nyuuzyou/wikis huggingface dataset:
Commensurability of Squares

== Theorem ==
=== Porism ===
== Proof ==
Let $A$ and $B$ be commensurable in length.
Then $A$ has to $B$ the ratio which a number has to a number.
Let $\dfrac A B = \dfrac C D$.
From Similar Polygons are composed of Similar Triangles: Porism:
:$\dfrac {A^2} {B^2} = \paren {\dfrac C D}^2 = \dfrac {C^2} {D^2}$
Let:
:$\dfrac {A^2} {B^2} = \dfrac {C^2} {D^2}$
Then it follows that:
:$\dfrac A B = \dfrac C D$
and from Magnitudes with Rational Ratio are Commensurable it follows that $A$ and $B$ are commensurable in length.
Let $A$ and $B$ be incommensurable in length.
Suppose:
:$\dfrac {A^2} {B^2} = \dfrac {C^2} {D^2}$
where $C$ and $D$ are integers.
Then $A$ and $B$ would be commensurable in length, which they are not.
So the square on $A$ and the square on $B$ do not have to one another the ratio which a square number has to a square number.
Let the square on $A$ not have to the square on $B$ not have the ratio which a square number has to a square number.
Suppose $A$ was commensurable in length with $B$.
Then the square on $A$ would have to the square on $B$ the ratio which a square number has to a square number, which it does not.
So $A$ is not commensurable in length with $B$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Commensurability of Squares on Proportional Straight Lines

== Theorem ==
=== Lemma ===
== Proof ==
Let $A$, $B$, $C$ and $D$ be four straight lines in proportion, such that:
:$A : B = C : D$
Using the lemma let the straight lines $E$ and $F$ be found such that:
: $A^2 = B^2 + E^2$
: $C^2 = D^2 + F^2$
As $A : B = C : D$ it follows from Similar Figures on Proportional Straight Lines that:
:$A^2 : B^2 = C^2 : D^2$
But:
: $E^2 + B^2 = A^2$
: $F^2 + D^2 = C^2$
Therefore:
:$E^2 + B^2 : B^2 = F^2 + D^2 : D^2$
and so by Magnitudes Proportional Compounded are Proportional Separated:
:$E^2 : B^2 = F^2 : D^2$
By Similar Figures on Proportional Straight Lines:
:$B : E = D : F$
But:
:$A : B = C : D$
Therefore from Equality of Ratios Ex Aequali:
:$A : E = C : F$
From Commensurability of Elements of Proportional Magnitudes it follows that:
:if $A$ is commensurable with $E$ then $C$ is also commensurable with $F$
and:
:if $A$ is incommensurable with $E$ then $C$ is also incommensurable with $F$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Commensurability of Squares on Proportional Straight Lines/Lemma

== Lemma to Commensurability of Squares on Proportional Straight Lines ==
== Proof ==
:
Let $AB$ and $C$ be two unequal straight lines.
Let $AB > C$.
Let the semicircle $ADB$ be described with $AB$ as the diameter.
Using Fitting Chord Into Circle, let $AD$ be fitted into $ADB$ equal to $C$.
Let $DB$ be joined.
From Relative Sizes of Angles in Segments, $\angle ADB$ is a right angle.
From Pythagoras's Theorem:
:$AB^2 = AD^2 + DB^2$
and so $AB^2$ is greater than $AD^2$, that is, $C^2$, by $DB^2$.
Conversely, given two straight lines $A$ and $B$ the same technique can be used to find the straight line the square of whose length equals the sum of the squares on $A$ and $B$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Commensurability of Sum of Commensurable Magnitudes

== Theorem ==
== Proof ==
Let $AB$ and $BC$ be commensurable magnitudes which are added together to make $AC$.
Since $AB$ and $BC$ are commensurable, then some magnitude $D$ will measure them both.
Since $D$ measures both $AB$ and $BC$, $D$ also measures $AC$.
That is, $D$ measures $AB$, $BC$ and $AC$.
Therefore from , $AC$ is commensurable with each of the magnitudes $AB$ and $BC$.
Let $AC$ be commensurable with $AB$.
Since $AC$ and $AB$ are commensurable, then some magnitude $D$ will measure them both.
Since $D$ measures both $AC$ and $AB$, $D$ also measures the remainder $BC$.
That is, $D$ measures $AB$, $BC$ and $AC$.
Therefore from , $AB$ and $BC$ are commensurable.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Commensurable/Examples/3 Feet and 2 Inches

== Example of Commensurability ==
The lengths $2$ inches and $3$ feet are commensurable.
== Proof ==
$3$ feet is $18$ times $2$ inches.
Hence $2$ inches is a common divisor of both $2$ inches and $3$ feet.
Hence the result by definition of commensurable.
== Sources ==
* 
* 
Category:Examples of Commensurability
Commensurable Magnitudes are Incommensurable with Same Magnitude

== Theorem ==
== Proof ==
Let $A$ and $B$ be magnitudes which are commensurable with each other.
Let $A$ be incommensurable with any other magnitude $C$.
Suppose $B$ is commensurable with $C$.
But $A$ is commensurable with $B$.
So from Commensurability is Transitive Relation it follows that $A$ is commensurable with $C$.
From that contradiction it follows that $B$ cannot be commensurable with $C$.
Hence the result.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Common Denominator/Examples/2, 3, 7

== Examples of Common Denominators ==
The fractions:
:$\dfrac 1 2$, $\dfrac 1 3$, $\dfrac 3 7$
have common denominators:
:$42$, $84$, $126$, $168$
and so on.
The lowest common denominator of $\dfrac 1 2$, $\dfrac 1 3$, $\dfrac 3 7$ is $42$.
== Sources ==
* 
* 
Category:Examples of Common Denominators
Common Denominator/Examples/Multiples of 12

== Examples of Common Denominators ==
Multiples of $12$ are all common denominators of:
:$\dfrac 1 2$, $\dfrac 1 4$, $\dfrac 1 6$
Hence these fractions can be expressed as:
:$\dfrac 6 {12}$, $\dfrac 3 {12}$, $\dfrac 2 {12}$
== Sources ==
* 
Category:Examples of Common Denominators
Common Divisor Divides Difference

== Theorem ==
Let $c$ be a common divisor of two integers $a$ and $b$.
That is:
:$a, b, c \in \Z: c \divides a \land c \divides b$
Then:
:$c \divides \paren {a - b}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 11.4$: The division algorithm
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$: $\mathbf D \, 4$
Category:Divisors
Category:Common Divisor Divides Difference
Common Divisor Divides Difference/Proof 1

== Theorem ==
== Proof ==
Let $c \divides a \land c \divides b$.
From Common Divisor Divides Integer Combination:
:$\forall p, q \in \Z: c \divides \paren {p a + q b}$
Putting $p = 1$ and $q = -1$:
:$c \divides \paren {a - b}$
Category:Common Divisor Divides Difference
Common Divisor Divides GCD

== Theorem ==
Let $a, b \in \Z$ such that not both of $a$ and $b$ are zero.
Let $c$ be any common divisor of $a$ and $b$.
That is, let $c \in \Z: c \divides a, c \divides b$.
Then:
:$c \divides \gcd \set {a, b}$
where $\gcd \set {a, b}$ is the greatest common divisor of $a$ and $b$.
== Proof ==
Let $d = \gcd \set {a, b}$.
Then $d \divides a$ and $d \divides b$ by definition.
Then from Bézout's Identity, $\exists u, v \in \Z: d = u a + v b$.
From Common Divisor Divides Integer Combination, $c \divides a \land c \divides b \implies c \divides u a + v b$ for all $u, v \in \Z$.
Thus $c \divides d$.
== Sources ==
* : $\S 12.4$: Highest common factors and Euclid's algorithm
Category:Greatest Common Divisor
Common Divisor Divides Integer Combination

== Corollary to Common Divisor in Integral Domain Divides Linear Combination ==
Let $c$ be a common divisor of two integers $a$ and $b$.
That is:
:$a, b, c \in \Z: c \divides a \land c \divides b$
Then $c$ divides any integer combination of $a$ and $b$:
:$\forall p, q \in \Z: c \divides \paren {p a + q b}$
=== Corollary ===
=== General Result ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\text I$: The Series of Primes: $1.1$ Divisibility of integers
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 2 \ (7)$
* : $1$ Divisibility: $1.1$ The Euclidean algorithm and unique factorization: $\text {(v)}$
Category:Divisors
Category:Integer Combinations
Category:Common Divisor Divides Integer Combination
Common Divisor Divides Integer Combination/Corollary

== Corollary to Common Divisor Divides Integer Combination ==
Let $c$ be a common divisor of two integers $a$ and $b$.
That is:
:$a, b, c \in \Z: c \divides a \land c \divides b$
Then:
:$c \divides \paren {a + b}$
=== Converse does not Hold ===
== Proof ==
From Common Divisor Divides Integer Combination:
:$\forall p, q \in \Z: c \divides \paren {p a + q b}$
The result follows by setting $p = q = 1$.
== Sources ==
* : $\S 11.4$: The division algorithm
Category:Common Divisor Divides Integer Combination
Common Divisor Divides Integer Combination/Corollary/Converse does not Hold

== Theorem ==
Let $a, b, c \in \Z$ be integers.
Let:
:$c \divides \paren {a + b}$
Then it is not necessarily the case that:
:$c \divides a \land c \divides b$
== Proof ==
Proof by Counterexample:
Let $a = 2, b = 4, c = 3$.
Then we have:
:$3 \divides \paren {2 + 4}$
but:
:$3 \nmid 2$
and:
:$3 \nmid 4$
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $3$
Category:Common Divisor Divides Integer Combination
Common Divisor Divides Integer Combination/General Result

== Theorem ==
Let $c$ be a common divisor of a set of integers $A := \set {a_1, a_2, \dotsc, a_n}$.
That is:
:$\forall x \in A: c \divides x$
Then $c$ divides any integer combination of elements of $A$:
:$\forall x_1, x_2, \dotsc, x_n \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_n x_n}$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$\forall x \in \set {a_1, a_2, \dotsc, a_n}: c \divides x \implies \forall x_1, x_2, \dotsc, x_n \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_n x_n}$
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\forall x \in \set {a_1, a_2}: c \divides x \implies \forall x_1, x_2 \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2}$
This is demonstrated to be true in Common Divisor Divides Integer Combination.
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\forall x \in \set {a_1, a_2, \dotsc, a_k}: c \divides x \implies \forall x_1, x_2, \dotsc, x_k \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k}$
from which it is to be shown that:
:$\forall x \in \set {a_1, a_2, \dotsc, a_k, a_{k + 1} }: c \divides x \implies \forall x_1, x_2, \dotsc, x_k, x_{k + 1} \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k + a_{k + 1} x_{k + 1} }$
=== Induction Step ===
This is the induction step:
Let :
:$\forall x \in \set {a_1, a_2, \dotsc, a_k, a_{k + 1} }: c \divides x$
We have that:
:$c \divides a_{k + 1} \implies \forall x_{k + 1} \in \Z: c \divides a_{k + 1} x_{k + 1}$
and we have that:
:$\forall x \in \set {a_1, a_2, \dotsc, a_k}: c \divides x \implies \forall x_1, x_2, \dotsc, x_k \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k}$
Let $x_1, x_2, \dotsc, x_k \in \Z$ be arbitrary.
Let $d = a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k$.
Then:
But $x_1, x_2, \dotsc, x_k \in \Z$ are arbitrary, and so:
:$\forall x \in \set {a_1, a_2, \dotsc, a_k, a_{k + 1} }: c \divides x \implies \forall x_1, x_2, \dotsc, x_k, x_{k + 1} \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k + a_{k + 1} x_{k + 1} }$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 2}: \forall x \in \set {a_1, a_2, \dotsc, a_n}: c \divides x \implies \forall x_1, x_2, \dotsc, x_n \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_n x_n}$
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor
Category:Common Divisor Divides Integer Combination
Common Divisor Divides Integer Combination/Proof 2

== Theorem ==
== Proof ==
== Sources ==
* : Chapter $3$: The Integers: $\S 10$. Divisibility: Theorem $16 \ \text{(iii)}$
* : $\text {2-2}$ Divisibility: Example $\text {2-4}$
* : $\S 11.4$: The division algorithm
Category:Common Divisor Divides Integer Combination
Common Divisor in Integral Domain Divides Linear Combination

== Theorem ==
Let $\struct {D, +, \times}$ be an integral domain.
Let $c$ be a common divisor of two elements $a$ and $b$ of $D$.
That is:
:$a, b, c \in D: c \divides a \land c \divides b$
Then:
:$\forall p, q \in D: c \divides \paren {p \times a + q \times b}$
=== Corollary ===
== Proof ==
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 26$. Divisibility: Theorem $49 \ \text{(iii)}$
Category:Divisibility
Common Factor Cancelling in Congruence

== Theorem ==
Let $a, b, x, y, m \in \Z$.
Let:
:$a x \equiv b y \pmod m$ and $a \equiv b \pmod m$
where $a \equiv b \pmod m$ denotes that $a$ is congruent modulo $m$ to $b$.
Then:
:$x \equiv y \pmod {m / d}$
where $d = \gcd \set {a, m}$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
We have that $d = \gcd \set {a, m}$.
From Law of Inverses (Modulo Arithmetic), we have:
: $\exists a' \in \Z: a a' \equiv d \pmod m$
Hence:
Then:
Hence the result.
== Examples ==
Category:Modulo Arithmetic
Category:Common Factor Cancelling in Congruence
Common Factor Cancelling in Congruence/Corollary 1

== Corollary to Common Factor Cancelling in Congruence ==
Let $a, b, x, y, m \in \Z$.
Let:
:$a x \equiv b y \pmod m$ and $a \equiv b \pmod m$
where $a \equiv b \pmod m$ denotes that $a$ is congruent modulo $m$ to $b$.
If $a$ is coprime to $m$, then:
:$x \equiv y \pmod m$
== Proof ==
Let $a \perp m$.
Then by definition of coprime:
:$\gcd \set {a, m} = 1$
The result follows immediately from Common Factor Cancelling in Congruence.
== Warning ==
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Law $\text{B}$
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $20$
Category:Common Factor Cancelling in Congruence
Common Factor Cancelling in Congruence/Corollary 1/Warning

== Theorem ==
Let $a, b, x, y, m \in \Z$.
Let:
:$a x \equiv b y \pmod m$ and $a \equiv b \pmod m$
where $a \equiv b \pmod m$ denotes that $a$ is congruent modulo $m$ to $b$.
Let $a$ not be coprime to $m$.
Then it is not necessarily the case that:
:$x \equiv y \pmod m$
== Proof ==
Proof by Counterexample:
Let $a = 6, b = 21, x = 7, y = 12, m = 15$.
We note that $\map \gcd {6, 15} = 3$ and so $6$ and $15$ are not coprime.
We have that:
Then:
But:
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $22$
Category:Common Factor Cancelling in Congruence
Common Factor Cancelling in Congruence/Corollary 2

== Corollary to Common Factor Cancelling in Congruence ==
Let $a, b, x, y, m \in \Z$.
Let:
:$a x \equiv a y \pmod m$
where $a \equiv b \pmod m$ denotes that $a$ is congruent modulo $m$ to $b$.
If $a$ is coprime to $m$, then:
:$x \equiv y \pmod m$
== Proof ==
From Common Factor Cancelling in Congruence: Corollary 1:
:$a x \equiv b y \pmod m$ and $a \equiv b \pmod m$
and:
:$a$ is coprime to $m$
then:
:$x \equiv y \pmod m$
The result follows immediately from setting $a = b$.
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Theorem $\text {4-3}$
Category:Common Factor Cancelling in Congruence
Common Factor Cancelling in Congruence/Examples/6 equiv 12 mod 2 does not lead to 3 equiv 6 mod 2

== Example of Common Factor Cancelling in Congruence ==
We have that:
:$6 \equiv 12 \pmod 2$
but:
:$3 \not \equiv 6 \pmod 2$
== Proof ==
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Example $\text {4-5}$
Category:Common Factor Cancelling in Congruence
Common Factor Cancelling in Congruence/Examples/6 equiv 12 mod 2 leads to 2 equiv 4 mod 2

== Example of Common Factor Cancelling in Congruence ==
We have that:
:$6 \equiv 12 \pmod 2$
and so:
:$2 \equiv 4 \pmod 2$
== Proof ==
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Example $\text {4-5}$
Category:Common Factor Cancelling in Congruence
Common Logarithm/Examples

== Examples of Common Logarithm ==
=== Common Logarithm: $\log_{10} 1000$ ===
=== Common Logarithm: $\log_{10} 0 \cdotp 01$ ===
=== Common Logarithm: $\log_{10} 2$ ===
=== Common Logarithm: $\log_{10} 3$ ===
=== Common Logarithm: $\log_{10} e$ ===
=== Common Logarithm: $\log_{10} \pi$ ===
=== Common Logarithm: $\log_{10} 2 \cdotp 36$ ===
=== Common Logarithm: $\log_{10} 23 \cdotp 6$ ===
=== Common Logarithm: $\log_{10} 236$ ===
=== Common Logarithm: $\log_{10} 2360$ ===
=== Common Logarithm: $\log_{10} 0 \cdotp 236$ ===
=== Common Logarithm: $\log_{10} 0 \cdotp 0236$ ===
=== Common Logarithm: $\log_{10} 6 \cdotp 573$ ===
=== Common Logarithm: $\log_{10} 657 \cdotp 3$ ===
=== Common Logarithm: $\log_{10} 0 \cdotp 06573$ ===
Category:Examples of Common Logarithms
Common Logarithm/Examples/0.0236

== Example of Common Logarithm ==
The common logarithm of $0 \cdotp 0236$ is:
:$\log_{10} 0 \cdotp 0236 = \overline 2 \cdotp 3729 = -1.6271$
== Proof ==
== Sources ==
* : Chapter $1$: Logarithms: Example 3.
Category:Examples of Common Logarithms
Common Logarithm/Examples/0.06573

== Example of Common Logarithm ==
The common logarithm of $0 \cdotp 06573$ is:
:$\log_{10} 0 \cdotp 06573 = \overline 2 \cdotp 8178 = -1 \cdotp 182$
== Proof ==
== Sources ==
* 
* 
Category:Examples of Common Logarithms
Common Logarithm/Examples/0.236

== Example of Common Logarithm ==
The common logarithm of $0 \cdotp 236$ is:
:$\log_{10} 0 \cdotp 236 = \overline 1 \cdotp 3729 = -0.6271$
== Proof ==
== Sources ==
* : Chapter $1$: Logarithms: Example 3.
Category:Examples of Common Logarithms
Common Logarithm/Examples/2

== Example of Common Logarithm ==
The common logarithm of $2$ is:
:$\log_{10} 2 \approx 0.30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots$
== Also see ==
* Logarithm Base 10 of 2 is Irrational
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.13$
* : $0 \cdotp 301 \, 029 \, 995 \, 663 \, 981 \ldots$
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: $(10)$
* : $0 \cdotp 30102 \, 99956 \, 63981 \ldots$
Category:Examples of Common Logarithms
Common Logarithm/Examples/2.36

== Example of Common Logarithm ==
The common logarithm of $2 \cdotp 36$ is:
:$\log_{10} 2 \cdotp 36 = 0 \cdotp 3729$
== Sources ==
* : Chapter $1$: Logarithms: Example 1.
Category:Examples of Common Logarithms
Common Logarithm/Examples/23.6

== Example of Common Logarithm ==
The common logarithm of $23 \cdotp 6$ is:
:$\log_{10} 23 \cdotp 6 = 1 \cdotp 3729$
== Proof ==
== Sources ==
* : Chapter $1$: Logarithms: Example 2.
Category:Examples of Common Logarithms
Common Logarithm/Examples/3

== Example of Common Logarithm ==
The common logarithm of $3$ is:
:$\log_{10} 3 = 0.47712 \, 12547 \, 19662 \, 43729 \, 50279 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.14$
Category:Examples of Common Logarithms
Common Logarithm/Examples/e

== Example of Common Logarithm ==
The common logarithm of Euler's number $e$ is:
:$\log_{10} e = 0 \cdotp 43429 \, 44819 \, 03251 \, 82765 \, 11289 \, 18916 \, 60508 \, 22943 \, 97005 \, 803 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.15$
* : $0,43429 44819 \ldots$
* : $0 \cdotp 434 \, 294 \, 481 \, 903 \, 251 \, 827 \, 651 \, 128 \, 918 \, 916 \, 605 \, 082 \, 294 \, 397 \, 005 \, 803 \ldots$
* : $0 \cdotp 43429 \, 44819 \, 03251 \, 82765 \, 11289 \, 18916 \, 60508 \, 22943 \, 97005 \, 803 \ldots$
Category:Examples of Common Logarithms
Category:Euler's Number
Common Logarithm/Examples/pi

== Example of Common Logarithm ==
The common logarithm of $\pi$ is:
:$\log_{10} \pi = 0.49714 \, 98726 \, 94133 \, 85435 \, 12683 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.16$
Category:Examples of Common Logarithms
Common Logarithm of Number in Scientific Notation

== Theorem ==
Let $n$ be a positive real number which is presented (possibly approximated) in scientific notation as:
:$n = a \times 10^d$
where:
:$1 \le a < 10$
:$d \in \Z$ is an integer.
Then:
:$\log_{10} n = \log_{10} a + d$
where:
:$0 \le \log_{10} a < 1$
== Proof ==
We are given that:
:$1 \le a < 10$
It follows from Range of Common Logarithm of Number between 1 and 10 that:
:$0 \le \log_{10} a < 1$
Category:Common Logarithms
Common Ratio in Integer Geometric Sequence is Rational

== Theorem ==
Let $\sequence {a_k}$ be a geometric sequence whose terms are all integers.
Then the common ratio of $\sequence {a_k}$ is rational.
== Proof ==
From Integers form Subdomain of Rationals it follows that $a_k \in \Q$ for all $0 \le k \le n$.
The result follows from Common Ratio in Rational Geometric Sequence is Rational.
Category:Geometric Sequences of Integers
Common Ratio in Rational Geometric Sequence is Rational

== Theorem ==
Let $\sequence {a_k}$ be a geometric sequence whose terms are rational.
Then the common ratio of $\sequence {a_k}$ is rational.
== Proof ==
Let $r$ be the common ratio of $\sequence {a_k}$.
Let $p, q$ be consecutive terms of $r$.
We have that:
:$p, q \in \Q$
Then, by definition of geometric sequence:
:$q = r p$
It follows that:
:$r = \dfrac q p$
From Rational Numbers form Field, $\Q$ is closed under division.
Thus $r \in \Q$ and hence the result.
Category:Geometric Sequences
Common Section of Bisecting Planes of Cube Bisect and are Bisected by Diagonal of Cube

== Theorem ==
== Proof ==
:
Let $AF$ be a cube.
Let the edges of the opposite faces $CF$ and $AH$ of $AF$ be bisected at the points $K, L, M, N, O, P, Q, R$.
Let the plane $KN$ be constructed through $K, L, M, N$.
Let the plane $OR$ be constructed through $O, P, Q, R$.
Let $US$ be the common section of $KN$ and $OR$.
Let $DG$ be the diameter of the cube $AF$.
It is to be demonstrated that:
:$UT = TS$
:$DT = TG$
Let $DU, UE, BS, SG$ be joined.
We have that $DO$ is parallel to $PE$.
Therefore from :
:$\angle DOU = \angle UPE$
We have that:
:$DO = PE$
:$OU = UP$
:$\angle DOU = \angle UPE$
So from :
:$\triangle DOU = \triangle PUE$
Therefore:
:$\angle OUD = \angle PUE$
So from :
:$DUE$ is a straight line.
For the same reason:
:$BSG$ is a straight line.
and:
:$BS = SG$
We have that:
:$CA = DB$ and $CA \parallel DB$
and:
:$CA = EG$ and $CA \parallel EG$
Therefore from:
:$DB = EG$ and $DB \parallel EG$
We have that the straight lines $DE$ and $BG$ join the endpoints of $DB$ and $EG$.
Therefore from :
:$DE \parallel BG$
Therefore from :
:$\angle EDT = \angle BGT$
Therefore from :
:$\angle DTU = \angle GTS$
Therefore from :
:$\triangle DTU = \triangle GTS$
Therefore:
:$DT = TG$
and
:$UT = TS$
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Cubes
Common Section of Planes Perpendicular to other Plane is Perpendicular to that Plane

== Theorem ==
== Proof ==
:
Let $AB$ and $BC$ be planes which are perpendicular to the plane of reference.
Let $BD$ be the common section of $AB$ and $BC$.
It is to be demonstrated that $BD$ is perpendicular to the plane of reference.
Suppose $BD$ is not perpendicular to the plane of reference.
From the point $D$, let:
:$DE$ be drawn in the plane $AB$ perpendicular to $AD$
:$DF$ be drawn in the plane $AC$ perpendicular to $CD$
We have that:
:$AB$ is perpendicular to the plane of reference
and:
:$DE$ is in the plane $AB$ perpendicular to $AD$.
Therefore from :
:$DE$ is perpendicular to the plane of reference.
Similarly it can be proved that $DF$ is also perpendicular to the plane of reference.
Therefore from $D$ we have two straight lines have been set up perpendicular to the plane of reference on the same side.
This contradicts .
Hence the result.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Planes
Common Section of Two Planes is Straight Line

== Theorem ==
== Proof ==
:
Let $AB$ and $BC$ be two distinct planes that cut one another.
Let $DB$ be their common section.
Suppose $DB$ were not a straight line.
Then let:
: the straight line segment $DEB$ be drawn in the planes $AB$
and:
: the straight line segment $DFB$ be drawn in the planes $BC$.
Thus the two straight line segments $DEB$ and $DFB$ have the same endpoints.
Thus $DEB$ and $DFB$ enclose an area, which is absurd.
Therefore $DEB$ and $DFB$ are not straight lines.
Hence the result.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Planes