 | 10,158 | # 1. What is 7.445 rounded to the nearest tenths? 2. What is 7.999 rounded to the nearest tenths? 3. What is \$5.68 rounded to the hearts ones? 4. What is 10.49 rounded to the nearest ones? 5. What is 2.499 rounded to the nearest hundredths? 6. What is 40.458 rounded to the nearest hundredths? 7. What is 5.4572 rounded to the nearest thousandths? 8. What is 45.0099 rounded to the nearest thousandths? 9. What is 9.56303 rounded to the nearest ten-thousandths? 10. What is 988.08055 rounded to the hearts ten-thousandths? 11. What is \$87.09 rounded to the besets tens? 12. What is 1,567.893 rounded to the nearest tens?
2
by Fashionista22435678
2014-09-17T16:28:33-04:00
When you are rounding to a certain place value, first identify the place value you are rounding to by underlining it. The digit to the immediate right of that number will determine whether you add 1 to the underlined digit or it remains the same.
For example....
1. 7.445 rounded to the nearest tenth. Underline the first 4 after the decimal and circle the 4 that is to the right. Answer this question....is the number that is circled 5 or greater? If the answer is YES, add one to the underlined digit. If the answer is NO, leave the underlined number the same and change all remaining digits to the right of it to ZERO. Most of the time, those ZEROs will simply vanish because they are unnecessary.....
7.445 = 7.4 rounded.
2. 7.999 rounded to the nearest tenth. This one is a bit trickier....but you'll still follow the same process. Underline the first 9 next to the decimal and circle the 9 immediately to the right. The answer to the question is yes, so you'll add one to the underlined digit. This will cause it to become a zero and you'll add 1 to your whole number. BUT....keep that 0 there because it's in the tenths place. All of the other digits will vanish.
7.999 = 8.0 rounded.
Follow that same process for the remainder of your questions and you will have mastered the art of ROUNDING!!!!
Hope that's clear and GOOD LUCK!
2014-09-19T07:15:17-04:00
Key: Always look at the back of what you are rounding to, to be able to round your numbers.
1.) 7.445 rounded to the nearest tenths is 7.4. How I got it was looking at what in the back of the number i'm rounding to.
2.) 7.999 rounded to the nearest tenths place is 8.0. How I got was looking at what in the back of the number i'm rounding to. The 9 in the hundredths place had pushed the 9 in the tenths place witch had also pushed the 7 in the ones place.( The tenths place is zero).
3.) \$5.68 rounded to the nearest hearts ones is \$6. How I got it was looking at what in the back of what i'm rounding to.
4.) 10.49 rounded to the nearest ones is 10. How I got it was looking at my tenths place. (The 4 couldn't push the number 0 to 1).
5.) 2.499 rounded to the nearest hundredths place is 2.50. How I got it was looking at what in the back of what i'm rounding to.
6.) 40.458 rounded to the nearest hundredths place is 40.46. How I got it was looking at what in the back of what i'm rounding to.
7.) 5.4572 rounded to the nearest thousandths place is 5.457. How I got it was looking at what in the back of what i'm rounding to.
8.) 45.0099 rounded to the nearest thousandths place is 45. How I got was looking at what in the back of what i'm rounding to. Because the 9 in the ten-thousandths place had moved the 9 i the thousandths place to a zero, it had cost all of the numbers from the tenths place to the ten-thousandths place to change into a zero.
9.) 9.56303 rounded to the nearest ten-thousandths place is 9.563.How I got it was looking at what in the back of what i'm rounding to. Because the three couldn't move up the 0, it is then 9.563.
10.) 988.08055 rounded to the hearts ten-thousandths place is 988.0806.How I got was looking at what in the back of what i'm rounding to.
11.) \$87.09 rounded to the besets tens place is \$9. How I got it was looking at what in the back of what i'm rounding to.( The seven had moved up the 8 to 9).
12.) 1,567.893 rounded to the nearest tens place is 1,570. How I got it was looking at what in the back of what i'm rounding to.( The 7 had rounded the 6 up to 7).
All underlined numbers are the regular number that is meant to be rounded. | 1,220 | 4,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-04 | latest | en | 0.904896 |
https://www.gradesaver.com/textbooks/math/statistics-probability/elementary-statistics-12th-edition/chapter-9-inferences-from-two-samples-9-2-two-proportions-beyond-the-basics-page-453/20 | 1,576,138,617,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540542644.69/warc/CC-MAIN-20191212074623-20191212102623-00260.warc.gz | 732,769,462 | 12,270 | ## Elementary Statistics (12th Edition)
We want to see if the mean is different to $\frac{1404}{2000}=0.702$, so we know $H_0:μ=0.702$ and $H1:μ\ne0.702$. We now use a Ti-84 calculator. To solve, go to "Stat," then "Tests," and then select "Stats." Doing this, we can find that the critical value and use a z-table calculator to find that $p\lt0.05$, hence the claim $p_1\ne p_2$ is supported. We use a ti-84 to find the confidence interval. To do this, we press “Stat” and then select “Tests.” Next, we select “2-PropZInt” to conduct the test. Doing this, we can find that the confidence interval goes through $0$, so we see that there is not a significant difference, meaning that $p_1=p_2$ is supported. Thus, we see that the two results are different. | 221 | 756 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-51 | latest | en | 0.908336 |
https://gateoverflow.in/866/gate-cse-2002-question-13 | 1,632,779,619,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058552.54/warc/CC-MAIN-20210927211955-20210928001955-00406.warc.gz | 314,754,883 | 28,245 | 3,907 views
1. In how many ways can a given positive integer $n \geq 2$ be expressed as the sum of $2$ positive integers (which are not necessarily distinct). For example, for $n=3$, the number of ways is $2$, i.e., $1+2, 2+1$. Give only the answer without any explanation.
2. In how many ways can a given positive integer $n \geq 3$ be expressed as the sum of $3$ positive integers (which are not necessarily distinct). For example, for $n=4$, the number of ways is $3$, i.e., $1+2+1, 2+1+1$ and $1+1+2$. Give only the answer without explanation.
3. In how many ways can a given positive integer $n \geq k$ be expressed as the sum of $k$ positive integers (which are not necessarily distinct). Give only the answer without explanation.
### Subscribe to GO Classes for GATE CSE 2022
1. $n= 2 \left(1+1\right),\;n=3\left(1+2, 2+1\right),\\n=4\left(1+3,3+1,2+2\right),\;n=5\left(1+4,4+1,2+3,3+2\right)$
so $x_1+x_2=n\;\text{and}\;x_1,x_2>{0}$ (no.of integral sol)
This is same as number of ways of putting $\left(n-2\right)$ (as we can't have $0$ for either $x_1$ or $x_2$) identical balls into two distinct bins, which is obtained by putting a divider across $\left(n-2\right)$ balls and taking all possible permutations with $\left(n-2\right)$ being identical. i.e., $\frac{(n-2 + 1)!}{(n-2)!} = (n-1).$
We can also use the following formula ,
$^{(n-2+2-1)}C_{(2-1)}=^{n-1}C_1.$
2. $n=3\left(1+1+1\right),\;n=4\left(1+1+2,1+2+1,2+1+1\right),\\ n=5\left(1+1+3,1+3+1,3+1+1,2+2+1,2+1+2,1+2+2\right)$
so $x_1+x_2+x_3=n\;\text{and}\;x_1,x_2,x_3>0$ (no.of integral sol)
Here, we can permute $\left(n-3\right)$ items with $2$ dividers which will give $\frac{(n-3 + 2)!}{(n-3)!2!}$
\begin{align}&=\frac{\left(n-1\right)!}{\left(n-1-2\right)!2!}\\\\&=\;^{n-1}C_2\end{align}
3. $^{(n-k+k-1)}C_{k-1}=^{n-1}C_{k-1}.$
edited by
reference :
### Theorem one
Suppose one has n objects (to be represented as stars; in the example below n = 7) to be placed into k bins (in the example k = 3), such that all bins contain at least one object; one distinguishes the bins (say they are numbered 1 to k) but one does not wish to distinguish the n stars (so configurations are only distinguished by the number of stars present in each bin; in fact a configuration is represented by a k-tuple of positive integers as in the statement of the theorem). Instead of starting to place stars into bins, one starts by placing the stars on a line:
★ ★ ★ ★ ★ ★ ★
Fig. 1: seven objects represented by stars
where the stars for the first bin will be taken from the left, followed by the stars for the second bin, and so forth. Thus the configuration will be determined once one knows what is the first star going to the second bin, and the first star going to the third bin, and so on. One can indicate this by placing k − 1 separating bars at some places between two stars; since no bin is allowed to be empty, there can be at most one bar between a given pair of stars:
★ ★ ★ ★ | ★ | ★ ★
Fig. 2: two bars give rise to three bins containing 4, 1, and 2 objects
Thus one views the n stars as fixed objects defining n − 1 gaps, in each of which there may or not be one bar (bin partition). One has to choose k − 1 of them to actually contain a bar; therefore there are $\tbinom {n-1}{k-1}$ possible configurations (see combination).
nice & simple......
$x_{1}+x_{2}= n$ where $x_{1},x_{2}> 0$ or $x_{1},x_{2} \geq 1$
So $x_{1}+1+x_{2}+1= n$
or, $x_{1}+x_{2}= n-2$
For those who r getting confused with n-2 .... like me :p
edited
*best explanation @ supromit roy*
In simple words, We can think of expressing an integer $n$ as a sum of $k$ positive integers as dividing a queue of stars using $k-1$ bars and the number of stars in any partition give that positive number.
$Example- \ n = 5 \ k =3$
$\star \star | \star \star | \star \rightarrow 5= 2+2+1$
In general, $n$ stars create $n-1$ gaps between them and we have to choose $k-1$ gaps for placing bars.
$\Large\star \_ \star\_ \star\_ \star\_ \star$
This can be done in $\binom{n-1}{k-1}$ ways.
I am trying to solve this using gen functions.
x1+x2=n . An integer is divided into two boxes where each box will contain a number.
LHS= $(x+x^{2}+x^{3}+x^{4}+x^{5}....)(x+x^{2}+x^{3}+x^{4}+x^{5}....)$
$x(1+x+x^{2}+x^{3}+x^{4}) \times x(1+x+x^{2}+x^{3}+x^{4})$
$x(1-x)^{^-1} \times x(1-x)^{^-1}$
$x{^2}(1-x)^{-2}$
$x^{2} \prod_{r=0}^{\propto }\binom{r+2-1}{2-1}x^{r}$
$x^{2} \prod_{r=0}^{\propto }(r+1)x^{r}$
$\prod_{r=0}^{\propto }(r+1)x^{r+2}$
RHS =$[x^{n}]$
solving r I get r=n-2.
So coefficient of x^{n} would be n-1
Is this correct way to do it as I have a doubt when I expand the gen function? Kindly help!!
We know that the no. of +ve integral solution to the equation $x_{1} + x_{2} + x_{3}+ ... + x_{k} = n$ is given by $\binom{n-1}{k-1}$.
(a) Let, $x_{1}$ and $x_{2}$ be the two positive integers (not necessarily distinct) such that their sum is equal to $n\ (n≥2)$.
Therefore, $x_{1} + x_{2} = n$.
Now, we have to find all such values of $x_{1}$ and $x_{2}$ that satisfy the above equation. In other words, the problem basically reduces to finding the no. of +ve integral solution to the above equation.
Therefore, the req. no. of ways is $\binom{n-1}{2-1} = \binom{n-1}{1}$.
(b) Let, $x_{1}$, $x_{2}$ and $x_{3}$ be the three positive integers (not necessarily distinct) such that their sum is equal to $n\ (n≥3)$.
Therefore, $x_{1} + x_{2} + x_{3} = n$.
Now, we have to find all such values of $x_{1}$, $x_{2}$ and $x_{3}$ that satisfy the above equation. In other words, the problem basically reduces to finding the no. of +ve integral solution to the above equation.
Therefore, the req. no. of ways is $\binom{n-1}{3-1} = \binom{n-1}{2}$.
(c) Let, $x_{1}$, $x_{2}$, $x_{3}$,...,$x_{k}$ be the $k$ positive integers (not necessarily distinct) such that their sum is equal to $n\ (n≥k)$.
Therefore, $x_{1} + x_{2} + x_{3}+ ... + x_{k} = n$.
Now, we have to find all such values of $x_{1}$, $x_{2}$, $x_{3}$,...,$x_{k}$ that satisfy the above equation. In other words, the problem basically reduces to finding the no. of +ve integral solution to the above equation.
Therefore, the req. no. of ways is $\binom{n-1}{k-1}$.
(a) Total no of ways = C(n-1 ,1) = n-1
(b)Total no of ways = C(n-1 ,n-3) = C(n-1 ,2) = (n-1)⨉(n-2) /2
(c)Total no of ways = C(n-k+k-1 ,n-k) = C(n-1 , n-k) = C(n-1 , k-1)
by
SIMPLY APPLY THE LOGIC:
x1+x2+x3+x4+...........xn=r where x1,x2.....xn>=0 apply c(n+r-1,r)
make similarly all the 3 cases in order that it satisfies the equation
a- x1+x2=n where x1,x2>=1 so make it x1+x2=n-2 now x1,x2>=0 so c(n-2+2-1,1)=n-1
b-x1+x2+x3=n where x1,x2,x3>=1 so x1+x2+x3=n-3 so c(n-3+3-1,2)=c(n-1,2)
c-x1+x2+.....xk=n where x1,x2,x3....xk>=n so x1+x2+x3+....xk=n-k=c(n-k+k-1,k-1)=c(n-1,k-1)
For Distinct k positive integers : (n+k-1)C(k-1)
For Non-Distinct k positive integers : (n-1)C(k-1)
Here, i solved for the generalized case i.e. part c of the question.
Here what we have to do is
a) Let the 2 numbers is x1 and x2
Now it is given that $x1+x2 \ge n$
x1 and x2 can be anything greater than 0 such that $x1+x2 \ge n$
minimum value of x1 and x2 is 1 so 1 is a value which is mandatory for x1 and x2
so, $x1+x2 \ge n$ can be expressed as 10000.... no. of 0s in n-2 times 1 acts as a partition between x1 and x2 . so, no. of
ways = $^{n-2+1}C_1 = n-1$
b) $x1+x2+x3\ge n$ for this it will be $^{n-3+2}C_{2} \implies ^{n-1}C_{2}$
c) similarly for k variables it will be $^{n-1}C_{k-1}$ | 2,777 | 7,503 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-39 | latest | en | 0.720177 |
https://mathspace.co/textbooks/syllabuses/Syllabus-1014/topics/Topic-20189/subtopics/Subtopic-265978/?textbookIntroActiveTab=overview | 1,637,966,922,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358074.14/warc/CC-MAIN-20211126224056-20211127014056-00041.warc.gz | 477,129,477 | 56,153 | Victorian Curriculum Year 10A - 2020 Edition
11.07 Investigating data with scatter plots
Lesson
### Interpreting bivariate data
When we have bivariate data, we want to determine what sort of relationship the two variables have. As the independent variable ($x$x-axis) changes notice how the dependent variable ($y$y-axis) tends to change. Just by observation, we may notice the following:
• A simple relationship: if the distribution of points appears to follow a trend either linear or non-linear depending on if the points appear to follow the shape of a line or not.
Consider being given an $x$x-value that doesn't correspond to any data point we have. Does the data set give us an idea of what $y$y-value that point should have to fit in with the rest of the data? If yes there might be a relationship. If not, there might be no relationship between the variables.
• Outliers: in a scatterplot, any data points that are very different from the other data points will be quite obvious especially if the rest of the points appear to have a relationship.
Causal relationships
Even when two variables have a relationship, it may not be a causal relationship. We cannot say for sure that a change in the value of $x$x causes $y$y to change or that the value of $y$y causes a corresponding value of $x$x even when a relationship is apparent. It may be that both $x$x and $y$y have a relationship with some other hidden variable, which creates an indirect relationship between $x$x and $y$y.
#### Practice questions
##### question 1
The scatter plot shows the relationship between sea temperature and the amount of healthy coral.
1. Which variable is the dependent variable?
Sea temperature
A
Level of healthy coral
B
Sea temperature
A
Level of healthy coral
B
2. Which variable is the independent variable?
Sea temperature
A
Level of healthy coral
B
Sea temperature
A
Level of healthy coral
B
##### question 2
The scatterplot shows the height and weight of six students. Use the plot to answer the questions.
1. How tall is the student who weighs $45$45kg?
2. How tall is the tallest student?
##### question 3
The price of ten houses is graphed against the house's land area below.
Describe the relationship between land area and house price in the data.
1. As land area increases, house price decreases.
A
As land area increases, house price increases.
B
Land area has no effect on house price.
C
As land area increases, house price decreases.
A
As land area increases, house price increases.
B
Land area has no effect on house price.
C
### Outcomes
#### VCMSP352
Use scatter plots to investigate and comment on relationships between two numerical variables
#### VCMSP353
Investigate and describe bivariate numerical data where the independent variable is time. | 628 | 2,810 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2021-49 | latest | en | 0.863996 |
null | null | null | null | null | null | Date: Mon, 12 Apr 2010 13:01:42 -0500 From: Sean Josh Subject: josh's summer job 14 This is an explicit fictional account of homoerotic behavior. If you are offended by such material or access to it is banned in your community read no further! The characters and situations portrayed are purely fictional and therefor(e) have no fear of catching or passing on STDs. If you are sexually active, play safe and use protection. Comments and criticism of this work are welcome. Email: Josh's Summer Job 14 The following morning Billy's phone rang. It was Josh's mother again. She seemed somewhat concerned. Josh hadn't been home, that didn't bother her, but he hadn't called and that was unusual. She'd tried his cell phone, but it was still off. Billy had little to tell her, but promised he'd have Josh call if he heard from him. He called Danny and picked him up on the way to Jamie's, but Josh was nowhere to be found. The truck was still parked in the back lot of the high school. They searched it carefully. Josh's calendar was in the pocket of his shirt. A small address book was clipped to it. The calendar detailed the fixed appointments, the address book held a broad number of customers and friends. By cross referencing the two they slowly began to see the differences between clients and customers. "Look here," Billy said. "He's got Mr. Clark down for lawn service twice a week all summer." "Yeah so, Officer Clark is anal about his lawn?" Danny said. "Clark's yard is no bigger than ours. He's got Clark down for a two hour appointment twice a week, and our yard is only an hour, and then we're on call. You've seen him work, he doesn't dawdle." "So how's that help?" "I don't know, at least we can figure out who his special clients are, maybe he went to one of them?" "And if he did what then? We knock on the door and ask, `Hi, we were wondering if Josh was here fucking you?' That'll work." "There are some names that are crossed out." "Okay, so people move, and hire different companies, what's that prove?" The different notations weren't the only thing he noticed. There were notations on the calendar for billing. He noticed that his family paid a lower rate than many of Josh's customers. >>>>>>>>>>>>>>>>>>> The room was pitch dark. Josh grunted in time with the machine that pounded his wide splayed ass. He had no idea how long he'd been here. His entire body ached. Sweat dripped from every pour and he was exhausted. The lights blazed to life. Voices, voices filled the room. They broke through the haze created by the persistent pulsing vibration ravaging his frame. A man in leather chaps appeared before him. A fat cock dangled between his thighs. "He's a beauty, where'd ya find him." The machine was removed. Even that simple act made Josh's body quiver. The hose was employed to wash away the crust of slime that covered his legs and the pommel horse. The welts on his back and ass still ached. It would be a week before his anus would recover from the pounding it received. "That is a perfect ass." "Help yourself. It's why I called ya." The chaps moved up and smeared his greasy cock across Josh's face. "Better be careful, he's not as well trained as he once was, he may bite." "You won't bite me will ya pretty boy." Josh accepted the offered cock. As he did a tongue slithered across his balls and up into his crack. It delved into his gaping hole. "I'm sorry Billy." "Jman, I wish you wouldn't ream `em so with that damn machine. We ain't all hung like you ya know." "We can fix that." The man he serviced had grabbed his hair and was beginning to fuck his face. His cock was thick, but of average length. He could manage the fat mushroom head popping in and out of this throat so long as he could focus. There was chuckling behind him. "What the fuck is that." "Made molds from all the attachments for the machine." "I'm sorry Billy" Something cold touched Josh's ass. The sensation broke his concentration and he gagged. "Yeah baby, take it." The man fucking his face growled. The cold sensation moved into his gaping anus. The ice made the tortured flesh shrink and tighten. It hurt just as much as when it was forced open, and the welts of the beatings he'd taken ached as his muscles clenched. Around the room the monitors played a variety of video; they all featured Josh. Josh at twelve, Josh at thirteen, at the beach, playing in the yard, being yelled at, spanked, restrained, strapped to the pommel horse, in the sling, tied to the wall. An X shaped cross constructed of two by tens was bolted to the wall in front of the pommel horse. He was tied to the cross. Thick ropes coiled tightly around his arms, legs and torso. A heavy weight hung from his balls and clamps were applied to his tits. A thick metal butt plug was shoved into his greasy hole. Wires trailed away from it and were attached to a set of batteries. The electrical current coursed through his gut forcing his ass to clamp tight around the intrusion. His limp cock was slid into a thick Plexiglas tube. A rubber collar on the device was pressed firmly into his hairless groin. The vacuum pump was engaged as his sizable endowment thickened and stretched until it pushed the tube away from his groin breaking the vacuum. `We're gonna need the big sleeve for this bitch.' `He's hung like you Jay. Gonna be a major league stud some day.' `Let him off the wall and he'll show ya some day.' They employed the vacuum pump until his cock reached monumental dimensions, and the festivities began. The boy's long thick pole was to be the focal point; the purpose to make him cum, keep him hard, extend orgasm. They licked and sucked and teased his aching cock. He'd no sooner fill one mouth with his adolescent spunk than another hungry cocksucker would stoop to continue the process. He raved, and groaned and begged for respite. They tortured him with pleasure, and when that no longer worked, they bound his genitals with cords and wires, slapping it, buffing it, grinding it until he raved incoherently. Occasionally he'd pass out, and they'd revive him with salts or the hose. Tears and sweat blurred his vision. He no longer focused on the noise around him. Not the taunts of his tormentors or his own incoherent cries and pleading. "Who's next?" A hairy middle aged man panted as he pulled free of Josh's sopping ass. The sound brought Josh to his senses. They'd fucked him with the ice sculpture until it melted away, and applied a metal butt plug hooked to electrodes to finish tightening his overworked hole. Once the process was complete they drew numbers and fucked him in turn. A river of cum oozed from his hole over his balls to streak his legs and the pommel horse. One of the old queens was slurping at the cum cocktail that poured from him. "Everybody get a chance to breed this pig?" "I'm sorry Billy" It had been his mantra through the endless haze of sex and torture that had encompassed him since arriving. They dragged him from the pommel horse. The icy pinpricks of the pressure hose cleared his vision and restored some semblance of reason. He was rinsed inside and out. They stuffed him into a small cage in the corner. There was room enough for him to sit with his knees drawn up. He remembered the cage being larger. Food and water were delivered. He gargled and spat, then wolfed the steaming bowl of food which proved to be very spicy. He gulped the water knowing it would do little to ease the fire and then fell into and uneasy sleep. The throbbing between his legs woke him. He was painfully erect, and understood immediately what had happened. The two old queens in the group sat watching. They smiled when he woke. "Jay said we could have you for a bit," said the heavy set man. "So we've decided to give you a choice, so long as you don't tell Jay." Josh didn't say a word, but watched them warily. "You'll be a good boy won't you?" Harold said dangling the key to his cage. Josh nodded. They set him free. He stood and stretched for what felt like the first time in days. "He's magnificent." The fat man cooed running a hand over Josh's torso. "What do you want?" "To fuck." "Fine how do you want me?" Josh said his eyes projecting a sorrow that they both perceived. "No, no, no dear boy, fuck us." Harold said feeling naughty. "In the sling." The fat man added. Josh grimaced as Harold stroked his aching cock. The fat man used a chair to mount the sling. The leather and beams groaned under his weight. Harold led Josh to him and greased his partner's hole. "No holding back now. We can see the man you are behind those gorgeous blue eyes." Josh grunted as the friction from the fat man's blubbery butt and tight hole increased the pain in his tender cock. He grabbed the chains from which the sling hung and pulled it toward him as he stiffened his hips and presented his drug induced hardon. The fat man grunted as the hard meat entered him. Harold took the opportunity to caress Josh's highly defined frame. The stress, physical demands, and dehydration had combined to leave his long lean frame defined to levels rarely seen. He traced the welts that marked his muscled back and ribs and caressed the boy's flexing ass. Once Josh was firmly in place he labored, slowly mastering his pain. The effort brought relief. Moving freely, his aching joints and muscle began to relax into the practiced motion. His mind settled, the semblance of control fed his recovery. He reared back and began delivering a brutal fucking. The sling swung wildly, the chains and leather creaking and rattling as he hammered into their occupant. The fat man grunted his approval. "That what you want?" Josh grinned viciously. He set his rhythm to meet the fat man's swinging ass, pounding relentlessly into the gaping aperture. Harold grinned as the young man he'd helped to demean seemed to be taking his revenge. He grew bolder and more determined. "He's a stud," the fat man cried. "It's so big, bigger than Jay." A fountain of cum shot up from between his chubby legs. Josh didn't seem to notice. He flailed into the clutching hole, his hips blurring, the sweat pouring down his powerful back. "Don't use him up," Harold complained. "Use me up?" Josh said pulling suddenly free of the fat man and turning on Harold. The angry red cock swaying before him looked more like weapon than a pleasurable tool. He stalked Harold. Padding soft and sure until the man had backed into the pommel horse. Fear and passion confused the expression on Harold's face. Josh pressed him against the horse. He was a head taller than the man. His stiff cock pressed into the man's belly and chest. "You scared Harold?" "Yes." "You should be." Josh grabbed him beneath the arms and flipped him face down onto the pommel horse. He fisted his erection and jammed home into the little man. Harold grunted and squealed as the thick pole shot into him. Josh fucked him with wild abandon, his strength mounting, his body reveling in the movement, his libido once more in control. Orgasm came, Harold discharging a spray of cum unlike any he'd experienced in years. Josh stepped back allow him to slump to the floor. The fat man was still nearby jerking off. Josh strode to him and took hold of his thinning hair and presented his shit slick tool. "Clean me faggot." The fat man rushed to comply. He was a consummate cocksucker. Josh had known him for years. He sucked Josh prodigious endowment tip to root leaving it gleaming in a coat of saliva and reminding Josh what good head felt like. Exhaustion was catching up with him, he could feel the strength that the drug and situation had given him flagging. He lifted an iron bar that was used for restraint; two cuffs had been welded to its wide spread ends. He swung it testing its weight. Harold scuttled away looking for sanctuary. The fat man simply stroked faster as his eyes widened. Josh swung the pole in a hard sideward stroke. It smashed the flywheel of the machine that had tormented him. Harold fled. The machine lay on its side on the floor. Josh swung the improvised cudgel until the device was a crumbled ruin. He turned to vent his anger on the electric fencing device to find his father, hands on hips, watching angrily from the doorway behind the cage. "Put down the fuckin' bar and get back in your cage." "Let me out or I'll split your skull." "You don't have the balls." "You should know," Josh said hurling the bar. The man dove behind the door. The bar crushed the case of the fencing device. It flashed and sputtered raining sparks across the room. A deep buzz filled the space followed by the smell of ozone. The lights flickered, and several of the monitors flashed before the electricity flailed completely. Emergency lighting came on line. "I'm gonna beat you `til you bleed." His father growled as he entered the room swinging a strap. Several shadows huddled in the doorway behind him. "I've missed this." Josh said distinctly. The man stopped. He drew several deep breaths slowly mastering his anger and stepped aside. "You always cost more than you're worth." He hissed as Josh passed. Getting out of his clothes gracefully was never difficult, getting back into them posed a far more difficult problem. Luckily Josh had been wearing little when he arrived. He slipped on his jeans and stepped into his sneakers. He handed Harold his sweat stained boxers, and turned to leave. The climb up these stairs was always a liberating experience. It was evening when he emerged into the air. He wasn't sure what day it was. His body and soul ached. He walked off into the shadows as he had arrived. The High School was close by, he hoped his truck was still there as he started on the lonely walk. | null | null | null | null | null | null | null | null | null |
https://www.physicsforums.com/threads/if-f-x-0-for-every-bounded-linear-map-f-is-x-0.479896/ | 1,606,412,433,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188899.42/warc/CC-MAIN-20201126171830-20201126201830-00267.warc.gz | 798,972,929 | 18,008 | # If f(x) = 0 for every bounded linear map f, is x = 0?
Suppose you're looking at a complex vector space X, and you know that, for some x in X, you have f(x) = 0 for every linear map on X. Can you conclude that x = 0? If so, how?
This seems easy, but I can't think of it for some reason.
(EDIT: Assume it holds for every CONTINUOUS (i.e., bounded) linear map on X. This may, or may not, make a difference.)
Last edited:
Fredrik
Staff Emeritus
Gold Member
The identity map is linear and bounded.
Well, you can take f the identity map on X. Then f(x)=0 means that x=0. But I'm quite sure that this is not what you meant...
Well the identity function is a bounded linear map, so yes.
If you mean bounded linear functional, this is still true by Hahn-Banach, because we can send x to ||x||, and send each vector of the form ax to a||x||, where a is in C. By Hahn-Banach, this extends to a bounded linear functional.
Well the identity function is a bounded linear map, so yes.
If you mean bounded linear functional, this is still true by Hahn-Banach, because we can send x to ||x||, and send each vector of the form ax to a||x||, where a is in C. By Hahn-Banach, this extends to a bounded linear functional.
Yes, perhaps I've phrased it badly. I'm talking about X being a normed space, and I'm talking about f(x) = 0 for every bounded linear functional $f: X \to \mathbb C$.
Perhaps I screwed up by writing "map" instead of "functional"...sorry, guys
Well the identity function is a bounded linear map, so yes.
If you mean bounded linear functional, this is still true by Hahn-Banach, because we can send x to ||x||, and send each vector of the form ax to a||x||, where a is in C. By Hahn-Banach, this extends to a bounded linear functional.
So, just following up...if one has $f(x) = 0$ for every bounded linear functional $f: X \to \mathbb C$, then we have $x = 0$, but one has to invoke Hahn-Banach to show that? Ok...no wonder I couldn't think of it
Fredrik
Staff Emeritus
Gold Member
I didn't think of it either, but it's a really nice solution. Definitely the simplest one. I have a more complicated one that only works for Hilbert spaces, if you're interested. It's based on the fact that there's a unique vector $x_f$ such that the map $y\mapsto\langle x_f,y\rangle$ is equal to f (the Riesz representation theorem). I'm also using that $(\ker f)^\perp=\mathbb Cx_f$.
Edit: I thought of a way to avoid the Hahn-Banach theorem, but it requires an inner product. Ebola's solution goes like this: Suppose that $x\neq 0$. Then $ax\mapsto a\|x\|$ is a bounded linear functional on the 1-dimensional subspace $\mathbb Cx$. Let's call this functional $f_0$. By the Hahn-Banach theorem, there's a linear functional $f:X\rightarrow\mathbb C$ such that $f|_{\mathbb Cx}=f_0$ and $\|f\|=\|f_0\|$. Since $f(x)=f_0(x)=\|x\|\neq 0$, we have a contradiction.
If there's an inner product, we don't have to invoke Hahn-Banach. We just define f explicitly: $f(y)=\langle x,y\rangle$.
Last edited:
Landau
It shouldn't come as a surprise that Hahn-Banach is needed. For an arbitrary normed vector space, it isn't even clear a priori that the dual spaces is non-trivial. If it were, then your condition would be satisfied trivially! You use Hahn-Banach to show that the dual space is non-trivial.
mathwonk
Homework Helper
this seems non trivial as hahn banach uses the axiom of choice as i recall.
Fredrik
Staff Emeritus
Gold Member
Yes, it's definitely not trivial if we're dealing with an arbitrary normed space or Banach space. However, if there's an inner product on the space, we can use the inner product instead of the Hahn-Banach theorem.
If we're dealing with an arbitrary normed space X, I guess we would have apply the argument based on the Hahn-Banach theorem to the completion of X.
The proof I've seen of the Hahn-Banach theorem relies on Zorn's lemma, which is equivalent to the AoC.
Landau
@Fredrik: Hahn-Banach is already a result about normed vector spaces, not necessarily complete. So no need to consider the completion.
Hahn-Banach is strictly weaker than AoC: in ZF, AoC implies H-B, but not the other way around. | 1,144 | 4,140 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-50 | latest | en | 0.951884 |
https://www.dailycodebuffer.com/return-matrix-in-spiral-order-java/ | 1,716,160,249,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00022.warc.gz | 637,335,515 | 21,259 | # Return matrix in Spiral order (Java)
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
For example, given the following matrix:
```[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
```
You should return [1,2,3,6,9,8,7,4,5].
### Java Solution 1
If more than one row and column left, it can form a circle and we process the circle. Otherwise, if only one row or column left, we process that column or row ONLY.
```public class Solution {
public ArrayList<Integer> spiralOrder(int[][] matrix) {
ArrayList<Integer> result = new ArrayList<Integer>();
if(matrix == null || matrix.length == 0) return result;
int m = matrix.length;
int n = matrix[0].length;
int x=0;
int y=0;
while(m>0 && n>0){
//if one row/column left, no circle can be formed
if(m==1){
for(int i=0; i<n; i++){
}
break;
}else if(n==1){
for(int i=0; i<m; i++){
}
break;
}
//below, process a circle
//top - move right
for(int i=0;i<n-1;i++){
}
//right - move down
for(int i=0;i<m-1;i++){
}
//bottom - move left
for(int i=0;i<n-1;i++){
}
//left - move up
for(int i=0;i<m-1;i++){
}
x++;
y++;
m=m-2;
n=n-2;
}
return result;
}
}
```
Similarly, we can write the solution this way:
```public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> result = new ArrayList<Integer>();
if(matrix==null||matrix.length==0||matrix[0].length==0)
return result;
int m = matrix.length;
int n = matrix[0].length;
int left=0;
int right=n-1;
int top = 0;
int bottom = m-1;
while(result.size()<m*n){
for(int j=left; j<=right; j++){
}
top++;
for(int i=top; i<=bottom; i++){
}
right--;
//prevent duplicate row
if(bottom<top)
break;
for(int j=right; j>=left; j--){
}
bottom--;
// prevent duplicate column
if(right<left)
break;
for(int i=bottom; i>=top; i--){
}
left++;
}
return result;
}
```
### Java Solution 2
We can also recursively solve this problem. The solution’s performance is not better than Solution 1. Therefore, Solution 1 should be preferred.
```public class Solution {
public ArrayList<Integer> spiralOrder(int[][] matrix) {
if(matrix==null || matrix.length==0)
return new ArrayList<Integer>();
return spiralOrder(matrix,0,0,matrix.length,matrix[0].length);
}
public ArrayList<Integer> spiralOrder(int [][] matrix, int x, int y, int m, int n){
ArrayList<Integer> result = new ArrayList<Integer>();
if(m<=0||n<=0)
return result;
//only one element left
if(m==1&&n==1) {
return result;
}
//top - move right
for(int i=0;i<n-1;i++){
}
//right - move down
for(int i=0;i<m-1;i++){
}
//bottom - move left
if(m>1){
for(int i=0;i<n-1;i++){
}
}
//left - move up
if(n>1){
for(int i=0;i<m-1;i++){ | 789 | 2,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-22 | longest | en | 0.335947 |
null | null | null | null | null | null | Growth and Development
“Development” and “growth” are sometimes used interchangeably in conversation, but in a botanical sense they describe separate events in the organization of the mature plant body.
Development is the progression from earlier to later stages in maturation, e.g. a fertilized egg develops into a mature tree. It is the process whereby tissues, organs, and whole plants are produced. It involves: growthmorphogenesis (the acquisition of form and structure), and differentiation. The interactions of the environment and the genetic instructions inherited by the cells determine how the plant develops.
Growth is the irreversible change in size of cells and plant organs due to both cell division and enlargement. Enlargement necessitates a change in the elasticity of the cell walls together with an increase in the size and water content of the vacuole. Growth can be determinate—when an organ or part or whole organism reaches a certain size and then stops growing—or indeterminate—when cells continue to divide indefinitely. Plants in general have indeterminate growth.
Differentiation is the process in which generalized cells specialize into the morphologically and physiologically different cells described in Table 1 . Since all of the cells produced by division in the meristems have the same genetic make up, differentiation is a function of which particular genes are either expressed orrepressed. The kind of cell that ultimately develops also is a result of its location: Root cells don't form in developing flowers, for example, nor do petals form on roots.
Mature plant cells can be stimulated under certain conditions to divide and differentiate again, i.e. to dedifferentiate. This happens when tissues are wounded, as when branches break or leaves are damaged by insects. The plant repairs itself bydedifferentiating parenchyma cells in the vicinity of the wound, making cells like those injured or else physiologically similar cells.
Plants differ from animals in their manner of growth. As young animals mature, all parts of their bodies grow until they reach a genetically determined size for each species. Plant growth, on the other hand, continues throughout the life span of the plant and is restricted to certain meristematic tissue regions only. This continuous growth results in:
• Two general groups of tissues, primary and secondary.
• Two body types, primary and secondary.
• Apical and lateral meristems.
Apical meristems, or zones of cell division, occur in the tips of both roots and stems of all plants and are responsible for increases in the length of the primary plant body as the primary tissues differentiate from the meristems. As the vacuoles of the primary tissue cells enlarge, the stems and roots increase in girth until a maximum size (determined by the elasticity of their cell walls) is reached. The plant may continue to grow in length, but no longer does it grow in girth. Herbaceous plants with only primary tissues are thus limited to a relatively small size.
Woody plants, on the other hand, can grow to enormous size because of the strengthening and protective secondary tissues produced by lateral meristems, which develop around the periphery of their roots and stems. These tissues constitute the secondary plant body. | null | null | null | null | null | null | null | null | null |
https://www.york.ac.uk/students/studying/manage/programmes/module-catalogue/module/MAT00053M/2022-23 | 1,685,257,318,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643585.23/warc/CC-MAIN-20230528051321-20230528081321-00306.warc.gz | 1,214,076,575 | 7,764 | Accessibility statement
# Partial Differential Equations I - MAT00053M
« Back to module search
• Department: Mathematics
• Module co-ordinator: Dr. Konstantin Ilin
• Credit value: 10 credits
• Credit level: M
• Academic year of delivery: 2022-23
## Related modules
• None
### Prohibited combinations
Pre-requisite modules: MSc Mathematical Sciences students: knowledge of Vector calculus and elementary complex function theory.
## Module will run
Occurrence Teaching cycle
A Autumn Term 2022-23
## Module aims
A partial differential equation (PDE) is a differential equation that contains an unknown function and its partial derivatives. PDEs are used to describe a wide range of natural processes. Examples include fluid mechanics, elasticity theory, electrodynamics, quantum mechanics, etc. PDEs also play an important role in other areas of mathematics such as analysis and differential geometry.
The aim of this course is to give an introduction to the basic properties of PDEs and to the basic analytical techniques to solve them.
## Module learning outcomes
At the end of the module students should:
• Be able to determine the type of a second order PDE
• Be able to solve simplest first order PDEs
• Understand what are well-posed initial (and/or boundary) value problems for classical PDEs such as the wave equation, the Laplace equation and the heat (diffusion) equation
• Know basic analytical techniques for solving the above classical equations
## Module content
Syllabus
• Introduction: what is a PDE, first-order linear PDEs, initial and boundary conditions, well-posed problems, types of second-order PDEs.
• Wave equation: d'Alembert’s formula, causality and energy, reflection of waves.
• Heat (diffusion) equation: maximum principle, heat equation on the whole line and on the half-line
• Laplace equation: maximum principle, Poisson’s formula, rectangular domain.
• Periodic solutions for wave and heat equations. (This topic is not taught in the H-level variant of this module.)
• Academic skills: the techniques taught are used in many areas of pure and applied mathematics.
• Graduate skills: through lectures, examples, classes, students will develop their ability to assimilate, process and engage with new material quickly and efficiently. They develop problem solving-skills and learn how to apply techniques to unseen problems. Students on this module will learn to work more independently and assimilate advanced material at a greater rate than those on the H-level variant.
## Assessment
Task Length % of module mark
Closed/in-person Exam (Centrally scheduled)
Partial Differential Equations I
2 hours 100
None
### Reassessment
Task Length % of module mark
Closed/in-person Exam (Centrally scheduled)
Partial Differential Equations I
2 hours 100
## Module feedback
Current Department policy on feedback is available in the student handbook. Coursework and examinations will be marked and returned in accordance with this policy. | 629 | 2,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-23 | latest | en | 0.840454 |
https://www.clutchprep.com/chemistry/practice-problems/143684/the-barometer-shown-here-measures-atmospheric-pressure-in-millimeters-of-mercury | 1,638,744,960,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363226.68/warc/CC-MAIN-20211205221915-20211206011915-00276.warc.gz | 761,499,871 | 32,668 | Pressure Units Video Lessons
Concept:
# Problem: The barometer shown here measures atmospheric pressure in millimeters of mercury (mmHg). According to this barometer, what is the atmospheric pressure P in inches of mercury (inHg)?Express your answer in inches of mercury using five significant figures.Air pressure is measured by a barometer. One type of barometer utilizes an evacuated, calibrated tube inverted in a pool of mercury. The pool of mercury is open to the atmosphere. As air pressure increases, the mercury will rise in the column until the weight of the mercury in the column equals the force exerted by the air on the surface of the pool. The difference in height between the top of the mercury in the column and level of mercury in the pool is directly proportional to air pressure and quantitatively measures air pressure. Such columns are often calibrated in units of millimeters and thus the pressure read from such a column is reported as mm of Hg, abbreviated mmHg, since the liquid in the column is mercury.Units of pressureThe column of a mercury barometer is calibrated in linear units (such as inches or millimeters) and thus the air pressure observed on such a barometer is reported in terms of these linear units. This measurement of pressure can be converted to other common pressure units. Some conversion factors are shown here.1 mmHg = 1 torr1 atm = 760 mmHg1 atm = 1.01325 bar1 Pa = 10−5 bar1 in = 2.54 cm1 cm = 10 mm
###### FREE Expert Solution
96% (297 ratings)
###### Problem Details
The barometer shown here measures atmospheric pressure in millimeters of mercury (mmHg). According to this barometer, what is the atmospheric pressure P in inches of mercury (inHg)?
Air pressure is measured by a barometer. One type of barometer utilizes an evacuated, calibrated tube inverted in a pool of mercury. The pool of mercury is open to the atmosphere. As air pressure increases, the mercury will rise in the column until the weight of the mercury in the column equals the force exerted by the air on the surface of the pool. The difference in height between the top of the mercury in the column and level of mercury in the pool is directly proportional to air pressure and quantitatively measures air pressure. Such columns are often calibrated in units of millimeters and thus the pressure read from such a column is reported as mm of Hg, abbreviated mmHg, since the liquid in the column is mercury.
Units of pressure
The column of a mercury barometer is calibrated in linear units (such as inches or millimeters) and thus the air pressure observed on such a barometer is reported in terms of these linear units. This measurement of pressure can be converted to other common pressure units. Some conversion factors are shown here.
1 mmHg = 1 torr
1 atm = 760 mmHg
1 atm = 1.01325 bar
1 Pa = 10−5 bar
1 in = 2.54 cm
1 cm = 10 mm | 641 | 2,866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-49 | latest | en | 0.894707 |
https://www.nagwa.com/en/videos/430185671327/ | 1,721,058,991,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00680.warc.gz | 798,018,274 | 34,578 | Question Video: Using Objects to Decompose Numbers up to 10 in More Than One Way | Nagwa Question Video: Using Objects to Decompose Numbers up to 10 in More Than One Way | Nagwa
# Question Video: Using Objects to Decompose Numbers up to 10 in More Than One Way Mathematics • First Year of Primary School
There are 10 fish. Fill in the numbers to find another way to make 10.
02:51
### Video Transcript
There are 10 fish. Fill in the numbers to find another way to make 10.
In this question, we can see two additions shown using pictures. And both additions make a total of 10. How do we know this? Well, because the first sentence tells us there are 10 fish, but also because we can see that the final picture in our number sentences or equations shows the number 10. Our first number sentence is complete. And if we look carefully, we can see that each picture’s labeled, but we can count the fish just to check. One, two, three, four, five, six, seven plus one, two, three equals one, two, three, four, five, six, seven, eight, nine, 10. Seven and three are a pair of numbers that go together to make 10.
And we could model this in different ways. For example, we could split up a ten frame to show seven plus three. Or if we had 10 counting beads on a string, we could move three to the other end, to show that a group of seven and a group of three go together to make 10.
Now, if we look at our second addition, we can see that some of the numbers are missing. What plus what equals 10? The question asks us to fill in the numbers to find this other way to make 10. How many fish are there in our first picture? Let’s start at the top and work our way down. One, two, three, four, five fish. Now, what do we add to five to make 10?
We could find the answer by counting the fish in the second picture. But let’s use our models to help. If we have five pink counters, how many orange counters are we going to need? We’ll need the same number of orange counters to make a row underneath. In other words, we’re going to need another five. Can we model five plus five using our beads? Yes, we can. And if we count the fish in our second picture from top to bottom, we have one, two, three, four, five fish altogether.
Just because the fish in the second picture make a different shape doesn’t mean there’s a different number of them. They’re just arranged differently. So, as well as seven plus three, five plus five equals 10. Our missing numbers are five and five.
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# There was a trial going on in a laboratory. Students were doing an
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There was a trial going on in a laboratory. Students were doing an experiment. This experiment showed that the probability of a bird flies away when the cage is opened is 7/10. It was also observed that the bird that flew away also comes back is 1/5.what is the probability that when the cage is opened a bird not only flies away but also comes back.
a. 7/10
b. 3/10
c. 1/5
d. 7/50
e. None of the above
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Originally posted by GeorgeKo111 on 16 Jul 2019, 11:51.
Last edited by GeorgeKo111 on 16 Jul 2019, 13:57, edited 1 time in total.
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16 Jul 2019, 13:23
Is it a DS question?
Posted from my mobile device
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16 Jul 2019, 13:58
Shrey9 wrote:
Is it a DS question?
Posted from my mobile device
no it is a PS.
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Re: There was a trial going on in a laboratory. Students were doing an [#permalink]
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16 Jul 2019, 14:34
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There was a trial going on in a laboratory. Students were doing an experiment. This experiment showed that the probability of a bird flies away when the cage is opened is 7/10. It was also observed that the bird that flew away also comes back is 1/5.what is the probability that when the cage is opened a bird not only flies away but also comes back.
In order for two probabilities to happen, we need to multiply them by each other: P1 * P2 = P
7/10 * 1/5 = 7/50
Re: There was a trial going on in a laboratory. Students were doing an [#permalink] 16 Jul 2019, 14:34
Display posts from previous: Sort by | 853 | 3,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-35 | longest | en | 0.950743 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-5-linear-functions-5-1-rate-of-change-and-slope-practice-and-problem-solving-exercises-page-296/22 | 1,534,912,548,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219469.90/warc/CC-MAIN-20180822030004-20180822050004-00467.warc.gz | 883,464,548 | 14,218 | ## Algebra 1
$-\frac{1}{3}$
To find the slope, we use the formula $m=\frac{y_2-y_1}{x_2-x_1}$. We plug in the $x$ and $y$ values from the given coordinates: $m=\frac{-4-(-3)}{5-2}$ We simplify above and below the fraction bar with subtraction: $m=\frac{-1}{3}$ We simplify the fraction by writing the negative sign out front: $-\frac{1}{3}$ | 116 | 341 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-34 | longest | en | 0.593109 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-12th-edition/chapter-1-section-1-3-applications-of-linear-equations-1-3-exercises-page-74/26 | 1,537,466,548,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156554.12/warc/CC-MAIN-20180920175529-20180920195929-00179.warc.gz | 765,768,808 | 12,936 | ## Intermediate Algebra (12th Edition)
expression; $\dfrac{8}{15}x+\dfrac{13}{2}$
Without an equal sign, the given is an $\text{ expression }.$ By combining like terms, the given expression, $\dfrac{1}{3}x+\dfrac{1}{5}x-\dfrac{1}{2}+7 ,$ simplifies to \begin{array}{l}\require{cancel} \left( \dfrac{1}{3}x+\dfrac{1}{5}x \right)+\left(-\dfrac{1}{2}+7\right) \\\\= \left( \dfrac{5}{15}x+\dfrac{3}{15}x \right)+\left(-\dfrac{1}{2}+\dfrac{14}{2}\right) \\\\= \dfrac{8}{15}x+\dfrac{13}{2} .\end{array} | 210 | 497 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-39 | latest | en | 0.460175 |
https://math.stackexchange.com/questions/2736933/how-many-ways-to-get-an-odd-number-of-each-color-in-each-bin | 1,563,716,560,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527000.10/warc/CC-MAIN-20190721123414-20190721145414-00371.warc.gz | 464,248,945 | 38,164 | How many ways to get an odd number of each color in each bin?
Suppose you have an odd number of white balls and the same number of blacks balls.
How many different ways are there of putting the balls into bins so that you have an odd number of each color in each bin?
For example, if you have $3$ white and $3$ black there are $2$ different ways. You either put them all in one bin or one white and one black in each of $3$ bins. For $5$ white and $5$ white balls there are $4$ different ways. These are:
(wwwwwbbbbb)
(wwwbbb)(wb)(wb)
(wwwb)(wbbb)(wb)
(wb)(wb)(wb)(wb)(wb)
• Generating function: $\sum\limits_{k=1}^\infty(x+x^3+x^5+x^7+\dots)^k(y+y^3+y^5+\dots)^k$ and looking at the coefficient of $x^ay^b$ where $a$ is the number of white balls and $b$ is the number of black balls. – JMoravitz Apr 14 '18 at 16:13
• @JMoravitz Does this give you 12 for $a = b = 7$? – Anush Apr 14 '18 at 17:28
• No, it gives much more than that. My initial comment assumes that the bins are each labeled. For 5 balls each I come to a total of 11 outcomes (one outcome each for one bin or five bins, and then with 3 bins it is broken into cases of which bin gets the three balls of each color for 9 more cases) If you intend the bins to be unlabeled, then my approach doesn't work. – JMoravitz Apr 14 '18 at 17:39
• For 7 black and 7 white, if my mental calculations are correct I get 63 outcomes. – JMoravitz Apr 14 '18 at 17:41
• Dropping the odd requirement gives OEIS A108469. – Christian Sievers Apr 14 '18 at 23:41
The cycle index $Z(S_n)$ of the symmetric group (multiset operator $\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{MSET}$) has $Z(S_0)=1$ and the recurrence
$$Z(S_n) = \frac{1}{n}\sum_{l=1}^n a_l Z(S_{n-l}).$$
Extracting coefficients from this Maple will produce
$$1, 2, 4, 12, 32, 85, 217, 539, 1316, 3146, 7374, 16969, 38387, 85452, \\ 187456, 405659, 866759, 1830086, 3821072, 7894447, 16148593, \\ 32723147, 65719405, 130871128, 258513076, 506724988, \ldots$$ where we have used memoization.
The repertoire here was $$f(W, B) = \sum_{p_1=0}^q \sum_{p_2=0}^q W^{2p_1+1} B^{2p_2+1},$$
the substitution $a_l = f(W^l, B^l)$ and the coefficient being extracted
$$\sum_{k=1}^{2q+1} [W^{2q+1}] [B^{2q+1}] Z(S_k)(f(W,B)).$$
The Maple code runs as follows.
X :=
proc(n, q, q1, q2)
option remember;
if n = 0 then
if q1 = 0 and q2 = 0 then
return 1;
else
return 0;
fi;
fi;
p2=0..floor((q2/l-1)/2)),
p1=0..floor((q1/l-1)/2)),
l=1..n)/n;
end;
R := q -> add(X(k, q, 2*q+1, 2*q+1), k=1..2*q+1);
• Thanks for this! What do you think the asymptotics might be? – Anush Apr 15 '18 at 20:30
More general, the coefficient of $x^my^n$ in
$$\prod_{i,j\geq 0}(1-x^{2i+1}y^{2j+1})^{-1}$$
tells you how many ways there are to distribute $m$ white and $n$ black balls into bins such that each bin contains an odd number of white and an odd number of black balls.
There is a correspondance between the partitions of a number into odd numbers and those of the same number into different numbers. The same idea gives a correspondance between the ways to put balls into bins according to the given rules and the ways to put the balls into bins such that no two bins have the same contents and, for each bin, the number of white balls and the number of black balls is divisible by the same power of two (they have the same $2$-adic valuation). For example, a bin with $6$ and $10$ balls is allowed, but a bin with $6$ and $12$ balls isn't.
Hence the generating function can also be expressed as
$$\prod (1+x^iy^j)$$
where the product is over all pairs of positive integers $i$ and $j$ such that $i$ and $j$ have the same $2$-adic valuation (i.e. there is an integer $k$ such that $(i,j)=2^k\cdot (r,s)$ for some odd integers $r$ and $s$).
• Can you explain the $-1$ exponent? – qwr Apr 15 '18 at 2:51
• @qwr Recall the geometric series $(1-z)^{-1}=1+z+z^2+\dots$ and let $z=x^ry^s$ for some odd $r$ and $s$. The corresponding factor will allow for any number of bins having $r$ white and $s$ black balls. – Christian Sievers Apr 15 '18 at 8:28
• What do you think the asymptotics might be if $m = n$? It seems it might be asymptotic to $\sqrt{2}^n$ but it's hard to tell. – Anush Apr 15 '18 at 20:28
• @Anush Extracting information about asymptotics from generating functions seems to be much more difficult in the multivariate case. From other series involving partitions I expect something more complicated. – Christian Sievers Apr 16 '18 at 11:27
• @Anush $f(501)/f(499)\approx 1.39$ – Christian Sievers Apr 17 '18 at 9:25 | 1,533 | 4,558 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-30 | latest | en | 0.832733 |
https://ca.answers.yahoo.com/question/index?qid=20200225110158AACZkhv | 1,586,132,238,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371611051.77/warc/CC-MAIN-20200405213008-20200406003508-00075.warc.gz | 384,659,735 | 19,326 | Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago
How do you write asinx+bcozx as a single trigonometric ratio?please show all steps.?
Relevance
• 1 month ago
Consider the right triangle in the diagram below.
In the right triangle, θ is an angle with opposite side b and adjacent side a.
By Pythagorean theorem, the hypotenuse = √(a² + b²)
Hence, cos(θ) = a/√(a² + b²) and sinθ = b/√(a² + b²)
Also tan(θ) = b/a, and thus θ = tan⁻¹(b/a)
Besides, consider the following trigonometric identity:
sin(A + B) = cos(B) sin(A) + sin(B) cos(A)
a sin(x) + b cos(x)
= √(a² + b²) * {[a/√(a² + b²)] sin(x) + [b/√(a² + b²)] cos(x)}
= √(a² + b²) * {cos(θ) sin(x) + sin(θ) cos(x)}
= √(a² + b²) sin[x + θ]
= √(a² + b²) sin[x + tan⁻¹(b/a)]
• 1 month ago
Think of a and b as the legs of another right triangle. I'm sure you meant a * sin(x) + b * cos(x) as well.
sqrt(a^2 + b^2) * (a * sin(x) / sqrt(a^2 + b^2) + b * cos(x) / sqrt(a^2 + b^2))
a / sqrt(a^2 + b^2) = cos(t)
b / sqrt(a^2 + b^2) = sin(t)
sqrt(a^2 + b^2) * (sin(x) * cos(t) + sin(t) * cos(x))
sqrt(a^2 + b^2) * sin(x + t)
• Nate1 month agoReport
so is that the same as say, Rsin(x+t)? | 457 | 1,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-16 | latest | en | 0.596083 |
https://biz.libretexts.org/Workbench/MGT_1010/06%3A_Book-_Growth_and_Competitive_Strategy_in_Three_Circles/6.09%3A_Summary-_Growth_Strategy_in_10_Steps/6.9.09%3A_Step_9-_Dynamics | 1,723,273,582,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00725.warc.gz | 97,340,723 | 29,749 | # 6.9.9: Step 9- Dynamics
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Step 9 involves an analysis of market dynamics depicted in the three circles. This step recognizes that markets are constantly moving, and in potentially predictable ways. Recall from Chapter 8 that changes in the market can be reflected one of two ways in the model. First, the circles move, often approaching one another as the offerings of the different competitors become more similar and customer needs become more well known as a product or service category matures. Second, though, is the flow of value through the circles, which helps illustrate the typical competitive innovation-imitation cycle of healthy markets. In the case of MedFactor and its drug OptiMod, the market dynamics analysis would suggest that once the firm is able to establish its unique Area A with doctors, there is a very real possibility that its advantages can be eroded over time as its competitors seek to imitate its unique advantages. A careful exploration of the forces that evolve the market toward commoditization is imperative, as patent protection is limited and other firms are likely to be aggressive in their imitation of a demonstrated competitive advantage.
6.9.9: Step 9- Dynamics is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. | 2,014 | 5,655 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-33 | latest | en | 0.19415 |
https://www.nagwa.com/en/videos/346126483810/ | 1,642,865,538,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303864.86/warc/CC-MAIN-20220122134127-20220122164127-00353.warc.gz | 962,748,558 | 23,705 | Question Video: Writing a Complex Number in Polar Form given Its Modulus and Principal Argument Mathematics • 12th Grade
Given that |π| = 5 and the argument of π is π = 2π + 2ππ, where π β β€, find π, giving your answer in trigonometric form.
01:40
Video Transcript
Given that the modulus of π is equal to five and the argument of π is π equals two π plus two ππ, where π is an integer, find π, giving your answer in trigonometric form.
When we write a complex number in trigonometric or polar form, we write it as π equals π multiplied by cos π plus π sin π, where π is known as the modulus of the complex number π and π is its argument. In polar form, π can be in degrees or radians whereas in exponential form, it does need to be in radians.
Letβs substitute what we know about the complex number π into this formula. The modulus of π was five and the argument was two π plus two ππ. So we get π equals five multiplied by cos of two π plus two ππ plus π sin of two π plus two ππ.
Next, we can recall what we know about the cosine and sine functions. They are periodic. That is to say, they repeat. And their period is two π radians. That means that for any value of π, sin of π plus some multiple of two π is equal to sin π. And cos of π plus some multiple of two π is equal to cos of π.
This means that cos of two π plus two ππ is simply cos of two π. And sin of two π plus two ππ is equal to sin of two π. Our complex number can therefore be written as five multiplied by cos of two π plus π sin of two π, in trigonometrical polar form. | 566 | 1,725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2022-05 | latest | en | 0.884649 |
https://www.physicsgoeasy.com/distance-and-displacement/ | 1,653,553,048,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00028.warc.gz | 1,087,103,687 | 33,013 | Home » Mechanics » What is Distance and displacement in physics?
What is Distance and displacement in physics?
In this article learn about distance and displacement in physics. Also learn about the difference between these two quantities distance and displacement.
Before learning what is distance and displacement do not forget to check our article on How to find position in physics
Get more articles in kinematics.
Now that we know how to describe the position of a particle moving along a straight line at a given time, we will now learn about distance and displacement.
In our day-to-day language, we use terms distance and displacement interchangeably but in physics, both terms have different meanings.
What is distance?
The Length of the actual path between the initial and final position of the moving object in the given time interval is known as the distance traveled by the object.
Distance Definition
To explain this further let us take our example where a man starts moving towards the right of the tree we are using as a reference point.
Now at t = 10 minute he is at 10 meters from the reference point O. Let’s call this point as A. He then starts to move towards left and at t = 8 minute he is 5 meters from the origin. Let’s call this point as B
So, distance traveled by a man going from O to A is equal to 10 meters and distance traveled in going from A to B is equal to 5 meters.
The total distance traveled is equal to OB. So,
$$OB = OC + AB$$
Putting in the values we get
$$OB= 10\,\, meter + 5\,\, meter =15 \,\, meters$$
So, here we are considering the actual length of the path traveled by the man between initial point O and final point B.
What is Displacement?
Let us now define the term displacement.
Displacement is the shortest distance between the initial position and the final position of a moving object in the given interval of time along with its direction.
Displacement Definition
Let us now find the displacement of the man moving from his initial position O to position A and then to position B. Here to find the displacement of the man from the initial position to final position we must have knowledge of
• How far the final position is from the initial position, which is nothing but the straight-line distance between initial and final positions.
• The direction of the final position as seen from the initial position.
Now going back to our example let us now find the displacement of man from his initial position.
Displacement is equal to straight line distance between points O and B which is equal to OB. Here O is the initial position and B is the final position.
From the above figure, displacement is equal to
$$OB= 5\,\,meters$$ towards the right of origin O.
This straight-line distance between initial and final positions of a particle is called the magnitude of the displacement
Also note that we have mentioned the direction in our example which is ‘towards the right’ or ‘right direction’ of the reference point or origin.
So, displacement has both magnitude and direction whereas distance traveled has magnitude but no direction.
How to find displacement in physics x
How to find displacement in physics
Difference between distance and displacement
Let us now look at the differences between distance and displacement.
The first one is the displacement of an object may be zero but the distance traveled by the object is never zero.
To explain this further again consider that in our example the man travels to point A, then to point B and finally returned to reference point O then
$$Distance = OA + OB + BO$$
$$= 10\,\, meters + 5\,\, meters + 5\,\, meter= 20\,\, meters$$
Now, displacement, in this case, would be zero since O is the initial as well as the final position of the man.
Again, if any object is moving along a circular path, starts at point A and return to point A then the total distance covered would be $$2 \pi r$$ but the net displacement of the object would be zero. So,
Distance traveled by a moving body cannot be zero but the final displacement of a moving body can be zero.
We have also discussed earlier that displacement of an object can be negative but distance traveled by an object can never be negative.
Another difference between distance and displacement is that
Distance is a a scalar quantity that is it only has magnitude and no direction is specified.
Whereas displacement is a vector quantity as it has both magnitudes and direction.
For example, if a car travels a distance of 60 Km then this ’60 kilometer’ is the distance traveled.
Again, if the car is moving in straight line towards west then the displacement of the car of the car would be expressed as ’50 kilometer towards west’.
Watch this video on position, distance and displacement for a clear idea.
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https://www.gradesaver.com/textbooks/math/algebra/college-algebra-7th-edition/chapter-p-prerequisites-section-p-8-solving-basic-equations-p-8-exercises-page-59/7 | 1,532,140,755,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592150.47/warc/CC-MAIN-20180721012433-20180721032433-00246.warc.gz | 914,449,712 | 12,987 | ## College Algebra 7th Edition
(a) Using $x=-2$: $4(-2)+7=9(-2)-3$ $-8+7=-18-3$ $-1=-21$ Which is False. (b) Using $x=2$: 4(2)+7=9(2)-3 8+7=18-3 15=15 Which is True. | 87 | 166 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-30 | longest | en | 0.435608 |
https://studytiger.com/free-essay/stoichiometry-of-a-precipitation-reaction/ | 1,726,217,170,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651510.65/warc/CC-MAIN-20240913070112-20240913100112-00839.warc.gz | 509,921,854 | 11,170 | || || Data Tables: Step 3: Show the calculation of the needed amount of Na2CO3 Convert 1. 0g of CaCl2-. 2H2O to moles of CaCl2-. 2H2O 1. 0g x 1 mole CaCl2-. 2H2O 147. 0 g CaCl2-. 2H2O = 0. 00680 moles CaCl2-. 2H2O The mole ratio is 1:1 Hence if we have 0. 00680 moles of CaCl2-. 2H2O we will as well need 0. 00680 moles of Na-2CO3 Convert moles of Na-2CO3 to grams of Na2CO3 = 0. 00680 moles Na-2CO3 x 105. 99g Na-2CO3 1 mole Na-2CO3 = 0. 72g This means that we need 0. 72g of Na-2CO3 to fully react with 1g of CaCl2-. H2O Step 4: Mass of weighing dish_0. 7___g Mass of weighing dish and Na2CO3__1. 4__g Net mass of the Na2CO3 __0. 7__g Step 6: Mass of filter paper __0. 7__g Step 10: Mass of filter paper and dry calcium carbonate__1. 2__g Net mass of the dry calcium carbonate_0. 5___g (This is the actual yield) Step 11: Show the calculation of the theoretical yield of calcium carbonate. The mole ration between CaCl2-. 2H2O and CaCO3 is 1:1 that means that if we have 0. 00680 moles of CaCl2-. 2H2O we will get 0. 00680 moles CaCO3
Convert the moles of CaCO3 to grams of CaCO3 = 0. 00680 moles CaCO3 x 100 g CaCO3 1 mole CaCO3 = 0. 68g CaCO3 Show the calculation of the percent yield. = Actual yield/Theoretical yield x 100 = 0. 5/0. 68 x 100 = 73. 5% Conclusion: The objective of the experiment is to predict the amount of product produced in a precipitation reaction using stoichiometry. Secondly, the experiment accurately measures the reactants and products of a reaction. Also, the experiment is to determine actual yield vs. theoretical yield and to calculate the percent yield.
For example in this experiment, we were able to predict that we need 0. 72g of Na-2CO3 to fully react with 1g of CaCl2-. 2H2O. Another example is that, we calculate the amount of theoretical yield of Calcium Carbonate to be 0. 68g and the percentage yield to be 73. 5%. The scientific principles involved here was that when two or more soluble substances in separate solutions are mixed together to form an insoluble compound they settles of a combined solution as a solid. The solid insoluble compound is called a precipitate.
For example in this experiment, we combined sodium carbonate and calcium chloride dehydrates to produce a precipitate of calcium carbonate. The formula mathematically is Na2CO3(aq) + CaCl2. 2H2-- = CaCO3(s) + 2NaCl(aq) + 2H2O. Sources of Error and ways to minimize them: There may still be some solid particles in the beaker thereby we will not be able to get the correct mass (quantity) of the Calcium Carbonate. To minimize the error we should use an instrument that can be able to scoop out the entire solid from the beaker.
Also if the water in the Calcium Carbonate is not properly dried, the net mass of the Calcium Carbonate can be extremely high. To solve this we must make sure the Calcium Carbonate is well dried. Error of approximation: the molar mass if not well approximated, can lead to an error in the calculation. To minimize this error the instruction should indicate how many decimal point or how significant figure to approximate to. I am highly impressed with the experiment. | 917 | 3,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-38 | latest | en | 0.824767 |
https://medium.com/@spiringwriter/limit-calculator-online-8d8080040068?source=author_recirc-----5b7ea37bb97----1---------------------9946b6dd_2600_4cf8_bf29_787850b6cd21-------&responsesOpen=true&sortBy=REVERSE_CHRON | 1,685,284,950,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643784.62/warc/CC-MAIN-20230528114832-20230528144832-00070.warc.gz | 451,610,406 | 33,468 | # Limit Calculator Online
--
Before I talk about calculating limits with uncertainty, I want to believe that you already have an understanding of what a limit is and how to calculate elementary limits. If there is no such understanding, first read the article “Limits. The concept of limits. Calculation of limits.”
We now turn to the consideration of limits with uncertainty.
There is a group of limits, when x is the Arrow of Infinity, and the function is a fraction, substituting in which the value x = Infinity will receive the uncertainty of the form Limits with uncertainty.
Example Limit Calculator Online
It is necessary to calculate the limit. Limits with uncertainty.
We use our rule №1 and substitute the Infinity in the function. As you can see, we get uncertainty. Limits with uncertainty.
In the numerator we find x in the highest degree, which in our case = 2:
Limits to uncertainty
Do the same with the denominator:
Limits to uncertainty
Here also the highest power = 2.
Next, you need to choose the highest of the two degrees found. In our case, the degree of the numerator and denominator coincide and = 2.
So, to uncover uncertainty Limits with uncertainty, we will need to divide the numerator and denominator by x to the highest degree, i.e. on x2:
Limits to uncertainty
There are also limits with another uncertainty — the type Limits with uncertainty. The difference from the previous case is that x tends not to infinity, but to a finite number.
Example.
It is necessary to calculate the limit limits with uncertainty.
Again, use rule number 1 and substitute the number -1 in place x:
Limits to uncertainty
We have obtained uncertainty Limits with uncertainty, for the disclosure of which it is necessary to factor the numerator and denominator, which in turn usually solves a quadratic equation or use abbreviated multiplication formulas.
In our case, we solve the equation:
Limits to uncertainty
Find the discriminant:
Limits to uncertainty
Limits to uncertainty.
If the root is not extracted the whole is most likely D calculated incorrectly.
Now we find the roots of the equation:
Limits to uncertainty
Limits to uncertainty
Substitute:
Limits to uncertainty
Numerator laid out.
In the denominator, we have x + 1, which is therefore the simplest factor.
Then our limit will look like:
Limits to uncertainty
x + 1 beautifully reduced:
Limits to uncertainty
Now we substitute the value -1 in the function instead of x and get:
2 * (- 1) — 5 = -2–5 = -7
Consider the basic provisions used in solving various kinds of problems with the limits:
The limit of the sum of 2 or more functions is equal to the sum of the limits of these functions:
Limits — Rules
The limit of a constant value is the most constant value:
Limits — Rules
For the sign of the limit you can make a constant coefficient:
Limits — Rules
The limit of the product of 2 or more functions is equal to the product of the limits of these functions (the latter must exist):
Limits — Rules
The limit of the ratio of 2 functions is equal to the ratio of the limits of these functions (in that case, if the limit of the denominator is not equal to 0:
Limits — Rules
The degree of the function under the limit sign applies to the very limit of this function (the degree must be a real number):
Limits — Rules
On this with the calculation of limits with uncertainty everything. Even in the article “Remarkable Limits: The First and Second Remarkable Limits” we separately consider an interesting group of limits. The article will insert another block to address most of the limits found in the vast learning space. | 752 | 3,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2023-23 | latest | en | 0.928029 |
http://www.brainpop.com/educators/community/lesson-plan/basic-geometry-a-tangled-web-game-lesson-plan/?bp-game=a-tangled-web | 1,429,921,877,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246645538.5/warc/CC-MAIN-20150417045725-00170-ip-10-235-10-82.ec2.internal.warc.gz | 381,885,549 | 17,823 | # Basic Geometry Lesson Plan: A Tangled Web Puzzle Game
In this lesson plan adaptable for grades 3 through 8, students use BrainPOP resources (including an interactive game) to learn about basic geometry concepts. Students will explore angles, parallel lines, polygons, triangles, circles, and more while applying their findings to real-life situations.
### Students will:
1. Understand basic geometry concepts related to angles, parallel lines, polygons, triangles, and circles.
2. Apply geometry understandings to real-life situations and virtual ones through online game play.
### Materials:
• Internet and BrainPOP access
• LCD projector/interactive whiteboard
• Computers for students to use when playing the game in pairs
• One protractor and level for each pair of students (optional)
• Copies of the Graphic Organizer (optional)
### Vocabulary:
angle; ray; degree; acute angle; right angle; obtuse angle; straight angle; line; parallel
### Preparation:
Familiarize yourself with game play for A Tangled Web. In this math game, you control a tiny robotic spider named Itzi, who has fallen from the clock that is his home. He must climb back up the clock, finding a safe path through an intricate maze of tangled webs and solving cunning angle puzzles as he goes.
To start the game, click ‘Play’ and then ‘OK’, then '1' to start on level one, stage one. Your ultimate goal is to guide Itzi safely to the Warp Gate. You don’t control Itzi’s movement directly; instead, you rotate the giant cog and then Itzi ‘rolls’ according to gravity. Dotted around each level are blue Glo-Flies. Itzi must collect all the blue Glo-Flies to activate the Warp Gate--only then can he enter it and travel to the next level. Some special gold Glo-Flies are tucked away in hard-to-reach places, and they're worth big bonus points if you find them.
At certain points, Itzi’s path to the Gate will be blocked by a colored line. To remove the line, you must click on the angle puzzle of the same color and input its value: If the angle value is correct, the line disappears and Itzi can pass. If the angle value is incorrect, the line remains in place and Itzi loses energy. Itzi’s energy is shown by a bar in the top-left of the screen – if it runs out, you fail the level. You also fail the level if you don’t reach the Gate before time runs out. But don’t worry: you can always try again! Itzi also loses energy if he hits the cog rim or bumps into dangerous obstacles, such as Laser Walls and Spike Balls.
You can click "Help" at the beginning of game play and during game play to access this information. More directions (including information about the game controls, notes function, stages and levels, and scoring) can be found on the Manga High site under "Instructions". Portions of this lesson plan have been adapted from the resources provided by Manga High.
### Lesson Procedure:
1. To build background knowledge, have students pair up and use the protractors and levels to measure the angles of several objects in the classroom. Discuss the different degrees using the Experiment as a guide and talking point. Alternatively, have students use the Graphic Organizer to look for angles around the classroom.
2. Play the Angles Movie for students. Turn on closed captioning to aid students in comprehension. You may wish to display the Review Quiz afterward and discuss the questions with students.
3. Project the Do It tab of the FYI features and talk about angles in relation to clock hands. Challenge students to work in pairs to determine the angles for the clock hands shown.
4. Tell students they will have the opportunity to practice and apply their understanding of angles and other geometric concepts through an interactive game featuring a spider and a clock. Project the game for the class to see.
5. Show students how to get the A Tangled Web game started and model game play through a few rounds. When you start a new level, demonstrate how to spend a few moments spotting its main mathematical features. Are there any triangles that will be useful? Any parallel lines? Any circles that will help you with your calculations? Show students how to decide the best order in which to calculate the angles. Often you need to work out other angles in order to calculate the angles needed to remove lines. Make good use of the Notes function and encourage students to do the same when they play.
6. Allow students to explore the game in pairs.
7. Bring the group back to a whole-class discussion. Talk about some of the strategies students used. Guide them to understand that the more correct angle values you add, the easier it will be to work out the angles you need. Sometimes it helps to imagine extra lines in your diagram to solve problems. You will need to remove some lines more than once. Remind students that once they’ve correctly found an angle, they can remove its line as often as they wish. Sometimes it’s useful to rest Itzi against a line you can remove; then, when you remove it, Itzi will start rolling. Use this control to your advantage when timing is important. Ask students how the in-game math help assisted them. Have volunteers share shortcuts and tricks that saved time, and encourage them to take risks and try new strategies during game play.
8. Review some of the math strategies used in the game. Remind students that if they can’t work out how to find an angle, they can use the in-game math help. You may want to ask volunteers to share information about angles and create a chart for students to reference. Record that angles making a straight line add up to 180°, angles meeting at a point add up to 360°, and angles in a triangle add up to 180°. In an isosceles triangle, two of the angles are equal. If you know one of the two equal angles, double it and then take the number you get away from 180° to find the other angle. You can also visit the Manga High site and click on "Improve Your Score" for more ideas on math strategies and the math concepts behind this game that could be recorded in your chart.
9. Allow students to explore the game a second time, either independently or in pairs. Encourage them to reference the chart as they play.
10. Use the Activity to assess what students have learned, or have them take the Quiz independently.
### Extension Activity:
Use the movies and activities from our other Geometry and Measurement Unit to further students' understanding of the concepts in this game. More math games are available in GameUp and on the MangaHigh.com website. | 1,375 | 6,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2015-18 | latest | en | 0.873235 |
http://mathtrench.com/Contents/Statistics/Confidence+Intervals/One+population+CI+with+unknown+standard+deviation-ID9 | 1,435,973,727,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375096293.72/warc/CC-MAIN-20150627031816-00186-ip-10-179-60-89.ec2.internal.warc.gz | 167,588,353 | 3,989 | # One population CI with unknown standard deviation
Assume you are the manager of a mustard seed factory in Colombia. Your company has received complaints that there is not enough mustard seed in your economy size packages.
You ask your supervisor and chief operating officer Juan Valdez to test the new mustard seed packaging machine you are installing. He runs a sample of 36 packages, with the results of package sizes in ounces:
Dataset two
A. Calculate a 95% confidence interval (CI) on the average weight of packaged mustard seed. Explain very carefully to the packaging workers what the 95% confidence interval numbers mean. Include your SPSS output. (1 points)
B. Now, let’s assume that the package claims that it contains 1.7 oz. of mustard seed. Some of the customers claim that there isn’t enough mustard seed in the 1.7 oz. economy size; corporate management is worried that there may be too much. Given the sample of 36, assuming that it is adequate, test the hypothesis that the average amount of mustard seed in the package meets the 1.7 oz. standard with, say, 95% confidence. Be sure to state the null and alternative hypothesis and which you support. Include your SPSS output. (2 points)
Solution: (a) We compute the 95% confidence interval for the population weight of packaged mustard seed with the aid of SPSS.
The 95% CI is (1.7244, 1.8123). This means that there is a 0.95 probability that the real population mean weight is contained by the interval (1.7244, 1.8123).
(b) We have to test the following null and alternative hypotheses:
We use a two-tailed t-test, using SPSS. The output is shown below:
The p-value of the test is p = 0.403, which means that we fail to reject the null hypothesis, at the 0.05 significance level. | 406 | 1,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2015-27 | latest | en | 0.916529 |
http://www.askmehelpdesk.com/math-sciences/calculus-672266.html | 1,369,473,640,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705884968/warc/CC-MAIN-20130516120444-00057-ip-10-60-113-184.ec2.internal.warc.gz | 331,988,366 | 13,735 | # Calculus
solve for x, e^x+1=0
ebaines Posts: 10,055, Reputation: 5539 Expert #2 Jun 18, 2012, 08:28 AM
Quote:
Originally Posted by VICTOR101010101 solve for x, e^x+1=0
Have you studied d'Moivre's theorem?
$
e^{i \theta} = \cos \theta + i \sin \theta
$
where $\theta$ is a real number. Here you need to find a value for $\theta$ such that $\cos \theta = -1$ and $\sin \theta = 0$.
cyber_i9 Posts: 9, Reputation: 1 New Member #3 Jun 19, 2012, 07:59 AM
well, i guess we can Manipulate the equation,
given: e^x+1=o
:- e^x= -1
:- taking log on both sides
:- x = -log 1 ( log and e cancel out)
therefore,
:- x = 0
hope it works, bcoz its just a Hunch!
ebaines Posts: 10,055, Reputation: 5539 Expert #4 Jun 19, 2012, 09:09 AM
Quote:
Originally Posted by cyber_i9 well, i guess we can Manipulate the equation, given: e^x+1=o :- e^x= -1
You made a mistake there. From e^x+1=0 you get e^x = -1, which leads to x = ln(-1). Hence the need to apply complex numbers.
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What is calculus
Calculus help [ 4 Answers ]
A charter flight charges a fare of $300 per person plus$10 per person for each unsold seat on the plane. The plane holds 100 passengers. The total revenue received can be modelled by R(x)=-10x squared 700x 30,000, where x represents the number of unsold seats. The maximum revenue is \$__ ,which... | 530 | 1,666 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 5, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2013-20 | latest | en | 0.895951 |
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### Author Topic: Hive Temperature Calculations (Read 6942 times)
#### BlueBee
• Galactic Bee
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##### Hive Temperature Calculations
« on: March 28, 2011, 09:21:21 PM »
For those of you who like a little math, I thought I would provide some equations for estimating the temperature inside your hives. Did I hear a collective sigh and a moan? There goes BB again with his heated hives? Not exactly. The math applies to wood hives too.
I agree with the crowd that bees don’t set out to warm up their entire hive, however depending upon the R value of the hive and air infiltrations, the bees heat CAN end up heating a hive rather substantially.
The hive temp equations are simplest if you calculate the hive temp based on a perfect R value and then adjust it later for air infiltration losses and solar gains. The equations for heat transfer were discovered by Newton 300 years ago (http://en.wikipedia.org/wiki/Convective_heat_transfer). Using modern building materials, the differential nature of Newton’s equation can be simplified to a simple energy relation equation we can all understand.
(Hive Surface Area/ R value) x (Temperature Inside – Temperature Outside) = BTUs of bee heat.
A full colony of bees can generate somewhere between 20 and 40 watts according to the literature I’ve seen and my casual observations. 1watt = 3.41 BTU/hour.
A wood hive has a R-value of about 1. A 2 deep hive has a surface area of about 14.5 sq feet. Plugging these numbers in the heat loss equation we have:
(14.5/R1) x Delta T = 20watts x 3.41
Delta T = 4.7F. Where Delta T is the temperature delta ABOVE the outside air temp.
A perfect wood hive with bees making 20 watts of heat would be at most 4.7F warmer than the outside air before accounting for air infiltration loses. When you take those into account, then the math agrees very closely to what people believe; “ bees don’t heat the hive, they only heat the cluster”.
Now let’s run those same numbers with a foam hive having R10 insulation.
14.5/R10 * DeltaT = 20watts x 3.41
DeltaT = 47F.
A perfectly insulated R10 foam hive will be up to 47F above the outside temperature if the bees are making 20watts of heat in the cluster. In this case, the bees DO end up heating the hive along with the cluster.
This winter it got down to -6F here. The math says a wood hive got down to -2F inside. The math says a foam hive got down to 41F inside. If you were a bee, which house would you prefer? My bees are all alive; they seemed to prefer 41F over -2F.
« Last Edit: March 28, 2011, 09:31:23 PM by BlueBee »
#### MTWIBadger
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##### Re: Hive Temperature Calculations
« Reply #1 on: March 29, 2011, 12:20:15 AM »
BlueBee,
What is your setup for heating your hive and insulating it? In a weak trapout hive, I used two 7 watt lights under a SBB set on the ground covered with 2 inch foam insulation. Only turned the light on when temps reached single digits or below and the hive is still alive. May expand lighting to all four of my hives next year.
#### BlueBee
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##### Re: Hive Temperature Calculations
« Reply #2 on: March 29, 2011, 12:42:13 AM »
MTWIBadger, you’re one step ahead of me typing! See my post “Bee Heater Construction” for some details of my setup. My goal is to size the volume of my foam hives to the number of bees in the hive in such a fashion that the heat from the bee cluster will be sufficient to keep the hive temperature between 40 and 60F without electrical heat.
In an insulated foam hive, the bees are not going to have to generate as much heat (equations above) to keep the queen warm, meaning they eat less honey which also means they have less waste products stored in their gut over winter. I figure that is probably a good thing too.
This season my hive volumes were larger than ideal and hence I ended up adding electrical heat to keep the nucs in the desired temp range. The nucs got between 10 and 20 watts because the hive volumes were bigger than I should have made them.
#### JRH
• New Bee
• Posts: 27
##### Re: Hive Temperature Calculations
« Reply #3 on: March 29, 2011, 10:43:59 AM »
(Hive Surface Area/ R value) x (Temperature Inside – Temperature Outside) = BTUs of bee heat.
A full colony of bees can generate somewhere between 20 and 40 watts according to the literature I’ve seen and my casual observations. 1watt = 3.41 BTU/hour.
A wood hive has a R-value of about 1. A 2 deep hive has a surface area of about 14.5 sq feet. Plugging these numbers in the heat loss equation we have:
(14.5/R1) x Delta T = 20watts x 3.41
Delta T = 4.7F. Where Delta T is the temperature delta ABOVE the outside air temp.
Now let’s run those same numbers with a foam hive having R10 insulation.
14.5/R10 * DeltaT = 20watts x 3.41
DeltaT = 47F.
A perfectly insulated R10 foam hive will be up to 47F above the outside temperature if the bees are making 20watts of heat in the cluster. In this case, the bees DO end up heating the hive along with the cluster.
These calculations all depend on WHERE in the hive you are measuring the heat and WHERE in the hive the cluster is located. I am speaking from a full winter's experience with temperature probes suspended about an inch down through the hole in the inner cover in two hives. (The hives are wooden and unheated. There is a 1" thick piece of styrofoam insulation inside the outer cover.) The probes show temperatures that ranged from 10 to 60 degrees higher than the outside ... hive temperatures as high as 60 degrees with outside temperatures a few degrees below zero on occaision. This difference in temperatures, I think, would result in unrealistically high heat output calculations.
As cluster size decreases over the course of the winter, the hive temperatures have decreased as well.
On 50 degree days, hive temperatures at the probe have been as high as 88 degrees.
« Last Edit: March 29, 2011, 03:24:17 PM by JRH »
#### BlueBee
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##### Re: Hive Temperature Calculations
« Reply #4 on: March 29, 2011, 03:43:49 PM »
JRH, congrats on the instrumentation, I’m still working on computerizing mine.
The math isn’t something I invented, that was done by Isaac Newton; the inventor of Calculus. Unless the laws of physics work differently in Vermont, the trick will be interpreting how your data fits Newton’s equations.
If you were reading 60F on 0F days, do you think the cluster may have been around your probe at that point? The cluster does move up during the course of winter. The bees in the cluster are activating their wing muscles to keep the cluster warm.
You noted on a warm 50 degree day reading 88F. I bet the bees like that! That is an advantage of a wood hive on a sunny day. There is so much heat hitting a hive on a sunny day that it can really add up. Solar radiation is on the order of 100watts per square foot. A 2 deep hive is likely to have about 2 sq feet incident to the incoming sunlight. That’s 200watts of heat in. About 10x what the bees can generate.
Newton’s equations are really differential in nature and to get an exact calculation of heat flows you would need to integrate over the entire surface of the hive using Calculus. In that case you would use the exact temp at each point on the inner surface of the hive for the math. The simplified equation used here is an approximation of that integration result. In this case I would guess the proper number to plug into the equation would be an average surface temp in your wood hive. Probably the inside surface temp half way up the hive would be appropriate.
Keep us posted, it’s always great to have real data.
#### JRH
• New Bee
• Posts: 27
##### Re: Hive Temperature Calculations
« Reply #5 on: March 30, 2011, 10:59:13 AM »
Unless the laws of physics work differently in Vermont, the trick will be interpreting how your data fits Newton’s equations.
I am not sure I questioned the equation. I did say that where the cluster is relative to the temperature probe makes a big difference in the observations.
Using Newton's equation, I get 255 watts for a 60 degree differential between the temperature at the probe of my indoor-outdoor thermometer. As I said previously, this sounds like a number that is higher than the likely output of a cluster of bees.
If someone has done a thorough and scientific study of the heat output of a cluster, I would be interested in reading it.
Regards,
Jeff Hills
Dorset, Vermont
#### BlueBee
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##### Re: Hive Temperature Calculations
« Reply #6 on: March 30, 2011, 02:29:25 PM »
I’ll say it again, Newton’s equation is differential in nature. To compute the exact temperature inside a hive, you would have to use calculus and integrate over the entire surface area. That would require more than 2 temperature probes. The best you can do with incomplete data is make some approximate calculations.
The equation I posted is an approximation of integrating Newton’s law over the surface of a hive using calculus. That means the temperatures you plug into the equation needs to be an average of the hive surface temperature; NOT the temperature above or inside the cluster.
An analogy to what you’re doing is measuring the temperature inside a house right above the furnace (or even inside the furnace) and claiming that is an average temperature for the entire house. Not so.
Here’s a good article (often posted) to start learning about the temperature profile in a winter cluster.
http://jeb.biologists.org/cgi/content/full/206/2/353
#### JRH
• New Bee
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##### Re: Hive Temperature Calculations
« Reply #7 on: March 31, 2011, 11:05:34 AM »
The equation I posted is an approximation of integrating Newton’s law over the surface of a hive using calculus.
The equation you posted and the general thrust of the post claimed to be a way to measure the heat output of the cluster in watts.
An analogy to what you’re doing is measuring the temperature inside a house right above the furnace (or even inside the furnace) and claiming that is an average temperature for the entire house.
That is not what I said. I said that your equation would result in radically different results depending on the location of the probe relative to the location of the cluster. Nothing more and nothing less.
#### BlueBee
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##### Re: Hive Temperature Calculations
« Reply #8 on: March 31, 2011, 11:37:15 AM »
Quote
The equation you posted and the general thrust of the post claimed to be a way to measure the heat output of the cluster in watts.
Actually the title of the post it is about computing the temperature inside a hive, not for computing the wattage of your bees. Really the point I was trying to get across is that bees DO end up heating up their environment in a foam hive whereas they do not do so in a wood hive. Basically confirming the standard Beeks mantra that “Bees don’t heat the hive, they only heat the cluster”.
Quote
I said that your equation would result in radically different results depending on the location of the probe relative to the location of the cluster. Nothing more and nothing less.
Fair enough, I can’t argue with that.
Bee Energy output in watts:
If you want to know the heat output of your bees in watts, there is probably a better way to do than the argumentative direction this post has gone. As you say, the location of your temp sensors throws off the equation here. There is a simple good way for finding the wattage if that is what you really want to know. I’ll make a new post later today detailing how you can do that.
#### JRH
• New Bee
• Posts: 27
##### Re: Hive Temperature Calculations
« Reply #9 on: March 31, 2011, 02:52:39 PM »
Really the point I was trying to get across is that bees DO end up heating up their environment in a foam hive whereas they do not do so in a wood hive. Basically confirming the standard Beeks mantra that “Bees don’t heat the hive, they only heat the cluster”.
The conclusion of your post is that it's warmer in a foam hive than it is in a wood hive. You say "they seemed to prefer 41F over -2F. " This implies that bees DO heat the hive. The observation that a foam hive holds the heat better than a wooden one is fairly obvious.
#### BlueBee
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##### Re: Hive Temperature Calculations
« Reply #10 on: March 31, 2011, 04:00:42 PM »
Jeff, I think we’re probably agreeing with each other but our written words aren’t exactly conveying that sentiment.
Quote
A perfectly insulated R10 foam hive will be up to 47F above the outside temperature if the bees are making 20watts of heat in the cluster. In this case, the bees DO end up heating the hive along with the cluster.
What I think I said above was this: The bees DO heat up an insulated hive, but do not appreciably heat up a wood hive. Hence the beeks mantra holds for wood hives, but NOT for foam hives.
Quote
The observation that a foam hive holds the heat better than a wooden one is fairly obvious.
You would think everybody knows that! However that was not evident in the forum this winter. You must have missed out in the discussions this winter about insulation. Many beeks don’t believe insulation warms up a hive. They believe the old mantra “Bees only heat the cluster” applies in all cases. At least that was my interpretations of their posts this winter.
#### BlueBee
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##### Re: Hive Temperature Calculations
« Reply #11 on: April 01, 2011, 06:51:12 PM »
The point of this post is not to prove insulated hives are warmer than wooden hive. I would hope that is fairly obvious to everybody by now. A better use of the equation is for designing wintering systems for next year or spring buildup. If you know how many watts your bees are generating (see my other post), you can design an enclosure of the proper volume and R-value to maintain your bees at about any temperature you want in your climate. How much insulation you would need can be figured out from the equation I posted. There is no need to make your bees suffer in hives that are 0F.
Take for example a 5 frame nuc. People know from reading the forums that you want to pack them in a small hive (nuc) not a big one. However how cold will that small colony get? We know small colonies will freeze out due to a lack of heater bees. What are you to do?
Well, using this math you can tune the thermal performance of the hive to the size of your colony to give them a better chance of thriving. You can make an informed decision rather to use a wood hive, a 1” foam hive, a 2” foam hive, or even adding electric heat if you want. The equation will predict the average temperature inside that hive. (before accounting for air infiltration losses and solar gains).
#### MarkF
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##### Re: Hive Temperature Calculations
« Reply #12 on: June 01, 2011, 04:22:58 PM »
Does anyone know at what the bees would be the most comfortable. What temperature would the bees use the least food, raise the most brood move freely in the hive.
Sting me once shame on you!
Sting me twice I guess I should have learned faster!
#### Grieth
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##### Re: Hive Temperature Calculations
« Reply #13 on: June 01, 2011, 04:45:07 PM »
Great discussion on heat. It doesn't get so cold where I am, so MarkF's question is more important to me, coupled with the calcs for heat. I would also like to know about cooling as it gets up to 40 centigrade here on the hottest summer days - would an insulated box (higher R value) potentially keep the excess heat in? I have noticed in warmer climates that roof insulation slows heating, but also seems to slow cooling in the evenings. I am interested in your thoughts.
"The time has come," the walrus said, "to talk of many things:
Of shoes and ships - and sealing wax - of cabbages and kings”
Lewis Carroll
#### derekm
• Field Bee
• Posts: 555
##### Re: Hive Temperature Calculations
« Reply #14 on: August 04, 2011, 05:52:35 PM »
Does anyone know at what the bees would be the most comfortable. What temperature would the bees use the least food, raise the most brood move freely in the hive.
from the scientific literature it appears to be around 34C. So the trick is to insulate so the bees can heat the hive to 34C if they want to and keep down to 34C as well, we can keep the bees in their "office environment"
Using that temp, the edothermic heat output( bees heater full on) and ectothermic ( bees just chillin) output for volume of bees, the dimensions of the hive, the thermal conductivity of the hive material, you can work out the temperature range the bees can control with resorting to clustering or fanning.
I've done that and i 've manage to confirm the obvious Bee's do well in 37" diameter tree with 12" hole in the middle and only need to cluster when it gets below freezing and only need to fan when it way above 20C.I'm using this to see what temperature I need to open vents on a 2" thick polyurethane hive, and what cold its good for.
If they increased energy bill for your home by a factor of 4.5 would you consider that cruel? If so why are you doing that to your bees?
#### rbinhood
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##### Re: Hive Temperature Calculations
« Reply #15 on: August 04, 2011, 06:38:02 PM »
While you are at it why don't you try and redesign the work of mother nature....build a new bee.
For thousands of years the humble bee has managed to survive and some may even say thrive in a natural state, they adapt to the changes of the seasons no matter what is thrown at them. Bees unlike humans do not need central heat and air or indoor plumbing in order to survive. If you continue down this path you are on, the bees will soon be like the human race.....a bunch of fat, lazy, unhealthy couch potatoes. They will be waiting lined up on the landing board for you to hand feed them every ounce of nectar and pollen they need to survive, thus you are going to create another welfare state.
Seriously....have you given any thought to how many problems you may be creating for the future of your bees? What kind of new virus and pest, possably some we have never had to contend with that you may be helping to create by not letting nature take its course.
Don't get me wrong I think your discussion is very interesting, but you have to admit mother nature did a real good job when she designed the honey bee....what else on the face of this earth can take a cow pattie and turn it into the sweetest thing on earth HONEY! ROFLMAO
Only God can make these two things.....Blood and Honey!
#### BlueBee
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##### Re: Hive Temperature Calculations
« Reply #16 on: August 04, 2011, 07:44:40 PM »
Did mother nature design the bees to live in ¾” thick wood boxes? Does a ¾” wood box emulate the thermal conditions in a tree cavity? How many natural hives use foundation? How many natural hives have 5.3mm fat cell bees? How natural is it to ship millions of southern bees to the northern USA every spring? What are the consequences of that?
The ‘don’t mess with mother nature’ genie came out of the bottle a LONG time ago!
What I’ve been trying to do is better emulate the thermal conditions in the bee’s NATURAL home (tree cavities) giving our northern bees a better chance to survive. I’m trying to head back towards nature, not away from it.
It might be bad for the economy, but I don’t enjoy buying new Southern bees every spring as much as some people do.
#### rbinhood
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##### Re: Hive Temperature Calculations
« Reply #17 on: August 04, 2011, 08:00:21 PM »
Are there not any cold hearty bees that will survive the northern winters.....not trying to be funny.....it is a serious question. If so why not stick to that stock and stay away from those southern gal's. I have purchased a few queens from northern breeders in the past and they did well here in the southeast but not as good as feral or mutts with no long linage with so called resistante traits toughted by many breeders.
Only God can make these two things.....Blood and Honey!
#### BlueBee
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##### Re: Hive Temperature Calculations
« Reply #18 on: August 04, 2011, 08:12:08 PM »
We do have plenty of cold hearty bees here in Michigan, but they live in trees; not thin wood boxes. We had 60% losses in cold wood boxes last winter. It gets pretty cold here in the winter :)
#### rbinhood
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##### Re: Hive Temperature Calculations
« Reply #19 on: August 04, 2011, 08:25:01 PM »
:lau:
You should just build them and igloo it would be warmer than an old thin wood box....lol
My girls would not know how to act if they could not pris around on Christmas day in their bikini and get some sun!
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## [link] 1. Introduction, Background, and Motivation
When many real-world problems are addressed or solved mathematically and computationally, the details of those problems are abstracted away until they can be represented directly as idealized mathematical structures (e.g., numbers, sets, trees, graphs, matrices, and so on). In this course, we will study a collection of such idealized mathematical objects: integers, groups, residues, and several others. We will see how these structures and their properties can be used for implementing useful computational solutions to problems such as random number generation, prime number generation, error correction, trusted and distributed storage, secure communication, and others.
In covering the material for this course, we will use the standard language and conventions for discussing these mathematical structures that have been developed by the community of mathematicians over the course of history. You will need to become familiar with these conventions in order to find, identify, and use the structures and techniques that have already been developed for representing and solving certain computational problems. At the same time, we will also learn how modern programming languages and programming paradigms can be used to implement these structures and algorithms both accessibly and efficiently.
The development and application of mathematics involves abstraction. A problem can be viewed at multiple levels of abstraction, and in developing mathematics humans have adopted a variety of techniques that allow them to successfully employ abstraction to study natural phenomena and solve problems.
symbolic abstract meaning concrete meaning inapplication domain 2 + 3 5 five objects {(1, 2), (1, 3)} acyclic graph file system {(1, 2), (2, 3), (3, 1)} graph with cycle network {(0,1), (1,2), (2,0)} permutation random number sequence
The above illustrates the different levels of abstraction that may exist for a given problem. We employ a language of symbols to denote certain abstract structures, which may correspond to actual structures in the world. A string of symbols corresponds to a particular abstract object. Notice that the actual object being modelled and the abstract structure behave the same way, and that this behavior implies certain rules about how we can manipulate the symbols without changing the object that they name. For example, we can represent the same graph using the two strings of symbols "{(1,2), (2,3), (3,1)}" and "{(1,2), (2,3), (3,1)}", or the same number of objects using "2 + 3", "3 + 2", "1 + 4", and so on.
In this course, we will begin to reviewing the terminology and concepts of logic, integer arithmetic, and set theory, which we will use throughout the course. We will then show that the algebraic properties of the integers also apply to congruence classes of integers (i.e., the properties of modular arithmetic operations), and we will derive and utilize theorems that have useful computer science applications (such as for generating random numbers and creating cryptographic protocols). We will then go further and show that some of the algebraic properties that hold in integer and modular arithmetic can also apply to any data structure, and we will study how to recognize and take advantage of these properties.
### [link] 1.2. Informal motivating example: random number generation
Let us informally consider the problem of generating a sequence of random positive integers. Random number generators are needed in many situations and applications, including:
• generating unique identifiers for database records, objects, etc.;
• generating a one-time pad for a simple encryption scheme;
• generating public and private keys for more sophisticated encryption and signature schemes;
• simulation and approximation methods that employ random sampling (Monte-Carlo, and so on).
Different applications will impose different requirements on what is and is not a sufficiently "random" sequence of number. Suppose we adopt the following method:
• n0 = a number in the range (inclusive) 0 to 5;
• ni = (2 ⋅ ni-1 + 1) mod 6.
We can consider another method:
• n0 = an initial seed integer 104 > n ≥ 103;
• ni = only the last four digits of ni-12.
Frequent repetition of a sequence may or may not be allowed in our given application. Does the above method produce repeating numbers? How often? For how many initial seeds? How do we choose a good seed? We can measure a physical process or component (a clock, a keyboard), but even under these circumstances we need a way to reason about the range of random values the measurement produces, and the range of random values the application requires. How do we begin to approach and formally characterize these aspects of the problem so that we are certain we are meeting the requirements imposed by the application?
One way to model a random number generation process is to view it is a permutation. In fact, there is more than one way to view the process as a permutation. We could simply count up from 0 to m and apply the same permutation to each 0 ≤ nm in order to produce the nth random number in the sequence. Is there an efficient way (i.e., using no more memory than O(log m)) to compute a random number from each n such that a number never repeats?
In this course we will learn about a variety of mathematical structures and their properties that will allow us to precisely specify the above problem and others like it, to identify what solutions are appropriate for such a problem, and to implement these solutions correctly and, where necessary, efficiently.
## [link] 2. Review of Logic with Sets, Relations, and Operators
In this section, we will review several abstract structures and associated properties (and the symbolic language used to represent them) that you should have already encountered in past courses. Simultaneously, we will review one way in which these structures can be implemented and manipulated within the modern programming language Python. As with most human languages that have developed organically over time, mathematics has a rich and often redundant vocabulary. We introduce many terms in this section that we will use consistently in this course. However, keep in mind that there are often other synonyms within mathematics and computer science for these structures.
### [link] 2.1. Formulas without quantifiers
Definition: A logical formula or formula is a string of symbols that follow a certain syntax. If the formula is written using a correct syntax, we can ask whether the formula is true or false. The symbols or, and, not, implies, and iff are logical operators.
formula meaning example of one possiblePython representation true always true `True` false always false `False` f1 and f2 only true if both f1 and f2 are true `True and False` f1 or f2 true if f1 or f2 (or both) are true `True or (False and True)` f1 implies f2 if f1 is true, then f2 must be true f1 is false, or f2 is true f1 is "less than or equal to" f2(if false is 0 and true is 1) `False <= True` f1 iff f2 f1 is true if and only if f2 is true f1 and f2 are eitherboth true or both false `True == False` ¬ f true if f is false `not (True or (False and True))` ( f ) true if f is true `(True and (not (False))` predicate example depends on the definitionof the predicate `isPrime(7)`
The following table may help with gaining a good intuition for the meaning of the implies operator.
meaning ofleft-hand side(premise) meaning ofright-hand side(conclusion) meaning ofentire formula comments true true true if the premise is true and the conclusionis true, the claim of implication is true;thus, the whole formula is true true false false if the premise is true but the conclusion isfalse, the conclusion is not impliedby the premise, so the claim of implicationis false; thus, the formula is false false true true if the conclusion is true on its own, it doesn't matterthat the premise is false, because anything impliesan independently true conclusion; thus, the claimof implication is true, and so is theentire formula false false true if we assume that a false premise is true, then "false"itself is "true"; in other words, falseimplies itself, so the formula is true
Example: Suppose we have the following formula involving two predicates the sun is visible and it is daytime:
the sun is visible ⇒ it is daytime
This formula might describe a property of our real-world experience of a person that is in a particular fixed location on the surface of the Earth. We could state that the above formula is always true (i.e., it is always an accurate description of the system it describes). For every possible assignment of values to each variable, the above formula is indeed accurate, in that it is true exactly in those situations that might occur on Earth, and false in any situation that cannot occur:
the sun is visible it is daytime meaning interpretation true true true a sunny day true false false false true true a cloudy day false false true nighttime
In particular, only one set of values causes the formula to be false: if the sun is in the sky, but it is not daytime. This is indeed impossible; all the others are possible (it may be day or night, or it may be cloudy during the day). The contrapositive of the formula is true if the formula is true:
¬(it is daytime) ⇒ ¬(the sun is visible)
Notice that the contrapositive of the above is a direct result of the fact that if the sun is visible it is daytime must be true, the rows in the truth table in which it is false must be ignored, and then the only possible row in the truth table in which it is daytime is false is the one in which the sun is visible is also false.
### [link] 2.2. Terms: integers and term operators that take integer inputs
Definition: A term is a string of symbols that represents some kind of mathematical structure. In our case, terms will initially represent integers or sets of integers. Terms may contain term operators. We can view these as functions that take terms as input and return terms as output. The term operators for terms that represent integers with which we will be working are +, -, ⋅, and mod.
term what it represents example of one possiblePython representation 0 0 `0` 1 1 `1` z1 + z2 the integer sum of z1 and z2 `3 + 4` z1 − z2 the integer difference of z1 and z2 `(1 + 2) - 4` z1 ⋅ z2 the integer product of z1 and z2 `3 * 5` z1 mod z2 the remainder of the integer quotient z1 / z2z1 - ⌊ z1/z2 ⌋ ⋅ z2 `17 % 5` z1z2 product of z2 instances of z1 `2**3``pow(2,3)`
### [link] 2.3. Formulas: relational operators and predicates dealing with integers
Definition: A term can only appear in a formula if it is an argument to a predicate. A few common predicates involving integers are represented using relational operators (e.g, ≤, ≥).
formula what it represents example of one possiblePython representation z1 = z2 true if z1 and z2have the same meaning;false otherwise `1 == 2` z1 < z2 true if z1 is less than z2;false otherwise `4 < 3` z1 > z2 true if z1 is greater than z2;false otherwise `4 > 3` z1 ≤ z2 true if z1 is less than or equal to z2;false otherwise `4 <= 3` z1 ≥ z2 true if z1 is greater than or equal to z2;false otherwise `4 >= 3` z1 ≠ z2 true if z1 is not equal to z2;false otherwise `4 != 3`
Example: We can define our own predicates as well. Notice that one way we can represent these in Python is by defining a function that returns a boolean result.
predicate definition example of one possiblePython representation P(x) iff x > 0 and x < 2 `def P(x): return x > 0 and x < 2` Q(x) iff x > 3 `Q = lambda x: x > 3`
formula what it represents example of one possiblePython representation P(1) true `P(1)` P(1) or P(2) true `P(1) or P(2)` Q(1) and P(1) false `Q(1) and Q(1)`
We will use the following predicates throughout the course.
Definition: For any x,y ℤ, x | y iff y/x ℤ. If x | y, we then say that x is a factor of y.
Definition: For any y ℤ, y is prime iff for any integer x where 2 ≤ x < y, it is not true that x | y. In other words, y is prime if its only factors are 1 and y (itself).
formula what it represents x | y y / x ∈ ℤ x divides y y is divisible by x y is an integer multiple of x y mod x = 0 x is a factor of y y is prime y > 1 andx | y implies x = 1 or x = y y > 1 andy is divisible only by 1 and itself
Example: We can define the divisibility and primality predicates in Python in the following way:
def divides(x,y):
return y % x == 0 # The remainder of y/x is 0.
def prime(y):
for x in range(2,y):
if divides(x,y):
return False
return True
Definition: For any x,y ℤ, x is a proper factor of y iff y/x ℤ and x < y.
### [link] 2.4. Terms: finite sets of integers, term operators that take set inputs, and set comprehensions
Definition: A finite set of integers is an unordered, finite collection of zero or more integers with no duplicates. The following are examples of terms the meaning of which is a finite set of integers (with the exception of the set size terms, the meaning of which is a positive integer).
term what it represents example of one possiblePython representation ∅ a set with no elements in it `set()` {1,2,3} {1,2,3} `{1,2,3}` {2,..,5} {2,3,4,5} `set(range(2,6))` { x | x ∈ {1,2,3,4,5,6}, x > 3 } {4,5,6} `{x for x in {1,2,3,4,5,6} if x > 3}` |{1,2,3,4}| 4 `len({1,2,3,4})`
The following are term operators on terms the meaning of which is a finite set of integers.
term what it represents example of one possiblePython representation S1 ∪ S2 {z | z ∈ ℤ, z ∈ S1 or z ∈ S2} `{1,2,3}.union({4,5})``{1,2,3} | {4,5}` S1 ∩ S2 {z | z ∈ ℤ, z ∈ S1 and z ∈ S2} `{1,2,3}.intersection({2,3,5})``{1,2,3} & {2,3,5}` |S| the number of elements in S `len({1,2,3})`
While the terms below do not represent finite sets of integers, we introduce the following two set terms in order to reference them throughout the notes.
Definition: Let ℤ be the set of all integers.
Definition: Let ℕ be the set of all positive integers, including 0.
term what it represents ℕ {0, 1, 2, ...} ℤ {..., -2, -1, 0, 1, 2, ...}
### [link] 2.5. Formulas: quantifiers over finite sets of integers
Definition: Suppose we define the following two Python functions that take predicates (or, more specifically, functions that represent predicates) as input.
def forall(S, P):
for x in S:
if not P(x):
return False
return True
def exists(S, P):
for x in S:
if P(x):
return True
return False
We could redefine the above using comprehensions. We will also introduce a `subset()` operation on sets.
def forall(X, P):
S = {x for x in X if P(x)}
return len(S) == len(X)
def exists(X, P):
S = {x for x in X if P(x)}
return len(S) > 0
def subset(X,Y):
return forall(X, lambda x: x in Y)
Then we can introduce the following definitions and corresponding Python examples.
formula what it represents example of one possiblePython representation 1 ∈ {1,2,3} true `1 in {1,2,3}` 4 ∈ {1,2,3} false `4 in {1,2,3}` ∀ x ∈ {1,2,3}, x > 0 and x < 4 true `forall({1,2,3}, lambda x: x > 0 and x < 4)` ∃ x ∈ {1,2,3}, x < 1 and x > 3 false `exists({1,2,3}, lambda x: x < 1 or x > 3)` ∀ x ∈ ∅, f true ∃ x ∈ ∅, f false
Notice that when we quantify over an empty set with a universal quantifier ∀, the formula is always true. When we quantify over an empty set with an existential quantifier, the formula is always false (since no element satisfying any formula could exist if no elements exist at all). We can see that the Python functions for these quantifiers are consistent with this interpretation.
Fact: Let X = {x1 , ..., xn} be a finite set and let P be a predicate that applies to a single integer argument. Then we have the following correspondences between quantifiers and logical operators:
∀ x ∈ X, P(x)
iff
P(x1) and P(x2) and P(x3) and ... and P(xn)
∃ x ∈ X, P(x)
iff
P(x1) or P(x2) or P(x3) or ... or P(xn)
Notice that if X is empty, the "base case" for ∀ must be true (since that is the identity of the and logical operator), while the "base case" for ∃ must be false (since that is the identity of the or logical operator).
Exercise: Implement Python functions that correspond to formulas which can be used to define each of the following statements about a set X and a predicate P.
• All the elements of a set X satisfy the predicate P.
# We provide two equivalent implementations.
def all(X, P):
return forall(X, P)
def all(X, P):
S = {x for x in X if P(x)}
return len(S) == len(X)
• None of the elements of a set X satisfy the predicate P.
# We provide two equivalent implementations.
def none(X, P):
return forall(X, lambda x: not P(x))
def none(X, P):
S = {x for x in X if P(x)}
return len(S) == 0
• At most one of the elements of a set X satisfy the predicate P.
def atMostOne(X, P):
S = {x for x in X if P(x)}
return len(S) <= 1
• At least one of the elements of a set X satisfy the predicate P.
# We provide two equivalent implementations.
def atLeastOne(X, P):
return exists(X, P)
def atLeastOne(X, P):
S = {x for x in X if P(x)}
return len(S) >= 1
Exercise: Use quantifiers to implement a Python function corresponding to the predicate p is prime for any integer p.
def prime(p):
return p > 1 and forall(set(range(2, p)), lambda n: p % n != 0)
### [link] 2.6. Formulas: predicates dealing with finite sets of integers
Definition: The following are examples of formulas that contain relational operators dealing with finite sets of integers.
formula what it represents example of one possiblePython representation 3 ∈ {1,2,3} true `3 in {1,2,3}` {1,2} ⊂ {1,2,3} true `subset({1,2}, {1,2,3})` {4,5} ⊂ {1,2,3} false `subset({4,5}, {1,2,3})`
Below are the general forms of formulas containing relational operators dealing with finite sets of integers.
formula what it represents z ∈ S true if z is an element of S; false otherwise S1 ⊂ S2 ∀ z ∈ S1, z ∈ S2 S1 = S2 S1 ⊂ S2 and S2 ⊂ S1
### [link] 2.7. Terms: set products and binary relations
Definition: The product of two sets X and Y is denoted X × Y and is defined to be the set of ordered pairs (x,y) for every possible combination of x X and y Y.
Example:
term what it represents example of one possiblePython representation {1,2} × {5,6,7} {(1,5),(1,6),(1,7),(2,5),(2,6),(2,7)} `{ (x,y) for x in {1,2} for y in {4,5,6,7} }`
Definition: A set R is a relation between the sets X and Y if RX × Y. We also say that a set R is a relation on a set X if RX × X.
Example: Suppose we have the sets X = {a, b, c} and Y = {D, E, F}. Then one possible relation between X and Y is {(a, D), (c, E)}. One possible relation on X is {(a, a), (a, b), (a, c), (b, b), (c, a)}.
### [link] 2.8. Formulas: predicates dealing with relations
There are several common properties that relations may possess.
predicate definition graphical example X × Y is the set product of X and Y X × Y = { (x,y) | x ∈ X, y ∈ Y } R is a relation between X and Y R ⊂ X × Y R is a function from X to YR is a (many-to-one) map from X to Y R is a relation between X and Y and ∀ x ∈ X, there is at most one y ∈ Y s.t. (x,y) ∈ R R is an injection from X to Y R is a relation between X and Y and ∀ y ∈ Y, there is at most one x ∈ X s.t. (x,y) ∈ R R is a surjection from X to Y R is a relation between X and Y and ∀ y ∈ Y, there is at least one x ∈ X s.t. (x,y) ∈ R R is a bijection between X and Y R is an injection from X and Y andR is a surjection from X and Y R is a permutation on X R ⊂ X × X and R is a bijection between X and X R is a reflexive relation on X R ⊂ X × X and∀ x ∈ X, (x,x) ∈ R R is a symmetric relation on X R ⊂ X × X and∀ x ∈ X, ∀ y ∈ X, (x,y) ∈ R implies (y,x) ∈ R R is a transitive relation on X R ⊂ X × X and ∀ x ∈ X, ∀ y ∈ X, ∀ z ∈ X, ((x,y) ∈ R and (y,z) ∈ R) implies (x,z) ∈ R R is an equivalence relation on XR is a congruence relation on X R ⊂ X × X and R is a reflexive relation on X and R is a symmetric relation on X and R is a transitive relation on X
Exercise: Define the set of all even numbers between 0 and 100 (inclusive). There are at least two ways we can do this:
evens = { 2 * x for x in set(range(0,51)) }
evens = { x for x in set(range(0,101)) if x % 2 == 0 }
Exercise: Implement a Python function that computes the set product of two sets `X` and `Y`.
def product(X, Y):
return { (x,y) for x in X for y in Y }
Exercise: Implement a Python function that takes a finite set of integers and builds the relation on that set correspondingto the operator relational operator ≤.
def leq(S):
return { (x, y) for x in S for y in S if x <= y }
Exercise: Implement a Python function that determines whether a relation `R` is a relation over a set `X`.
# Using our definition of subset().
def relation(R, X):
return subset(R, product(X, X))
# Using the built-in set implementation.
def relation(R, X):
return R.issubset(product(X, X))
Exercise: One property of relations that is studied in other subject areas within computer science and mathematics is asymmetry. We say that R is an asymmetric relation on a set X if:
∀ x ∈ X, ∀ y ∈ X, (x,y) ∈ R implies ¬((y,x) ∈ R)
One example of an asymmetric relation is the "less than" relation on integers, usually represented using the < relational operator. How can we write a Python function that takes as its input a relation `R` and a set `X` and determines whether that relation is asymmetric? Recall that we can represent the implication logical operator using the Python operator `<=`.
def isAsymmetric(X, R):
return relation(R,X) and forall(X, lambda a: forall(X, lambda b: ((a,b) in R) <= (not ((b,a) in R))))
### [link] 2.9. Terms: set quotients and quotient maps
Given an equivalence relation on a set, we can partition that set into a collection of distinct subsets, called equivalence classes, such that all the elements of each subset are equivalent to one another.
Definition: For any set X and equivalence relation R on X, let the quotient set of X with respect to R, denoted X/R, be defined as:
X/R
=
{{y | y ∈ X, (x,y) ∈ R} | x ∈ X}
Exercise: Implement a Python function that takes two inputs (a set X and an equivalence relation R on that set), and outputs the quotient set X/R.
def quotient(X,R):
return {frozenset({y for y in X if (x,y) in R}) for x in X}
Below, we evaluate the above function on an example input.
>>> quotient({1,2,3,4}, {(1,1),(2,2),(3,3),(2,3),(3,2),(4,4)})
{frozenset({4}), frozenset({2, 3}), frozenset({1})}
Definition: For a set X and a relation R over X, the relation that relates each x X to its equivalence class in X under R is called the quotient map. The function is typically denoted using [ ... ]. That is, [x] is the equivalence class of x under R.
Exercise: Implement a Python function that takes two inputs (a set X and an equivalence relation R on that set), and outputs the quotient map taking each element x X to its corresponding equivalence class [x] X/R.
def quotientMap(X,R):
return {(x, frozenset({y for y in X if (x,y) in R})) for x in X}
Exercise: Determine whether {(x,y) | x ℤ, y ℤ, (x + y) mod 2 = 0} is an equivalence relation.
It is reflexive because any integer plus itself must be even (the sum of two odd numbers is even, and the sum of two even numbers is even). Because of commutativity of addition, the relation is symmetric. Finally, it is transitive because if x + y is even and y + z is even, then x and z must either both be even, or they must both be odd, so x + z must be even, as well. Thus, the relation is reflexive, symmetric, and transitive, so it is an equivalence relation.
Example: Let X be a set of humans. Let R be the following relation RX × X:
R
=
{ (x, y) | x ∈ X, y ∈ X, x is y or x is a relative of y }
Then R is an equivalence relation (we assume everyone is related to themselves, and that if two people are both related to the same person, then they are themselves related). Furthermore, the quotient set X/R is a separation of the humans in X into families of relatives. No one in any equivalence class (a.k.a., a family) in X/R is related to anyone in any other equivalence class, and everyone in each equivalence class in X/R is related to everyone else in that equivalence class. Thus, |X/R| is the number of distinct families of humans in the set X. More generally, we can view the quotient set X/R as the separation of X into the largest number of groups such that no two relatives are separated into separate groups.
We can illustrate this with a Python example. Suppose we have the following relation on the set `{'Alice', 'Bob', 'Carl', 'Dan', and 'Eve'}`:
R = {\
('Alice''Alice'), ('Bob''Bob'), ('Carl''Carl'), ('Dan''Dan'), ('Eve''Eve'),\
('Alice''Carl'), ('Carl''Alice'), ('Dan''Eve'), ('Eve''Dan')\
}
We can then compute the set of families:
families = quotient({'Alice''Bob''Carl''Dan''Eve'}, R)
### [link] 2.10.Assignment #1: Logic, Integers, Sets, and Relations
In this assignment you will define Python functions that represent various constructs. You must submit a single Python source file named `hw1/hw1.py`. Please follow the gsubmit directions.
Your file may not import any modules or employ any external library functions associated with integers and sets (unless the problem statement explicitly permits this). You may include in your module any function that is defined in the lecture notes.
Solutions to each of the programming problem parts below should fit on 1-3 lines. You will be graded on the correctness, concision, and mathematical legibility of your code. The different problems and problem parts rely on the lecture notes and on each other; carefully consider whether you can use functions from the lecture notes, or functions you define in one part within subsequent parts.
1. Implement a Python function `cube(n)` that takes a single positive integer argument `n` and returns `True` if and only if `n` is a perfect cube (i.e., there exists an integer `k` such that `k`3 = `n`). You cannot use loops, and you must use `exists()` to receive credit. Efficiency is not important.
>>> cube(27)
True
>>> cube(12)
False
2. Implement a Python function `properPrimeFactors(n)` that takes a single positive integer argument `n` and returns a finite set of integers containing all the prime proper factors of that number.
>>> properPrimeFactors(9)
{3}
>>> properPrimeFactors(14)
{2,7}
3. Implement a Python function `shared(S)` that takes any finite set of positive integers `S` and returns a relation over `S` in which any two numbers in `S` that share at least one prime proper factor are related. If a number has no proper prime factors, then it shares no proper prime factors with any other number (including itself).
>>> shared({4,5,6,7,8,9})
{(4,4), (4,6), (4,8), (6,4), (6,6), (6,8),
(6,9), (8,4), (8,6), (8,8), (9,6), (9,9)}
>>> shared({9,14,28})
{(9,9), (14,14), (14,28), (28,14), (28,28)}
>>> shared({11,12})
{(12,12)}
4. Determine which of the three properties of an equivalence relation (reflexivity, symmetry, and transitivity) always apply to any result of `shared()` (assuming its input is a finite set of positive integers). Define three veriables in your code; each will represent a counterexample input (if it exists) that leads to an output that does not have the corresponding property:
reflexive = ?
symmetric = ?
transitive = ?
If a property always applies to any output of `shared()`, simply assign `None` to the corresponding variable. If a property does not apply to all outputs of `shared()`, assign an input set to the corresponding variable that returns an output relation that does not satisfy that property. For example, if we were doing the same thing for the property of asymmetry, we would choose a value for `asymmetric` such that:
>>> isAsymmetric(asymmetric, shared(asymmetric))
False
1. Solve the following problems involving relations. Recall that using Python, we are representing relations as sets of tuples (i.e., pairs such as `(1,2)`).
1. Implement a function `isSymmetric()` that takes as its input two arguments: a set `X` and a relation `R`. The function should return `True` if the relation `R` is a symmetric relation on `X`, and it should return `False` otherwise.
isSymmetric({1,2}, {(1,1), (2,2), (2,1), (1,2)}) == True
isSymmetric({1,2,3}, {(1,2), (2,1), (3,3)}) == True
isSymmetric({'a','b','c'}, {('a','a'), ('b','b'), ('a','c')}) == False
2. Implement a function `isTransitive()` that takes as its input two arguments: a set `X` and a relation `R`. The function should return `True` if the relation `R` is a transitive relation on `X`, and it should return `False` otherwise.
3. Implement a function `isEquivalence()` that takes as its input two arguments: a set `X` and a relation `R`. The function should return `True` if the relation `R` is an equivalence relation on `X`, and it should return `False` otherwise. Hint: read the notes in this section carefully and reuse functions that have already been defined.
isEquivalence({1,2,3}, {(1,1), (2,2), (3,3)}) == True
isEquivalence({1,2,3}, {(1,2), (2,1), (3,3)}) == False
isEquivalence({1,2}, {(1,1), (2,2), (1,2), (2,1)}) == True
isEquivalence({0,3,6}, {(0,3), (3,6), (0,6), (3,0), (6,3), (6,0)}) == False
2. You should include the following definition in your module:
def quotient(X, R):
return {frozenset({y for y in X if (x,y) in R}) for x in X}
Solve the following problems by defining appropriate relations.
1. Include the following definitions in your code:
X1 = {"a","b","c","d","e"}
R1 = ?
Define the variable `R1` above so that `R1` is an equivalence relation over `X1`, and so that the following will evaluate to `True`:
>>> isEquivalence(X1,R1)
True
>>> quotient(X1,R1) == {frozenset({"a""c"}), frozenset({"b""d"}), frozenset({"e"})}
True
2. Include the following definitions in your code:
X2 = {0,1,2,3,4,5,6,7,8,9}
R2 = ?
Define the variable `R2` above so that `R2` is an equivalence relation over `X2`, and so that the following will evaluate to `True`:
>>> isEquivalence(X2,R2)
True
>>> quotient(X2,R2) == {frozenset({0,4,8}), frozenset({1,5,9}), frozenset({2,6}), frozenset({3,7})}
True
3. Include the following definitions in your code:
X3 = set(range(-25,26))
R3 = ?
Define the variable `R3` above so that `R3` is an equivalence relation over `X3`, and so that the following will evaluate to `True`:
>>> isEquivalence(X3,R3)
True
>>> quotient(X3,R3) == {frozenset(range(-25,0)), frozenset({0}), frozenset(range(1,26))}
True
You must use a set comprehension to define `R3`. Solutions that use an explicitly defined set will receive no credit.
Modular arithmetic can be viewed as a variant of integer arithmetic in which we introduce a congruence (or equivalance) relation on the integers and redefine the integer term operators so that they are defined on these congruence (or equivalance) classes.
### [link] 3.1. Terms: congruence classes in ℤ/mℤ, term operators, and relations
Definition: For any m ℤ, define:
mℤ
=
{x ⋅ m | x ∈ ℤ}
Definition: For any m ℤ, define:
k + mℤ
=
{k + (x ⋅ m) | x ∈ ℤ}
Exercise: Show that the relation R = {(x,y) | x ℤ, y ℤ, x mod 17 = y mod 17} is an equivalence relation.
Definition: For any given m ℤ, define:
ℤ/mℤ
=
ℤ/{(x,y) | x ∈ ℤ, y ∈ ℤ, x mod m = y mod m}
Example: How do we determine whether 7 2 + 5ℤ is a true formula? We can expand the notation 2 + 5ℤ into its definition:
7
2 + 5ℤ
7
{ 2 + 5 ⋅ z | z ∈ ℤ }
Thus, if 7 is in the set of elements of the form 2 + 5 ⋅ z, then we must be able to solve the following equation on integers for z:
7
=
2 + 5 ⋅ z
5
=
5 ⋅ z
1
=
z
Since we can solve for z, it is true that 7 2 + 5ℤ.
Informally and intuitively, we could think of the structure of the above set as a logical consequence of letting all multiples of m be equivalent to 0. That is, if 0 = m = 2m = ..., then 1 = m + 1 = 2m + 1 = ..., and so on.
term what it represents zz mod mz + mℤ {z + (a ⋅ m) | a ∈ ℤ} c1 + c2 {(x + y) | x ∈ c1, y ∈ c2} c1 − c2 {(x − y) | x ∈ c1, y ∈ c2} c1 ⋅ c2 {(x ⋅ y) | x ∈ c1, y ∈ c2} cz c ⋅ ... ⋅ c c! c ⋅ (c-1) ⋅ (c-2) ⋅ ... ⋅ 1
formula what it represents c1 ≡ c2 true only if c1 ⊂ c2 and c2 ⊂ c1, i.e., set equalityapplied to the congruence classes c1 and c2;false otherwise
### [link] 3.2. Algebra of congruence classes
We use the familiar symbols +, -, ⋅, and 0, 1, 2, 3, 4, ... to represent operations on congruence classes. When these symbols are used to represent operations on integers, they have certain algebraic properties. This allows us, for example, to solve equations involving integers and variables, such as in the example below (in which we add the same integer to both sides, use associativity of + and commutativity of ⋅, and cancel 2 on both sides of the equation):
2 ⋅ x − 3
=
1
(2 ⋅ x − 3) + 3
=
1 + 3
2 ⋅ x
=
4
2 ⋅ x
=
2 ⋅ 2
x ⋅ 2
=
2 ⋅ 2
x
=
2
Do the operations on congruence classes, represented by the operators +, -, and ⋅, also share the familiar algebraic properties of the corresponding operations on integers? In many cases they do, but in some cases these properties only apply under specific circumstances.
Example: Suppose we write the term 3 + 4 ≡ 2 where 2, 3, and 4 are congruence classes in ℤ/5ℤ. What is the meaning of this term? First, note the following equivalence.
{ x + y | x ∈ ℤ, y ∈ ℤ} = {z | z ∈ ℤ }
Now, we expand the definitions of congruence classes and the operation + on congruence classes below.
3 + 4
(3 + 5ℤ) + (4 + 5ℤ)
=
{3 + a ⋅ 5 | a ∈ ℤ} + {4 + b ⋅ 5 | b ∈ ℤ}
=
{(x + y) | x ∈ {3 + a ⋅ 5 | a ∈ ℤ}, y ∈ {4 + b ⋅ 5 | b ∈ ℤ}}
=
{(3 + a ⋅ 5) + (4 + b ⋅ 5) | a ∈ ℤ, b ∈ ℤ}
=
{(3 + 4) + (a ⋅ 5) + (b ⋅ 5) | a ∈ ℤ, b ∈ ℤ}
=
{2 + 5 + (a ⋅ 5) + (b ⋅ 5) | a ∈ ℤ, b ∈ ℤ}
=
{2 + (1 + a + b) ⋅ 5 | a ∈ ℤ, b ∈ ℤ}
=
{2 + c ⋅ 5 | c ∈ ℤ}
2 + 5ℤ
2
Fact: The set ℤ/mℤ is closed under the operation represented by +.
Fact: It is the case that ℤ/mℤ = {0,...,m-1} where 0,...,m-1 are congruence classes, and thus, |ℤ/mℤ| = m.
Fact: The addition operation on congruence classes in ℤ/mℤ represented by + is commutative, associative, and has the additive identity 0 + mℤ (a.k.a., mℤ, or simply 0).
Fact: The multiplication operation on congruence classes in ℤ/mℤ represented by ⋅ is commutative, associative, and has the multiplicative identity 1 + mℤ (a.k.a., 1).
property definition ℤ/mℤ is closed under + ∀ x,y ∈ ℤ/mℤ, x + y ∈ ℤ/mℤ + is commutative on ℤ/mℤ ∀ x,y ∈ ℤ/mℤ, x + y ≡ y + x + is associative on ℤ/mℤ ∀ x,y,z ∈ ℤ/mℤ, (x + y) + z ≡ x + (y + z) + has a (left and right) identity 0 in ℤ/mℤ ∀ x ∈ ℤ/mℤ, 0 + x ≡ x and x + 0 ≡ x ℤ/mℤ has inverses with respect to + ∀ x ∈ ℤ/mℤ, (m - x) + x ≡ 0 ℤ/mℤ is closed under ⋅ ∀ x,y ∈ ℤ/mℤ, x ⋅ y ∈ ℤ/mℤ ⋅ is commutative on ℤ/mℤ ∀ x,y ∈ ℤ/mℤ, x ⋅ y ≡ y ⋅ x + is associative on ℤ/mℤ ∀ x,y,z ∈ ℤ/mℤ, (x ⋅ y) ⋅ z ≡ x ⋅ (y ⋅ z) + has a (left and right) identity 1 in ℤ/mℤ ∀ x ∈ ℤ/mℤ, 1 ⋅ x ≡ x and x ⋅ 1 ≡ x ⋅ distributes across + in ℤ/mℤ ∀ x,y,z ∈ ℤ/mℤ, x ⋅ (y + z) ≡ (x ⋅ y) + (x ⋅ z)
In the rest of this subsection, we derive some familiar algebraic properties for congruence classes. We derive some of these properties from the properties of the divisibility predicate (i.e., for any x, y ℤ, x | y iff y/x ℤ). These properties will allow us to use algebra to solve equations involving congruence classes in ℤ/mℤ.
It is worth considering why we choose to work with the set of congruence classes ℤ/mℤ = {0 + mℤ, 1 + mℤ, 2 + mℤ, ..., (m-1) + mℤ} and operations over it rather than simply working with equations involving integer variables and the modulus operator. Modular arithmetic textbooks can be written (and such textbooks exist) in which the techniques covered in these notes are used to solve integer equations of the form f(x) mod m = g(x) mod m for some functions f and g. Some of the reasons for using the set of congruence classes ℤ/mℤ include:
• it is often possible to find the unique solution to an equation over ℤ/mℤ, while equations over ℤ involving the modulus operation may have infinitely many solutions;
• the set ℤ/mℤ is finite, so there is always a finite number of possible solutions to test, even if this is very inefficient, while equations over the integers involving modulus have an infinite range of possible solutions to test;
• the set ℤ/mℤ is a group and is a prototypical example of an algebraic structure, and gaining experience with algebraic structures is one of the purposes of this course, as algebraic structures are ubiquitous in computer science and its areas of application.
Fact: Given an equation involving congruence classes, we are allowed to add the same value to both sides. In other words, for any congruence classes a, b, c ℤ/mℤ, ab implies a + cb + c:
a
b (mod m)
a + c
a + c
To see that this is true, we can simply appeal to algebraic facts about integers:
a
=
b
a + c
=
b + c
(a + c) mod m
=
(b + c) mod m
a + c
b + c (mod m)
Thus, the two congruence classes contain the same elements, so they are equivalent.
Fact: For any congruence classes a, b, c ℤ/mℤ, ab implies a - cb - c. We can adjust the argument for + in the following way:
a
=
b
a - c
=
b - c
(a - c) mod m
=
(b - c) mod m
a - c
b - c (mod m)
We saw that we can add and subtract from both sides of an equation involving congruence classes. Can we also "divide" both sides by the same factor (or "cancel" that factor) in such an equation?
Example: Consider the following sequence of equations within ℤ/2ℤ:
4
6
2 ⋅ 2
2 ⋅ 3
2
3
Clearly, 2 ≢ 3 (mod 2) since the left-hand side is even and the right-hand side is odd. Thus, cancelling 2 on both sides of the equation in the above case is not correct. On the other hand, we have the following:
10
6
2 ⋅ 5
2 ⋅ 3
5
3
In the above case, cancelling 2 on both sides led to a true equation.
It seems that we cannot always "divide" by the same factor on both sides, but we can do so under certain conditions. In order to characterize at least some of the cases in which this is possible, we need a few preliminary facts.
Fact: For any a, m ℤ, a mod m = 0 iff we have that m | a. Thus, the following are all equivalent (i.e., all three are true at the same time):
m
|
a
a mod m
=
0
a
0 (mod m)
0
a (mod m)
We can derive the fact that m | a iff a mod m = 0 as follows. If a mod m = 0 then by definition of mod we have:
a - ⌊ a/m ⌋ ⋅ m
=
0
a
=
⌊ a/m ⌋ ⋅ m
a / m
=
⌊ a/m ⌋
a / m
ℤ
m
|
a
If m | a then by definition of m | a we have:
m
|
a
a / m
ℤ
a / m
=
⌊ a/m ⌋
a
=
⌊ a/m ⌋ ⋅ m
a - ⌊ a/m ⌋ ⋅ m
=
0
a mod m
=
0
Fact: For any a, b, c ℕ, if c|a then c|(ab).
Because c|a, it must be that a/c ℤ. But then we have that:
(a ⋅ b) / c = (a / c) ⋅ b
Since (a / c) ℤ and b ℤ, (a / c) ⋅ b ℤ and (ab) / c ℤ. Thus, c|(ab).
Fact: In ℤ/mℤ, Multiplying by the 0 + mℤ congruence class yields the 0 + mℤ congruence class. For any a, b, m ℕ, if a ≡ 0 (mod m) then ab ≡ 0 (mod m).
We can show this as follows:
a
0 (mod m)
m
|
a
m
|
(a ⋅ b)
a ⋅ b
0 (mod m)
Thus, 0 ℤ/mℤ behaves with respect to multiplication over ℤ/mℤ much the same way that 0 ℤ behaves with respect to multiplication over ℤ.
Example: For any a, b, m ℕ, it is not necessarily the case that just because ab ≡ 0 (mod m), either a or b must be the congruence class 0 ℤ/mℤ. For example, let m = 6, a = 4, and b = 9. Then we have:
4 ⋅ 9
0 (mod 6)
(2 ⋅ 2) ⋅ (3 ⋅ 3)
0 (mod 6)
2 ⋅ (2 ⋅ 3) ⋅ 3
0 (mod 6)
2 ⋅ 6 ⋅ 3
0 (mod 6)
However, we have that:
6
4
4
0 (mod 6)
6
9
9
0 (mod 6)
Thus, the congruence class 0 ℤ/mℤ does not always behave the same way that 0 ℤ behaves.
Fact (Euclid's lemma): For any a, b, p ℤ, if p is prime and p | (ab), then it must be that p|a or p|b (or both).
Fact: For any congruence classes a, b, c ℤ/pℤ, if c is not divisible by p then acbc implies ab.
We can derive the above fact by using the following steps:
a ⋅ c
b ⋅ c
(a ⋅ c) - (b ⋅ c)
0
((a ⋅ c) - (b ⋅ c)) mod p
=
0
((a - b) ⋅ c) mod p
=
0
p
|
((a - b) ⋅ c)
By Euclid's lemma, the fact that c is not divisible by p requires that a - b must be divisible by p. Thus:
p
|
(a - b)
(a - b) mod p
=
0
a - b
0
a
b
Example: Solve the following equation for all possible congruence classes x ℤ/3ℤ:
6 ⋅ x
0 (mod 3)
Since 6 ≡ 0 mod 3, we can rewrite the equation as follows:
0 ⋅ x
0 (mod 3)
Thus, any congruence class x {0 + 3ℤ, 1 + 3ℤ, 2 + 3ℤ} is a solution to the equation.
Example: Solve the following equation for all possible congruence classes x ℤ/5ℤ:
2 ⋅ x
0 (mod 5)
We know that 2 ⋅ 0 ≡ 0 (mod 5), so we can rewrite the above by substituting the right-hand side of the equation:
2 ⋅ x
2 ⋅ 0 (mod 5)
We can now cancel 2 on both sides of the equation using the cancellation law because 5 is prime and 2 ≢ 0 (mod 5):
x
0 (mod 5)
Thus, the only solution is the single congruence class x = 0 + 5ℤ.
Example: Solve the following equation for all possible congruence classes x ℤ/11ℤ:
3 ⋅ x + 5
6 (mod 11)
We can begin by subtracting 5 from both sides:
3 ⋅ x
1 (mod 11)
We can then see that 12 ≡ 1 (mod 11), so we can substitute 1 with 12 on the right-hand side:
3 ⋅ x
12 (mod 11)
We can then rewrite 12 as 3 ⋅ 4:
3 ⋅ x
3 ⋅ 4 (mod 11)
Since 11 is prime and 3 ≢ 0 (mod 11), we can cancel the 3 on both sides to solve the problem:
x
4 (mod 11)
Thus, the only solution is the single congruence class x = 4 + 11ℤ.
Example: Let a ℤ be any integer. Solve the following equation for all possible congruence classes x ℤ/7ℤ:
a + 3 ⋅ x
6 - 6 ⋅ a (mod 7)
We can begin by adding 6 ⋅ a to both sides:
(a + 6 ⋅ a) + 3 ⋅ x
6 (mod 7)
Now we can add a + 6 a to obtain 7 ⋅ a:
7 ⋅ a + 3 ⋅ x
6 (mod 7)
We know that 7 ≡ 0 (mod 7), so we know that for any a ℤ, 7 ⋅ a ≡ 0 (mod 7). Thus, we can substitute the term 7 ⋅ a with 0:
0 + 3 ⋅ x
6 (mod 7)
3 ⋅ x
6 (mod 7)
3 ⋅ x
3 ⋅ 2 (mod 7)
Since 7 is prime and 3 ≢ 0 (mod 7), we can cancel 3 on both sides:
x
2 (mod 7)
Thus, the only solution is the single congruence class x = 2 + 7ℤ.
Exercise: Solve the following equation for all possible congruence classes x ℤ/13ℤ:
4 ⋅ x − 2
10 (mod 13)
We can add 2 to both sides to obtain:
4 ⋅ x
12 (mod 13)
At this point, since 13 is prime and 4 is not equivalent to the congruence class 0 + 13ℤ (i.e., it is not a multiple of 13), we know a solution must exist. Since 12 is a multiple of 4, we can cancel 4 on both sides:
4 ⋅ x
4 ⋅ 3 (mod 13)
x
3
Thus, we have a unique solution x = 3 + 13ℤ, which we can also denote using the symbol 3.
Exercise: Let a ℤ be any integer. Solve the following equation for all possible congruence classes x ℤ/19ℤ. Hint: notice that 17 + 19 = 36.
6 ⋅ x − 11
6 (mod 19)
We can add11 to both sides to obtain:
6 ⋅ x
17 (mod 19)
At this point, we know a solution must exist because 19 is prime and 6 is not a multiple of 19 (so it is not in 0 + 19ℤ and not equivalent to the congruence class 0 (mod 19)). However, it is not immediately obvious that we can cancel a multiple of 6 on the right-hand side. What we do know is that there must exist an integer in 17 + 19ℤ that will allow us to cancel. We can list the members of this set:
17 + 19ℤ
=
{..., 17, 36, 55, ...}
We note that 36 is a multiple of 6 and that 17 ≡ 36 (mod 19), so we can cancel:
6 ⋅ x
36 (mod 19)
6 ⋅ x
6 ⋅ 6
x
6
Thus, we have a unique solution x = 6 + 19ℤ, which we can also denote using the symbol 6.
### [link] 3.3. Multiplication by a congruence class as a permutation
We have seen that in certain situations, it is possible to cancel on both sides of an equation involving congruence classes. While Euclid's lemma made this possible, we might be interested in finding other ways to understand why cancelling is possible in this particular situation. In fact, the alternative explanation is useful in its own right because it can be applied to the practical problem of generating random numbers.
Let us consider the situations in which we can cancel on both sides in an equation involving integers. Suppose a,b,c ℤ, and:
a ⋅ c
=
b ⋅ c
It is possible to cancel in the above equation exactly when the operation of multiplication by c is invertible. In particular, if c is 0, then ac = 0, and all information about a is lost (likewise for bc = 0). So, if c = 0, the operation of multiplication by c is not invertible (i.e., multiple inputs map to the same output, namely 0, so multiplication by c = 0 is not a bijection), and it is not possible to cancel c on both sides. In all other situations where c ≠ 0, the operation is invertible (we can simply perform integer division by c on ac and bc). This raises a natural question: does the ability to cancel congruence classes on both sides of a congruence class equation also imply that the operation of multiplying by the congruence class that can be cancelled on both sides is an invertible operation? The answer is "yes".
For a prime p, multiplication by a congruence class in ℤ/pℤ corresponds to an invertible relation, also known as a bijection or a permutation.
Fact: For any p ℕ, for any a {1,...,p-1}, if p is prime then the following relation R is a permutation from {0, 1,...,p-1} to ℤ/pℤ (the non-zero congruence classes in ℤ/pℤ):
R
=
{ (0, (0 ⋅ a) mod p), (1, (1 ⋅ a) mod p), (2, (2 ⋅ a) mod p), ..., (p-1, ((p-1) ⋅ a) mod p) }
=
{ (i, (i ⋅ a) mod p) | i ∈ {0,...,p-1} }
Recall that R is a permutation if R is a bijection. In order to be a bijection, R must be both an injection and a surjection.
To show that R is an injection, suppose that it is not. We will derive a contradiction from this assumption, which will tell us that the assumption must be false.
If it is not injective, then there exist distinct i {0,...,p-1} and j {0,...,p-1} where without loss of generality j < i such that:
i
j
(i ⋅ a) mod p
=
(j ⋅ a) mod p
But the above implies the following:
(i ⋅ a) mod p
=
(j ⋅ a) mod p
((i ⋅ a) - (j ⋅ a)) mod p
=
0 mod p
((i - j) ⋅ a) mod p
=
0 mod p
p
|
(i - j) ⋅ a
By Euclid's lemma, the above implies that p must divide either (ij) or a. But also know that:
• because a < p, p does not divide a;
• because p > i - j > 0, p cannot divide (ij).
Alternatively, notice that in (ia) mod p = (ja) mod p , we should be able to simply divide both sides of the equation by a because p is prime; however, this contradicts our initial assumption!
Since assuming that distinct i and j can be mapped to the same element when they are multiplied by a leads to a contradiction, it must be that this is not possible. Thus, no two distinct i and j map to the same result, so R is an injection from {0,...,p-1} to ℤ/pℤ and we have that:
|{0,...,p-1}|
=
|ℤ/pℤ|
Thus, since R maps to at least p distinct elements, and |ℤ/pℤ| has at most p elements, R must map to every element in ℤ/pℤ, so it is also a surjection by the Pigeonhole principle.
Since R is both an injection and a surjection from {1,...,p-1} to ℤ/pℤ - {0}, it must be a bijection, and thus a permutation.
Example: Consider 2 ℤ/5ℤ. We can write out the results of multiplying all the congruence classes in ℤ/5ℤ by the congruence class 2:
2 ⋅ 0
0 (mod 5)
2 ⋅ 1
2 (mod 5)
2 ⋅ 2
4 (mod 5)
2 ⋅ 3
1 (mod 5)
2 ⋅ 4
3 (mod 5)
Notice that each congruence class in ℤ/5ℤ appears exactly once as a result.
### [link] 3.4. Generating random numbers
Suppose we wish to automatically generate a sequence of "random" numbers using an algorithm. Before we can implement an algorithm and determine whether it solves our problem, we must first determine what constitutes an acceptable "random" sequence.
Example: Suppose we want to find a way to generate a "random" sequence v of positive integers. Assume we have only one requirement.
Requirement 1: The sequence v has m distinct positive integers between 0 and m-1, where vi is the ith element in the sequence.
In this case, a relation R ⊂ ℕ × ℤ/mℤ that is a permutation would be sufficient. One such relation is:
v0
=
0 mod m
vi
=
(vi-1 + 1) mod m
R0
=
{(i, vi) | i ∈ {0,...,m-1}}
Notice that the second term in (x, x mod m) is in this case the congruence class modulo m that corresponds to x. The relation R0 is indeed a permutation, but it does not satisfy our intuitive notion of a random sequence because it simply counts from 0 to m − 1, so we impose another requirement.
Requirement 2: The sequence v must not be the trivial sequence (0,...,m-1).
Suppose we propose the following relation:
v0
=
0
vi
=
(vi-1 + 2) mod m
R1
=
{(i, vi) | i ∈ {0,...,m-1}}
Notice that we can redefine R1 above more concisely:
R1
=
{(i, (0 + 2 ⋅ i) mod m) | i ∈ {0,...,m-1}}
Does R1 always satisfy both requirements? Suppose that m is even. Then we have that there exists j {0,...,m-1} such that 2 ⋅ j = m. But this means that 2 ⋅ j ≡ 0, so 2 ⋅ (j+1) ≡ 2 ⋅ j + 2 ⋅ 1 ≡ 2 ⋅ 1 ≡ 2 and so on. This means that R1 is not injective, so the first requirement is not met when m is even. Suppose we define R2 to be a variant of R1 parameterized by some b {0,...,m-1}:
R2
=
{(i, (0 + b ⋅ i) mod m) | i ∈ {0,...,m-1}}
What conditions can we impose on b and m so that they satisfy both requirements?
After examining the permutation we can obtain by multiplying all the congruence classes in some set ℤ/pℤ by a particular a ℤ/pℤ, we might wonder if we can use this fact to implement a random number generator. One immediate benefit of this approach is that this approach would satisfy several conditions that we might associate with a "good" algorithm for generating random numbers:
• the "state" of the algorithm is easy to store: it consists of a single congruence class in ℤ/pℤ, which can be represented using an integer;
• the sequence that is generated will contain exactly one instance of all the numbers in the chosen range {0,...,p-1};
• the sequence that is generated can, at least in some cases, be a non-trivial sequence that might appear "random".
Fact: If m is prime and b {2,...,m-1}, then R2 satisfies both requirements.
We know this is true because in this case, R is a permutation, so it satisfies Requirement 1. Furthermore, element v1 = b, so v is never the trivial sequence. Thus, Requirement 2 is satisfied.
Algorithm: The following is one possible implementation of a simple random number generation algorithm.
1. inputs: upper bound (prime) p ℕ, seed a {0,...,p-1}, index i {0,...,p-1}
1. return (ai) mod p
Exercise: What are some drawbacks (or unresolved issues) with building random sequences by choosing a prime m and some a {2,...,m-1}?
The biggest limitation is that the range must be a prime number. This will present a number of difficulties:
• if a non-prime upper bound is needed, it would be necessary to scale down the results from the smallest prime number that is greater than the desired upper bound, and this would eliminate the bijective property of the generator (multiple indices might map to the same random number);
• where do we even find a prime p efficiently that is close to the desired upper bound?
Regarding the second point above: we will see that an efficient, commonly used algorithm for finding prime numbers in a given range actually uses a random number generator as a subroutine. This circular dependency would certainly present a problem, so it would be ideal if we could generate a random number without relying on the ability to generate prime numbers within a certain range.
### [link] 3.5. Greatest common divisor and related facts
It is actually possible to generalize Euclid's lemma so that it does not rely on prime numbers existing at all. In order to do so, however, we must first introduce concepts that make it possible to reason about a particular relationship between numbers that is similar to the property of primality, but is less restrictive.
Definition: For any two x, y ℤ, we define the greatest common divisor, denoted gcd(x,y), as the greatest integer z ℤ such that z | x and z | y. Equivalently, we can define it as the maximum of a set:
gcd(x,y)
=
max{z | z ∈ ℤ, z | x, z | y}
We can also define it recursively (not that z | 0 for all z ℤ because 0/z ℤ):
gcd(x,0)
=
x
gcd(x,y)
=
gcd(y, x mod y)
To see why the recursive definition of gcd works, consider two cases. If x < y, then the two inputs are simply reversed. This ensures that the first input x is eventually larger than the second input y. If xy and they share a greatest common divisor a, then we have for n = ⌊ x/y ⌋ that:
y
=
y' ⋅ a
x
=
x' ⋅ a
x mod y
=
x - (n ⋅ y)
=
(x' ⋅ a) - (n ⋅ y)
=
x' ⋅ a - ((n ⋅ y') ⋅ a)
=
(x' - n ⋅ y') ⋅ a
Notice that (x' - ny') ⋅ a < x' ⋅ a, but that the new smaller value is still a multiple of a, so the greatest common divisor of this value and y is still a.
Example: Consider the number 8 and 9. The factors of 8 are 1, 2, 4, and 8, while the factors of 9 are 1, 3, and 9. Thus, the maximum of the numbers in the intersection {1,2,4,8} ∩ {1,3,9} is 1, so we have that gcd(8, 9) = 1.
Example: We can implement the inefficient algorithm for the greatest common divisor using Python in the following way:
def gcd(x, y):
return max({z for z in range(0, min(x,y)) if x % z == 0 and y % z == 0})
Exercise: Consider the following relation:
{ (x, y) | gcd(x,y) ≠ 1 }
Is this an equivalence relation?
The relation is reflexive because gcd(x,x) = x, and it is symmetric because gcd(x,y) = gcd(y,x). However it is not transitive. For example, gcd(4,6) = 2 and gcd(6,9) = 3, but gcd(4,9) = 1. Note also that we can define such a relation for some range using Python:
{(x,y) for x in S for y in S if gcd(x,y) != 1}
Fact: For any x ℤ, y ℤ, x | y iff gcd(x,y) = x.
Definition: For any x ℤ, y ℤ, x and y are relatively prime, relative primes, and coprime iff gcd(x,y) = 1.
Fact (Euclid's lemma generalization): For any a, b, c ℕ, if a | (bc) and a and b are relatively prime, then it must be that a | c.
Fact: If m, a ℕ and x, y ℤ/mℤ where gcd(a, m) = 1 (i.e., a and m are coprime), and suppose we have that:
x
y (mod m)
Then it must be that:
a ⋅ x
a ⋅ y (mod m)
Notice that we can prove the above by contradiction. If we instead suppose that axay (mod m), then because a and m are coprime, by Euclid's generalized theorem, we can canel a on both sides of the equation to obtain xy (mod m).
Fact: For any a ℕ and m ℕ, if gcd(a,m) = 1, then {(i, (ia) mod m) | i {0,...,m-1}} is a permutation.
The above can be proven by noticing that if gcd(a,m) = 1, then a does not divide m and m does not divide a. Notice that in this fact in which p was required to be prime, the fact that p is prime was not used in isolation; only the coprime relationship between p and a was required.
Using the generalization of Euclid's lemma, it is now possible to address the drawback we observed in our initial random number generating algorithm. We can now accept any upper bound m, not just a prime upper bound, and there is no need for either the algorithm or the user to find a prime p before generating random numbers. However, we have a new problem: how to do we obtain a non-trivial coprime for any given m?
Fact: Suppose that we have some m ℕ, and that we choose some b {2,...,m-1} such that b > m/2. Then it is guaranteed that:
gcd(m, b)
<
b
To see why, consider that if gcd(m, b) = b, this would mean that there exists some k ≥ 2 such that bk = m, and this would mean that:
b ⋅ 2
m
b
m/2
This contradicts our assumption that b > m/2, so it must be that gcd(m, b) < b. We can then further conclude that:
b / gcd(m, b)
>
1
Thus, if gcd(m, (b / gcd(m, b))) = 1, this provides a way to find a number that is greater than 1 and coprime with m. However, this is not guaranteed to work every time because it may still be that gcd(m, (b / gcd(m, b))) > 1. Under those conditions, the options would be to try a different b, or to use a different technique.
Fact: Suppose that m ℕ is an odd positive integer. Then for any k ℕ, gcd(m, 2k) = 1. This is because if m had any factors of 2, it would be even.
Fact: For any m ℤ where m ≥ 2, gcd(m,m+1) = 1.
We can prove this fact by contradiction. Suppose there exists a factor z > 1 of m and m+1. In other words, gcd(m,m+1) > 1. Then we have that:
z ⋅ a
=
m
z ⋅ b
=
m+1
(z ⋅ b) - (z ⋅ a)
=
m+1-m
z ⋅ (b- a)
=
1
z ⋅ (b- a)
=
1/z
If z > 1 then 1/z ∉ ℤ, so (ba) ∉ ℤ. Since b-a ℤ, this is a contradiction, so it must be that gcd(m,m+1) = z = 1.
Fact: Suppose that m ℕ is of the form 2km' for some odd m' and some k ≥ 1. Then m' ⋅ 2 and m have exactly the same prime factors, which means (m' ⋅ 2) − 1 and m share no factors, so gcd(m, (m' ⋅ 2) − 1) = 1.
Algorithm: The following algorithm uses this fact to generate a new coprime. In the worst case, it runs in a linear amount of time in the length of the bit representation of m, and it may in some cases return m − 1 as a result. Note that the operations below (e.g., multiplication ⋅ and subtraction −) are on congruence classes in ℤ/mℤ and not on integers.
1. inputs: positive integer m
1. p := any number in {3,...,m-1}
2. while p − 1 and m are not coprime
1. p := p ⋅ gcd(p − 1, m)
3. return p − 1
At this point, we can define an improved random number generation algorithm that works for any upper bound.
Algorithm: The following is another variant of a simple random number generation algorithm.
1. inputs: upper bound m ℕ, index i {0,...,m-1}
1. a := number in {2,...,m-1} s.t. a and m are coprime (always the same a for an m)
2. return (ai) mod m
This algorithm has a more subtle flaw: poor choices of a (e.g., very small values such as 2) result in a very predictable "random" sequence. It is preferable to choose an a that is coprime with the upper bound m, and that falls somewhere between the middle and the upper quarter of the range {0,...,m-1} (i.e., between 0.5 ⋅ m and 0.75 ⋅ m). However, even this is not ideal, and common random number generators found in standard libraries (such as the linear congruential generator) use a slightly different fact about permutations that results in sequences that appear more "random". This fact and related concepts and results are presented in the latter portions of these notes.
Algorithm: In this variant, the algorithm attempts to find a coprime that is as close as possible to (4/7) ⋅ m. The value 4/7 is chosen in an ad hoc manner in this example. Other values in the range between 1/2 and 3/4 might also produce "nice"-looking results.
1. inputs: upper bound m ℕ, index i {0,...,m-1}
1. b := number in {2,...,m-1} s.t. b and m are coprime
2. for possible powers k in the range 1 to the bit length of m
1. a := a power bk of b that is as close as possible to ((4/7) ⋅ m)
3. return (ai) mod m
We introduce another useful fact about greatest common divisors. Notice that the above fact suggests one possible way to build a prime number generator: for a product p of known prime numbers, it is guaranteed that p+1 shares no factors with p, so any prime factors it has must be new.
Exercise: Solve the following problems using the algebraic facts you know about the gcd operation.
• Find gcd(18,42).
• Find gcd(21000, 2100).
• For a positive even integer a ℤ, find gcd(a/2, a - 1).
• Suppose that for some a ℤ/mℤ, the set {ia mod m | i {1,...,m-1}} contains every number in the set {1,...,m-1}. What is gcd(a, m)?
Exercise: Solve the following equation for all possible congruence classes x ℤ/16ℤ:
9 ⋅ x + 2
4 (mod 16)
We can subtract 2 from both sides to obtain:
9 ⋅ x
2 (mod 16)
At this point, since 9 and 16 are coprime, we know a solution must exist. We can list all the members of 2 + 16ℤ:
2 + 16ℤ
=
{..., 2, 18, 34, ...}
We note that 18 is a multiple of 9, so we can cancel 9 on both sides:
9 ⋅ x
18 (mod 16)
9 ⋅ x
9 ⋅ 2
x
2
Thus, we have a unique solution x = 2 + 16ℤ, which we can also denote using the symbol 2.
Exercise: Solve the following equation for all possible congruence classes x ℤ/15ℤ:
30 ⋅ x
14 (mod 15)
We note that 30 = 15 ⋅ 2, so 15|30, which means:
30
0 (mod 15)
Thus, we can write the equation as:
0 ⋅ x
14 (mod 15)
0
14
There is no possible congruence class x ℤ/15ℤ that will make the above formula true; the formula 0 ≡ 14 is always false in ℤ/15ℤ. Thus, there are no solutions to the original equation.
### [link] 3.6. Generating prime numbers
Many applications require the generation of new primes. We have already seen a simple example in which generating new random sequences required prime numbers. Another important class of applications with this requirement are cryptographic schemes and protocols. In this section, we consider the problem of generating prime numbers, and in particular, random prime numbers in a particular range.
Algorithm: There exists a simple algorithm that is guaranteed to generate a new prime number distinct from any of its inputs, but it is not efficient.
1. inputs: set of primes {p1 ,... , pn}
1. n := p1 ⋅ ... ⋅ pn + 1
2. F := factors of n
3. return any element in F
The above algorithm must return a new prime distinct from any of the primes p1 ,... , pn. To see why, consider the following:
P
=
p1 ⋅ ... ⋅ pn
gcd(P, P + 1)
=
1
There are two possibilities: P+1 is prime, or P+1 is not prime.
• If P+1 is prime, then P > pi for all i {1,...,n}, so P+1 is a new prime.
• If P+1 is not prime, it cannot share any factors with P since gcd(P, P + 1) = 1, so no factors of P+1 are in the set {p1 ,... , pn}. But it must have factors, so any of these factors will be different from the primes in the input set {p1 ,... , pn}.
Thus, the algorithm is a guaranteed method for generating new primes. It also constitutes a proof that there are infinitely many primes. Unfortunately, this algorithm is impractical because the new primes produced by it grow exponentially as the set of primes {p1 ,... , pn} is extended with new primes returned by the algorithm.
In practice, most algorithms that need to generate large primes for commercial applications simply choose a range of numbers and filter out non-primes using some efficient algorithm that does not provide an absolute guarantee that the numbers that remain are all prime. As long as it is not too likely that the generated number is not a prime, this may be sufficient.
Example: Suppose we want to generate a d-digit prime number (in decimal representation). The prime number theorem states that for a given N, the number of primes in the range {2,...,N} is about N/(ln(N)). We can roughly estimate the number of primes with d-digit decimal representations using the following formula:
(10d+1-1 / ln(10d+1-1)) - (10d / ln(10d))
For d = 8, this value is about 4,780,406, so we can roughly say that the chances that an 8-digit number chosen at random (here we are ignoring the details of what distribution is used) is prime are about:
4,780,406/((109 - 1) - 108) ≈ 5.5/100
Algorithm: Suppose we defined the following algorithm for generating a prime with a d-digit representation.
1. inputs: d
1. do
1. n := any number from {10d, ..., 10d+1-1}
while n is not prime
2. return n
Assuming we were choosing numbers "well" with respect to their distribution (we are being imprecise here), we could optimistically hope that for d = 8, the above algorithm would only need to check for primality about 20 times (since roughly 1 out of every 20 numbers it tries should be a prime).
It remains to define an algorithm for checking whether an arbitrary input m ℕ is prime. We could check every number k between 2 and ⌊ √(m) ⌋ to see if it is a factor of m. However, ⌊ √(m) ⌋ still grows exponentially in the representation size of m. For example, for an n-bit input, an integer m in {0,...,2n-1} which must have a representation size of at least n bits, we have the following exponential running time:
√(m)
=
√(2n)
=
2n/2
=
(21/2)n
1.42n
If we only consider primes and not any of their multiples (i.e., we apply the Sieve of Eratosthenes to the set {2,...,⌊ √(m) ⌋}), we can decrease the number of times we check the divisibility of m. However, we would need to do a lot of extra work to filter out the multiples of primes. Modern algorithms such as ECPP run in polynomial time, in practice it is currently difficult to implement a version of these algorithms that runs quickly enough for many applications.
Algorithm: Given the above considerations, we introduce a modified algorithm.
1. inputs: d
1. do
1. n := any number from {10d-1, ..., 10d-1}
while n is not probably prime
2. return n
It remains to define a subroutine for checking whether a number is probably prime (for some appropriate definition of "probably") that is very efficient.
### [link] 3.7. Detecting probable prime numbers
In this subsection, we consider the problem of defining a very efficient algorithm to check whether a positive integer m ℕ is prime. In fact, the algorithm we consider will be detectors of some, but not all, composite numbers.
Fact: For any n ℕ, n is composite iff n > 1 and it is not the case that n is prime.
That is, the algorithms we consider recognize prime numbers but with false positives. They only guarantee that there are no false negatives (i.e., if the algorithm outputs that its input is composite, then it is indeed composite; otherwise, the number may or may not be prime and we call it probably prime because we were not able to detect that it is composite). First, consider how an algorithm for checking primality that never has a "false" output behaves:
algorithm input algorithm output meaning description actually a composite number(this is not known at time of input) composite the input is composite true negative actually a prime number(this is not known at time of input) prime the input is prime true positive
Compare the above table to the following table describing three possible conditions (and one forbidden condition) for an algorithm that detects probable primes.
algorithm input algorithm output meaning description actually a composite number(this is not known at time of input) composite the input isdefinitely composite true negative actually a composite number(this is not known at time of input) probably prime the input is eithercomposite or prime false positive actually a prime number(this is not known at time of input) probably prime the input is eithercomposite or prime true positive actually a prime number(this is not known at time of input) composite impossible false negative(we will not consider algorithmsthat return such outputs)
Below is a comparison of the outputs of four possible probable prime algorithms on inputs in the range {2,...,10} ⊂ ℕ.
inputnumber perfectalgorithm perfect probableprime algorithm less accurateprobable primealgorithm very inaccurateprobable primealgorithm 2 prime probablyprime probablyprime probablyprime 3 prime probablyprime probablyprime probablyprime 4 composite composite probablyprime probablyprime 5 prime probablyprime probablyprime probablyprime 6 composite composite composite probablyprime 7 prime probablyprime probablyprime probablyprime 8 composite composite probablyprime probablyprime 9 composite composite composite probablyprime 10 composite composite probablyprime probablyprime
Algorithm: We now define our first algorithm for testing whether a number is probably prime.
1. inputs: m ℕ, k
1. repeat k times:
1. a := a random number from {2,...,m-1}
2. if a | m then return composite
2. return probably prime
Notice that the above algorithm will never say that a prime number is actually composite. If it does not find a factor of m because it did not run for sufficiently many iterations, then it will indicate that m is probably prime. Thus, it will have no false negatives (i.e., an incorrect output indicating a prime number is composite).
Algorithm: We now define another algorithm for testing whether a number is probably prime.
1. inputs: m ℕ, k
1. repeat k times:
1. a := a random number from {2,...,m-1}
2. if a | m then return composite
3. if gcd(a,m) ≠ 1 then return composite
2. return probably prime
The above algorithm is interesting because by using the gcd operation, we get more value out of each random number we try. In fact, the gcd operation runs in polynomial time but tells us if the intersection between the two sets of factors (the factors of a and the factors of m) contains any numbers. Checking this intersection using the naive approach would take exponential time.
The above algorithm is somewhat problematic if we want to have a good idea of how to set k given our desired level of confidence in the output. For example, how high should k be so that the probability that we detect a composite is more than 1/2? If we require that k ≈ √(m) to be sufficiently confident in the output, we might as well use the brute force method of checking every a {2,..., ⌊ √(m) ⌋}.
To define a more predictable testing approach for our algorithm, we derive a theorem that is frequently used in applications of modular arithmetic (in fact, this fact underlies the prime number generators found in many software applications).
Fact (Fermat's little theorem): For any p ℕ, for any a {0,...,p-1}, if p is prime then it is true that:
ap-1
1 (mod p)
We have already shown that if p is a prime then R defined as below is a permutation:
R
=
{ (1, (1 ⋅ a) mod p), (2, (2 ⋅ a) mod p), ..., (p-1, ((p-1) ⋅ a) mod p) }
=
{ (i, (i ⋅ a) mod p) | i ∈ {1,...,p-1} }
Next, to make our notation more concise, note that:
1 ⋅ 2 ⋅ ... ⋅ p-1
=
(p - 1)!
(1 ⋅ a) ⋅ (2 ⋅ a) ⋅ ... ⋅ ((p-1) ⋅ a)
=
ap-1 (p - 1)!
Recall that p is prime, so p does not divide (p - 1)!. Thus, we can divide by (p - 1)! both sides of the following equation:
ap-1 (p - 1)!
1 ⋅ (p - 1)!
ap-1
1
We now have derived the statement of Fermat's little theorem.
Fact: A number p ℕ is prime iff p > 1 and for all a {1,...,p-1}, ap-1 mod p = 1.
If we negate the statement above, we can define when a number is composite (i.e., when it is not prime) in a way that suggests a straightforward algorithm.
Definition: A number m ℕ is composite iff m > 1 and there exists a {1,...,m-1} such that am-1 mod m ≠ 1. In this case, a is a Fermat witness to the compositeness of m.
Definition: If for composite m ℕ and a {1,...,m-1}, we have that am-1 mod m = 1, then a is a Fermat liar and m is a pseudoprime with respect to a.
Algorithm (Fermat primality test): We now extend our algorithm. The following algorithm can be used to test whether a number is probably prime.
1. inputs: m ℕ, k
1. repeat k times:
1. a := a random number from {2,...,m-1}
2. if a | m then return composite
3. if gcd(a,m) ≠ 1 then return composite
4. if am-1 mod m ≠ 1 then return composite
2. return probably prime
If m is a prime, the above algorithm will always return probably prime.
For any given candidate a in the above algorithm, if the first test fails and gcd(a,m) ≠ 1 then a is a factor of m. Thus, in the worst case, the first is gcd(a,m) = 1 for all k instances of a that we consider. How many of these k instances must pass the second test before we are confident that m is prime? In fact, for most composite numbers m, k can be very low.
Fact: If for a composite m ℤ there is at least one Fermat witness a {2,...,m-1} such that gcd(a,m) = 1, then at least half of all a such that gcd(a,m) = 1 are Fermat witnesses.
Suppose that a is a Fermat witness and a1,...,an are distinct Fermat liars. Then for every Fermat liar we have that:
(a ⋅ ai)n-1
an-1 ⋅ ain-1
an-1
But a is a Fermat witness, so an-1 mod m ≠ 1. Thus, (aai)n-1 mod m ≠ 1, so aai is also Fermat witness. Since there is a witness for every liar, there are at least as many witness as liars, so at least half the values are witnesses. How many numbers m have at least one Fermat witness? Equivalently, how many numbers have no Fermat witnesses?
Definition: For any m ℤ, if m has no coprime Fermat witnesses, then m is a Carmichael number, also known as a Fermat pseudoprime.
The distribuation of Carmichael numbers is high enough that the Fermat primality test is usually not used in favor of slightly more complex tests for probable primes. However, those tests follow a similar principle. The Fermat primality test is used in some deployed software applications (such as PGP).
for the chosena we have... what it means probability of this occurringif m is a non-Carmichael composite a | m a is a non-trivial factor of m,so m is composite (# integers in {2,...,m-1} that are factors with m) / (m-2) gcd(a,m) ≠ 1 m and a have a non-trivial factor,so m is composite (# integers in {2,...,m-1} that share factors with m) / (m-2) am-1 mod m ≠ 1 a is a Fermat witnessthat m is composite at least 1/2
We can consider a particular example input for the primality test to see how each successive check in the algorithm can extract valuable information about whether the input is composite. The following table is for m = 15.
m = 15 and a = ... 2 3 4 5 6 7 8 9 10 11 12 13 14 a | m PP C PP C PP PP PP PP PP PP PP PP PP gcd(a,m) ≠ 1 PP C PP C C PP PP C C PP C PP PP am-1 mod m = ... 4 9 1 10 6 4 4 6 10 1 9 4 1
We can now summarize all the facts and algorithms we have introduced and how their relationships allow us to construct a prime number generator.
Euclid'slemmageneralization ⇐ multiples of coprimea in ℤ/mℤ are apermutation ⇑ Fermat'slittletheorem randomnumbergenerator ⇒ gcd(m,m+1) = 1 ⇑ ⇑ greatestcommondivisoralgorithm ⇐ Fermatprimalitytest ⇐ probableprimedetector ⇑ probableprimegenerator
### [link] 3.8.Assignment #2: Modular Arithmetic, Random Numbers, and Primes
In this assignment you will solve several equations, and you will define a collection of Python functions that will allow you to generate probable prime numbers of any size. You must submit a single Python source file named `hw2/hw2.py`. Please follow the gsubmit directions.
You may import the following library functions in your module:
from fractions import gcd
from math import log
You may also use the built-in `pow()` function, which can compute modular exponents efficiently (as in, ak mod n can be written in Python as `pow(a,k,n)`), the `abs()` function for computing the absolute value of an integer, and the `//` for integer division (you should avoid using `/` because it does not work for very large integers). Your file may not import any other modules or employ any external library functions associated with integers and sets unless explicitly permitted to do so in a particular problem.
Solutions to each of the programming problem parts below should be fairly concise. You will be graded on the correctness, concision, and mathematical legibility of your code. The different problems and problem parts rely on the lecture notes and on each other; carefully consider whether you can use functions from the lecture notes, or functions you define in one part within subsequent parts.
1. Solve the following equations using step-by-step equational reasoning, and list each step. You must list all solutions (zero or more) for x. You may need to use some automation to perform some of the steps (e.g., to check whether a number is prime or whether two numbers are coprime). Your solutions for this problem should appear as comments, delimited using `'''`...`'''`, in `hw2.py`. You may use the `=` ascii character to represent the ≡ relational operator on congruence classes.
1. 4 ⋅ x + 1 ≡ 9 (mod 17)
2. x ≡ − 6 (mod 13)
3. 40 ⋅ x ≡ 5 (mod 8)
4. 3 ⋅ x + 1 ≡ 1 (mod 3)
5. 5 ⋅ x + 7 ≡ 13 (mod 29)
6. 1 + 2 ⋅ x ≡ 2 (mod 10)
7. 17 ⋅ x + 11 ≡ 300 (mod 389)
8. 650472472230302 ⋅ x ≡ 1 (mod 8910581811374)
9. 48822616 ⋅ x ≡ 14566081015752 (mod 3333333333333333333333333)
2. Implement a function `closest(t, ks)` that takes two arguments: a target integer `t` and a list of integers `ks`. The function should return the integer `k` in `ks` that is closest to `t` (i.e., the integer `k` in `ks` that minimizes the absolute value of the difference |`t``k`| between the two numbers). This will serve as a helper function for subsequent problems in this assignment.
>>> closest(5, [1,3,4,9,10])
4
>>> closest(8, [1,3,4,9,10])
9
3. Implement the following Python functions. These functions take advantage of the generalized Euclid's lemma to make it possible to generate a random number within a specified range. Your implementations must be extremely efficient, and must handle very large inputs, as shown in the examples below. Implementations that perform exhaustive, exponential-time searches will receive no credit.
1. Implement a function `findCoprime(m)` that takes a single positive integer argument `m` and returns an integer `b` where `b` > 1 and `b` is coprime with `m`. Your implementation does not need to return exactly the same answers as you see in the example outputs. However, the output generated by your implementation must be coprime with the input. Hint: use facts about coprime numbers (e.g., this, this, this, this, and/or this); the efficiency of the probable prime generator you must assemble in the last problem will depend on the coprimes your algorithm finds not being on the "edges" of the range.
>>> findCoprime(10)
7
>>> findCoprime(100)
63
>>> findCoprime(872637825353262)
545398640845789
>>> findCoprime(2**200)
1004336277661868922213726307713226626576376871114245522063361
>>> gcd(findCoprime(2**100000), 2**100000)
1
2. Implement a function `randByIndex(m, i)` that takes two positive integer arguments: `m` represents the upper bound of random numbers to be generated, and `i` represents an index specifying which random number in a the sequence should be generated. You may assume `m` ≥ 4 and that 1 ≤ `i``m` − 1. The function must return the `i`th "random" number in a permutation of the numbers {0, ..., `m` − 1}. Your implementation does not need to return exactly the same answers as you see in the example outputs. However, the output generated by your implementation must produce a permutation when used in a comprehension, as in the examples below.
>>> [randByIndex(10, i) for i in {0,1,2,3,4,5,6,7,8,9}]
[0, 7, 4, 1, 8, 5, 2, 9, 6, 3]
>>> [randByIndex(77, i) for i in range(0,76)]
[ 0, 48, 19, 67, 38, 9, 57, 28, 76, 47, 18, 66,
37, 8, 56, 27, 75, 46, 17, 65, 36, 7, 55, 26,
74, 45, 16, 64, 35, 6, 54, 25, 73, 44, 15, 63,
34, 5, 53, 24, 72, 43, 14, 62, 33, 4, 52, 23,
71, 42, 13, 61, 32, 3, 51, 22, 70, 41, 12, 60,
31, 2, 50, 21, 69, 40, 11, 59, 30, 1, 49, 20,
68, 39, 10, 58]
>>> randByIndex(2**200, 2**99+1)
1004336277661868922213726307713860451876490985814993873666049
Hints:
1. do not use floating point numbers and only use the integer division operator `//` when dividing;
2. you can use the `m.bit_length()` method to efficiently obtain the bit length of an integer.
4. Implement a function `probablePrime(m)` that takes a single integer argument `m` where `m` >= 1. The function should return `True` if `m` is probably prime, and `False` otherwise. Your code should employ the Fermat primality test by generating some number of random witnesses in the appropriate range and using them to test the primality of `m`. You will need to determine what is a reasonable number of potential witnesses to test. Implementations that perform an exponentially large exhaustive search, even if the algorithm is mathematically correct, will not earn full credit.
>>> probablePrime(31)
True
>>> probablePrime(107)
True
>>> probablePrime(230204771)
True
>>> probablePrime(10738019798475862873464857984759825354679201872)
False
5. Implement a function `makePrime(d)` that takes a single integer argument `d` where `d` >= 1 and returns a probably prime number that has exactly `d` digits. Your implementation should be sufficiently efficient to produce an output for `d` = `100` in a reasonably short amount of time. Implementations that perform an exponentially large exhaustive search, even if the algorithm is mathematically correct, will not earn full credit.
>>> makePrime(2)
47
>>> makePrime(100)
3908330587430939367983163094172482420761782436265274101479718696329311615357177668931627057438461519
To better understand multiplicative inverses, we first review the definition of an additive inverse.
Fact: For any m ℕ, every element in the set ℤ/mℤ has an inverse with respect to addition defined over ℤ/mℤ (i.e., an additive inverse). Consider any x ℤ/mℤ. Then px ℤ/mℤ and
x + (p − x)
p (mod p)
0
We denote by −x the additive inverse of x.
Example: What is the additive inverse of 2 ℤ/5ℤ?
The additive inverse is 5 − 2 = 3, since 2 + 3 mod 5 = 0.
There is more than one way to compute multiplicative inverses; in this subsection, we will present facts that will help us build algorithms for computing multiplicative inverses.
Definition: Given a positive integer m ℕ and a congruence classes x ℤ/mℤ, suppose there exists a congruence class y ℤ/mℤ such that:
x ⋅ y
1 (mod m)
Then we say that y is the multiplicative inverse of x in ℤ/mℤ. We usually denote the multiplicative inverse of x as x-1 (as is often done for multiplicative inverses over the integers, i.e., 2-1 = 1/2).
Fact: Let p ℕ be a prime number, and let a ℤ/pℤ. Then we know by Fermat's little theorem that:
ap-1
1 (mod p)
But we can factor the above to get:
a ⋅ ap-2
1 (mod p)
Thus, the multiplicative inverse of a ℤ/pℤ is:
a-1
ap-2 (mod p)
Note that:
a ⋅ a-1
a ⋅ ap-2 (mod p)
ap-1
1
Example: What is the multiplicative inverse of 2 ℤ/5ℤ? We can compute it as follows:
2-1
25-2 (mod 5)
23
8
3
We can check to confirm that this is true:
2 ⋅ 3
6 (mod 5)
1
### [link] 3.10. Chinese remainder theorem (CRT) and applications
In previous sections we presented facts that allowed us to solve certain individual equations with solution spaces corresponding to sets of congruence classes such as ℤ/mℤ. It is also possible to solve systems of equations over sets of congruence classes.
Theorem (Chinese remainder theorem): The Chinese remainder theorem (CRT) states that given primes p1,...,pk ℕ, for any a1,...,ak ℤ there exists a solution x ℤ to the system of equations:
x mod p1
=
a1
x mod pk
=
ak
We can also state the theorem in terms of congruences. Given primes p1,...,pk ℕ, for any a1 ℤ/p1ℤ, ..., ak ℤ/pkℤ there exists a unique solution x ℤ/(p1 ⋅ ... ⋅ pk)ℤ to the system of equations:
x
a1 (mod p1)
x
ak (mod pk)
In other words, all the solutions to the first system above are from the same congruence class of ℤ/(p1 ⋅ ... ⋅ pk)ℤ. The theorem applies even if p1,...,pk are only relatively prime or coprime.
Example: Solve the following system of equations for the unique solution x ℤ/10ℤ:
x
3 (mod 5)
x
0 (mod 2)
We can list the integers corresponding to each congruence class and find the unique integer in {0, ..., 2 ⋅ 5 - 1} that is in both lists:
3 + 5ℤ
=
{..., 3, 8, 13, 18, 23, 28, ...}
0 + 2ℤ
=
{..., 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, ...}
We can compute the intersection, which should contain all the integers that satisfy both equations:
(3 + 5ℤ) ∩ (0 + 2ℤ)
=
{..., 8, 18, 28, ...}
This appears to be the congruence class 8 + 10ℤ. Thus, we have the unique solution:
x
8 (mod 10)
The Chinese remainder theorem has many applications in a variety of contexts. In this section we present the following algorithms, which all rely on the ability to solve systems of equations involving congruence classes.
efficientmodulararithmetic ⇓ Chineseremaindertheorem ⇐ CRT solver ⇐ rangeambiguityresolution ⇑ Shamir secretsharing protocol
Fact: Given a, a ℤ, if a + b {0,...,m-1}, then it is true that
(a mod m) + (b mod m)
=
(a + b) mod m
Likewise, if a ⋅ b {0,...,m-1}, then it is true that
(a mod m) ⋅ (b mod m)
=
(a ⋅ b) mod m
Example (efficient modular arithmetic): Suppose we want to perform a large number of arithmetic operations in sequence. The operations could be specified as a program that operates on a single variable and performs a sequence of variable updates that correspond to arithmetic operations, such as the example below.
x
:=
3
x
:=
x + 6
x
:=
x ⋅ 2
x
:=
x − 17
x
:=
x + 1
Suppose that over the course of the computation, x might become very large (e.g., 0 ≤ x ≤ 21000000000). However, we have an additional piece of information: once the sequence of operations ends, we know that 0 ≤ x < 1024.
Given our additional information about the range of the final output, we do not need to store 1000000000 bit numbers in order to perform the computation and get a correct result. It is sufficient to instead perform all the operations in ℤ/1024ℤ:
x
:=
3 mod 1024
x
:=
(x + 6) mod 1024
x
:=
(x ⋅ 2) mod 1024
x
:=
(x − 17) mod 1024
x
:=
(x + 1) mod 1024
The above will produce the same result in ℤ/1024ℤ, and we will only need 1024 bits at any single point in the computation to store each intermediate result.
Example (efficient distributed modular arithmetic): Suppose that, as in the previous example, we want to perform a large number of arithmetic operations in sequence on large integers (e.g., in ℤ/270ℤ). However, our resources may be limited. For example, we may only have a collection of processors that can each perform arithmetic on relatively small integers (e.g., in ℤ/28ℤ). Is it possible for us to perform this computation using these processors, and is it possible for us to speed up the computation by running the processors in parallel? We may assume that a single arithmetic computation in ℤ/2nℤ running on a single processor takes n time steps to perform.
Suppose we have ten processors that can perform arithmetic computations in ℤ/28ℤ or any smaller space (such as ℤ/27ℤ). We can approach this problem by choosing a collection of primes p1,...,p10 such that 27 < pi < 28, which implies that:
p1 ⋅ ... ⋅ p10
>
27 ⋅ ... ⋅ 27
270
We can then perform the sequence of computations modulo each of the primes pi to obtain ten results a1, ..., a10. Once we obtain the results, we can apply the Chinese remainder theorem to obtain x:
x
=
a1 mod p1
x
=
a10 mod p10
Since the product of the primes is greater than 270, the unique solution x to the above system of equations will be the correct result of the computation. Since the processors were running in parallel, the computation was about 10 times faster than it would have been if we had performed the computation in ℤ/270ℤ (or in sequence using a single processor that can perform computations in ℤ/28ℤ).
Example (variant of range ambiguity resolution): Suppose we want to build a radar or other sensing device that sends signals out and listens for reflections of those signals in order to detect the distances of obstacles in the environment. The device has a clock that counts up from 0, one integer per second. If the device sends a single signal out that travels at 1 km per second at time 0 and receives a response in 12 seconds at time 12, it knows that the distance to the object and back is 12 km.
However, what if we cannot wait 12 seconds or more? For example, the obstacle may be moving quickly and we want to constantly update our best guess of the distance to that object. We would need to send signals more frequently (for example, every 5 seconds). But then if an object is 12 seconds away, we would have no way to tell when running in a steady state which of the signals we just received.
However, we can obtain some information in this scenario. Suppose we send a signal every 5 seconds, only when the clock's timer is at a multiple of 5. Equivalently, imagine the clock counts up modulo 5 (i.e., 0,1,2,3,4,0,1,2,3,4,0,...) and we only send signals when the clock is at 0. What information can we learn about the object's distance in this scenario? If the distance to the object and back is d, then we would learn d mod 5, because we would get the signal back when the clock is at 0, 1, 2, 3, or 4.
We can use multiple instances of the above device (each device using its own distinct frequency for sending signals) to build a device that can check for obstacles more frequently while not giving up too much accuracy. Pick a collection of primes p1,..., pn such that their product is greater than the distance to any possible obstacle (e.g., if this is a ship or plane, we could derive this by considering the line of sight and the Earth's curvature). Take n instances of the above devices, each with their own clock that counts in cycles through ℤ/piℤ and sends out a signal when the clock is at 0. Running in a steady state, if at any point in time the known offsets are a1,...,an, we would know the following about the distance d to an obstacle:
d
a1 (mod p1)
d
an (mod pn)
We can then use the Chinese remainder theorem to derive the actual distance d < p1 ⋅ ... ⋅ pn.
Protocol (Shamir secret sharing): Suppose there are N participants and we want to divide some secret information among them into N parts so that any k or greater number of participants can reconstruct the secret information, but no subset of fewer than k participants can reconstruct it. Let s ℤ be the secret information. Collect a set of randomly chosen relatively prime integers M = {m1,...,mN} such that:
• the product of any collection of at least k integers in M is greater than s;
• the product of any collection of k-1 integers in M is less than s.
Give each participant i {1,...,N} the value s mod mi. Now, any number of participants nk can use the Chinese remainder theorem to solve for s.
Note: There are many alternative ways to implement Shamir secret sharing. Consider the following example using curve-fitting. We choose some large m ℤ, and then randomly select integers c1,...,ck ℤ/mℤ. We then use these integers as coefficients in a polynomial:
f(x)
=
s + c1 x + c2 x2 + ... + ck xk
Each participant i {1,...,N} is given f(i). Any k participants can now use curve-fitting techniques or techniques for solving collections of equations (e.g., computing the reduced row echelon form of a matrix) to determine all the coefficients of f and, thus, solve for s.
### [link] 3.11. Solving systems of equations with CRT solutions using multiplicative inverses
The Chinese remainder theorem guarantees that a unique solution exists to particular systems of equations involving congruence classes. But can these solutions be computed automatically and efficiently? In fact, they can. However, computing such solutions requires the ability to compute multiplicative inverses in ℤ/mℤ.
greatestcommondivisoralgorithm Fermat'slittletheorem Euler'stheorem ⇑ ⇑ ⇑ Bézout'sidentity ⇐ extendedEuclideanalgorithm ⇐ algorithm forfindingmultiplicativeinverses ⇒ Euler'stotientfunction φ ⇑ Chineseremaindertheorem ⇐ CRT solverfor twoequations ⇑ induction ⇐ CRT solverfor nequations
How is computing multiplicative inverses related to solving systems of equations that have solutions according to CRT? Consider the following example.
Example: Suppose we want to solve the following system of equations:
x
1 (mod 5)
x
0 (mod 4)
The above two equations are constraints on the integers that can be in the congruence class x. One way to state these constraints in English is: "x must be a multiple of 4, and x must be in 1 + 5ℤ". But then we can rewrite the above as a single equation:
4 ⋅ y
1 (mod 5)
Then, we only need to solve for y, and let x = 4 ⋅ y. What is y? The above equation implies that y is the multiplicative inverse of 4 in ℤ/5ℤ. Thus, we can compute:
y
4-1 (mod 5)
45-2 (mod 5)
43
64
4
Thus, the multiplicative inverse of 4 in ℤ/5ℤ is itself. Plugging y into x = 4 ⋅ y gives us 16. Thus, we know by CRT that we have our unique solution in ℤ/(4 ⋅ 5)ℤ = ℤ/20ℤ:
x
16 (mod 20)
The above example suggests that we can solve certain pairs of equations with CRT solutions if one of the congruence classes is 0 and the other 1. What if the other congruence class is not 1?
Example: Suppose we want to solve the following system of equations for x ℤ/15ℤ:
x
4 (mod 5)
x
0 (mod 3)
We can observe that we want some x that is a multiple of 3 and is in 4 + 5ℤ. We can set x = 3 ⋅ y for some y, and then we want to solve the following for y ℤ/5ℤ:
3 ⋅ y
4 (mod 5)
Using Fermat's little theorem, we can compute the multiplicative inverse of 3 ℤ/5ℤ:
35-1
1 (mod 5)
35-1 ⋅ 3-1
1 ⋅ 3-1
3(5-1)-1
3-1
35-2
3-1
33
3-1
27
3-1
2
3-1
Thus, we know that the multiplicative inverse of 3 ℤ/5ℤ is 2, and we have that 2 ⋅ 3 ≡ 1 (mod 5). Notice that 4 ≡ 4 ⋅ 1 (mod 5):
3 ⋅ y
4 (mod 5)
3 ⋅ y
4 ⋅ 1 (mod 5)
Since 1 ≡ 2 ⋅ 3 (mod 5), we can substitute:
3 ⋅ y
4 ⋅ (3 ⋅ 2) (mod 5)
We can cancel 3 on both sides using Euclid's lemma (since 5 is prime) or Euclid's generalized lemma (since 3 and 5 are coprime):
y
4 ⋅ 2 (mod 5)
y
8
y
3
Since we originally set x = 3 ⋅ y, we can now substitute and solve for x ℤ/15ℤ:
x
3 ⋅ y (mod 15)
x
3 ⋅ 3
x
9
x
9
Thus, x ≡ 9 (mod 15) is a solution to our equation. We can confirm this:
9
4 (mod 5)
9
0 (mod 3)
Example: Note that what we actually did in the previous example when we cancelled 3 ℤ/5ℤ on both sides is that we multiplied both sides by the multiplicative inverse of 3 ℤ/5ℤ. Suppose we knew that the multiplicative inverse of 3 ℤ/5ℤ is 2. We can use this information to help us solve the following equation:
3 ⋅ x
2 (mod 5)
We multiply both sides by 3-1 ≡ 2 (mod 5):
3-1 ⋅ 3 ⋅ x
3-1 ⋅ 2 (mod 5)
x
2 ⋅ 2
x
4
Notice that we have now reduced the problem of solving an equation with a single coefficient before x into the problem of finding the multiplicative inverse of the coefficient.
Example: Suppose we want to solve the following system of equations:
x
0 (mod 11)
x
4 (mod 7)
The above equations require that x ℤ/77ℤ be divisible by 11, and that x 4 + 7ℤ. Since x is divisible by 11, it is a multiple of 11, so we want to find x = 11 ⋅ y where:
11 ⋅ y
4 (mod 7)
To solve the above, it is sufficient to multiply both sides of the equation by 11-1 (mod 7). Since 11 ≡ 4 (mod 7), it is sufficient to find 4-1 (mod 7).
11-1
4-1 (mod 7)
47-2
45
1024
2
Thus, we can multiply both sides to obtain:
11 ⋅ y
4 (mod 7)
11-1 ⋅ 11 ⋅ y
11-1 ⋅ 4
y
2 ⋅ 4
y
8
y
1
Thus, we have:
x
11 ⋅ y (mod 77)
x
11 ⋅ 1
x
11 (mod 77)
Fact: Suppose we are given two unequal prime numbers p, q ℕ, and the following two equations:
x
1 (mod p)
x
0 (mod q)
This implies x must be a multiple of q, so rewrite x = qy. Then we have:
q ⋅ y
1 (mod p)
Thus, we can solve for q-1 by computing:
q-1
qp-2 (mod p)
Then we have:
q-1 ⋅ q ⋅ y
q-1 ⋅ 1 (mod p)
y
q-1 (mod p)
x
q ⋅ y (mod (p ⋅ q))
Notice that qy is indeed a solution to the original system because:
q ⋅ y
1 (mod p)
because y ≡ q-1 (mod p);
q ⋅ y
0 (mod q)
because q ⋅ y is a multiple of q.
Fact: Suppose we are given two unequal prime numbers p, q ℕ, and the following two equations where a ℤ/pℤ:
x
a (mod p)
x
0 (mod q)
This implies x must be a multiple of q and a multiple of a, so rewrite x = aqy. Then we have:
a ⋅ q ⋅ y
a (mod p)
As in the previous fact, the above works if y = q-1 (mod p), so compute:
y
=
q-1 (mod p)
x
a ⋅ q ⋅ y (mod (p ⋅ q))
Notice that aqy is indeed a solution to the original system because:
a ⋅ q ⋅ y
a (mod p)
because y ≡ q-1 (mod p);
a ⋅ q ⋅ y
0 (mod q)
because a ⋅ q ⋅ y is a multiple of q.
Fact: Suppose we are given two unequal prime numbers p, q ℕ, and the following two equations where a ℤ/pℤ and b ℤ/qℤ:
x
a (mod p)
x
b (mod q)
Suppose we instead solve the following two systems:
x1
a (mod p)
x1
0 (mod q)
x2
0 (mod p)
x2
b (mod q)
Notice that x1 + x2 is a solution to the original system because:
x1 + x2
x1 + 0 (mod p)
a + 0 (mod p)
a (mod p)
x1 + x2
0 + x2 (mod q)
0 + b (mod q)
b (mod q)
We know how to solve the above two systems separately:
x1
a ⋅ q ⋅ q-1 (mod (p ⋅ q))
x2
b ⋅ p ⋅ p-1 (mod (p ⋅ q))
Thus, we have the solution to the original system:
x
x1 + x2 (mod (p ⋅ q))
We have shown that we can solve a system of equations with a solution according to CRT if the moduli in the equations are both prime. What if the moduli are merely coprime? So far, we only needed a way to compute multiplicative inverses of numbers modulo a prime, and Fermat's little theorem was sufficient for this purpose. However, if the moduli are not prime, we need some other method to compute multiplicative inverses.
Fact: For any m ℕ, an x ℤ/mℤ has an inverse with respect to multiplication defined over ℤ/mℤ (i.e., a multiplicative inverse) iff gcd(x,m) = 1.
Fact (Bezout's identity): For any two integers x ℤ, y ℤ where x ≠ 0 or y ≠ 0, let z = gcd(x,y). Then there exist a ℤ and b ℤ such that:
a ⋅ x + b ⋅ y
=
z
Fact: For any two integers x ℤ, y ℤ where x ≠ 0 or y ≠ 0, and gcd(x,y) = 1, there exist a ℤ and b ℤ such that:
a ⋅ x + b ⋅ y
=
1
This fact is a special case of Bézout's identity (i.e., the case in which gcd(x,y) = 1).
Example: Suppose we have s,t ℤ such that:
5 ⋅ s + 3 ⋅ t
=
1
We can then do the following:
− 5 ⋅ t + (5 ⋅ s + 3 ⋅ t) + 5 ⋅ t
=
1
(5 ⋅ s − 5 ⋅ t) + (3 ⋅ t + 5 ⋅ t)
=
1
5 ⋅ (s − t) + 8 ⋅ t
=
1
Thus, we have converted a instance of Bézout's identity for 5 and 3 into an instance of Bézout's identity for 5 and 8.
We can repeat the above as many times as we want. Suppose we instead want Bézout's identity for 3 and 13. We can do the following:
− 5 ⋅ 2 ⋅ t + (5 ⋅ s + 3 ⋅ t) + 5 ⋅ 2 ⋅ t
=
1
(5 ⋅ s − 5 ⋅ 2 ⋅ t) + (3 ⋅ t + 5 ⋅ 2 ⋅ t)
=
1
5 ⋅ (s − 2 ⋅ t) + 13 ⋅ t
=
1
Fact: For any two integers a, b, s, t ℤ, suppose we have that:
a ⋅ s + b ⋅ t
=
1
Let us assume that a > b and that a mod b = r (in other words, for some k,
a mod b
=
r
b ⋅ k + r
=
a
Then we have that:
a ⋅ s + b ⋅ t
=
1
− b ⋅ k ⋅ s + (a ⋅ s + b ⋅ t) + b ⋅ k ⋅ s
=
1
(a ⋅ s − b ⋅ k ⋅ s) + (b ⋅ (t + k ⋅ s))
=
1
(a − b ⋅ k) ⋅ s + b ⋅ (t + k ⋅ s)
=
1
r ⋅ s + b ⋅ (t + k ⋅ s)
=
1
(a mod b) ⋅ s + b ⋅ (t + k ⋅ s)
=
1
Thus, for any instance of Bézout's identity for a and b and a > b, there must exist an instance of Bézout's identity for a mod b and b.
The above fact suggests that if we want to find the s and t coefficients for an equation as + bt = 1 given a > b, we should try finding Bézout's identity for a mod b and b. But notice that:
a mod b
<
b
The above implies that the problem of finding the coefficients for an instance of Bézout's identity can be reduced to a smaller version of the problem: find Bézout's identity for a mod b and b can then be reduced further to finding b mod (a mod b) and a mod b. At this point, we have a strictly smaller instance of the problem:
a mod b
<
b
<
a
b mod (a mod b)
<
b
<
a
Thus, we can use recursion; the recursive algorithm that solves this problem is called the extended Euclidean algorithm, and is a modification of the recursive algorithm that computes the gcd of two numbers.
Algorithm (extended Euclidean algorithm): The collection of equations considered in the Chinese remainder theorem can be solved constructively (i.e., in a way that provides a concrete solution and not just a proof that a solution exists) by applying an extended version of the greatest common divisor algorithm. We provide the definition of the algorithm below.
1. extended Euclidean algorithm: x ℤ, y
1. if y = 0
1. (s,t) := (1, 0)
2. return (s,t)
2. otherwise
1. (s,t) := extended Euclidean algorithm(y, x mod y)
2. return (t, s - (⌊ x/y ⌋ ⋅ t) )
Given two inputs x ℤ, y ℤ, the extended Euclidean algorithm returns two integers u, v such that
u ⋅ x + v ⋅ y
=
gcd(x,y)
We can check that the above is indeed a solution to xa (mod m). Consider the following:
u ⋅ m + v ⋅ n
=
1
v ⋅ n
=
1 - u ⋅ m
v ⋅ n
1 (mod m)
Furthermore, we have that:
((u ⋅ m) ⋅ b) mod m
=
0
Then, we can conclude:
((u ⋅ m) ⋅ b + (v ⋅ n) ⋅ a) mod m
=
0 + ((v ⋅ n) ⋅ a) mod m
=
0 + (1 ⋅ a) mod m
=
a mod m
Using a similar argument, we can show that the solution is also equivalent to b (mod m).
Example: Suppose we want to find the multiplicative inverse of 49 in ℤ/100ℤ and the multiplicative inverse of 100 in ℤ/49ℤ. We run the extended Euclidean algorithm on the inputs 49 and 100 to obtain the following instance of Bézout's identity:
(-24) ⋅ 100 + 49 ⋅ 49
=
1
We can use the above to find the multiplicative inverse of 49 in ℤ/100ℤ:
(-24) ⋅ 100 + 49 ⋅ 49
=
1
(-24) ⋅ 100 + 49 ⋅ 49
1 (mod 100)
49 ⋅ 49
1 (mod 100)
Thus, 49-1 = 49 in ℤ/100ℤ. We can also find the multiplicative inverse of 100 in ℤ/49ℤ (also known as 2 ℤ/49ℤ):
(-24) ⋅ 100 + 49 ⋅ 49
=
1
(-24) ⋅ 100 + 49 ⋅ 49
1 (mod 49)
-24 ⋅ 100
1 (mod 49)
25 ⋅ 100
1 (mod 49)
Thus, 100-1 = 25 in ℤ/49ℤ.
Example: Suppose we want to solve the following system:
x
23 (mod 100)
x
31 (mod 49)
We use the extended Euclidean algorithm to find that:
(-24) ⋅ 100 + 49 ⋅ 49
=
1
This tells us that −24 is the inverse of 100 in ℤ/49ℤ and that 49 is the inverse of 49 in ℤ/100ℤ. Thus, to build 31 in ℤ/49ℤ, we need:
31
1 ⋅ 31 (mod 49)
(100 ⋅ 100-1) ⋅ 31
(100 ⋅ -24) ⋅ 31
To build 23 in ℤ/100ℤ, we need:
23
1 ⋅ 23 (mod 100)
(49 ⋅ 49-1) ⋅ 23
(49 ⋅ 49) ⋅ 23
Then the solutions to the system are in the congruence class:
x
(100 ⋅ -24) ⋅ 31 + (49 ⋅ 49) ⋅ 23 (mod (100 ⋅ 49))
-19177 mod 4900
423
Algorithm: Suppose we are given a collection of equations of the following form such that m1,...,mk are all pairwise coprime.
x
a1 (mod m1)
x
ak (mod mk)
Let C be the set of these equations, where Ci is the ith equation. The following algorithm can be used to find a solution for this system of equations.
1. solve system of equations: C is a set of constraints xai mod mi
1. while |C| > 1
1. remove two equations Ci and Cj from C and solve them to obtain a new equation xc (mod mimj)
2. add the new equation to C
2. return the one equation left in C
### [link] 3.12. More practice with CRT
Example: Solve the following equation for x ℤ/5ℤ by multiplying both sides by the appropriate multiplicative inverse:
3 ⋅ x
2 (mod 5)
Example: Solve the following system of equations for x ℤ/35ℤ by finding multiplicative inverses of 5 ℤ/7ℤ and 7 ℤ/5ℤ:
x
4 (mod 5)
x
2 (mod 7)
Example: Suppose you know that 7-1 ≡ 3 (mod 10). Solve the following system of equations:
x
0 (mod 2)
x
1 (mod 5)
x
3 (mod 7)
We first solve the first two equations. We know x must be a multiple of 2 that is in 1 + 5ℤ. Thus, we set x = 2 ⋅ y and we solve:
x
1 (mod 5)
2 ⋅ y
1 (mod 5)
Then we know that:
y
2-1 (mod 5)
25-2
23
8
3
Thus, x ≡ 2 ⋅ 3 ≡ 6 (mod (2 ⋅ 5)). This leaves two equations:
x
6 (mod 10)
x
3 (mod 7)
We first find 10-1 (mod 7):
10-1
107-2 (mod 7)
105
5
We now have 10-1 ≡ 5 (mod 7) and 7-1 ≡ 3 (mod 10), so now we can express 6 as a multiple of 7 in ℤ/10ℤ, and we can express 3 as a multiple of 10 in ℤ/7ℤ:
x
6 (mod 10)
6 ⋅ 1
6 ⋅ (7 ⋅ 7-1)
x
3 (mod 7)
3 ⋅ 1
3 ⋅ (10 ⋅ 10-1)
Thus, we add these two terms to obtain our solution in ℤ/(7 ⋅ 10)ℤ:
x
6 ⋅ (7 ⋅ 7-1) + 3 ⋅ (10 ⋅ 10-1) (mod 70)
6 ⋅ (7 ⋅ 3) + 3 ⋅ (10 ⋅ 5)
6 ⋅ 21 + 3 ⋅ 50
126 + 150
276
66
Example: Suppose we have a single processor that can perform arithmetic operations (addition, subtraction, multiplication, and modulus) on integers that can be represented with at most 11 bits (211 = 2048). On this processor, a single arithmetic operation can be performed in 11 time steps. We also have three other processors that can perform arithmetic on integers that can be represented with at most 4 bits (24 = 16). Each of these processors can perform an arithmetic operation on 4-bit integers in 4 time steps.
For example, suppose we want to perform 1000 arithmetic operations on 11-bit integers. Using a single processor, this would require:
1000 ⋅ 11 = 11,000 time steps
If we use three coprime numbers 13, 14, and 15, and we use each of the three 4-bit processors to perform these operations modulo 13, 14, and 15 in parallel, 1000 operations would require:
1000 ⋅ 4 = 4,000 time steps
Note that 13 ⋅ 14 ⋅ 15 = 2730, and that 2730 > 2048, so:
13 ⋅ 14 ⋅ 15
>
211
Suppose it takes 1400 time steps to solve a system of three congruence equations of the following form:
x
a (mod 13)
x
b (mod 14)
x
c (mod 15)
If we want to perform the computations as quickly as possible and we can use either the 11-bit processor or the three 4-bit processors, how many operations k would we need to perform before we decided to switch from the the 11-bit processor to the 4-bit processors?
To solve this problem, we can first write down two function describing the time cost of the two strategies for k operations:
f(k)
=
k ⋅ 11
g(k)
=
k ⋅ 4 + 1400
Notice that f(1) < g(1), so if we only wanted to perform k = 1 operation, we would prefer to use the 11-bit processor. The point at which we would want to switch would be when f(k) ≥ g(k), so we can find the point at which the two functions intersect:
f(k)
g(k)
k ⋅ 11
k ⋅ 4 + 1400
k ⋅ 7
1400
k
200
Thus, if k ≥ 200, we want to use the 4-bit processors.
Example: Suppose we are using echolocation to measure the distance to a wall that is at most 15 distance units away. We have two devices that emit sounds at two different frequencies. One device emits sound every 3 seconds, while the other device emits a sound every 11 seconds. Suppose we hear the following:
• the device that emits a sound every 3 seconds hears a response 2 seconds after each time it emits a sound;
• the device that emits a sound every 11 seconds hears a response 4 seconds after each time it emits a sound.
If sound travels one distance unit per second, how far away is the wall?
Example: Suppose Alice, Bob, and Eve are using the Shamir secret sharing protocol to store a combination for a lock; all three participants would need to work together to retrieve the secret lock combination in ℤ/60ℤ. They are each given the following equations:
Alice:
x
1 (mod 3)
Bob:
x
3 (mod 4)
Eve:
x
2 (mod 5)
1. What is the lock combination?
2. The lock only permits anyone to try two incorrect combinations before locking down completely and becoming inaccessible. Suppose Eve has a chance to steal either Bob's secret information or Alice's secret information, but she can only choose one. Whose information should she steal in order to unlock the lock?
Example: Suppose we want to store a number n between 0 and 500,000 on a collection of 5-bit memory regions. However, we want to make sure that if any one of the memory regions is turned off, we can still recover the number exactly, without any missing information or errors. How many memory regions will we need to use? Note that 323 = 32,768 and 324 = 1,048,576.
We will need five memory regions. Suppose we choose five coprime numbers less than or equal to 25 = 32: 25,27,29,31, and 32. Note that any product of four of these numbers is greater than 500,000 because 25 ⋅ 27 ⋅ 29 ⋅ 31 = 606,825. Then with five 5-bit memory regions, we can store each of the following five values in one region:
n mod 25
n mod 27
n mod 29
n mod 31
n mod 32
If any of the above values are lost, it is still possible to recover n by solving a system with four equations. In the worst case, the product of the moduli would be 606,825 > 500,000.
Example: Suppose we make the following simplifications: for every t years,
• when the Earth revolves around the sun, it travels a circumference of 1 unit, at a rate of 1 ⋅ t (once per year);
• when the asteroid Ceres revolves around the sun, it travels a circumference of 5 units;
• when the planet Jupiter revolves around the sun, it travels a circumference of 11 units.
Suppose that on June 21st, 2000, the Earth, Ceres, and Jupiter all align (i.e., one can draw a straight line through all three). Next, suppose that it is June 21st of some year between 2000 and 2055. At this time, there is no alignment. However, Jupiter aligned with earth on June 21st two years ago, and Ceres aligned with Earth on June 21st three year ago. What year is it?
### [link] 3.13. Euler's totient function, Euler's theorem, and applications
Definition: For any input m ℕ, define Euler's totient function φ by:
φ(m)
=
|{k | k ∈ {1,...,m}, gcd(k,m) = 1}|
Example: Compute φ(15).
φ(15)
=
|{k | k ∈ {1,...,15}, gcd(k,15) = 1}|
=
|{1,2,4,7,8,11,13,14}|
=
8
Example: Suppose p ℕ is a prime number. What is φ(p)?
φ(p)
=
|{k | k ∈ {1,...,p}, gcd(k,p) = 1}|
=
|{1,2,3,...,p-1}|
=
p-1
Example: What is φ(15)?
φ(15)
=
|{k | k ∈ {1,...,15}, gcd(k,15) = 1}|
=
15 - |{k | k ∈ {1,...,15}, gcd(k,15) ≠ 1}|
=
15 - |{3,6,9,12,15} ∪ {5,10,15}|
=
15 - |{3,6,9,12}| - |{5,10}| - |{15}|
=
15 - (5-1) - (3-1) - 1
=
15 - 5 - 3 + 1 + 1 - 1
=
15 - 5 - 3 + 1
=
(3 ⋅ 5) - 5 - 3 + 1
=
(3-1) ⋅ (5-1)
=
2 ⋅ 4
=
8
Fact: For any x ℕ and y ℕ, if gcd(x,y) = 1 then:
φ(x) ⋅ φ(y)
=
φ(x ⋅ y)
Example: Suppose p ℕ and q ℕ are prime numbers. What is φ(pq)?
φ(p ⋅ q)
=
φ(p) ⋅ φ(q)
=
(p-1) ⋅ (q-1)
Fact: For any prime p ℕ, we have that:
φ(pk)
=
pk - pk-1
Fact: For any a ℕ and m ℕ, if am-1 mod m = 1 then:
am-1 mod m
=
1
am-1
=
1 + k ⋅ m
1
=
gcd(1 + k ⋅ m, k ⋅ m)
=
gcd(am-1, k ⋅ m)
=
gcd(a, k ⋅ m)
=
gcd(a, m)
Thus, a and m are coprime.
Example: Suppose m ℕ is a Carmichael number. At most how many Fermat liars does m have?
Fact: We can use φ to provide a formula for the probability that the Fermat primality test will detect that a Carmichael number m ℕ is actually composite. It is approximately:
(m - φ(m)) / m
To be more precise (since we do not check 0 or 1 in our actual implementation), it is:
((m - 3) - φ(m)) / (m - 3)
Unfortunately, Euler's totient function does not in general have a better upper bound than f(m) = m.
Example: How many elements of ℤ/mℤ have a multiplicative inverse in ℤ/mℤ? Since an x ℤ/mℤ has an inverse iff gcd(x,m) = 1. Thus, the set of such x is exactly the set {x | x {1,...,m}, gcd(k,m) = 1}. But this is the definition of φ(m). Thus, there are φ(m) elements in ℤ/mℤ that have a multiplicative inverse.
Theorem (Euler's theorem): For any m ℕ and a ℤ/mℤ, if gcd(m,a) = 1 then we have that:
aφ(m) mod m
=
1
Notice that if m is a prime number, then φ(m) = m-1. Then for any a ℤ/mℤ, gcd(a,m) = 1 and am-1 mod m = 1. This is exactly the statement of Fermat's little theorem. Thus, Euler's theorem is a generalization of Fermat's little theorem.
Fact: For any m ℕ and a ℤ/mℤ, if gcd(m,a) = 1 then for any i ℤ/φ(m)ℤ such that i ≡ 0 we have that
ai mod m
=
1
This is because:
i
0 (mod φ(m))
=
k ⋅ φ(m)
aφ(m) ⋅ k mod m
=
(aφ(m))k mod m
=
1k mod m
=
1 mod m
Fact: For any p ℕ, if p is prime and a ℤ/pℤ then for any k ℤ we have that:
ak mod p
=
a(k mod (p-1)) mod p
Fact: For any m ℕ and a ℤ/mℤ, if gcd(m,a) = 1 then for any k ℤ we have that:
ak mod m
=
a(k mod φ(m)) mod m
Example: We can compute the integer value 238 mod 7 as follows because 7 is prime:
238
238 mod (7-1) (mod 7)
238 mod 6
22
4
Since the final operation in the integer term is a modulus operation, the congruence class 4 is also exactly the integer result of the term.
Example: We can compute 4210000000 mod 5 as follows because 5 is prime:
4210000000
4210000000 mod (5-1) (mod 5)
=
4210000000 mod 4
=
40
=
1
Example: We can compute 48100+ 3 mod 15 as follows because gcd(4,15) = 1:
4(8100 + 3)
4(8100 + 3) mod φ(15) (mod 15)
=
4(8100 + 3) mod ((5-1) ⋅ (3-1))
=
4(8100 + 3) mod 8
=
43
=
64
=
4
Example: Compute 56603 mod 7.
Fact: For any m ℕ and a ℤ/mℤ where gcd(m,a) = 1, we can use the Euler's theorem to find the inverse of a. Notice that:
aφ(m) mod m
=
1
(aφ(m)-1 ⋅ a) mod m
=
1
Thus, aφ(m)-1 mod m is the multiplicative inverse of a in ℤ/mℤ.
Example: Find the multiplicative inverse of 52 in ℤ/7ℤ.
It is sufficient to notice that 56 ≡ 1 (mod 7), so 52 ⋅ 54 ≡ 1, so 54 is the inverse of 52 in ℤ/7ℤ.
Example: We can find the multiplicative inverse of 3 in ℤ/22ℤ using the following steps. We first compute φ(22) = 10.
φ(22)
=
φ(11 ⋅ 2)
=
φ(11) ⋅ φ(2)
=
(11 − 1) ⋅ (2 − 1)
=
10 ⋅ 1
=
10
Next, we compute the inverse using Euler's theorem.
3-1
3φ(22) − 1 (mod 22)
310 − 1
39
33 ⋅ 33 ⋅ 33
5 ⋅ 5 ⋅ 5
25 ⋅ 5
3 ⋅ 5
15
Definition: For m ℕ, We define (ℤ/mℤ)* to be the following subset of ℤ/mℤ:
(ℤ/mℤ)*
=
{ a | a ∈ ℤ/mℤ, a has an inverse in ℤ/mℤ }
Example: Does 11 have an inverse in ℤ/22ℤ (i.e., is it true that 11 (ℤ/22ℤ)*)?
Example: Compute |(ℤ/35ℤ)*|.
|(ℤ/35ℤ)*|
=
|{ a | a ∈ ℤ/35ℤ, a has an inverse in ℤ/35ℤ }|
=
|{ a | a ∈ ℤ/35ℤ, gcd(a,35) = 1 }|
=
|{ a | a ∈ ℤ/35ℤ, gcd(a,35) = 1 }|
=
φ(35)
=
φ(5 ⋅ 7)
=
φ(5) ⋅ φ(7)
=
4 ⋅ 6
=
24
Fact: For any m ℕ, (ℤ/mℤ)* is closed under multiplication modulo m. That is, for any a ℤ/mℤ and b ℤ/mℤ, if there exist a-1 ℤ and b-1 ℤ then (ab) has an inverse (a-1b-1). We can use the commutativity of multiplication to show this:
(a ⋅ b) ⋅ (a-1 ⋅ b-1)
(a ⋅ a-1) ⋅ (b ⋅ b-1)
1 ⋅ 1
1
### [link] 3.14.Assignment #3: Multiplicative Inverses, CRT, and Efficient Computation
In this assignment you will solve several equations, and you will define a collection of Python functions for finding inverses of congruence classes, for solving systems of equations using the Chinese remainder theorem, and for employing CRT solutions. You must submit a single Python source file named `hw3/hw3.py`. Please follow the gsubmit directions.
You may import the following library functions in your module (you may not need all these functions for this assignment depending on how you approach the problems, but they may be used):
from math import floor
from fractions import gcd
You may also use the following built-in functions:
• the `pow()` function can compute modular exponents efficiently (as in, ak mod n can be written in Python as `pow(a,k,n)`);
• the `sum()` function returns the sum of a list of integers (e.g., `sum(1,2,3,4)` returns `10`).
Your file may not import any other modules or employ any external library functions associated with integers and sets unless explicitly permitted to do so in a particular problem.
Solutions to each of the programming problem parts below should be fairly concise. You will be graded on the correctness, concision, and mathematical legibility of your code. The different problems and problem parts rely on the lecture notes and on each other; carefully consider whether you can use functions from the lecture notes, or functions you define in one part within subsequent parts.
1. Solve the following equations using step-by-step equational reasoning, and list each step. For this problem, you must use Fermat's little theorem and/or Euler's theorem to compute inverses of congruence classes. Your solutions for this problem should appear as comments, delimited using `'''`...`'''`, in `hw3.py`. You may use the `=` ascii character to represent the ≡ relational operator on congruence classes.
1. Solve the following equation for x ℤ/5ℤ:
8 ⋅ x ≡ 2 (mod 5)
2. Solve the following system of equations for x ℤ/35ℤ:
x
1 (mod 7)
x
3 (mod 5)
3. Let p and q be unequal prime numbers. If p-1 ≡ 5 (mod q) and q-1 ≡ 3 (mod p), find the solution x ℤ/(pq)ℤ to the following system of equations (your solution should be in terms of p and q):
x
4 (mod p)
x
2 (mod q)
4. Solve the following system of equations for a unique congruence class x, and specify the range of congruence classes in which this system of equations has a unique solution:
2 ⋅ x
3 (mod 5)
x
6 (mod 14)
2. Implement the following Python functions for computing multiplicative inverses of congruence classes.
1. Implement a function `invPrime(a, p)` that takes two integers `a` and `p` > 1 where `p` is prime. The function should return the multiplicative inverse of `a` ℤ/`p`ℤ (if `a` ≡ 0, it should return `None`). Your solution must use Fermat's little theorem.
>>> [invPrime(i, 7) for i in range(0,7)]
[None, 1, 4, 5, 2, 3, 6]
>>> [invPrime(i, 13) for i in range(1,13)]
[1, 7, 9, 10, 8, 11, 2, 5, 3, 4, 6, 12]
2. Include the following definition in your code. This non-recursive implementation of the extended Euclidean algorithm avoids a stack overflow error on large inputs.
def egcd(a, b):
(x, s, y, t) = (0, 1, 1, 0)
while b != 0:
k = a // b
(a, b) = (b, a % b)
(x, s, y, t) = (s - k*x, x, t - k*y, y)
return (s, t)
Given two inputs `a` and `b`, `egcd(a, b)` returns a solution `(s, t)` to the following instance of Bézout's identity:
`a` ⋅ `s` + `b` ⋅ `t`
=
`gcd(a, b)`
Using `egcd()`, implement a function `inv(a, m)` that takes two integers `a` and `m` > 1. If `a` and `m` are coprime, it should return the multiplicative inverse of `a` ℤ/`m`ℤ. If `a` and `m` are not coprime, it should return `None`.
>>> [inv(i, 13) for i in range(1,13)]
[1, 7, 9, 10, 8, 11, 2, 5, 3, 4, 6, 12]
>>> [inv(i, 8) for i in range(1,8)]
[1, None, 3, None, 5, None, 7]
3. Implement the following Python functions for solving certain systems of equations involving congruence classes.
1. Implement a function `solveOne(c, a, m)` that takes three integers `c`, `a`, and `m` ≥ 1. If `c` and `m` are coprime, the function should return the solution x {0, ..., `m`-1} to the following equation:
`c` ⋅ x
`a` (mod `m`)
If `c` and `m` are not coprime, the function should return `None`.
>>> solveOne(3, 4, 7)
6
>>> solveOne(1, 5, 11)
5
>>> solveOne(2, 3, 8)
None
2. Implement a function `solveTwo(e1, e2)` that takes two tuples `e1` and `e2` as inputs, each of the form `(c, a, m)` (i.e., containing three integer elements). Each tuple `(c, a, m)` corresponds to an equation of the form:
`c` ⋅ x
`a` (mod `m`)
Thus, the two tuples, if we call them `(c, a, m)` and `(d, b, n)`, correspond to a system of equations of the form:
`c` ⋅ x
`a` (mod `m`)
`d` ⋅ x
`b` (mod `n`)
The function `solveTwo()` should return the unique solution x to the above system of equations. If either equation cannot be solved using `solveOne()`, or `n` and `m` are not coprime, the function should return `None`.
>>> solveTwo((3, 4, 7), (1, 5, 11))
27
3. Implement a function `solveAll(es)` that takes a list of one or more equations, each of the form `(c, a, m)`. The list corresponds to the system of equations (assume all the mi are mutually coprime):
c1 ⋅ x
a1 (mod m1)
ck ⋅ x
ak (mod mk)
The function `solveAll()` should return the unique solution x to the above system of equations. If any individual equation cannot be solved using `solveOne()`, or if the moduli are not all mutually coprime, the function should return `None`.
>>> solveAll([(3,4,7), (1,5,11)])
27
>>> solveAll([(5,3,7), (3,5,11), (11,4,13)])
856
>>> solveAll([(1,2,3), (7,8,31), (3,5,7), (11,4,13)])
7109
>>> solveAll([(3,2,4), (7,8,9), (2,8,25), (4,4,7)])
554
4. Suppose we represent the sum of a collection of exponentiation operations as a list of tuples, where each tuple contains two integers: the base and the exponent. For example, the list `[(2,4),(3,5),(-6,3)]` represents the sum of powers 24 + 35 + (−6)3.
1. Implement a function `sumOfPowers(nes, ps)` that takes a list of one or more tuples `nes` (i.e., `nes` is of the form `[(a1,n1),...,(ak,nk)]`) as its first argument, and a list of one or more primes `ps` (i.e., of the form `[p1,...,pm]`) as its second argument. The function should return the correct result of the sum of powers as long as the following is true (e.g., on a computer with unlimited memory and time):
0
`a1``n1` + ... + `ak``nk`
<
`p1` ⋅ ... ⋅ `pm`
You may assume the second list contains distinct prime numbers. You may not assume that the numbers in the first input list have any particular patterns or relationships; they can be in any order, they can be of any size, and they may or may not share factors. Your implementation must work efficiently on very large inputs (e.g., with computations like 229999999999999999999999999999999996, as presented in the examples below).
>>> sumOfPowers([(2,3), (5,6)], [3,5,7,11,13,17,19,23,29]) == 2**3 + 5**6
True
>>> primes =[\
15481619,15481633,15481657,15481663,15481727,15481733,15481769,15481787
,15481793,15481801,15481819,15481859,15481871,15481897,15481901,15481933
,15481981,15481993,15481997,15482011,15482023,15482029,15482119,15482123
,15482149,15482153,15482161,15482167,15482177,15482219,15482231,15482263
,15482309,15482323,15482329,15482333,15482347,15482371,15482377,15482387
,15482419,15482431,15482437,15482447,15482449,15482459,15482477,15482479
,15482531,15482567,15482569,15482573,15482581,15482627,15482633,15482639
,15482669,15482681,15482683,15482711,15482729,15482743,15482771,15482773
,15482783,15482807,15482809,15482827,15482851,15482861,15482893,15482911
,15482917,15482923,15482941,15482947,15482977,15482993,15483023,15483029
,15483067,15483077,15483079,15483089,15483101,15483103,15483121,15483151
,15483161,15483211,15483253,15483317,15483331,15483337,15483343,15483359
,15483383,15483409,15483449,15483491,15483493,15483511,15483521,15483553
,15483557,15483571,15483581,15483619,15483631,15483641,15483653,15483659
,15483683,15483697,15483701,15483703,15483707,15483731,15483737,15483749
,15483799,15483817,15483829,15483833,15483857,15483869,15483907,15483971
,15483977,15483983,15483989,15483997,15484033,15484039,15484061,15484087
,15484099,15484123,15484141,15484153,15484187,15484199,15484201,15484211
,15484219,15484223,15484243,15484247,15484279,15484333,15484363,15484387
,15484393,15484409,15484421,15484453,15484457,15484459,15484471,15484489
,15484517,15484519,15484549,15484559,15484591,15484627,15484631,15484643
,15484661,15484697,15484709,15484723,15484769,15484771,15484783,15484817
,15484823,15484873,15484877,15484879,15484901,15484919,15484939,15484951
,15484961,15484999,15485039,15485053,15485059,15485077,15485083,15485143
,15485161,15485179,15485191,15485221,15485243,15485251,15485257,15485273
,15485287,15485291,15485293,15485299,15485311,15485321,15485339,15485341
,15485357,15485363,15485383,15485389,15485401,15485411,15485429,15485441
,15485447,15485471,15485473,15485497,15485537,15485539,15485543,15485549
,15485557,15485567,15485581,15485609,15485611,15485621,15485651,15485653
,15485669,15485677,15485689,15485711,15485737,15485747,15485761,15485773
,15485783,15485801,15485807,15485837,15485843,15485849,15485857,15485863]
>>> sumOfPowers(\
[(2,29999999999999999999999999999999996)\
,(-8,9999999999999999999999999999999999)\
,(2,29999999999999999999999999999999996)\
,(7,7),(-13,3)], primes)
821346
2. Extra credit: Modify your implementation of `sumOfPowers()` so that it can handle inputs even if the exponents themselves are extremely large. You must use Euler's theorem to accomplish this; you may not assume that any particular patterns will exist in the bases or exponents.
>>> sumOfPowers([(2,10**1000000 + 1), (-2,10**1000000 + 1), (3,3)], primes)
27
## [link] Review #1. Properties, Algorithms, and Applications of Modular Arithmetic
This section contains a comprehensive collection of review problems going over the course material covered until this point. Many of these problems are an accurate representation of the kinds of problems you may see on an exam.
Exercise: For some a ℕ, suppose that a-1 ℤ/21ℤ and a-1 ℤ/10ℤ (that is, a has an inverse in ℤ/21ℤ, and it also has an inverse in ℤ/10ℤ). Determine whether or not a has an inverse in ℤ/210ℤ. Explain why or why not. Hint: use gcd.
If a has an inverse in ℤ/10ℤ and ℤ/21ℤ, then gcd(a,10) = 1 and gcd(a,21) = 1. Since gcd(10,21) = 1, a shares no factors with the product 10 ⋅ 21 = 210, so gcd(a, 210) = 1. Thus, a must have a multiplicative inverse in ℤ/210ℤ.
Exercise: Bob is trying to implement a random number generator. However, he's distracted and keeps making mistakes while building his implementation.
1. Bob begins his algorithm by generating two coprime numbers a and m such that gcd(a,m) = 1. However, he mixes them up and defines the following computation:
[ (i ⋅ m) mod a | i ∈ {1,...,a-1} ]
Is Bob going to get a permutation? Why or why not?
Yes, Bob will still get a permutation because gcd(a,m) = 1, which means m mod a is also coprime with m (we can see this because this is exactly what the extended Euclidean algorithm computes before making a recursive call, or by observing that if m mod a shares a factor with a, so must m).
2. Bob notices part of his mistake and tries to fix his algorithm; he ends up with the following:
[ (i ⋅ m) mod m | i ∈ {1,...,m-1} ]
How many distinct elements does the list he gets in his output contain?
It contains exactly one element: 0 (mod m), since all multiples of the congruence class m are equivalent to 0 (mod m).
3. Bob notices his algorithm isn't returning a permutation, but he mixes up a few theorems and attempts the following fix:
[ (i ⋅ am-1) mod m | i ∈ {1,...,m-1} ]
Bob tests his algorithm on some m values that are prime numbers. How many elements does the set he gets in his output contain?
It contains all the elements {1,...,m-1}, generated in ascending order, since by Fermat's little theorem, am-1 ≡ 1 (mod m) if m is prime.
4. Bob doesn't like the fact that his permutation doesn't look very random, so he moves the i term to the exponent:
[ (ai ⋅ (m-1)) mod m | i ∈ {1,...,m-1} ]
Bob tests his algorithm on some m values that are prime numbers. How many elements does the set he gets in his output contain?
It contains exactly one element: 1 (mod m), since by the consequences of Fermat's little theorem and Euler's theorem, if m is prime we have:
ak
ak mod φ(m) (mod m)
ak mod (m-1) (mod m)
Exercise: Suppose you have the following instance of Bézout's identity: 2 ⋅ 3 + (−1) ⋅ 5 = 1. Solve the following system of equations:
x
2 (mod 3)
x
3 (mod 5)
Given 3-1 ≡ 2 (mod 5), and 5-1 ≡ −1 ≡ 2 (mod 3), the formula for the solution is:
x
2 ⋅ (5 ⋅ 5-1) + 3 ⋅ (3 ⋅ 3-1) (mod (3 ⋅ 5))
2 ⋅ (5 ⋅ 2) + 3 ⋅ (3 ⋅ 2)
20 + 18
38 (mod 15)
8 (mod 15)
Exercise: Solve the following system of equations:
x
2 (mod 7)
x
3 (mod 5)
To demonstrate an alternative but inefficient method to find a solution, we can list the positive members of the equivalence classes 2 ℤ/7ℤ and 3 ℤ/5ℤ that are less than 7 ⋅ 5 = 35. The one element that appears in both lists is the unique solution in ℤ/35ℤ to the above system.
2 + 7ℤ
=
{..., 2, 9, 16, 23, 30, ...}
3 + 5ℤ
=
{..., 3, 8, 13, 18, 23, 28, 33, ...}
Exercise: Determine the size of the following set:
{x | x ∈ ℤ/(11 ⋅ 13)ℤ, x ≡ 5 mod 11, x ≡ 7 mod 13}
By the Chinese remainder theorem, we know there exists exactly one solution in ℤ/(11 ⋅ 13)ℤ to the following system of equations:
x
5 (mod 11)
x
7 (mod 13)
Thus, the size of the set is 1.
Exercise: For a given y ℤ/(pq)ℤ where p and q are distinct primes, how many solutions does the following system of equations have:
x
y2 (mod p)
x
y2 (mod q)
The value y2 ℤ/(pq)ℤ is a constant with respect to x, so by the Chinese remainder theorem, there is exactly one solution x ℤ/(pq)ℤ to the above system.
Exercise: Determine the size of the following set:
{x | x ∈ ℤ/(11 ⋅ 13)ℤ, s ∈ ℤ/11ℤ, t ∈ ℤ/13ℤ, x ≡ s mod 11, x ≡ t mod 13 }
Consider the following set. Notice that the right-hand side of the comprehension is exactly the same as that of the above. The only difference is that the left-hand side x in the above expression has been replaced with (x, s, t).
{(x, s, t) | x ∈ ℤ/(11 ⋅ 13)ℤ, s ∈ ℤ/11ℤ, t ∈ ℤ/13ℤ, x ≡ s mod 11, x ≡ t mod 13 }
By the Chinese remainder theorem, we know that exactly one tuple (x, s, t) for each x ℤ/(11 ⋅ 13)ℤ will satisfy the conditions in the comprehension, because for each distinct pair (s, t), exactly one x ℤ/(11 ⋅ 13)ℤ will be a solution to the system of equations:
x
s (mod 11)
x
t (mod 13)
Thus, the conditions in the comprehension will be satisfied at least once for every x ℤ/(11 ⋅ 13)ℤ. Thus, the entire set is exactly the set ℤ/(11 ⋅ 13)ℤ, and it is the case that
|ℤ/(11 ⋅ 13)ℤ|
=
11 ⋅ 13
=
143
Exercise: Suppose that n ℕ is even and n/2 − 1 is odd. Determine the size of the following set:
{i ⋅ (n/2 - 1) mod n | i ∈ {0,...,n-1} }
We know that gcd(n/2 − 1, n/2) = 1. Since n/2 − 1 is odd, 2 is not a factor of n/2 − 1, so gcd(n/2 − 1, n) = 1. Thus, n/2 − 1 and n are coprime. Thus, the above set must be a permutation.
Exercise: For any n ℕ, let a ℤ/nℤ have an inverse a-1 ℤ/nℤ. Determine the size of the following set:
{ (a ⋅ i) mod n | i ∈ ℤ/nℤ }
If a has an inverse in ℤ/nℤ (whether or not n is prime), then gcd(a, n) = 1, so the above set must be a permutation.
Exercise: Let p be a prime number. Compute the set size |ℤ/pℤ - (ℤ/pℤ)*|.
Since (ℤ/pℤ)* is the set of elements of ℤ/pℤ that have multiplicative inverses, it is the set of elements a ℤ/pℤ such that gcd(a, p) = 1. However, because p is prime, all of the elements in ℤ/pℤ except 0 have this property (since gcd(p, 0) = p; recall that 0/p ℤ and p|0). Thus,
ℤ/pℤ - (ℤ/pℤ)*
=
{0}
Thus, |ℤ/pℤ - (ℤ/pℤ)*| = 1.
Alternatively, we know that |ℤ/pℤ| = p and |(ℤ/pℤ)*| = φ(p) = p − 1, and p − (p − 1) = 1.
Exercise: In a game, you win if you can guess correctly whether a large number n is prime in under a minute (if you are wrong, you win nothing and you lose nothing). You are given a handheld calculator that can only perform addition, subtraction, multiplication, division, exponentiation, and modulus (the calculator can represent arbitrarily large numbers, and can provides quotients to any precision). Describe one strategy you can use to give yourself a high probability of winning.
If you pick a few random a {2,...,n-1} and compute an-1 mod n, if the result is ever greater than 1 (i.e., the Fermat primality test) then a would be a witness that n is composite. You would then be able to say with certainty that p is not prime, winning the game. Once time is about to run out, you should guess that the number is prime.
Exercise: Suppose that n ℕ. Compute the following:
534 ⋅ n + 1 mod 11
We can compute the exponent 34 ⋅ n + 1 modulo φ(11) by Euler's theorem, because gcd(5,11) = 1. Likewise, we can compute the exponent within the exponent, 4 ⋅ n + 1, modulo φ(φ(11)) because gcd(3,φ(11)) = 1. Thus, we have:
φ(11)
=
11 - 1
=
10
φ(10)
=
φ(5) ⋅ φ(2)
=
(5-1) ⋅ (2-1)
=
4
534 ⋅ n + 1 mod 11
=
534 ⋅ n + 1 mod φ(φ(11)) mod φ(11) mod 11
=
534 ⋅ n + 1 mod 4 mod 10 mod 11
=
531 mod 10 mod 11
=
53 mod 11
=
(25 ⋅ 5) mod 11
=
(3 ⋅ 5) mod 11
=
15 mod 11
=
4 mod 11
Exercise: Suppose we make the following simplifications:
• the Earth rovolves around the sun once per year;
• the asteroid Ceres rovolves around the sun every 5 years;
• the planet Jupiter revolves around the sun every 11 years.
Suppose that on June 21st, 2000, the Earth, Ceres, and Jupiter all align (i.e., one can draw a straight line through all three).
1. Which two of these objects will align again on June 21st, and in which year?
The next time Earth and Ceres align will be in five years, in 2005, as x = 5 is the smallest non-zero solution in ℕ to the following system (the first equation represents alignment with Earth; the second equation represents alignment with Ceres):
x
0 mod 1
x
0 mod 5
2. How many years will pass before all three align again?
The smallest non-zero solution in ℕ to the following system is x = 55, so they will all align again in 2055:
x
0 mod 1
x
0 mod 5
x
0 mod 11
3. Suppose that it is June 21st of some year between 2000 and 2055. At this time, there is no alignment. However, Jupiter aligned with earth on June 21st four years ago, and Ceres aligned with Earth on June 21st one year ago. What year is it?
The following system of equations captures the situation. Since the solution must be in ℤ/55ℤ, we can find a unique solution using the Chinese remainder theorem. By inspection of the elements of 4 + 11ℤ = {4, 15, 26, ...}, the solution is x = 26.
x
0 mod 1
x
1 mod 5
x
4 mod 11
Alternatively, we can use the formula; since x ≡ 0 (mod 1) is true for all x, there are actually only two equations:
x
1 mod 5
x
4 mod 11
Since 11 ≡ 1 (mod 5), 11-1 ≡ 1 (mod 5); we also have:
5-1
59 (mod 11)
254 ⋅ 5
34 ⋅ 5
5 ⋅ 3 ⋅ 5
4 ⋅ 5
9
Then the solution is:
x
1 ⋅ (11 ⋅ 1) + 4 ⋅ (5 ⋅ 9) (mod (5 ⋅ 11))
11 + 180
191
26
Exercise: Suppose there exist two devices, where one can either produce or consume exactly 2 units of power and another can either produce or consume exactly 7 units of power:
• device A: +/− 2 units
• device B: −/+ 7 units
Suppose we want to produce exactly 1 unit of power using a combination of some number of A devices and B devices. Is this possible?
Yes, this is possible because 2 and 7 are coprime, so there exists an instance of Bézout's identity of the form 2 ⋅ s + 7 ⋅ t = 1.
The following is a breakdown of what you should be able to do at this point in the course (and of what you may be tested on in an exam). Notice that many of the tasks below can be composed. This also means that many problems can be solved in more than one way.
• problems that can be solved
• generate a permutation of a set {1,...,m-1}
• generate a "random" number using permutations
• find the greatest common divisor of two relatively small integers
• check if a number is prime
• using an exhaustive search
• using the Fermat primality test
• generate a random number in a certain range
• generate a random prime in a certain range
• use random primes to...
• to share information requiring cooperation using Shamir secret sharing
• to store information on unreliable storage devices (using Shamir secret sharing)
• to perform many arithmetic operations x1 ⊕ ... ⊕ xn for some operator ⊕ in sequence more efficiently (using CRT)
• if the range of the final output is known
• if working in parallel along multiple distinct pi in ℤ/piℤ, and then performing CRT once is less expensive than working in ℤ/nℤ the whole time
• solving equations and performing computations
• solve a system of linear equations where each equation is modulo some ni and the coefficient is coprime with the moduli
• derive a system of equations from a word problem
• rotating objects
• objects that generate/consume different amounts of power
• Shamir secret sharing and applications
• solve a system with additive and multiplicative inverses
• solve generalized cases (gcd(n,m) > 1 and/or c > 1 in cxa (mod n))
• solve a system of three or four equations
• compute exponents modulo n efficiently
• using Euler's theorem and φ (when possible)
• using the efficient repeated-squaring method
• compute multiplicative inverses ℤ/n
• using φ(n)
• using a given output from the extended Euclidean algorithms or (equivalently) an instance of Bézout's identity
• recognize when you cannot solve problems efficiently
• computing φ(n) for prime n or composite n with no powers of primes in its factorization
• factoring n for an arbitrary n
## [link] 4. Computational Complexity of Modular Arithmetic Algorithms
### [link] 4.1. Definition of computational problems and their complexity
Below, we review a small set of definitions and facts from complexity theory. We will only use these facts as they relate to problems in modular arithmetic and abstract algebra. A course on computational complexity theory would go into more detail.
Definition: Informally, for some formula f, we call a statement of the following form a problem:
• "Given x, find y such that f(x, y) is true."
In the above, x can be viewed as the input describing the problem, and y can be viewed as the solution to the problem.
Definition: The computational complexity of a problem refers to the running time of the most efficient algorithm that can solve the problem.
### [link] 4.2. Complexity of algorithms for solving tractable problems
In this subsection we consider the running time of efficient algorithms for performing common arithmetic operations (addition, subtraction, multiplication, exponentiation, and division). We consider the complexity of these arithmetic operations on each of the following domains:
• unbounded positive integers;
• integers modulo 2n;
• integers modulo k for some k ℕ.
All of our arithmetic algorithms will operate on bit string representations of positive integers. A bit string representation such as
an-1...a0
is defined to represent the integer
2n-1 ⋅ an-1 + ... + 20 ⋅ a0
Below are some specific examples:
111
=
22 ⋅ 1 + 21 ⋅ 1 + 20 ⋅ 1
1101
=
23 ⋅ 1 + 22 ⋅ 1 + 21 ⋅ 0 + 20 ⋅ 1
10
=
21 ⋅ 1 + 20 ⋅ 0
Since the operations we consider usually take two arguments, we will follow the following conventions:
• the first (left-hand side) input is x, an n-bit integer;
• the second (right-hand side) input is y, an m-bit integer.
Thus, x ≤ 2n - 1 and y ≤ 2m - 1.
Algorithm: There exists an algorithm that can compute the sum of an n-bit integer x and an m-bit integer y in time O(max(n,m)+1). The size of the output is O(max(n,m)+1).
1. addition of unbounded positive integers: n-bit integer x, m-bit integer y
1. r (a bit vector to store the result)
2. c := 0 (the carry bit)
3. for i from 0 to max(n,m)-1
1. r[i] := (x[i] xor y[i]) xor c
2. c := (x[i] and y[i]) or (x[i] and c) or (y[i] and c)
4. r[max(n,m)+1] := c
5. return r
How can we use the addition algorithm to implement multiplication? One approach for multiplying two positive integers x, y ℕ is to do repeated addition of y (repeating the addition operations x times). However, if x is an n-bit integer, this would require up to 2n-1 addition operations, which would take exponential time in the representation size of the input x.
A more efficient approach is to use the representation of x as a sum of powers of 2, and to apply the distributive property. Suppose x is represented as the binary bit string an-1...a0. Then we have:
x ⋅ y
=
(an-1 ⋅ 2n-1 + ... + a1 ⋅ 21 + a0 ⋅ 20) ⋅ y
=
an-1 ⋅ 2n-1 ⋅ y + ... + a1 ⋅ 21 + a0 ⋅ 20 ⋅ y
Notice that we have now rewritten multiplication as n-1 addition operations. The only other problem is how to multiply y by powers of 2. We can do so simply by appending a 0 to the bit string representation of y. Suppose y is represented as the binary bit string bn-1...b0. Then we have:
2 ⋅ y
=
2 ⋅ bn-1...b1b0
=
2 ⋅ (bn-1 ⋅ 2n-1 + ... + b1 ⋅ 21 + b0 ⋅ 20)
=
bn-1 ⋅ 2n + ... + b1 ⋅ 22 + b0 ⋅ 21
=
(bn-1 ⋅ 2n + ... + b1 ⋅ 22 + b0 ⋅ 21) + 0 ⋅ 20
=
bn-1...b1b00
Thus, our algorithm only needs to depend on addition, and on shifting bit strings left by one (a.k.a., appending a 0 to the bit string at the position of the least significant bit).
Algorithm: There exists an algorithm that can compute the product of an n-bit integer x and an m-bit integer y in time O(n ⋅ (max(n,m)+1+n)) or O(max(n,m)2). The size of the output is O(n+m) (because the shift left for the 21 case does not contribute to the final result, the m-bit integer is shifted left at most n-1 times, but there may still be a carried bit on the last addition operation performed).
1. multiplication of unbounded positive integers: n-bit integer x, m-bit integer y
1. r (a bit vector to store the result)
2. for i from 0 to n-1
1. if x[i] is 1
1. r := r + y (using unbounded integer addition)
2. shift the bits of y left by one bit (i.e., multiply y by 2)
3. return r
Algorithm: There exists an algorithm that can compute the exponentiation xy of an n-bit integer x and an m-bit integer y in time O(n ⋅ 2m). The size of the output is O(n ⋅ 2m). Notice that this means that for unbounded integer outputs, the algorithm runs in exponential time.
1. exponentiation of unbounded positive integers: n-bit integer x, m-bit integer y
1. r (a bit vector to store the result)
2. for i from 0 to m-1
1. if y[i] is 1
1. r := rx (using unbounded integer multiplication)
2. x := xx (using unbounded integer multiplication)
3. return r
Algorithm: There exists an algorithm that can compute the integer quotient ⌊ x / y ⌋ of an n-bit integer x and an m-bit integer y in time O((nn) + (n ⋅ (2 ⋅ n))) or O(n2).
1. integer division of unbounded positive integers: n-bit integer x, m-bit integer y
1. if y > x
1. return 0
2. for i from 0 to n-1
1. shift y left by one bit
3. r (a bit vector to store ⌊ x / y ⌋ ⋅ y)
4. q (a bit vector to store the integer quotient)
5. p := 2n-1 (to keep track of the current power of 2)
6. for i from 0 to n-1
1. if r+y < x
1. r := r+y (using unbounded integer addition)
2. q := q+p (using unbounded integer addition)
2. shift y right by one bit
3. shift p right by one bit
7. return q
Algorithm: There exists an algorithm that can compute x mod y of an n-bit integer x and an m-bit integer y in time O(n2). This is accomplished by first performing integer division, then an integer multiplication, and then a subtraction. This corresponds to the formula for the modulus operation:
x mod y
=
x - ⌊ x/y ⌋ ⋅ y
When we consider the operations above as operating on integers modulo 2n (with results also in 2n), this corresponds to simply dropping any bits beyond the n least-significant bits when performing the computation.
Fact: There exists an algorithm that can compute the sum of two n-bit integers x and y in time O(n). The size of the output is O(n).
Fact: There exists an algorithm that can compute the product of two n-bit integers x and y in time O(n2). The size of the output is O(n).
Fact: There exists an algorithm that can compute xy for two n-bit integers x and y in time O(n3). The size of the output is O(n).
Fact: The recursive algorithm for gcd (and the extended Euclidean algorithm) makes O(log (max(x,y))) recursive calls on an integer inputs x ℕ and y ℕ. Notice that this means that the number of recursive calls is linear, or O(max(n,m)), for inputs consisting of an n-bit integer x and an m-bit integer y.
To see the above, consider the following fact: for any a ℕ, b ℕ, if ba then a mod b < (1/2) ⋅ a. Consider the two possibilities for a and b:
• if b ≤ (1/2) ⋅ a, then ⌊ a / b ⌋ > 1, so:
(a mod b)
< b
(1/2) ⋅ a
• if b > (1/2) ⋅ a, then ⌊ a / b ⌋ = 1, so:
a mod b
= a - ⌊ a/b ⌋ ⋅ b
= a - 1 ⋅ b
= a - b
< a - ((1/2) ⋅ a)
< (1/2) ⋅ a
Thus, every time a mod b is computed in the algorithms, size of the second paramter is halved. Since every other invocation switches the two parameters, both parameters are halved. Thus, the number of invocations or iterations for an input m is log(m).
Fact: The recursive algorithm for the extended Euclidean algorithm on inputs consisting of an n-bit integer x and an m-bit integer y runs in time O(max(n,m) ⋅ (2 ⋅ max(n,m)2 + max(n,m))), or O(max(n,m)3). The number of recursive calls is about max(n,m), and each recursive call involves an integer division, a multiplication, and a subtraction.
If all inputs and outputs are integers that can be represented with at most n bits, the running time is then O(n3).
Fact: The following problem can be solved in polynomial time: given x (ℤ/nℤ)*, compute x-1. This can be reduced to running the extended Euclidean algorithm, which has a polynomial running time.
If all inputs and outputs are integers that can be represented with at most n bits, the running time is then O(n3).
Fact: There exists an O(max(n,m)3 + (n+m)2) algorithm that can solve the following system of two equations (for n-bit integers x,x' and m-bit integers y,y') using the Chinese remainder theorem:
s
x' (mod x)
s
y' (mod y)
This algorithm calls the extended Euclidean algorithm on x and y, and then performs four multiplications modulo (xy).
If all inputs and outputs are integers that can be represented with at most n bits, the running time is then O(n3).
Exercise: Multiply the following two numbers (represented in binary) using the multiplication algorithm presented in lecture: 1101101.
We can proceed by repeatedly shifting the first input left (corresponding to multiplication by 2), and multiplying each shifted version by a corresponding digit from the second input:
1101 ⋅ 1 + 11010 ⋅ 0 + 110100 ⋅ 1
=
1101 + 110100
=
1000001
### [link] 4.3. Complexity of (probably) intractable problems
In the previous section we saw that addition, subtraction, multiplication, exponentiation, and division (both integer division, modulus, and multiplication by multiplicative inverses) can all be computed efficiently (i.e., in polynomial time) both over integers and over congruence classes. It is also possible to efficiently compute roots and logarithms of integers (we omit proofs of this fact in this course). However, no efficient algorithms are known for computing roots and logarithms of congruence classes.
Definition: A problem can be solved in polynomial time iff there exists for some constant c an algorithm that solves all instances of the problem in time O(nc). The set of all problems that can be solved in polynomial time is called P, and if a problem can be solved in polynomial time, we say that the problem is in P.
Definition: A problem can be solved in exponential time iff there exists an algorithm that solves all instances of the problem in time O(2n).
Definition: There exists a polynomial-time reduction from a problem A to a problem B iff there exists a polynomial-time algorithm that can convert any instance of problem A into an instance of problem B (i.e., convert an input for A into an input for B, and convert the output from B into an output from A).
A polynomial-time reduction from one problem to another can be viewed as two separate polynomial-time algoritms: a conversion algorithm that takes inputs to problem A and invokes a solver for problem B some polynomial number of times, and a conversion algorithm that takes all the outputs obtained from the solver for problem B and assembles and/or converts them into outputs for problem A.
solver forproblem B ⇒⇒⇒ conversionfrom output(s) fromB to output from A ⇑⇑⇑ ⇓ conversionfrom input forA to input(s) for B ⇐ solver forproblem A
We can summarize the above diagram by simply saying that problem A reduces to problem B.
problem B ⇐ problem A
We have already seen examples of such reductions. For example, a CRT solver for two equations makes a single call to the extended Euclidean algorithm. Thus, there exists a polynomial-time reduction from the problem of solving a two-equation system using CRT to the problem of computing multiplicative inverses.
findingmultiplicativeinverses ⇐ solving two-equationsystems using CRT
Fact: If there exists a polynomial-time reduction from problem A to problem B, and problem A is not in P (i.e., there exists no polynomial-time algorithm to solve A), then problem B must not be in P, either.
To see why B cannot be in P, we can present a proof by contradiction. Suppose that there does exist a polynomial-time algorithm to solve problem B. Then the polynomial-time reduction from A to B can invoke a polynomial-time algorithm. But then the reduction and algorithm for B working together will constitute a polynomial-time algorithm to solve A. Then it must be that A is in P. But this contradicts the fact that A is not in P, so no such polynomial-time algorithm for B could exist.
The above fact allows us to make conclusions about the computational complexity of certain problems based on their relationships (in terms of implementation) to other problems.
problem Bpremise:can be solved inpolynomial timeB ∈ P ⇐ problem Aconclusion:can be solved inpolynomial timeA ∈ P problem Bconclusion:cannot be solved inpolynomial timeB ∉ P ⇐ problem Apremise:cannot be solved inpolynomial timeA ∉ P
Intuitively, we can imagine that if problem A is "attached to" (i.e., depends on) problem B, an "easy" problem B will "pull" A down into the set of easily solvable problems P, while a "difficult" problem A will "pull" problem B into the set of hard-to-solve problems.
Conjecture (factoring): The following problem is not in P: given any integer n ℕ where n = pq and p and q are prime, find p and q.
Fact: Suppose that n = pq for two primes p ℕ and q ℕ. Given only n and φ(n), it is possible to compute p and q. Consider the following:
φ(n)
=
(p − 1) ⋅ (q − 1)
φ(n)
=
p ⋅ q − p − q + 1
φ(n)
=
n - p − q + 1
φ(n) - n
=
- p − q + 1
φ(n) - n - 1
=
− p − q
Thus, it is sufficient to solve the following system of equations for p and q:
n
=
p ⋅ q
φ(n) - n - 1
=
− p − q
Example: Suppose that n = 15 and φ(n) = 8. Factor n.
We can plug n and φ(n) into the system of equations derived in the applicable fact:
15
=
p ⋅ q
8 − 15 − 1
=
− p − q
With two equations and two unknowns, we can now solve for p and q:
8 − 15 − 1
=
− p − q
p
=
15 − 8 + 1 − q
=
8 − q
15
=
(8 − q) ⋅ q
0
=
− q2 + 8q − 15
0
=
q2 − 8q + 15
At this point, we use the quadratic equation:
q
=
1/2 ⋅ (8 ± √(64 − 4(1)(15)))
q
=
1/2 ⋅ (8 ± √(4))
q
=
1/2 ⋅ (8 ± 2)
q
{3, 5}
{p, q}
=
{3, 5}
Conjecture (computing φ): The following problem is not in P: given any integer n ℕ where n = pq and p and q are prime, find φ(n).
If we can compute φ(n), then we can compute p and q. If computing φ(n) were any easier than factoring n (e.g., if we had a polynomial-time algorithm for computing φ(n)), then our claim about the hardness of factoring n would be a contradiction. In other words, factoring n can be reduced to solving φ(n).
computing φ(n)conclusion:cannot be solved inpolynomial timecomputing φ(n) ∉ P ⇐ factoring nconjecture:cannot be solved inpolynomial timefactoring ∉ P
The above fact (i.e., that if factoring n is not in P, then neither is computing φ(n)) holds for arbitrary n, not just a product of two primes. However, the proofs in those cases are more sophisticated [Shoup].
Suppose we are given the following equation:
xy
=
z
There are three computational questions we could ask about the above equation:
• given x and y, compute z (this is the exponentiation operation);
• given x and z, compute y (this is the logarithm operation, since we have logx z = y in an equivalent notation);
• given y and z, compute x (this is the yth root operation, since we have y√(z) = x in an equivalent notation).
We have efficient algorithms for computing all three of the above if x, y, and z are all integers or real numbers. Suppose we instead consider the following equation for some n ℕ:
xy
z (mod n)
In other words, we can interpret the equation as a congruence of equivalence classes in ℤ/nℤ. In this case, we already know that the first operation (exponentiation) has an efficient implementation because exponentiation and modulus are both efficient operations. However, we believe that the other two operations (computation of logarithms and roots of congruence classes) are computationally difficult (no polynomial-time algorithm exists to compute solutions).
Conjecture (RSA problem): The following problem is not in P: compute m given only the following:
n
=
p ⋅ q
for two primes p and q in ℕ
e
ℤ/φ(n)ℤ
where e ≥ 3
c
=
me mod n
for an unknown m ∈ ℤ/nℤ
Notice that the RSA problem is analogous to computing the eth root of c in ℤ/nℤ:
e√(c) (mod n)
=
e√(me) (mod n)
=
m (mod n)
Note also that this can be accomplished by first finding φ(n) and then computing the inverse of e, but this is as difficult as factoring n, and we assume that is not in P. Is there another way to compute m? We do not know, but we assume that there is no other faster (i.e., polynomial-time) way to do so.
Conjecture (discrete logarithm assumption): The following problem is not in P: compute e given only the following:
n
ℕ
m
{1,...,n − 1}
c
=
me mod n
for an unknown e ∈ ℕ
Notice that this is analogous to computing the logarithm of a value c in ℤ/nℤ with respect to a known base m:
logm (c)
=
logm (me)
=
e
Note that the RSA problem requires that e ≥ 3, so it does not include the problem of computing square roots. The last intractable problem we will consider in this section is the problem of computing square roots of congruence classes within ℤ/nℤ for some n ℕ. We examine this problem separately from the RSA problem because it is possible to reduce factoring directly to the problem of computing square roots (while there is currently no known deterministic polynomial-time reduction from the factoring problem to the RSA problem).
Before we formally define the problem of computing square roots in ℤ/nℤ, we must first introduce some concepts and facts. This is because the problem of computing square roots in ℤ/nℤ is different from the problem of computing square roots of integers in ℕ, and this difference is likely what makes it computationally more difficult.
Definition: Given some n ℕ and some y ℤ/nℤ, we say that y is a quadratic residue in ℤ/nℤ if there exists x ℤ/nℤ such that x2y.
Example: Let us find the quadratic residues in ℤ/7ℤ:
02 ≡ 0 (mod 7)
12 ≡ 1 (mod 7)
22 ≡ 4 (mod 7)
32 ≡ 2 (mod 7)
42 ≡ 2 (mod 7)
52 ≡ 4 (mod 7)
62 ≡ 1 (mod 7)
The quadratic residues in ℤ/7ℤ are 0, 1, 2, and 4. Notice that 3, 5, and 6 are not quadratic residues in ℤ/7ℤ. Thus, the equations x2 ≡ 3, x2 ≡ 5, and x2 ≡ 6 have no solution in ℤ/7ℤ.
Example: Consider the set of congruence classes ℤ/5ℤ. We have that:
02 ≡ 0 (mod 5)
12 ≡ 1 (mod 5)
22 ≡ 4 (mod 5)
32 ≡ 4 (mod 5)
42 ≡ 1 (mod 5)
Notice that 2 and 3 are not quadratic residues in ℤ/5ℤ. Thus, neither x2 ≡ 2 nor x2 ≡ 3 have solutions in ℤ/5ℤ.
Fact: Given some n ℕ and some y ℤ/nℤ, if y and n are coprime and y is a non-zero quadratic residue in ℤ/nℤ then there exist at least two a,b ℤ/nℤ such that ab, a2y, and b2y.
Note that this is analogous to square roots in ℤ (since √(z) ℤ and −√(z) ℤ are both square roots of z ℤ if they exist).
We can prove this fact in the following way: suppose that y is a quadratic residue. Then there exists at least one x ℤ/nℤ such that:
x2 mod n
=
y
But this means that (nx) ℤ/nℤ is such that:
((n − x)2) mod n
=
(n2 − (2 ⋅ n ⋅ x) + x2) mod n
=
x2 mod n
=
y mod n
Thus, x and (nx) are both roots of y.
Example: It is the case that 4 ℤ/5ℤ is a quadratic residue in ℤ/5ℤ, with two roots 2 and 3:
22 mod 5
=
4
32 mod 5
=
9 mod 5
=
4
Example: Consider 0 ℤ/3ℤ. We have that:
02 ≡ 0 (mod 3)
12 ≡ 1 (mod 3)
22 ≡ 1 (mod 3)
Thus, x2 ≡ 0 has exactly one solution in ℤ/3ℤ.
Fact: Let p ℕ be a prime such that p mod 4 = 3, and suppose that y ℤ/pℤ. Then y has either 0, 1, or 2 roots in ℤ/pℤ.
Example: Suppose we want to solve the following equation for x ℤ/7ℤ:
x2
3 (mod 7)
Suppose we start by squaring both sides:
x4
32 (mod 7)
We can then use Euler's theorem to add any multiple of φ(7) to the exponent:
x4
32 ⋅ 1 (mod 7)
x4
32 ⋅ 3φ(7)
x4
32 + φ(7)
Since 7 is prime, φ(7) must be even, so 2 + φ(7) is also even. Thus, we can divide the exponent by 2 on both sides.
x2
3(2 + φ(7))/2
Furthermore, since 7 ≡ 3 (mod 4), we know that 2 + φ(7) is a multiple of 4. Thus, we can actually divide both exponents by 4:
x
3(2 + φ(7))/4
Thus, we have found x as a power of the original quadratic residue 3.
Fact: Let p ℕ be a prime such that p mod 4 = 3, and suppose that y ℤ/pℤ is a quadratic residue with two roots in ℤ/pℤ. Then we can compute the roots using the following formula:
x ≡ ± y(p+1)/4 (mod p)
In fact, if the modulus n is not prime, there may exist more than two roots of a value in ℤ/nℤ.
Example: It is the case that 1,-1,6,-6 ℤ/35ℤ are all square roots of 1 ℤ/35ℤ:
12 mod 35
=
1
(-1)2 mod 35
=
342 mod 35
=
1156 mod 35
=
((33 ⋅ 35)+1) mod 35
=
1 mod 35
62 mod 35
=
36 mod 35
=
1 mod 35
(-6)2 mod 35
=
292 mod 35
=
841 mod 35
=
((24 ⋅ 35)+1) mod 35
=
1 mod 35
Example: Suppose we are given an instance of the congruent squares problem where y = 2 and n = 15. We want to find x ℤ/15ℤ such that x ≢ ± y but x2y2 ≡ 22 ≡ 4. Notice that we have that:
y
2 mod 3
y2
22 mod 3
1 mod 3
(3-y)2
12 mod 3
1 mod 3
Notice also that we have that:
y
2 mod 5
y2
22 mod 5
4 mod 5
(5-y)2
32 mod 5
4 mod 5
Thus, the square roots of 4 in ℤ/3ℤ are 1 and 2, and the square roots of 4 in ℤ/5ℤ are 2 and 3. We can then apply the Chinese remainder theorem to every pair of combinations:
r1
1 mod 3
r1
2 mod 5
r1
7 mod 15
r2
2 mod 3
r2
2 mod 5
r2
2 mod 15
r3
1 mod 3
r3
3 mod 5
r3
13 mod 15
r4
2 mod 3
r4
3 mod 5
r4
8 mod 15
Thus, x = 8 and x = 7 are solutions to x ≢ ± 2 and x2 ≡ 4.
Fact (Hensel's lemma): Let p ℕ be a prime number greater than 2, and let k ℕ be any positive integer (i.e., k ≥ 1). Suppose that x and p are coprime, and that x ℤ/pkℤ can be squared to obtain some quadratic residue r ℤ/pkℤ:
x2 ≡ r (mod pk)
We can compute y ℤ/pk+1ℤ such that:
y2 ≡ r (mod pk+1)
We compute it as follows. First, we compute c using the following formula:
c
x-1 ⋅ 2-1 ⋅ ((r - x2) / pk) (mod p)
Then, we have that:
y
x + c ⋅ pk
To see why Hensel's lemma is true, suppose that we have that:
x2
r (mod pk)
Notice that if it is possible to "lift" x to a root of r in ℤ/pk+1ℤ, the only possibility is that this new root y has an additional multiple of pk. Thus, it must be that for some integer multiple c, we have:
y
=
x + c ⋅ pk
We can then substitute:
y2
r (mod pk+1)
(x + (c ⋅ pk))2
r (mod pk+1)
But we can simplify the above equation:
(x + (c ⋅ pk))2
r (mod pk+1)
x2 + (2 ⋅ x ⋅ c ⋅ pk) + (c2 ⋅ p2k)
r (mod pk+1)
But notice that the third term on the left-hand side in the above equation is equivalent to the congruence class 0 + pk+1ℤ:
c2 ⋅ p2k
0 (mod pk+1)
Thus, we have:
x2 + (2 ⋅ x ⋅ c ⋅ pk)
r (mod pk+1)
(x2 − r) + (2 ⋅ x ⋅ c ⋅ pk)
0
The above can be rewritten using the divisibility predicate as:
pk+1
|
(x2 − r) + (2 ⋅ x ⋅ c ⋅ pk)
Thus, we can divide both sides of the above relationship by pk to obtain:
p
|
(x2 − r)/(pk) + (2 ⋅ x ⋅ c)
We can then rewrite the above as an equation of congruence classes:
(x2 − r)/(pk) + (2 ⋅ x ⋅ c)
0 (mod p)
(2 ⋅ x ⋅ c)
− (x2 − r)/pk
c
x-1 ⋅ 2-1 ⋅ (− (x2 − r)/pk)
c
x-1 ⋅ 2-1 ⋅ ((r − x2)/pk)
Thus, we have derived the formula in Hensel's lemma.
Example: To better understand Hensel's lemma, we can derive the lemma for a particular example. Let us start with the following equation:
42
2 (mod 7)
Suppose we want to find y ℤ/72ℤ such that:
y2
2 (mod 49)
We know that the difference between 4 and y must be a multiple of 7, so we write:
y
=
4 + 7 ⋅ c
Then we proceed:
y2
2 (mod 49)
(4 + 7 ⋅ c)2
2 (mod 49)
42 + (2 ⋅ 7 ⋅ c ⋅ 4) + (49 ⋅ c2)
2 (mod 49)
42 + (2 ⋅ 7 ⋅ c ⋅ 4)
2 (mod 49)
We simplify further to compute c:
(42 - 2) + (2 ⋅ 7 ⋅ c ⋅ 4)
0 (mod 49)
14 + (2 ⋅ 7 ⋅ c ⋅ 4)
0 (mod 49)
The above can be rewritten using the divisibility predicate as:
49 | 14 + (2 ⋅ 7 ⋅ c ⋅ 4)
7 | 2 + (2 ⋅ c ⋅ 4)
We can again rewrite the above as an equation of congruence classes:
2 + (2 ⋅ c ⋅ 4)
0 (mod 7)
2 ⋅ c ⋅ 4
− 2 (mod 7)
2 ⋅ c ⋅ 4
5 (mod 7)
c
2-1 ⋅ 4-1 ⋅ 5 (mod 7)
c
4 ⋅ 2 ⋅ 5 (mod 7)
c
5 (mod 7)
Thus, we have:
y
4 + 7 ⋅ 5 (mod 49)
y
39 (mod 49)
Since 49 − 39 = 10, we have:
y
± 10 (mod 49)
y2
2 (mod 49)
Example: We want to find both solutions y ℤ/121ℤ to:
y2
5 (mod 121)
Since 121 = 112, we have p = 11, k = 1, and r = 5. We begin by finding x ℤ/11ℤ such that:
x2
5 (mod 11)
Since 11 3 + 4ℤ, we can use an explicit formula:
x
± 5(11+1)/4 (mod 11)
± 53
± 3 ⋅ 5
± 4
Thus, it is sufficient to lift the solution 4 ℤ/11ℤ to a solution in ℤ/121ℤ using Hensel's lemma. We compute c:
c
x-1 ⋅ 2-1 ⋅ ((r − x2)/pk) (mod p)
4-1 ⋅ 2-1 ⋅ ((5 − 16)/11) (mod 11)
4-1 ⋅ 2-1 ⋅ (− 1)
3 ⋅ 6 ⋅ (− 1)
(− 18)
4
Thus, we have:
y
4 + 4 ⋅ 11 (mod 121)
4 + 44
48
Thus, we have the solution
y
± 48 (mod 121)
Example: The distance travelled by an object that is at rest at time t = 0 and then immediately begins accelerating at 4 meters/second2 (i.e., the speed of the object increases by the quantity 4 meters/second every second) can be defined in terms of time in second t as:
d
=
1/2 ⋅ 4 ⋅ t2
We might expect an object to behave this way if it is being pulled by gravity, or if it is using a stable propulsion engine (e.g., a rocket).
Suppose we are using a range ambiguity resolution technique to track the distance the object has travelled. If it a particular moment, we know that the distance from the object is in the congruence class 10 + 11ℤ, what can we say about the amount of time t that has elapsed since the object started moving?
Since the distance is in 10 + 11ℤ, we can say:
d
10 (mod 11)
1/2 ⋅ 4 ⋅ t2
10
2 ⋅ t2
10
We know that 2-1 ≡ 6 (mod 11), so we multiply both sides of the above equation to obtain:
t2
60 (mod 11)
t2
5 (mod 11)
Thus, we can compute:
t
5(11+1)/4 (mod 11)
t
53
t
3 ⋅ 5
t
4 (mod 11)
Thus, we can say that the amount of time that has elapsed is in 4 + 11ℤ.
Example: Solve the following system of equations for x ℤ/21ℤ (find all solutions):
x2
1 (mod 3)
x2
1 (mod 7)
We know that there is exactly one solution y ℤ/21ℤ to the following system:
y
1 (mod 3)
y
1 (mod 7)
The solution is simply y = 1, and since there is only one solution, this is the only possibility. Thus, we are looking for all the solutions to the following equation:
x2
1 (mod 21)
Since 3 mod 4 = 7 mod 4 = 3, we know that there are two solutions to each of the following equations:
x2
1 (mod 3)
x2
1 (mod 7)
The solutions are as follows:
x
1 (mod 3)
x
2 (mod 3)
x
1 (mod 7)
x
6 (mod 7)
Taking every pair of combinations with one solution from ℤ/3ℤ and one solution from ℤ/7ℤ, we get:
x1
1 (mod 3)
x1
1 (mod 7)
x1
1 (mod 21)
x2
2 (mod 3)
x2
1 (mod 7)
x2
8 (mod 21)
x3
1 (mod 3)
x3
6 (mod 7)
x3
13 (mod 21)
x4
2 (mod 3)
x4
6 (mod 7)
x4
20 (mod 21)
Example: How many solutions x ℤ/(33 ⋅ 35)ℤ does the following system of equations have:
x2
4 (mod 33)
x2
4 (mod 35)
We know that each of the following equations have two solutions (2 and -2 in the respective sets). Notice that 4 mod 3 = 1.
x2
1 (mod 3)
x2
4 (mod 11)
x2
4 (mod 5)
x2
4 (mod 7)
Thus, there are two possible choices for each of the variables r1 {-2,2}, r2 {-2,2}, r3 {-2,2}, r4 {-2,2}, so there are 2 ⋅ 2 ⋅ 2 ⋅ 2 = 24 = 16 possible systems of the form:
x
r1 (mod 3)
x
r2 (mod 11)
x
r3 (mod 5)
x
r4 (mod 7)
Each system has a unique solution because the tuple (r1, r2, r3, r4) is unique, so there are 16 solutions for x in ℤ/(33 ⋅ 35)ℤ. Alternatively, we could break the problem down into two subproblems. First, we solve the following equation:
x2
4 (mod 33)
We obtain four distinct solutions (r1, r2, r3, r4) in ℤ/33ℤ. Next, we solve the following equation:
x2
4 (mod 35)
We then have four distinct solutions in (s1, s2, s3, s4) ℤ/35ℤ. Since gcd(33,35) = 1, we can then take any combination of solutions ri and si and set up the system:
x
ri (mod 33)
x
si (mod 35)
There will be exactly one solution to each of the above systems. There are 42 = 16 distinct systems, so there will be 16 distinct solutions.
We can summarize everything we know about computing square roots of congruence classes as follows. Suppose we want to find all solutions to the equation x2a (mod n) for some n ℕ and some a ℤ/nℤ.
• If n is a prime p, the possibilities are:
• a is not a quadratic residue in ℤ/pℤ, so there are no solutions to the equation,
• a ≡ 0, in which case x ≡ 0 is the one and only solution to the equation,
• a is a quadratic residue in ℤ/pℤ, so there are exactly two solutions to the equation, ± x ℤ/pℤ.
• If n is prime power pk+1 and a is coprime with p, the possibilities are:
• a is not a quadratic residue in ℤ/pkℤ, so it is not a quadratic residue in ℤ/pk+1ℤ;
• a is a quadratic residue in ℤ/pkℤ, and both square roots of a in ℤ/pkℤ can be "lifted" to ℤ/pk+1ℤ using Hensel's lemma.
• If n is a product of two coprime numbers k and m, then there is a solution in ℤ/nℤ for every possible combination of y and z such that:
y2
a mod k
z2
a mod m
Each combination corresponds to a solution x ℤ/nℤ defined using CRT as:
x
y mod k
x
z mod m
Let us consider the problem of finding all of the square roots of a member of ℤ/nℤ. Notice that this problem is analogous to computing all the square roots of y in ℤ/nℤ:
√(y)
=
± x
The main difference is that the number of square roots may be greater than 2. This problem is believed to be computationally difficult (i.e., no algorithm in P exists that can solve the problem). In fact, even finding just one additional square root is believed to be computationally difficult.
Conjecture (congruent squares): The following problem is not in P: given n = pq for two primes p and q in ℕ and y ℤ/nℤ, find an x ℤ/nℤ such that x2y2 but x ≢ ± y.
Factoring can be reduced to finding congruent squares. Suppose we want to factor n. We find x and y such that:
x2 mod n
=
y2 mod n
0 mod n
=
(x2 − y2) mod n
=
((x + y) ⋅ (x − y)) mod n
n
|
(x + y) ⋅ (x − y)
Since n cannot divide (x+y) (because x ≢ ± y, so x + yn), and it cannot divide (x-y) (since (x+y) < n), and (x-y) ≠ 0 (since x ≢ ± y), it must be that n shares factors with both (x+y) and (x-y). Thus, it must be that either gcd(n,x + y) or gcd(n,x - y) is a non-trivial factor of n, and this can be computed efficiently.
The following diagram summarizes the relationships between the problems that are conjectured to be intractable (i.e., not in P). Each directed edge represents that there exists a polynomial-time reduction from the source problem to the destination problem. All of the nodes in the graph are conjectured to be not in P.
congruent squares(square roots ofcongruence classes) ⇑ ⇑ computing φ(n)for n = p ⋅ q ⇐⇒ factoringn = p ⋅ q ⇑ ⇑ RSA problem(eth roots ofcongruence classes) discrete logarithm(logarithms ofcongruence classes)
### [link] 4.4. Applications of intractability
The computational intractability of certain problems in modular arithmetic makes it possible to address some practical security issues associated with implementing communication protocols. In particular, it helps address two common problems:
• parties must communicate over a public communications channel, so everything they send is visible both to their receiver and to anyone that may be eavesdropping;
• parties trying to communicate cannot physically meet to agree on shared secret information before communicating.
Protocol (hard-to-forge identification with meeting): Suppose Alice and Bob know that Alice will need to send Bob a single message at some point in the future. However, it is possible that Eve might try to impersonate Alice and send a message to Bob while pretending to be Alice.
In order to help Bob confirm that a message is truly from Alice (or to determine which message is from Alice given multiple messages), Alice and Bob meet in person and agree on a secret identifier s. When Alice decides to send a message m to Bob, she will send (m, s). Bob can then compare s to his own copy of s and confirm the message is from Alice.
Eve's only attack strategy is to try and guess s. As long as Alice and Bob choose s from a very large range of integers, the probability that Eve can guess s correctly is small.
A major flaw in the above identification protocol is that Alice and Bob must first meet in person to agree on a secret. Can Alice and Bob agree on a secret without meeting in person?
Protocol (hard-to-forge identification without meeting): Suppose Alice and Bob know that Alice will need to send Bob a single message at some point in the future. Alice prepares for this by doing the following:
• choose two large primes p and q at random;
• compute n = pq;
• send the public identifier n to Bob over a public/non-secure communication channel.
When Alice is ready to send her message m to Bob, Alice will send (m, (p, q)), where (p, q) is the private identifier. Bob can confirm that pq = n, at which point he will know Alice was the one who sent the message.
Conjecture (forging identification): The following problem is not in P: in the previously defined protocol, given a public identifier n, compute the private identifier (p, q).
If it were possible to quickly (i.e., in polynomial time) compute the private identifier, then it would be easy to forge the identity of a sender by recording their public identifier n. However, suppose that this forging computation could be used as a subprocedure in a factoring algorithm. Then it would be possible to implement an efficient factoring algorithm. In other words, factoring n can be reduced to forging a private identifier. Thus, the problem of forging a private identifier must not be in P (i.e., the fastest algorithm that exists for forging an identity is not in P, which means there is no polynomial-time for forging an identity).
forging privateidentifier for nconclusion:cannot be solved inpolynomial timeforging ∉ P ⇐ factoring nconjecture:cannot be solved inpolynomial timefactoring ∉ P
Note that the reduction also works in the other direction: an algorithm for factoring can be used for forging an identity. However, this proves nothing about the difficulty of forging! Just because forging can be reduced to an inefficient algorithm for factoring does not mean that there does not exist some other algorithm for forging that does not rely on factoring.
alternativeefficientalgorithm ⇐ forging privateidentifier for n ⇒ factoring n
The above identification protocol improves over the previous protocol because it allows Alice and Bob to agree on an identifier for Alice without meeting in person. Its security relies on the fact that it is unlikely that Eve is capable of forging identifiers because her ability to forge would solve a problem that we believe is very difficult to solve.
However, the protocol still has many other flaws. For example, Eve could preempt Alice and send her own signature n before Alice has a chance to send her signature to Bob. A more thorough examination of such protocols is considered in computer science courses focusing explicitly on the subject of cryptography.
Protocol (Diffie-Hellman key exchange): We introduce the Diffie-Hellman key exchange protocol. This protocol is useful if two parties who cannot meet physically want to agree on a secret value that only they know.
• Public key generation (performed by one party):
1. Randomly choose a public large prime number p ℕ and an element g ℤ/pℤ.
• Private key generation (performed by both parties):
1. Party A randomly chooses a secret a ℤ/φ(p)ℤ.
2. Party B randomly chooses a secret b ℤ/φ(p)ℤ.
• Protocol:
1. Party A computes (ga mod p) and sends this public value to party B.
2. Party B computes (gb mod p) and sends this public value to party A.
3. Party A computes (gb mod p)a mod p.
4. Party B computes (ga mod p)b mod p.
5. Since multiplication over ℤ/φ(p)ℤ is commutative, both parties now share a secret gab mod p.
This protocol's security only relies on the discrete logarithm assumption.
It is not known whether the discrete logarithm problem is related to the factoring problem. Factoring can be reduced using a probabilistic approach to the discrete logarithm problem modulo pq.
Protocol (RSA protocol): We introduce the RSA public-key cryptographic protocol. This protocol is useful in many scenarios, such as the following:
• a sender wants to send the receiver a secret message over a public channel;
• a receiver wants to allow any number of senders to send him messages over a public channel, and the receiver does not yet know who the senders will be.
This protocol can also be used to prove the identity of the receiver.
• Key generation (performed by the receiver):
1. Randomly choose two secret prime numbers p ℕ and q ℕ of similar size.
2. Compute a public key value n = pq.
3. Compute the secret value φ(n) = (p-1) ⋅ (q-1).
4. Choose a public key value e {2,...,φ(n)-1} such that gcd(e, φ(n)) = 1.
5. Compute the secret private key d = e-1 mod φ(n)
• Protocol (encryption and decryption): There are two participants: the sender and the receiver.
1. The sender wants to send a message m {0,...,n-1} where gcd(m,n) = 1 to the receiver.
2. The receiver reveals the public key (n,e) to the sender.
3. The sender computes the ciphertext (encrypted message) c = me mod n.
4. The sender sends c to the receiver.
5. The receiver can recover the original message by computing m = cd mod n.
The above encryption-decryption process works because for some k ℤ:
e ⋅ d
1 (mod φ(n))
=
1 + φ(n) ⋅ k
(me)d mod n
=
(m1 + φ(n) ⋅ k) mod n
=
(m ⋅ (mφ(n) ⋅ k)) mod n
=
(m ⋅ (mφ(n) ⋅ k)) mod n
=
(m ⋅ mφ(n)) mod n
=
(m ⋅ 1) mod n
=
m mod n
Besides the message m, there are three pieces of secret information that an eavesdropper cannot know in order for the encryption to provide any privacy:
• p and q
• φ(n)
• d = e-1
Notice that if an eavesdropper knows p and q where n = pq, the eavesdropper can easily compute φ(n) (which was supposed to be private). If the eavesdropper can compute φ(n), then they can use the extended Euclidean algorithm to compute the inverse d = e-1 of the public key value e. They can then use d to decrypt messages.
Suppose the eavesdropper only knows φ(n). Then the eavesdropper can compute d and decrypt any message. In fact, the eavesdropper can also recover p and q.
Protocol (Rabin cryptosystem): We introduce the Rabin cryptosystem protocol. It is similar to the RSA scheme, but it does not rely on the difficulty of the RSA problem.
• Key generation (performed by the receiver):
1. Randomly choose two secret prime numbers p ℕ and q ℕ of similar size.
2. Compute a public key value n = pq.
• Protocol (encryption and decryption): There are two participants: the sender and the receiver.
1. The sender wants to send a message m {0,...,n-1} to the receiver.
2. The receiver reveals the public key n to the sender.
3. The sender computes the ciphertext (encrypted message) c = m2 mod n.
4. The sender sends c to the receiver.
5. The receiver can recover the original message by computing √(c) in ℤ/pℤ and ℤ/qℤ, and then finding the four solutions to the following system by using the Chinese remainder theorem:
m
√(c) mod p
m
√(c) mod q.
Notice that the receiver must guess which of the square roots corresponds to the original message. Also notice that it is not a good idea to encrypt messages in the ranges {0, ..., √(n)} and {n − √(n), ..., n − 1} because it is easy to decrypt such messages by computing the integer square root √(c) of c and then confirming that √(c)2c (mod n).
The following diagram summarizes the relationships between algorithms that might break each of the protocols presented in this section, and existing problems that are believed not to be in P. Thus, our conjectures imply that all of the problems below are not in P.
breakingRabin encryption ⇐ congruent squares(square roots ofcongruence classes) ⇑ ⇑ finding RSAsecret key ⇐ computing φ(n)for n = p ⋅ q ⇐⇒ factoringn = p ⋅ q ⇑ ⇑ decrypting individualRSA messages ⇐ RSA problem(eth roots ofcongruence classes) breakingDiffie-Hellman ⇐ discrete logarithm(logarithms ofcongruence classes)
Example: Bob decides to create his own online currency BobCoin. Bob knows that in order for BobCoin to be successful, it needs to be possible to make more BobCoins as more and more people start using them. However, he also does not want rapid inflation to occur. Thus, Bob issues BobCoins according to the following protocol:
• every day, Bob chooses two new random primes p and q;
• Bob computes n = pq, and then discards p and q;
• Bob posts n online for everyone to see;
• at any time on that day, anyone can submit a factor f of n;
• if f is a factor of n, Bob issues that person a BobCoin, invalidates n permanently so that no one else can use it, and generates a new n.
1. Why is it okay for Bob to discard p and q?
Because Bob can easily check whether a submitted f actually is a factor by computing gcd(f, n), or simply dividing n by f.
2. Suppose that Bob always posts numbers n that have 100 digits, and it takes the fastest computer one year to factor a 100-digit number through trial and error. If Alice wants to earn a BobCoin in one day, how many computers will Alice need to run in parallel to earn a BobCoin?
Since it takes a 365 days to try all possible k factors of a 100-digit number, in one day one computer can try 1/365 ⋅ k factors. Thus, Alice would need 365 computers that she could run for one day, with each computer looking through a different region of factors.
3. Suppose Bob wants to issue a complimentary BobCoin coupon to a group of 100 people. However, he wants to make sure that they can use their BobCoin coupon only if at least 20 out of those 100 people agree that the coupon should be redeemed for a BobCoin. How can Bob accomplish this?
Bob can use Shamir secret sharing to accomplish this. Bob can take one of the factors f of n when he generates n, and choose 100 distinct primes such that the product of any 20 of the primes is greater than f while the product of any 19 of these primes is less than f. Bob can then distribute f mod m for each prime modulus m to each of the 100 people. If any 20 (or greater) of these 100 people agree to redeem the coupon, then can use CRT to reconstruct f and submit it to Bob for redemption.
### [link] 4.5.Assignment #4: Intractable Problems in Modular Arithmetic
In this assignment you will solve several equations, and you will define a collection of Python functions that demonstrate the relationships between intractable problems in modular arithmetic. You must submit a single Python source file named `hw4.py` (submitted to the location `hw4/hw4.py`). Please follow the gsubmit directions.
You may import the following library functions in your module (you may not need all these functions for this assignment depending on how you approach the problems, but they may be used):
from math import floor
from fractions import gcd
Your file may not import any other modules or employ any external library functions associated with integers and sets unless explicitly permitted to do so in a particular problem. Solutions to each of the programming problem parts below should be fairly concise. You will be graded on the correctness, concision, and mathematical legibility of your code. The different problems and problem parts rely on the lecture notes and on each other; carefully consider whether you can use functions from the lecture notes, or functions you define in one part within subsequent parts.
1. Solve the following equations using step-by-step equational reasoning, and list each step. Your solutions for this problem should appear as comments, delimited using `'''`...`'''`, in `hw4.py`. You may use the `=` ascii character to represent the ≡ relational operator on congruence classes.
1. Solve the following equation for all solutions x ℤ/23ℤ:
x2
3 (mod 23)
2. Solve the following equation for all solutions x ℤ/43ℤ:
x2
25 (mod 43)
3. Solve the following equation for all solutions x ℤ/41ℤ (note that you cannot use the explicit formula for square roots in this case):
3 ⋅ x2
7 (mod 41)
4. Solve the following equation for all solutions x ℤ/49ℤ:
x2
11 (mod 49)
5. Solve the following equation for all solutions x ℤ/21ℤ:
(8 ⋅ x2) + 4
6 (mod 21)
6. Using congruence classes, explain why the following equation involving integers cannot have any integer solutions x ℤ:
x5 + x + 1
=
256
2. Implement the following Python functions for factoring a positive integer. These algorithms must be efficient (i.e., they must run in polynomial time).
1. Implement a function `factorsFromPhi(n, phi_n)` that takes two integers `n` and `phi_n`. You may assume (i.e., you do not need to verify in your code) that `n` is a product of two distinct positive prime numbers and that `phi_n` = φ(`n`). The function should return both prime factors of `n` as a tuple.
>>> factorsFromPhi(77, 60)
(7, 11)
>>> factorsFromPhi(14369648346682547857, 14369648335605206820)
(9576890767, 1500450271)
2. Implement a function `factorsFromRoots(n, x, y)` that takes three integers `n`, `x`, and `y`. You may assume that `n` is a product of two distinct positive prime numbers, and that the following is true:
`x`2
`y`2 (mod `n`)
`x`
± `y` (mod `n`)
The function should return both prime factors of `n` as a tuple.
>>> factorsFromRoots(35, 1, 6)
(5, 7)
>>> factorsFromRoots(14369648346682547857, 12244055913891446225, 1389727304093947647)
(1500450271, 9576890767)
3. Consider the following problem (call it "φ-four"): "given an integer n = pqrs that is the product of four distinct primes, find φ(n)." Notice that the four distinct primes are not known in the specification of the problem. It is only known that n is the product of some four distinct primes.
Given our assumptions about the intractability of various problems (i.e., that certain problems are not in P), you must show that φ-four is also not in P. To do so, you must implement a function `phiFromPhiFour(n)` for the existing intractable problem of computing φ that uses a solver for φ-four (call it `phi_four()`) as a subroutine. Note: remember that the four primes must be distinct; your reduction must work for all possible instances of the intractable problem.
3. In this problem you will implement the three component algorithms of the RSA cryptographic protocol described in lecture.
1. Define a Python function `generate(k)` that takes a single integer input `k` and returns a tuple `(n,e,d)` corresponding to the public values `n` and `e` and private key `d` in the RSA cryptographic protocol. The output `n` must be the product of two distinct, randomly chosen `k`-digit primes. You may import and use the Python random number generator (`from random import random` or `from random import randint`), and you may want to reuse the extended Euclidean algorithm implementation provided in the previous homework assignment.
2. Define a Python function `encrypt(m, t)` that takes two inputs: an integer `m` and a tuple `(n,e)` representing an RSA public key. It should return a single integer: the RSA ciphertext `c`.
3. Define a Python function `decrypt(c, t)` that takes two inputs: an integer `c` representing the ciphertext and a tuple of two integers `(n,d)` representing an RSA private key. It should decrypt `c` and return the original message `m`.
4. In this problem you will implement an algorithm for computing all the square roots of a congruence class in ℤ/nℤ, given a complete factorization of n into its distinct prime factor powers (assuming all the prime factors are in 3 + 4ℤ).
1. Implement a Python function `sqrtsPrime(a, p)` that takes two arguments: an integer `a` and a prime number `p`. You may assume that `a` and `p` are coprime. If `p` is not in 3 + 4ℤ or `a` has no square roots in ℤ/`p`ℤ, the function should return `None`. Otherwise, it should return the two congruence classes in ℤ/`p`ℤ that solve the following equation:
x2
`a` (mod `p`)
>>> sqrtsPrime(2, 7)
(3, 4)
>>> sqrtsPrime(5, 7) # 5 has no square roots
None
>>> sqrtsPrime(5, 17) # 17 mod 4 =/= 3
None
>>> sqrtsPrime(763472161, 5754853343)
(27631, 5754825712)
2. Implement a Python function `sqrtsPrimePower(a, p, k)` that takes three arguments: an integer `a`, a prime number `p`, and a positive integer `k`. You may assume that `a` and `p` are coprime. If `p` is not in 3 + 4ℤ or `a` has no square roots in ℤ/`p``k`ℤ, the function should return `None`. Otherwise, it should return the congruence classes in ℤ/`p``k`ℤ that solve the following equation:
x2
`a` (mod `p``k`)
>>> sqrtsPrimePower(2, 7, 2)
(10, 39)
>>> sqrtsPrimePower(763472161, 5754853343, 4)
(27631, 1096824245608362247285266960246506343570)
3. Implement a Python function `sqrts(a, pks)` that takes two arguments: an integer `a` and a list of tuples `pks` in which each tuple is a distinct positive prime number paired with a positive integer power. You may assume that `a` and `n` are coprime. You may assume that all the primes in `pks` are in 3 + ℤ/4ℤ (if any are not, the function should return `None`). Let `n` be the product of all the prime powers in the list `pks`. Then the function should return a set of all the distinct square roots of `a` in ℤ/`n`ℤ that are solutions to the following equation:
x2
`a` (mod `n`)
Your implementation must be efficient (i.e., it may not iterate over all possible values in ℤ/`n`ℤ to look for square roots), and it must work for any positive number of entries in `pks`.
>>> sqrts(2, [(7,4)])
{235, 2166}
>>> sqrts(1, [(7,1), (11,1)])
{1, 76, 43, 34}
>>> sqrts(1, [(7,1), (11,1), (3,1)])
{1, 76, 43, 34, 155, 188, 197, 230}
>>> sqrts(1, [(7,2), (11,1), (3,2)])
{1, 197, 881, 1079, 3772, 3970, 4654, 4850}
>>> sqrts(1, [(7,1), (11,1), (3,1), (19,1)])
{1, 265, 419, 1198, 1310, 1462, 1616, 1882, 2507, 2773, 2927, 3079, 3191, 3970, 4124, 4388}
>>> sqrts(76349714515459441, [(1500450271,3), (5754853343,2)])
{276314521,
111875075121925861006908948065990250824231090118,
50900491283175338098734392241768315796265192809,
60974583838750522908174555824221935028242211830}
You may find the following helper function useful in implementing your solution.
def combinations(ls):
if len(ls) == 0:
return [[]]
else:
return [ [x]+l for x in ls[0] for l in combinations(ls[1:]) ]
This function takes a list of lists as its input and returns a list of all possible combinations of one element from each list.
>>> combinations([[1,2], ['a','b'], ['X','Y']])
[[1, 'a''X'], [1, 'a''Y'], [1, 'b''X'], [1, 'b''Y'], [2, 'a''X'], [2, 'a''Y'], [2, 'b''X'], [2, 'b''Y']]
5. Suppose you are supplied with an efficient implementation of an algorithm for breaking the Rabin cryptographic protocol. Below, we provide a mockup of an efficient implementation of such an algorithm that only work on a few sample inputs (since we do not believe efficient algorithms for breaking this protocol exists, we must cheat in this way in order to test our code). You may assume that the below algorithm works on all inputs, not just those provided in the fake implementation.
# Efficiently computes m from (pow(m,2,n), n).
def decryptMsgRabin(c, n):
input_output = {\
(14, 55): 17,\
(12187081, 8634871258069605953): 7075698730573288811,\
(122180308849, 16461679220973794359): 349543,\
(240069004580393641, 19923108241787117701): 489968371\
}
return input_output[(c, n)]
Implement an efficient function `roots(a, n)` that takes two integers `a` and `n` and returns all four square roots of `a` in ℤ/`n`ℤ as a tuple. You may assume that `n` is the product of two distinct positive prime numbers. Your algorithm in this problem must work on all possible inputs under the assumption that `decryptMsgRabin()` also works on all inputs (not just the fake inputs handled by the definitions above) and that an encrypted message c is never represented by the same congruence class as the original, unencrypted message m or − m (i.e., the implementation follows the guideline regarding good ranges for messages at the end of the definition of the Rabin protocol).
>>> roots(12187081, 8634871258069605953)
(3491, 8634871258069602462, 1559172527496317142, 7075698730573288811)
In the previous sections, we studied a specific algebraic structure, ℤ/nℤ, as well as its operations (e.g., addition, multiplication), and its properties (commutativity, associativity, and so on). There exist many other algebraic structures that share some of the properties of ℤ/nℤ. In fact, we can create a hierarchy, or even a web, of algebraic structures by picking which properties of ℤ/nℤ we keep and which we throw away.
Showing that some new algebraic structure is similar or equivalent to another, more familiar structure allows us to make inferences about that new structure based on everything we already know about the familiar structure. In computer science, the ability to compare algebraic structures using their properties is especially relevant because every time a programmer defines a new data structure and operations on that data structure, they are defining an algebraic structure. Which properties that algebraic structure possesses determines what operations can be performed on it, in what order they can be performed, how efficiently they can be performed, and how they can be broken down and reassembled.
Recall that a permutation on a set X is a bijective relation between X and X (i.e., a subset of the set product X × X). Since a permutation is a bijective map (i.e., a function), we can reason about composition of permutations (it is just the composition of functions). Thus, we can study sets of permutations as algebraic structures under the composition operation o.
Notice that for any set X of finite size n, we can relabel the elements of X to be {0,...,n-1} (that is, we can define a bijection between X and {0,...,n-1}). Thus, we can study permutations on {0,...,n-1} without loss of generality. We will adopt the following notation for permutations:
[a1,...,an]
Where a1,...,an is some rearrangement of the integers from 0 to n-1. For example, the identity permutation on n elements would be:
[0,1,2,3,4,5,...,n-1]
Definition: Any permutation that swaps exactly two elements is called a swap. Examples of swaps are [0,3,2,1], [1,0], and [0,6,2,3,4,5,1].
Definition: Any permutation that swaps exactly two adjacent elements is called an adjacent swap. Examples of adjacent swaps are [0,1,3,2], [1,0,2,3,4], and [0,1,3,2,4,5,6].
Definition: Define Sn to be the set of all permutations of the set {0,...,n-1}.
Example: The set of permutations of {0,1} is S2 = {[0,1], [1,0]}.
Example: The set of permutations of {0,1,2} is S3 = {[0,1,2], [0,2,1], [1,0,2], [1,2,0], [2,0,1], [2,1,0]}.
Fact: The set Sn contains n! permutations.
Suppose we want to construct a permutation [a1,...,an] using the elements in {0,...,n-1}, where we are only allowed to take each element in the set once and assign it to an unassigned entry ai. Then for the first slot, we have n possibilities; for the second, we have n-1 possibilities. For the third, we have n-2 possibilities, and so on until we have only one possibility left. Thus, the number of possible permutations we can make is:
n!
=
n ⋅ (n-1) ⋅ (n-2) ⋅ ... ⋅ 2 ⋅ 1
Definition: Define the set Cn to be the set of all cyclic permutations on n elements. Any permutation that performs a circular shift on elements is a cyclic permutation (also known as a cyclic shift permutation, a circular shift permutation, or just a shift permutation). Examples of shifts are [6,7,0,1,2,3,4,5], [2,3,4,0,1], and [4,0,1,2,3].
Definition: Define the set Mn to be the set of all multiplication-induced permutations on n elements. Any permutation on n elements that corresponds to multiplication by some coprime a < n is called a multiplication-induced permutation. Examples of such permutations include [0,2,4,1,3] (corresponding to multiples 2 ⋅ i for ascending i in ℤ/5ℤ).
Example: The set of multiplication-induced permutations on 6 elements (i.e., permutations of {0,1,2,3,4,5}) is the collection of permutations of the form [a ⋅ 0 mod 6, a ⋅ 1 mod 6, a ⋅ 2 mod 6, a ⋅ 3 mod 6, a ⋅ 4 mod 6, a ⋅ 5 mod 6] for each a that is coprime with 6. Thus, a {1,5}, and so we have:
M6 = {[0,1,2,3,4,5], [0,5,4,3,2,1]}.
Example: The set of multiplication-induced permutations on 7 elements (i.e., permutations of {0,1,2,3,4,5,6}) is:
M7 = {[0,1,2,3,4,5,6], [0,2,4,6,1,3,5], [0,3,6,2,5,1,4], [0,4,1,5,2,6,3], [0,5,3,1,6,4,2], [0,6,5,4,3,2,1]}.
Note that |M7| = φ(7) = 6 because there are 6 possible a ℤ/7ℤ that are coprime with 7.
### [link] 5.2. Isomorphisms: Equivalence of Algebraic Structures
An algebraic structure is a set together with a binary operator over that set. All algebraic structures are closed under their binary operation.
Definition: Let S be a set, and let ⊕ be a binary operator. Let closure(S,⊕) be the closure of S under ⊕. We can define the set closure(S,⊕) in the following way:
closure(S, ⊕)
=
{ x1 ⊕ x2 | x1,x2 ∈ S } ∪ { x1 ⊕ (x2 ⊕ x3) | x1,x2,x3 ∈ S } ∪ { (x1 ⊕ x2) ⊕ x3 | x1,x2,x3 ∈ S } ∪ ...
Alternatively, we could define it in the following way using recursion:
closure0(S, ⊕)
=
S
closuren(S, ⊕)
=
{ x ⊕ y | x,y ∈ (closuren-1(S, ⊕) ∪ ... ∪ closure0(S, ⊕)}
closure(S, ⊕)
=
closure0(S, ⊕) ∪ closure1(S, ⊕) ∪ closure2(S, ⊕) ∪ ...
Notice that if a set S is finite, there is a natural way to algorithmically list all elements in closure(S, ⊕) by starting with the elements in S and "building up" all the elements in each of the closure_i(S, ⊕) subsets.
The concept of an isomorphism between two algebraic structures captures the fact that two structures are not only the same size, but that the two structures have the same internal "structure" with respect to their respective operations. Isomorphisms are useful because they allow us to learn more about a structure by studying the structure isomorphic to it. They can also be useful because if two structures are isomorphic, we can perform computations in one structure instead of another structure (e.g., because it is more secure, more efficient, and so on) while obtaining the same final result.
Fact: Let A be an algebraic structure with operator ⊕ and let B be an algebraic structure with operator ⊗. We say that A is isomorphic to B, which we denote as (A,⊕) ≅ (B,⊗) or simply AB, if the following conditions hold:
• there exists a bijection (i.e., a bijective relation) between A and B, which we denote using = ;
• for all a, a' A and b,b' B, if a = b and a' = b' then aa' = bb'.
Another way to state the definition is to write it in terms of a bijective map m between A and B:
• there exists a bijective map m between A and B;
• for all a, a' A, m(aa') = m(a) ⊗ m(a').
In other words, an isomorphism is a bijection that preserves (or respects) the binary operations on the two sets: if any true equation involving elements from A and the operator ⊕ is transformed by replacing all elements of a with their corresponding elements m(b) B and by replacing all instances of ⊕ with ⊗, the resulting equation is still true.
Example: Consider the set of permutations on two elements S2 and the set of congruence classes ℤ/2ℤ. It is true that (S2,o) ≅ (ℤ/2ℤ,+), where o is composition of permutations and where + is addition of congruence classes in ℤ/2ℤ. The following table demonstrates the bijection:
S2 ℤ/2ℤ [0,1] 0 [1,0] 1
The following table demonstrates that the bijection above respects the two operations.
S2 ℤ/2ℤ [0,1] o [0,1] = [0,1] 0 + 0 = 0 [0,1] o [1,0] = [1,0] 0 + 1 = 1 [1,0] o [0,1] = [1,0] 1 + 0 = 1 [1,0] o [1,0] = [0,1] 1 + 1 = 0
Another way to demonstrate the above is to show that the "multiplication tables" (though the operation need not be multiplication) for the two operators are exactly the same (i.e., the entries in the multiplication table all correspond according to the bijection).
+o 0[0,1] 1[1,0] 0[0,1] 0[0,1] 1[1,0] 1[1,0] 1[1,0] 0[0,1]
Fact: For any positive integer n ℕ, (ℤ/nℤ,+) ≅ (Cn, o).
Example: Consider the set of cyclic permutations on three elements C3 and the set of congruence classes ℤ/3ℤ. It is true that (C3,o) ≅ (ℤ/3ℤ,+).
C3 ℤ/3ℤ [0,1,2] o [0,1,2] = [0,1,2] 0 + 0 = 0 [0,1,2] o [1,2,0] = [1,2,0] 0 + 1 = 1 [0,1,2] o [2,0,1] = [2,0,1] 0 + 2 = 2 [1,2,0] o [0,1,2] = [1,2,0] 1 + 0 = 1 [1,2,0] o [1,2,0] = [2,0,1] 1 + 1 = 2 [1,2,0] o [2,0,1] = [0,1,2] 1 + 2 = 0 [2,0,1] o [0,1,2] = [2,0,1] 2 + 0 = 2 [2,0,1] o [1,2,0] = [0,1,2] 2 + 1 = 0 [2,0,1] o [2,0,1] = [1,2,0] 2 + 2 = 1
Example: To compute the composition of two permutations [45,46,...,49,0,1,2,...,44] o [3,4,5,...,49,0,1,2], it is sufficient to recognize that [45,46,...,49,0,1,2,...,44] corresponds to 45 ℤ/50ℤ, and [3,4,5,...,49,0,1,2] corresponds to 3 ℤ/50ℤ. Thus, since 45 + 3 = 48, the result of the composition must be [48,49,0,1,2,...,47].
45 + 3
=
48
[45,46,...,49,0,1,2,...,44] o [3,4,5,...,49,0,1,2]
=
[48,49,0,1,2,...,47]
Fact: For any prime p ℕ, (ℤ/φ(p)ℤ,+) ≅ ((ℤ/pℤ)*, ⋅).
Note that |ℤ/φ(p)ℤ| = φ(p) = |(ℤ/pℤ)*|.
Example: Consider the set ℤ/2ℤ with the addition operation + modulo 2, and the set (ℤ/3ℤ)* together with the multiplication operation ⋅ modulo 3. It is true that (ℤ/2ℤ, +) ≅ ((ℤ/3ℤ)*,⋅).
(ℤ/2ℤ, +) ((ℤ/3ℤ)*, ⋅) 0 + 0 = 0 1 ⋅ 1 = 1 0 + 1 = 1 1 ⋅ 2 = 2 1 + 0 = 1 2 ⋅ 1 = 2 1 + 1 = 0 2 ⋅ 2 = 1
Example: Consider the set ℤ/2ℤ with the addition operation + modulo 2, and the set (ℤ/6ℤ)* together with the multiplication operation ⋅ modulo 6. It is true that (ℤ/2ℤ, +) ≅ ((ℤ/6ℤ)*,⋅). Note that (ℤ/6ℤ)* = {1,5}, because only 1 and 5 in the range {0,...5} are coprime with 6.
(ℤ/2ℤ, +) ((ℤ/6ℤ)*, ⋅) 0 + 0 = 0 1 ⋅ 1 = 1 0 + 1 = 1 1 ⋅ 5 = 5 1 + 0 = 1 5 ⋅ 1 = 5 1 + 1 = 0 5 ⋅ 5 = 1
Isomorphisms need not be defined between different sets. It is possible to define an isomorphism between a set and itself that has non-trivial, interesting, and even useful characteristics.
Fact: For any n ℕ and any a ℤ/nℤ where a is coprime with n, (ℤ/nℤ,+) ≅ (ℤ/nℤ,+) under the bijection that relates x ℤ/nℤ with ax ℤ/nℤ. This is because for any x,y ℤ/nℤ, we have:
a ⋅ (x + y)
a ⋅ x + a ⋅ y
Fact: For any n ℕ and any e ℤ/φ(n)ℤ where e is coprime with φ(n), ((ℤ/nℤ)*,⋅) ≅ ((ℤ/nℤ)*,⋅) under the bijection that relates x (ℤ/nℤ)* with xe (ℤ/nℤ)*. This is because for any x, y (ℤ/nℤ)*, we have:
(x ⋅ y)e
xe ⋅ ye (mod n)
Example: Consider the set ℤ/3ℤ with the addition operation + modulo 3, and another instance of the set ℤ/3ℤ with the addition operation + modulo 3. The following is a bijection between (ℤ/3ℤ, +) and (ℤ/3ℤ, +):
ℤ/3ℤ ℤ/3ℤ 0 0 1 2 2 1
This bijection is an isomorphism:
ℤ/3ℤ ℤ/3ℤ 0 + 0 = 0 0 + 0 = 0 0 + 1 = 1 0 + 2 = 2 0 + 2 = 2 0 + 1 = 1 1 + 0 = 1 2 + 0 = 2 1 + 1 = 2 2 + 2 = 1 1 + 2 = 0 2 + 1 = 0 2 + 0 = 2 1 + 0 = 1 2 + 1 = 0 1 + 2 = 0 2 + 2 = 1 1 + 1 = 2
If we only consider algebraic structures with particular algebraic properties, we can actually show that there is only one algebraic structure of a particular size (i.e., there is only one "isomorphism class" of algebraic structures having that size).
Fact: Suppose we have an algebraic structure (A, ⊕) with two elements in which the elements in the set must have inverses, one of them must be an identity, and ⊕ is associative. Without loss of generality, let's label the two elements a and b, and let a be the label of the identity element. Because a is the identity, we must have:
a ⊕ a
=
a
a ⊕ b
=
b
b ⊕ a
=
b
The identity is its own inverse, so a-1 = a. The only question that remains is to determine what bb must be. If we have bb = b, then we must ask what the inverse of b can be (since it isn't b itself, as we would then have bb = a). But then the only option is a. That would mean that ba should be a (since b and its inverse should yield the identity element). But this contradicts the equations we already derived above. So it must be that b is its own inverse:
a ⊕ b
=
a
Thus, there can be only one distinct algebraic structure (in terms of its "multiplication table") having two elements; it's the algebraic structure isomorphic to (A, ⊕), as well as all the other algebraic structures isomorphic to it: (S2, o), (C2, o), (ℤ/2ℤ, +), ((ℤ/3ℤ)*, ⋅), ((ℤ/6ℤ)*, ⋅), and so on.
Example (partial homomorphic encryption supporting addition): Suppose that for some n ℕ, Alice wants to store a large number of congruence classes b1,...,bk ℤ/nℤ in Eve's database (perhaps Alice will generate or collect these over a long period of time). Alice does not have enough memory to store the congruence classes herself, but she does not want to reveal the congruence classes to Eve.
What Alice can do before she stores anything in Eve's database is to pick some secret a ℤ/nℤ that is coprime with n. Then, every time Alice needs to store some b in Eve's database, Alice will instead send Eve the obfuscated value (ab) mod n. Since a is coprime with n, there exists a-1 ℤ/nℤ. Thus, if Alice retrieves some obfuscated data entry c from Eve's database, she can always recover the original value by computing (a-1c) mod n, because:
a-1 ⋅ (a ⋅ b)
b (mod n)
Furthermore, Alice can ask Eve to compute the sum (modulo n) of all the entries in the database (or any subset of them). Suppose that Alice has stored obfuscated versions of b1,...,bk ℤ/nℤ in Eve's database. Then if Eve computes the sum of all the obfuscated entries stored in her database, she will get:
a ⋅ b1 + ... + a ⋅ bk
=
a ⋅ (b1 + ... + bk) (mod n)
Thus, if Alice asks Eve for the sum of all the obfuscated entries in the database, Alice can recover the actual sum of the original entries that she stored in the database because:
a-1 ⋅ (a ⋅ (b1 + ... + bk))
=
b1 + ... + bk (mod n)
In this way, Alice has avoided having to store and add all the database entries, while preventing Eve finding out the actual entries, or their sum.
Example (partial homomorphic encryption supporting multiplication): Suppose that for some n ℕ, Alice wants to store a large number of congruence classes b1,...,bk (ℤ/nℤ)* in Eve's database. Alice does not have enough memory to store the congruence classes herself, but she does not want to reveal the congruence classes to Eve.
We assume that Alice knows or can easily compute φ(n), while Eve does not know and cannot compute it (perhaps Alice generated n using a method similar to the one in RSA encryption protocol).
What Alice can do before she stores anything in Eve's database is to pick some secret e ℤ/φ(n)ℤ that is coprime with φ(n). Since Alice knows e and φ(n), she can compute e-1 using the extended Euclidean algorithm.
Then, every time Alice needs to store some b in Eve's database, Alice will instead send Eve the encrypted value be mod n. If Alice retrieves some encrypted data entry c from Eve's database, she can always recover the original value by computing (ce-1}) mod n, because by Euler's theorem:
(be)e-1
be ⋅ e-1
b (mod n)
Furthermore, Alice can ask Eve to compute the product (modulo n) of all the entries in the database (or any subset of them). Suppose that Alice has stored encrypted versions of b1,...,bk ℤ/nℤ in Eve's database. Then if Eve computes the product of all the encrypted entries stored in her database, she will get:
b1e ⋅ ... ⋅ bke
=
(b1 ⋅ ... ⋅ bk)e (mod n)
Thus, if Alice asks Eve for the product of all the encrypted entries in the database, Alice can recover the actual product of the original entries that she stored in the database because:
((b1 ⋅ ... ⋅ bk)e)e-1
=
b1 ⋅ ... ⋅ bk (mod n)
In this way, Alice has avoided having to store and multiply all the database entries, while preventing Eve from finding out the actual entries, or their product. Furthermore, because it is believed that factoring n, computing φ(n), and solving the RSA problem is computationally difficult, it is highly unlikely that Eve can decrypt the database entries or the result.
We know by Euler's theorem and the algebraic properties of exponents that for any ℤ/nℤ, any b (ℤ/nℤ)*, and any x, y ℤ/φ(n)ℤ the following identity must hold:
bx ⋅ by
bx + y (mod n)
b(x + y) mod φ(n) (mod n)
We might naturally ask whether there might be an isomorphism between (ℤ/φ(n)ℤ, +) and ((ℤ/nℤ)*, ⋅). In fact, sometimes there is (although more often it is an isomorphism between a subset of (ℤ/nℤ)* and ℤ/kℤ for k | φ(n)).
Given the above, we might ask whether it might be possible to create a homomorphic encryption protocol using an isomorphism of the form (ℤ/φ(n)ℤ, +) ≅ ((ℤ/nℤ)*, ⋅) in which Alice can encrypt her data x and y by computing bx (mod n) and by (mod n). This should be secure because it is believed that no efficient algorithms for computing discrete logarithms exist. Then, in order to have Eve compute a sum of the data values x and y, Alice can ask Eve to compute the product bxby on her end, which is equivalent to bx + y.
However, there is a flaw in this protocol: Alice has no way to retrieve x + y from bx + y because that requires computing a discrete logarithm, as well. Thus, an isomorphism of the form (ℤ/φ(n)ℤ, +) ≅ ((ℤ/nℤ)*, ⋅) would not necessarily give us a practical homomorphic encryption protocol.
Fact: Given a set of possible data values ℤ/nℤ (e.g., integers within a certain range), any compression algorithm for elements in ℤ/nℤ must be a bijection and a permutation (since it must be invertible in order for decompression to be possible). As a result, it must necessarily expand the representation size of some elements.
Example: Suppose that we are working with elements in ℤ/11ℤ = {0,1,2,3,4,5,6,7,8,9,10}. Suppose we define an algorithm that compresses 10 ℤ/11ℤ into an element in ℤ/11ℤ with a smaller representation size. One example of such an element is 1, since:
10 ⋅ 10
1 (mod 11)
Thus, one possible implementation of a compression algorithm is a function that takes any x ℤ/11ℤ and returns (10 ⋅ x) mod 11. Since 10 has an inverse in ℤ/11ℤ, this function is invertible, so decompression is possible (simply multiply by 10 again). However, this will necessarily expand the representation of at least one value: 1 ℤ/11ℤ:
10 ⋅ 1
10 (mod 11)
Note that this cannot be avoided because multiplying all the elements in ℤ/11ℤ by 10 amounts to a permutation, so at least one compressed element must be 10.
### [link] 5.3. Generators of Algebraic Structures
Because an algebraic structure (A,⊕) often consists of a set of objects that can be "built up" using the binary operator ⊕ from a smaller, possibly finite, collection of generators GA, it is often easier to reason about an algebraic structure by first reasoning about its generators, and then applying structural induction.
Fact: Let W be the set of swap permutations on n elements. Then W is a set of generators for the set of permutations Sn, which can be defined as:
S
=
closure(W, o)
Fact: Let A be the set of adjacent swap permutations on n elements. Then A is a set of generators for the set of permutations Sn, which can be defined as:
S
=
closure(A, o)
Fact: The set ℤ/nℤ has a single generator 1 ℤ/nℤ with respect to addition + modulo n:
ℤ/nℤ
=
closure({1}, +)
Fact: If a and n are coprime, then a is a generator for ℤ/nℤ with respect to addition + modulo n:
ℤ/nℤ
=
closure({a}, +)
Fact: Suppose we have two algebraic structures A and B where for some operator ⊕:
A
=
closure({a}, ⊕)
B
=
closure({b}, ⊕)
If the generator a can be expressed in terms of b, i.e., a = b ⊕ ... ⊕ b, then it must be that:
A
B
Furthermore, if in addition to the above, the generator b can be expressed in terms of a, i.e., b = a ⊕ ... ⊕ a, then it must be that:
B
A
This would then imply (by basic set theory):
A
=
B
### [link] 5.4.Assignment #5: Algebraic Structures and Isomorphisms
In this assignment you will solve several problems involving algebraic structures and isomorphisms, and you will define a collection of Python functions that exploit isomorphisms between algebraic structures. You must submit a single Python source file named `hw5.py` (submitted to the location `hw5/hw5.py`). Please follow the gsubmit directions.
You may import the following library functions in your module (you may not need all these functions for this assignment depending on how you approach the problems, but they may be used):
from math import floor
from fractions import gcd
from random import randint
from urllib.request import urlopen
Your file may not import any other modules or employ any external library functions associated with integers and sets unless explicitly permitted to do so in a particular problem. Solutions to each of the programming problem parts below should be fairly concise. You will be graded on the correctness, concision, and mathematical legibility of your code. The different problems and problem parts rely on the lecture notes and on each other; carefully consider whether you can use functions from the lecture notes, or functions you define in one part within subsequent parts.
1. Solve the following equations using step-by-step equational reasoning, and list each step. Your solutions for this problem should appear as comments, delimited using `'''`...`'''`, in `hw5.py`.
1. Compute the following composition of permutations (the result should be a permutation):
[3,5,4,0,1,2] o [4,5,3,1,2,0] = ?
2. Compute the following composition of permutations (the result should be a permutation):
[71,72,...,99,0,1,2,3,4,...,70] o [11,12,...,99,0,1,2,3,4,...,10] = ?
3. Let p S7 be a swap permutation on 7 elements. Compute the following (your result should be an explicit permutation):
p o p o p o p = ?
4. Let p, q C11 be two distinct cyclic permutation on 11 elements. Compute the following (your result should be an explicit permutation):
p o q o p-1 o q-1 = ?
5. Define the isomorphism between the two algebraic structures (C2, o) and ((ℤ/4ℤ)*, ⋅). You must write down a bijection that specifies how each element in C2 corresponds to an element of (ℤ/4ℤ)*, and you must write down four pairs of corresponding equation that show that the behavior of o on elements of C2 is the same as the behavior of ⋅ on corresponding elements of (ℤ/4ℤ)*.
6. Define the isomorphism between the two algebraic structures (closure({2 + 15ℤ}, ⋅), ⋅) and (ℤ/4ℤ, +) (where ⋅ in the first case refers to multiplication modulo 15). You must write down a bijection that specifies how each element in closure({2 + 15ℤ}, ⋅) corresponds to an element of ℤ/4ℤ, but you do not need to write out all the equations.
7. Explain why there can be no isomorphism between the two algebraic structures (S6, o) and (ℤ/6ℤ, +).
2. Implement the following Python functions for working with permutations. Any code that does not conform exactly to the specified requirements will receive no credit.
1. Implement a function `permute(p, l)` that takes two arguments: a permutation `p` (represented as a Python list of integers) and a list `l` of the same length as the permutation. It should return the list after it has been permuted according to the permutation.
>>> permute([2,1,0], ['a','b','c'])
['c','b','a']
2. Implement a function `C(k, m)` that takes two integers `k` and `m` where `k` < `m` and returns the cyclic permutation in Cm that shifts all elements up by k.
>>> C(1, 4)
[1, 2, 3, 0]
3. Implement a function `M(a, m)` that takes two coprime integers `a` and `m` where `0` < `a` < `m` and returns the multiplication-induced permutation in Mm that corresponds to multiplication by a modulo m:
>>> M(2, 5)
[0, 2, 4, 1, 3]
4. Implement a function `sort(l)` that takes a list of integers of some length n and returns:
• a cyclic permutation p Cn that will sort the list into ascending order, if it exists;
• a multiplication-induced permutation p Mn that will sort the list into ascending order, if it exists;
• `None` otherwise.
>>> sort([38,16,27])
[1,2,0]
>>> permute(sort([38,49,16,27]), [38,49,16,27])
[16,27,38,49]
>>> sort([1, 13, 4, 17, 6, 23, 9])
[0, 2, 4, 6, 1, 3, 5]
>>> sort([0, 17, 4, 21, 8, 25, 12, 29, 16, 3, 20, 7, 24, 11, 28,\
15, 2, 19, 6, 23, 10, 27, 14, 1, 18, 5, 22, 9, 26, 13])
[0, 23, 16, 9, 2, 25,18, 11, 4, 27, 20, 13, 6, 29, 22,
15, 8, 1, 24, 17, 10, 3, 26, 19, 12, 5, 28, 21, 14, 7]
3. In this problem you will implement a secure and correct multiplication algorithm by using an untrusted, unreliable third-party web service to perform the actual multiplication. The web service consists of a PHP script at `http://cs-people.bu.edu/lapets/235/unreliable.php` (e.g., the result of ((2 ⋅ 3 ⋅ 4) mod 7) can be obtained at `http://cs-people.bu.edu/lapets/235/unreliable.php?n=7&data=2,3,4`). You can use the Python function below to invoke this script (this Python code needs to run on a computer connected to the internet).
def unreliableUntrustedProduct(xs, n):
url = 'http://cs-people.bu.edu/lapets/235/unreliable.php'
return int(urlopen(url+"?n="+str(n)+"&data="+",".join([str(x) for x in xs])).read().decode())
To test your code on a service that does not sometimes return incorrect answers, you can also use the URL `http://cs-people.bu.edu/lapets/235/reliable.php`.
1. Implement a function `privateProduct(xs, p, q)` that takes three inputs: a non-empty list of integers `xs`, a prime `p`, and another distinct prime `q`. The function must compute the product modulo `p` of all the integers in the list `xs` (assuming the web service performs its job correctly, which it may sometimes not do):
`privateProduct(xs, p, q)`
=
`(xs[0] * xs[1] * ... * xs[len(xs)-1]) % p`
However, your implementation may not multiply any of the integers in `xs` on its own; it must use `unreliableUntrustedProduct()` to do so, and at the same time it must not send the actual integers in `xs` over the web or reveal them to the web service. Your implementation should leak no information about the entries in `xs` or the product obtained by multiplying them to anyone (unless they can solve an intractable problem in modular arithmetic).
To solve this problem, use the extra prime `q` to create public and private RSA keys, and then encrypt all the integers in `xs` (you will need to encrypt them modulo `n` instead of modulo `p` because RSA needs a composite modulus). Next, use `unreliableUntrustedProduct()` to compute the product of the RSA-encrypted integers modulo `n` (i.e., take advantage of this isomorphism as in this homomorphic encryption example). Finally, your `privateProduct()` implementation should decrypt the result and return the product modulo `p`.
2. Implement a function `validPrivateProduct(xs, p, q)` that takes three inputs: a non-empty list of integers `xs`, a prime `p`, and another distinct prime `q`. The function must always correctly compute the product modulo `p` of all the integers in the list `xs`:
`validPrivateProduct(xs, p, q)`
=
`(xs[0] * xs[1] * ... * xs[len(xs)-1]) % p`
As before, your implementation may not multiply any of the integers in `xs` on its own; it must use `unreliableUntrustedProduct()` to do so, and at the same time it must not send the actual integers in `xs` over the web or reveal them to the web service.
Start with your solution to part (a) and extend it by choosing a random value `r` in ℤ/`q`ℤ. For each integer `xs[i]` in the list, convert the pair `(xs[i], r)` into a value in ℤ/`n`ℤ via the CRT isomorphism. Then encrypt all these ℤ/`n`ℤ values using RSA and use `unreliableUntrustedProduct()` to compute their product. Finally, your `validPrivateProduct()` implementation should decrypt the result and convert it back to an answer modulo `p` via the opposite direction of the CRT isomorphism.
To check if the answer you obtained from `unreliableUntrustedProduct()` is actually correct, determine whether the decrypted value modulo `q` has the expected value (i.e., is it really `r``len(xs)` modulo `q`). If this check fails, keep repeating the entire process from the beginning until it succeeds.
4. Implement a function `isomorphism(A, B)` that takes two tuples `A` and `B` as inputs. Each tuple consists of two entries: the first is an ordered list of elements from an algebraic structure, and the second is a function on elements of that binary structure. Thus, the tuple `A` represents an algebraic structure (A, ⊕), and the tuple `B` represents an algebraic structure (B, ⊗). You may assume that the list of elements is closed under the operation represented by the function. You may also assume that the bijection between the sets A and B is already provided by the order of the elements in each list (i.e., the ith entry in the first list corresponds to the ith entry in the second). The `isomorphism()` function should return `True` if there is indeed an isomorphism between (A, ⊕) and (B, ⊗), and `False` otherwise.
>>> plusMod2 = lambda x,y: (x + y) % 2
>>> A = ([0,1], plusMod2)
>>> B = ([C(0,2), C(1,2)], permute)
>>> isomorphism(A, B)
True
>>> plusMod4 = lambda x,y: (x + y) % 4
>>> multMod8 = lambda x,y: (x * y) % 8
>>> A = ([0,1,2,3], plusMod4)
>>> B = ([1,3,5,7], multMod8)
>>> isomorphism(A, B)
False
5. Extra credit: In a comment, explain why there cannot be an isomorphism between (ℤ/4ℤ, +) and ((ℤ/8ℤ)*, ⋅), even though the two sets are the same size because |(ℤ/8ℤ)*| = φ(8) = 23 − 22 = 4 = |ℤ/4ℤ|.
### [link] 5.5. Isomorphisms and Linear Equations of Congruence Classes
Fact: Let n be a positive integer. Then if + represents addition modulo n, we have:
ℤ/nℤ
closure({1}, +)
In other words, 1 is a generator for ℤ/nℤ with respect to +.
Fact: Let a and n be coprime positive integers. Then if + represents addition modulo n, we have:
ℤ/nℤ
closure({a}, +)
In other words, a can be a single generator for ℤ/nℤ with respect to +. This is equivalent to a fact we have already seen.
Fact: Let a and n be any two positive integers. Then if + represents addition modulo n, we have:
ℤ/(n/gcd(n,a))ℤ
closure({a}, +)
closure({gcd(n,a)}, +)
Example: Consider n = 6 and a = 4. Then we have g = gcd(4,6) = 2. We have:
closure({4}, +)
=
{4 ⋅ 0, 4 ⋅ 1, 4 ⋅ 2}
=
{0, 4, 2}
=
closure({2}, +)
Note that:
4
=
2 + 2 (mod 6)
2
=
4 + 4 (mod 6)
Thus, 2 can be expressed using the generator 4, and 4 can be expressed using the generator 2.
Fact (linear congruence theorem): Suppose that for a positive integer n and two congruence classes a ℤ/nℤ and b ℤ/nℤ where g ≡ gcd(a,n), we are given the following equation:
a ⋅ x
b (mod n)
Since a and n are not coprime, we cannot solve the above equation. Furthermore, if b ∉ closure({a}, +), we know the equation cannot be solved. Since closure({a}, +) = closure({g}, +), the equation can only be solved if b closure({g}, +). In other words, the equation can only be solved if g|b.
Note that if g ≡ gcd(n,m) and b closure({g}, +), we have:
n
=
n' ⋅ g
a
=
a' ⋅ g
b
=
b' ⋅ g
(a' ⋅ g) ⋅ x
(b' ⋅ g) (mod (n' ⋅ g))
We can then rewrite the above as:
(a' ⋅ g) ⋅ x
=
(b' ⋅ g) + k ⋅ (n' ⋅ g)
We can divide both sides of the above equation by g:
a' ⋅ x
=
b' + k ⋅ n'
We can convert the above equation back into an equation of congruence classes:
a' ⋅ x
b' mod n'
At this point a' and n' are coprime, we can compute a'-1 (mod n') and multiply both sides by it to find our solution x:
a'-1 ⋅ a' ⋅ x
a'-1 ⋅ b' mod n'
x
a'-1 ⋅ b' mod n'
Example: Solve the following equation for x ℤ/8ℤ, or explain why no solution exists:
2 ⋅ x
3 (mod 8)
Since 3 ∉ closure({2}, +), there is no solution x.
Example: Solve the following equation for x ℤ/24ℤ, or explain why no solution exists:
16 ⋅ x
7 (mod 24)
If we attempt to apply the linear congruence theorem, we will find that gcd(16,24) = 8, and 7 ∉ closure({8},+), which means there is no solution.
Example: Solve the following equation for x ℤ/15ℤ, or explain why no solution exists:
9 ⋅ x
6 (mod 15)
We know that gcd(9,15) = 3, so we can apply the linear congruence theorem:
3 ⋅ x
2 (mod 5)
We can now multiply both sides by 3-1 ≡ 2 (mod 5) to obtain the solution:
3-1 ⋅ 3 ⋅ x
3-1 ⋅ 2 (mod 5)
x
4 (mod 5)
### [link] 5.6. Isomorphisms and the Chinese Remainder Theorem
Fact (Chinese remainder theorem isomorphism): Let n and m be coprime positive integers. Let ℤ/nℤ × ℤ/mℤ be the set product of ℤ/nℤ and ℤ/mℤ, and let ⊕ be an operation on ℤ/nℤ × ℤ/mℤ defined as follows:
(a,b) ⊕ (c,d)
=
(a + c,b + d)
Then it is true that (ℤ/nℤ × ℤ/mℤ, ⊕) ≅ (ℤ/(mn)ℤ, +). The bijective relationship in this isomorphism is as follows:
(a mod n, b mod m) ∈ ℤ/nℤ × ℤ/mℤ
corresponds to
(a ⋅ (m ⋅ m-1) + b ⋅ (n ⋅ n-1)) ∈ ℤ/(m ⋅ n)ℤ
In other words, given (a, b), we can map it to a ⋅ (mm-1) + b ⋅ (nn-1), and given some c = a ⋅ (mm-1) + b ⋅ (nn-1) from ℤ/(nm)ℤ, we can map it back to ℤ/nℤ × ℤ/mℤ using (c mod n, c mod m).
Example (partial homomorphic encryption with validation): Suppose that for some p ℕ, Alice wants to store a large number of congruence classes b1,...,bk (ℤ/pℤ)* in Eve's database. Alice does not have enough memory to store the congruence classes herself, but she does not want to reveal the congruence classes to Eve. Alice also wants Eve to compute the product of the congruence classes for her as in this example. However, because Alice does not actually have the numbers Eve is multiplying, Alice has no way to know that the product Eve returns to her corresponds to the actual product; perhaps Eve is saving money and cheating by returning a random number to Alice.
To address this, Alice first chooses a new prime q (distinct from p). She then computes n = pq and φ(n) = (p − 1) ⋅ (q − 1), finds e (ℤ/φ(n)ℤ)* and computes de-1 (mod φ(n)). This will allow Alice to encrypt things before storing them in Eve's database. However, at this point Alice will not simply encrypt her congruence classes b1, ..., bk.
Instead, Alice will first choose a single random value r ℤ/qℤ. Then, Alice will map each of her values (bi, r) ℤ/pℤ × ℤ/qℤ via the CRT isomorphism to some values ci ℤ/nℤ. Alice will then encrypt the values ci by computing cie (mod n), and will submit these values to Eve's database. Now, whenever Alice can ask Eve to compute the following product:
c1e ⋅ ... ⋅ cke
(c1 ⋅ ... ⋅ ck)e (mod n)
If Eve returns the product (c1 ⋅ ... ⋅ ck)e to Alice, Alice can decrypt it by computing:
((c1 ⋅ ... ⋅ ck)e)d
c1 ⋅ ... ⋅ ck (mod n)
Next, Alice can compute (c1 ⋅ ... ⋅ ck) mod p to retrieve the actual product in ℤ/pℤ. However, Alice also wants to make sure that Eve actually multiplied all the entries. Alice can do so by computing:
(c1 ⋅ ... ⋅ ck) mod q
r ⋅ ... ⋅ r (mod q)
rk (mod q)
Alice can quickly compute rk (mod q) and compare it to (c1 ⋅ ... ⋅ ck) mod q. This gives Alice some confidence (but not total confidence) that Eve actually computed the product because it ensures that Eve really did multiply k distinct values provided by Alice.
What is one way that Eve can still cheat and save money under these circumstances (if she knows that Alice is using this validation method)? Is there any way Alice can counter this (hint: what if Alice chooses different values r for each entry)?
Fact: Let n and m be positive integers. and let g = gcd(n,m). Let ℤ/(n/g)ℤ × ℤ/(m/g)ℤ × ℤ/gℤ be a set product, and let ⊕ be an operation on ℤ/(n/g)ℤ × ℤ/(m/g)ℤ × ℤ/gℤ defined as follows:
(a,b,c) ⊕ (x,y,z)
=
(a + x,b + y, c + z)
Then it is true that (ℤ/(n/g)ℤ × ℤ/(m/g)ℤ × ℤ/gℤ, ⊕) ≅ (ℤ/((mn)/g)ℤ, +).
Example: Consider the set ℤ/2ℤ × ℤ/3ℤ with the operation ⊕, and the set ℤ/6ℤ together with the operation +. It is true that (ℤ/2ℤ × ℤ/3ℤ, ⊕) ≅ (ℤ/6ℤ, +). The bijection is specified below.
ℤ/2ℤ × ℤ/3ℤ ℤ/6ℤ (0,0) 0 (0,1) 4 (0,2) 2 (1,0) 3 (1,1) 1 (1,2) 5
The isomorphism is demonstrated below.
ℤ/2ℤ × ℤ/3ℤ ℤ/6ℤ (0,0) ⊕ (0,0) = (0,0) 0 + 0 = 0 (0,0) ⊕ (0,1) = (0,1) 0 + 4 = 4 (0,0) ⊕ (0,2) = (0,2) 0 + 2 = 2 (0,0) ⊕ (1,0) = (1,0) 0 + 3 = 3 (0,0) ⊕ (1,1) = (1,1) 0 + 1 = 1 (0,0) ⊕ (1,2) = (1,2) 0 + 5 = 5 (0,1) ⊕ (0,0) = (0,1) 4 + 0 = 4 (0,1) ⊕ (0,1) = (0,2) 4 + 4 = 2 (0,1) ⊕ (0,2) = (0,0) 4 + 2 = 0 (0,1) ⊕ (1,0) = (1,1) 4 + 3 = 1 (0,1) ⊕ (1,1) = (1,2) 4 + 1 = 5 (0,1) ⊕ (1,2) = (1,0) 4 + 5 = 3
ℤ/2ℤ × ℤ/3ℤ ℤ/6ℤ (0,2) ⊕ (0,0) = (0,2) 2 + 0 = 2 (0,2) ⊕ (0,1) = (0,0) 2 + 4 = 0 (0,2) ⊕ (0,2) = (0,1) 2 + 2 = 4 (0,2) ⊕ (1,0) = (1,2) 2 + 3 = 5 (0,2) ⊕ (1,1) = (1,0) 2 + 1 = 3 (0,2) ⊕ (1,2) = (1,1) 2 + 5 = 1 (1,0) ⊕ (0,0) = (1,0) 3 + 0 = 3 (1,0) ⊕ (0,1) = (1,1) 3 + 4 = 1 (1,0) ⊕ (0,2) = (1,2) 3 + 2 = 5 (1,0) ⊕ (1,0) = (0,0) 3 + 3 = 0 (1,0) ⊕ (1,1) = (0,1) 3 + 1 = 4 (1,0) ⊕ (1,2) = (0,2) 3 + 5 = 2
ℤ/2ℤ × ℤ/3ℤ ℤ/6ℤ (1,1) ⊕ (0,0) = (1,1) 1 + 0 = 0 (1,1) ⊕ (0,1) = (1,2) 1 + 4 = 5 (1,1) ⊕ (0,2) = (1,0) 1 + 2 = 3 (1,1) ⊕ (1,0) = (0,1) 1 + 3 = 4 (1,1) ⊕ (1,1) = (0,2) 1 + 1 = 2 (1,1) ⊕ (1,2) = (0,0) 1 + 5 = 0 (1,2) ⊕ (0,0) = (1,2) 5 + 0 = 5 (1,2) ⊕ (0,1) = (1,0) 5 + 4 = 3 (1,2) ⊕ (0,2) = (1,1) 5 + 2 = 1 (1,2) ⊕ (1,0) = (0,2) 5 + 3 = 2 (1,2) ⊕ (1,1) = (0,0) 5 + 1 = 0 (1,2) ⊕ (1,2) = (0,1) 5 + 5 = 4
Since 1 and 5 are generators for ℤ/6ℤ with respect to +, the corresponding elements (1,1) and (1,2) are generators for ℤ/2ℤ × ℤ/3ℤ.
Fact: Suppose that for two positive integers n and m where g ≡ gcd(n,m) and two congruence classes a ℤ/nℤ and b ℤ/mℤ, we are given the following system of equations:
x
a (mod n)
x
b (mod m)
We then know that:
n
=
n' ⋅ g
m
=
m' ⋅ g
But this means that:
x
a (mod (n' ⋅ g))
x
b (mod (m' ⋅ g))
The above equations can be converted into facts about divisility:
x
=
a + k ⋅ (n' ⋅ g)
x
=
b + l ⋅ (m' ⋅ g)
But note that:
x
=
a + (k ⋅ n')) ⋅ g
x
=
b + (l ⋅ (m')) ⋅ g
The above implies:
x
=
a (mod g)
x
=
b (mod g)
Since xx, it must be that:
a
=
b (mod g)
Thus, a solution x exists for the system of equations only if ab (mod (gcd(n,m)).
Fact: Suppose that for two positive integers n and m where g ≡ gcd(n,m) and two congruence classes a ℤ/nℤ and b ℤ/mℤ, we are given the following system of equations:
x
a (mod n)
x
b (mod m)
Note that because g ≡ gcd(n,m), we have:
n
=
n' ⋅ g
m
=
m' ⋅ g
To find a solution, first determine whether ab (mod g), and compute r = a (mod g). Then set:
x = y + r
We can now solve the following system for y:
y + r
a (mod n)
y + r
b (mod m)
We substruct r from both sides in both equations. We now have g | a-r and g | b-r, since r was the remainder when dividing a and b by g:
y
(a − r) (mod n)
y
(b − r) (mod m)
Now set:
y = g ⋅ z
We can now solve the following system for y:
g ⋅ z
(a − r) (mod (n' ⋅ g))
g ⋅ z
(b − r) (mod (m' ⋅ g))
Using the linear congruence theorem on both equations, we get:
z
(a − r)/g (mod n')
z
(b − r)/g (mod m')
We know that n' and m' are coprime, so we can solve for z using the usual method for solving systems of equations with coprime moduli. Once we find z, we can compute a solution x to the original system of equations:
x
g ⋅ z + r (mod ((n ⋅ m) / g))
Example: Suppose we want to solve the following system of equations:
x
1 (mod 6)
x
3 (mod 8)
First, we compute gcd(6,8) = 2. Then we check that 1 ≡ 3 (mod 2). Since this is true, we know we can find a solution. We proceed by subtracting 1 from both sides of both equations:
x − 1
0 (mod 6)
x − 1
2 (mod 8)
We can now apply the linear congruence theorem to both equations:
x − 1 2
0 (mod 3)
x − 1 2
1 (mod 4)
We can now solve the above system of equations using the usual CRT solution computation because the moduli are now coprime:
x − 1 2
0 ⋅ (4 ⋅ 4-1) + 1 ⋅ (3 ⋅ 3-1)
0 + 1 ⋅ (3 ⋅ 3)
9
We now compute x:
x − 1 2
9
x − 1
18
x
19
Since the range of unique CRT solutions with coprime moduli is ℤ/((6 ⋅ 8)/gcd(6,8))ℤ = ℤ/24ℤ, the congruence class solution is:
x
19 (mod 24)
greatestcommondivisoralgorithm Fermat'slittletheorem Euler'stheorem ⇑ ⇑ ⇑ Bézout'sidentity ⇐ extendedEuclideanalgorithm ⇐ algorithm forfindingmultiplicativeinverses ⇒ Euler'stotientfunction φ ⇑ Chineseremaindertheorem ⇐ CRT solverfor twoequations ⇑ linearcongruencetheorem ⇐ generalCRT solverfor twoequations ⇑ induction ⇐ generalCRT solverfor nequations
formula for3+4ℤprimes ⇐ square rootsmodulo p ⇑ Hensel's lemma ⇐ square rootsmodulo pk ⇑ CRT solverfor twoequations ⇐ square rootsmodulo n ⋅ m
Example: Suppose we want to solve the following equation for any congruence classes in ℤ/6ℤ that solve it:
4 ⋅ x + 3 ⋅ x2
2 (mod 6)
One approach is to use the Chinese remainder theorem and split the problem into two equations by factoring 6:
4 ⋅ x + 3 ⋅ x2
2 (mod 2)
4 ⋅ x + 3 ⋅ x2
2 (mod 3)
We can now simplify each of the above:
3 ⋅ x2
0 (mod 2)
4 ⋅ x
2 (mod 3)
We can simplify each further:
x2
0 (mod 2)
x
2 (mod 3)
We know that the only solution to the first is x ≡ 0 (mod 2), so we have now obtained the following system of equations; we can use CRT to find the unique solution modulo 6:
x
0 (mod 2)
x
2 (mod 3)
x
2 (mod 6)
We can check that, indeed, 4 ⋅ 2 + 3 ⋅ 22 ≡ 20 ≡ 2 (mod 6).
Example: Solve the following equation for all congruence classes in ℤ/7ℤ that satisfy it:
x2 − 3 ⋅ x
0 (mod 7)
### [link] 5.7. Abstract Algebraic Structures and Axioms
Algebraic structures can be categorized according to the additional algebraic properties they possess. In this section we introduce terminology for several common types of algebraic structure.
Definition: Let G be a set, and let (A, ⊕) be an algebraic structure where A = closure(G, ⊕). In this case, we call A a magma and we call G the generating set or the set of generators of A.
In other words, a set A is a magma under operator ⊕ if:
• A is closed under ⊕ (i.e., for all x, y A, xy A).
If (A, ⊕) possesses no other algebraic properties, we say (A, ⊕) is a free magma, or that (A, ⊕) is strictly a magma.
Given any finite set of elements S, a closure over that set with respect to an operator that has no other algebraic properties will be a magma. Furthermore, it will correspond to the set of binary trees in which leaves are elements from S.
Example: Suppose we have the algebraic structure (closure({a, b}, ⊕), ⊕) where ⊕ has no other algebraic properties. This structure is then a magma, and every element in closure({a, b}, ⊕) corresponds to a binary tree in which every non-leaf node is labelled ⊕ and every leaf node is either a or b.
Definition: We call an algebraic structure (A, ⊕) a semigroup under ⊕ if:
• A is closed under ⊕;
• ⊕ is associative on A (for all x,y,z A, the equation (xy) ⊕ z = x ⊕ (yz) is true).
If (A, ⊕) possesses no other algebraic properties, we call it a free semigroup or we say it is strictly a semigroup.
Definition: We call an algebraic structure (A, ⊕) a monoid under ⊕ if:
• A is closed under ⊕;
• ⊕ is associative on A;
• A contains an identity (which we call 1) in A (where for all x A, the equations 1x = x and x1 = x are always true).
If (A, ⊕) possesses no other algebraic properties, we call it a free monoid or we say it is strictly a monoid.
Example: Let S be the set of all strings, let + represent string concatenation, and let `""` be the empty string. Then (S, +) is an algebraic structure that is a monoid because string concatenation is associative and `""` is the identity with respect to string concatenation.
Example: Let L be the set of all lists, let + represent list concatenation, and let `[]` be the empty list. Then (L, +) is an algebraic structure that is a monoid because list concatenation is associative and `[]` is the identity with respect to list concatenation.
Definition: We call an algebraic structure (A, ⊕) a group under ⊕ if:
• A is closed under ⊕;
• ⊕ is associative on A;
• A contains an identity;
• A has inverses with respect to ⊕ (for all x A, there exists x-1 A such that xx-1 = 1 and x-1x = 1 are always true).
If (A, ⊕) possesses no other algebraic properties, we call it a free group or we say it is strictly a group.
Definition: We call an algebraic structure (A, ⊕) an abelian group under ⊕ if:
• A is closed under ⊕;
• ⊕ is associative on A;
• A contains an identity;
• A has inverses with respect to ⊕;
• ⊕ is commutative on A (i.e., for all x,y A, the equation xy = yx is always true).
Fact: Define Sn to be the set of permutations of the set {0,...,n-1}. Together with the binary composition operation o on permutations, Sn is a group, and we call it the symmetric group on n element:
• Sn is closed under composition of permutations (since Sn contains all of them);
• composition of functions (including permutations) is associative;
• there is an identity permutation [0,1,2,3,4,5,...,n-1];
• every permutation p Sn has an inverse p-1 Sn such that p o p-1 = [0,1,...,n-1].
Notice that Sn is not commutative.
Fact: The set of cyclic permutations Cn is an abelian group (i.e., it is a commutative group).
Fact: The set of multiplication-induced permutations Mn is an abelian group (i.e., it is a commutative group).
Fact: For any n ℕ, the algebraic structures (ℤ/nℤ, +) and ((ℤ/nℤ)*, ⋅) are abelian groups.
Example: Recall that a magma is an algebraic structure with a set that is closed under a binary operation, and a free magma is such an algebraic structure with no additional algebraic properties. An example of a Python data structure that corresponds to such a structure is the set of nested lists. An implementation of the binary operation for such a data structure might be:
def op(x, y):
return [x, y];
Notice that the above operation is not associative, has no identity or inverses, and is not commutative.
>>> op(1,2) == op(2,1)
False
>>> op(1,op(2,3)) == op(op(1,2),3)
False
Example: A free commutative magma is an algebraic structure with no algebraic properties except commutativity. An example of a Python data structure that corresponds to this algebraic structure is the set of nested sets. An implementation of the binary operation for such a data structure might be:
def op(x, y):
return {x, y};
Notice that the above operation is not associative, has no identity or inverses, but is commutative.
>>> op(1,2) == op(2,1)
True
>>> op(3,op(1,2)) == op(op(2,1),3)
True
>>> op(1,op(2,3)) == op(op(1,2),3)
False
Example: A free monoid is an algebraic structure with no algebraic properties except associativity and the existence of an identity. An example of a Python data structure that corresponds to this algebraic structure is the set of lists (not nested lists). An implementation of the binary operation for such a data structure might be:
def op(x, y):
return x + y # List concatenation done using +.
Notice that the above operation is associative and has the identity `[]`.
>>> op([1],op([2],[3])) == op(op([1],[2]),[3])
True
>>> op([1],op([2],[3])) == op(op([1],[2]),[3])
True
>>> op([], [1,2,3]) == [1,2,3]
True
Example: Suppose a customer wants us to store a very large individual element x from each of the following sets in a distributed manner (i.e., we want to store it in parts on multiple computers). We assume that the customer does not care if we store (and return upon retrieval) an equivalent element. What additional annotation information will we need in order to reconstruct the element? What information can we eliminate, thus optimizing our use of storage space?
• x A1, where A1 is a free magma;
• x A2, where A2 is a monoid;
• x A3, where A3 is a group and has an element y A3 s.t. yx takes much less space to store than x;
• x A4, where A4 is a commutative semigroup.
The additional annotation information is as follows:
• an element of a free magma is a binary tree in which every node has ordered children, so for any node whose children we want to store separately, we would need a unique pair of ordered identifiers for each child subtree;
• an element x in a monoid is assembled using an associative operation, and we would not need to store anything about the grouping/hierarchization of the individual generators found in an element x, so it would be sufficient to use an integer to represent each separate portion (reassembly would involve sorting the pieces in ascending order by the integer associated with them, and then gluing them back together using the associative operation);
• whenever we need to store a copy of x, we should instead store yx and return y-1 ⋅ (yx) = x to the user when we need to retrieve the stored element;
• since a commutative semigroup's operation is both associative and commutative, we can group all the like generators in x together and simply store how many of each there are (for example, if x = abaabbacc, we would only store 4a3b2c).
The following table lists the correspondences between the parts of an algebraic structure and how they might correspond to a concrete data structure.
abstract algebraic structure concrete data structure element in set of generators S individual character, string, or data object set of generators S base cases; alphabet; data containers operator ⊕ constructor for node with children; operator/function/method on data items element in closure of S under ⊕ individual data object closure of S under ⊕ set of all possible data objects
Suppose a concrete data structure corresponds to an algebraic structure with certain properties. The following table summarizes how those properties can inform or constrain implementations of algorithms or applications that operate on the concrete data sttructure.
magma commutativemagma semigroup commutativesemigroup monoid group abeliangroup typicaldatastructures binary treesunder nodeconstructor binary treeswith unorderedbranches undernode constructor lists withlist concatenation;strings withstring concatenation sets withduplicates lists withlist concatenationand empty list;strings withstring concatenationand empty string distributedcomputation must operate ondata in itsoriginalordering andhierarchization must operate ondata in itsoriginalhierarchization must operate overoriginal ordering can operate ondata in anyorder can employ identityas a base case;can ignoreidentity entries can "cancel" pairs ofadjacent inverses can "cancel" allelements thatcan be pairedwith an inverse distributedstorage must store originalorderingand hierarchization must store originalhierarchization must store original ordering can store in anyorder no need tostore identities compression common subtrees common hierarchies run-length encoding distinct elementswith quantities can ignoreidentity entries can "cancel" pairs ofadjacent inverses;can applyinvertibletransformations can "cancel" allelements thatcan be pairedwith an inverse example/test caseenumeration can enumerate examples/test cases automatically provingimplementationworks correctlyfor all inputs can show this by structural induction
Example: Suppose we have an algebraic structure (ℕ, max) consisting of the set ℕ and the binary operator max. Which algebraic properties does this algebraic structure satisfy?
• ℕ is closed under max
• max is associative on ℕ
• max is commutative on ℕ
Notice that there are no identities or inverses. Thus, (ℕ, max) is a commutative semigroup.
Suppose that we have a distributed database (i.e., different parts of the database are on different computers) containing records that indicate the ages of individual people. These age values are represented as elements in ℕ. How can we compute the age of the oldest individual that has a record in the database?
Because max is associative and commutative over A, each computer can compute max over the records that are stored on it. Once each computer has computed its maximum, it can broadcast it to the other computers. One computer can then be designated to receive these maxima and take their maximum to obtain the final result.
Example: Consider the following set A:
A
=
{unknown} ∪ {(a,n) | a ∈ ℕ, n ∈ ℕ}
Consider the following definition of a binary operator ⊕ over A: given (a,n) and (b,m):
• if gcd(n,m) = 1, then (a, n) ⊕ (b, m) = (c, (nm)) where c is the unique solution to the system of equations:
c
a (mod n)
c
b (mod m)
• if gcd(n,m) ≠ 1, then (a, n) ⊕ (b, m) = unknown;
• for any (a,n) A, unknown ⊕ (a, n) = unknown;
• for any (a,n) A, (a, n) ⊕ unknown = unknown.
Is A an algebraic structure? What are its algebraic properties? What does this tell us about how we can implement an algorithm that performs this operation over many elements in A?
Yes, A is an algebraic structure in which the operation ⊕ is associative, commutative, and has (1,0) as the identity element (because every integer is a solution to x ≡ 0 (mod 1)). Thus, A is at least a commutative monoid.
The fact that the operation is commutative and associative tells us that we do not need to worry about the order in which we perform the operation on a collection of elements (we can perform the operation on any subset of the elements in any grouping, then put this solution together with other solutions assembled from other subsets of the overall set of elements). This means we can distribute the computation of the operation across many computers if we wish. The identity element allows us to easily include a base case in an iterative or recursive algorithm for computing the operation over a collection of elements.
Example: Determine which properties each of the following algebraic structures satisfy.
1. The generating set G = {∅, {1}, {2}, {3}} with the union operation ∪ as the binary operator.
We know that closure(G, ∪) is the set of subsets of {1,2,3}:
closure(G, ∪)
=
{∅, {1}, {2}, {3}, {1,2}, {2,3}, {1,3}, {1,2,3}}
Thus, we know from set theory that the union operation on sets is associative and commutative. There is an identity element ∅. There are no inverses.
2. For positive integers a,n ℕ where gcd(a,n) = 1, generating set G = {a} with integer addition modulo n.
The closure is the entire set {0,..., n − 1} because:
{a, a + a, a + a + a, ..., a + ... + a}
=
{(i ⋅ a) mod n | i ∈ {1 ,..., n}}
Integer addition modulo n is associative and commutative, has identity 0, and for every x ℤ/nℤ, the inverse is n-x ℤ/nℤ.
### [link] 5.8. Subgroups and the Hidden Subgroup Problem
If we look at subsets of the elements in a group, we might find that certain subsets are closed under the operator for that group. These subsets are called subgroups, and the concept of a subgroup can be very useful when studying and using groups.
Definition: Let A be a group under the operator ⊕. We say that B is a subgroup of A if BA, B is closed under ⊕, and B is a group.
Example: The following are all the subgroups of ℤ/4ℤ under addition + modulo 4:
• {0}, because all terms of the form 0, 0+0, 0+0+0, and so on are equivalent to 0;
• {0,2}, since closure({0,2}, +) = {0,2};
• {0,1,2,3} = ℤ/4ℤ.
The following are all the subgroups of ℤ/6ℤ under addition + modulo 6
• {0}, because all terms of the form 0, 0+0, 0+0+0, and so on are equivalent to 0;
• {0,2,4}, since closure({2}, +) = closure({0,2,4}, +) = {0,2,4};
• {0,3}, since closure({3}, +) = closure({0,3}, +) = {0,3};
• {0,1,2,3,4,5} = closure({1}, +) = ℤ/6ℤ.
Fact: Given some n ℕ and some factor f ℕ such that f|n, then (closure({f}, +), +) is a subgroup of (ℤ/nℤ, +), and it is isomorphic to the group (ℤ/(n/f)ℤ, +).
Example: The following are all the non-trivial subgroups of ℤ/6ℤ under addition + modulo 6, together with their corresponding isomorphic group:
• ({0,2,4}, +) ≅ (ℤ/(6/2)ℤ) ≅ (ℤ/3ℤ, +);
• ({0,3}, +) ≅ (ℤ/(6/3)ℤ) ≅ (ℤ/2ℤ, +).
The notion of a subgroup allows us to introduce an alternative definition for prime numbers.
Definition: Given an integer p ℕ where p > 1, we say that p is prime if the only subgroups of (ℤ/pℤ, +) are the trivial subgroups ({0}, +) and (ℤ/pℤ, +).
Conjecture (hidden subgroup problem): The following problem is not in P: given a group (A, ⊕), find a non-trivial subgroup of A (non-trivial means not the subgroup that contains only the identity, ({I}, ⊕), and not the subgroup consisting of the entire group, (A, ⊕)).
Often, we are interested in a more restricted versions of this problem, which are also not believed to be in P:
• finding a non-trivial subgroup of (ℤ/nℤ, +);
• finding a non-trivial subgroup of ((ℤ/nℤ)*, ⋅).
Example: Suppose that for some n ℕ, we are given a ℤ/nℤ such that gcd(a, n) = 1. We know from Euler's theorem that aφ(n) ≡ 1 (mod n). However, φ(n) is not necessarily the smallest exponent of a that will yield 1.
For example, consider 3 ℤ/8ℤ. Even though φ(8) = 23 - 22 = 4 and 34 ≡ 1 (mod 8), it is also true that 32 ≡ 9 ≡ 1 (mod 8).
Thus, given some a ℤ/nℤ such that gcd(a, n) = 1, the problem of determining the smallest r such that ar ≡ 1 (mod n) amounts to finding the smallest subgroup of ((ℤ/nℤ)*, ⋅) that contains a.
Algorithm (Shor's algorithm): Shor's algorithm relies on the ability of a quantum computer to find the smallest r > 0 such that ar ≡ 1 (mod n). It takes an arbitrary integer n as its input and finds a non-trivial factor of n with high probability.
1. Shor's algorithm: n
1. while r is odd or a(r/2) ≡ −1 (mod n)
1. choose a random a ℤ/n
2. if gcd(a, n) > 1, return gcd(a, n)
3. otherwise, find the smallest r > 0 such that ar ≡ 1 (mod n)
2. since we have exited the loop, then r is even and a(r/2) ≢ −1 (mod n)
3. thus, a(r/2) is a non-trivial root of ar
4. return gcd(a(r/2) ± 1, n), which is a non-trivial factor by this fact
Example (arithmetic with unbounded error and bounded unreliability): Suppose you need to perform a sequence of k addition operations in ℤ/15ℤ, but all the addition operators ⊕ modulo n available to you are error-prone. To add two numbers a, b modulo n accurately, you must perform the computation ab at least n times (because up to ⌈ n/2 ⌉ of those attempts will result in an arbitrarily large error).
This means that to perform k addition operations modulo 15, it will be necessary to perform every operation 15 times, for a total of k ⋅ 15 operations modulo 15. If each addition operation modulo n takes about log2 n steps, this would mean that k operations would take:
k ⋅ 15 ⋅ 4 steps
Assuming that performing CRT to find a solution in ℤ/15ℤ takes 10,000 steps, determine how you can use CRT to speed up the computation of these k addition operations, and for what minimum k this would be advantageous.
Example: Suppose we want to find all the solutions in ℤ/6ℤ to the following equation:
2 ⋅ x
4 (mod 6)
Using the linear congruence theorem, we can find the unique solution modulo 3:
x
2 (mod 3)
The solutions modulo 6 will be those congruence classes in ℤ/6ℤ whose integer representatives are members of 2 + 3ℤ = {..., 2, 5, 8, 11, 14, ...}. These are 2 + 6ℤ and 5 + 6ℤ, since 2 ≡ 2 (mod 3) and 5 ≡ 2 (mod 3). Note that:
2 + 3ℤ
=
2 + 6ℤ ∪ 5 + 6ℤ
{..., 2, 5, 8, 11, 14, 17, 20, ...}
=
{..., 2, 8, 14, 20, ...} ∪ {..., 5, 11, 17, ...}
### [link] 5.9. Structural Induction and Direct Product of Algebraic Structures
Suppose an algebraic structure A with operator ⊕ has a generating set. Since any element in A must be either a genrator or some finite number of generators combined using the operator ⊕, we can use induction to prove facts (i.e., true formulas) about all the elements in A.
Fact (fundamental theorem of arithmetic): Consider a commutative semigroup A with generating set S = { p | p ℕ, p is prime } and an operator ⊗ that is associative and commutative. Let ℕ be the commutative semigroup of positive integers { n | n ℕ, n ≥ 2 } with the integer multiplication operator. The semigroup A is isomorphic to ℕ:
(A, ⊗)
(ℕ, +)
Equivalently, we can say that every natural number has a unique prime factorization (notice that the algebra A is actually the set of prime factorizations, with ⊗ corresponding to multiplication of prime factorizations).
Example: We have introduced the following definition of the totient function φ.
φ(m)
=
|{k | k ∈ {1,...,m}, gcd(k,m) = 1}|
We have also introduced the following properties of the totient function φ.
φ(p)
=
p - 1
if p is prime
φ(pk)
=
pk - pk-1
if p is prime
φ(n ⋅ m)
=
φ(n) ⋅ φ(m)
if gcd(n,m) = 1
Is the set of three equations above equivalent to our one-line definition of φ? That is, if we know the factorization of any input n > 1 to φ, can we compute φ exactly using the three equations above?
The answer is "yes". Consider the algebraic structure consisting of the set {2,3,4,...} and the operation ⋅ (integer multiplication). Then the set of generators of A is the set of prime numbers. Thus, any element x A is a product of some collection of not necessarily distinct prime numbers p1 ⋅ ... ⋅ pk. If we write down n as this product, we can always break down our computation of φ(n) into smaller computations until we reach the base case.
### [link] 5.10.Assignment #6: Generalized CRT, Data Structures, and More Isomorphisms
In this assignment you will solve several problems using the generalized Chinese remainder theorem and other techniques and facts you have learned in this course. You must submit a single file named `hw6.py` (submitted to the location `hw6/hw6.py`). Please follow the gsubmit directions.
For the programming parts, you may import the following library functions in your module (you may not need all these functions for this assignment depending on how you approach the problems, but they may be used):
from math import floor
from fractions import gcd
from random import randint
Your file may not import any other modules or employ any external library functions associated with integers and sets unless explicitly permitted to do so in a particular problem.
1. Solve the following equations using step-by-step equational reasoning, and list each step.
1. Solve the following system of equations and find a unique congruence class solution (including the set of congruence classes from which the congruence class is drawn) if it exists; if no solution exists, explain why not:
x
7 (mod 21)
x
21 (mod 49)
2. Solve the following system of equations and find a unique congruence class solution if it exists; if no solution exists, explain why not:
x
11 (mod 14)
x
18 (mod 21)
3. Solve the following system of equations and find all congruence class solutions if any exist; if no solution exists, explain why not:
x2
4 (mod 35)
3 ⋅ x
15 (mod 21)
4. Solve the following system of equations and find a unique congruence class solution if it exists; if no solution exists, explain why not:
x
10 (mod 12)
x
2 (mod 16)
5. Suppose that you have a standard 12-hour analog clock. You also have an alarm that rings every 16 hours. At some point in the last two days, the alarm rang at exactly midnight. At this moment, the analog clock reads 6 (you do not know if it is AM or PM), and it has been 14 hours since the alarm last rang. How many hours have passed since the alarm rang at exactly midnight, and what time is it? Set up an appropriate equation or system of equations and use step-by-step equational reasoning to find the solution.
6. Bob needs to reserve virtual machines on a cloud computing service to solve a number of problems before a deadline. He has two options: reserve some number of virtual machines that can each solve 12 problems before the deadline, or reserve some number of virtual machines that can each solve 15 problems before the deadline:
• with virtual machines that can each solve 12 problems, Bob will end up using all the virtual machines to their full capacity except one, which will have only 1 problem to solve;
• with virtual machines that can each solve 15 problems, Bob will end up using all the virtual machines to their full capacity except one, which will have only 7 problems to solve.
Answer both of the following questions:
• Assuming that Bob has at most 60 problems to solve, how many problems does Bob have?
• If Bob can only reserve batches of machines that all have the same capacity (capacities are always 2 or greater), and the the fastest machine can only solve 19 problems before the deadline, is there anything Bob can do to avoid having at least one machine not used to its full capacity?
2. Implement the following Python functions for solving systems of equations involving congruence classes.
1. Implement a function `solveOne(c, a, m)` that takes three integers `c`, `a`, and `m` ≥ 1. If it exists, the function should return the solution x {0, ..., `m`-1} to the following equation:
`c` ⋅ x
`a` (mod `m`)
If no solution exists, the function should return `None`. The function must work correctly for all possible equations (you should use the linear congruence theorem).
>>> solveOne(1, 2, 3)
2
>>> solveOne(3, 4, 7)
6
>>> solveOne(1, 5, 11)
5
>>> solveOne(2, 3, 8)
None
>>> solveOne(6, 2, 8)
3
2. Implement a function `solveTwo(e1, e2)` that takes two tuples `e1` and `e2` as inputs, each of the form `(c, a, m)` (i.e., containing three integer elements). Each tuple `(c, a, m)` corresponds to an equation of the form:
`c` ⋅ x
`a` (mod `m`)
Thus, the two tuples, if we call them `(c, a, m)` and `(d, b, n)`, correspond to a system of equations of the form:
`c` ⋅ x
`a` (mod `m`)
`d` ⋅ x
`b` (mod `n`)
The function `solveTwo()` should return the unique solution x to the above system of equations. If either equation cannot be solved using `solveOne()`, the function should return `None`.
>>> solveTwo((3, 4, 7), (1, 5, 11))
27
>>> solveTwo((1, 1, 6), (1, 3, 8))
19
>>> solveTwo((1, 0, 6), (1, 3, 8))
None
3. Implement a function `solveAll(es)` that takes a list of one or more equations, each of the form `(c, a, m)`. The function should return the unique solution x to the system of equations represented by the list of equations. If the system of equations has no solution, the function should return `None`.
>>> solveAll([(1,2,3)])
2
>>> solveAll([(3,4,7), (1,5,11)])
27
>>> solveAll([(5,3,7), (3,5,11), (11,4,13)])
856
>>> solveAll([(1,2,3), (7,8,31), (3,5,7), (11,4,13)])
7109
>>> solveAll([(3,2,4), (7,8,9), (2,8,25), (4,4,7)])
554
>>> solveAll([(1, 1, 6), (1, 3, 8)])
19
>>> solveAll([(1, 0, 6), (1, 3, 8)])
None
3. Suppose you are given the following function (which simulates the component of Shor's algorithm that can run efficiently on a quantum computer). Given n ℕ and a ℤ/nℤ, it returns the smallest non-zero congruence class r ℤ/φ(n)ℤ such that ar ≡ 1 (mod n).
def quantum(a, n):
return [pow(a,k,n) for k in range(1,n)].index(1) + 1
Implement a function `factor(n)` that finds a non-trivial factor of a composite number input `n` by calling `quantum()`. Solutions that use exhaustive search will receive no credit.
4. In this problem you will implement a reliable addition algorithm for ℤ/256ℤ by using an unreliable addition algorithm for ℤ/256ℤ. You may not use the addition operator `+` anywhere in your solutions to this problem. You may assume you are given access to a function `plus256unreliable(x, y)` that returns an answer that is at most 4 away from the true sum modulo `256` (assume there is no chance of this error causing the answer to wrap around):
| `plus256unreliable(x, y)` − (`(x + y) % 256`) |
<
`4`
You may use the following Python simulation of this unreliable function in order to test your code:
from random import randint
def plus256unreliable(x, y):
r = randint(0,7) - 4
return (min(255, max(0, ((x + y)%256) + r)))
1. Implement a Python function `plus16(x, y)` that reliably returns `(x + y) % 16` with 100% accuracy. You may use `//`, `*`, `plus256unreliable()`, and numerical constants, but you may not use anything else in your definition. Hint: you can call `plus256unreliable()` more than once.
2. Implement a Python function `plus256(x, y)` that reliably returns `(x + y) % 256` with 100% accuracy. Your solution must use `solveAll()`, and you are allowed to use `%`, but you may not use the addition or subtraction operators: choose four appropriate prime or mutually coprime moduli, perform the addition operations modulo those moduli, then restore the original result using `solveAll()`.
5. Extra credit: Suppose you are working with a commutative magma (A, ⊕) where the set of elements is A = closure({a, b, c, d, e, f, g}, ⊕) and the commutative (but not associative) operator is ⊕. Add the following functions to your Python file; the first function implements the operator, and the second function randomly rearranges the elements in a given array.
def oplus(x, y):
return (x, y)
from random import shuffle
def rearrange(l):
r = l[:]
shuffle(r)
return r
1. Implement a Python function `decompose()` that decomposes a given element in the algebra into a list in which each generator from {a, b, c, d, e, f, g} that occurs in the input element is paired with exactly one integer. The integer should be chosen so that `reassemble()` below can work correctly. Your output should look something like the following (the full output is not revealed as it is part of the problem):
>>> oplus(oplus("a""b"), "c")
(("a","b"), "c")
>>> decompose(oplus(oplus("a""b"), "c"))
[("a", ???), ("b", ???), ("c", ???)]
2. Implement a Python function `reassemble()` that takes a randomly rearranged output from `decompose()` and reassembles an equivalent element within the algebraic structure.
>>> rearrange([("a", ???), ("b", ???), ("c", ???)])
[("c", ???), ("a", ???), ("b", ???)]
>>> reassemble([("c", ???), ("a", ???), ("b", ???)])
(("a","b"), "c")
6. Extra extra credit: The group S2C2 ≅ ℤ/2ℤ is the only group containing two elements; there is no way to build a group containing two elements that is not isomorphic to these. Likewise, C3 ≅ ℤ/3ℤ is the only group containing three elements. Both of these groups are cyclic.
The smallest possible non-cyclic group is called the Klein group, and is defined to be V = (closure({a, b, c}, ⊕), ⊕), where a is an identity and:
b ⊕ b
=
a
c ⊕ c
=
a
(b ⊕ c) ⊕ (b ⊕ c)
=
a
Note that V is a group, so bc = cb. Determine how many distinct elements there are in the group (i.e., provide an exhaustive set of distinct elements and informally or formally explain why all possible terms involving ⊗, a, b, and c must equal one of these elements).
## [link] Review #2. Algebraic Structures and their Properties
This section contains a comprehensive collection of review problems going over all the course material. Many of these problems are an accurate representation of the kinds of problems you may see on an exam.
Exercise: Suppose we have the following polynomial in the integers (i.e., all operations are arithmetic operations):
6 x1001 + 2 x600 + 1
Prove that the arithmetic expression above is always divisible by 3.
We first notice that the above is simply asking if all possible congruence classes x ℤ/3ℤ are solutions to the equation:
6 x1001 + 2 x600 + 1
0 (mod 3)
We can use Euler's theorem to simplify the exponents in the above equation, since φ(3) = 2:
6 x1001 mod 2 + 2 x600 mod 2 + 1
0 (mod 3)
6 x1 + 2 x0 + 1
0
6 x + 2 + 1
0
6 x + 0
0
0 ⋅ x
0
0
0
The above shows that, indeed, for any x ℤ/3ℤ, the expression yields a result in 0 + 3ℤ, so the original arithmetic expression is always divisible by 3 for all integers x.
Exercise: Suppose you have n ℕ and a congruence class a ℤ/nℤ such that gcd(a, n) = 1. Compute in terms of n (and only n) the congruence class corresponding to the following term:
0 ⋅ a + 1 ⋅ a + 2 ⋅ a + ... + (n − 1) ⋅ a
As we did when we proved Fermat's little theorem, we notice that the multiples of a are just a permutation of {0,1,...,n-1} = ℤ/nℤ. This means that:
0 ⋅ a + 1 ⋅ a + 2 ⋅ a + ... + (n − 1) ⋅ a
0 + 1 + ... + n-1 (mod n)
We know that there is a closed formula for the summation, which we can use:
0 + 1 + ... + n-1
((n ⋅ (n − 1)) / 2) (mod n)
Exercise: Find all x ℤ/29ℤ that satisfy the following:
y2
16 (mod 29)
x2
y (mod 29)
We first solve for all possible y. Since 29 > 16, we have that ± 4 are the square roots of 16 in ℤ/29ℤ. Thus, we have two solutions for y:
y
{4, 29 − 4}
y
{4, 25}
We want to find all x that satisfy the system, so we need to solve for x for each possible y. Thus, we have:
y
4 (mod 29)
x2
4 (mod 29)
x
± 2 (mod 29)
x
{2, 29 − 2}
x
{2, 27}
We also have:
y
25 (mod 29)
x2
25 (mod 29)
x
± 5 (mod 29)
x
{5, 29 − 5}
x
{5, 24}
Thus, the possible solutions for x are:
x
{2, 5, 24, 27}
Notice also that the problem could have been stated as:
x4
16 (mod 29)
Exercise: Solve the following problems.
1. Consider the following two circular shift permutations:
[8,9,0,1,2,3,4,5,6,7]
[5,6,7,8,9,0,1,2,3,4]
How many of each would you need to compose to obtain the permutation [9,0,1,2,3,4,5,6,7,8]?
These permutations specified are in C10, and we know that C10 ≅ ℤ/10ℤ. Notice that [8,9,0,1,2,3,4,5,6,7] C10 can correspond to 8 ℤ/10ℤ, that [5,6,7,8,9,0,1,2,3,4] C10 can correspond to 5 ℤ/10ℤ, and that [9,0,1,2,3,4,5,6,7,8] C10 can correspond to 9 ℤ/10ℤ. By Bézout's identity we have that:
8 ⋅ 2 + 5 ⋅ (-3)
=
1
8 ⋅ 2 + 5 ⋅ 7
1 (mod 10)
8 ⋅ (9 ⋅ 2) + 5 ⋅ (9 ⋅ 7)
9 ⋅ 1 (mod 10)
8 ⋅ 8 + 5 ⋅ 3
9 (mod 10)
Thus, we would need to compose 8 instances of the first permutation and 3 instances of the second permutation to obtain [9,0,1,2,3,4,5,6,7,8].
2. Rewrite the permutation [3,4,0,1,2] as a composition of adjacent swap permutations.
We can simply run the bubble sort algorithm on the above permutation and record the permutation that represents each swap. This leads to the following sequence:
p1
=
[0,2,1,3,4]
p2
=
[0,1,3,2,4]
p3
=
[0,1,2,4,3]
p4
=
[1,0,2,3,4]
p5
=
[0,2,1,3,4]
p6
=
[0,1,3,2,4]
Thus, by applying the above sequence of permutations to [0,1,2,3,4] in reverse order, we can obtain [3,4,0,1,2], so the factorization of [3,4,0,1,2] into adjacent swap permutations is:
[3,4,0,1,2]
=
p1 o p2 o p3 o p4 o p5 o p6 o [0,1,2,3,4]
Exercise: Suppose we want to perform k exponentiation operations (e.g., if k = 4, we want to compute (((ab)c)d)e) modulo 21. Assume the following:
• a single exponentiation operation modulo 21 takes 213 = 9261 steps;
• a single exponentiation operation modulo 3 takes 33 = 27 steps;
• a single exponentiation operation modulo 7 takes 73 = 343 steps;
• an exponentiation operation modulo 3 and an exponentiation operation modulo 7 together take 343 + 27 = 370 steps;
• solving a two-equation system for two values, a modulo 3 and b modulo 7, takes 8000 steps using CRT;
• we can either compute the exponentiation sequence directly modulo 21, or we can split it into two sequences of computations (one modulo 3, the other modulo 7) and then recombine using CRT at the end.
1. What is the number of steps needed to perform k exponentiations modulo 21?
f(k)
=
9261 ⋅ k
2. What is the number of steps needed to perform k exponentiations modulo 3 and k exponentiations modulo 7, then to recombine using CRT?
f(k) = 370 ⋅ k + 8000
Exercise: Find solutions to the following problems.
1. Explain why the following polynomial has no integer solutions (Hint: you only need to evaluate the polynomial for two possible values of x):
x4 + x2 + 3
=
0
If the above equation has an integer solution x, then it must have an integer solution modulo 2, since we can take the modulus of both sides:
x4 + x2 + 3
=
0
(x4 + x2 + 3) mod 2
=
0 mod 2
x4 + x2 + 3
0 (mod 2)
Thus, we have using logic that:
(exists solution in ℤ)
(exists solution modulo 2)
(no solution modulo 2)
(no solution in ℤ)
Note that this only works in one direction. A solution modulo 2 does not necessarily imply that there is an integer solution.
We see that for x = 0 and x = 1, the left-hand side is odd. The right-hand side is 0, so it is always even. Thus, no integer solution exists.
2. Find at least one solution x ℤ/10ℤ to the following system of equations (you must use Bézout's identity):
6 ⋅ y + 5 ⋅ x - 1
0 (mod 10)
x2
y (mod 10)
Since 5 and 6 are coprime, we can find a solution to the first equations. One such solution is:
y
1 (mod 10)
x
-1 (mod 10)
9 (mod 10)
We also have that 92 ≡ 81 ≡ 1 (mod 10). Thus, since 12 ≡ 1 (mod 10), both equations are satisfied by this solution.
Exercise: Find solutions to the following problems.
1. Suppose you want to send some s ℤ/nℤ to Alice and Bob, but you want to ensure that the only way Alice and Bob can retrieve s is if they work together. What two distinct pairs (s1, p1) and (s2, p2) would you send to Alice and Bob, respectively, so that they would need to work together to recover s?
You would need to send (s mod p, p) to Alice and (s mod q, q) to Bob where s < pq and p and q are distinct and coprime.
2. Suppose Bob is generating a public RSA key; he chooses a very large prime p, and then he chooses q = 2. Why is this not secure?
Bob must share his public key (n, e). Since n = p ⋅ 2, n is even. This can immediately be seen by looking at the last bit of n (which will be 0, since n mod 2 = 0). It is then easy to recover the secret value p.
3. Suppose Alice and Bob use Shamir secret sharing to share a password s to a lock that is not protected from brute force attacks (i.e., anyone can keep trying different passwords until they unlock it). Alice holds s mod p and Bob holds s mod q, where s < pq. However, suppose that Bob happens to be using q = 2, and Alice knows this. What can Alice do to quickly break the lock?
If Alice holds a and Bob holds b, the system of equations that would be set up to recover s is as follows:
x
a (mod p)
x
b (mod 2)
Notice that there are only two possibilities for Bob's value b ℤ/2ℤ. Thus, Alice could set up two systems, one for b = 0 and another for b = 1, solve both, and try both secrets s on the lock.
Exercise: Suppose that Alice, Bob, Carl, and Dan are sharing a secret s using Shamir secret sharing, where each participant is assigned a distinct modulus n that is coprime to everyone else's modulus. Each participant is holding a part of the secret s mod n, and the secret can be recovered by any two participants. However, Eve has sabotaged the value stored by one one the participants. Below are the values currently stored by everyone; one of them is corrupted.
• Alice: nAlice = 3 and (s mod 3) = 2
• Bob: nBob = 4 and (s mod 4) = 3
• Carl: nCarl = 5 and (s mod 5) = 2
• Dan: nDan = 7 and (s mod 7) = 4
1. Which participant's stored value s mod n has Eve sabotaged?
Any two participants can recover the secret s by setting up a system with two equations and solving the system using CRT. We list all of the possible combinations and solve for s to find which participant's value has been corrupted.
• Alice and Bob:
s
2 (mod 3)
s
3 (mod 4)
s
11 (mod 12)
• Alice and Carl:
s
2 (mod 3)
s
2 (mod 5)
s
2 (mod 15)
• Alice and Dan:
s
2 (mod 3)
s
4 (mod 7)
s
11 (mod 21)
• Bob and Carl:
s
3 (mod 4)
s
2 (mod 5)
s
7 (mod 20)
• Bob and Dan:
s
3 (mod 4)
s
4 (mod 7)
s
11 (mod 28)
• Carl and Dan:
s
2 (mod 5)
s
4 (mod 7)
s
32 (mod 35)
Since three of the participants can consistently recover the same secret s = 11, and all pairs of participants in which Carl is present do not yield s = 11 and are inconsistent with one another, it must be Carl's data that has been sabotaged.
2. What is the correct secret value s?
Since three of the six possible pairings result in s = 11, and the other three are inconsistent and all involve Carl, it must be that s = 11 was the original uncorrupted secret.
3. What's the number of different shared secret values these four participants can store (assuming they use the same moduli, and require that any two members should be able to recover the secret).
Since any two participants must be able to recover s, it must be that s < nm for every possible pair n and m. The smallest two values are 3 and 4, so 3 ⋅ 4 - 1 = 11, where 11 ℤ/(3 ⋅ 4)ℤ, is the largest possible s that can be receovered.
4. Suppose you want to store an n-bit number s. You want to store it in a way that makes it possible to recover s even if one of the bits is corrupted. How can you accomplish this using at most approximately 2 ⋅ n bits?
Choose four distinct coprime numbers m1, m2, m3, and m4 such that s is less than the product of every pair of these, but such that the product of any pair is not much larger than s (e.g., m1 and m2 can be stored in the same number of bits as s). Then store (s mod m1, s mod m2, s mod m3, s mod m4) using about 2 ⋅ n bits.
If any individual bit is corrupted, this corrupts at most one of the four values stored. As with Alice, Bob, and Dan, the original value can still be recovered.
## [link] Appendix A. Using gsubmit
In this course, you will submit your assignments using `gsubmit`. This section reproduces and extends some of the instructions already made available by the BU Computer Science Department.
### [link] A.1. Register for a CS account
You must obtain a CS account to use the `csa` machines maintained by the CS Dept. You will need to physically visit the undergraduate computing lab located at 730 Commonwealth Avenue, on the third floor in room 302.
You will need an SCP or SFTP client (such as WinSCP for Windows or CyberDuck for OS X) to copy files from your local computer to your `csa` home directory. If you are using Windows, you will also need an SSH client (such as PuTTY).
### [link] A.3. Submitting assignments using gsubmit
A typical workflow can be described as follows.
1. You assemble your assignment solution file(s) on your own computer or device.
local
device
hw1
hw1.py
your `csa2`/`csa3`
home directory
your `gsubmit`
directory for CS 235
2. You log into `csa2` or `csa3` using an SCP or SSH client and create a directory for your submission in your CS account home directory. Note that in the examples below `%>` represents a terminal prompt, which may look different on your system.
%> cd ~
%> mkdir hw1
local
device
hw1
hw1.py
your `csa2`/`csa3`
home directory
hw1
your `gsubmit`
directory for CS 235
3. If you have not already done so (e.g., if you were using an SSH client in the previous step), you log into `csa2` or `csa3` using an SCP client and copy your completed file(s) into that directory.
local
device
hw1
hw1.py
your `csa2`/`csa3`
home directory
hw1
hw1.py
your `gsubmit`
directory for CS 235
4. If you have not already done so, you log into `csa2` or `csa3` using an SSH client and run the `gsubmit` commands to copy the files from your CS account home directory to the `gsubmit` directories to which the course staff has access.
%> cd ~
%> gsubmit cs235 hw1
local
device
hw1
hw1.py
your `csa2`/`csa3`
home directory
hw1
hw1.py
your `gsubmit`
directory for CS 235
hw1
hw1.py
5. To view your submitted files, you can use the following command:
%> gsubmit cs235 -ls
To look at a file that has already been submitted, you can use:
%> gsubmit cs235 -cat hw1/hw1.py
After grades are posted (normally, this will be announced on the mailing list and in lecture), you can check your grade using:
The Python programming language will be among the languages we use in this course. This language supports the object-oriented, imperative, and functional programming paradigms, has automatic memory managememt, and natively supports common high-level data structures such as lists and sets. Python is often used as an interpreted language, but it can also be compiled.
The latest version of Python 3 can be downloaded at: https://www.python.org/downloads/. In this course, we will require the use if Python 3, which has been installed on all the CS Department's undergraduate computing lab machines, as well as on `csa2/csa3`.
### [link] B.2. Assembling a Python module
The simplest Python program is a single file (called a module) with the file extension `.py`. For example, suppose the following is contained within a file called `example.py`:
# This is a comment in example.py.
# Below is a Python statement.
print("Hello, world.")
Assuming Python is installed on your system, to run the above program from the command line you can use the following (you may need to use `python3`, `python3.2`, `python3.3`, etc. depending on the Python installation you're using). Note that in the examples below `%>` represents a terminal prompt, which may look different on your system.
%> python example.py
Hello, world.
If you run Python without an argument on the command line, you will enter Python's interactive prompt. You can then evaluate expressions and execute individual statements using this prompt; you can also load and execute a Python module file:
%> python
Python 3.2 ...
Hello, world.
>>> x = "Hello." # Execute an assignment statement.
>>> print(x) # Execute a print statement.
Hello.
>>> x # Evaluate a string expression.
'Hello.'
>>> 1 + 2 # Evaluate a numerical expression.
3
### [link] B.3. Common data structures (i.e., Python expressions)
Python provides native support for several data structures that we will use throughout this course: integers, strings, lists, tuples, sets, and dictionaries (also known as finite maps). In this subsection, we present how instances of these data structures are represented in Python, as well as the most common operations and functions that can be applied to these data structure instances.
• Booleans consist of two constants: `True` and `False`.
• The usual logical operations are available using the operators `and`, `or`, and `not`.
>>> True # A boolean constant.
True
>>> False # A boolean constant.
False
>>> True and False or True and (not False) # A boolean expression.
True
• Integers are written as in most other programming languages (i.e., as a sequence of digits).
• The usual arithmetic operations are available using the operators `+`, `*`, `-`, and `/`. The infix operator `//` represents integer division, and the infix operators `**` represents exponentiation. Negative integers are prefixed with the negation operator `-`.
• The usual relational operators `==`, `!=`, `<`, `>`, `<=`, `>=` are available.
• The `int()` function can convert a string that looks like an integer into an integer.
>>> 123 # An integer constant.
True
>>> 1 * (2 + 3) // 4 - 5 # An integer expression.
-4
>>> 4 * 5 >= 19 # A boolean expression involving integers.
True
>>> int("123") # A string being converted into an integer
123
• Strings are delimited by either `'` or `"` characters. Strings can be treated as lists of single-character strings. Another way to look at this is that there is no distinction between a character and a string: all characters are just strings of length 1. Multiline strings can be delimited using `"""` or `'''` (i.e., three quotation mark characters at the beginning and end of the string literal).
• The empty string is denoted using `''` or `""`.
• Two strings can be concatenated using `+`.
• The function `len()` returns the length of a string.
• Individual characters in a string can be accessed using the bracketed index notation (e.g., `s[i]`). These characters are also strings themselves.
>>> 'Example.' # A string.
'Example.'
>>> "Example." # Alternate notation for a string.
'Example.'
>>> len("ABCD") # String length.
4
>>> "ABCD" + "EFG" # String concatenation.
'ABCDEFG'
>>> "ABCD"[2] # Third character in the string.
'C'
• Lists are similar to arrays: they are ordered sequences of objects and/or values. The entries of a list can be of a mixture of different types, and lists containing one or more objects are delimited using `[` and `]`, with the individual list entries separated by commas. Lists cannot be members of sets.
• The empty list is denoted using `[]`.
• Two lists can be concatenated using `+`.
• The function `len()` returns the length of a list.
• Individual entries in a list can be accessed using the bracketed index notation (e.g., `a[i]`).
• To check if a value is in a list, use the `in` relational operator.
>>> [1,2,"A","B"] # A list.
[1, 2, 'A''B']
>>> [1, 2] + ['A','B'] # Concatenating lists.
[1, 2, 'A''B']
>>> len([1,2,"A","B"] ) # List length.
4
>>> [1,2,"A","B"][0] # First entry in the list.
1
>>> 1 in [1, 2] # List containment check.
True
• Tuples are similar to lists (they are ordered, and can contain objects of different types), except they are delimited by parentheses `(` and `)`, with entries separated by commas. The main distinction between lists and tuples is that tuples are hashable (i.e., they can be members of sets).
• The empty tuple is denoted using `()`.
• A tuple containing a single object `x` is denoted using `(x, )`.
• Two tuples can be concatenated using `+`.
• A tuple can be turned into a list using the `list()` function.
• A list can be turned into a tuple using the `tuple()` function.
• The function `len()` returns the length of a tuple.
• Individual entries in a tuple can be accessed using the bracketed index notation (e.g., `t[i]`).
• To check if a value is in a tuple, use the `in` relational operator.
>>> (1,2,"A","B") # A tuple.
(1, 2, 'A''B')
>>> (1,) # Another tuple.
(1,)
>>> (1, 2) + ('A','B') # Concatenating tuples.
(1, 2, 'A''B')
>>> list((1, 2, 'A','B')) # A tuple being converted into a list.
[1, 2, 'A''B']
>>> tuple([1, 2, 'A','B']) # A list being converted into a tuple.
(1, 2, 'A''B')
>>> len((1,2,"A","B")) # Tuple length.
4
>>> (1,2,"A","B")[0] # First entry in the tuple.
1
>>> 1 in (1, 2) # Tuple containment check.
True
• Sets are unordered sequences that cannot contain duplicates. They are a close approximation of mathematical sets. Sets cannot be members of sets.
• The empty set is denoted using `set()`.
• The methods `.union()` and `.intersect` correspond to the standard set operations.
• A list or tuple can be turned into a set using the `set()` function.
• A set can be turned into a list or tuple using the `list()` or `list()` function, respectively.
• The function `len()` returns the size of a set.
• To access individual entries in a set, it is necessary to turn the set into a list or tuple.
• To check if a value is in a set, use the `in` relational operator.
>>> {1,2,"A","B"} # A set.
{1, 2, 'A''B'}
>>> ({1,2}.union({3,4})).intersection({4,5}) # Set operations.
{4}
>>> set([1, 2]).union(set(('A','B'))) # Converting a list and a tuple to sets.
{'A', 1, 2, 'B'}
>>> len({1,2,"A","B"}) # Set size.
4
>>> 1 in {1,2,"A","B"} # Tuple containment check.
True
• Frozen sets are like sets, except they can be members of other sets. A set can be turned into a frozen set using the `frozenset()` function.
>>> frozenset({1,2,3}) # A frozen set.
frozenset({1, 2, 3})
>>> {frozenset({1,2}), frozenset({3,4})} # Set of frozen sets.
{frozenset({3, 4}), frozenset({1, 2})}
• Dictionaries are unordered collections of associations between some set of keys and some set of values. Dictionaries are also known as finite maps.
• The empty dictionary is denoted using `{}`.
• The list of keys that the dictionary associates with values can be obtained using `list(d.keys())`.
• The list of values that the dictionary contains can be obtained using `list(d.values())`.
• The function `len()` returns the number of entries in the dictionary.
• Individual entries in a dictionary can be accessed using the bracketed index notation (e.g., `d[key]`).
>>> {"A":1, "B":2} # A dictionary.
{'A': 1, 'B': 2}
>>> list({"A":1, "B":2}.keys()) # Dictionary keys.
['A''B']
>>> list({"A":1, "B":2}.values()) # Dictionary values.
[1, 2]
>>> len({"A":1, "B":2}) # Dictionary size.
2
>>> {"A":1, "B":2}["A"] # Obtain a dictionary value using a key.
1
### [link] B.4. Function, procedure, and method invocations
Python provides a variety of ways to supply parameter arguments when invoking functions, procedures, and methods.
• Function calls and method/procedure invocations consist of the function, procedure, or method name followed by a parenthesized, comma-delimited list of arguments. For example, suppose a function or procedure `example()` is defined as follows:
def example(x, y, z):
print("Invoked.")
return x + y + z
To invoke the above definition, we can use one of the following techniques.
• Passing arguments directly involves listing the comma-delimited arguments directly between parentheses.
>>> example(1,2,3)
Invoked.
6
• The argument unpacking operator (also known as the `*`-operator, the scatter operator, or the splat operator) involves providing a list to the function, preceded by the `*` symbol; the arguments will be drawn from the elements in the list.
>>> args = [1,2,3]
>>> example(*args)
Invoked.
6
• The keyword argument unpacking operator (also known as the `**`-operator) involves providing a dictionary to the function, preceded by the `**` symbol; each named paramter in the function definition will be looked up in the dictionary, and the value associated with that dictionary key will be used as the argument passed to that parameter.
>>> args = {'z':3, 'x':1, 'y':2}
>>> example(**args)
Invoked.
6
• Default parameter values can be specified in any definition. Suppose the following definition is provided.
def example(x = 1, y = 2, z = 3):
return x + y + z
The behavior is then as follows: if an argument corresponding to a parameter is not supplied, the default value found in the definition is used. If an argument is supplied, the supplied argument value is used.
>>> example(0, 0)
3
>>> example(0)
5
>>> example()
6
Python provides concise notations for defining data structures and performing logical computations. In particular, it support a comprehension notation that can be used to build lists, tuples, sets, and dictionaries.
• List comprehensions make it possible to construct a list by iterating over one or more other data structure instances (such as a list, tuple, set, or dictionary) and performing some operation on each element or combination of elements. The resulting list will contain the result of evaluating the body for every combination.
>>> [ x for x in [1,2,3] ]
[1, 2, 3]
>>> [ 2 * x for x in {1,2,3} ]
[2, 4, 6]
>>> [ x + y for x in {1,2,3} for y in (1,2,3) ]
[2, 3, 4, 3, 4, 5, 4, 5, 6]
It is also possible to add conditions anywhere after the first `for` clause. This will filter which combinations are actually used to add a value to the resulting list.
>>> [ x for x in {1,2,3} if x < 3 ]
[1, 2]
>>> [ x + y for x in {1,2,3} for y in (1,2,3) if x > 2 and y > 1 ]
[5, 6]
• Set comprehensions make it possible to construct a set by iterating over one or more other data structure instances (such as a list, tuple, set, or dictionary) and performing some operation on each element or combination of elements. The resulting list will contain the result of evaluating the body for every combination. Notice that the result will contain no duplicates because the result is a set.
>>> { x for x in [1,2,3,1,2,3] }
{1, 2, 3}
• Dictionary comprehensions make it possible to construct a dictionary by iterating over one or more other data structure instances (such as a list, tuple, set, or dictionary) and performing some operation on each element or combination of elements. The resulting dictionary will contain the result of evaluating the body for every combination.
>>> { key : 2 for key in ["A","B","C"] }
{'A': 2, 'C': 2, 'B': 2}
### [link] B.6. Other useful built-in functions
The built-in function `type()` can be used to determine the type of a value. Below, we provide examples of how to check whether a given expression has one of the common Python types:
>>> type(True) == @bool
True
>>> type(123) == int
True
>>> type("ABC") == str
True
>>> type([1,2,3]) == list
True
>>> type(("A",1,{1,2})) == tuple
True
>>> type({1,2,3}) == set
True
>>> type({"A":1, "B":2}) == dict
True
### [link] B.7. Common Python definition and control constructs (i.e., Python statements)
A Python program is a sequence of Python statements. Each statement is either a function definition, a variable assignment, a conditional statement (i.e., `if`, `else`, and/or `elif`), an iteration construct (i.e., a `for` or `while` loop), a `return` statement, or a `break` or `continue` statement.
• Variable assignments make it possible to assign a value or object to a variable.
x = 10
It is also possible to assign a tuple (or any computation that produces a tuple) to another tuple:
(x, y) = (1, 2)
• Function and procedure definitions consist of the `def` keyword, followed by the name of the function or procedure, and then by one or more arguments (delimited by parentheses and separated by commas).
def example(a, b, c):
return a + b + c
• Conditional statements consist of one or more branches, each with its own boolean expression as the condition (with the exception of `else`). The body of each branch is an indented sequence of statements.
def fibonacci(n):
# Computes the nth Fibonacci number.
if n <= 0:
return 0
elif n <= 2:
return 1
else:
return fibonacci(n-1) + fibonacci(n-2)
• Iteration constructs make it possible to repeat a sequence of statements over and over. The body of an iteration construct is an indented sequence of statements.
• The while construct has a boolean expression as its condition (much like `if`). The body is executed over and over until the expression in the condition evaluates to `False`, or a `break` statement is encountered.
def example1(n):
# Takes an integer n and returns the sum of
# the integers from 1 to n-1.
i = 0
sum = 0
while i < n:
sum = sum + i
i = i + 1
return sum
def example2(n):
# Takes an integer n and returns the sum of
# the integers from 1 to n-1.
i = 0
sum = 0
while True:
sum = sum + i
i = i + 1
if i == n:
break
return sum
• The for construct makes it possible to repeat a sequence of statements once for every object in a list, tuple, or set, or once for every key in a dictionary.
def example3(n):
# Takes an integer n and returns the sum of
# the integers from 1 to n-1.
sum = 0
for i in range(0,n):
sum = sum + i
return sum
def example4(d):
# Takes a dictionary d that maps keys to
# integers and returns the sum of the integers.
sum = 0
for key in d:
sum = sum + d[key]
return sum | 87,263 | 282,925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-13 | latest | en | 0.943745 |
https://www.physicsforums.com/threads/spring-problem-help.191545/ | 1,513,335,308,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948568283.66/warc/CC-MAIN-20171215095015-20171215115015-00787.warc.gz | 792,846,952 | 17,882 | # Spring Problem, help!
1. Oct 15, 2007
### bulldog23
1. The problem statement, all variables and given/known data
A moving 1.3 kg block collides with a horizontal spring whose spring constant is 491 N/m.
A) The block compresses the spring a maximum distance of 5.0 cm from its rest postion. The coefficient of kinetic friction between the block and the horizontal surface is 0.49. What is the work done by the spring in bringing the block to rest? REMEMBER: Work has a sign.
B) How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring?
C) What is the speed of the block when it hits the spring?
2. Relevant equations
W=F_0*d
3. The attempt at a solution
I am unsure how to approach this problem. If someone could please walk me through it, that would help a lot! My teacher gave us this problem, but we haven't even learned this stuff yet. So please help me out!
Last edited: Oct 15, 2007
2. Oct 16, 2007
### Hootenanny
Staff Emeritus
Let's take question (A) first. So, when the block collides with the srping it has some kinetic energy, when the block stops it has no kinetic energy. Some of this kinetic energy will be used to do work against friction, what do you think the rest of this kinetic energy will be used for?
3. Oct 16, 2007
### bulldog23
Is the rest of the kinetic energy used to hold the spring in place?
4. Oct 16, 2007
### Hootenanny
Staff Emeritus
Not hold in place, but to compress the spring yes. So, do you know the expression for the potential energy stored in a compressed spring?
5. Oct 16, 2007
### bulldog23
Is it F=kx?
6. Oct 16, 2007
### Hootenanny
Staff Emeritus
Close, thats the force required to compress a spring by xm; how about the energy stored?
7. Oct 16, 2007
### bulldog23
PE=1/2kx^2?
8. Oct 16, 2007
### Hootenanny
Staff Emeritus
Sounds good to me. So, how much potential energy is stored after the block collides with the spring?
9. Oct 16, 2007
### bulldog23
So then do I plug in the 491 N/m for k and 5 m for x? If I do that I get 6137.5 J.
10. Oct 16, 2007
### Hootenanny
Staff Emeritus
The value is correct, but what about the sign?
11. Oct 16, 2007
### bulldog23
It should be negative because it is opposing the force of the block, right?
12. Oct 16, 2007
### bulldog23
So is that all you have to do for part A?
13. Oct 16, 2007
### Hootenanny
Staff Emeritus
Yes, since the force extered by the spring is in the opposite direction to the direction in which the block is moving, the work is negative.
Yup. Now for part (B) you do exactly the same, but this time for friction.
14. Oct 16, 2007
### bulldog23
It says that the answer for part A is wrong though. Doesn't the coefficient of the kinetic friction play into the problem somewhere?
15. Oct 16, 2007
### Hootenanny
Staff Emeritus
Is this a Webassign problem? Try rounding your answer to 3sf. And no, the spring is still compressed the same amount regardless of friction, therefore, the work done by the spring will be the same.
16. Oct 16, 2007
### bulldog23
I tried it again and it said that it was wrong. We must have gone wrong somewhere...
17. Oct 16, 2007
### Hootenanny
Staff Emeritus
Wait, I see what's happened, we've used x = 5m, when actually x = 5cm = 0.05m
I can't believe I missed that.
18. Oct 16, 2007
### bulldog23
So then the answer should be -.61375 J ? So then how do I do the same thing with friction for part B?
Last edited: Oct 16, 2007
19. Oct 16, 2007
### Hootenanny
Staff Emeritus
Yes, but be careful, web assign is notoriously pedantic about accuracy and rounding, I would say use 3sf or the accuracy it tells you to use in the question.
20. Oct 16, 2007
### bulldog23
Alright, it accepted the answer. I am lost when it comes to Part B | 1,078 | 3,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-51 | longest | en | 0.914081 |
http://mathhelpforum.com/number-theory/18415-divisor-print.html | 1,495,766,156,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608622.82/warc/CC-MAIN-20170526013116-20170526033116-00382.warc.gz | 227,635,605 | 2,963 | # divisor
• Sep 3rd 2007, 04:03 AM
chibuike1
divisor
For every positive even integer n, the function H(n) is defined to be the product of all even integers from 2 to n. What is the lowest possible prime factor of H(100) =1.
• Sep 3rd 2007, 04:23 AM
topsquark
Quote:
Originally Posted by chibuike1
For every positive even integer n, the function H(n) is defined to be the product of all even integers from 2 to n. What is the lowest possible prime factor of H(100) =1.
Do you mean what is the lowest possible prime factor of H(100) - 1?
-Dan
• Sep 3rd 2007, 05:40 AM
topsquark
Quote:
Originally Posted by chibuike1
For every positive even integer n, the function H(n) is defined to be the product of all even integers from 2 to n. What is the lowest possible prime factor of H(100) =1.
$H(100) = 2 \cdot 4 \cdot 6 \cdot ~ ... ~ \cdot 98 \cdot 100$
$= 2^{50} \cdot (1 \cdot 2 \cdot 3 \cdot ~ ... ~ \cdot 49 \cdot 50 )$
So H(100) - 1 cannot be divisible by any prime less than 53, the first prime which does not appear on this list. I'm not a number theorist so there may be a way to prove that this is not divisible by a prime higher than this.
-Dan
• Sep 3rd 2007, 07:19 AM
chibuike1
Math solution reqd
Sorry the right question is:
For every positive even integer n, the function H(n) is defined to be the product of all even integers from 2 to n. What is the lowest possible prime factor of H(100) + 1.
• Sep 3rd 2007, 09:16 PM
topsquark
Quote:
Originally Posted by chibuike1
Sorry the right question is:
For every positive even integer n, the function H(n) is defined to be the product of all even integers from 2 to n. What is the lowest possible prime factor of H(100) + 1.
From my knowledge base, the answer would be the same as the one I gave for H(100) - 1.
-Dan | 546 | 1,782 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-22 | longest | en | 0.876567 |
http://assignmentgrade.com/?page=2 | 1,585,766,763,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505826.39/warc/CC-MAIN-20200401161832-20200401191832-00320.warc.gz | 17,420,411 | 11,216 | ### Pls help me i need help fast
So if we add all the shirts together we get 48. So starting with the 1st one. 8/48 if the chance Jack selects a blue shirt. If we simplify 8/48 we get 1/6. So the answer is The Probability of Jack selecting a blue shirt
ANSWERED AT 01/04/2020 - 01:46 PM
QUESTION POSTED AT 01/04/2020 - 01:46 PM
### A 7.5 kg block is placed on a table. If its bottom surface area is 0.6 m2, how much pressure does the block exert on the tabletop? A. 73.5 Pa B. 122.5 Pa C. 367.5 Pa D. 226.5 Pa
A 7.5 kg block is placed on a table. If its bottom surface area is 0.6 m2, how much pressure does the block exert on the tabletop?
A. 367.5 Pa
B. 122.5 Pa
C. 73.5 Pa
D. 226.5 Pa
ANSWERED AT 01/04/2020 - 01:46 PM
QUESTION POSTED AT 01/04/2020 - 01:46 PM
### What is the sine ratio of angle E? A - 5/3 B - 3/5 C - 5/4 D - 4/5
The correct option is B.
Step-by-step explanation:
In a right angled triangle
In the given triangle DEF, the opposite sides of angle E is DF=6 units and hypotenuse EF=10 units.
The sine ratio of angle E is
The sine ratio of angle E is 3/5, therefore the correct option is B.
ANSWERED AT 01/04/2020 - 01:46 PM
QUESTION POSTED AT 01/04/2020 - 01:46 PM
### Communist Soviets claimed that capitalism
Created an unfair divide between rich and poor
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
### Explain how the election of 1860 was the final blow that initiated the start of the Civil War.
Following Lincoln's election, and before he even took the oath of office, seven states immediately seceeded from the Union. Those actions forced the president's hand. He couldn't restore the Union without a war. the majority of people did not vote for Lincoln, and the south as a whole detested him because of his policies
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
### How does magma form at divergent tectonic plate boundaries?
I got this from NationalGeographic.org
"Decompression melting involves the upward movement of Earth's mosty-solid mantle..."The rifting movement causes the buoyant magma below to rise and fill the space of the lower pressure. The rock then cools into new crust."
I hope this helps you!!!:)
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
### Kevin and some college friends are hiking on a trail that is 70 feet above sea level they hike into a canyon that is 14 feet below sea level. how many feet have Kevin and his friends descended
Im not sure of the answer but try quizlet
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
Mes Arc ZWX (clockwise) = mes Arc ZW + mes Arc WX
mes Arc ZW =180° & mes Arc WX=45° (given)
Then mes Arc ZWX (clockwise) = 180°+45° = 225°
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
### Please someone come thru for me idk how to get these answers
I can barley see can you take another one please
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
### How will our Sun die?
Our Sun is in the main sequence star stage.
From there our Sun will evolve to be a red giant.
When it evolves many many years from now it will engulf all of the inner planets. But don't worry, that won't happen in our lifetime!
From there the red giant (sun) will die. The death of a red giant is known as a planetary nebula. Though at the center of the nebula there is still a burning core.
When the debris flies away from the planetary nebula, our Sun will turn into a white dwarf, then the core will burn out which gives us a black dwarf.
I've provided an image of a good life cycle of the star, if you follow along with the brief description I've given you it should make sense!
If you have any questions let me know! I hope this helps!
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
### Bruce buys a trampoline prices at $54. if the sales tax is 3 2/5% how much tax will bruce pay$1.836 round off to $1.84 tax on$54 product with tax of 3.4%
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
### Fill all of this out please
The answer of 2 is a
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
### Sarah roller skates 18.5km in 3 1/2 hours. what is her average speed in kilo per hour?
5.2857 km/hrs... formula: speed = distance/time
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
### What's the diameter of a circle with a radius of 12
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
### If you combine 370.0 mL of water at 25.00 °C and 130.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.
Even at STP, water can exist under 0 deg C. it quite is stated as sub-cooling. there is an significant concern in aviation stated as rime icing. Rime is a white or milky opaque granular deposit of ice. It occurs whilst supercooled water droplets strike an merchandise at temperatures at or under freezing.
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
### Using examples, how can you look at a substance and determine if it is a pure substance or a mixture?
A pure substance is something that is not mixed with anything like like a jar of rice or a jar of marbles
A mixture is something that is mixed together like if you made lemonade and you mixed lemon juice and water and sugar together it's a mixture and another example is if you mix sand and rocks together it's a mixture I hope I helped you
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
### What is the area of the figure below?
192 ft.
We can start by finding the area of the whole shape. 18*12= 216
Then we can see that the shape cut out has a length of 6 and the missing 8 from the 10. so 6*8= 48 and we cut 48 in half since the small shape was cut in half and get 24. So 216-24= 192ft
If you don't get it tell me, it is hard to explain.
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
### What kind of bond is not formed by the differences in electronegativity
It's nonpolar covalent
ANSWERED AT 01/04/2020 - 01:45 PM
QUESTION POSTED AT 01/04/2020 - 01:45 PM
### A train travels 92 miles in 11 1/2 hours. what is its average speed in miles per hour?
The answer is 8 miles per hour
ANSWERED AT 01/04/2020 - 01:44 PM
QUESTION POSTED AT 01/04/2020 - 01:44 PM
### I need help asap!! I don't really get it bc I don't know if you include the \$12 or not.
12+3x=30
Subtract 12 from both sides
3x=18
Divide by 3 both side
X=6
ANSWERED AT 01/04/2020 - 01:44 PM
QUESTION POSTED AT 01/04/2020 - 01:44 PM
### Read the statement: Both Jerry in "President Cleveland, Where Are You?" and Squeaky in "Raymond's Run" reach a point in where they realize that __________. Which best completes the statement and identifies a recurring theme? a.giving up collecting cards is something that everyone must do eventually b.the most important thing in life is achieving one's individual goals c.selling something of great value to one's rivals is sometimes necessary d.helping others is more important than helping oneself
The correct answer is D) helping others is more important than helping oneself.
Both, Jerry in “President Cleveland, Where Are You?” and Squeaky in “Raymond’s Run” reach a point in where they realize that helping others is more important than helping oneself.
In “President Cleveland, Where Are You? Jerry is an eleven-year-old boy and the narrator of the story, which is the actions Jerry has to do in order to help his family. On the other hand, Squeaky is the nickname of Hazel Elizabeth Deborah Parker, the narrator in the story of “Raymond’s Run”, a great and fast runner. Both characters, Jerry in “President Cleveland, Where Are You?” and Squeaky in “Raymond’s Run” reach a point in where they realize that helping others is more important than helping oneself.
ANSWERED AT 01/04/2020 - 01:44 PM
QUESTION POSTED AT 01/04/2020 - 01:44 PM
### What is an assessment mean?
The evaluation or estimation of the nature, quality, or ability of someone or something.
ANSWERED AT 01/04/2020 - 01:44 PM
QUESTION POSTED AT 01/04/2020 - 01:44 PM
### Compare shield volcanoes and composite volcanoes.
When a shield volcano eruptions they are less explosive than composite volcanoes. Composite volcanoes are steep and shield volcanoes are more sloping. Shield volcanoes also have low viscosity and fast flowing magma and the composite volcano's magma is low viscosity and is slow flowing. Shield volcanoes magma is also hot and composite volcanos magma is cooler.
Hope This Helps!
ANSWERED AT 01/04/2020 - 01:44 PM
QUESTION POSTED AT 01/04/2020 - 01:44 PM
### What is the distance between a point on one wave and the same point on the next cycle of the wave called
It is called the wavelength
ANSWERED AT 01/04/2020 - 01:44 PM
QUESTION POSTED AT 01/04/2020 - 01:44 PM
### To improve the typical American diet of the average adult, one should do what?
decrease fat and salt
Explanation:
Adult Americans often consume many processed, precooked, and fast foods. This type of diet is often very harmful because of the high fat and salt content. To improve the average American diet for the average adult, salt and sugar should be lowered in the foods they eat.
Used sparingly in culinary preparations based on fresh or minimally processed foods, these ingredients contribute to diversifying and tasting the food without being nutritionally unbalanced. However, salt and fats contain high levels of some nutrients that can be harmful to health, such as sodium (which is the basis of table salt) and saturated fats (found in fats in cheeses, butter and some vegetable oils. used for the female firms). Fats are six times more calories per gram than rice, beans and other grains, and twenty times more than vegetables. High salt content is considered to be a major factor in the development of high blood pressure, which affects 90% of Americans over a lifetime and is related to heart disease and stroke,
ANSWERED AT 01/04/2020 - 01:44 PM
QUESTION POSTED AT 01/04/2020 - 01:44 PM
Brown belly,floundering legs,had trouble getting up
i think he turned into a spider.
ANSWERED AT 01/04/2020 - 01:44 PM
QUESTION POSTED AT 01/04/2020 - 01:44 PM
### Work the following calculations: A) How many moles are in 23 grams of oxygen (Z=8)? B) What is the mass of 2.5 moles of nitrogen (Z=7)?
THE MASS OF 2.5 IS THE AMOUNT OF THE LINER EQUITATION SO DIVIDE 2.5 INTO 23 GRAMS AND MULTIPLY YOUR PROFIT BRO BTW IM NOT A COP SO STAY COOL MY FRIEND
ANSWERED AT 01/04/2020 - 01:44 PM
QUESTION POSTED AT 01/04/2020 - 01:44 PM
### A ball has a radius of 5 inches. What is the volume it can hold
A ball is a Sphere sooooo......
And when you enter the radius.....
ANSWERED AT 01/04/2020 - 01:44 PM
QUESTION POSTED AT 01/04/2020 - 01:44 PM
### Monique says that three and one fourth g/cm3 is an Outlier. Is she right or wrong
She is right. Monique is right
ANSWERED AT 01/04/2020 - 01:44 PM
QUESTION POSTED AT 01/04/2020 - 01:44 PM
### A change in which factor will affect the rate of reaction only when gases are involved?
I believe it is the pressure
ANSWERED AT 01/04/2020 - 01:44 PM
QUESTION POSTED AT 01/04/2020 - 01:44 PM | 3,313 | 11,386 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-16 | latest | en | 0.876927 |
https://www.simplylogical.studio/2022/08/25-horses-puzzle-find-fastest-in.html | 1,721,367,227,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00878.warc.gz | 840,319,869 | 36,202 | # 25 Horses Puzzle - Find The Fastest In Minimum race || Google Interview Puzzles
Puzzle Details
There are 25 horses among which you need to find out the fastest 3 horses. You can conduct a race among at most 5 to find out their relative speed. At no point, you can find out the actual speed of the horse in a race. Find out how many races are required to get the top 3 horses.
Solution
Let the horses be numbered as H1, H2, H3, H4, H5, ..., H24, H25. Now First arrange 5 races among horses as :
Race 1: H1, H2, H3, H4, H5
(let's say that H1 comes first in the race then H2 then H3 then H4 and then H5 and similarly happens for the next races ...)
Race 2: H6, H7, H8, H9, H10
Race 3: H11, H12, H13, H14, H15
Race 4: H16, H17, H18, H19, H20
Race 5: H21, H22, H23, H24, H25
Now arrange the race between winners of all the above races as :
Race 6: H1, H6, H11, H16, H21
So, Now we have H1, H6, and H11 as the top 3 winners of Race 6. So, there are some points that can be noticed through race 6:
So, Now we have H1, H6, and H11 as the top 3 winners of Race 6. So, there are some points that can be noticed through race 6:
1) As H16 comes fourth in the race, so we can be sure that horses (H17, H18, H19, H20) cannot acquire top 3 positions rather they can be at any position from the fifth one to the last position. So, In our final race, we do not need to include H16, H17, H18, H19, and H20 as we are interested in only the top 3 fastest horses.
2) As H21 comes fifth in the race, we can be sure that horses (H22, H23, H24, H25) cannot acquire top 3 positions rather they can be at any position between the sixth one to the last position. So, In our final race, we do not need to include H21, H22, H23, H24, and H25 as we are interested in only the top 3 fastest horses.
3) As H11 comes third in the race, we can be sure that horses (H12, H13, H14, H15) cannot acquire top 3 positions rather they can be at any position from the fourth one to the last position. So, In our final race, only H11 will participate, and the rest of all (H12, H13, H14, and H15) need not participate as we are interested in only the top 3 fastest horses.
4) As H6 comes second in the race, we can be sure that only H7 could have the possibility of third fastest horse, and horses (H8, H9, H10) cannot acquire top 3 positions rather they can be in the any position between the fourth one to the last position. So, In our final race, only H6 and H7 will participate, and the rest of all (H8, H9, and H10) need not participate as we are interested in only the top 3 fastest horses.
5) As H1 wins race 1 and also wins the race with the winner of all rest of the races, so, till now we have found that H1 is the fastest horse among all. So, we do not need to race H1 again with any other horse.
6) As H1 comes first in the race, so we can be sure that only H2 and H3 could have the possibility of second and third fastest horse respectively and horses (H4, H5)cannot acquire top 3 positions rather they can be at any position between the fourth one to the last position. So, In our final race, only H4 and H5 will participate, and the rest of all (H8, H9, and H10) need not participate as we are interested in only the top 3 fastest horses. So, for our final race, we have H2, H3, H6, H7, and H11 as the competitors for second and third positions. So, we have,
Race 7: H2, H3, H6, H7, H11 Let's say H2 and H3 come at first and second position respectively. So, finally, we have the top 3 fastest horses in just 7 races among 25 horses as: H1 > H2 > H3 | 1,050 | 3,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-30 | latest | en | 0.946302 |
https://www.homeandlearn.co.uk/excel2007/excel2007s4p1.html | 1,620,804,643,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991685.16/warc/CC-MAIN-20210512070028-20210512100028-00416.warc.gz | 843,200,789 | 4,124 | # The Excel SUM Function
In earlier parts of this course, you used addition formula quite a lot. You saw that the basic way to add things up was by doing this:
=A1 + B1 + C1
You've also used the in-built SUM function:
=SUM(A1:C1)
Whichever of these two you used, the answer was the same - Excel will add up whatever numbers you have in the cells A1, B1, and C1. The two methods above are adding up consecutive cells. But what if you want to add up the following, non-consecutive cells: A1, B1, C1, D9?
Well, you can combine the two methods. So you can do it like this:
= Sum(A1:C1) + D9
or you can do it like this:
= Sum(A1:C1, D9)
For the first method, just type a plus sign after your SUM function, followed by the cell you want to include:
= Sum(A1:C1) + D9
You can include as many other cells as you like:
= Sum(A1:C1) + D9 + E12 + G25
You can even use another SUM function:
= Sum(A1:C1) + SUM(G1:H1)
The second method to add up non-consecutive cells starts in the same way: use a SUM function, and separate your consecutive cells with a colon:
= Sum(A1:C1)
To include the non consecutive cells, type a comma, followed by the cell you want to include:
= Sum(A1:C1, D9)
You can include other cells, as well:
= Sum(A1:C1, D9, E12, G25)
The thing to note is that all the cells are between the round brackets of the SUM function. Excel knows that SUM means to add up, so it sees each cell reference separated by commas, and then includes them in the addition.
To give you some practice, try this exercise.
Exercise
Create a simple spreadsheet with the number 3 in cells A1, B1, C1 and D1. Enter another number 3 in cell A2. Use one of the non consecutive addition formulas above to add up the values in all five cells. Your spreadsheet will then look like this, once you have the correct formula:
In the picture above, cell A4 displays the correct answer.
### Selecting Non Consecutive Cells
Another way to select non-consecutive cells for your SUM functions is by holding down the CTRL key on your keyboard, and then left click in the cell you want to add. Try this:
• Click inside a different cell in your spreadsheet (B4, for example). Then click inside the formula bar at the top
• Now type the following into the formula bar (Don't forget to add the colon at the end):
=SUM(A1:
• The cell A1 will be highlighted on the spreadsheet. It will have sizing handles, so that you can stretch the selection
• Hold your left mouse button over the bottom right blue square, and drag to cell D1. You spreadsheet should look like this:
Excel will add the cells to your formula. But it will also add a colon after D2. We don't want this, because a colon means "add up a range of cells". So delete the colon and type a comma instead.
Now that you have the cells A1 to D1 selected, hold down the left CTRL key on your keyboard. Keep it held down, and click inside cell A2 with your left mouse button:
The cell A2 is highlighted, in the image above. Excel will add this to your formula.. To finish off, add the right bracket ). Then press the enter key on your keyboard.
Using this method, you can add as many individual cells as you want for your formula.
Exercise
On a new sheet, enter the number 3 in the following cells: A1, B1, C1, D1, E1. Then type a 3 in the cells A3, C3 and E3. Using non-consecutive addition, display your answer in cell A5. The finished spreadsheet will then look like ours below:
The answer to the addition, 24, is displayed in A5. Only one Sum function was used here, with the other cells separated by commas.
Adding up shouldn't cause you too many problems. The tricky part is selecting all the cells that you want to include. In the next part, we'll at multiplication.
<--Back to the Excel Contents Page | 938 | 3,762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-21 | longest | en | 0.935688 |
https://www.physicsforums.com/threads/derivative-of-sinc-z-in-the-complex-plane.709560/ | 1,701,693,689,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100529.8/warc/CC-MAIN-20231204115419-20231204145419-00378.warc.gz | 1,050,276,575 | 18,910 | # Derivative of sinc(z) in the complex plane
• nateHI
In summary, the question asks to find the derivative and the maximal region in which the complex function f(z)=sin(z)/z is analytic. The derivative is found to be (-z^-2)sin(z)+cos(z)/z, and the maximal region of convergence is the entire complex plane, since the function has no singularities and is holomorphic on \mathbb{C}. The Cauchy-Riemann equations are used to determine the analyticity of the function, and while they can be challenging to work with, they are important in understanding the strong condition for analycity and the power of the Laurent series representation of complex functions.
## Homework Statement
$z=x+iy; f(z)=sin(z)/z$
find f'(z) and the maximal region in which f(z) is analytic.
## Homework Equations
The sinc function is analytic everywhere.
## The Attempt at a Solution
Writing f(z) as $(z^{-1})sin(z)$ and differentiating with respect to z using the chain rule I get...
$(-z^{-2})sin(z)+cos(z)/z$
However, this seems to simple since the context of the chapter of the book this problem comes from is cauchy-riemann. I would suspect I need to put f(z) in the form Re{f(z)}=u(x,y) and Im{f(z)}=v(x,y). Then $df/dz$ would be $du/dx+idv/dx$. If that is the case then I'm in trouble because the I can't separate the imaginary part from the real part of $sinc(z)$.
You can always remove the complex value from the denominator:
1. Assume an expression like F(z)/G(z)
2. Multiply it by unity of the form G*(z)/G*(x) where * is the complex conjugation operator.
3. Your expression now looks like F(z)G*(z)/[G(z)G*(z)] = F(z)G*(z)/|G(z)|^2.
OK thanks. Now I'm getting somewhere.
So now I have the following:
$sin(z)/z=\frac {1}{|z|^2}[z^*sin(z)]$
$=\frac{1}{(|z|^2)}[(x-iy)\frac{(e^z-e^{-z})}{2i}]$
$=\frac{-i}{(2*|z|^2)}[(x-iy)(e^z-e^{-z})]$
$=\frac{-i}{(2*|z|^2)}[xe^{(x+iy)}-xe^{-(x+iy)}-iye^{(x+iy)}+iye^{-(x+iy)}]$
I'm not really sure what to do now. I can't group the e^(iy)'s. I'm guessing I'm missing some sort of trick or something.
Thanks!
Now go back to the definitions of sin(z); definitions are your friend. :-)
$=\frac{-i}{(2*|z|^2)}[xe^{(x+iy)}-xe^{-(x+iy)}-iye^{(x+iy)}+iye^{-(x+iy)}]$
$=\frac{1}{|z|^2}[x*sin(z)-y*cos(z)]$
I'm not sure how this helps me though because I still have imaginary parts in both terms because of the z. Or, is this when I use cos(z)=Re{e^{iz}} ? That seems like a stretch but it's the only connection I'm making atm.
In your reconstruction the final line is incorrect. That cos(z) is wrong.
In any case you want to find the derivative ... if your goal is to find the real and imaginary parts you can find them either before or after the derivative. But you must carry out the Cauchy-Riemann tests as well ...
We've been on a side issue: how to remove the complex values from the denominator. You probably don't want to do that at the start -
OK, if we ignore my attempt at a solution, how would you suggest I attack the original question?
Just take the Laurent series expansion of $\sin z/z$ and show that its convergence radius is infinity and that it has no singularities anywhere.
Since you are saying I only need to find the maximal region of convergence and show that there are no singularities in order to solve; can I assume you think my original solution to finding the derivative was correct? Specifically
(−z^-2)sin(z)+cos(z)/z
nateHI said:
. If that is the case then I'm in trouble because the I can't separate the imaginary part from the real part of $sinc(z)$.
If you muscle through it you can. You mean you can't separate the real and imaginary part of the expression:
$$\frac{e^{i(x+iy)}-e^{-i(x+iy)}}{2i(x+iy)}$$
Bet you can. You know, convert the exponents to sines and consines, multiply top and bottom by conjugate of the denominator, bingo bango.
I don't know, what you are allowed to assume. For me the most simple solution is to look for the Laurent expansion around 0 and showing that the series is convergent on the entire complex plane, showing that the continuation of the function to the entire complex plane
$$f(z)=\begin{cases} \frac{\sin z}{z} & \text{for} \quad z \neq 0,\\ 1 & \text{for} \quad z=0 \end{cases}$$
in fact is an entire function, i.e., holomorphic on $\mathbb{C}$.
Your derivative was correct for $z \neq 0$.
Since my derivative was correct, I'm wondering if I can always simply pretend my complex function is a real function and differentiate as usual then go back and consider any singularities, branch cuts, etc...
$sin(z)/z=\frac {1}{|z|^2}[z^*sin(z)]$
$=\frac{1}{(|z|^2)}[(x-iy)\frac{(e^{iz}-e^{-iz})}{2i}]$
$=\frac{-i}{(2*|z|^2)}[(x-iy)(e^{iz}-e^{-iz})]$
$=\frac{-i}{(2*|z|^2)}[xe^{i(x+iy)}-xe^{-i(x+iy)}-iye^{i(x+iy)}+iye^{-i(x+iy)}]$
$=\frac{-i}{(2*|z|^2)}[xe^{ix-y}-xe^{-ix+y}-iye^{ix-y}+iye^{-ix+y}]$
$=\frac{-i}{(2*|z|^2)}[xe^{-y}e^{ix}-xe^ye^{-ix}-iye^{-y}e^{ix}+iye^ye^{-ix}]$
$=\frac{-i}{(2*|z|^2)}[xe^{-y}(cosx+isinx)-xe^y(cosx-isinx)-iye^{-y}(cosx+isinx)+iye^y(cosx-isinx)]$
$=\frac{1}{(2*|z|^2)}[-ixe^{-y}(cosx+isinx)+ixe^y(cosx-isinx)-ye^{-y}(cosx+isinx)+ye^y(cosx-isinx)]$
$=\frac{1}{(2*|z|^2)}[ -ixe^{-y}cosx +xe^{-y}sinx +ixe^ycosx +xe^{y}sinx +ye^{-y}cosx -iye^{-y}sinx +ye^ycosx -iye^ysinx]$
$=\frac{1}{(2*|z|^2)}[ +xe^{-y}sinx +ye^{-y}cosx +xe^{y}sinx +ye^{y}cosx -ixe^{-y}cosx -iye^{-y}sinx +ixe^ycosx -iye^ysinx ]$
$u(x,y)=e^{-y}(xsinx+ycosx)+e^y(xsinx+ycosx)$
$v(x,y)=-e^{-y}(xcosx+ysinx)+e^y(xcosx-ysinx)$
I was missing an 'i' in the exponent at the very beggining. I'm almost there I think.
Last edited:
What is this good for?
vanhees71 said:
What is this good for?
In theory, putting f(z) in the form u(x,y)+iv(x,y) allows you to use cauchy-riemann to determine if f(z) is analytic. However, in practice, attempting this has driven me to drinking...so it is good for something!
nateHI said:
In theory, putting f(z) in the form u(x,y)+iv(x,y) allows you to use cauchy-riemann to determine if f(z) is analytic. However, in practice, attempting this has driven me to drinking...so it is good for something!
That's the best response possible.
while this type of question is rather pedantic it is important for the student to become familiar with the Cauchy-Riemann equations in order to appreciate that they are a very strong condition for analycity (which suggests that only a small class of functions are analytic) and these topics are often taught prior to being introduced to Laurent series or power series representations of complex functions.
Furthermore, by asking students to solve problems in this manner before they learn the Laurent series the student develops an appreciation for the power of the Laurent series and the fact that they are infinitely differentiable.
EDIT: Furthermore, the Cauchy-Riemann equations prove useful in proving several important theorems and corollaries that will follow. | 2,090 | 6,849 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2023-50 | latest | en | 0.893403 |
dentisttlv.com | 1,597,507,389,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740929.65/warc/CC-MAIN-20200815154632-20200815184632-00242.warc.gz | 29,606,387 | 6,701 | מרפאת שיניים
ברזאני 4, תל-אביב
03-6414195
What variety mean in math? What exactly is a function in math? What exactly is a function in math?
For this one, keep in mind that the function called the square root is definitely an object which has a name, and when it really is being expressed, you may place the name inside the square brackets. It’s named a function because of its formula. If you’d like to express the square root with regards to a product, the formula is then X=R3.
What do you believe will be the equation for the square root? R3. In this case, the square root is called the derivative. So the square root may essay help be described by two formulas.
The very first formula can be represented by the quantity (R3). Then the derivative is represented by the word the. In case you replace the word the using the word differentiation, you will get the equation for the square root. This would include things like a solution, a worth plus a time. This would be a time which is different from the original.
Here is what exactly is the equation for the square root? Now it truly is an ordinary item among the product and also the value.
How would you describe what is the square root with regards to a time? Effectively, it’s the derivative of your time with respect to the value, and it’s the product from the two values.
So the time would be the derivative https://new.trinity.edu/ on the value, which can be normally equal to the item on the time plus the value. As a result, for those who add these items up, you get the value. When you take the derivative of the value, you get the time. For that reason, you’ve got to take the time into account if you are obtaining the derivative. So the square root is an expression for time. But what’s a time?
Time is a price at which we can repeat an action or possibly a method. When the action or approach is continuous, then the time will generally be the identical.
You can compare the time to a point on the course of action. A worth may be thought of as a point in a procedure.
In this case, the factor which has an impact around the time could be the worth. Once you measure time, you will be measuring how the process is going. So to acquire the worth, you have got to add the time for you to the value. | 528 | 2,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2020-34 | latest | en | 0.937429 |
https://www.physicsforums.com/threads/destructive-interference.273922/ | 1,579,769,591,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250609478.50/warc/CC-MAIN-20200123071220-20200123100220-00147.warc.gz | 1,032,452,735 | 16,445 | # Destructive interference
## Homework Statement
Two identical loudspeakers are located at points A and B, 2.00 m apart. The loudspeakers are driven by the same amplifier and produce sound waves with a frequency of 784 Hz. Take the speed of sound in air to be 344 m/s . A small microphone is moved out from point B along a line perpendicular to the line connecting A and B
http://img179.imageshack.us/img179/1504/yf1644vy5.jpg [Broken]
a)At what distances from B will there be destructive interference
b)At what distances from B will there be constructive interference
c)If the frequency is made low enough, there will be no positions along the line BC at which destructive interference occurs. How low must the frequency be for this to be the case?
λ=v/f
## The Attempt at a Solution
λ=v/f=344/784=0.4388 m
okay so first I labelled the distance from A to C as dA and from B-C as dB.
http://e.imagehost.org/0070/Capture.jpg [Broken]
and to see when its destructive or constructive you would input odd or even numbers respectivley. Now I still have a wrong answer with this method and I suspect its the way I derived dB. Can anyone help me figure out whats wrong. As always any help is appreciated.
Last edited by a moderator:
Related Introductory Physics Homework Help News on Phys.org
Doc Al
Mentor
Now I still have a wrong answer with this method and I suspect its the way I derived dB.
okay for destructive interference I use odd numbers for 3 and this needs to be done 5 times with values of 2 sig figs in ascending order.
for n=1 db=9m n=3 db=2.7m n=5 db=1.3 n=7 db=0.53m n=9 db=0.028 then entering in ascending order answer is
0.028,0.53,1.3,2.7,9.0
Doc Al
Mentor
I just checked the first couple (n = 1, 3) and your answers look good to me.
Okay I got the right answer thanks for the help. | 477 | 1,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-05 | longest | en | 0.91774 |
https://www.examveda.com/the-ghaziabad-hapur-meerut-emu-and-the-meerut-hapur-ghaziabad-emu-start-at-the-same-time-from-ghaziabad-and-meerut-and-proceed-towards-each-other-at-87821/ | 1,721,075,858,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00274.warc.gz | 664,767,673 | 11,463 | Examveda
# The Ghaziabad - Hapur - Meerut EMU and the Meerut - Hapur - Ghaziabad EMU start at the same time from Ghaziabad and Meerut and proceed towards each other at 16 km/hr and 21 km/hr respectively. When they meet, it is found that one train has traveled 60 km more than the other . The distance between two stations is?
A. 440 km
B. 444 km
C. 445 km
D. 450 km
\eqalign{ & {\text{At the time of meeting ,}} \cr & {\text{let the distance travelled by the}} \cr & {\text{first train be }}x{\text{ km}}{\text{.}} \cr & {\text{Then distance travelled by the }} \cr & {\text{second train is (}}x{\text{ + 60) km}} \cr & \therefore \frac{x}{{16}} = \frac{{x + 60}}{{21}} \cr & \Rightarrow 21x = 16x + 960 \cr & \Rightarrow 5x = 960 \Rightarrow x = 192 \cr & {\text{Hence,}} \cr & {\text{distance between two stations}} \cr & {\text{ = (192 + 192 + 60) km}} \cr & {\text{ = 444 km}}{\text{.}} \cr} | 307 | 901 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-30 | latest | en | 0.767677 |
http://mathhelpforum.com/algebra/212436-positions-points-intersecting-circles-difficult-print.html | 1,526,813,831,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863277.18/warc/CC-MAIN-20180520092830-20180520112830-00373.warc.gz | 184,514,358 | 4,065 | Positions of points in intersecting circles (DIFFICULT)
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• Feb 2nd 2013, 03:06 AM
NatalieSv94
Positions of points in intersecting circles (DIFFICULT)
Hi! here is the task I was talking about in my introductory thread I just posted a few minutes ago. I got this task yesterday and I have no idea how to even start with it. Here is an image of the task. I read it 100 times but I still don't know what to do! (Crying)
Attachment 26817
• Feb 2nd 2013, 04:46 AM
Paze
Re: Positions of points in intersecting circles (DIFFICULT)
You should post this here: Trigonometry
To get you started, try reading this: Intersecting Circles
• Feb 2nd 2013, 05:22 AM
ILikeSerena
Re: Positions of points in intersecting circles (DIFFICULT)
Hi NatalieSv94! :)
The triangle OPA is an isosceles triangle.
Furthermore the triangle OP'A is also an isosceles triangle.
Moreover this second triangle has a common angle with the first, meaning that these two triangles are similar.
Can you say anything about what that means for OP'?
• Feb 2nd 2013, 08:15 AM
NatalieSv94
Re: Positions of points in intersecting circles (DIFFICULT)
I dunno (Worried). OP'/AO=AP'/PO=AO/AP ? Is that right? And then, what am I supposed to do?
• Feb 2nd 2013, 11:36 AM
ILikeSerena
Re: Positions of points in intersecting circles (DIFFICULT)
That is good! :D
So suppose r=1 and OP=2, as your problem states.
Then AO, which is the radius, is also 1.
What is OP' then?
• Feb 3rd 2013, 12:14 AM
NatalieSv94
Re: Positions of points in intersecting circles (DIFFICULT)
It must be 1/2! Right? :D
Thank you so much ;)
• Feb 3rd 2013, 01:58 AM
ILikeSerena
Re: Positions of points in intersecting circles (DIFFICULT)
Yep! It is 1/2!
Can you find the general statement?
• Feb 3rd 2013, 02:50 AM
NatalieSv94
Re: Positions of points in intersecting circles (DIFFICULT)
OP'=1/OP :) Right?
• Feb 3rd 2013, 02:50 AM
NatalieSv94
Re: Positions of points in intersecting circles (DIFFICULT)
OP'=1/OP :) Right?
• Feb 3rd 2013, 02:51 AM
ILikeSerena
Re: Positions of points in intersecting circles (DIFFICULT)
Right!
That is for r=1.
So the next challenge is, what if r=2?
• Feb 3rd 2013, 02:57 AM
NatalieSv94
Re: Positions of points in intersecting circles (DIFFICULT)
I'll find it! OMG! I'll find it! it's OP'=r2/OP !!!! Is that right?
• Feb 3rd 2013, 03:00 AM
ILikeSerena
Re: Positions of points in intersecting circles (DIFFICULT)
Yep, that is right! (Wait)
• Feb 3rd 2013, 03:04 AM
NatalieSv94
Re: Positions of points in intersecting circles (DIFFICULT)
OMG! Thank you sooooooooooooooooooooo much! :) You're a great teacher you know!
In the last part when it says Discuss the scope and/or limitations of the general statement. I know that OP' approaches infinity as r approaches infinity and OP' approaches 0 as OP approaches infinity. Is that all? Or is there any situation in which OP' is not defined or the general statement is not working? (I know that it's not defined if OP is 0, but ...)
• Feb 3rd 2013, 03:13 AM
ILikeSerena
Re: Positions of points in intersecting circles (DIFFICULT)
Thanks! (Blush)
The only real limitation is that OP may not be zero, so the point P may not be chosen to coincide with the point O.
EDIT: Hold on! See below.
• Feb 3rd 2013, 03:15 AM
NatalieSv94
Re: Positions of points in intersecting circles (DIFFICULT)
Thanks again. ;)
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last | 1,069 | 3,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-22 | latest | en | 0.898537 |
https://api-project-1022638073839.appspot.com/questions/what-is-the-equation-of-the-line-perpendicular-to-y-23x-that-passes-through-1-6 | 1,718,533,818,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00649.warc.gz | 77,117,361 | 6,109 | # What is the equation of the line perpendicular to y=-23x that passes through (-1,-6) ?
Dec 27, 2015
The slope of a perpendicular line is always the negative reciprocal of the other line's slope.
#### Explanation:
If the slope of y = -23x is -23, the slope of the perpendicular line is $\frac{1}{23}$.
y - (-6) = $\frac{1}{23}$(x - (-1)
y = $\frac{1}{23} x$ + $\frac{1}{23}$ - 6
y = $\frac{1}{23} x$ - $\frac{137}{23}$
y = $\frac{1}{23} x$ - $\frac{137}{23}$ is the equation of the line perpendicular to y = -23x and that passes through (-1, -6).
Hopefully you understand now! | 192 | 588 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-26 | latest | en | 0.78342 |
http://www.acadblock.com/physical-chemistry/change-in-ph-df2m/ | 1,709,218,023,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474843.87/warc/CC-MAIN-20240229134901-20240229164901-00733.warc.gz | 32,513,990 | 7,239 | # change in pH
An ammonia ammonium chloride buffer has a pH value of 9 with[NH3]-0.25.By how much will the pH will change if 75 mL 0.1M KOH be added to 200 mL buffer solution.Kb=2*10-5
1357
Manish Shankar ·
pH=9 so [OH-]=10-5
NH3+H2O [eq] NH4++OH-
0.25 x 10-5
x*10-5/0.25=Kb=2*10-5
x=0.5
NH3 + H2O [eq] NH4+ + OH-
mmoles 0.25*200 0.5*200 10-5*200+0.1*75~7.5
let the whole of OH- is consumed
NH3 + H2O [eq] NH4+ + OH-
mmoles 50+7.5 100-7.5 0
mmoles 57.5-x 92.5+x x
total volume = V=275 ml
(x/275)((92.5+x)/275)/((57.5-x)/275)=Kb
let x be very small
x*92.5/57.5=275Kb
x=3.42*10-3
[OH-]=x/275=1.24*10-5M
pOH=4.9 and pH=9.1
so pH changes by 0.1
24
eureka123 ·
thanx sir....... | 344 | 687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-10 | latest | en | 0.487026 |
http://answers.google.com/answers/threadview/id/749355.html | 1,701,834,419,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.13/warc/CC-MAIN-20231206031946-20231206061946-00731.warc.gz | 3,424,237 | 3,188 | View Question
Q: current day value of cash flow ( Answered , 0 Comments )
Question
Subject: current day value of cash flow Category: Business and Money Asked by: coffey5-ga List Price: \$14.88 Posted: 25 Jul 2006 09:43 PDT Expires: 24 Aug 2006 09:43 PDT Question ID: 749355
```I have an investment that pays me \$19.50 per year for the next 40 years. The 19.50 increases 3% per year as a cost of living increase. I want to sell the cash flow and the buyer expects a 10% return. What is the current day value? How muc is it worth if the buyer want 8% or 12% return respectively?```
```Hi!! This is a Growing Annuity case and you need to find the present value of it according to the required discount rate, to solve this problem you need to use the following formula: {1 - [(1+g)/(1+r)]^t} PV = A * ------------------------- (r-g) where: PV = Present Value of the growing annuity A = Initial annuity value r = Interest rate g = Growth rate t = number of time periods In this case (10% rate): A = \$19.5 r = 0.1 g = 0.03 t = 40 Then: PV = \$19.5 * {1 - [(1+0.03)/(1+0.1)]^40} / (0.1-0.03) = = \$19.5 * {0.92792551} / 0.07 = = \$258.49 For the 8% case: PV = \$19.5 * {1 - [(1+0.03)/(1+0.08)]^40} / (0.08-0.03) = = \$19.5 * {0.849845356} / 0.05 = = \$331.44 For the 12% case: PV = \$19.5 * {1 - [(1+0.03)/(1+0.12)]^40} / (0.12-0.03) = = \$19.5 * {0.964943538} / 0.09 = = \$209.07 For references see: "GROWING ANNUITIES" by Albert L. Auxier and John M. Wachowicz, Jr.: http://web.utk.edu/~jwachowi/growing_annuity.pdf "Growing Annuity": http://www.pitt.edu/~schlinge/fall99/example_growth.htm Search strategy: "present value" "Growing Annuity" I hope this helps you. Feel free to request for a clarification if you need it. Regards, livioflores-ga``` | 600 | 1,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-50 | latest | en | 0.839045 |
null | null | null | null | null | null | Adolescent Medicine
Sometimes it's difficult to see your child as anything but that: a child. Yet, in many ways, teens today are growing up faster than ever. They learn about violence and sex through the media and their peers, but they rarely have all the facts. That's why it's so important for you to talk to your kids about sex, particularly sexually transmitted diseases (STDs).
Teens are one of the groups most at risk for contracting STDs. You can help your kids stay safe by talking to them and sharing some important information about STDs and prevention.
Before you tackle this sensitive subject, however, it's important to make sure you not only know what to say, but how and when to say it.
Timing Is Everything
It's never too late to talk to your kids about STDs, even if they're already teens. A late talk is better than no talk at all. But the best time to start having these discussions is some time during the preteen years.
Of course, the exact age varies from child to child: Some kids are more aware of sex at age 9 than others are at age 11. You'll need to read your child's cues.
No matter how old your child is, if he or she starts having questions about sex, it's a good time to talk about STDs.
Questions are a good starting point for a discussion. When kids are curious, they're often more open to hearing what their parents have to say.
But not all kids ask their parents questions about sex. One way to initiate a discussion is to use a media cue, like a TV program, a movie, or an article in the paper, and ask what your child thinks about it.
Another way is to use the human papillomavirus (HPV) vaccine as a starting point for a conversation. The HPV vaccine is recommended for preteen girls (and also boys), and has the best chance of protecting against infection if the series of shots is given before someone becomes sexually active.
The surest way to have a healthy dialogue is to establish lines of communication early on. If parents aren't open to talking about sex or other personal subjects when their kids are young, kids will be a lot less likely to seek out mom or dad when they're older and have questions.
Spend time talking with your kids from the beginning and it'll be much easier later to broach topics like sex because they'll feel more comfortable sharing thoughts with you.
Tips for Talking
To make talking about STDs a little easier for both you and your kids:
• Be informed. STDs can be a frightening and confusing subject, so it may help if you read up on STD transmission and prevention. You don't want to add any misinformation and being familiar with the topic will make you feel more comfortable. If kids ask for information that you're not sure about, find out the answer from a reliable source and get back to them.
• Ask what your kids already know about STDs and what else they'd like to learn. Remember, though: Kids often already know more than you realize, although much of that information could be incorrect. Parents need to provide accurate information so their kids can make the right decisions and protect themselves.
• Ask what your kids think about sexual scenarios on TV and in movies and use those fictional situations as a way to talk about safe sex and risky behavior.
• Encourage your kids to raise any fears, questions, or concerns they have.
• Make your kids feel that they're in charge of this talk, not you, by getting their opinions on whatever you discuss. If you let their questions lead the way, you'll have a much more productive talk than if you stick to an agenda or give a lecture.
• Explain that the only sure way to remain STD-free is to not have sex or intimate contact with anyone outside of a committed, monogamous relationship, such as marriage. However, those who are having sex should always use condoms to protect against STDs, even when using another method of birth control. Most condoms are made of latex, but both male and female condoms made of polyurethane are available for people with a latex allergy.
Common Questions About STDs
• What is an STD? An STD is a sexually transmitted disease.
• How does someone catch one? These infections and diseases are spread from one individual to another during anal, oral, or vaginal sex. They also can be spread by fingers or objects after they have touched genitals or body fluids.
• What do STDs do to a person's body? The type of STD determines what kinds of symptoms, if any, someone has. Some STDs cause virtually no symptoms, whereas others can cause the person to have discharge from the vagina or penis, sores, or pain.
But even when there are no symptoms, if STDs are untreated, they can lead to damage to the internal organs and may cause long-term health problems, like infertility or cancer. This is why anyone who has had any type of sex (vaginal, oral, or anal) needs to be tested for STDs regularly.
• Are STDs curable or do you have them forever? Some STDs like chlamydia and gonorrhea can be cured with antibiotics, but some infections — like herpes or HIV — have no cure.
• Are people who catch STDs somehow bad? Getting an STD does not mean that someone is a bad person, just that he or she needs to learn how to prevent future infections.
• Can you tell that someone has an STD just by looking at him or her? People often may not even know that they're infected themselves. Although there may be visible signs around the genitals with certain kinds of STDS, like genital warts and herpes, most of the time, there is no way to look at someone and know that he or she has an STD.
Answering any of these questions or others as openly as possible is the best approach. It's up to you to gently correct any misinformation your kids may have learned. And always answer questions honestly without being overly dramatic.
It can be tough, but try not to be too emotional or preachy. You want your kids to know that you're there to support and help, not condemn.
Finding Reliable Information
Communicating with your kids may not be simple, but it's necessary. If you're always available to talk, discussions will come easier. Literature from your doctor's office or organizations like Planned Parenthood can provide answers.
And websites like discuss STDs and sex in teen-friendly language. Viewing them together can help you and your kids start talking.
Your child's school can be an information resource. Find out when sexuality will be covered in health or science class and read the texts that will be taught. The PTA may even offer sessions about talking to teens where you can share tips and experiences with other parents.
And don't shy away from discussing STDs or sex out of fear that talking will make kids want to have sex. Informed teens are not more likely to have sex; but when they do become sexually active they are more likely to practice safe sex.
If you try these tactics and still don't feel comfortable talking about STDs, make sure your kids can talk to someone who will have accurate information: a doctor, counselor, school nurse, teacher, or another family member.
Kids and teens need to know about STDs, and it's better that they get the facts from someone trustworthy instead of discovering them on their own.
Reviewed by: Larissa Hirsch, MD
Date reviewed: August 2011 | null | null | null | null | null | null | null | null | null |
https://boazcommunitycorp.org/763-determinant-properties.html | 1,623,674,556,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612154.24/warc/CC-MAIN-20210614105241-20210614135241-00409.warc.gz | 142,048,009 | 9,942 | # Determinant Properties
The determinants associated with square order matrices no have the following properties:
P1) When all elements of a row (row or column) are null, the determinant of this matrix is null. Example:
P2) If two rows of an array are equal, then their determinant is null. Example:
P3) If two parallel rows of a matrix are proportional, then their determinant is null. Example:
P4) If the elements of a row in a matrix are linear combinations of the corresponding elements of parallel rows, then their determinant is null. Examples:
P5) Jacobi's theorem: The determinant of a matrix does not change when we add to the elements of a row a linear combination of the corresponding elements of parallel rows. Example:
Replacing the 1st column with the sum of that same column twice the 2nd, we have:
Next: Properties (Part 2) | 188 | 849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-25 | latest | en | 0.810656 |
https://boards.straightdope.com/t/math-is-cool/205488 | 1,611,778,215,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704832583.88/warc/CC-MAIN-20210127183317-20210127213317-00569.warc.gz | 241,431,465 | 8,493 | # Math is cool
There’s just something inherently beautiful about stuff like
111,111,111 x 111,111,111 = 12,345,678,987,654,321
Trigonometry, too. It’s so priceless that the sun and moon appear to be exactly the same size from here, despite their differences in diameter and distance.
How about these equations ? Notice that the “left-side” numbers contain no zeroes.
8 * 125 =             1,000
64 * 15,625 =       1,000,000
512 * 1,953,125 = 1,000,000,000
screams
covers eyes
I did NOT just read that.
backs away
1/27= .037037…
1/37= .027027…
[From Charles Fleischer’s old Moleeds routine.]
and the digits of multiples of 3 add up to be a multiples of 3
and the digits of multiples of 9 add up to be a multiples of 9
10[sup]3[/sup] + 9[sup]3[/sup] = 1,729
12[sup]3[/sup] + 1[sup]3[/sup] = 1,729
1,729 is the smallest number that can be expressed as the sum of two perfect cubes in two different ways. That’s pretty cool.
There are no natural numbers which are not interesting. If there were, there would be a least one, and that would make it interesting.
e[sup]i*pi[/sup]=-1
1/243=0.004115226337448559670781893004115…
Today in physics, the professor was talking about equations depicting motion. He put one equation up that I recognized from high school physics, then did the integral of it and came up with another equation I recognized. Having just taken integral calculus last year, I first realized how the second equation was created, and it made my day. I was like “Whoa, now I know where that equation comes from, and I know how it came to be!” Seriously, it completely awed me and it seemed like the most awesome thing ever. Even though it’s pretty mundane compared to the rest of physics and math.
So anyway, math is cool.
Philosophy is written in this grand book—I mean the universe—
which stands continually open to our gaze, but it cannot be understood unless one first learns to comprehend the language and interpret the characters in which it is written. It is written in the language of mathematics, and its characters are triangles, circles, and other geometrical figures, without which it is humanly impossible to understand a single word of it. : Galileo Galilei
I thank God everyday for the base ten system.
3 is a magic number.
Yes it is.
Do you remember when math actually clicked in your head and the “confusion” vanished. I remember all through school. Clear up to graduating from high school, struggling with mathematical concepts. I did well in school, straight A’s, honors and all that crap but it was tough. My math teachers apparently didn’t understand it well enough to make sense of it, I guess.
I retaught myself a lot of math after high school. When I went to college at 30…damn, math started getting tough again. I remember working through some trig functions in physics while taking an advanced calc class…and a programming course one summer semester (at the same time). :eek:
It was like a switch went off or ON in this case and it just started making sense. After college I went on to teach math at an alternative school for a few years. Actually I taught all the courses offered between grades 6-12 but specialized in math.
These kids were mostly 14-18 y/o boys that had been expelled from school for ??? drugs, weapons etc…You know?
Most were basically NON-FUNCTIONAL in math, some actually counted on their fingers!
There is nothing as great as seeing a kid snap to a concept, especially math. Talk about a confidence builder…a person who is not afraid of math, that used to be, can do anything they want.
My lil’ bro…is an ace mechanic, but was alway shy of the math.
I sat him down one weekend and taught him some algebra/geometry/ little calc…he got IT when we converted cu"/torque/horsepower back and forth on different motors. Since then, he loves math.
Sorry ‘bout the rant. Gotta go build some stairs today…
I’ve got three sets of stringers for a set of steps on a deck. They are 2"x12" treated boards. The stairs will have a 6" rise per 12" deep step. The deck is 4 1/2’ off the ground, the ground slopes downhill at 10degrees. The angle of the stringers/steps will be 30degrees from the deck to where it meet the ground.
How long will the 2"x12" boards/stringers have to be in order to reach the deck AND how many steps will it take?
I’m gonna go start in a few minutes…be back this evening. See if ya got it right.
Nope, 'cause I’m almost 33, and it still hasn’t happened. I’m a pretty smart guy in a lot of respects, but numbers have always been tough for me. Actually, being essentially math-illiterate never really bothered me much until fairly recently. I’ve been learning how to play poker, and quickly figuring out pot odds when you have trouble with basic math is a bitch. I’m seriously thinking about getting a ‘Math For Dummies’ type book just to help with that. Anyone recommend one?
I agree that math is cool… I just wish I were better at it. Pre-calc was rough and it went downhill from there.
I’ll accept that if you explain a little more what you mean. The math they attempted to teach me in school certainly seemed to only be about numbers (save for geometry, which at its base was about numbers as well). But as I said, it’s not a subject I was ever good at, so I’m certainly willing to learn I’m wrong. Exactly how do you mean it’s not about numbers?
A lot of mathmatics doesn’t really involve numbers at all, e.g. Euclidian geometry could probably be tackled without ever dealing with a single number. A lot of other mathematics uses numbers, but is less dependent on them than, oh…let’s say, literature is dependent on the printed word. The study of calculus uses numbers as a means to provide concrete examples as a means to work through & better understand the material, but to really understand calculus, and therefore a huge chunk of the universe we live in, doesn’t actually require any numbers at all. Indeed, economists have proven (& discovered) some very important things using mathmatics without really relying on numbers at all, e.g. the tragedy of the commons.
You may not be good at arithmetic. But I’m willing to bet that it is for no particularly good reason. Unfortunately we live in a society that is more or less mathmatically illiterate, so that even the most basic study is seen as a Herculean task of no real value. That poor attitude rubs off on us and we pass it on in turn, IMO.
If you have time, check out Mathematics: The Science of Patterns as an overview of what subjects are really covered by math.
Yep, mathematics is cool.
(1/1)+(1/4)+(1/9)+(1/16)+(1/25)+(1/36)+… [the numbers are squares]
equals
(4/3)(9/8)(25/24)(49/48)(121/120)(169/168)… [numerators are squares of prime numbers. denomenators are one less.]
equals
(pi^2)/6
Now, that is just freaky. | 1,669 | 6,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-04 | latest | en | 0.939776 |
https://www.esaral.com/q/write-the-following-cubes-in-expanded-form-i-2-x13-ii-2-a-3-b3iii-leftfrac32-x1right3-iv-leftx-frac23-yright3 | 1,722,725,454,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640380725.7/warc/CC-MAIN-20240803214957-20240804004957-00444.warc.gz | 618,854,807 | 11,579 | Write the following cubes in expanded form: <br/><br/>(i) $(2 x+1)^{3}$ <br/><br/>(ii) $(2 a-3 b)^{3}$<br/><br/>(iii) $\left[\frac{3}{2} x+1\right]^{3}$ <br/><br/>(iv) $\left[x-\frac{2}{3} y\right]^{3}$
Solution:
It is known that,
$(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
and $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$
(i) $(2 x+1)^{3}=(2 x)^{3}+(1)^{3}+3(2 x)(1)(2 x+1)$
$=8 x^{3}+1+6 x(2 x+1)$
$=8 x^{3}+1+12 x^{2}+6 x$
$=8 x^{3}+12 x^{2}+6 x+1$
(ii) $(2 a-3 b)^{3}=(2 a)^{3}-(3 b)^{3}-3(2 a)(3 b)(2 a-3 b)$
$=8 a^{3}-27 b^{3}-18 a b(2 a-3 b)$
$=8 a^{3}-27 b^{3}-36 a^{2} b+54 a b^{2}$
(iii) $\left[\frac{3}{2} x+1\right]^{3}=\left[\frac{3}{2} x\right]^{3}+(1)^{3}+3\left(\frac{3}{2} x\right)(1)\left(\frac{3}{2} x+1\right)$
$=\frac{27}{8} x^{3}+1+\frac{9}{2} x\left(\frac{3}{2} x+1\right)$
$=\frac{27}{8} x^{3}+1+\frac{27}{4} x^{2}+\frac{9}{2} x$
$=\frac{27}{8} x^{3}+\frac{27}{4} x^{2}+\frac{9}{2} x+1$
(vi) $\left[x-\frac{2}{3} y\right]^{3}=x^{3}-\left(\frac{2}{3} y\right)^{3}-3(x)\left(\frac{2}{3} y\right)\left(x-\frac{2}{3} y\right)$
$=x^{3}-\frac{8}{27} y^{3}-2 x y\left(x-\frac{2}{3} y\right)$
$=x^{3}-\frac{8}{27} y^{3}-2 x^{2} y+\frac{4}{3} x y^{2}$ | 655 | 1,166 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-33 | latest | en | 0.175213 |
http://slideplayer.com/slide/4386813/ | 1,506,013,776,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687833.62/warc/CC-MAIN-20170921153438-20170921173438-00347.warc.gz | 289,039,800 | 20,915 | # The Practice of Statistics Third Edition Chapter 6: Probability and Simulation: The Study of Randomness Copyright © 2008 by W. H. Freeman & Company Daniel.
## Presentation on theme: "The Practice of Statistics Third Edition Chapter 6: Probability and Simulation: The Study of Randomness Copyright © 2008 by W. H. Freeman & Company Daniel."— Presentation transcript:
The Practice of Statistics Third Edition Chapter 6: Probability and Simulation: The Study of Randomness Copyright © 2008 by W. H. Freeman & Company Daniel S. Yates
Two computer simulations of tossing a balanced coin 100 times
A Head A four An Ace
Ex. What is the sample space for the roll of a die? s= {1,2,3,4,5,6} All equal probability What is the sample space for the roll of a pair of dice? s={2,3,4,5,6,7,8,9,10,11,12} different probability
Events, Sample Spaces and Probability An event is a specific collection of sample points: –Event A: Observe an even number on the roll of a single die. Often represented by a venn diagram.
Tree diagrams help to determine the sample space Ex. An experiment consists of flipping a coin and tossing a die.
Ex. For your dinner you need to choose from six entrees, eight sides, and five desserts. How many different combinations of entrée, side and dessert are possible? 6x8x5 = 240
Unions and Intersections Compound Events Made of two or more other events Union A B Either A or B, or both, occur Intersection A B Both A and B occur
10 Venn diagrams for (a) event (not E), (b) event (A & B), and (c) event (A or B)
McClave: Statistics, 11th ed. Chapter 3: Probability 12 The Additive Rule and Mutually Exclusive Events Events A and B are mutually exclusive (disjoint) if A B contains no sample points.
Complementary Events The complement of any event A is the event that A does not occur, A C. A: {Toss an even number} A C : {Toss an odd number} B: {Toss a number ≤ 4} B C : {Toss a number ≥ 5} A B = {1,2,3,4,6} [A B] C = {5} (Neither A nor B occur)
Complementary Events
What is the probability of rolling a five when a pair of dice are rolled?
P(A) = 4/36 =.111 =11.1%
Independent Events – two events are independent if the occurrence of one event does not change the probability that the other occurs. Ex. What is the probability of rolling a 7 each of three consecutive roles of a pair of dice? P(rolling 7) = 6/36 = 1/6 ; P( three 7s) = (1/6) 3 =0.5%
If A and B are mutually exclusive,
Ex. Event A = { a household is prosperous; Inc. > 75 k} Event B = { a household is educated; completed college} P(A) = 0.125P(A and B) = 0.077 P(B) = 0.237 What is the probability the household is either prosperous or educated? P (A or B) = P(A) + P(B) – P(A and B) = 0.125 + 0.237 – 0.077 = 0.285
Conditional probability - The probability of the occurrence of an event given that another event has occurred. ♣ Notation – P(B|A) > probability of B given A
Ex. A die has sides 1,2,3 painted red and sides 4,5,6 painted blue. Event A = {Roll Blue}Event B = {Roll Red} Event C = {Roll even}Event D = {Roll Odd} What is the probability that you roll an even given we rolled a blue? What is the probability of rolling a red, even? What is the probability that you roll an odd given you rolled a blue?
General Multiplication rule can be extended for any number of events. P(A and B and C) = P(A) x P(B|A) x P(C|A and B)
Example: Deborah and Matthew have applied to become a partner in their law firm. Over lunch one day they discuss the possibility. They estimate that Deborah has a 70% chance of becoming partner while Matthew has a 50% chance. They also believe that there about a 30% chance that both of them will be chosen. a)What is the probability that either Deborah or Matthew become partner? b)What is the probability that either Deborah or Matthew become partner but not both? c)What is the probability that only Mathew becomes partner? d)What is the probability that neither one becomes partner? e)Are the events Independent? a)P(D U M) = P(D) + P(M) – P(D∩M) = 0.7 + 0.5 – 0.3 = 0.9 b)P(D∩M c ) U P(D c ∩M) = P(D∩M c ) + P(D c ∩M) = 0.4 + 0.2 = 0.6 c) P(D c ∩M) = P(M) x P(D c M) = 0.2 d) P(D c ∩M c ) = 0.1 Event D = {Deborah is made partner} Event M = {Matthew is made partner} e)P(D∩M) = P(D) x P(M) if independent 0.3 ≠ 0.7 x 0.5; No, they are not independent
Example: 29% of internet users are 1-20 years old, 47% are 21- 50 years old and 24% are 51-90 years old. 47% of 1- 20 year old users chat, 21% of 21-50 year old users chat and 7% of 51-90 year old users chat a)What is the probability that a 19 year old internet user does not chat? b)What is the probability that an internet user chats? c)If we know an internet user chats, what is the probability that they are 60 years old? a) P(C c ∩ A 1 ) = P (A 1 ) x P(C c A 1 ) = 0.29 x 0.53 = 0.1537 b) P(C) = P(C ∩ A 1 ) + P(C ∩ A 2 ) + P(C ∩ A 3 ) 0.1363 +0.0987 + 0.0168 = 0.2518 c) P(A 3 C) = P(A 3 ∩ C) / P(C) = 0.0168/(0.1363 +0.0987 + 0.0168) = 0.0667 Event A 1 = {person is 1-20 years old} Event A 2 = {person is 21-50 years old} Event A 3 = {person is 51-90 years old} Event C = {person chats online}
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http://mathoverflow.net/feeds/question/43925 | 1,369,164,214,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700477029/warc/CC-MAIN-20130516103437-00027-ip-10-60-113-184.ec2.internal.warc.gz | 171,999,224 | 5,858 | What are the polynomial relations between these characteristic 2 "thetas" ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:23:26Z http://mathoverflow.net/feeds/question/43925 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43925/what-are-the-polynomial-relations-between-these-characteristic-2-thetas What are the polynomial relations between these characteristic 2 "thetas" ? paul Monsky 2010-10-28T01:47:32Z 2011-10-16T08:44:49Z <p>Suppose $\ell=2m+1$, $m>0$. Define $[i]$ in $\mathbb{Z}/2\mathbb{Z}[[x]]$ to be $$\sum_{n\equiv i\mod l} x^{n^2}.$$ Note that $[0]=1$, and that $[i]=[j]$ whenever $\ell$ divides $i+j$ or $i-j$.</p> <p>Now let $u_1,...,u_m$ be indeterminates over $\mathbb{Z}/2\mathbb{Z}$, and $f$ be the homomorphism $\mathbb{Z}/2\mathbb{Z}[u_1,...,u_m]\to \mathbb{Z}/2\mathbb{Z}[[x]]$ taking $u_i$ to $[i]$. Using the theory of modular forms I think I can show that the kernel, $P$, of $f$ is a dimension 1 prime ideal.</p> <p>Question 1: What is the genus of (a non-singular projective model) of the curve corresponding to $P$?</p> <p>Examples: When $\ell=5$ the curve one desingularizes is $x^5+y^5+xy+(xy)^2=0$, and the genus is 0.</p> <p>When $\ell=7$, the curve has the following affine plane model of degree 14: $\sum x^iy^j=0$ where $(i,j)$ runs over the 10 pairs $(14,0)$, $(12,1)$, $(10,2)$, $(7,7)$, $(6,4)$, $(5,8)$, $(5,1)$, $(4,5)$, $(1,10)$ and $(0,14)$. (Perhaps someone with access to Singular or time on their hands can work out the genus?).</p> <p>When $\ell=9$ the curve has an affine plane model of degree 27; this time one gets the 20 pairs $(27,0)$, $(24,3)$, $(21,6)$, $(20,1)$, $(15,3)$, $(13,2)$, $(12,15)$, $(12,6)$, $(11,10)$, $(11,1)$, $(9,18)$, $(9,9)$, $(7,17)$, $(6,21)$, $(5,16)$, $(5,7)$, $(4,20)$, $(4,11)$, $(1,23)$ and $(0,27)$.</p> <p>One has the following curious but easily proved relations between the various $[i]$. Let $a$,$b$,$c$,$d$,$e$,$f$ be $[i]$,$[j]$,$[2i]$,$[2j]$,$[i+j]$,$[i-j]$. Then $d(a^4)+c(b^4)+cd+(ef)^2=0$. Each such identity gives rise to a "quintic relation" lying in $P$. (I used these relations to get the curves in the above examples). Let $J$ be the ideal contained in $P$ that is generated by these quintic relations.</p> <p>Rather vague Question 2: What can be said about $J$? For example: Are all the minimal primes of $J$ of dimension 1? If so, what are the associated primes other than $P$? Is $J$ a radical ideal?</p> <p>Examples: When $\ell=5$, $J=P$, and I believe the same holds when $\ell=7$. But when $\ell=9$ one needs to add the element $a(b^2)+b(c^2)+c(a^2)+d+(d^2)+(d^3)$, where $a$,$b$,$c$,$d$ are $u_1$,$u_2$,$u_4$,$u_3$ to $J$ in order to get $P$. Let $K$ be the ideal $(a+ad,b+bd,c+cd,ab+c^2,ac+b^2,bc+a^2)$. Then $K$ is the intersection of three dimension 1 primes, and I believe that $J$ is the intersection of $P$ and $K$.</p> <p>@sleepless--I hope you like this orthography better.</p> <p>EDIT: Here are answers to question 1 when l=9 and l=11. (As I explained in a comment the genus is 3 when l=7. It now appears that it's 10 when l=9 and 26 when l=11). Remarkably when l=3,5,7,9, or 11 the genus is the same as the genus of the compactification of the quotient of the upper half-plane by the principal congruence group, Gamma(l). I doubt that this is a coincidence, and am interested in what experts in the theory of characteristic p modular forms have to say. </p> <p>Suppose first l=9. Extend the constant field from Z/2 to its algebraic closure,K. Let C in affine 4-space be the zero-locus of P, and L/K be the function field of C. P is generated by the "quintic relations" together with ab^2+bc^2+ca^2+d+d^2+d^3, where a,b,c,d are the coordinate functions u1,u2,u4 and u3. It follows that P is stabilized by the linear automorphisms (a,b,d,c)-->(b,c,d,a) and (a,b,d,c)-->(ua,ub,d,uc) with u^3=1. These automorphisms generate an order 9 group, G, which acts on L; let L_0 be the fixed field. It can be shown that L_0 is generated over K by abc and d and that (abc)^3=d^7+d^8+d^9. So L_0/K has genus 1. We now use Riemann-Hurwitz to calculate the genus, g, of L/K. (Since G has odd order, L/L_0 is tamely ramified).</p> <p>The quintic relations all vanish on the line a=b=c=0. It follows that C has 3 points on this line; they are (0,0,d,0) with d+d^2+d^3=0. Each of these points is an ordinary triple point, and G permutes the branches at each of these points in a size 3 orbit. All the other orbits of G acting on the places of the function field L/K (including the places at infinity) are of size 9. Riemann-Hurwitz now tells us that 2g-2=9(2-2)+(9-3)+(9-3)+(9-3), so that g=10.</p> <p>When l=11, one can argue in like manner. Now P is generated by the quintic relations, and the similar group G, acting on L/K, has order 55. I think one can again show that the genus of L_0/K is 1; this is the one thing I haven't checked completely. Now C sits in affine 5-space, the origin is an ordinary singular point of multiplicity 5, and G permutes the branches at the origin in a size 5 orbit. All other orbits of G acting on the places of L/K are of size 55 and Riemann-Hurwitz tells us that 2g-2=55(2-2)+(55-5), so that g=26.</p> http://mathoverflow.net/questions/43925/what-are-the-polynomial-relations-between-these-characteristic-2-thetas/68668#68668 Answer by paul Monsky for What are the polynomial relations between these characteristic 2 "thetas" ? paul Monsky 2011-06-23T22:23:40Z 2011-07-01T16:14:42Z <p>In my question I remarked that when l=2m+1 a modular form argument might be used to show that the field generated by [1], ... ,[m] over Z/2 has transcendence degree 1. Below I give an elementary proof that the transcendence degree is 1 when l is prime, based on the quintic relations. The proof is too long to be a comment or edit, so I'm posting it as an answer.</p> <p>Lemma----Let F be a field of characteristic 2 and n-->a_n be a function Z/l-->F satisfying:</p> <p>(1) a_0 =1</p> <p>(2) a_i =a_-i</p> <p>(3) The sum of (a_2i)(a_j)^4, (a_2j)(a_i)^4, (a_2i)(a_2j) and (a_(i+j)a_(i-j))^2 is 0.</p> <p>Then if a_1 =0, each of a_2, ... ,a_(l-1) is 0.</p> <p>The proof proceeds in 3 steps: First I claim that if a_2 is 0 then a_1 is 0. For suppose the contrary. Since a_2l is 1, there is an odd positive r with a_2r non-zero. Take such an r as small as possible; since a_2 =0,r>1 and so (r+1)/2 is less than r. Taking i=r and j=1 in (3) we find that (a_2r)(a_1)^4 is the square of (a_(r+1))(a_(r-1)). So a_(r+1) and a_(r-1) are non-zero. But one of (r+1)/2, (r-1)/2 is odd. This contradicts the minimality of r.</p> <p>Observe that if a_2s is 0 then a_s is 0. To see this note that s is not 0 in Z/l, and apply the result of the paragraph above to the function (s)(i)-->a_i.</p> <p>Suppose finally that a_r and a_s are non-zero while a_(r+s) is 0. Then a_((r+s)/2) is 0. Applying (3) with i=(r+s)/2 and j=(r-s)/2 we find that the square of (a_r)(a_s) is 0, a contradiction. So the n in Z/l with a_n non-zero form a subgroup of the additive group, completing the proof.</p> <p>Theorem---Let K be an algebraic closure of Z/2 and T be the subring of K[[x]] generated over K by all the [i]. Then the only prime ideal of the affine domain T that contains [1] is the maximal ideal ([1], ... ,[m]).</p> <p>Note first that T is generated by [1], ... ,[m]. So the ideal of T generated by these elements is indeed maximal. Let I be a prime ideal that contains [1], and F be the field of fractions of T/I. Consider the function Z/l-->F taking the congruence class i+lZ to the image of [i] in T/I. This function clearly satisfies (1) and (2) of the Lemma. The quintic relations show that it satisfies (3) as well. As the function takes 1+lZ to 0, we find by the Lemma that it takes 2+lZ, ... , m+lZ to 0 as well. So I contains each of [1], ... ,[m].</p> <p>The result I mentioned is an immediate consequence. For let X be the irreducible algebraic set in affine m-space over K corresponding to T. The Theorem tells us that the intersection of X with one of the coordinate hyperplanes consists of the origin alone. So X has dimension 1 and T has transcendence degree 1 over K.</p> <p>EDIT: Let X be as in the paragraph above. In my question I showed that X is contained in the zero-locus of a set of "quintic relations". In my comment below I showed that this zero-locus imbeds in projective m-space in such a way that it has only finitely many "points at infinity". So all its irreducible components are of dimension 0 or 1. Let Y_i be the 1-dimensional irreducible components and Y their union, so that X is one of the Y_i. All this goes through even when l is composite. But when l is prime I now think I can prove more:</p> <p>A)... Each Y_i passes through the origin and is the image of X under a certain permutation of coordinates. (The permutation corresponds to the permutation of the theta-series given by [i]-->[ri] for some r prime to l).</p> <p>B)... deg(Y)= l(l-1)(l+1)/24. So the degree of X divides this number.</p> <p>I believe that X is the only component of Y. To prove this amounts to showing that if S is the subring of Z/2[[x]] generated by the theta-series, then for each r prime to l there is an automorphism of S taking [i] to [ri] for each i. (This conjecture has a modular forms feel. Can anyone provide a proof of it?)</p> http://mathoverflow.net/questions/43925/what-are-the-polynomial-relations-between-these-characteristic-2-thetas/78258#78258 Answer by paul Monsky for What are the polynomial relations between these characteristic 2 "thetas" ? paul Monsky 2011-10-16T08:44:49Z 2011-10-16T08:44:49Z <p>Felipe Voloch referred me to two 1959 papers of Igusa in v. 81 of Amer. J. of Math. pages 453-475 and 561-577. Results from these papers and techniques I've developed on MO give an answer to my question when l is prime.</p> <p>As I suggested in the edit to the question, the genus is (l-3)(l-5)(l+2)/24. More is true. In the first of the above-cited papers Igusa constructs, for each prime p and each N prime to p, a "field of modular functions of level N", finite and Galois over k(j) where k is the algebraic closure of Z/p. When N is a prime, l, he shows that this field has Galois group PSL_2(Z/l) over k(j) and is the splitting field of the "invariant transformation equation" Phi(X,j). In Lemma 2 of the second paper cited above he shows that this symmetric 2 variable Phi is the mod p reduction of the classical modular equation. I use these results to show that the field of my question, generated over the algebraic closure, k, of Z/2 by the theta series, identifies with Igusa's field of modular functions of level l and characteristic 2.</p> <p>A key observation(the key observation according to Kevin Buzzard--it allows me to pass from modular functions that appear to be of even level to ones evidently of level l) is that Phi(1/G,1/F)=0 where F=x+x^9+x^25+... and G=F(x^l). To see this recall Jacobi's identity (1-q)(1-q^2)(1-q^3)...=1-3q+5q^3-7q^6+9q^10... where the exponents are the triangular numbers. Raising to the power 8, multiplying by q, and reducing mod 2 shows that the mod 2 reduction of the Fourier expansion of Delta(z) is F(q). Since j(z) is the quotient of (E_4(z))^3 by Delta(z), the mod 2 reduction of the Fourier expansion of j(z) is 1/F(q), while that of j(lz) is 1/G(q). This together with the result from Igusa's second paper gives the observation.</p> <p>It now suffices to show that my field is the field generated over the algebraic closure, k, of Z/2 by G together with the l+1 conjugates of F over k(G). In various answers to other MO questions I've sketched a proof that my field admits PSl_2(Z/l) as an automorphism group, that it contains all of the above elements, and that the elements of PSL_2 all fix G. I now look at G sitting inside the fixed field of PSL_2, and show that it has exactly one zero (counted with multiplicity) in that field. So the fixed field is precisely k(G). It follows that my field is generated over k(G) by F and its k(G)-conjugates concluding the proof.</p> | 3,837 | 12,035 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2013-20 | latest | en | 0.624654 |
https://www.uen.org/core/displayLinks.do?courseNumber=5600&standardId=70381&objectiveId=70388 | 1,680,002,875,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00784.warc.gz | 1,152,188,596 | 15,593 | Secondary Mathematics I
Strand: GEOMETRY - Congruence (G.CO)
Experiment with transformations in the plane. Build on student experience with rigid motions from earlier grades (Standards G.CO.1-5). Understand congruence in terms of rigid motions. Rigid motions are at the foundation of the definition of congruence. Reason from the basic properties of rigid motions (that they preserve distance and angle), which are assumed without proof. Rigid motions and their assumed properties can be used to establish the usual triangle congruence criteria, which can then be used to prove other theorems (Standards G.CO.6-8). Make geometric constructions (Standards G.CO.12-13).
Standard G.CO.6
Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms of rigid motions to decide if they are congruent.
• 3D Transmographer
This lesson contains an applet that allows students to explore translations, reflections, and rotations.
• Are the Triangles Congruent?
The purpose of this task is primarily assessment-oriented, asking students to demonstrate knowledge of how to determine the congruency of triangles.
• Building a tile pattern by reflecting hexagons
This task applies reflections to a regular hexagon to construct a pattern of six hexagons enclosing a seventh: the focus of the task is on using the properties of reflections to deduce this seven hexagon pattern.
• Building a tile pattern by reflecting octagons
This task applies reflections to a regular octagon to construct a pattern of four octagons enclosing a quadrilateral: the focus of the task is on using the properties of reflections to deduce that the quadrilateral is actually a square.
• GEOMETRY - Congruence (G.CO) - Sec Math I Core Guide
The Utah State Board of Education (USBE) and educators around the state of Utah developed these guides for the Secondary Mathematics I - Congruence (G.CO).
• Geometry in Tessellations
In this lesson students will learn about lines, angles, planes, and experiment with the area and perimeter of polygons.
• Introduction to the Materials (Math 1)
Introduction to the Materials in the Mathematics One of the The MVP classroom experience begins by confronting students with an engaging task and then invites them to grapple with solving it. As students ideas emerge, take form, and are shared, the teacher orchestrates the student discussions and explorations towards a focused mathematical goal. As conjectures are made and explored, they evolve into mathematical concepts that the community of learners begins to embrace as effective strategies for analyzing and solving problems.
• Module 6: Transformations & Symmetry - Student Edition (Math 1)
The Mathematics Vision Project, Secondary Math One Module 6, Transformations and Symmetry, builds on students experiences with rigid motion in earlier grades to formalize the definitions of translation, rotation, and reflection.
• Module 6: Transformations & Symmetry - Teacher Notes (Math 1)
The Mathematics Vision Project, Secondary Math One Module 6 Teacher Notes, Transformations and Symmetry, builds on students experiences with rigid motion in earlier grades to formalize the definitions of translation, rotation, and reflection.
• Module 7: Congruence, Construction & Proof - Student Edition (Math 1)
The Mathematics Vision Project, Secondary Math One Module 7, Congruence, Construction, and Proof, begins by developing constructions as another tool to be used to reason about figures and to justify properties of shapes. Individual constructions are not taught for the sake of memorizing a series of steps, but rather to reason using known properties of shapes such as circles.
• Module 7: Congruence, Construction & Proof - Teacher Notes (Math 1)
The Mathematics Vision Project, Secondary Math One Module 7 Teacher Notes, Congruence, Construction, and Proof, begins by developing constructions as another tool to be used to reason about figures and to justify properties of shapes. Individual constructions are not taught for the sake of memorizing a series of steps, but rather to reason using known properties of shapes such as circles.
• Properties of Congruent Triangles
The goal of this task is to understand how congruence of triangles, defined in terms of rigid motions, relates to the corresponding sides and angles of these triangles.
• Reflections and Equilateral Triangles
This activity is one in a series of tasks using rigid transformations of the plane to explore symmetries of classes of triangles, with this task in particular focusing on the class of equilaterial triangles.
• Reflections and Equilateral Triangles II
This task examines some of the properties of reflections of the plane which preserve an equilateral triangle: these were introduced in ''Reflections and Isosceles Triangles'' and ''Reflection and Equilateral Triangles I''.
• Reflections and Isosceles Triangles
This activity is one in a series of tasks using rigid transformations of the plane to explore symmetries of classes of triangles, with this task in particular focussing on the class of isosceles triangles.
• Tessellations: Geometry and Symmetry
Students can explore polygons, symmetry, and the geometric properties of tessellations in this lesson.
• Translations, Reflections, and Rotations
Students are introduced to the concepts of translation, reflection and rotation in this lesson plan.
• Visual Patterns in Tessellations
In this lesson students will learn about types of polygons and tessellation patterns around us.
• When Does SSA Work to Determine Triangle Congruence?
The triangle congruence criteria, SSS, SAS, ASA, all require three pieces of information. It is interesting, however, that not all three pieces of information about sides and angles are sufficient to determine a triangle up to congruence. In this problem, we considered SSA. Also insufficient is AAA, which determines a triangle up to similarity. Unlike SSA, AAS is sufficient because two pairs of congruent angles force the third pair of angles to also be congruent.
• Why Does ASA Work?
The two triangles in this problem share a side so that only one rigid transformation is required to exhibit the congruence between them. In general more transformations are required and the "Why does SSS work?'' and "Why does SAS work?'' problems show how this works.
• Why does SAS work?
For these particular triangles, three reflections were necessary to express how to move from ABC to DEF. Sometimes, however, one reflection or two reflections will suffice. Since any rigid motion will take triangle ABC to a congruent triangle DEF, this shows the remarkable fact that any rigid motion of the plane can be expressed as one reflection, a composition of two reflections, or a composition of three reflections.
• Why does SSS work?
This particular sequence of transformations which exhibits a congruency between triangles ABC and DEF used one translation, one rotation, and one reflection.
http://www.uen.org - in partnership with Utah State Board of Education (USBE) and Utah System of Higher Education (USHE). Send questions or comments to USBE Specialist - Lindsey Henderson and see the Mathematics - Secondary website. For general questions about Utah's Core Standards contact the Director - Jennifer Throndsen.
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# Recreational Math I: Magic Squares: the “really good” kind – Part 3
Notice that to show the rules for making these kind of magic squares, I used only odd-ordered square matrices as examples. What about matrices of even numbers of rows and columns? The rules for these vary.
This is a small part of a 1514 engraving by Albrecht Durer, called Melancholia. The author of the article that houses this graphic asserts that there are 32 possible 4x4 magic squares with the famous pun "1514" in the same position as above. This magic square has a symmetry in the numbers, as explained below.
The famous Durer magic square, with the year of the engraving cleverly made a part of a magic square, has a certain organization in its construction, as well as a certain symmetry. The numbers are constructed, in sequence:
```_ 3 2 _ _ 3 2 _
_ _ _ _ 5 _ _ 8
_ _ _ _ ====> _ 6 7 _
4 _ _ 1 4 _ _ 1
_ 3 2 _ 16 3 2 13
5 10 11 8 5 10 11 8
9 6 7 12 ====> 9 6 7 12
4 _ _ 1 4 15 14 1```
So, you start from the bottom right and proceed in a horseshoe to the top then the bottom left. The next diagram places the numbers 5-8 in a pattern that is left-to-right u-shape. Then the same u-shape for the numbers 9-12 from left to right, except this time it’s upside-down. Finally ending as we started, the same horseshoe shape (except right side up) from right to left.
Durer’s square has many things about it, apart from its magic number (34) which works on all the attendant diagonals, rows and columns. The middle 4 squares add up to 34 (10 + 11 + 6 + 7 = 34); the four corners add to 34 (16 + 13 + 4 + 1 = 34), and all corner foursomes add to 34: (16 + 3 + 5 + 10); (2 + 13 + 11 + 8); (9 + 6 + 4 + 15); and (7 + 12 + 14 + 1).
The numbers at the ends of the two middle rows add to 34:
```16 3 2 13
5 10 11 8
9 6 7 12
4 15 14 1
```
and the numbers at the tops and bottoms of the two middle columns add to 34:
```16 3 2 13
5 10 11 8
9 6 7 12
4 15 14 1
```
If we take another symmetrical combination: a rightward-slanting rectangle whose corners are 2, 8, 9, and 15, these also add to 34:
```16 3 2 13
5 10 11 8
9 6 7 12
4 15 14 1
```
The leftward-slanting rectangle, whose corners are 5, 3, 12, and 14 also add to 34:
```16 3 2 13
5 10 11 8
9 6 7 12
4 15 14 1```
Starting from 2 and proceeding in an “L”-shape to the left to the number 5, and continuing counter-clockwise in the same manner gets us the corners of a tilted square whose numbers 2, 5, 15, and 12, add to 34:
```16 3 2 13
5 10 11 8
9 6 7 12
4 15 14 1```
Starting from “3” and doing likewise yields the numbers 3, 9, 14, and 8, also adding to 34:
```16 3 2 13
5 10 11 8
9 6 7 12
4 15 14 1```
And what about this talk about “symmetry”? By this, we mean that we may take pairs of numbers at the start and end of any row, and they add up to the same number in a symmetrical place elsewhere. 16 + 3 = 4 + 15, taking the top and bottom of the first and second column. Likewise can be done for the last two columns: 2 + 13 = 14 + 1. The middle two rows have the same property: 5 + 10 = 9 + 6; and 11 + 8 = 7 + 12. On a larger scale, the sums of the middle two rows of columns 1 and 2 are the same as the tops and bottoms of columns 3 and 4: 11 + 8 = 7 + 12 = 16 + 3 = 4 + 15. Likewise, the sums of the middle two rows of columns 3 and 4 are the same as the tops and bottoms of columns 1 and 2: 2 + 13 = 14 + 1 = 5 + 10 = 9 + 6. These two groups of symmetrical numbers are illustrated below in red and green:
```16 3 2 13
5 10 11 8
9 6 7 12
4 15 14 1```
The sums of 15 (green) and 18 (red) across each row form this pattern
```16 3 2 13
5 10 11 8
9 6 7 12
4 15 14 1```
The downward symmetry is also interesting. Here, the sum of 25 is in gold and the sum of 9 is in blue. In the process, we can discern the patterns that we used to construct the square in the first place:
```16 3 2 13
5 10 11 8
9 6 7 12
4 15 14 1```
This is an incredible amount of magic, but if you follow the order of filling (horseshoes are right-to-left, u-shapes are left-to-right, along with the peculiar pattern of filling u’s and horseshoes), there really are four possible patterns that have these “hyper-magic” qualities, but you lose the “1514” idea in two of them:
``` 8 11 10 5 12 7 6 9 4 15 14 1
13 2 3 16 1 14 15 4 12 6 7 9
1 14 15 4 13 2 3 16 5 11 10 8
12 7 6 9 8 11 10 5 16 3 2 13
```
Of course, you could reverse all of the numbers in the rows of the first two squares to get your “1514” back.
Every time I look at that darned Durer square, I keep seeing more patterns. I think there comes a point where one has to leave the remaining observations up to the reader.
There is yet another 4×4 square, and with it we can increase the magic, if that can even be conceivable after all I have said. But there is a square with even more magic than the Durer square. R. J. Reichmann mentioned it in his book “The Fascination of Numbers”, first published in 1957. The square could be constructed like this:
```- - 3 - - - 3 6 - 10 3 6 15 10 3 6
4 - - - ====> 4 5 - - ====> 4 5 - 9 ====> 4 5 16 9
- - 2 - - - 2 7 - 11 2 7 14 11 2 7
1 - - - 1 8 - - 1 8 - 12 1 8 13 12
```
This square has all the magic of the Durer square and then some. One thing this new square has over the Durer square is that any four numbers in square formation will add to 34, from anywhere in the square. These include the foursomes:
```10 3 16 9 11 2 4 5
5 16 2 7 8 13 14 11
```
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A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is:
A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively?
A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's speed against the current is:
A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream.
In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then, A wins by:
A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him by 8 seconds. The speed of B is:
In a 100 m race, A can give B 10 m and C 28 m. In the same race B can give C:
The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
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This Test consists of 10 questions on Speed, Time & Distance for BAT/GRE/GMAT or any other entrance exam. All questions consists of 4 options & there is a time limit of 15 minutes to complete these 10 questions. There will be more upcoming tests so that students can prepare well for there targeted exams. Best of Luck.
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# If 3x + 4 : X + 5 is the Duplicate Ratio of 8 : 15, Find X. - ICSE Class 10 - Mathematics
#### Question
If 3x + 4 : x + 5 is the duplicate ratio of 8 : 15, find x.
#### Solution
(3x + 4)/(x + 5) = (8)^2/(15)^2
⇒ (3x + 4)/(x + 5) -64 /( 225)
⇒ 675x + 900 = 64x + 320
⇒ 611x = -580
⇒ x = - (580)/(611)
Is there an error in this question or solution?
#### APPEARS IN
Selina Solution for Selina ICSE Concise Mathematics for Class 10 (2018-2019) (2017 to Current)
Chapter 7: Ratio and Proportion (Including Properties and Uses)
Ex.7A | Q: 26
#### Video TutorialsVIEW ALL [1]
Solution If 3x + 4 : X + 5 is the Duplicate Ratio of 8 : 15, Find X. Concept: Ratios.
S | 265 | 697 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-30 | longest | en | 0.613833 |
http://science.answers.com/Q/What_is_8_over_24_as_a_percent | 1,544,718,395,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824912.16/warc/CC-MAIN-20181213145807-20181213171307-00429.warc.gz | 253,495,145 | 45,202 | # What is 8 over 24 as a percent?
Would you like to merge this question into it?
#### already exists as an alternate of this question.
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#### exists and is an alternate of .
The Hanky rocks answer is wrong. 1/3 is the fraction, NOT the percentage. As anyone with basic mathematical knowledge will tell you, 1/3, as a percentage, is 33.33... (repeating) %.
1/3
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# What is 100 percent over 8?
100 percent over 8 = 1 /8 or 0.125 = 100% /8 = 1 /8 or 0.125
# What is the percent for 8 over 12?
To work it out, you would do (8/12)x(100). So the answer is 67%. By the way / means divide
# What is 3 over 8 written as a percent?
I believe it's 37.5%. The result of 3 over 8 written as a percent is 37.5 percent. Therational approximation is 3/8.
# How do you turn 5 over 8 into a percent?
Type 5 divided by 8 into your calculator. This should give you 0.625.. Then, you can do one of two things.. You can multiply by one hundred like this: 0.625 times 100 = 62.5% or . You can move the decimal place two places to the right. 0.625 â 62.5. This gives you 62.5 also. . This means t (MORE)
# What percent of 3 over 8 is 5 over 6?
5 over 6 is more than 3 over 8 so it is 222% of it.
# What is 7 over 8 plus 13 over 24?
7/8 + 13/24 = (7*3)/(8*3) + 13/24 = 21/24 + 13/24 = 34/24 = 17/12 or 1 5/12
# What is 8 over 26 as a percent?
Expressed as a percentage, rounded to two decimal places, eight twenty-sixths is equal to 8/26 x 100 = 30.77 percent.
# What is the percent of 8 over 12?
% rate = 66.67% = 8 /12 * 100% = 0.6667 * 100% = 66.67%
# What is 21 over 24 divided by 7 over 8?
21/24 = 7/8 So the question can be reworded as 7/8 divided by 7/8 and since any non-zero number divided by itself is 1, the answer is 1.
# How do you change 7 over 8 to a percent?
you have to try to find how 8 can go into either 10 or 100, but it can't so u must divide 7 by 8, u will come out with a decimal 7/8 8 divided by 7 8 goes into 70 8 times (64) you then have 6 left so add the zero and 8 goes into 60 7 times 60-56= 4 so bring down 0 and 8 goes into 40 5 times (MORE)
# Is 3 over 8 266 percent yes or no?
3 is 37.5% of 8, you are thinking of the percentage of 8 in 3 (266%).
# Find 5 over 8 of 24 equals?
5/8 of 24. First divide 24 x 8 to see what 1/8th is. Then multiply by 5 to get 5/8th's.
# What is the reduced fraction of 8 over 24?
8 over 24=8/24 8 and 24's greatest common factor is 8, so this is equivalent to: (8/8)/(24/8)=1/3
# 24 over 30 to a percent?
24/30 = x/100 cross-multiply: 30x = 2400 divide by 30: x = 80 Answer is 80%
# 8 percent of a number is 24 what is the number?
Let the required number be z, then 8% of z =24 i.e. 8z/100 = 24 On solving we get z = 300 300 is the answer to the question.
# What is the fraction 8 over 25 as a percent?
To convert a fraction (or decimal) to percentage, multiply by 100, thus: 8/25 = 8/25 x 100% = 32% .
# 3 over 8- 5 over 24?
Okay - make both denominators the same, 3/8 = 9/24. 9/24 - 5/24 = 4/24 or 1/6
# What is the percent form of 3 over 8?
To convert a fraction (or decimal) to a percentage, multiply by 100: 3 / 8 = 3 / 8 x 100 % = 37 1 / 2 % (as a fraction) or 37.5% (in decimal form) .
# What is 2 over 8 as a percent?
percentage = 25% % rate: = 2 /8 * 100% = 0.25 * 100% = 25 %
# What is 8 over 1 of a percent?
percentage = 800% % rate: = 8 /1 * 100% = 8 * 100% = 800 %
# Which is bigger 9 over 40 or 24 percent?
9 over 40 = 0.225 = 22.5% So, 9 over 40 is smaller than 24% AKA 24% is bigger than 9 over 40.
# What is 8 over 13 as a percent?
Divide 8 by 13 then move the decimal two places to the right to find percent: 8 ÷ 13 = .6153846 = 61.53846%
# How do you simplify 8 over 24?
Both the numerator (8) and the denominator (24) are evenlydivisible by 8. Divide both by 8 and get: â
# Is 8 over 30 equivalent to 9 over 24?
No. You can solve this by converting both to the lcd (Least common denominator) 4/15=8/30 3/8=9/24 They are not equal, so the two expressions are not equivalent.
# What is 24 over 40 as a percent?
24/40 reduces to 6/10 because the numerator and denominator both are divisible by the number 4. 6/10 translates to .6 in decimal form which is 60%
# What is the fraction 8 over 24 in reducing to the lowest amount?
It's 1/3 (just keep dividing the numerator and denominator by 2, until you can no longer divide it by 2).
# How is 20 over 24 equal to 7 over 8?
It's not. 20 / 24 is 4.76% less than 7 / 8 . 21 / 24 = 7 / 8
# How do you write 8 over 8 as a percent?
The first thing to do when converting into a percentage is to find out what decimal the fraction calculates to. In this case we have 8/8 which is equal to 1. The next step is to note that a percent is that number out of 100. Thus to convert from decimal to percentage you have to multiply by 100. In (MORE)
# What is 18 over 24 as a percent?
18 / 24 = 0.75 Converting decimal to a percentage: 0.75 * 100 = 75%
# Is 7 over 8 greater than or less than 24 over 25?
7 / 8 is less than 24 / 25 . Either convert the fractions to equivalent fractions with a common denominator and compare the numerators, or convert them to decimals and compare directly. I prefer fractions: lcm of 8 & 25 is 200. 7 / 8 = 175 / 200 24 / 25 = 192 / 200 175 < 192 â (MORE)
# What is 22 over 24 minues 3 over 8?
The easiest way to solve fractions is to make them have the same denominator (lower number)... Multiply both the 3 and the 8 by 3, and you get the fraction 9/24. Your sum is now 22 / 24 - 9 / 24 - Now you simply subtract the 9 from 22 which equals 13. Your answer, therefore is 13/24
# What is 5 over 8 into a percent?
5 divided by eight is 0.625 in a decimal. The answer in percent is 62.5%
# What is 24 divided by 8 over 3?
Do you mean 24/(8/3)? Be careful how you write it because 24/(8/3) and (24/8)/3 are two completely different values. We know that when a number is divided by a fraction or a fraction is divided by a number, we can invert the denominator and then multiply that by the numerator. Therefore, 24/(8 (MORE)
# Is 24 percent of 12 greater than 25 percent of 8?
Yes. 24% of 12 = 24% x 12 = 0.24 x 12 = 2.88 25% of 8 = 25% x 8 = 0.25 x 8 = 2 Therefore, 24% of 12 > 25% of 8
# What is 3 over 8 percent as a decimal?
0.375% Hint: Next time you want to find out a decimal as a fraction, simply enter it into Google. That way, you'll be able to get an answer instantly whenever you have a computer and an internet connection available, and you'll be totally clueless when you don't. A lot like your present situation (MORE)
# What is 24 reduced by 8 percent?
24 reduced by 8% = 24 - (8% x 24) = 24 - (0.08 x 24) = 24 - 1.92 = 22.08
# What is 8 over 11 as a percent?
8 over 11 as a percent = 72.73% % rate: = 8 /11 * 100% = 0.7273 * 100% = 72.73%
# How do you turn 8 over 15 into a percent?
It's really very easy. First, 8 over 15 simply means 8 divided by 15. This equates to 0.533. Next, you simplymultiply that by 100 to get the percent: 0.533 x 100 = 53.3%
# What is a 8 over 200 fraction converted as a percent?
% rate = 4%. = 8 /200 * 100%. = 0.04 * 100%. = 4%.
# What should be subtracted from 4 over 8 to get to 5 over 24?
Subtract 7 /24 from 4 /8 to get 5 /24 . 4/ 8 * 3/ 3 = 12/ 24. 12/ 24 - 5/ 24 = 7/ 24.
# What is the reduced fraction of 24 over 8?
reduced fractio n of 24 /8 = 3 /1. 24 divided by 8 = 3. 8 divided by 8 = 1.
# How do you convert 8 over 110 into a percent?
% rate = 7.27% . = 8 /110 * 100%. = 0.0727 * 100%. = 7.27%.
# What should be subtracted from 7 over 8 to get 5 over 24?
Suppose x needs to be subtracted. That is 7/8 - x = 5/24 Rearranging, 7/8 - 5/24 = x so 21/24 - 5/24 = x = 16/24 = 2/3 Suppose x needs to be subtracted. That is 7/8 - x = 5/24 Rearranging, 7/8 - 5/24 = x so 21/24 - 5/24 = x = 16/24 = 2/3 Suppose x needs to be subtracted. That is 7/8 - x = 5/2 (MORE)
# What is 24 over 52 as a percent?
The easiest way to turn a fraction into a percentage is to make the fraction as small as possible. 24/52 in it's smallest form is 6/13. 6/13 = 46.154%
# What is 8 over 24 a fraction?
8 over 24 equals 1 over 3. An easy way to reduce fractions is to look at a multiplication table across the multiples of 1 and 3.
# What is 3 over 24 as a percent?
To make 3/24 into a percent, divide three by twenty four. You then move the decimal point two places to get a number out of 100. The answer is 12.5 percent.
# How do you write 67 over 8 as a percent?
First, divide 67 by 8 to get 8.375. Then multiply that by 100 toget the percent. The answer is " 67 is 837.5% of 8 ".
# Which is bigger 7 over 8 1 over 2 or 24 over 25?
7/8 = 0.875 1/2 = 0.5 24/25 = 0.96 So 24/25 is the biggest of the 3. 7/8 is also bigger - it is bigger than 1/2.
# What is 5 over 8 minus 5 over 24?
The answer to the problem of 5/8 minus 5/24 is 5/12. In decimalform it is 0.41666.
# Is 16 over 24 greater than 4 over 8?
16/24 = 2/3 4/8 = 1/2 1/2 < 2/3 So, yes, 16 over 24 is greater than 4 over 8. | 3,164 | 9,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-51 | latest | en | 0.963259 |
https://www.bartleby.com/questions-and-answers/4-blood-calcium-levels-are-measured-in-mgdl.-in-a-group-of-patients-over-30-you-find-the-following-m/e153b4b8-28cc-43fd-ba1a-6d2b47120003 | 1,582,936,663,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148163.71/warc/CC-MAIN-20200228231614-20200229021614-00277.warc.gz | 644,052,050 | 29,388 | # 4) Blood calcium levels are measured in mg/dL. In a group of patients over 30 you find the following: μ = 9.3 mg/dL, and σ = 2.35 mg/dL.a) Give the blood calcium values for the middle 50% of patients. (Hint/comment: if you want the middle 50%, how many percent go in each tail?)b) Give the blood calcium values for the middle 60% of patients.c) Give the 80th percentile.d) Are you surprised by the answers to (b) and (c)? Explain. If you're not sure what's going on, draw some pictures of the normal curves.
Question
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4) Blood calcium levels are measured in mg/dL. In a group of patients over 30 you find the following: μ = 9.3 mg/dL, and σ = 2.35 mg/dL.
a) Give the blood calcium values for the middle 50% of patients. (Hint/comment: if you want the middle 50%, how many percent go in each tail?)
b) Give the blood calcium values for the middle 60% of patients.
c) Give the 80th percentile.
d) Are you surprised by the answers to (b) and (c)? Explain. If you're not sure what's going on, draw some pictures of the normal curves.
check_circle
Step 1
It is provided that the sample size, n is 30, mean is 9.3mg/dL and the standard deviation is 2.35 mg/dL.
Step 2
(a) The 50% in the middle leaves the other 25% in each tail, that is, 25th and 75th percentile. The z-value is obtained using Excel function =NORMSINV (percentile). Therefore, the corresponding z-values are obtained as ±0.6745 respectively.
Hence, the middle 50% of the patient’s blood calcium values is,
Step 3
(b) The 60% in the middle leaves the other 20% in each tail, that is, 20th and 80th percentile. The z-value is obtained using Excel function =NORMSINV (percentile). Therefo...
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Prob 1.2 8/18/04 PROBLEM 1.2 PROBLEM STATEMENT: If an object weighs 170 lbf on Earth, find its mass in lbm and kg and its weight in Newtons. GIVEN: The object’s weight in English units (170 lbf). FIND: The mass in lbm and kg and the weight in Newtons. ASSUMPTIONS: g= 32.2 ft/s 2 = 9.8 m/s 2 GOVERNING RELATIONS: . c c 22 mg W g lbm ft kg m g3 2 2 1 sl b f sN = ⋅⋅ == PROPERTY DATA: Not applicable QUANTITATIVE SOLUTION: . . 2 c 2 c Wg mg 170lbf 32 2lbm ft s lbf Wm 1 7 0 gg 32 2ft s =⇒ = = = l b m . ./ .. . 2 2 c 170lbm m7 7 1 1 k g 2 2046lbm kg mg 77 11kg 9 8m s W7 g 1kg m s N = 5 5 7 N DISCUSSION OF RESULTS: This problem illustrates the importance of using the units conversion factor g c . Note that, even though g c has the numerical value of 1 in SI units, the units of weight would not be correct if one just wrote W=mg. In English units, g c has a numerical value of 32.2, but its units are not those of g (m/s 2 ). Always make a clear distinction between g and g c and their units. PROBLEM 1.4 PROBLEM STATEMENT: A spacecraft of dry weight 50,000 lbf leaves Earth with 180,000 lbf of fuel on board flies to a planet where the acceleration due to gravity is 12 ft/s 2 . During the flight to the planet, 2/3 of the fuel is consumed. How much thrust does the rocket need to insure lift-off from the planet? DIAGRAM DEFINING SYSTEM AND PROCESS: f W = -120,000 lbf W r = 50,000 lbf W f = 180,000 lbf
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Prob 1.2 8/18/04 GIVEN: W r = 50,000 lbf, W f = 180,000 lbf, a=12 ft/s 2
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#### Resources tagged with Cubes similar to Next Size Up:
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### There are 36 results
Broad Topics > 3D Geometry, Shape and Space > Cubes
### Next Size Up
##### Stage: 2 Challenge Level:
The challenge for you is to make a string of six (or more!) graded cubes.
### Dicey
##### Stage: 2 Challenge Level:
A game has a special dice with a colour spot on each face. These three pictures show different views of the same dice. What colour is opposite blue?
### Cube Drilling
##### Stage: 2 Challenge Level:
Imagine a 4 by 4 by 4 cube. If you and a friend drill holes in some of the small cubes in the ways described, how many will not have holes drilled through them?
### Three Cubed
##### Stage: 2 Challenge Level:
Can you make a 3x3 cube with these shapes made from small cubes?
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##### Stage: 2 Challenge Level:
Imagine a 3 by 3 by 3 cube made of 9 small cubes. Each face of the large cube is painted a different colour. How many small cubes will have two painted faces? Where are they?
### A Puzzling Cube
##### Stage: 2 Challenge Level:
Here are the six faces of a cube - in no particular order. Here are three views of the cube. Can you deduce where the faces are in relation to each other and record them on the net of this cube?
### Four Layers
##### Stage: 1 and 2 Challenge Level:
Can you create more models that follow these rules?
### Cubic Conundrum
##### Stage: 2 Challenge Level:
Which of the following cubes can be made from these nets?
### Holes
##### Stage: 1 and 2 Challenge Level:
I've made some cubes and some cubes with holes in. This challenge invites you to explore the difference in the number of small cubes I've used. Can you see any patterns?
### Triple Cubes
##### Stage: 1 and 2 Challenge Level:
This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.
### Green Cube, Yellow Cube
##### Stage: 2 Challenge Level:
How can you paint the faces of these eight cubes so they can be put together to make a 2 x 2 cube that is green all over AND a 2 x 2 cube that is yellow all over?
### Construct-o-straws
##### Stage: 2 Challenge Level:
Make a cube out of straws and have a go at this practical challenge.
##### Stage: 2 Challenge Level:
Make a cube with three strips of paper. Colour three faces or use the numbers 1 to 6 to make a die.
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##### Stage: 2 Challenge Level:
How many models can you find which obey these rules?
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##### Stage: 3 Challenge Level:
In the game of Noughts and Crosses there are 8 distinct winning lines. How many distinct winning lines are there in a game played on a 3 by 3 by 3 board, with 27 cells?
### Drilling Many Cubes
##### Stage: 3 Challenge Level:
A useful visualising exercise which offers opportunities for discussion and generalising, and which could be used for thinking about the formulae needed for generating the results on a spreadsheet.
### Changing Areas, Changing Volumes
##### Stage: 3 Challenge Level:
How can you change the surface area of a cuboid but keep its volume the same? How can you change the volume but keep the surface area the same?
### Counting Triangles
##### Stage: 3 Challenge Level:
Triangles are formed by joining the vertices of a skeletal cube. How many different types of triangle are there? How many triangles altogether?
### Nine Colours
##### Stage: 3 and 4 Challenge Level:
You have 27 small cubes, 3 each of nine colours. Use the small cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of every colour.
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##### Stage: 3 Challenge Level:
A 3x3x3 cube may be reduced to unit cubes in six saw cuts. If after every cut you can rearrange the pieces before cutting straight through, can you do it in fewer?
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##### Stage: 3 Challenge Level:
Find all the ways to cut out a 'net' of six squares that can be folded into a cube.
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##### Stage: 3 Challenge Level:
Imagine you are suspending a cube from one vertex (corner) and allowing it to hang freely. Now imagine you are lowering it into water until it is exactly half submerged. What shape does the surface. . . .
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##### Stage: 2 and 3
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##### Stage: 1, 2 and 3
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
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Here are four cubes joined together. How many other arrangements of four cubes can you find? Can you draw them on dotty paper?
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Investigate the number of faces you can see when you arrange three cubes in different ways.
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### How Many Dice?
##### Stage: 3 Challenge Level:
A standard die has the numbers 1, 2 and 3 are opposite 6, 5 and 4 respectively so that opposite faces add to 7? If you make standard dice by writing 1, 2, 3, 4, 5, 6 on blank cubes you will find. . . .
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Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube made from 27 unit cubes so that the surface area of the remaining solid is the same as the surface area of the original 3 by 3 by 3. . . .
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##### Stage: 2, 3 and 4
A description of how to make the five Platonic solids out of paper.
### Painting Cubes
##### Stage: 3 Challenge Level:
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In a three-dimensional version of noughts and crosses, how many winning lines can you make?
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Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
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##### Stage: 3 and 4 Challenge Level:
Toni Beardon has chosen this article introducing a rich area for practical exploration and discovery in 3D geometry | 1,698 | 7,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-04 | longest | en | 0.917024 |
https://www.chegg.com/homework-help/algebra-2-student-edition-1st-edition-chapter-ep12.3-solutions-9780078279997 | 1,558,630,071,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257316.10/warc/CC-MAIN-20190523164007-20190523190007-00190.warc.gz | 725,511,799 | 45,588 | Solutions
Algebra 2, Student Edition
# Algebra 2, Student Edition (1st Edition) Edit edition Solutions for Chapter EP12.3
We have solutions for your book!
Chapter: Problem:
Step-by-step solution:
Chapter: Problem:
• Step 1 of 4
Suppose a jar contains 3 red, 4 green, and 5 orange marbles and three marbles are drawn at random without replacing.
It is required to find.
• Step 2 of 4
There are 12 marbles in total. Choose 3 marbles from these 12 marbles.
Since the order is not important of choosing marbles. Therefore, 3 marbles from 12 marbles can be chosen in ways.
So total no. of ways
Since, therefore,
Therefore, 3 marbles from 12 marbles can be chosen in ways.
• Step 3 of 4
Choose 3 green marbles from 4 available green marbles.
This can be chosen inways.
So no. of ways of choosing 3 green
Since, therefore,
Therefore, 3 green marbles from 4 available green marbles can be chosen in 4 ways.
• Step 4 of 4
Then,is given by,
Therefore, the required probability is.
Corresponding Textbook
Algebra 2, Student Edition | 1st Edition
9780078279997ISBN-13: 0078279992ISBN: | 304 | 1,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-22 | latest | en | 0.833316 |
https://www.physicsforums.com/threads/analyzing-convergence-of-1-n-ln-n-p-and-1-n-ln-n-ln-ln-n-p.143839/ | 1,716,197,397,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058254.21/warc/CC-MAIN-20240520080523-20240520110523-00237.warc.gz | 835,636,567 | 15,900 | # Analyzing Convergence of 1/n(ln n)^p and 1/((n(ln n)(ln (ln n)))^p
• jkh4
In summary, the question is for what value of p>0 does the series 1/(n(ln n ) ^p) and 1/((n(ln n)(ln (ln n))))^p converges and for what value does it diverges, with a=1 and b=infinity for the integral signs. The integral test and ratio test can be used to determine the convergence or divergence of the series, with the criterion \sum a_n converges iff \sum 2^n a_{2^n} converges being particularly useful for series involving ln(n). It is also important to note that the series should be monotonically decreasing for this method to work.
jkh4
Hi, I just wondering if someone can help me on this question.
1) 1/(n(ln n ) ^p)
2) 1/((n(ln n)(ln (ln n))))^p
Both are b = infinity and a = 1 for the integral signs.
The question is for what value of p>0 does the series converges and for what value does it diverges.
Thank you !
There was a similar question in the HW section recently. Use the integral test.
Do you know about the criterion $\sum a_n$ converges iff $\sum 2^n a_{2^n}$ converges? It is usefull for series involving ln(n) because ln(2^n)=nln(2) !
Then, for the values of p, use the ratio test (a_{n+1}/a_n).
Last edited:
quasar987 said:
Do you know about the criterion $\sum a_n$ converges iff $\sum 2^n a_{2^n}$ converges?
It's important to keep in mind this only works when the series is monotonically decreasing (which this one is). Otherwise, just take the terms to be all 1's except for 0's when n is a power of 2.
## Question 1: What is the purpose of analyzing the convergence of 1/n(ln n)^p and 1/((n(ln n)(ln (ln n)))^p)?
The purpose of analyzing the convergence of these two sequences is to understand their behavior as n approaches infinity. This information can be useful in various fields of science, such as statistics, physics, and computer science.
## Question 2: How do you determine the convergence of a sequence?
To determine the convergence of a sequence, we look at its limit as n approaches infinity. If the limit exists and is a finite number, the sequence is said to be convergent. If the limit does not exist or is infinite, the sequence is said to be divergent.
## Question 3: What is the relationship between the values of p and the convergence of 1/n(ln n)^p and 1/((n(ln n)(ln (ln n)))^p)?
The values of p play an important role in determining the convergence of these two sequences. If p is greater than 1, both sequences will converge. If p is equal to or less than 1, both sequences will diverge. Additionally, as p increases, the rate of convergence also increases.
## Question 4: How do you analyze the convergence of 1/n(ln n)^p and 1/((n(ln n)(ln (ln n)))^p)?
To analyze the convergence of these sequences, we can use various mathematical techniques such as the ratio test, comparison test, or the integral test. These tests help us determine the behavior of the sequence as n approaches infinity.
## Question 5: What are the practical applications of understanding the convergence of 1/n(ln n)^p and 1/((n(ln n)(ln (ln n)))^p)?
Understanding the convergence of these sequences can have practical applications in fields such as data analysis, optimization problems, and computational algorithms. It can also help in predicting the behavior of a system or process as it approaches a large number of iterations.
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#### Resources tagged with Combinatorics similar to Square Ratio:
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Broad Topics > Decision Mathematics and Combinatorics > Combinatorics
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##### Stage: 4 Challenge Level:
To win on a scratch card you have to uncover three numbers that add up to more than fifteen. What is the probability of winning a prize? | 2,114 | 8,610 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2016-40 | longest | en | 0.896715 |
https://math.stackexchange.com/questions/3468789/modifications-in-the-schwarz-christoffel-formula | 1,716,133,307,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00572.warc.gz | 336,853,322 | 35,281 | # Modifications in the Schwarz-Christoffel formula
I am reading Ahlfors's Complex Analysis book, chapter $$6.2.2$$; The Schwarz-Christoffel Formula, and I have some questions in the exercises.
Firstly, the Schwarz-Christoffel formula is given as follows:
Theorem. The functions $$z=F(w)$$ which map the unit disk $$|w|<1$$ conformally onto polygons with (interior) angles $$\alpha_k \pi$$ ($$k=1,...,n; 0<\alpha_k<2$$) are of the form $$F(w)=C \int _0 ^w (w-w_1)^{-\beta_1}...(w-w_n)^{-\beta_n} dw +C'$$, where $$\beta_k=1-\alpha_k$$, the $$w_k$$ are points on the unit circle, and $$C,C'$$ are constants.
$$(1)$$ Show that the $$\beta_k$$ may be allowed to be $$-1$$. What is the geometric interpretation?
$$(2)$$ If the vertex of the polygon is allowed to be at $$\infty$$, what modification does the formula undergo? If in this context $$\beta_k=1$$, what is the polygon like?
In $$(1)$$, if the $$\beta_k=-1$$, then what do I have to justify? Also, if $$\beta_k=1$$ then $$\alpha_k=2$$ so the polygon would have a slit; is this the desired geometric interpretation?
Next, in $$(2)$$, I don't have any idea of modification. If $$\beta_k=1$$ in this case, then I see that the polygon would be a half-plane. Am I right?
I am having a hard time with these. Any help will be appreciated. Thanks!
## 1 Answer
You're correct about the first question. The formula still works for $$\alpha_k = 2 \pi$$ and gives two fully or partially overlapping segments as part of the boundary.
If $$\alpha_k \leq 0$$, then $$w_k$$ is mapped to $$\infty$$ since $$(w - w_k)^{\alpha_k/\pi - 1}$$ is a non-integrable singularity. If the angle between two lines at $$\infty$$ is defined as minus the angle at their finite intersection point, the formula also works for polygons with a vertex (or several vertices) at $$\infty$$.
Since we have to distinguish between inner and outer angles, the angle at $$\infty$$ for parallel rays (a $$\Pi$$-shaped boundary) is either $$0$$ or $$-2 \pi$$. The angle at $$\infty$$ for antiparallel rays (a $$Z$$-shaped boundary) is $$-\pi$$.
All this is exactly the same for the mapping from the upper half-plane to a polygon, but in that case we can eliminate one of the factors by choosing $$w_k = \infty$$. | 650 | 2,234 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 36, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-22 | latest | en | 0.873671 |
http://www.educator.com/mathematics/geometry/pyo/tangents.php | 1,508,523,163,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824293.62/warc/CC-MAIN-20171020173404-20171020193404-00404.warc.gz | 425,899,998 | 68,706 | Start learning today, and be successful in your academic & professional career. Start Today!
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### Tangents
• Tangent Theorems:
• If a line is a tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency
• In a plane, if a line is perpendicular to a radius of a circle at the endpoint on the circle, then the line is a tangent of the circle
• Know what common external tangents and common internal tangents are using geometric pictures
• Tangent segments: If two segments from the same exterior point are tangent to a circle, then they are congruent
• Circumscribed Polygons: A polygon is circumscribed about a circle if each side of the polygon is tangent to the circle
### Tangents
Determine whether the following statement is true or false.
If AB ⊥CD, B is on circle A, then CD is the tangent of circle A.
True.
Find a common internal tangent and a common external tangent.
• Internal tangent:AB
• External tangent:CD
Internal tangent:AB
External tangent:CD
AB and CD are internal tangents, write two pairs of congruent segments.
OD ≅ OA , OB ≅ OC .
Determine whether the following statement is true or false.
A pentagon is circumscribed about a cirle if each side of the pentagon is tangent to the circle.
True.
Determine whether the following statement is true or false.
A tangent of a circle is always perpendicular to the radius drawn to the point of tangency.
True.
CD and BD are tangents of circle A, AC = 3, AD = 5, find BD.
• AC ⊥BD
• CD = √{AD2 − AC2} = 4
• CD ≅ BD
BD = CD = 4.
CG = 18, EF = 10, find the perimeter of ∆CEG.
• GF = BG, BC = CD, DE = EF
• C = CB + CD + DE + EF + GF + BG
• C = 2CB + 2EF + 2BG
• C = 2CG + 2EF
• C = 2*18 + 2*10
C = 56.
Determine whether the following statement is true or false.
No common internal tangent can be drawn for two concentric circles.
True
Determine whether the following statement is true or false.
If two circles are concentric, at least one common external tangent can be drawn.
False
Fill in the blank with sometimes, never or always.
For a circle, the area of its inscribed polygon is _____ smaller than the area of its circumscribed polygon.
Always
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
### Tangents
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Tangent Theorems 0:04
• Tangent Theorem 1
• Tangent Theorem 1 Converse
• Common Tangents 1:34
• Common External Tangent
• Common Internal Tangent
• Tangent Segments 3:08
• Tangent Segments
• Circumscribed Polygons 4:11
• Circumscribed Polygons
• Extra Example 1: Tangents & Circumscribed Polygons 5:50
• Extra Example 2: Tangents & Circumscribed Polygons 8:35
• Extra Example 3: Tangents & Circumscribed Polygons 11:50
• Extra Example 4: Tangents & Circumscribed Polygons 15:43
### Transcription: Tangents
Welcome back to Educator.com.0000
For this next lesson, we are going to go over tangents.0001
Now, remember: tangents are lines that intersect the circle at exactly one point.0006
If a line is tangent to a circle, and there is a radius that is also touching that same point, then the radius and this tangent are perpendicular.0013
This point that they meet at--this is called the point of tangency.0030
If the radius and a tangent meet at that point, then they are perpendicular.0047
In a plane, if a line is perpendicular to a radius of a circle with endpoint on the circle, then the line is a tangent of the circle.0056
So, it is just the converse of this theorem.0063
So again, a tangent, we know, touches the circle at one point; and if the radius is also right there,0066
at the point of tangency, then those two are perpendicular.0075
And then, the converse is that, if a line is perpendicular to a radius of the circle, then the line is a tangent.0080
Common tangents: now again, a tangent has to be intersecting the circle at one point.0096
We have the same tangent touching two different circles; it is intersecting two different circles;0103
this tangent right here is intersecting this circle at this point, and intersecting this circle at this point; then that is a common tangent,0111
because two circles are sharing the same tangent, so it is a common tangent.0118
Now, a common tangent can be either external or internal; we know that external means outside, and internal means inside.0124
So, when the common tangent (the tangent that the two circles share) is on the outsides of the two circles, then it is an external tangent.0132
This is another one; this one is also on the outside, so it is a common external tangent.0144
And then, the shared tangents for this one are internal, because it is crossing through between them.0151
See how there is nothing here; it is just on the outside, just making a wall with them.0160
But here, it is crossing in between them; that is internal, in between--one on that side, and then the other one on the other side of the circle.0168
These are common internal tangents; external tangents, and common internal tangents.0180
Tangent segments: here is a tangent that is intersecting the circle at this point, point B;0191
and this tangent is intersecting at point C; so we have two tangents.0197
And those tangents intersect at a point outside the circle, right there; they intersect at point A.0205
Then, this tangent segment and this tangent segment are congruent.0216
It can't be the whole tangent, because it is going on forever; this segment, from that point of tangency0227
to the point where they intersect--that part, that segment right there, is going to be congruent to this segment right here for this tangent.0235
Remember that they are congruent.0249
Circumscribed polygons: we learned about inscribed polygons; inscribed polygons are when you have a polygon0254
that is inside the circle, with all of the vertices, all of the endpoints, touching the circle.0261
But this one is on the outside; "circumscribed" means that a polygon is on the outside of the circle,0267
so that each side of the polygon is tangent to the circle.0278
See, look at this one: this is tangent to this, because it is intersecting the circle at one point.0283
This side is tangent to this circle; tangent, tangent, tangent.0289
That is circumscribed polygons; I can also say that this circle is inscribed in the pentagon, or the pentagon is circumscribed about the circle.0296
Just remember that "inscribed" is inside; so whatever you say is inside--you have to use the word "inscribed."0311
The circle is inscribed in the pentagon, and the pentagon is circumscribed about the circle.0319
The same thing here: we have a triangle; we have a circle that is inscribed; or I can say that the triangle is circumscribed about the circle.0330
And again, the sides of each polygon are tangent to the circle.0340
OK, our examples: the first one: Triangle ABC is circumscribed about the circle; if the perimeter of the triangle is 80, find DC.0354
You are probably going to get a lot of problems like this, where you are going to have circumscribed polygons; and they all have to do with tangents.0365
If you look at this right here, this side of this triangle, it is tangent to this circle, because it is intersecting the circle at that point.0379
The same thing happens here: it is intersecting at point E and intersecting at point D.0389
All of these sides are tangent to the circle; now, that one theorem that says that,0395
if you have two tangents that intersect at an exterior point (we know that this right here and this right here0403
are two tangents of the circle that intersect at point B; therefore) these are congruent.0414
And then, the same thing happens here: this tangent and this tangent segment intersect at point A; therefore, this part and this part are congruent.0422
And then, the same thing happens for those; so for this here, it is like we have three pairs of congruent segments.0434
And so, if they give us the perimeter (they tell us that the perimeter is 80), well, if this side is 8, then this side also has to be 8;0448
if this side is 12, then this side has to be 12; if this side is x, then this side also has to be x.0459
And then, to find the perimeter, we know that we have to just add up all of the sides;0470
so then, 8 + 8 is 16, plus 12 + 12 is 24, plus x + x is 2x; that is all going to add up to 80.0472
This right here is 40, plus 2x equals 80; 2x...we will subtract 40, and then divide the 2; so x is 20.0489
And what are they asking for? DC...well, that is x; so I can say DC is 20.0504
OK, again, they want us to find DC; we have a tangent here; AC is a tangent, so this is the point of tangency.0518
And the radius is also at the point of tangency right there--the endpoint of that radius.0534
Therefore, this radius and this tangent are perpendicular; that was the first theorem that we went over.0541
This radius and this tangent are perpendicular, because they are both intersecting at the point of tangency.0552
Well, if this is a right angle, then I see here that I have a right triangle.0560
They are asking for DC, but I want to first find BC; I am going to find BC first, because I know that BC is a side of the triangle,0567
and I can use a triangle to find the unknown side.0578
And then, from there, I can look for DC.0582
In order to find the missing side of a right triangle, I use the Pythagorean theorem.0586
This would be 52 + 122 =...let's make that BC2.0591
5 squared is 25, plus 144, equals BC squared; this is 169 equals BC squared; therefore, BC is 13.0602
Now, if BC, this whole thing, is 13, and I just want to find DC, well, do I know BD?0622
If I know BD, then I can just subtract that from 13 and get DC; but how do I find DB?0633
Well, BA is a radius with a measure of 5; isn't BD also a radius?0641
If BD is a radius, and we know that all radii have the same measure, if this is 5, then BD has to be 5.0650
Then, I just subtract 13 from 5, and I get 8; so if this is 5, then this has to be 8; I can say that DC is 8.0658
So again, the tangent line and the radius are perpendicular, because they meet at the point of tangency.0675
That gives me a right triangle; and then, I use the Pythagorean theorem to find the whole unknown side, BC.0682
If they ask me for BC, then that would be the answer; but they are asking for DC.0692
So, I found that BD, since that is a radius, is 5, the same as this; and then, subtract it from the whole thing, and I get 8 as DC.0697
OK, the next one: Find the value of x.0712
Now, for this one, here is a tangent; here is a tangent; and then, here is a segment0716
from the center of the circle all the way out to that external point, that point of intersection.0728
Then, here is the radius; all I need to look for is x.0736
I know that this tangent and this tangent are congruent, because if a tangent and a tangent meet at the same point, then they are congruent.0745
So, I can just make this and this equal to each other.0757
This segment and this radius mean nothing; they don't mean anything to me; that is all you need.0759
Now, you might get a problem similar to this, where they are asking for this segment here.0765
Or they give you this segment, and they ask for the tangent segment.0774
So, in that case, you can make this radius meet at that point of tangency so that it will be perpendicular.0778
And then, this radius will have a measure of 6, and then you can just work with that there.0789
But for this problem, we don't need that; here, we just need to make these tangents equal to each other.0794
It is going to be 2x - 7 = x + 3; to solve, I can subtract the x, so that would be x; I can add the 7, so x = 10.0800
And then, for this one, the radius is 12; they are asking for this whole side of the square.0821
We know that it is a square, because all sides are perpendicular, and each side is tangent to the circle, so it has to be a square.0829
Now, you can think of this two ways: if you just think of this in a very simple way,0844
this side is the same as the diameter of a circle, because it is from one end to the other end, and that is x.0853
If the radius is 12, then we know that the whole thing, the diameter, is 24.0863
You can also look at this as tangents; and I am explaining this both ways, even though we know that that could be the easiest way to solve,0868
because you might have different versions of this kind of problem, where you have a square circumscribed about a circle.0878
And just keep in mind that these are all tangent; that is tangent here, and so each of these are going to be congruent to each other.0887
Then, remember that tangent segments that meet on an external point are congruent to the same thing here:0896
tangent, tangent, congruent, congruent, congruent, congruent.0901
Whatever you need to be able to find whatever is that that problem is asking for...0906
For our problem here, I can just make x become 24, because if this is 12, then this is 12, and the whole thing is 24.0914
If that is x, then this also is going to be x, so x here is 240928
And the fourth example: A regular hexagon is circumscribed about a circle; the radius is 10; find the measure of each side.0945
"Regular" means that all of the sides of the hexagon are congruent, and all of the angles are congruent.0956
It is equilateral, and it is equiangular.0968
Now, if all of the sides are congruent, and we know that they are all tangent to the circle, then, first of all,0973
let's draw the radius first, and then we will go on from there: if I am going to draw a radius, I could draw it anywhere.0988
They are all going to be 10; but I want to draw it so that it is to the point of tangency, because that helps me out.0999
We learned a theorem on that today; so I want to just draw the radius like that--and what do we know about that?1005
This radius and this tangent segment are perpendicular; and this is 10.1016
Now, I am looking for this whole side; so let's see, that is all we have to work with; all that is given is the radius.1023
So, how would we solve this problem? Well, since you know that it is a regular hexagon, you know that each part is broken up1038
into congruent parts, into congruent segments or congruent sections of the circle.1053
So, if I were to just draw out each radius to each of the points of tangency, then what would each angle measure be?1060
I have this right here; I have the radius; the radius has a measure of 10; I don't have anything else.1088
So, in that case, I know that this is a right angle; but still, if I want to use the Pythagorean theorem, I still need a second side.1096
So, that one chapter on right triangles...if you want to use the Pythagorean theorem, you need two of the three sides.1105
If we don't have another side, then we have to have an angle measure, at least.1117
So, I can use this circle to find my angle measures, since they are all divided up into equal sections of the circle, because it is a regular hexagon.1124
Each angle measure is going to be 360 (because that is the whole thing), divided by 1, 2, 3, 4, 5, 6.1137
So, 360 degrees, divided by 6, is going to be 60; that means that this is 60; that means that this is 60 degrees; 60 degrees; 60 degrees; and so on.1147
If this is 60, now, that doesn't help me too much; it helps me, but then I need a triangle.1164
So, I am going to draw, from this point right here, another segment to the middle, like that; that way, I have a right triangle.1173
I'll draw this triangle right here; this is the triangle I am going to work with.1187
Now, we found that this whole thing is 60--not that each of these is 60, but the whole thing; this was each of these sections.1198
This is 60; then, this has to be 30; now, if I want to draw that triangle out again, just so that it is easier to see,1206
it is a right triangle; this is 30 degrees; this is 10; what is this angle measure here? This is 60.1223
Now, if you want, you can go ahead and use Soh-cah-toa; and we know that we use Soh-cah-toa when we have angles and sides.1233
The Pythagorean theorem we only use when we have sides; that is it--only sides--nothing to do with angles.1244
But it is all for right triangles, of course, but only when it comes to sides.1250
Soh-cah-toa, we use when we have a combination of angles and sides.1256
We can use Soh-cah-toa; now, we are looking for this right here, x, because if we find this,1261
then we can just multiply that by 2, and we will find the whole side.1268
Or we can use special right triangles; now, special right triangles are either 45-45-90 triangles or 30-60-90 triangles.1274
If you have a 30-60-90 triangle, this is n; the side opposite the 60-degree angle is going to be n√3;1294
and then, the side opposite the 90 is going to be 2n.1307
This, the side opposite 30, is going to be n; the side opposite the 60 is going to be n√3; and this is going to be 2n.1315
Now, we want to look for this here; what are we given?1330
We are given that the side opposite the 60 is 10, so that means I can make 10 equal to n√3.1334
Again, in a 30-60-90 triangle, the side opposite the 30 is n; the side opposite the 60 is n√3; and the side opposite the 90 is 2n.1347
This is the rule for special right triangles; now, what is given?--this side right here, the side opposite the 60; they gave us that it is 10.1358
So, in that case, you can make 10 equal to n√3, and that will help you; and then, you just solve for n.1367
So, here I divide this by √3; n = 10/√3.1375
Now, here, we have a radical in the denominator; I don't like to have (you shouldn't have) a radical in the denominator.1384
So, you want to go ahead and rationalize it; that means that we want to make it so that that denominator doesn't have the square root.1395
I can multiply this by itself, √3/√3; now, I have to multiply the top and the bottom by √3,1404
because this √3/√3 becomes just 1; this is 1; √3/√3 = 1.1415
So, I have to multiply this by this, so that the radical will go away.1423
This will become 10√3/3; so, this side right here, n, is 10√3/3, which is also x; that is this side.1430
Now, I want to find the measure of each side; that means that each side is this whole thing right here.1450
I can take 10√3/3, and I am going to multiply it by 2 times 2; and then, I am going to get 20√3/3, and that is my answer.1457
Now, if you want, you can use your calculator; you could just make that into a decimal...or that should be fine, too.1479
Again, just to explain how we did the problem: it is a regular hexagon, so I know that,1488
if I draw a radius to each point of tangency, then the circle will be divided up into equal congruent angle measures,1499
which makes each of these 60; this whole thing is 60.1509
Then, since I want a triangle, because with triangles, I have a lot to work with; I drew a segment from this point all the way to the center.1514
And then, since this whole thing is 60, and this is 30, this became a 30-60-90 triangle, which is also a special right triangle.1529
The side opposite 30 is n; opposite 60 is n√3; opposite 90 is 2n.1540
So then, I want to look for the side opposite the 30; this is given to me, so make that equal to n√3; solve for n.1546
That is this side right here; multiply that by 2 to get this whole side, and that is 20√3/2.1560
That is it for this lesson; thank you for watching Educator.com.1572 | 5,401 | 19,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2017-43 | latest | en | 0.806695 |
http://stackoverflow.com/questions/2151162/find-the-most-points-enclosed-in-a-fixed-size-circle?answertab=active | 1,433,312,796,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1433195036630.10/warc/CC-MAIN-20150601214356-00050-ip-10-180-206-219.ec2.internal.warc.gz | 171,778,664 | 20,016 | # Find the most points enclosed in a fixed size circle
This came up when a friend talked about a programming competition, and we wondered what the best approach was:
Given a list of points, find the centre of a circle of predetermined size that covers the most points. If there are several such circles, its only important to find one of them.
Example input: 1000 points, in a 500x500 space, and a circle of 60 diameter.
-
My best approach so far is:
Every circle containing points must have a left-most point. So it makes a list of all the points to the right of a point that are potentially within the bounds of a circle. It sorts the points by x first, to make the sweep sane.
It then sorts them again, this time by the number of neighbours to the right that they have, so that the point with the most neighbours get examined first.
It then examines each point, and for each point to the right, it computes a circle where this pair of points is on the left perimeter. It then counts the points within such a circle.
Because the points have been sorted by potential, it can early-out once it's considered all the nodes that might potentially lead to a better solution.
``````import random, math, time
from Tkinter import * # our UI
def sqr(x):
return x*x
class Point:
def __init__(self,x,y):
self.x = float(x)
self.y = float(y)
self.left = 0
self.right = []
def __repr__(self):
return "("+str(self.x)+","+str(self.y)+")"
def distance(self,other):
return math.sqrt(sqr(self.x-other.x)+sqr(self.y-other.y))
def equidist(left,right,dist):
u = (right.x-left.x)
v = (right.y-left.y)
if 0 != u:
r = math.sqrt(sqr(dist)-((sqr(u)+sqr(v))/4.))
theta = math.atan(v/u)
x = left.x+(u/2)-(r*math.sin(theta))
if x < left.x:
x = left.x+(u/2)+(r*math.sin(theta))
y = left.y+(v/2)-(r*math.cos(theta))
else:
y = left.y+(v/2)+(r*math.cos(theta))
else:
theta = math.asin(v/(2*dist))
x = left.x-(dist*math.cos(theta))
y = left.y + (v/2)
return Point(x,y)
class Vis:
def __init__(self):
self.frame = Frame(root)
self.canvas = Canvas(self.frame,bg="white",width=width,height=height)
self.canvas.pack()
self.frame.pack()
self.run()
def run(self):
self.count_calc0 = 0
self.count_calc1 = 0
self.count_calc2 = 0
self.count_calc3 = 0
self.count_calc4 = 0
self.count_calc5 = 0
self.prev_x = 0
self.best = -1
self.best_centre = []
for self.sweep in xrange(0,len(points)):
self.count_calc0 += 1
if len(points[self.sweep].right) <= self.best:
break
self.calc(points[self.sweep])
self.sweep = len(points) # so that draw() stops highlighting it
print "BEST",self.best+1, self.best_centre # count left-most point too
print "counts",self.count_calc0, self.count_calc1,self.count_calc2,self.count_calc3,self.count_calc4,self.count_calc5
self.draw()
def calc(self,p):
for self.right in p.right:
self.count_calc1 += 1
if (self.right.left + len(self.right.right)) < self.best:
# this can never help us
continue
self.count_calc2 += 1
assert abs(self.centre.distance(p)-self.centre.distance(self.right)) < 1
count = 0
for p2 in p.right:
self.count_calc3 += 1
if self.centre.distance(p2) <= radius:
count += 1
if self.best < count:
self.count_calc4 += 4
self.best = count
self.best_centre = [self.centre]
elif self.best == count:
self.count_calc5 += 5
self.best_centre.append(self.centre)
self.draw()
self.frame.update()
time.sleep(0.1)
def draw(self):
self.canvas.delete(ALL)
# draw best circle
for best in self.best_centre:
outline="red")
# draw current circle
if self.sweep < len(points):
outline="pink")
# draw all the connections
for p in points:
for p2 in p.right:
self.canvas.create_line(p.x,p.y,p2.x,p2.y,fill="lightGray")
# plot visited points
for i in xrange(0,self.sweep):
p = points[i]
self.canvas.create_line(p.x-2,p.y,p.x+3,p.y,fill="blue")
self.canvas.create_line(p.x,p.y-2,p.x,p.y+3,fill="blue")
# plot current point
if self.sweep < len(points):
p = points[self.sweep]
self.canvas.create_line(p.x-2,p.y,p.x+3,p.y,fill="red")
self.canvas.create_line(p.x,p.y-2,p.x,p.y+3,fill="red")
self.canvas.create_line(p.x,p.y,self.right.x,self.right.y,fill="red")
self.canvas.create_line(p.x,p.y,self.centre.x,self.centre.y,fill="cyan")
self.canvas.create_line(self.right.x,self.right.y,self.centre.x,self.centre.y,fill="cyan")
# plot unvisited points
for i in xrange(self.sweep+1,len(points)):
p = points[i]
self.canvas.create_line(p.x-2,p.y,p.x+3,p.y,fill="green")
self.canvas.create_line(p.x,p.y-2,p.x,p.y+3,fill="green")
width = 800
height = 600
points = []
# make some points
for i in xrange(0,100):
points.append(Point(random.randrange(width),random.randrange(height)))
# sort points for find-the-right sweep
points.sort(lambda a, b: int(a.x)-int(b.x))
# work out those points to the right of each point
for i in xrange(0,len(points)):
p = points[i]
for j in xrange(i+1,len(points)):
p2 = points[j]
if p2.x > (p.x+diameter):
break
if (abs(p.y-p2.y) <= diameter) and \
p.distance(p2) < diameter:
p.right.append(p2)
p2.left += 1
# sort points in potential order for sweep, point with most right first
points.sort(lambda a, b: len(b.right)-len(a.right))
# debug
for p in points:
print p, p.left, p.right
# show it
root = Tk()
vis = Vis()
root.mainloop()
``````
-
Unless I've missed something obvious I think there is a simple answer.
For a rectangular area MxN, number of points P, radius R:
• Initialise a map (e.g. 2D array of int) of your MxN area to all zeroes
• For each of your P points
• increment all map points within radius R by 1
• Find map element with maximum value - this will be the centre of the circle you are looking for
This is O(P), assuming P is the variable of interest.
-
That works for an integer grid, but if the point coordinates are real values, you might have an issue. – Mark Bessey Jan 28 '10 at 0:25
(Original poster) Reminds me of one of my most unjust downvotes: stackoverflow.com/questions/244452/… :) – Will Jan 28 '10 at 6:54
@Mark - good point - I think the same technique can probably still be applied if we think of each element in the map as a "bin", but this may still leave some edge cases that we won't find using this method. – Paul R Jan 28 '10 at 8:34
How about using a clustering algorithm to identify the cluster of points. Then ascertain the cluster with the maximum number of points. Take the mean point of the cluster having the maximum points as the center of your circle and then draw the circle.
MATLAB supports implementation of k-means algorithm and it gives back a 2-d array (a matrix to be precise) of cluster means and corresponding cluster ids.
One well known flip side of k-means is deciding on k(number of clusters) before hand. This can be resolved though - one can learn the value of k from the data points. Please check this paper.
I hope this helps.
cheers
-
Very quick idea, not necessarily right one:
• for each point P you calculate a "candidate covering area" - a continuum of points where the center of its covering circle might be. Naturally it is also a circle of diameter D with the center in P.
• for each point P you intersect its candidate covering area with the corresponding areas of other points. Some of the candidate covering areas may intersect with the P's and with each other. For each intersection you count the number of intersected areas. A figure which is intersected by the most of candidate areas is a candidate area for the center of a covering circle which covers P and as many of other points as possible.
• find a candidate area with the highest number of intersections
Seems to be N^2 complexity, provided that calculating intersections of circle shaped areas is easy
-
So the question is: how do we efficiently calculate/store intersections of circles? :) – BlueRaja - Danny Pflughoeft Jan 28 '10 at 21:20 | 2,054 | 7,709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2015-22 | latest | en | 0.811077 |
https://stemgeeks.net/hive-163521/@dkmathstats/algebra-rationalize-the-denominator | 1,606,995,902,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141727627.70/warc/CC-MAIN-20201203094119-20201203124119-00645.warc.gz | 475,232,022 | 100,757 | # Algebra - Rationalize The Denominator
in STEMGeeks18 days ago
Hi everyone. This mathematics post is focused on rationalizing the denominator. It is an algebra technique that is used to remove square roots from the denominator of a fraction.
Math text rendered with QuickLatex.com.
Pixabay Image Source
## Short Review Of Exponent Laws and Square Roots
Given two numbers with a common base and different exponents being multiplied together, they can be combined into a single base where the different exponents are added together.
As an example if you have , this can be rewritten into a single base and a single exponent as . The exponents of 2 and 7 are added together to obtain the sum of 9.
In general, you have:
where `b` is the number base, `m` is the first number exponent and `n` is the second number exponent.
Square Root As The Exponent Of Half
A concept that some math students forget or are not aware of is that the square root as an exponent is one half.
The square root of a product such as `xy` is the same as the square root of `x` multiplied by the square root of `y`. This is based on the exponent laws.
In summary for this section, the result worth remembering is:
Pixabay Image Source
## No Square Roots On The Bottom
When it comes to mathematics formatting there are some unspoken mathematics style rules. It is common to prefer no square root values in the denominator of fractions.
We are not there in rationalizing the denominator and removing square roots yet. This simple of trick of "multiplying by one" to the fraction is a nice way of removing square root values from the bottom of fractions.
Example One
Sometimes you may obtain answers such as . The square root of 3 is on the denominator. To make this fraction into more a "valid" answer we multiplying by "one" by multiplying top and bottom by the square root of 3.
The result from earlier is applied to the bottom of the fraction.
The square root of 9 is 3. Another way is to use exponents instead of square roots on the denominator.
Note that:
Pixabay Image Source
## The Use Of Conjugates In Rationalizing The Denominator
In the earlier section there was only on term in the square root. What if there two terms (binomial) in a square root? How can we remove square roots of binomials?
We do the multiply by one trick along with the use of conjugates.
The Conjugate Of A Binomial
If you have a binomial of the form `a + bz`, the conjugate would be `a - bz`. The conjugate of `a - bz` would be `a - (-bz)` or `a + bz`.
As one example, the conjugate of 5 + 7x would be 5 - 7x.
The conjugate of 3 - x is 3 + x.
Rationalizing The Denominator With Conjugates
This rationalizing the denominator algebra technique will be shown through an example.
To remove the square root below, multiply by "one" with the conjugate of x plus the square root of 2 on the numerator and denominator.
The numerator of this fraction is just x minus the square root of 2. The bottom is a product of two binomials. In fact, this product of binomials is the factored version of the difference of squares factoring technique.
In the bottom of the part of the denominator the middle two terms out of the four cancel each other out. The last term of two square root of two behind the negative becomes 2 (with square root exponent laws).
In summary you have:
Another Example
This next example features two square roots in a binomial. What if we wanted to remove the two square roots on the bottom of the fraction
To rationalize the denominator, we still use conjugates. The conjugate of the square root of 5 minus the square root of 7 is . This conjugate is used when multiplying the fraction by "one".
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Unformatted text preview: j ) = ij. Each 26 CHAPTER 2. DISCRETE-TYPE RANDOM VARIABLES 4/36 p (k) Y 3/36 2/36 1/36 k 5 10 20 15 25 30 35 Figure 2.3: The pmf for the product of two fair dice of the 36 possible values of (X1 , X2 ) has probability 1/36, so we have E [Y ] = = = 1 36 6 6 ij i=1 j =1 1 36 6 i i=1 (21)2 36 6 = j j =1 7 2 2 = 49 = 12.25. 4 Variance and standard deviation Suppose you are to be given a payment, with the size of the payment, in some unit of money, given by either X or by Y, described as follows. The random 1 999 variable X is equal to 100 with probability one, whereas pY (100000) = 1000 and pY (0) = 1000 . Would you be equally happy with either payment? Both X and Y have mean 100. This example illustrates that two random variables with quite different pmfs can have the same mean. The pmf for X is concentrated on the mean value, while the pmf for Y is considerably spread out. The variance of a random variable X is a measure of how spread out the pmf of X is. Letting µX = E [X ], the variance is defined by: Var(X ) = E [(X − µX )2...
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# Factors for Data Analysis: Margin of Error and Null Hypothesis
✅ Paper Type: Free Essay ✅ Subject: Data Analysis ✅ Wordcount: 1893 words ✅ Published: 8th Feb 2020
1. What does the margin of error tell us about a sample taken from a large population? How does the confidence level for a sample outcome differ from the sample’s margin of error?
First, the margins of error and sample size are considered an inverse relationship (Rumsey, n.d.). Second, increasing the sample size furthermore than what you already may have will give you a reduced return since the increases accuracy will be negligible (Rumsey, n.d.).
A notion, indispensable in estimating the chances that a sample is accurate is considered the margin of error (Carey, 2011). The population is defined as the average value of a variable, where the reference class is a population interest, and the sample can be defined as the same; however, the reference class is a sample from the population. When taking samples from a larger population one must keep in mind that the larger the population, the greater the chances the results took will be accurate (Carey, 2011). In lamest terms, the margin of error decreases and as the sample size increases (Carey, 2011). The aforementioned relationship is called an inverse because both, the margin of error and sample size, move in opposite directions.
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The population is gauged by two important statistics, the margin of error and confidence level. The confidence level can be defined as the percentage of all samples, which can be expected to include the true population parameter or in lamest terms, how confident we are in a given margin of error. Both margins of error and confidence level are closely related. Both depend on the sample size and not the population size. As known, with the increase of the sample size, the margin of error decreases. However, as the population increases, the margin of error increases as well. Lastly, as the confidence level increases so does the margin or error increases.
2. What is a null hypothesis in causal research and what does it mean to say a study has failed to reject the null hypothesis?
A hypothesis is defined as a theory or speculation, which is based on insufficient evidence that lends it to further experimentation and testing (Gonzalez, n.d.). However, the null hypothesis can be defined as a hypothesis, which says there is no statistical significance between the two variables in the hypothesis (Gonzalez, n.d.). The following are a couple of examples of the null hypothesis. Question: do teens access the Internet more than adults using a cellular phone? Null hypothesis: age, teens or adults, has zero effect when a cellular phone is used to access the Internet. Question: does taking a daily dose of aspirin to reduce your chances of a heart attack? Null hypothesis: taking a daily dose of aspirin does not reduce your chances of a heart attack.
However, in causal research, unable to establish a causal link is often seen as a failure to reject the null hypothesis. Furthermore, an experiment, which establishes a successful large enough difference in levels of effect between control and experimental groups, will often be said to reject the null hypothesis (Carey, 2011). With the aforementioned stated there must be null and alternative hypotheses. If it is concluded, during the experimentation and testing, that a null hypothesis is rejected; then we must be inclined to accept the alternative hypothesis (Taylor, 2018). However, if the null hypothesis were not rejected, then we would say that we accept the null hypothesis (Taylor, 2018).
In a nutshell, researchers are attempting to provide enough evidence to prove and accept the alternative hypothesis, not the null hypothesis (Taylor, 2018). The null hypothesis is considered to be a true statement until evidence proves otherwise (Taylor, 2018).
3. How do randomized, prospective and retrospective studies differ from one another? What are the major advantages and disadvantages of each type of study?
As mentioned in A Beginner’s Guide to Scientific Method (Carey, 2011) randomized causal studies can be explained as a selection of subjects, dividing the selected into two groups, and administering the suspected causal agent to the selected subjects of one of the two groups. One advantage of randomized causal studies is that they are capable of providing strong evidence because it enables us to control other potential causal factors. Another advantage is that subjects are chosen prior to being exposed to the suspected cause, combined with being indiscriminately divided into control and experimental groups. The aforementioned allows controlling for extraneous cause factors. However, one of the biggest disadvantages of randomized causal studies is that they tend to be time consuming and expensive.
As mentioned in A Beginner’s Guide to Scientific Method (Carey, 2011) prospective causal studies can be explained as two groups of subjects, one that is the experimental group and already has the suspected causal factor while the other group does not. A huge advantage of prospective causal studies versus the abovementioned study is that they require less direct manipulations of the experimental subjects, therefore, are considered easier and less expensive to conduct. Furthermore, they provide the most accurate analysis of the effect (Why randomize?” n.d.) A downfall of prospective causal study is although if properly conducted it can indicate a strong causal link; however, the link would not be as strong as what a randomized causal study would provide.
As mentioned in A Beginner’s Guide to Scientific Method (Carey, 2011) retrospective causal studies can be explained as a beginning with two groups, control, and experimental groups; however, both are comprised of subjects who do and do not have the effect in question. Advantages of retrospective causal studies are similar to randomized causal studies in which they can be conducted fairly quickly and inexpensive. A few limitations of retrospective causal studies are they provide no way of estimating the level of differences of the effect being experimented. Furthermore, retrospective causal studies provide weak evidence of a causal link.
4. Describe each of the fallacies listed below and make up an example of each.
a) False anomalies: As mentioned in A Beginner’s Guide to Scientific Method (Carey, 2011) false anomalies are defined as omitting facts intentionally, which would show how something is not as bizarre as it may seem. For example, many of the conspiratorial theories, which surrounded the events of September 11, 2001, were considered false anomalies (Carey, 2011). One of the events, which could not be explained as many people felt as if a plane did not crash into the Pentagon due to the fact of little wreckage; however, an American Airlines fuselage was located in the front lawn of the Pentagon.
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b) Questionable arguments by elimination: As mentioned in A Beginner’s Guide to Scientific Method (Carey, 2011) questionable arguments by elimination is defined as having one true alternative by eliminating the possibility of the other alternative, which may be false. An example of this is telepathy. Case studies in the past have shown people to somehow receive unexplained messages from one person to another. Although telepathy may not be the scientific answer other explanations are far-fetched and are eliminated.
c) Illicit causal inference: As mentioned in A Beginner’s Guide to Scientific Method (Carey, 2011) illicit casual inference is defined as saying one thing caused another when there is clearly only a correlation. For example, students who sit in the front of the class tend to get better grades than those who sit in the back of the class; however, the answer might just be that students who sit in front of the class are more motivated to do better than those who sit in the back of the class (Carey, 2011).
d) Unsupported analogies and similarities: As mentioned in A Beginner’s Guide to Scientific Method (Carey, 2011) are defined as showing similarities between one speculative theory and a different well-established theory in science. An example of this would be luminiferous ether. Physicists hypothesized of certain similarities between light and sound; however, after much experimentation, they were able to conclude there was no there were no such similarities thus establishing light as a well-understood phenomenon (Carey, 2011).
e) Untestable explanations and predictions: As mentioned in A Beginner’s Guide to Scientific Method (Carey, 2011) untestable explanations is defined as presenting a theory, which by definition alone, cannot be tested. Furthermore, untestable predictions can be seen as providing a prediction, which cannot be tested or explained. Predictions should be testable, allowing the prediction to either be verified or invalidated (Brennan, 2017).
f) Empty jargon can be defined as a phrase or word, which changes or loses its meaning when used with people who do not understand the meaning of the phrase or word used. An example of empty jargon can be seen at the hospital. Doctors and nurses use of medical jargon to explain an issue their patient may be having is sometimes more confusing rather than telling their patient they have the flu or a simple cold.
References:
• Brennan, J. (2017). What is a testable prediction? Sciencing. Retrieved from
• https://sciencing.com/testable-prediction-8646215.html
• Carey, S. S. (2011). A beginner’s guide to scientific method (4th ed.). Boston, MA: Wadsworth, Cengage Learning.
• Gonzales, K. (n.d.). What is a null hypothesis? Study. Retrieved on October 12, 2018 from https://study.com/academy/lesson/what-is-a-null-hypothesis-definition-examples.html
• Rumsey, D. (n.d.) How sample size affects the margin of error. Dummies. Retrieved on October 12, 2018 from https://www.dummies.com/education/math/statistics/how-sample-size affects-the-margin-of-error/
• Taylor, C. (2018). Why say “fail to reject” in a hypothesis test? Thought Co. Retrieved from https://www.thoughtco.com/fail-to-reject-in-a-hypothesis-test-3126424
• Why randomize? (n.d.). Yale University. Retrieved on October 10, 2018 from https://isps.yale.edu/node/16697
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http://math.stackexchange.com/questions/783141/how-do-you-solve-this-permutation-problem-evaluate-np-1 | 1,469,550,911,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824995.51/warc/CC-MAIN-20160723071024-00063-ip-10-185-27-174.ec2.internal.warc.gz | 154,032,174 | 16,785 | # How do you solve this permutation problem? Evaluate $\,{}_nP_1?$
Evaluate ${}_nP_1$? I believe the answer is $n$. Am I correct?
-
Well, do you know the formula for $_n P_r$? $$_nP_r=\frac{n!}{(n-r)!}$$ Just plug in the values of $n$ and $r$ into this formula. In this case your $n$ still equals $n$, and $r$ equals $1$. Therefore: $$_nP_1=\frac{n!}{(n-1)!}$$ $$=\frac{n(n-1)(n-2)\dots 3\cdot 2\cdot 1}{(n-1)(n-2)\dots 3\cdot 2\cdot 1}$$ $$=n$$ $$\color{green}{\boxed{_nP_1=n}}$$ Hope I helped | 193 | 497 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2016-30 | latest | en | 0.599446 |
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What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
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##### Stage: 1 and 2 Challenge Level:
Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game.
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##### Stage: 2 Challenge Level:
Use the interactivity to investigate what kinds of triangles can be drawn on peg boards with different numbers of pegs.
### Centred Squares
##### Stage: 2 Challenge Level:
This challenge, written for the Young Mathematicians' Award, invites you to explore 'centred squares'.
### Spirals, Spirals
##### Stage: 2 Challenge Level:
Here are two kinds of spirals for you to explore. What do you notice?
### Oddly
##### Stage: 2 Challenge Level:
Find the sum of all three-digit numbers each of whose digits is odd.
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##### Stage: 1 and 2 Challenge Level:
How many different journeys could you make if you were going to visit four stations in this network? How about if there were five stations? Can you predict the number of journeys for seven stations?
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##### Stage: 2 Challenge Level:
This activity involves rounding four-digit numbers to the nearest thousand.
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##### Stage: 1 Challenge Level:
Stop the Clock game for an adult and child. How can you make sure you always win this game?
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##### Stage: 1 and 2 Challenge Level:
Strike it Out game for an adult and child. Can you stop your partner from being able to go?
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##### Stage: 2 Challenge Level:
Can you make dice stairs using the rules stated? How do you know you have all the possible stairs?
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##### Stage: 2 Challenge Level:
This task follows on from Build it Up and takes the ideas into three dimensions!
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##### Stage: 2 Challenge Level:
Can you find all the ways to get 15 at the top of this triangle of numbers?
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##### Stage: 2 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
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##### Stage: 2 Challenge Level:
Watch this video to see how to roll the dice. Now it's your turn! What do you notice about the dice numbers you have recorded?
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##### Stage: 2 Challenge Level:
What happens when you round these three-digit numbers to the nearest 100?
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##### Stage: 1 Challenge Level:
This challenge is about finding the difference between numbers which have the same tens digit.
### Sums and Differences 1
##### Stage: 2 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### Sums and Differences 2
##### Stage: 2 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
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##### Stage: 2 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
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##### Stage: 2 Short Challenge Level:
Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread?
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##### Stage: 2 and 3
The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails.
### Calendar Calculations
##### Stage: 2 Challenge Level:
Try adding together the dates of all the days in one week. Now multiply the first date by 7 and add 21. Can you explain what happens?
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##### Stage: 1 Challenge Level:
Ben and his mum are planting garlic. Use the interactivity to help you find out how many cloves of garlic they might have had.
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##### Stage: 1 Challenge Level:
This is a game for two players. Can you find out how to be the first to get to 12 o'clock?
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How can you arrange these 10 matches in four piles so that when you move one match from three of the piles into the fourth, you end up with the same arrangement?
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##### Stage: 2 Challenge Level:
Put the numbers 1, 2, 3, 4, 5, 6 into the squares so that the numbers on each circle add up to the same amount. Can you find the rule for giving another set of six numbers?
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##### Stage: 2 Challenge Level:
Can you dissect an equilateral triangle into 6 smaller ones? What number of smaller equilateral triangles is it NOT possible to dissect a larger equilateral triangle into?
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In each of the pictures the invitation is for you to: Count what you see. Identify how you think the pattern would continue.
### Games Related to Nim
##### Stage: 1, 2, 3 and 4
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
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Frances and Rishi were given a bag of lollies. They shared them out evenly and had one left over. How many lollies could there have been in the bag?
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An investigation that gives you the opportunity to make and justify predictions.
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Can you continue this pattern of triangles and begin to predict how many sticks are used for each new "layer"?
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In this calculation, the box represents a missing digit. What could the digit be? What would the solution be in each case?
### Round the Two Dice
##### Stage: 1 Challenge Level:
This activity focuses on rounding to the nearest 10.
### Nim-like Games
##### Stage: 2, 3 and 4 Challenge Level:
A collection of games on the NIM theme | 1,993 | 8,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-47 | latest | en | 0.909267 |
https://wiingy.com/resources/calculators/millimetre-to-inch/ | 1,718,370,156,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861546.27/warc/CC-MAIN-20240614110447-20240614140447-00303.warc.gz | 568,196,142 | 36,757 | #FutureSTEMLeaders - Wiingy's \$2400 scholarship for School and College Students
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Length
Convert Millimetre to Inch
Written by Rahul Lath
Updated on: 18 Aug 2023
Convert Millimetre to Inch
Conversion of Millimetre to Inch: Millimetres and inches are used to measure the length in the metric system. A length of 25.4 millimetres will count up to 1 inch. This article will give you the necessary information on how to convert millimetres to inches.
Millimeters
mm
Inches
in
What Is a Millimetre?
A Millimetre is a unit of length that is equal to one-thousandth of a metre in the International System of Standards (SI). It is denoted as ‘mm’ in short.
Millimetre = 1/1000 metres
Common Usage:
• Used to measure the size of the dial of a wristwatch.
• Used to measure the size of a diamond.
• Used to measure the size of a coin.
What Is an Inch?
Inch is a unit of measurement for length and is equal to 25.4 millimetres and 2.54 centimetres. It is denoted as ‘in’ or simply by the symbol “. So,
1 inch = 25.4 millimetres
1 inch = 2.54 centimetres
Common Usage:
• Used to measure the length of the paperclip.
• Used to measure the length of a standard sewing pin.
• Used to measure the length of a standard sewing pin.
How to Convert Millimetre to Inch?
To convert millimetre to inch, multiply the given millimetres by 0.03937 to get the value in inches. As
1 Millimetre = 0.0393701 Inches
The formula to convert millimetres to inches is given below,
X Millimetres = X × 0.03937 Inches
where X is the value in millimetres.
Examples of the Conversion of Millimetre to Inch
Problem 1: Convert 50 millimetres to inches.
Solution 1:
Step 1: The given value is 50 millimetres.
Step 2: To convert millimetres to inches, substitute the given values at the required places in the conversion formula.
X Millimetres = X × 0.03937 Inches
Hence,
50 millimetres = 50 × 0.03937
= 1.9685 in
Therefore, 50 millimetres is 1.9685 inches.
Problem 2: Convert 98.6 millimetres to inches.
Solution 2:
Step 1: The given value is 98.6 millimetres.
Step 2: To convert millimetres to inches, substitute the given values at the required places in the conversion formula.
X Millimetres = X × 0.03937 Inches
Hence,
98.6 millimetres = 98.6 × 0.03937
= 3.881882 in
Therefore, 98.6 millimetres is 3.881882 inches.
Problem 3: Convert 300 millimetres to inches.
Solution 3:
Step 1: The given value is 300 millimetres.
Step 2: To convert millimetres to inches, substitute the given values at the required places in the conversion formula.
X Millimetres = X × 0.03937 Inches
Hence,
300 millimetres = 300 × 0.03937
= 11.811 in
Therefore, 300 millimetres is 11.811 inches.
Problem 4: Convert 1500 millimetres to inches.
Solution 4:
Step 1: The given value is 1500 millimetres.
Step 2: To convert millimetres to inches, substitute the given values at the required places in the conversion formula.
X Millimetres = X × 0.03937 Inches
Hence,
1500 millimetres = 1500 × 0.03937
= 59.055 in
Therefore, 1500 millimetres is 59.055 inches.
Problem 5: Convert 685 millimetres to inches.
Solution 5:
Step 1: The given value is 685 millimetres.
Step 2: To convert millimetres to inches, substitute the given values at the required places in the conversion formula.
X Millimetres = X × 0.03937 Inches
Hence,
685 millimetres = 685 × 0.03937
= 26.96845 in
Therefore, 685 millimetres is 26.96845 inches.
FAQs on the Conversion of Millimetre to Inch
What size is a millimetre?
A millimetre is a unit of length in the International System of Standards (SI), which is equal to one-thousandth of a metre. It is denoted as ‘mm’ in short.
Millimetre = 1/1000 metres
What size is 1 mm in inches?
1 mm is 0.0393701 inches.
Is 20 mm same as 1 inch?
20 mm is 0.787402 inches, which is almost 1 inch.
How much is a mm?
1 Millimetre = 1/1000 metres and 0.0393701 inches.
How does 1 mm compare to 1?
One millimetre is 0.1 centimetre and 1 centimetre is 10 millimetres.
Related conversion posts
We hope this article on the conversion of millimetre to inch is helpful to you. If you have any queries related to this post, ping us through the comment section below and we will get back to you as soon as possible.
References
The SI Base Units
Written by
Rahul Lath
Reviewed by
Arpit Rankwar
Share article on | 1,249 | 4,288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-26 | latest | en | 0.822483 |
https://home-garden.blurtit.com/3818683/how-many-bricks-used-in-100-sq-ft-wall | 1,582,683,895,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146176.73/warc/CC-MAIN-20200225233214-20200226023214-00129.warc.gz | 396,463,490 | 11,633 | # How many bricks used in 100 sq. Ft. Wall?
A brick with a 2" by 8" face, surrounded by 1/4" of mortar will need to be replicated 776 times to have an area of 100 ft^2. The number of bricks actually required depends on their size, the edge finishing, the mortar thickness, breakage, the nature of the cuts, and probably several other factors, including the number of bricks thick the wall is.
thanked the writer.
Brick retaining walls are usually 2 walls with a brick row on top. ... Your wall 25 x 3 = 75 x 2 (the back wall) = 150 sq feel x (how many of your brick per sq foot) + 100 bricks for top row = total brick x 10% waste factor
thanked the writer. | 178 | 657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-10 | latest | en | 0.942851 |
https://socratic.org/questions/what-is-the-distance-between-the-following-polar-coordinates-3-15pi-12-5-3pi-8#291983 | 1,653,350,415,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562106.58/warc/CC-MAIN-20220523224456-20220524014456-00465.warc.gz | 607,918,798 | 6,467 | What is the distance between the following polar coordinates?: (3,(-15pi)/12), (5,(-3pi)/8)
Jul 25, 2016
$x = \sqrt{{3}^{2} + {5}^{2} - 2 \cdot 3 \cdot 5 \cdot \cos \left(\frac{- 15 \pi}{12} - \frac{- 3 \pi}{8}\right)} \approx 7.86$
Explanation:
Polar coordinates are written as $\left(r , \theta\right)$, where $r$ is the distance from the origin and $\theta$ is the angle with respect to the positive x-axis and the origin.
If we plot $\left(3 , \frac{- 15 \pi}{12}\right)$ and $\left(5 , \frac{- 3 \pi}{8}\right)$ on a graph, we notice we can form a triangle with the origin, $\left(0 , 0\right)$, as such:
We know the length of two of the sides of the triangle, since the $r$ value of the coordinates tell us. We can also calculate the angle $\alpha$, since we have the angles of the two points.
$| {\theta}_{1} - {\theta}_{2} | = | \frac{- 15 \pi}{12} - \frac{- 3 \pi}{8} | = \frac{7 \pi}{8}$
If the result was larger than $\pi$, we would simply subtract it from $2 \pi$ to get the angle inside of the triangle.
Finally, we know from trigonometry that if we know two sides and the angle in between, we can calculate the length of the other side using the Law of Cosines:
${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cos \left(\alpha\right)$
we can use this law to arrive at an equation for the distance between two polar points, $\left({r}_{1} , {\theta}_{1}\right)$ and $\left({r}_{2} , {\theta}_{2}\right)$
x = sqrt(r_1^2 + r_2^2 - 2r_1r_2cos(theta_1 - theta_2)
or in this case,
$x = \sqrt{{3}^{2} + {5}^{2} - 2 \cdot 3 \cdot 5 \cdot \cos \left(\frac{7 \pi}{8}\right)} \approx 7.86 \square$
p.s. note the lack of absolute value signs in the final equation. That's because $\cos \left(x\right) = \cos \left(- x\right)$ | 585 | 1,733 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 18, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2022-21 | latest | en | 0.796041 |
http://www.thescienceforum.com/physics/41675-help-speed-problem-please.html | 1,591,052,927,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419639.53/warc/CC-MAIN-20200601211310-20200602001310-00099.warc.gz | 212,652,123 | 13,138 | ## View Poll Results: What is answer the following question?
Voters
1. You may not vote on this poll
• Option A
0 0%
• Option B
0 0%
• Option C
1 100.00%
• Option D
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# Thread: Help in speed problem please
1. hello i need help in solving this
Q)If the speed of a vehicle increases by 2 m/s,its kinetic energy is doubled,then original speed of the vehicle is
a)(sqrt of 2 )+1 m/s
b)2(sqrt of 2 -1)m/s
c)2(sqrt of 2 + 1 )m/s
d)sqrt of 2 (sqrt of 2 + 1)m/s
sqrt is square root of
2.
3. What have you got so far?
4. I found the answer as 2(sqrt of 2 + 1 )m/s i just want you to cross check it
5. Originally Posted by Rajnish Kaushik
I found the answer as 2(sqrt of 2 + 1 )m/s
Not useful.
HOW did you arrive at that answer?
6. i just used the identity 1/2*m*(v+2)^2 = 1/2*m*v^2 * 2
7. 1/2*m*(v+2)^2 = 1/2*m*v^2 * 2 ie KE=PE
8. And did you check it?
E.g.:
2(sqrt of 2 + 1 )m/s = (2(1.414+1) = 4.828 m/ s,
Square that = 23.3.
New value = 6.828 (i.e. adding 2 m/ sec).
6.8282 = 46.6.
46.6/23.3 = 2.
Confirmed.
9. Originally Posted by Dywyddyr
And did you check it?
E.g.:
2(sqrt of 2 + 1 )m/s = (2(1.414+1) = 4.828 m/ s,
Square that = 23.3.
New value = 6.828 (i.e. adding 2 m/ sec).
6.8282 = 46.6.
46.6/23.3 = 2.
Confirmed.
So i was right yeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee eeeeeeeeeeeeeeeee
10. I've always found that plugging in the numbers and actually calculating them helps.
11. Originally Posted by Dywyddyr
I've always found that plugging in the numbers and actually calculating them helps. | 567 | 1,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2020-24 | latest | en | 0.889329 |
https://www.exactlywhatistime.com/days-before-date/june-26/6-days | 1,713,893,262,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818732.46/warc/CC-MAIN-20240423162023-20240423192023-00895.warc.gz | 677,070,401 | 6,362 | # Thursday June 20, 2024
## Calculating 6 days before Wednesday June 26, 2024 by hand
This page helps you figure out the date that is 6 days before Wednesday June 26, 2024. We've made a calculator to find the date before a certain number of days before a specific date. If you want to change the question on this page, you have two choices: you can change the URL in your browser's address bar, or go to our days before specific date calculatorto type in a new question or days from a specific date if you want to add 6 days.
But for all you time sickos out there who want to calculate 6 days before Wednesday June 26, 2024 - here's how you do it:
1. Start with the Input Date (Wednesday June 26, 2024): Write it down! I can't stress this enough
2. Count in Weeks: Recognize that 6 days is approximately 0.8571428571428571 weeks. Count forward 0.8571428571428571 weeks (1.2 work weeks) from the input date. This takes you to .
3. Add Remaining Days: Since you've counted 0.8571428571428571 weeks, you only need to add the remaining days to reach Thursday June 20, 2024
4. Use Mental Math: If Wednesday June 26, 2024 is a Thursday, then compare that to if 6 is divisible by 7. That way, you can double-check if June 20 matches that Thursday.
## Thursday June 20, 2024 Stats
• Day of the week: Thursday
• Month: June
• Day of the year: 172
## Counting 6 days backward from Wednesday June 26, 2024
Counting backward from today, Thursday June 20, 2024 is 6 before now using our current calendar. 6 days is equivalent to:
6 days is also 144 hours. Thursday June 20, 2024 is 47% of the year completed.
## Within 6 days there are 144 hours, 8640 minutes, or 518400 seconds
Thursday Thursday June 20, 2024 is day number 172 of the year. At that time, we will be 47% through 2024.
## In 6 days, the Average Person Spent...
• 1288.8 hours Sleeping
• 171.36 hours Eating and drinking
• 280.8 hours Household activities
• 83.52 hours Housework
• 92.16 hours Food preparation and cleanup
• 28.8 hours Lawn and garden care
• 504.0 hours Working and work-related activities
• 463.68 hours Working
• 758.88 hours Leisure and sports
• 411.84 hours Watching television
## Famous Sporting and Music Events on June 20
• 1975 "Jaws", based on the book by Peter Benchley, directed by Steven Spielberg and starring Roy Scheider is released
• 1987 1st Rugby World Cup Final, Eden Park, Auckland: New Zealand fly-half Grant Fox lands 4 penalties, a conversion and drop goal as the All Blacks beat France, 29-9 | 689 | 2,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-18 | latest | en | 0.928419 |
https://www.physicsforums.com/threads/likelyhood-ratio-test-hypotheses-and-normal-distribution.216366/ | 1,716,302,215,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00224.warc.gz | 829,626,876 | 16,520 | # Likelyhood ratio test hypotheses and normal distribution
• Hummingbird25
In summary, the conversation discusses using the normal distribution and likelihood function to test the null hypothesis that two sample means are equal. If the common variance is known, the z-test for equality of means should be used. If the common variance is not known, it should be calculated from the combined sample and then the t-test for equality of means should be used.
Hummingbird25
## Homework Statement
Given the normal distribution
$$X_{ij} \sim N(\mu_i, \omega^2)$$ where i = 1,2 and j = 1,...,n
deduce that $$H_{0\mu}: \mu_1 = \mu _2$$
## The Attempt at a Solution
Do I take in the Likelyhood function here?
and use it to analyse the case?
Sincerely Hummingbird
p.s. I have reading in Wiki that the Null hypo is rejected by the likehood ratio test, could be what I am expected to show here?
Last edited:
Hummingbird25 said:
I have reading in Wiki that the Null hypo is rejected by the likehood ratio test, could be what I am expected to show here?
You are taking 2 samples from 2 different normal distributions, where sample size is n for each sample. You are supposed to calculate the sample averages then test the null hyp. using a z-test, assuming their common variance $\omega^2$ is known (given).
If you don't know the variance you'll need to estimate it from the combined sample, then use a t-test for equality of means (assuming equal variances and equal sample sizes).
Last edited:
EnumaElish said:
You are taking 2 samples from 2 different normal distributions, where sample size is n for each sample. You are supposed to calculate the sample averages then test the null hyp. using a z-test, assuming their common variance $\omega^2$ is known (given).
If you don't know the variance you'll need to estimate it from the combined sample, then use a t-test for equality of means (assuming equal variances and equal sample sizes).
The sample average of the two norm distributions is that
$$\overline{x} = \frac{\sum_{i=1}^{2}f_i}{n}$$??
Sincerely
Hummingbird
Last edited:
No.
$$\overline{x_1} = \frac{\sum_{j=1}^{n}x_{1j}}{n}$$
Same for i = 2.
Hello again EnomaElish and thank you,
$$\overline{x_1} = \frac{\sum_{j=1}^{n}x_{1j}}{n}$$
$$\overline{x_2} = \frac{\sum_{j=1}^{n}x_{2j}}{n}$$
Then I say by the z-test then the null hypotheses is rejected if the variance isn't given since the samples aren't drawn from the same population.
But the null hypotheses is accepted if the means are equal which can be tested using the student t-test.
Is this it?
Sincerely Hummingbird
Is the variance given, or assumed known?
Yes, the variance is given (or the problem assumes it is known).
What you need to do: use the z test for equality of means to determine whether or not the two means are equal. (You are not supposed to use the t test in this case.)
No, the variance is not given (nor does the problem assume the variance is known).
What you need to do: calculate the common variance from the combined sample. Then use the t test for equality of means to determine whether or not the two means are equal. (You are not supposed to use the z test in this case.)
## What is a likelihood ratio test?
A likelihood ratio test is a statistical method used to compare two or more statistical models. It is used to determine whether one model fits the data significantly better than another.
## How is a likelihood ratio test used to test hypotheses?
A likelihood ratio test is used to compare the likelihood of the data under the null hypothesis (H0) to the likelihood of the data under the alternative hypothesis (HA). If the likelihood under HA is significantly greater than under H0, then the null hypothesis is rejected, indicating that the alternative hypothesis is a better fit for the data.
## What is a normal distribution?
A normal distribution, also known as a Gaussian distribution, is a probability distribution that is symmetric and bell-shaped. It is commonly used to model continuous data in many fields, and is characterized by its mean and standard deviation.
## How is a normal distribution related to a likelihood ratio test?
A likelihood ratio test assumes that the data follows a normal distribution, meaning that the data is continuous and symmetric. This assumption is necessary for the test to accurately compare the likelihood of the data under different statistical models.
## What are the assumptions of a likelihood ratio test?
The main assumptions of a likelihood ratio test are that the data follows a normal distribution, and that the models being compared are nested (i.e. one model is a special case of the other). Additionally, the sample size should be large enough for the test to be valid, and the observations should be independent of each other.
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7K | 1,293 | 5,423 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-22 | latest | en | 0.937515 |
http://www.enotes.com/homework-help/find-area-between-line-y-x-y-x-2-222887 | 1,477,120,953,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718840.18/warc/CC-MAIN-20161020183838-00513-ip-10-171-6-4.ec2.internal.warc.gz | 435,455,071 | 10,241 | # Find the area between the line y= x and y = x^2
Asked on by jude69
hala718 | High School Teacher | (Level 1) Educator Emeritus
Posted on
y=x and y = x^2
First we need to find the points of intersection that bounded area:
==> y = y
==> x^2 = x
==> x^2 - x = 0
==> x(x-1) = 0
Then x = 0 and x= 1
Then we will find the area between x^2 x = 0, and x= 1
==> We know that the area is:
A1 = intg y = intg x dx = x^2 /2
==> A1 = (1/2 - 0) = 1/2
== A2 = intg y= intg x^2 = x^3/3
==> A2 = ( 1/3- 0 ) = 1/3
Then the area is:
A = A1 - A2 = 1/2 - 1/3 = 1/6
Then the area is 1/6 square units
neela | High School Teacher | (Level 3) Valedictorian
Posted on
To find the area betwen y = x and x^2.
The intersection points of y = x and y = x^2 is given by:
x= x^2.
Or x-x^2 = 0.
Or x(1-x) = 0.
So x= 0 , Or 1-x = 0 , Or x = 1.
So the area between curves is to be found from x= 0 to x = 1.
If we draw the graph , y = x is above x = x^2 from x= 0 to x = 1 .
Therefore area between y = x and y =x^2 is given by:
Area = Integral (x-x^2)dx from x= 0 to x = 1.
Area = {(x^2/2 -x^3/3 at x= 1} - {(x^2/2 -x^3/3 at x= 0}
Area = { 1/2-1/3}- 0
Area = (3-2)/6 = 1/6.
Therefore the area between the curves = 1/6 sq units.
We’ve answered 317,420 questions. We can answer yours, too. | 552 | 1,305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2016-44 | latest | en | 0.884096 |
http://www.physicspages.com/2016/05/03/entropy-changes-in-macroscopic-systems/ | 1,500,964,695,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425082.56/warc/CC-MAIN-20170725062346-20170725082346-00060.warc.gz | 516,759,025 | 17,878 | # Entropy changes in macroscopic systems
Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 3.10 – 3.11.
The original definition of entropy was in terms of heat flow, rather than the multiplicity of states in a system. If an amount of heat ${Q}$ flows into a substance at a constant temperature ${T}$, the change in entropy is
$\displaystyle \Delta S=\frac{Q}{T} \ \ \ \ \ (1)$
If heat flows out of the substance, then ${Q}$ is negative and the system loses entropy. In order for a system to maintain a constant temperature when it gains or loses heat, it must differ from the types of systems we’ve considered up to now. One possibility is that the amount of heat gained is very small compared to the existing internal energy of the system so that the added heat makes a negligible difference to the temperature. Such a system is called a heat reservoir. Another example is a phase change, as when ice melts into liquid water, as during a phase change, heat is gained or lost and the substance doesn’t change its temperature.
Example 1 Suppose a 30 g ice cube at ${0^{\circ}\mbox{ C}=273\mbox{ K}}$ is placed on a table in a room at ${25^{\circ}\mbox{ C}=298\mbox{ K}}$. The ice will first melt into water, still at 273 K (because it’s a phase change), then the water will warm up to 298 K. All of this heat is transferred from the air in the room, which we can consider to be a heat reservoir at a constant temperature of 298 K. The changes in entropy are then:
The latent heat of fusion of water at 273 K is ${334\mbox{ J g}^{-1}}$, so the amount of heat required to melt the ice is
$\displaystyle Q=334\times30=10020\mbox{ J} \ \ \ \ \ (2)$
Since it occurs at a constant temperature, the entropy change of the water is
$\displaystyle \Delta S_{1}=\frac{Q}{T}=\frac{10020}{273}=36.7\mbox{ J K}^{-1} \ \ \ \ \ (3)$
As the water warms up, it absorbs heat, but the temperature varies. We can then use our relation between entropy and heat capacity, along with the fact that the specific heat capacity of water is roughly constant over the liquid range at ${1\mbox{ cal g}^{-1}\mbox{K}^{-1}=4.181\mbox{ J g}^{-1}\mbox{K}^{-1}}$ to get
$\displaystyle \Delta S_{2}$ $\displaystyle =$ $\displaystyle C_{V}\int_{T_{i}}^{T_{f}}\frac{1}{T}dT\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle C_{V}\ln\frac{T_{f}}{T_{i}}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(4.181\times30\right)\ln\frac{298}{273}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 11.0\mbox{ J K}^{-1} \ \ \ \ \ (7)$
Thus the total entropy increase of the water is
$\displaystyle \Delta S_{H_{2}O}=47.7\mbox{ J K}^{-1} \ \ \ \ \ (8)$
The total amount of heat transferred to the water from the room’s air is
$\displaystyle Q=10020+4.181\times30\times\left(298-273\right)=13155.75\mbox{ J} \ \ \ \ \ (9)$
This happens at a constant room temperature of 298 K so the entropy lost by the room is
$\displaystyle \Delta S_{room}=-\frac{13155.75}{298}=-44.1\mbox{ J K}^{-1} \ \ \ \ \ (10)$
The net entropy change of the universe is therefore
$\displaystyle \Delta S=\Delta S_{H_{2}O}+\Delta S_{room}=+3.6\mbox{ J K}^{-1} \ \ \ \ \ (11)$
which is positive, as required by the second law of thermodynamics.
Example 2 To draw a bath (in the days before hot and cold running water, presumably) we mix 50 litres of hot water at ${55^{\circ}\mbox{ C}=328\mbox{ K}}$ with 25 litres of cold water at ${10^{\circ}\mbox{ C}=283\mbox{ K}}$. The final temperature of the water is the weighted average:
$\displaystyle T_{f}=\frac{50\times328+25\times283}{75}=313\mbox{ K}\left(=40^{\circ}\mbox{ C}\right) \ \ \ \ \ (12)$
To find the entropy change, we can use 5. The hot water cools down and thus loses heat, so its entropy change is
$\displaystyle \Delta S_{hot}$ $\displaystyle =$ $\displaystyle 4.181\times5\times10^{4}\times\ln\frac{313}{328}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -9786\mbox{ J K}^{-1} \ \ \ \ \ (14)$
The entropy gained by the cold water is
$\displaystyle \Delta S_{cold}$ $\displaystyle =$ $\displaystyle 4.181\times2.5\times10^{4}\times\ln\frac{313}{283}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle +10532\mbox{ J K}^{-1} \ \ \ \ \ (16)$
Thus the net entropy change is
$\displaystyle \Delta S=+746\mbox{ J K}^{-1} \ \ \ \ \ (17)$
which is again positive.
## 6 thoughts on “Entropy changes in macroscopic systems”
1. Pingback: Entropy and heat | Physics pages
2. Pingback: Heat engines | Physics pages
3. Tabitha Booth-Seay
On line 9, wouldn’t the Q= -(334 J/g * 30 g + 30 g * 4.184 J/gK * (298-273)) = -13155.8 J
making Sroom = -13115.8 J / 298 K = -44.1 J/K ?
Without adding in the temp change on line 9, the units would not work out properly. | 1,492 | 4,798 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 42, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2017-30 | latest | en | 0.896833 |
https://analystprep.com/cfa-level-1-exam/quantitative-methods/t-distribution/ | 1,539,781,433,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511173.7/warc/CC-MAIN-20181017111301-20181017132801-00030.warc.gz | 608,894,926 | 39,383 | ## T-distribution
The student’s t-distribution is a bell-shaped probability distribution symmetrical about its mean. It is considered the best distribution to use for the construction of confidence intervals when:
1. Dealing with small samples of less than 30 elements.
2. The population variance is unknown.
3. The distribution involved is either normal or approximately normal.
In the absence of outright normality of a given distribution, the t-distribution may still be appropriate for use if the sample size is large enough such that the central limit theorem can be applied, in which case the distribution is considered approximately normal.
The t-statistic, also called the t-score is given by:
t = (x – μ)/(S/√n)
Where:
x is the sample mean,
μ is the population mean,
S is the sample standard deviation,
n is the sample size
The t-distribution allows us to analyze those distributions that are not perfectly normal. It has the following properties:
1. It has a mean of zero.
2. Its variance = v/(v/2), where v represents the number of degrees of freedom and v ≥ 2.
3. The variance is greater than 1 at all times, although it’s very close to one when there are many degrees of freedom. With a large number of degrees of freedom, the t-distribution resembles the normal distribution.
4. Its tails are fatter than those of the normal distribution, indicating more probability in the tails.
## T-distribution: The Degrees of Freedom
The t-distribution, just like several other distributions, has only one parameter: the degrees of freedom. The number of degrees of freedom refers to the number of independent observations (total number of observations less 1). i.e.
v = n-1
Hence, a sample of 10 observations/elements would be analyzed by the use of a t-distribution with 9 d.f. Similarly a 6 d.f. distribution would be used for a sample size of 7 observations.
### Notations
It is standard practice for statisticians to use tα to represent the t-score that has a cumulative probability of (1 – α). Therefore, if we were to be interested in at-score having 0.9 cumulative probability, α would be equal to 1 – 0.9 = 0.1. We would denote the statistic as t0.1.
However, the value of tα depends on the number of degrees of freedom. For example,
t0.05, 2 = 2.92 where the second subscript (2) represents the number of d.f and,
t0.05, 20 = 1.725
### Important Relationships
tα = -t1 – α and t1 – α = -tα
The above relationships are true because the t-distribution is symmetrical about the mean.
The t-distribution has thicker tails relative to the normal distribution.
The shape of the t-distribution is dependent on the number of degrees of freedom so that as the number of d.f. increases, the distribution becomes more ‘spiked’ and its tails become thinner.
The table below represents one-tailed confidence intervals and various probabilities for a range of degrees of freedom.
Describe properties of Student’s t-distribution and calculate and interpret its degrees of freedom.
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CFA Institute does not endorse, promote, review, or warrant the accuracy of the products or services offered by prep courses. | 751 | 3,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-43 | longest | en | 0.927479 |
https://excelpunks.com/2014/11/24/transfer-data-formulas-formats-or-even-perform-mathematical-operations-in-a-flash/ | 1,685,601,874,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647614.56/warc/CC-MAIN-20230601042457-20230601072457-00012.warc.gz | 283,493,601 | 34,735 | # Transfer data, formulas, formats or even perform mathematical operations in a flash!
Basics 5: Paste Special
Time to learn: 20 minutes
The Paste Special function allows you to easily transfer data, formulas, formats or even perform mathematical operations just by using good old copy and paste!
This aspect of Excel seldom gets much attention as most do not understand how it works. That is going to change today.
Remember that you always start with some data you need to copy first and then proceed to paste it in another location in a special format that you decide on – that is the principle of the Paste Special function.
Here we have 12 colleagues grouped into 3 teams and their call records, including how many deals they have closed.
Mathematical Operations
Step 1:
Let’s find out how many calls on average it takes to close a deal for each of them. Let’s create a column called ‘Average Calls/Deal’.
Step 2:
Highlight cells C2 to C13 – this cell reference contains the list of the number of calls made per person.
Now, with your cursor still within the cell reference C2 to C13, perform a right click and click on ‘Copy’. You can also use the Ctrl + C shortcut.
You should immediately see a line of running dashes around your selection – this is Excel’s way of saying that the cells have been selected for copying.
Step 3:
Paste the selection onto the ‘Average Calls/Deal’ column by right clicking on cell F2 (the first cell of the column ‘Average Calls/Deal’) and selecting ‘Paste’. You can also use the Ctrl + V shortcut.
Step 4:
Copy the cells D2 to D13 – this cell reference contains the list of deals closed by each person. With your cursor still within the cell reference D2 to D13, perform a right click and click on ‘Copy’. You can also use the Ctrl + C shortcut.
Ensuring that the cells in D2 to D13 are still selected, right click on the cell F2 and select ‘Paste Special’.
Step 5:
You will see an option box like this.
Under ‘Operation’, click on ‘Divide’ and click ‘OK’.
Now, you will see that all your colleagues’ call numbers have been divided by the number of deals they have closed per person.
Tip:
You would have noticed in the option box earlier under ‘Operation’ that there are other mathematical functions like ‘Add’, ‘Subtract’ and ‘Multiply’. That is exactly what it does. By copying and then using the ‘Paste Special’ function, you can add, subtract, multiply or divide a selection of numbers in a cell destination by the values you have copied instantly.
Paste Special
This section deals with the more interesting aspects of ‘Paste Special’.
Notice in the ‘Paste Special’ option box under ‘Paste’, the 6 options – ‘All’, ‘Formulas’, ‘Values’, ‘Formats’, ‘Comments’ and ’Validation’.
I’ve created column titles for each one of these options to show you how they work.
Now, I’ve done a few things with the column ‘Average Calls/Deal’.
• Notice that when the cells are selected, they have a formula ‘=C2/D2’ – all of them have that, referencing cells in the ‘Calls Made’ and ‘Deals Closed’ columns and dividing the ‘Calls Made’ by the ‘Deals Closed’.
• Notice that the cell and text colour have been changed.
• There is also a comment in cell F2.
You can insert comments into cells by right clicking on the cell and selecting ‘Insert Comment’. Comments are free text boxes that you can use to contain information about a cell without affecting its contents.
• In cell F3, I have also created a data validation list.
You can get more information on how to do this under Aesthetics 1: Dropdown Lists (Data Validation). This is just to illustrate the Paste Special function, though.
Step 1:
To illustrate what the 6 options, ‘All’, ‘Formulas’, ‘Values’, ‘Formats’, ‘Comments’ and ’Validation’ do, we will copy the whole cell selection from F2 to F13 and copy and Paste Special on each column, using one of the Paste functions.
All
Notice that every aspect has been copied over – formulas, formats, the comment and the data validation.
So what’s with the error messages?
Notice that when we copied over the cells F2 to F13, their contents contained the formulas ‘=C2/D2’? Now, the formula has moved 1 complete column to the right to cell ‘=D2/E2’ or ’=D3/E3’.
Why?
Because when we did a copy and Paste Special ‘All’ 1 column to the right, all the cell references moved as well in exactly the same way.
This is part of how Excel ‘intelligently’ tries to assist us. Don’t worry about this, though, as we expected this to occur.
Formulas
Notice that only the formulas within each cell have been copied. Notice also that the formulas now indicate ‘=E2/F2’.
As with Paste Special ‘All’, whenever formulas are copied, they are moved exactly the same number of rows or columns and in the same direction.
Values
Now, notice that only the values in the cells have been copied over?
There are no formulas, formatting, comments or data validation lists. This is for when you just want the results of the cells copied, nothing more.
Formats
Observe that only the aesthetic formatting has been copied.
The cells are grey and if you type in any text, they will be red, like the original list. If you have applied Conditional Formatting to the cells, they will be carried over as well.
Notice that only the comment has been copied over.
Nothing more.
Validation
Notice that only the data validation list has been copied over.
Nothing more.
The remaining options are variations of those we have looked at.
All using Source Theme copies the theme of your cells, should you have applied one.
You can access the Themes function by going to the ‘Page Layout’ tab and selecting ‘Themes’. This changes the overall look of your Excel spreadsheets.
All except borders is similar to All, except that the borders are not copied.
Column widths only copies the exact width of the original cells – more aesthetic than anything.
Formulas and number formats – copies only the formulas (remember that the cell references move accordingly) and the formatting of the contents of the cells.
Values and number formats – copies only the value of the contents of the cell; not its formula, and the formatting of the contents of the cells. | 1,378 | 6,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-23 | latest | en | 0.906166 |
https://www.gradesaver.com/textbooks/math/geometry/elementary-geometry-for-college-students-5th-edition/chapter-8-review-exercises-page-398/14b | 1,547,828,278,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660175.18/warc/CC-MAIN-20190118151716-20190118173716-00479.warc.gz | 791,395,347 | 11,694 | Elementary Geometry for College Students (5th Edition)
150 $cm^{2}$
sides of the triangle = 15 cm , 20cm , 25 cm. by pythagoras Theorem $15^{2}$ +$20^{2}$ = $25^{2}$ HENCE its a Right angled Triangle area of a right angled triangle = $1\div$2 $\times$15 $\times$20 = 150 $cm^{2}$ | 96 | 280 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2019-04 | latest | en | 0.821747 |
https://www.jiskha.com/questions/121073/what-is-the-base-number-for-729-625-as-expressed-as-a-power | 1,553,554,800,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204461.23/warc/CC-MAIN-20190325214331-20190326000331-00518.warc.gz | 795,712,470 | 4,965 | # math
what is the base number for 729, 625 as expressed as a power?
1. 👍 0
2. 👎 0
3. 👁 56
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Simplify the following number. ∛729 I think it is ∛9, but I do not know. Please help, and check my answer. I will appreciate it if you guys can answer this as soon as possible. It is part of a RATIONAL EXPONENTS question. The | 694 | 2,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-13 | latest | en | 0.926032 |
http://www.jiskha.com/display.cgi?id=1281718926 | 1,496,078,744,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612502.45/warc/CC-MAIN-20170529165246-20170529185246-00113.warc.gz | 713,071,535 | 4,343 | # college math
posted by on .
I do not understand a problem from a text book or how to solve the problem for the answer. Could shomeone show me the steps (show work) on how to solve this question.
The sum of the intergers from 1 through n is n(n+1)/2. the sum of the squares of the intergers from 1 through n is n(n+1)(2n+1)/6. The sum of the cubes of the integers from 1 through n is n squared(n+1)squared/4. Use the appropriate expressions to find the following values.
(A). The sum of the integers from 1 through 30.
(B) The sum of the squares of the integers from 1 through 30
(C) The sum of the cubes of integers from 1 through 30.
(D) The square of the sum of the integers from 1 though 30.
(E) The cube of the sum of integers from 1 through 30.
I's sure its much easier than what I am making it out to be. I tried working the problems but am lost in what it' asking.
• college math - ,
They have given you the formulas for each of the question for A, B, and C
I will do B
Sum of squares from 1 to n
= n(n+1)(2n+1)/6
so the sum of the squares from 1 to 30, n=30
just plug it in
sum of squares = 30(31)(61)/6 = 9455
do A and C the same way
Once you have the answer for A,
square that result for D
cube the result for A to get E
• college math - ,
During a walk, Dave completed the first two miles at a pace of 3mph. He walked the next two miles at a pace of 4.5 mph. Would his walk take more, less or the samr time if he walked the entire distance of four miles at a steady pace of 3.5 mph? | 430 | 1,505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2017-22 | latest | en | 0.947538 |
https://plfa.inf.ed.ac.uk/Quantifiers/ | 1,669,694,597,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710685.0/warc/CC-MAIN-20221129031912-20221129061912-00730.warc.gz | 510,517,984 | 15,754 | ```module plfa.part1.Quantifiers where
```
This chapter introduces universal and existential quantification.
## Imports
```import Relation.Binary.PropositionalEquality as Eq
open Eq using (_≡_; refl)
open import Data.Nat using (ℕ; zero; suc; _+_; _*_)
open import Relation.Nullary using (¬_)
open import Data.Product using (_×_; proj₁; proj₂) renaming (_,_ to ⟨_,_⟩)
open import Data.Sum using (_⊎_; inj₁; inj₂)
open import plfa.part1.Isomorphism using (_≃_; extensionality)
open import Function using (_∘_)
```
## Universals
We formalise universal quantification using the dependent function type, which has appeared throughout this book. For instance, in Chapter Induction we showed addition is associative:
``+-assoc : ∀ (m n p : ℕ) → (m + n) + p ≡ m + (n + p)``
which asserts for all natural numbers `m`, `n`, and `p` that `(m + n) + p ≡ m + (n + p)` holds. It is a dependent function, which given values for `m`, `n`, and `p` returns evidence for the corresponding equation.
In general, given a variable `x` of type `A` and a proposition `B x` which contains `x` as a free variable, the universally quantified proposition `∀ (x : A) → B x` holds if for every term `M` of type `A` the proposition `B M` holds. Here `B M` stands for the proposition `B x` with each free occurrence of `x` replaced by `M`. Variable `x` appears free in `B x` but bound in `∀ (x : A) → B x`.
Evidence that `∀ (x : A) → B x` holds is of the form
``λ (x : A) → N x``
where `N x` is a term of type `B x`, and `N x` and `B x` both contain a free variable `x` of type `A`. Given a term `L` providing evidence that `∀ (x : A) → B x` holds, and a term `M` of type `A`, the term `L M` provides evidence that `B M` holds. In other words, evidence that `∀ (x : A) → B x` holds is a function that converts a term `M` of type `A` into evidence that `B M` holds.
Put another way, if we know that `∀ (x : A) → B x` holds and that `M` is a term of type `A` then we may conclude that `B M` holds:
```∀-elim : ∀ {A : Set} {B : A → Set}
→ (L : ∀ (x : A) → B x)
→ (M : A)
-----------------
→ B M
∀-elim L M = L M
```
As with `→-elim`, the rule corresponds to function application.
Functions arise as a special case of dependent functions, where the range does not depend on a variable drawn from the domain. When a function is viewed as evidence of implication, both its argument and result are viewed as evidence, whereas when a dependent function is viewed as evidence of a universal, its argument is viewed as an element of a data type and its result is viewed as evidence of a proposition that depends on the argument. This difference is largely a matter of interpretation, since in Agda a value of a type and evidence of a proposition are indistinguishable.
Dependent function types are sometimes referred to as dependent products, because if `A` is a finite type with values `x₁ , ⋯ , xₙ`, and if each of the types `B x₁ , ⋯ , B xₙ` has `m₁ , ⋯ , mₙ` distinct members, then `∀ (x : A) → B x` has `m₁ * ⋯ * mₙ` members. Indeed, sometimes the notation `∀ (x : A) → B x` is replaced by a notation such as `Π[ x ∈ A ] (B x)`, where `Π` stands for product. However, we will stick with the name dependent function, because (as we will see) dependent product is ambiguous.
Show that universals distribute over conjunction:
```postulate
∀-distrib-× : ∀ {A : Set} {B C : A → Set} →
(∀ (x : A) → B x × C x) ≃ (∀ (x : A) → B x) × (∀ (x : A) → C x)
```
Compare this with the result (`→-distrib-×`) in Chapter Connectives.
#### Exercise `⊎∀-implies-∀⊎` (practice)
Show that a disjunction of universals implies a universal of disjunctions:
```postulate
⊎∀-implies-∀⊎ : ∀ {A : Set} {B C : A → Set} →
(∀ (x : A) → B x) ⊎ (∀ (x : A) → C x) → ∀ (x : A) → B x ⊎ C x
```
Does the converse hold? If so, prove; if not, explain why.
#### Exercise `∀-×` (practice)
Consider the following type.
```data Tri : Set where
aa : Tri
bb : Tri
cc : Tri
```
Let `B` be a type indexed by `Tri`, that is `B : Tri → Set`. Show that `∀ (x : Tri) → B x` is isomorphic to `B aa × B bb × B cc`. Hint: you will need to postulate a version of extensionality that works for dependent functions.
## Existentials
Given a variable `x` of type `A` and a proposition `B x` which contains `x` as a free variable, the existentially quantified proposition `Σ[ x ∈ A ] B x` holds if for some term `M` of type `A` the proposition `B M` holds. Here `B M` stands for the proposition `B x` with each free occurrence of `x` replaced by `M`. Variable `x` appears free in `B x` but bound in `Σ[ x ∈ A ] B x`.
We formalise existential quantification by declaring a suitable inductive type:
```data Σ (A : Set) (B : A → Set) : Set where
⟨_,_⟩ : (x : A) → B x → Σ A B
```
We define a convenient syntax for existentials as follows:
```Σ-syntax = Σ
infix 2 Σ-syntax
syntax Σ-syntax A (λ x → B) = Σ[ x ∈ A ] B
```
This is our first use of a syntax declaration, which specifies that the term on the left may be written with the syntax on the right. The special syntax is available only when the identifier `Σ-syntax` is imported.
Evidence that `Σ[ x ∈ A ] B x` holds is of the form `⟨ M , N ⟩` where `M` is a term of type `A`, and `N` is evidence that `B M` holds.
Equivalently, we could also declare existentials as a record type:
```record Σ′ (A : Set) (B : A → Set) : Set where
field
proj₁′ : A
proj₂′ : B proj₁′
```
Here record construction
``````record
{ proj₁′ = M
; proj₂′ = N
}``````
corresponds to the term
``⟨ M , N ⟩``
where `M` is a term of type `A` and `N` is a term of type `B M`.
Products arise as a special case of existentials, where the second component does not depend on a variable drawn from the first component. When a product is viewed as evidence of a conjunction, both of its components are viewed as evidence, whereas when it is viewed as evidence of an existential, the first component is viewed as an element of a datatype and the second component is viewed as evidence of a proposition that depends on the first component. This difference is largely a matter of interpretation, since in Agda a value of a type and evidence of a proposition are indistinguishable.
Existentials are sometimes referred to as dependent sums, because if `A` is a finite type with values `x₁ , ⋯ , xₙ`, and if each of the types `B x₁ , ⋯ B xₙ` has `m₁ , ⋯ , mₙ` distinct members, then `Σ[ x ∈ A ] B x` has `m₁ + ⋯ + mₙ` members, which explains the choice of notation for existentials, since `Σ` stands for sum.
Existentials are sometimes referred to as dependent products, since products arise as a special case. However, that choice of names is doubly confusing, since universals also have a claim to the name dependent product and since existentials also have a claim to the name dependent sum.
A common notation for existentials is `∃` (analogous to `∀` for universals). We follow the convention of the Agda standard library, and reserve this notation for the case where the domain of the bound variable is left implicit:
```∃ : ∀ {A : Set} (B : A → Set) → Set
∃ {A} B = Σ A B
∃-syntax = ∃
syntax ∃-syntax (λ x → B) = ∃[ x ] B
```
The special syntax is available only when the identifier `∃-syntax` is imported. We will tend to use this syntax, since it is shorter and more familiar.
Given evidence that `∀ x → B x → C` holds, where `C` does not contain `x` as a free variable, and given evidence that `∃[ x ] B x` holds, we may conclude that `C` holds:
```∃-elim : ∀ {A : Set} {B : A → Set} {C : Set}
→ (∀ x → B x → C)
→ ∃[ x ] B x
---------------
→ C
∃-elim f ⟨ x , y ⟩ = f x y
```
In other words, if we know for every `x` of type `A` that `B x` implies `C`, and we know for some `x` of type `A` that `B x` holds, then we may conclude that `C` holds. This is because we may instantiate that proof that `∀ x → B x → C` to any value `x` of type `A` and any `y` of type `B x`, and exactly such values are provided by the evidence for `∃[ x ] B x`.
Indeed, the converse also holds, and the two together form an isomorphism:
```∀∃-currying : ∀ {A : Set} {B : A → Set} {C : Set}
→ (∀ x → B x → C) ≃ (∃[ x ] B x → C)
∀∃-currying =
record
{ to = λ{ f → λ{ ⟨ x , y ⟩ → f x y }}
; from = λ{ g → λ{ x → λ{ y → g ⟨ x , y ⟩ }}}
; from∘to = λ{ f → refl }
; to∘from = λ{ g → extensionality λ{ ⟨ x , y ⟩ → refl }}
}
```
The result can be viewed as a generalisation of currying. Indeed, the code to establish the isomorphism is identical to what we wrote when discussing implication.
Show that existentials distribute over disjunction:
```postulate
∃-distrib-⊎ : ∀ {A : Set} {B C : A → Set} →
∃[ x ] (B x ⊎ C x) ≃ (∃[ x ] B x) ⊎ (∃[ x ] C x)
```
#### Exercise `∃×-implies-×∃` (practice)
Show that an existential of conjunctions implies a conjunction of existentials:
```postulate
∃×-implies-×∃ : ∀ {A : Set} {B C : A → Set} →
∃[ x ] (B x × C x) → (∃[ x ] B x) × (∃[ x ] C x)
```
Does the converse hold? If so, prove; if not, explain why.
#### Exercise `∃-⊎` (practice)
Let `Tri` and `B` be as in Exercise `∀-×`. Show that `∃[ x ] B x` is isomorphic to `B aa ⊎ B bb ⊎ B cc`.
## An existential example
Recall the definitions of `even` and `odd` from Chapter Relations:
```data even : ℕ → Set
data odd : ℕ → Set
data even where
even-zero : even zero
even-suc : ∀ {n : ℕ}
→ odd n
------------
→ even (suc n)
data odd where
odd-suc : ∀ {n : ℕ}
→ even n
-----------
→ odd (suc n)
```
A number is even if it is zero or the successor of an odd number, and odd if it is the successor of an even number.
We will show that a number is even if and only if it is twice some other number, and odd if and only if it is one more than twice some other number. In other words, we will show:
`even n` iff `∃[ m ] ( m * 2 ≡ n)`
`odd n` iff `∃[ m ] (1 + m * 2 ≡ n)`
By convention, one tends to write constant factors first and to put the constant term in a sum last. Here we’ve reversed each of those conventions, because doing so eases the proof.
Here is the proof in the forward direction:
```even-∃ : ∀ {n : ℕ} → even n → ∃[ m ] ( m * 2 ≡ n)
odd-∃ : ∀ {n : ℕ} → odd n → ∃[ m ] (1 + m * 2 ≡ n)
even-∃ even-zero = ⟨ zero , refl ⟩
even-∃ (even-suc o) with odd-∃ o
... | ⟨ m , refl ⟩ = ⟨ suc m , refl ⟩
odd-∃ (odd-suc e) with even-∃ e
... | ⟨ m , refl ⟩ = ⟨ m , refl ⟩
```
We define two mutually recursive functions. Given evidence that `n` is even or odd, we return a number `m` and evidence that `m * 2 ≡ n` or `1 + m * 2 ≡ n`. We induct over the evidence that `n` is even or odd:
• If the number is even because it is zero, then we return a pair consisting of zero and the evidence that twice zero is zero.
• If the number is even because it is one more than an odd number, then we apply the induction hypothesis to give a number `m` and evidence that `1 + m * 2 ≡ n`. We return a pair consisting of `suc m` and evidence that `suc m * 2 ≡ suc n`, which is immediate after substituting for `n`.
• If the number is odd because it is the successor of an even number, then we apply the induction hypothesis to give a number `m` and evidence that `m * 2 ≡ n`. We return a pair consisting of `suc m` and evidence that `1 + m * 2 ≡ suc n`, which is immediate after substituting for `n`.
This completes the proof in the forward direction.
Here is the proof in the reverse direction:
```∃-even : ∀ {n : ℕ} → ∃[ m ] ( m * 2 ≡ n) → even n
∃-odd : ∀ {n : ℕ} → ∃[ m ] (1 + m * 2 ≡ n) → odd n
∃-even ⟨ zero , refl ⟩ = even-zero
∃-even ⟨ suc m , refl ⟩ = even-suc (∃-odd ⟨ m , refl ⟩)
∃-odd ⟨ m , refl ⟩ = odd-suc (∃-even ⟨ m , refl ⟩)
```
Given a number that is twice some other number we must show it is even, and a number that is one more than twice some other number we must show it is odd. We induct over the evidence of the existential, and in the even case consider the two possibilities for the number that is doubled:
• In the even case for `zero`, we must show `zero * 2` is even, which follows by `even-zero`.
• In the even case for `suc n`, we must show `suc m * 2` is even. The inductive hypothesis tells us that `1 + m * 2` is odd, from which the desired result follows by `even-suc`.
• In the odd case, we must show `1 + m * 2` is odd. The inductive hypothesis tell us that `m * 2` is even, from which the desired result follows by `odd-suc`.
This completes the proof in the backward direction.
#### Exercise `∃-even-odd` (practice)
How do the proofs become more difficult if we replace `m * 2` and `1 + m * 2` by `2 * m` and `2 * m + 1`? Rewrite the proofs of `∃-even` and `∃-odd` when restated in this way.
```-- Your code goes here
```
#### Exercise `∃-+-≤` (practice)
Show that `y ≤ z` holds if and only if there exists a `x` such that `x + y ≡ z`.
```-- Your code goes here
```
## Existentials, Universals, and Negation
Negation of an existential is isomorphic to the universal of a negation. Considering that existentials are generalised disjunction and universals are generalised conjunction, this result is analogous to the one which tells us that negation of a disjunction is isomorphic to a conjunction of negations:
```¬∃≃∀¬ : ∀ {A : Set} {B : A → Set}
→ (¬ ∃[ x ] B x) ≃ ∀ x → ¬ B x
¬∃≃∀¬ =
record
{ to = λ{ ¬∃xy x y → ¬∃xy ⟨ x , y ⟩ }
; from = λ{ ∀¬xy ⟨ x , y ⟩ → ∀¬xy x y }
; from∘to = λ{ ¬∃xy → extensionality λ{ ⟨ x , y ⟩ → refl } }
; to∘from = λ{ ∀¬xy → refl }
}
```
In the `to` direction, we are given a value `¬∃xy` of type `¬ ∃[ x ] B x`, and need to show that given a value `x` that `¬ B x` follows, in other words, from a value `y` of type `B x` we can derive false. Combining `x` and `y` gives us a value `⟨ x , y ⟩` of type `∃[ x ] B x`, and applying `¬∃xy` to that yields a contradiction.
In the `from` direction, we are given a value `∀¬xy` of type `∀ x → ¬ B x`, and need to show that from a value `⟨ x , y ⟩` of type `∃[ x ] B x` we can derive false. Applying `∀¬xy` to `x` gives a value of type `¬ B x`, and applying that to `y` yields a contradiction.
The two inverse proofs are straightforward, where one direction requires extensionality.
Show that existential of a negation implies negation of a universal:
```postulate
∃¬-implies-¬∀ : ∀ {A : Set} {B : A → Set}
→ ∃[ x ] (¬ B x)
--------------
→ ¬ (∀ x → B x)
```
Does the converse hold? If so, prove; if not, explain why.
#### Exercise `Bin-isomorphism` (stretch)
Recall that Exercises Bin, Bin-laws, and Bin-predicates define a datatype `Bin` of bitstrings representing natural numbers, and asks you to define the following functions and predicates:
``````to : ℕ → Bin
from : Bin → ℕ
Can : Bin → Set``````
And to establish the following properties:
``````from (to n) ≡ n
----------
Can (to n)
Can b
---------------
to (from b) ≡ b``````
Using the above, establish that there is an isomorphism between `ℕ` and `∃[ b ] Can b`.
We recommend proving the following lemmas which show that, for a given binary number `b`, there is only one proof of `One b` and similarly for `Can b`.
``````≡One : ∀ {b : Bin} (o o′ : One b) → o ≡ o′
≡Can : ∀ {b : Bin} (cb cb′ : Can b) → cb ≡ cb′``````
Many of the alternatives for proving `to∘from` turn out to be tricky. However, the proof can be straightforward if you use the following lemma, which is a corollary of `≡Can`.
``proj₁≡→Can≡ : {cb cb′ : ∃[ b ] Can b} → proj₁ cb ≡ proj₁ cb′ → cb ≡ cb′``
```-- Your code goes here
```
## Standard library
Definitions similar to those in this chapter can be found in the standard library:
```import Data.Product using (Σ; _,_; ∃; Σ-syntax; ∃-syntax)
```
## Unicode
This chapter uses the following unicode:
``````Π U+03A0 GREEK CAPITAL LETTER PI (\Pi)
Σ U+03A3 GREEK CAPITAL LETTER SIGMA (\Sigma)
∃ U+2203 THERE EXISTS (\ex, \exists)`````` | 4,925 | 15,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-49 | latest | en | 0.789223 |
https://hectorpefo.github.io/2019-02-08-Century-Product/ | 1,571,061,275,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653247.25/warc/CC-MAIN-20191014124230-20191014151730-00541.warc.gz | 495,881,144 | 4,571 | Given any three random integers — X, Y and Z — what are the chances that their product is divisible by 100?
## Solution
Reading the question charitably (since “random integer” has no specific meaning), there will be an answer if there is a limit for a uniform distribution of positive integers up to some number $N$. But we can ignore that technicality, and make do with the idealization that since every second, fourth, fifth, and twenty-fifth integer are divisible by $2, 4, 5,$ and $25$, the chances of getting a random integer divisible by those numbers are $1/2$, $1/4$, $1/5$, and $1/25$.
The product $XYZ$ is divisible by $100$ if and only if it has at least two factors of $2$ and at least two factors of $5$. The chance that it has at least two factors of $2$ is $1$ minus the chance that it has exactly zero factors of $2$ (which is $(1/2)^3$ or $1/8$) and minus the chance that it has exactly one factor of $2$. It has exactly one factor of two if one of the three numbers itself has exactly one factor of $2$ and the other two are odd. A number has exactly one factor of $2$ if it is one of the half of all even numbers that is not divisible by $4$; so $1/2$ times $1/2$, or $1/4$ of all numbers have this property. So:
The calculation of the chance that $XYZ$ is divisible by $25$ is similar. In this case, the chance that a number has exactly one factor of $5$ is the chance that it has at least one (which is $1/5$) minus the chance that it has at least two (which is $1/25$). So:
These two probabilities are independent, because every twenty-fifth number divisible by $4$ is also divisible by $25$ (and vice versa). Therefore the probability that $XYZ$ is divisible by both $4$ and $25$, that is, that it is divisible by $100$, is their product, which is exactly $.1243$. | 471 | 1,792 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2019-43 | longest | en | 0.968845 |
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» Technology
Jedi Ambassador Shuttle (Light Freighter)
Navigational Stats:
Hyperspeed: 4
Sublight Speed: 45 MGLT
Max Speed: 450 km/h
Manoeuvrability: 4.00
Sensors: 3
ECM: 0
Cargo Stats:
Weight: 90 T
Volume: 720 m³
Weight Cap: 3 T
Volume Cap: 20 m³
Max Passengers: 4
Party Slot Size: 4.00
Hull Stats:
Length: 10 m
Hull: 30
Deflector Shields: 50
Ionic Capacity: 25
115,145 Credits
Landing CapacityFlight Grade Repulsorlifts
Required Raw Materials:
Quantum (Armour): 9
Meleenium (Durasteel): 88
Ardanium (Fuel Canisters): 24
Rudic (Electronics): 55
Rockivory (Antigrav Units / Mechanical Parts): 21
Varmigio (Hyperdrives): 74
Lommite (Transparisteel): 11
Durelium (Hyperdrives): 25
Developed during the clone wars specifically for use by Jedi ambassadors, the Eta-class Shuttle, more commonly called the Jedi Ambassador Shuttle, was engineered to provide reliable transportation and act as an official symbol of peaceful negotiation by the Galactic Republic, the ruling government of the time.
The shuttle's design consists of a forward mounted command pod, a small cabin and three large wings.Designed specifically for purely diplomatic purposes, the Jedi Ambassador Shuttle carried no armaments and was only lightly shielded, which made the bold statement that neither the ship nor its passengers should be perceived as a threat. The shuttle was small enough to be a fighter, but the complete lack of fighting power and the passenger count saw the ship designated a freighter instead. The shuttle's hyper and sublight speeds left little to be desired and the cargo capacity meant passengers could only bring the most essential of items and a very limited amount of supplies. Jedi were often unencumbered by personal goods and so more priority was given to installing other systems like sensors. The mid-range sensor package meant that the crew, often defenseless, could monitor and avoid trouble or use it to gather useful information during official investigations and reports.
Today, the Jedi Ambassador Shuttle still sees use on diplomatic missions, or acting as small passenger shuttles on some worlds. No longer produced, this relic of the Clone Wars has become somewhat of a collector's item.
Public Custom Images: | null | null | null | null | null | null | null | null | null |
null | null | null | null | null | null | Susan Clark argues with another protester about the Affordable Healthcare Act outside the U.S. Supreme Court on June 28, 2012 in Washington, D.C.
The Supreme Court is set to hear arguments tomorrow in what is considered the highest profile case of this term. Sebelius v. Hobby Lobby Stores touches on some of the most hot-button political and legal questions in the country right now, including the Affordable Care Act, abortion and the impact of the Supreme Court’s Citizens United ruling. The case involves two for-profit companies objecting to contraception requirements under the Affordable Care Act. The question the court will consider: Do corporations have the same religious rights as individuals? Diane and her guests preview the arguments and discuss the political and legal implications at stake.
• Mark Rienzi professor of constitutional law, Catholic University of America and senior counsel, Becket Fund for Religious Liberty.
• 10:06:53
MS. DIANE REHMThanks for joining us. I'm Diane Rehm. For the first time since 2012, the Affordable Care Act is back in front of the Supreme Court. Two corporations are arguing they have religious objections to providing certain forms of contraception to their female employees, a requirement under the Affordable Care Act.
• 10:07:17
MS. DIANE REHMHere to preview tomorrow's arguments in the case: Joan Biskupic -- she's editor for legal affairs for Reuters -- Mark Rienzi, senior counsel at the Becket Fund for Religious Liberty, and, joining us from the studios of NPR West in Los Angeles, Sandra Fluke. She's democratic candidate for the California Senate District 26. I hope you'll join us in this most important discussion. Give us a call at 800-433-8850. I'd like to hear your views so you can email us to Follow us on Facebook or Twitter. And welcome to all of you.
• 10:08:09
MR. MARK RIENZIGood morning, Diane.
• 10:08:11
MS. JOAN BISKUPICMorning, Diane.
• 10:08:11
MS. SANDRA FLUKEThank you.
• 10:08:12
REHMGood to have you with us. Joan Biskupic, people refer to this as the Hobby Lobby case, which sounds light but is so important. Tell us why.
• 10:08:28
BISKUPICIt really doesn't get any bigger than this. The law that's being challenged here is the Obama-sponsored healthcare law that everybody knows was passed four years ago, just about this week. Two years ago, in the biggest case of the term at that point, the Supreme Court upheld most of it. It implicates religious rights, reproductive rights. It's got just a little bit of everything.
• 10:08:56
BISKUPICAnd the two sides are sides that represent pro and con on abortion rights issues, too. So we have lots going on here. And at the center of it are two for-profit companies run by people with deeply-felt religious beliefs. One is the Hobby Lobby company based in Oklahoma that you mentioned in which the family-run operation -- it is for-profit.
• 10:09:22
BISKUPICBut they say their religious beliefs should entitle them to an exemption from the contraceptive mandate of the Obama-sponsored healthcare law based on two things: one, a religious freedom restoration act statute that was a 1993 law adopted by Congress, but also the corporations say that they have First Amendment religious free exercise rights that also would allow it to be exempt from this requirement that they provide healthcare coverage that includes contraceptives -- four disputed contraceptives within several that would actually interfere with implantation of the embryo.
• 10:10:07
BISKUPICThere's a lot of science involved here that will also come into play, but mainly what it comes down to, Diane, is a question of whether these for-profit corporations can be covered by a religious exemption in federal statute and the constitution.
• 10:10:20
REHMThat they could actually exercise freedom of religious rights.
• 10:10:29
BISKUPICExactly. And this also implicates, to add one more layer, the Citizens United ruling of 2010 where the Supreme Court said that corporations have free speech rights in the campaign finance regulation area, so this is another layer of that. And actually a lot of people say, oh, well, you know, are corporations really people? But, frankly, in the law, corporations are, for legal purposes, people to an extent, but the Supreme Court has never said that for-profit corporations could be entitled to this, which makes it an even more momentous case now.
• 10:11:02
REHMJoan Biskupic, she's editor in charge of legal affairs at Reuters News. Turning to you, Mark Rienzi, why should these two for-profit companies, Hobby Lobby and Conestoga Wood, why should they be exempt from the contraceptive mandate?
• 10:11:29
RIENZIBecause Congress and the constitution say so. The bottom line is that, when you open a family business in the United States, you don't give up your religious freedom rights. You don't, when you open up a for-profit doctor's office, have to start immediately performing abortions, and you don't, when you run a for-profit pharmacy, have to start compounding drugs to make -- for lethal injections. There's no rule that says, when you make money, you give up your religious exercise. The Green family exercise their religion when they run their business.
• 10:11:57
REHMHow do they do so?
• 10:11:59
RIENZIIn all sorts of ways. They pay their employees roughly twice the minimum wage to start, $14 an hour to start. They close their business early every evening so that their workers can be home with their families. They close entirely on Sundays. They refuse to sell items related to alcohol or sublease their stores to liquor stores in bad neighborhoods because they think it's bad for people.
• 10:12:19
RIENZISo the Greens give up millions of dollars every year because they exercise religion while they're running their business. This is a case that's utterly unnecessary though because the government has so many other ways to achieve its goals here that there's really no reason to force the Greens to violate their religion.
• 10:12:33
REHMMark Rienzi, he's professor of constitutional law at Catholic University. He's senior counsel for the Becket Fund for Religious Liberty. He's one of the attorneys for Hobby Lobby. And turning to you, Sandra Fluke, you see it differently. Tell us how.
• 10:12:59
FLUKEI do. I want to actually just back up one step and make sure that everyone who's listening understands exactly which type of organizations we're talking about in this Hobby Lobby case, as well as the Conestoga Wood Specialties case that is its partner at the Supreme Court. First of all, this is not about houses of worship, churches or synagogues or mosques. Those types of organizations are not required under the Affordable Care Act to provide contraception for their employees.
• 10:13:25
FLUKEThey're completely exempt. And then, even a step beyond that, religiously-affiliated non-profits, like universities or hospitals, have also been offered a policy that balances their concerns against the health and welfare of their employees and students. So those types of organizations are removed from the process. They don't need to use their money or their resources to arrange for insurance coverage of contraception.
• 10:13:53
FLUKEInstead, the insurance company works directly with employees and students to make sure that they have access to the comprehensive healthcare that they need. So that's already in place in the policy and not being argued in these cases. This is about a third set of organizations, for-profit corporations that have no religious identity in the corporation itself but simply have owners who happen to have religious beliefs, absolutely, deep-held, sincere religious beliefs, but the beliefs of the owners and not a part of the mission of the organization or its identity.
• 10:14:30
REHMAnd if I could interrupt you for a moment, you heard Mark refer, however, to the practices of Hobby Lobby, for example, not selling anything related to alcohol, closing the stores on Sundays, being located in non-alcohol related areas. Doesn't that change the picture slightly?
• 10:15:01
FLUKEI don't think that it does because I agree that anyone who opens a business and goes into the marketplace does not give up their own religious rights and concerns, but they don't gain the rights to impose them upon others. So those types of business decisions that don't have a negative impact on their employees are completely within the rights of these corporations and their owners.
• 10:15:25
FLUKEBut where we come to the sticking point is when we need to try to balance the need for affordable health insurance access and a number of other rights for both employees and customers of business against those of the owners of the corporation. And that's where, in our society, we have to find the right balance that protects the religious beliefs of individuals but not of corporations against the health and welfare of employees and customers and the public.
• 10:15:53
REHMSandra Fluke, she's democratic candidate for California Senate District 26. She is a social justice attorney in Los Angeles. If you'd like to join us, 800-433-8850. Joan Biskupic, will the solicitor general make an argument about women's health and paying for effective birth control?
• 10:16:28
BISKUPICYes, because there are two questions here. First, the question is, can a for-profit corporation actually assert a religious free exercise right? But then, once that's addressed, the second question is whether the government has a compelling interest here to nonetheless burden those rights. And what comes into that question are some of the things that Sandra Fluke just raised in terms of women's access to healthcare, women's access to contraceptives.
• 10:17:00
BISKUPICAnd the government makes a very strong argument in its brief about the importance of healthcare for women employees and the importance of birth control and how much various percentages of unwanted pregnancies and health issue questions that necessarily the government wants to raise to say, yes, we have a compelling interest to require this kind of coverage.
• 10:17:25
BISKUPICSo those kinds of health issue questions will come up. I imagine, given that the justices have now extended the oral arguments to a total of 90 minutes, rather than 60 minutes, we're going to see a whole range. We're going to see a lot on corporate rights, religious rights, and then get down to the nitty-gritty of what this law requires of employers and what the benefits might be for employees.
• 10:17:49
REHMHow unusual is it that they've extended the arguments to 90 minutes?
• 10:17:55
BISKUPICIt's very unusual. They only do that maybe once or twice a term, and I think it's a signal of the many faceted elements that come into this. And just so your listeners know, the two people arguing are two people who we've seen up there before, Solicitor General Don Verrilli opposing Paul Clement, a former solicitor general under George W. Bush.
• 10:18:18
REHMJoan Biskupic, editor in charge of legal affairs at Reuters News. Short break. When we come back, we'll talk further, take your calls, your email. I look forward to hearing from you.
• 10:19:59
REHMWelcome back. We're talking about a case that's coming before the Supreme Court tomorrow, two corporations arguing that they have religious objections to providing certain forms of contraception to their female employees, which is a requirement under the Affordable Care Act. Here in the studio, Mark Rienzi. He is with the Becket Fund for Religious Liberty, one of the lawyers for Hobby Lobby. Joan Biskupic is editor in charge of legal affairs at Reuters News. She's written biographies on Sandra Day O'Connor and Antonin Scalia.
• 10:20:51
REHMJoining us by ISDN from Los Angeles is Sandra Fluke. She's Democratic candidate for California Senate District 26. She's a social justice attorney in Los Angeles. And, Mark Rienzi, returning to you, it's not all contraceptives that Hobby Lobby and Conestoga would object to. Explain which contraceptives are the problem and why.
• 10:21:27
RIENZISure. You're right. Hobby Lobby, for a long time, has happily provided generous benefits, including coverage for most contraceptive methods. They're evangelical Christians. They're perfectly fine with standard contraception. Their objection is only to four drugs and devices that, according to the government -- this is conceded by the government -- may interfere with implantation of an already fertilized egg. And for the Greens, they understand that to be abortion.
• 10:21:48
REHMOK. First of all, give me the four that they are providing now.
• 10:21:55
RIENZIThey are not providing them now because the lower courts have ruled that they do have the right at issue here.
• 10:22:00
• 10:22:00
RIENZIBut four drugs are -- drugs and devices are Ella, which is sometimes called the week-after pill, Plan B, which is sometimes called the morning-after pill, and two types of intrauterine devices.
• 10:22:11
REHMAll right. And which are they providing?
• 10:22:15
RIENZIThey are providing all of the other forms of -- the pill, for example, and all the different variants of the pill. They have no objection to providing that.
• 10:22:23
REHMDo they provide condoms?
• 10:22:25
RIENZII don't know if condoms are ever -- condoms are not part of this mandate. I frankly don't know if condoms are generally part of an insurance package.
• 10:22:32
REHMInteresting. Interesting.
• 10:22:34
RIENZIMay I just say that...
• 10:22:35
• 10:22:35
RIENZI...the point is that the Greens, when they opened the family business and they started earning money, didn't give up the right to run that business according to moral, ethical, or religious beliefs. And if you look around our society, we actually see businesses do that all the time. We saw CVS a couple of weeks ago say they won't sell cigarettes anymore. It's bad for the bottom line, but they're doing it because they have a moral belief about selling cigarettes to people. That's something that's harmful for people.
• 10:22:59
RIENZIStarbucks, Whole Foods, Chipotle, companies, you know, on every corner exercise moral and ethical beliefs while they earn money. The government's view here that, if you're earning money, you can't do that, if you're earning money, you can't act on these other values is actually a bad view for society. I don't think most of us want businesses around the country operating only based on the bottom line.
• 10:23:21
RIENZIAnd it's such an extreme view that, in this case, the government is even saying -- not just Hobby Lobby and Conestoga, but Mardel Christian, which is a related company to Hobby Lobby, it's a Christian bookstore, right. All the signs say Mardel Christian in big letters on the front, and they sell Christian books and Bibles. The idea that that business is not exercising religion is actually a little bit absurd.
• 10:23:41
REHMYou -- CVS and other stores in regard to cigarettes, but isn't that actually related to public health because the surgeon general and others have said they're really bad for your health?
• 10:23:59
RIENZIYeah, and CVS legally could make money by selling them. And if corporations only sought after profits, they'd keep selling them. But CVS has made an ethical decision that it's bad to do that. Chipotle made a decision not to partner with the Boy Scouts in Utah last year because they didn't like the Boy Scouts' stance on gay scout masters. That's clearly not an exercise of bottom line. They said it didn't match with their values. Companies do that all the time. It's a great thing that companies do that all the time. They shouldn't be prohibited from doing it just because it's religious.
• 10:24:27
REHMJoan Biskupic, you say this case is really about abortion.
• 10:24:33
BISKUPICWell, it's about many things, but I think the abortion culture wars are definitely in the backdrop of this. This is an issue that we've had around since 1973 when the Supreme Court first declared a constitutional right for a woman to end a pregnancy. And it's an issue that just has not subsided in America. I mean, both sides feel very strong -- strong about where they're at.
• 10:25:00
BISKUPICAnd we saw that on the anniversary of Roe just recently in January. And the two sides that have joined this case to back either Hobby Lobby or back the federal government have very strong views on abortion rights. And, in fact, part of what comes into play with the four contraceptive methods that Mark mentioned are ones that are likened by these religious owners to abortion because they could interfere with the implantation of an embryo. So it's -- that's definitely in the background, and so is the idea of who decides.
• 10:25:34
BISKUPICI know that many of the women who will be out in front of the court tomorrow as part of demonstrations that we'll see really have tried to cast this in terms of equal rights for women, women being able to make decisions on reproductive rights that they'd be able to make for any kind of employer, whether it be for-profit or nonprofit corporations, so I think that's in the background. And we've had abortion cases marching toward the Supreme Court, but we haven't had a full-fledged abortion rights one for about five years now.
• 10:26:03
REHMAnd what about the relationship to Citizens United?
• 10:26:10
BISKUPICOK. To remind your listeners, in January of 2010, the Supreme Court ruled in Citizens United vs. Federal Election Commission that corporations and labor unions actually had speech rights that would -- that were being infringed on by the federal government in campaign finance laws. And as part of that ruling, the court said that these corporations would have the same kind of speech rights that individuals would have.
• 10:26:39
BISKUPICAnd so that -- there was a lot in that ruling, but the way it's been encapsulated in terms of the public view was the idea of corporations, you know, being people. But many times before that, the Supreme Court had actually allowed fairly robust speech rights for corporations. It's just that in the campaign finance sphere, the court had allowed government, because of its own interests in fighting corruption in politics, to have regulations.
• 10:27:09
BISKUPICNow there are many more precedents on the books for free speech rights for corporations than there are for any kind of religious rights, which is why the stakes in this case are frankly much higher. And you saw great public reaction in January of 2010 to Citizens United. And I think if the Supreme Court were to rule that there are special religious freedom rights here for corporations, you would probably see either equal or greater reaction.
• 10:27:34
REHMSandra Fluke, you've made the point that this is really a slippery slope not just with providing healthcare but in lots of other areas. Talk about that.
• 10:27:50
FLUKEYes, absolutely. I want to go back for just a moment to address the four types of contraception...
• 10:27:55
• 10:27:55
FLUKE...that these particular companies are objecting to. So their concern that these are types of birth control that could cause an abortion is not backed up by scientific fact and not what the medical community believes. But the facts of that part are not really the point. The point is that we're turning over the decision-making power on questions like that to owners of corporations.
• 10:28:20
FLUKEThis is your boss being able to make decisions about which types of medical procedures should or should not be covered on insurance based on their personal beliefs and not on medical or scientific fact. And for these particular corporations, there are four types of birth control that they're objecting to. For other corporations -- and there are a whole slew of cases in the courts right now -- it's all birth control.
• 10:28:46
FLUKEBut this goes even beyond just reproductive rights. It's about really any type of health insurance coverage, any type of healthcare service covered by insurance because there are owners of corporations who have religious beliefs who object to HIV or AIDS treatment, believing that -- and, you know, I'm paraphrasing someone else's belief here -- that that HIV and AIDS is the vengeance of a vengeful god visited upon sinners.
• 10:29:14
FLUKEThere are also religious beliefs who -- folks with religious beliefs that object to blood transfusions or to mental healthcare. So this could be about one's boss, one owner of a corporation, deciding that any part of health insurance coverage could be limited based purely on their personal belief. And then it even goes beyond health insurance coverage because there are a lot of nondiscrimination provisions and other types of protections for employees and for customers and the public, which could also be said to be in contrast -- or in conflict, excuse me, with a corporation's owner's belief.
• 10:29:54
FLUKEWe've seen in the past cases where employers have said that they didn't want to pay men and women equally because that conflicted with their religious beliefs. So these are the types of discrimination that could come into play if we said -- and this is what the court would be saying -- if we said that certain corporations and certain owners of corporations don't have to follow the same laws as the rest of us because of their religious beliefs.
• 10:30:20
REHMAll right. And to you, Mark, couldn't Hobby Lobby and other corporations continue to practice their own religious beliefs without somehow being heavily burdened by providing contraception of all types to employees?
• 10:30:45
RIENZINo, not at all because the government says to them, you will pay for the abortion drugs and devices, or we'll fine you massive, hundreds of millions of dollars a year in fines.
• 10:30:53
REHMSo they could opt out -- if they opted out. Is that correct?
• 10:30:58
RIENZIThey have to pay millions and millions of dollars in fines.
• 10:31:00
REHMIn fines.
• 10:31:01
RIENZIThey've have to cut all of their -- you know, they have thousands of employees. They'd have to cut those employees off from healthcare. It would be surprising to find out that the government thinks that's the right approach. If I could just go back to the slippery slope that was just described, a couple things about it.
• 10:31:15
RIENZIOne, RFRA's actually been the law -- the Religious Freedom Restoration Act has actually been the law for 20-plus years. And the government acknowledges that sole proprietorships and partnerships and lots of businesses actually can exercise religious exercise rights. But we've never seen the parade of horribles that was just described.
• 10:31:32
RIENZIAnd there's a reason for that. And the reason is that the Religious Freedom Restoration Act doesn't say religion always wins. It says religion wins unless the government can satisfy a test of proving it's got a compelling interest. Any time anybody's made the sort of claims that were just described, as I don't want to pay women equally to men or something like that, the government always wins those cases because it actually can prove it's got a compelling interest and this is the least restrictive means.
• 10:31:55
RIENZIThe difference here -- the problem for the government here is that it's so obvious that there are other ways to get these drugs to people without forcing the Greens to be involved. And to take just one, we have the healthcare exchanges open now, right. A few years ago, you might've said, well, how is the government ever going to get insurance to people?
• 10:32:12
RIENZIWell, we don't have to ask that question anymore because we have health insurance exchanges. And so if the government thinks that the policy offered by Hobby Lobby is insufficient in some way, well, that's OK. The government runs exchanges on which they provide people with insurance.
• 10:32:25
REHMSo you're saying employees of Hobby Lobby could go elsewhere.
• 10:32:30
RIENZIOf course. If they see the policy and say, gosh, you know, I may like this job, but I want more in my policy, it's a free country. Hobby Lobby doesn't stop anybody from getting a different policy. And if the government thinks it should be subsidized, the government can subsidize it. But there's no need to force Hobby Lobby and the Greens to pay for the abortion drug.
• 10:32:46
REHMIsn't there one flaw in your argument, which is that Roe v. Wade is and remains the law of the land?
• 10:32:57
RIENZII'm glad you brought up Roe v. Wade, and it's not a flaw at all. It's actually -- it shows why the argument is right. What Roe v. Wade says is that the government can't stop you from getting an abortion. But there were cases several years after that about whether that means you can make the government pay for your abortion. And the court said no. The court said that right in Roe is that the government has to let you do it, but you can't make the government fund it if they choose not to fund it.
• 10:33:17
RIENZIWell, here, we're not even talking about government funding it. We're talking about going out and finding private citizens and private businesses that disagree with it and saying, I'm going to make you be involved. The Rose v. Wade right has nothing to do with that. It doesn't include that. And, frankly, I think the court would be rather surprised if the government comes out and claims that the right includes something like that. It just doesn't...
• 10:33:34
REHMAre these actually abortion-inducing medications that we're talking about, Joan?
• 10:33:43
BISKUPICWell, I'm not equipped to say just the extent scientifically, and that is something that both sides have been backed by medical folks on. I don't think the court's going to get into it either, frankly.
• 10:33:56
REHMAnd you're listening to "The Diane Rehm Show." I'm going to open the phones here, 800-433-8850. First let's go to Brian in York, Penn. You're on the air.
• 10:34:14
• 10:34:15
• 10:34:15
BRIANI'm just calling in regards to -- I think it comes back to the fundamental issue, what we constitute life. I think this president and Sandra seems to think that the country wasn't founded on the constitution. And in the constitution, we give people rights. And it's -- to think that something's not a life unless it comes out of the womb -- once it comes out of the womb, then we have to give it constitutional rights -- it comes down to personal responsibility for reproduction.
• 10:34:51
BRIANIf you're going to have unprotected sex, there's chances of having a pregnancy, even if you have protected sex. A condom costs 50 cents. You shouldn't have to make a business pay for something. Practicing a religious belief is a lifestyle. It's a way of life.
• 10:35:06
REHMAll right. All right. Thanks for your call. And to you, Joan Biskupic.
• 10:35:13
BISKUPICWell, I think your first caller just demonstrated why there's so much emotion around this case. It really just brings in a lot of other elements that will not be present tomorrow in that marble courtroom. But it's why people are so energized over this case, the idea of, you know, when life begins, do some of these drugs actually stop that. What is the effectiveness of the IUD? Does the IUD cause more problems than it helps?
• 10:35:39
BISKUPICIn fact, I think it's the Guttmacher brief that points up the effectiveness of different kinds of contraceptives and why some women would turn to these that might be more controversial. But I think that, again, the caller just proves a point of why this case is getting so much attention.
• 10:35:55
REHMAll right. To Tim in Palm Desert, Calif. You're on the air.
• 10:36:02
TIMGood morning, Diane, and thank you. I'm honored to join Sandra Fluke in this discussion. As I see it, the employer is trying to assert an ownership claim to an aspect of the employee's life. The employer cannot own the employee. Slavery was abolished a long time ago, and the employer does not own the healthcare or the insurance. Any contribution the employer makes is in lieu of wages. She earns it every day she shows up for work. If the CEO doesn't like contraception, he's not asked to use it.
• 10:36:38
REHMAll right. Thanks for your call. I wonder about your comments, Mark.
• 10:36:44
RIENZISure. I think both callers, as Joan points out, show that there are, you know, heated and real disagreements about abortion in this country, and that's no surprise to anyone. Hobby Lobby and the Greens' point is simply, hey, that's a controversial issue, and they don't want to be involved in somebody's use of certain drugs. And it seems to me that, in a free country, they ought to be allowed to go about their lives and go about their business without the government forcing them to be involved in what they understand to be abortions. In a diverse place, that ought to not be such a big deal.
• 10:37:13
REHMMark Rienzi, he's professor of constitutional law at Catholic University. He's senior counsel for the Becket Fund for Religious Liberty. He's one of the attorneys for Hobby Lobby. Short break here. We'll be right back.
• 10:39:58
REHMAnd welcome back. We'll go right back to the phones. Let's go to Rob in Raleigh, N.C. Rob, you're on the air.
• 10:40:10
ROBHi, how are you today?
• 10:40:11
• 10:40:13
ROBI was just wondering about the rights of the employees and how it affects them, if Hobby Lobby chooses to force their religious beliefs on their employees, and an employee doesn't have the same religious belief.
• 10:40:28
REHMSandra Fluke, do you want to answer that?
• 10:40:32
FLUKEI think that gets right at the heart of the problem, that we can't look just at the religious beliefs of owners of the corporation. We have to understand that employees and customers, in other types of legal situations, have their own set of religious and moral beliefs and their own needs in their lives and need to be able to be the ones making their own personal decisions on these types of healthcare matters, rather than have their boss make those decisions for them.
• 10:40:59
FLUKEBut what I want to come back to is actually on the reproductive rights concerns, specifically how this type of decision, along with several other attacks on reproductive rights we've seen recently, box in employees and women specifically on how they can really actually access their rights, make those rights a reality because, yes, we have Roe v. Wade on the books. We have the right to access safe and legal abortion free of government interference. But is that something that women can actually act upon? And can they actually get other types of reproductive healthcare that they need?
• 10:41:38
FLUKEBecause, in addition to this attack that we're seeing at the Supreme Court of saying that employers should be able to refuse to provide comprehensive reproductive healthcare coverage on insurance, we've also seen that the exchanges in many states are restricting which types of coverage you can buy on the exchanges.
• 10:42:00
FLUKESo individual people, taking their own money and buying insurance on the exchange, they're not able to access insurance that covers all of the healthcare that they might require. We've also seen the government refusing to use federal and, in many cases, state dollars to provide certain types of reproductive healthcare to women who access their insurance that way. And then we've seen massive attacks on government funding for family planning and reproductive healthcare clinics.
• 10:42:28
FLUKESo this is about an attempt to really surround women from many sides and cut off their access to being able to afford these types of healthcare because those who are opposed to reproductive rights realize that Roe v. Wade is strong -- it's under constant attack, but it is a constitutional right -- and that the way to get to be able to actually control women's decisions on these matters is to cut off their ability to access those services so that we have a right on paper but not a right in reality.
• 10:42:57
• 10:42:58
BISKUPICYou know, she raises a couple points, and I'm wondering how they're going to play out in the courtroom tomorrow. Because what Sandra's getting to -- and what Mark had gotten to before in terms -- is kind of the second point of, what's the least restrictive means for the government to carry out its policy? And how effective might these health exchanges be as an alternative to forcing Hobby Lobby and Conestoga Wood to provide this kind of contraceptive insurance.
• 10:43:23
BISKUPICI'm not sure how much the government's going to want to get down into the nitty-gritty of what women can access or not. The government lawyers are making a couple claims about why this is a compelling interest for them. And it is -- you know, it's the public health issue, the equality issue for women, but also the question of that this is a comprehensive healthcare scheme that's supposed to apply to all employers.
• 10:43:48
BISKUPICThere are some exceptions. And I'm sure some of the Justices are going to say, you've already made some exceptions, how can you not make exceptions here? But the point that the recent dialog has gotten to is sort of, what would women's access be like and what are the restrictions? And, really, how can the government achieve its purpose for public health in an alternative way? And I think those are going to be where the rubber might meet the road, frankly.
• 10:44:14
• 10:44:15
RIENZIAnd the good news about that question is that the court doesn't have to guess at what it would look like because they can see it, because the government has exempted plans covering tens of millions of people just because the policies are grandfathered. You know, if you like your healthcare, you can keep it, was the promise. And that means that plans that were in effect before 2010, which don't cover these drugs, are perfectly legal. And those plans can cover tens of millions of people. So the government knows exactly what it would look like.
• 10:44:40
RIENZIIt would look like people can have plans, and if there's something that they don't like in their plan, they can either go to the exchanges -- and Sandra said that the exchanges don't cover some reproductive healthcare. The some reproductive healthcare that she's talking about is just surgical abortion. Every policy on every exchange has to cover the drugs and devices here. So there's no question that if the government thinks people need access to these drugs and devices that they can go to exchanges and they can get them.
• 10:45:04
RIENZIAnd if the government thinks the policies on its exchanges are too expensive, they can certainly subsidize it if they want to. But one broader point about that, the discussion about the exchanges not covering surgical abortion is, I think, a relevant one because it points to the fact that the government's arguments here and the arguments we're hearing from the left here would also support a flat-out surgical abortion mandate -- in other words, a requirement that every employer pay for surgical and late-term abortions.
• 10:45:29
RIENZIThe same argument would justify the same mandate. Well, it's reproductive healthcare, and, sorry, you gave up your right to exercise religion or to have a conscience about everything when you went into business. There's no dividing line between the government's argument in this case and a flat-out abortion mandate.
• 10:45:43
REHMSo you've got abortion. You've got the Affordable Care Act. How do you think that the debate the justices will have with lawyers is going to affect the outcome of the Affordable Care Act, Joan?
• 10:46:01
BISKUPICWell, it really shouldn't touch the Affordable Care Act itself, beyond this mandate. You know, there's just so much still happening with that 2010 law and, you know, all sorts of controversy in Congress and still in the states about how it's going to play out. But there's only one small provision that's at issue here. So however the justices rule on this contraceptive mandate, it will not -- it will -- it could undermine the government's overall effort, but it will not actually say anything broader about the Affordable Care Act itself.
• 10:46:34
REHMAll right. To Jody in Silver Spring, Md. Hi, you're on the air.
• 10:46:40
JODYThank you so much for taking my call.
• 10:46:41
• 10:46:42
JODYI wanted to make a few points. One is that I think language is critical here. Mr. Rienzi continued to say a remark about Hobby Lobby giving or providing contraceptives. They are not giving or providing anything. They are offering either an employee-paid or an employer-paid or a shared-paid insurance plan. And it is the individual that opts whether or not to access a certain kind of primary preventative care. So this is not about Hobby Lobby giving anyone anything. Employees earn their healthcare insurance.
• 10:47:19
JODYThe second is that Ms. Biskupic and Mr. Rienzi both are conflating this with abortion. And it has nothing to do with abortion. None of the methods, IUDs or emergency contraception have anything to do with abortion. That is now a scientifically proven fact. And every major public health association will tell you that. You can look on their websites and find the most recent science. I think it is irresponsible for a reporter or anyone else to continue to give weight to a small group of renegade people who want to conflate scientific fact with belief.
• 10:48:03
REHMAll right. Thanks.
• 10:48:05
RIENZITwo points on the claimed scientific fact there. One is the federal government agrees with the Greens here. This is in footnote five of their brief, but they say that these drugs and devices may prevent the egg from attaching and planting in the womb. They say it about the drugs. They say it about the devices. So, in this case, the Greens and the federal government agree as to the way the drugs work.
• 10:48:28
RIENZISecondly, in terms of the science, I would just point the listener and all your listeners to the American Journal of Obstetrics and Gynecology from 2011 which has an article saying that 57 percent of OB/GYNs nationwide believe that pregnancy begins at conception. And that makes sense. When people are trying to have a baby, they say they're trying to conceive, that it begins at conception. And the claimed scientific consensus that the caller just talked about is something that only 28 percent of OB/GYNs actually believe.
• 10:48:55
• 10:48:56
FLUKEWell, Diane, can I jump in on that actually?
• 10:48:57
REHMSure. Go ahead.
• 10:48:59
FLUKESo I don't think it's fair to say that the federal government and the Greens agree in this case. The federal government's footnote says that these birth control methods may -- emphasis on may -- prevent implantation. That's not the same thing as saying that these are methods of abortion because they're not. And that is the clear scientific consensus.
• 10:49:22
FLUKEThis actually, you know, it's a very minor point, but it goes back to a problematic study that came about when one of these drugs was introduced to the market. And the science at this point is much clearer. But this just comes back to my overall earlier point. We do not want our employers being the ones who decide what is or is not abortion, what is or is not the right medical choice for us to have insurance coverage of. This is the problem.
• 10:49:49
• 10:49:50
RIENZIThe employer in this case is simply trying to be out of this decision. The employer doesn't want anything to do with the employee's decision about whether or not to terminate a pregnancy. It is the government that is dragging the employer in and saying, you must be here, and you must pay for it. And the government's doing it in a way that's totally unnecessary, given that the government provides insurance on the exchanges to millions of people.
• 10:50:09
REHMBut isn't it because the employers are defining these medications as being abortion drugs?
• 10:50:22
RIENZIThey view them as causing abortions the way, again, most OB/GYNs, according to the American Journal of Obstetrics and Gynecology, have the same view of when pregnancy and when life begins.
• 10:50:30
REHMBut don't others disagree?
• 10:50:32
RIENZIOh, sure. There's a disagreement. And that just goes back to my...
• 10:50:34
• 10:50:35
RIENZI...original point, which is, sure, abortion's controversial. There are disagreements. People have disagreements, and therefore the government...
• 10:50:40
REHMAnd as to when life when life begins.
• 10:50:43
RIENZIAnd as to when life begins, although scientifically I think you will never find an embryology book that will tell you anything other than human life starts at fertilization. But, ultimately, the point is it's a controversial subject. I agree. But on a controversial subject, the government shouldn't be dragging unwilling people in and saying, you have to be part of this, and if you don't, we'll crush your business. That's not the way to live in a pluralistic society.
• 10:51:01
• 10:51:02
BISKUPICYeah, I just want to make sure the caller knows that my references to abortion had to do with the culture wars that are surrounding this. Reuters has written extensively about the medical and scientific dispute in this and where all the briefs have come in and where the science is. And it's not that we're asserting in any way that these do cause abortion. I'm saying that, in terms of the cultural arguments that -- and I think it was demonstrated today -- that you've got this overlay over fights over abortion rights that have infected this whole argument.
• 10:51:30
REHMAll right. To John in York, Penn. You're on the air.
• 10:51:35
JOHNGood morning. Thanks for taking my call.
• 10:51:37
• 10:51:38
JOHNI guess a lot of my comments are going to be like the last caller. I apologize for that. But I think we all have the right to our individual choices in terms of medical care and how we proceed with it. I think the root of the problems is that we've expected the employer and the government to take too active of a role in this whole process. And I understand the need to provide this from, you know, from an individual choice. But I don't see where we have the right as an employer or a government to say, this is the way you must do it.
• 10:52:08
• 10:52:10
RIENZIAnd I would agree with that. I don't think the Greens and Hobby Lobby are trying to tell anybody, this is the way you must do it. They're simply saying, there are certain things that, as a matter of their faith, they can't be involved in. And, honestly, I don't know what the caller does for a living or your listeners do for a living, but I seriously doubt there are many people listening who, when they go to work, say, well, there are things that I believe are deeply wrong. But from 9:00 to 5:00, I work for the company, so I'll do those deeply wrong things. And it's just the company acting. It's not really me.
• 10:52:34
RIENZII don't think people actually approach their jobs like that. At least I hope they don't approach their jobs like that. And the same is true for the Greens. There are just some things they can't do. Buying abortion drugs for people is one of them. And, again, in a world where there are so many other ways to get them, this is really a controversy that shouldn't even happen.
• 10:52:47
REHMJoan, I'm interested in the definition of an abortion drug versus a contraceptive.
• 10:52:56
BISKUPICWell, part of...
• 10:52:57
FLUKEDiane, can I actually just add on that last point before we move on?
• 10:53:00
REHMAll right, go ahead.
• 10:53:03
FLUKEI just want to point out that the problem here is that the government is not dragging employers in on this one reproductive healthcare decision. The problem is that, as a society, we have made a decision that our health insurance coverage for the vast majority of people in this country is going to be provided through their employers and/or their educational institution. And that means they're involved, period.
• 10:53:26
REHMAll right.
• 10:53:27
FLUKEYou can't take particular healthcare decisions about one drug or one type of operation or procedure and say that that's one where the employer should be able to step away from that decision. If we have an employee- and employment-based health insurance system, if we don't have single payer, then this is how we are going to have to counter these decisions.
• 10:53:45
REHMAll right. And you're listening to "The Diane Rehm Show." Joan.
• 10:53:51
BISKUPICWell, I can tell you that the Supreme Court has never defined when life begins. And I'm not going to try, for sure. And the justices will not, in this case. And, in fact, there are -- as I said, there's just a lot going on.
• 10:54:03
REHMSo how are they going to...
• 10:54:05
BISKUPICBecause the actual question...
• 10:54:06
REHM...argue and decide this case?
• 10:54:09
BISKUPICRight. This conversation that we're all having now will be much more interesting than what's going to happen from -- between the two lawyers at the lectern. But it will implicate a lot of what we've been talking about in terms of reproductive rights and exactly what these devices and pills actually do. But the question...
• 10:54:24
REHMBut does it come down to whether a corporation can exercise religious rights and impose those rights on its employees?
• 10:54:39
BISKUPICYes. But the court will start out with the presumption that these are valid religious rights and that these two parties have good reason to believe -- that they have sincere beliefs to object to these devices and drugs. So they won't have -- they don't have to prove in the courtroom that these drugs and devices do what they believe they do, even though the science might say otherwise.
• 10:55:01
REHMRight, right.
• 10:55:02
BISKUPICThey're going to go right to the much more legalistic but very important question about whether the Religious Freedom Restoration Act and the Constitution's Free Exercise clause would it allow them an exemption as for-profit corporations. Now, I've spent a lot of time saying about how big this is, how important this is.
• 10:55:21
BISKUPICBut there are ways that the court could actually go narrower and actually look at the owners of these companies and look at their religious beliefs rather than -- as individuals rather than as corporations. So there are ways that they can -- the justices can go narrower than we've been speaking. But the one thing I want to make clear is they will not define when life begins in this case.
• 10:55:41
• 10:55:42
RIENZII agree. That's exactly right. They won't define it. They don't have to because the government has conceded that the drugs act the way the Greens say they act. Ultimately, this case is about whether the government can pursue sort of an aggressive scorched-earth policy that says, this is not an issue on which you're allowed to disagree or opt out or say, hey, I want to stay away from that. It's the same aggressive policy that had the administrating dragging the Little Sisters of the Poor to the Supreme Court trying to make them sign forms about it.
• 10:56:07
RIENZIAnd it's part of a broader approach of saying, it will no longer be OK to say, hey, that's controversial. I want nothing to do with it. And instead, the government will say, you're going to be part of it, and you're going to be involved. And if you won't, we'll crush you with fines. That's the government's approach here, and it's clearly wrong.
• 10:56:22
REHMMark Rienzi, he's professor of constitutional law at Catholic University. He is one of the lawyers for Hobby Lobby. Joan Biskupic, editor in charge of legal affairs at Reuters. And Sandra Fluke, she's Democratic candidate for California Senate District 26, social justice attorney in L.A. I'm going to be fascinated to hear those arguments tomorrow, as I'm sure our listeners will. Thanks for listening today, and I'm Diane Rehm.
• 10:57:07
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