nyuuzyou/wikis huggingface dataset:
Rate of Change of Cartesian Coordinates of Cycloid

== Theorem ==
Let a circle $C$ of radius $a$ roll without slipping along the x-axis of a cartesian plane at a constant speed such that the center moves with a velocity $\mathbf v_0$ in the direction of increasing $x$.
Consider a point $P$ on the circumference of this circle.
Let $\tuple {x, y}$ be the coordinates of $P$ as it travels over the plane.
Then the rate of change of $x$ and $y$ can be expresssed as:
where $\theta$ is the angle turned by $C$ after time $t$.
== Proof ==
Let the center of $C$ be $O$.
, let $P$ be at the origin at time $t = t_0$.
By definition, $P$ traces out a cycloid.
From Equation of Cycloid, $P = \tuple {x, y}$ is described by:
:$(1): \quad \begin{cases} x & = a \paren {\theta - \sin \theta} \\
y & = a \paren {1 - \cos \theta} \end{cases}$
Let $\tuple {x_c, y_c}$ be the coordinates of $O$ at time $t$.
We have that:
:$y_c$ is constant: $y_c = a$
From Body under Constant Acceleration: Velocity after Time
:$x_c = \mathbf v_0 t$
as the acceleration of $O$ is zero.
But $x_c$ is equal to the length of the arc of $C$ that has rolled along the $x$-axis.
By Arc Length of Sector:
:$x_c = a \theta$
and so:
Thus:
and:
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Problem $4 \ \text{(a)}$
Category:Cycloids
Rate of Loss of Mass of Fluid per Unit Volume

== Theorem ==
Let $B$ be a body of fluid.
The rate of loss of mass of fluid from $B$ per unit volume is given by:
:$\rho \operatorname {div} \mathbf v$
where:
:$\rho$ denotes the density of $B$
:$\mathbf v$ denotes the velocity of $B$
:$\operatorname {div}$ denotes the divergence operator.
== Proof ==
== Sources ==
* 
* 
Category:Hydrodynamics
Ratio Equals its Multiples

== Theorem ==
That is:
:$a : b \implies ma = mb$
== Proof ==
Let $AB$ be the same multiple of $C$ that $DE$ is of $F$.
:
So as many magnitudes as there are in $AB$ equal to $C$, so many are there also in $DE$ equal to $F$.
Let $AB$ be divided into the magnitudes $AG, GH, HB$ equal to $C$.
Let $DE$ be divided into the magnitudes $DK, KL, LE$ equal to $F$.
Then the number of magnitudes $AG, GH, GB$ is the same as the number of magnitudes in $DK, KL, LE$.
We have that $AG = GH = HB$ and $DK = KL = LE$.
So from Ratios of Equal Magnitudes it follows that $AG : DK = GH : KL = HB : LE$.
Then from Sum of Components of Equal Ratios $AG : DK = AB : DE$.
But $AG = C$ and $DK = F$.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Ratio Test

== Theorem ==
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a series of real numbers in $\R$, or a series of complex numbers in $\C$.
Let the sequence $\sequence {a_n}$ satisfy:
:$\ds \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} } = l$
:If $l > 1 $, then $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges.
:If $l < 1 $, then $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.
== Proof ==
From the statement of the theorem, it is necessary that $\forall n: a_n \ne 0$; otherwise $\size {\dfrac {a_{n + 1} } {a_n} }$ is not defined.
Here, $\size {\dfrac {a_{n + 1} } {a_n} }$ denotes either the absolute value of $\dfrac {a_{n + 1} } {a_n}$, or the complex modulus of $\dfrac {a_{n + 1} } {a_n}$.
=== Absolute Convergence ===
Suppose $l < 1$.
Let us take $\epsilon > 0$ such that $l + \epsilon < 1$.
Then:
:$\exists N: \forall n > N: \size {\dfrac {a_n} {a_{n - 1} } } < l + \epsilon$
Thus:
By Sum of Infinite Geometric Progression, $\ds \sum_{n \mathop = 1}^\infty \paren {l + \epsilon}^n$ converges.
So by the the corollary to the comparison test, it follows that $\ds \sum_{n \mathop = 1}^\infty \size {a_n}$ converges absolutely too.
=== Divergence ===
Suppose $l > 1$.
Let $\epsilon > 0$ such that $l - \epsilon > 1$.
Then, for sufficiently large $N$, for all $k \ge N$ we have:
Hence, for all $n \ge N$:
But:
:$\paren {l - \epsilon}^{n - N} \size {a_{N} } \to \infty$
as $n \to \infty$.
So, by the Divergence Test, $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges.
== Warning ==
== Also known as ==
== Sources ==
* : $\S 6.17$
* : Appendix: $\S 18.5$: Tests for convergence of series
* : $\S 1.2.2$: Summary of convergence tests: Theorem $1.6$
* 
* 
* 
Category:Ratio Test
Category:Convergence Tests
Category:Named Theorems
Ratio Test/Warning

== Ratio Test: Warning ==
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a series of real numbers in $\R$, or a series of complex numbers in $\C$.
Let the sequence $\sequence {a_n}$ satisfy:
:$\ds \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} } = l$
If $l = 1$, the ratio test provides no information on whether $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely, converges conditionally, or diverges.
If $\size {\dfrac {a_{n + 1} } {a_n} } \to \infty$ as $n \to \infty$, then of course $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges.
== Sources ==
* : $\S 6.17$
* : $\S 1.2.2$: Summary of convergence tests: Theorem $1.6$
* 
* 
* 
Category:Ratio Test
Ratio between Consecutive Highly Composite Numbers Greater than 2520 is Less than 2

== Theorem ==
The ratio between $2$ consecutive highly composite numbers both greater than $2520$ is less than $2$.
== Proof ==
 $n$ and $m$ are consecutive highly composite numbers such that:
:$2520 < n < m$
:$m / n \ge 2$
By definition of highly composite:
:$\map \tau m > \map \tau n$
and, , $m$ is the smallest such integer.
We have that:
:$\map \tau {2 n} > \map \tau n$
so it follows that $m \le 2 n$, otherwise $m$ would not be the smallest such integer.
So from $m / n \ge 2$ and $m \le 2 n$, it follows that $m = 2 n$.
We have that $2520$ is a special highly composite number.
The prime decomposition of $2520$ s given by:
:$2520 = 2^3 \times 3^2 \times 5 \times 7$
We have that $n$ is a highly composite number such that $n > 2520$.
As $2520$ is a special highly composite number, $2520$ is a divisor of $n$.
Thus $n$ can be expressed as:
:$n = 2^a \times 3^b \times 5^c \times 7^d \times 11^e \times 13^f \times r$
where:
:$a \ge 3$
:$b \ge 2$
:$c \ge 1$
:$d \ge 1$
:$r$ is a possibly vacuous product of prime numbers strictly greater than $13$.
We have that $n$ and $m = 2 n$ are consecutive highly composite numbers.
Hence it follows that:
:$\map \tau {3 n / 2} \le \map \tau n$
and:
:$\map \tau {4 n / 3} \le \map \tau n$
otherwise $3 n / 2$ or $4 n / 3$ would be highly composite numbers between $n$ and $2 n$.
Then:
and:
This leads to:
It has already been established that:
:$a \ge 3$
:$b \ge 2$
so it is now possible to state:
Suppose:
:$(1): \quad f \ge 1$
Then:
So $f = 0$ and so by Prime Decomposition of Highly Composite Number $r = 1$.
Thus:
:$n = 2^3 \times 3^2 \times 5^c \times 7^d \times 11^e$
where $c = 1$ or $c = 2$.
Suppose $c = 2$.
Then:
So $c = 1$ and so by Prime Decomposition of Highly Composite Number:
:$n = 2^3 \times 3^2 \times 5 \times 7$
or:
:$n = 2^3 \times 3^2 \times 5 \times 7 \times 11$
Thus it has been established that these are the only possible values of $n$ greater than $2520$ which may fit the criteria for $n$ and $2 n$ to be consecutive highly composite numbers.
But the first of these is $2^3 \times 3^2 \times 5 \times 7 = 2520$ which fails through not being greater than $2520$.
Thus we consider:
:$n = 2^3 \times 3^2 \times 5 \times 7 \times 11 = 27 \, 720$
We have that:
It is now necessary to show that there are no highly composite numbers between $27 \, 720$ and $2 \times 27 \, 720 = 55 \, 440$.
That is, all numbers $n$ such that $27 \, 720 < n < 55 \, 440$ are to be shown to have $\map \tau n < 96$.
But:
So $45 \, 360$ has a higher $\tau$ than $27 \, 720$ and so the next higher highly composite number than $27 \, 720$ is less than twice it.
The result follows by Proof by Contradiction.
== Sources ==
* 
Category:Highly Composite Numbers
Ratio of 2016 to Aliquot Sum

== Theorem ==
$2016$ has the property that its ratio to its aliquot sum is $4 : 9$.
== Proof ==
The aliquot sum of an integer $n$ is the integer sum of the aliquot parts of $n$.
That is, the aliquot sum of $2016$ is the divisor sum of $2016$ minus $2016$.
Thus:
== Historical Note ==
== Sources ==
* : Chapter $\text {II}$. Formulas for the Number and Sum of Divisors, Problems of Fermat and Wallis
Category:2016
Category:Aliquot Sums
Ratio of 2016 to Aliquot Sum/Historical Note

== Historical Note on Ratio of 2016 to Aliquot Sum ==
 reports in his $1919$ work that this result was noted by , in response to 's 
observation that $360$ has the same property.
== Sources ==
* : Chapter $\text {II}$. Formulas for the Number and Sum of Divisors, Problems of Fermat and Wallis
Category:Historical Notes
Ratio of 360 to Aliquot Sum

== Theorem ==
$360$ has the property that its ratio to its aliquot sum is $4 : 9$.
== Proof ==
The aliquot sum of an integer $n$ is the integer sum of the aliquot parts of $n$.
That is, the aliquot sum of $360$ is the divisor sum of $360$ minus $360$.
Thus:
== Also see ==
* Ratio of 2016 to Aliquot Sum
== Historical Note ==
== Sources ==
* : Chapter $\text {II}$. Formulas for the Number and Sum of Divisors, Problems of Fermat and Wallis
Category:360
Category:Aliquot Sums
Ratio of Areas of Equiangular Parallelograms

== Theorem ==
== Proof ==
Let $\Box AC, \Box CF$ be equiangular parallelograms having $\angle BCD = \angle ECG$.
Let them be arranged so that $BC$ is in a straight line with $CG$.
Then $DC$ is in a straight line with $CE$.
:
Let the parallelogram $DG$ be completed.
Let a straight line $K$ be set out.
Using Construction of Fourth Proportional Straight Line, let $L$ and $M$ be constructed such that:
:$K : L = BC : CG$
:$L : M = DC : CE$
Then $K : L$ and $L : M$ are the same as the ratios of the sides.
But $K : M$ is compounded of $K : L$ and $L : M$.
So $K : M$ is the ratio compounded of the ratios of the sides.
From Areas of Triangles and Parallelograms Proportional to Base:
: $BC : CG = \Box AC : \Box CH$
From Equality of Ratios is Transitive:
: $K : L = \Box AC : \Box CH$
Similarly, from Areas of Triangles and Parallelograms Proportional to Base:
: $DC : CE = \Box CH : \Box CF$
Since we have $DC : CE = L : M$, it follows from Equality of Ratios is Transitive that:
: $L : M = \Box CH : \Box CF$
Since we have:
:$K : L = \Box AC : \Box CH$
:$L : M = \Box CH : \Box CF$
it follows that:
:$K : M = \Box AC : \Box CF$
But $K : M$ is the ratio compounded of the ratios of the sides.
Hence the result.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Area of Parallelogram
Ratio of Areas of Similar Triangles

== Theorem ==
That is, the ratio of the areas of the similar triangles is the square of the ratio of the corresponding sides.
== Proof ==
Let $\triangle ABC$ and $\triangle DEF$ be similar, such that $\angle ABC = \angle DEF$ and $AB : BC = DE : EF$ such that $BC$ corresponds to $EF$.
:
Let $BG$ be constructed such that $EF : BG = BC : EF$, and join $AG$.
From Proportional Magnitudes are Proportional Alternately $AB : DE = BC : EF$.
So from Equality of Ratios is Transitive $AB : DE = EF : BG$.
So in $\triangle ABC$ and $\triangle DEF$ the sides about the equal angles are reciprocally proportional.
From Sides of Equiangular Triangles are Reciprocally Proportional‎, the area of $\triangle ABG$ equals the area of $\triangle DEF$.
Now we have that $BC : EF = EF : BG$.
So from $BC$ has to $BG$ a ratio duplicate to that which $CB$ has to $EF$.
But from Areas of Triangles and Parallelograms Proportional to Base, $CB : BG = \triangle ABC : \triangle ABG$.
So $\triangle ABC$ has to $\triangle ABG$ a ratio duplicate to that which $BC$ has to $EF$.
But $\triangle ABC = \triangle DEF$.
So $\triangle ABC$ has to $\triangle DEF$ a ratio duplicate to that which $BC$ has to $EF$.
== Porism ==
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Ratio of Areas of Similar Triangles
Category:Areas of Triangles
Ratio of Commensurable Magnitudes

== Theorem ==
== Proof ==
Let $A$ and $B$ be commensurable magnitudes.
By definition of commensurable, some magnitude will measure them both.
Let $C$ be a common measure of $A$ and $B$.
Let $D$ be the number of times $C$ is the measure of $A$.
Let $E$ be the number of times $C$ is the measure of $B$.
Since:
:$C$ measures $A$ according to the number of units of $D$
and
:the unit measures $D$ according to the number of units of $D$
it follows that the unit measures $D$ the same number of times
$C$ measures $A$.
From Ratios of Fractions in Lowest Terms:
:$\dfrac C A = \dfrac 1 D$
and:
:$\dfrac A C = \dfrac D 1 = D$
Similarly, since:
:$C$ measures $B$ according to the number of units of $E$
and
:the unit measures $E$ according to the number of units of $E$
it follows that the unit measures $E$ the same number of times
$C$ measures $B$.
From Ratios of Fractions in Lowest Terms:
:$\dfrac C B = \dfrac 1 E$
So by Equality of Ratios Ex Aequali:
:$\dfrac A B = \dfrac D E$
Hence the result.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Ratio of Consecutive Fibonacci Numbers

== Theorem ==
For $n \in \N$, let $f_n$ be the $n$th Fibonacci number.
Then:
:$\ds \lim_{n \mathop \to \infty} \frac {f_{n + 1} } {f_n} = \phi$
where $\phi = \dfrac {1 + \sqrt 5} 2$ is the golden mean.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Historical Note ==
== Sources ==
* : $5$
* : $\S 1.2.8$: Fibonacci Numbers
* : $5$
* 
* 
* 
* 
Category:Ratio of Consecutive Fibonacci Numbers
Category:Fibonacci Numbers
Category:Golden Mean
Ratio of Consecutive Fibonacci Numbers/Historical Note

== Historical Note on Ratio of Consecutive Fibonacci Numbers ==
This result appears first to have been published by in $1560$, but the work it appeared in was not well-known and subsequent mathematicians appear to have been unaware of it.
This result was observed by , but not established rigorously by him:
:It is so arranged that the two lesser terms of a progressive series added together constitute the third ... and so on to infinity, as the same proportion continues unbroken. It is impossible to provide a perfect example in round numbers. However ... Let the smalllest numbers be $1$ and $1$, which you must imagine as unequal. Add them, and the sum will be $2$; add this to $1$, result $3$; add $2$ to this, and get $5$; add $3$, get $8$ ... as $5$ is to $8$, so $8$ is to $13$, approximately, and as $8$ is to $13$, so $13$ is to $21$, approximately.
However, it was in $1753$ who first explicitly stated that the Ratio of Consecutive Fibonacci Numbers tend to a limit.
== Sources ==
* : $5$
* 
* : $\S 1.2.8$: Fibonacci Numbers
* : $5$
Category:Historical Notes
Ratio of Consecutive Fibonacci Numbers/Proof 1

== Theorem ==
== Proof ==
Let:
Then:
Recall the Euler-Binet Formula:
:$f_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
Let $n \ge 1$.
It immediately follows that:
From the definition of $\alpha$:
:$\size \alpha > 1$
Therefore:
Category:Ratio of Consecutive Fibonacci Numbers
Ratio of Consecutive Fibonacci Numbers/Proof 2

== Theorem ==
== Proof ==
From Continued Fraction Expansion of Golden Mean: Successive Convergents, the $n$th convergent of the continued fraction expansion of $\phi$ is:
:$C_n = \dfrac {f_{n + 1} } {f_n}$
The result follows from Continued Fraction Expansion of Irrational Number Converges to Number Itself.
Category:Ratio of Consecutive Fibonacci Numbers
Ratio of Consecutive Fibonacci Numbers/Proof 3

== Theorem ==
== Proof ==
Let:
Then:
This sequence represents the iterated dynamical system $a_{n + 1} = \map g {a_n}$ with initial condition $a_1 = 1$.
We have that:
:$\size {\map {g'} \phi} = \dfrac 1 {\phi^2} < 1$
Hence the golden mean $\phi$ is an attracting fixed point of $g$.
Additionally, since the interval $I := \closedint {\dfrac 3 2} 2$ is positively invariant under $g$
and the derivative of $g$ is bounded above by $\dfrac 1 {\paren {3 / 2}^2} < 1$,
then $g$ is a contraction mapping on $I$. 
Thus, by the Banach Fixed-Point Theorem, $I$ is in the region of attraction of $\phi$.
Because $a_2 = 2 \in I$ is in the orbit of the dynamical system, the iterations converge to $\phi$. 
Therefore,
:$\ds \lim_{n \mathop \to \infty} \dfrac {f_{n + 1} } {f_n} = \lim_{n \mathop \to \infty} a_n = \phi$
Category:Ratio of Consecutive Fibonacci Numbers
Ratio of Consecutive Lucas Numbers

== Theorem ==
For $n \in \N$, let $L_n$ be the $n$th Lucas number.
Then:
:$\ds \lim_{n \mathop \to \infty} \frac {L_{n + 1} } {L_n} = \phi$
where $\phi = \dfrac {1 + \sqrt 5} 2$ is the golden mean.
== Proof ==
== Sources ==
* : $11$
* : $11$
Category:Lucas Numbers
Category:Golden Mean
Ratio of Lengths of Arms of Pentagram

== Theorem ==
Consider a pentagram.
:
Let $AC$ be the length of one of the lines which span the pentagram and define it.
Let $B$ be one of the points where $AC$ intersects one of the other such lines such that $AB > AC$.
Then:
:$\dfrac {AC} {AB} = \phi$
where $\phi$ denotes the golden mean.
== Proof ==
Follows directly from Straight Lines Subtending Two Consecutive Angles in Regular Pentagon cut in Extreme and Mean Ratio.
== Historical Note ==
== Sources ==
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* 
* 
Category:Golden Mean
Category:Pentagrams
Ratio of Lengths of Sides of Cube and Regular Icosahedron in Same Sphere

== Theorem ==
== Proof ==
:
Let a regular dodecahedron, a regular icosahedron and a cube be inscribed in a given sphere.
From :
:the circle which circumscribes the regular pentagon which is the face of the regular dodecahedron is the same size as the circle which circumscribes the equilateral triangle which is the face of the regular icosahedron.
Let $AHB$ be the circle which circumscribes both that regular pentagon and that equilateral triangle.
Let $C$ be the center of the circumscribing circle.
Let $CB$ be a radius of $AHB$.
Let $CB$ be divided at $D$ in extreme and mean ratio where $CD$ is the greater segment.
Then from:
:
and the converse of:
:
it follows that:
:$CD$ is the side of a regular decagon which has been inscribed in $AHB$.
Let $E$ be the side of the equilateral triangle that is the face of the regular icosahedron.
Let $F$ be the side of the regular pentagon which is the face of the regular dodecahedron.
Let $G$ be the side of the square which is the face of the cube.
From :
:if $G$ is divided in extreme and mean ratio, the greater segment is equal to $F$.
Thus:
Hence the result.
== Sources ==
* : The So-Called Book $\text{XIV}$, by Hypsicles
Category:Cubes
Category:Regular Icosahedra
Ratio of Number to Reversal which is Multiple

== Theorem ==
Take a (strictly) positive integer $n$, written in conventional decimal notation.
Let $m$ be the reversal of $n$.
Let $m = k n$ where $k$ is an integer.
Then $k$ is either $4$ or $9$.
== Proof ==
=== Existence ===
$8712 = 4 \times 2178$
$9801 = 9 \times 1089$
=== Uniqueness ===
Write $n = \sqbrk {a \dots b}$.
Then $m = \sqbrk {b \dots a} = k n$.
Comparing the leftmost digit, we must have $k a \le b < k \paren {a + 1}$.
Hence $k a < 10$.
Comparing the rightmost digit, we must have $a \equiv k b \pmod {10}$.
We only consider $2 \le k < 10$ since the case $k = 1$ is trivially satisfied by palindromes.
For $k \ge 5$, we must have $a = 1$.
$1 \equiv k b \pmod {10}$ has solutions only for $k = 7$ or $9$.
For $k = 7$, we must have $b = 3$.
However $3 < 7 \times 1$.
Hence $k = 9$.
For $k = 3$, $a \le 3$.
From $a \equiv 3 b \pmod {10}$ we have:
:$\tuple {a, b} = \tuple {1, 7}, \tuple {2, 4}, \tuple {3, 1}$
However none of the above satisfy the condition:
:$3 a \le b < 3 \paren {a + 1}$
For $k = 2$, $a \le 4$.
From $a \equiv 2 b \pmod {10}$, both sides must be even.
Hence $a = 2$ or $4$, giving $b = 1$ or $6$ or $2$ or $7$.
However none of the above satisfy the condition:
:$2 a \le b < 2 \paren {a + 1}$
Hence the only possible values of $k$ are $4$ and $9$.
== Also note that ==
* $2178 = 2 \times 1089$
== Sources ==
* : $4$
* : $1089$
* : $4$
* : $1089$
Category:Recreational Mathematics
Category:Reversals
Category:1089
Category:2178
Ratio of Products of Sides of Plane Numbers

== Theorem ==
== Proof ==
Let $a$ and $b$ be plane numbers.
Let $a = c d$ and $b = e f$.
From we can find $g, h, k$ such that:
:$c : e = g : h$
:$d : f = h : k$
Let $d e = l$.
We also have that $a = c d$.
So from :
:$c : e = a : l$
But we also have:
:$c : e = g : h$
and so:
:$g : h = a : l$
Since:
:$e d = l$
:$e f = b$
from :
:$d : f = l : b$
But we also have:
:$d : f = h : k$
and so:
:$h : k = l : b$
But we have:
:$g : h = a : l$
and so from :
:$g : k = a : b$
But $g : k$ is the ratio compounded of the ratios of the sides of $a$ and $b$.
Therefore $a : b$ is the ratio compounded of the ratios of their sides.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Euclidean Number Theory
Ratio of Proton Moment to Nuclear Magneton

== Theorem ==
The ratio of the proton moment to the nuclear magneton is given by:
== Proof ==
We have:
== Sources ==
* : $2$. Physical Constants and Conversion Factors: Table $2.3$ Adjusted Values of Constants
::which gives the mantissa as $2 \cdotp 792 \, 782$, with an uncertainty of $\pm 17$ corresponding to the $2$ least significant figures
Category:Proton Moment
Category:Nuclear Magneton
Ratio of Proton Moment to Nuclear Magneton Corrected for Diamagnetism

== Theorem ==
The ratio of the proton moment to the nuclear magneton, adjusted for diamagnetism, $\mathrm {H_2 O}$, is given by:
== Proof ==
== Sources ==
* : $2$. Physical Constants and Conversion Factors: Table $2.3$ Adjusted Values of Constants
Category:Proton Moment
Category:Nuclear Magneton
Ratio of Sizes of Surfaces of Cube and Regular Icosahedron in Same Sphere

== Theorem ==
== Proof ==
:
Let a regular dodecahedron, a regular icosahedron and a cube be inscribed in a given sphere.
From :
:the circle which circumscribes the regular pentagon which is the face of the regular dodecahedron is the same size as the circle which circumscribes the equilateral triangle which is the face of the regular icosahedron.
Let $ABC$ be the circle which circumscribes both that regular pentagon and that equilateral triangle.
Let $CD$ be the side of the equilateral triangle.
Let $AC$ be the side of the regular pentagon.
Let $E$ be the center of the circumscribing circle.
Let $EF$ and $EG$ be perpendiculars dropped from $E$ to $CD$ and $AC$ respectively.
Let $EG$ be produced past $G$ to meet the circumference of $ABC$ at $B$.
Let $BC$ be joined.
Let $H$ be equal to the side of the cube.
It is to be demonstrated that the ratio of the surface of the regular dodecahedron to the surface of the regular icosahedron is $H : CD$.
From :
:$EB$ equals the side of the regular hexagon that can be inscribed in $ABC$.
From :
:$EB + BC$ is divided at $B$ in extreme and mean ratio where $BE$ is the greater segment.
From :
:$EG = \dfrac {EB + BC} 2$
From (indirectly):
:$EF = \dfrac BE 2$
So:
:$EG : EF = \paren {EB + BC} : BE$
Therefore if $EG$ is divided in extreme and mean ratio, the greater segment is $EF$.
But from :
:if $H$ is divided in extreme and mean ratio, the greater segment is equal to $CA$.
So:
We have:
From :
:(surface of dodecahedron) : (surface of icosahedron) = $CA \cdot EG : FE \cdot CD$
Hence the result.
== Sources ==
* : The So-Called Book $\text{XIV}$, by Hypsicles
Category:Cubes
Category:Regular Icosahedra
Ratio of Sizes of Surfaces of Regular Dodecahedron and Regular Icosahedron in Same Sphere

== Theorem ==
== Proof ==
: $\qquad$ 
What is meant by its perpendicular is a perpendicular from the center of the circle circumscribed around it to the side of it.
In the above diagram:
From :
:the surface of the dodecahedron is $30$ times its perpendicular times its side.
From :
:the surface of the icosahedron is $30$ times its perpendicular times its side.
Hence the result.
== Sources ==
* : The So-Called Book $\text{XIV}$, by Hypsicles
Category:Regular Dodecahedra
Category:Regular Icosahedra