nyuuzyou/wikis huggingface dataset:
Perfect Number which is Sum of Equal Powers of Two Numbers

== Theorem ==
$28$ is the only perfect number which is the sum of equal powers of exactly $2$ positive integers:
:$28 = 1^3 + 3^3$
== Proof ==
Let $N$ be an even perfect number which is the sum of equal powers of exactly $2$ positive integers.
By Theorem of Even Perfect Numbers, write:
:$N = 2^{p - 1} \paren {2^p - 1} = a^n + b^n$
where $p, 2^p - 1$ are prime, $a, b, n \in \Z_{>0}$, $n > 1$.
, let $a \le b$.
Immediately we see that $a \ne b$:
 $a = b$.
We have:
:$2^{p - 1} \paren {2^p - 1} = 2 a^n$
In particular:
:$\paren {2^p - 1} \divides a^n$
By Prime Divides Power:
:$\paren {2^p - 1}^n \divides a^n = 2^{p - 2} \paren {2^p - 1}$
which leads to a contradiction, since $n > 1$.
Next, since:
:$2^p - 1 \equiv 3 \pmod 4$
by Integer as Sum of Two Squares, $N$ is not a sum of two squares.
This leads to $n$ being odd, and $n \ge 3$.
By Sum of Two Odd Powers:
:$\paren {a + b} \divides \paren {a^n + b^n} = N$
Suppose $\paren {2^p - 1} \divides \paren {a + b}$.
We have:
If $(*)$ holds, this is a contradiction.
Since $b > a > 0$, we have:
:$b \ge 2$
Hence for $n \ge 5$:
:$b^n \ge b^5 \ge 8 b^2 \ge 4 b^2$
so $(*)$ holds.
For $n = 3$ and $b \ge 4$:
:$b^3 \ge 4 b^2$
so $(*)$ holds.
We only need to inspect the cases:
:$n = 3$, $2 \le b \le 3$, $a^3 + b^3$ is even.
This leads to the only case:
:$1^3 + 3^3 = 28$
Now suppose that $\paren {2^p - 1} \nmid \paren {a + b}$.
Then $a + b = 2^k$ for some $k \le p - 1$.
As $0 < a < b$, we have:
:$a + b \ge 3$
Hence $a, b$ are either both odd or both even.
Suppose they are both odd.
By Sum of Two Odd Powers:
:$\ds a^n + b^n = \paren {a + b} \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j$
Each $\paren {-1}^j a^{n - j - 1} b^j$ is odd, and there are $n$ of them.
Therefore $\ds \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j$ is odd.
The only odd divisors of $N$ are $1$ and $2^p - 1$.
Since $a^n + b^n > a + b$:
:$\ds \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j = 2^p - 1$
Therefore $a + b = 2^{p - 1}$.
We have:
If $(*)$ holds, this is a contradiction.
Since $b > a > 0$, we have:
:$b \ge 2$
Hence for $n \ge 5$:
:$b^n \ge b^5 \ge 8 b^2$
so $(*)$ holds.
For $n = 3$ and $b \ge 8$:
:$b^3 \ge 8 b^2$
so $(*)$ holds.
We only need to inspect the cases:
:$n = 3$, $2 \le b \le 7$, both $a$ and $b$ are odd, $a + b$ is a power of $2$.
This leads to the cases:
Finally, suppose they are both even.
Let $2^m$ be the largest power of $2$ that divides both $a$ and $b$, where $m \ge 1$.
Write $s = \dfrac a {2^m}$ and $t = \dfrac b {2^m}$.
Then:
:$s + t = 2^{k - m}$
:$s^n + t^n = 2^{p - 1 - m n} \paren {2^p - 1}$
At least one of $s, t$ is odd.
If the other is even, then $2^{k - m}$ is odd.
This forces $2^{k - m} = 1 = s + t$, which contradicts $a, b > 0$.
Hence $s, t$ are both odd.
By our previous analysis:
:$s + t = 2^{p - 1 - m n}$
Hence:
== Sources ==
* : $28$
* : $28$
Category:Perfect Numbers
Category:28
Perfectly Normal Space is Completely Normal Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a perfectly normal space.
Then $T$ is also a completely normal space.
== Proof ==
Let $T = \struct {S, \tau}$ be a perfectly normal space.
From the definition:
:$T$ is a perfectly $T_4$ space
:$T$ is a $T_1$ (Fréchet) space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Additional Separation Properties
Category:Completely Normal Spaces
Category:Perfectly Normal Spaces
Perimeter of Circle

== Theorem ==
The perimeter $C$ of a circle with radius $r$ is given by:
:$C = 2 \pi r$
== Proof ==
By definition, the perimeter is the length of the circumference of the circle.
Let $C$ be the perimeter.
Then:
== Sources ==
* : $\S 4$: Geometric Formulas: Circle of Radius $r$: $4.12$
* 
* 
* : $\S 7$: Geometric Formulas: Circle of Radius $r$: $7.12.$
* 
* 
* : Appendix $1$: Areas and volumes
* : $\S 1$: What Is Curvature? The Euclidean Plane
* 
* 
* : Appendix $1$: Areas and volumes
Category:Circles
Category:Perimeter Formulas
Perimeter of Ellipse

== Theorem ==
Let $K$ be an ellipse whose major axis is of length $2 a$ and whose minor axis is of length $2 b$.
The perimeter $\PP$ of $K$ is given by:
:$\PP = 4 a \map E e$
where:
:$\ds \map E e = \int_0^{\pi / 2} \sqrt{1 - e^2 \sin^2 \theta} \rd \theta$ is the complete elliptic integral of the second kind
:$e = \dfrac {\sqrt {a^2 - b^2} } a$ is the eccentricity of $K$.
=== Approximation ===
== Proof ==
Let $K$ be aligned in a cartesian plane such that:
:the major axis of $K$ is aligned with the $x$-axis
:the minor axis of $K$ is aligned with the $y$-axis.
Then from Equation of Ellipse in Reduced Form: parametric form:
:$x = a \cos \theta, y = b \sin \theta$
Thus:
From Arc Length for Parametric Equations, the length of one quarter of the perimeter of $K$ is given by:
Since $\cos \theta = \map \sin {\dfrac \pi 2 - \theta}$ we can write for any real function $\map f x$:
:$\ds \int_0^{\pi / 2} \map f {\cos \theta} \rd \theta = \int_0^{\pi / 2} \map f {\map \sin {\frac \pi 2 - \theta} } \rd \theta$
So substituting $t = \dfrac \pi 2 - \theta$ this can be converted to:
justifying the fact that $\cos$ can be replaced with $\sin$ in $(1)$ above, giving:
:$\ds \PP = 4 a \int_0^{\pi / 2} \sqrt {1 - k^2 \sin^2 \theta} \rd \theta$
== Sources ==
* : $\S 4$: Geometric Formulas: Ellipse of Semi-major Axis $a$ and Semi-minor Axis $b$: $4.23$
* : $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems: Problem $5$
* 
* : $\S 7$: Geometric Formulas: Ellipse of Semi-major Axis $a$ and Semi-minor Axis $b$: $7.23.$
Category:Perimeter of Ellipse
Category:Ellipses
Category:Perimeter Formulas
Category:Complete Elliptic Integral of the Second Kind
Perimeter of Ellipse/Approximation

== Theorem ==
Let $K$ be an ellipse whose major axis is of length $2 a$ and whose minor axis is of length $2 b$.
The perimeter $\PP$ of $K$ can be approximated by the formula:
:$\PP \approx 2 \pi \sqrt {\dfrac {a^2 + b^2} 2}$
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Ellipse of Semi-major Axis $a$ and Semi-minor Axis $b$: $4.23$
* : $\S 7$: Geometric Formulas: Ellipse of Semi-major Axis $a$ and Semi-minor Axis $b$: $7.23.$
Category:Perimeter of Ellipse
Perimeter of Parallelogram

== Theorem ==
Let $ABCD$ be a parallelogram whose side lengths are $a$ and $b$.
The perimeter of $ABCD$ is $2 a + 2 b$.
== Proof ==
:
By Opposite Sides and Angles of Parallelogram are Equal it follows that:
:$AB = CD$
:$BC = AD$
The perimeter of $ABCD$ is $AB + BC + CD + AD$.
But $AB = CD = a$ and $BC = AD = b$.
Hence the result.
== Sources ==
* : $\S 4$: Geometric Formulas: Geometric Formulas: Parallelogram of Altitude $h$ and Base $b$: $4.4$
* : $\S 7$: Geometric Formulas: Parallelogram of Altitude $h$ and Base $b$: $7.4.$
Category:Parallelograms
Category:Perimeter Formulas
Perimeter of Rectangle

== Theorem ==
Let $ABCD$ be a rectangle whose side lengths are $a$ and $b$.
The perimeter of $ABCD$ is $2 a + 2 b$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 4$: Geometric Formulas: Rectangle of Length $b$ and Width $a$: $4.2$
* : $\S 7$: Geometric Formulas: Rectangle of Length $b$ and Width $a$: $7.2.$
* 
* 
Category:Perimeter of Rectangle
Category:Rectangles
Category:Perimeter Formulas
Perimeter of Rectangle/Proof 1

== Theorem ==
== Proof ==
:
From Rectangle is Parallelogram, $ABCD$ is a parallelogram.
By Opposite Sides and Angles of Parallelogram are Equal it follows that:
:$AB = CD$
:$BC = AD$
The perimeter of $ABCD$ is $AB + BC + CD + AD$.
But $AB = CD = a$ and $BC = AD = b$.
Hence the result.
Category:Perimeter of Rectangle
Perimeter of Regular Polygon

== Theorem ==
Let $P$ be a regular $n$-sided polygon whose side length is $b$.
Then the perimeter $L$ of $P$ is given by:
:$L = n b$
== Proof ==
By definition, an $n$-sided polygon has $n$ sides.
By definition, a regular polygon has sides all the same length.
By definition, the perimeter of a polygon is the total length of all its sides.
Thus $P$ has $n$ sides all of length $b$.
Hence $L = n b$.
== Sources ==
* : $\S 4$: Geometric Formulas: Regular Polygon of $n$ Sides Each of Length $b$: $4.10$
* : $\S 7$: Geometric Formulas: Regular Polygon of $n$ Sides Each of Length $b$: $7.10.$
Category:Regular Polygons
Category:Perimeter Formulas
Perimeter of Regular Polygon by Circumradius

== Theorem ==
Let $P$ be a regular $n$-gon.
Let $C$ be a circumcircle of $P$.
Let the radius of $C$ be $r$.
Then the perimeter $\PP$ of $P$ is given by:
:$\PP = 2 n r \sin \dfrac \pi n$
== Proof ==
:
From Regular Polygon is composed of Isosceles Triangles, let $\triangle OAB$ be one of the $n$ isosceles triangles that compose $P$.
Then $\PP$ is equal to $n$ times the base of $\triangle OAB$.
Let $d$ be the length of one side of $P$.
Then $d$ is the length of the base of $\triangle OAB$.
The angle $\angle AOB$ is equal to $\dfrac {2 \pi} n$.
Then:
:$d = 2 r \sin \dfrac \pi n$
So:
== Sources ==
* : $\S 4$: Geometric Formulas: Regular Polygon of $n$ Sides Inscribed in Circle of Radius $r$: $4.18$
* : $\S 7$: Geometric Formulas: Regular Polygon of $n$ Sides Inscribed in Circle of Radius $r$: $7.18.$
Category:Circumradii
Category:Regular Polygons
Category:Perimeter Formulas
Perimeter of Regular Polygon by Inradius

== Theorem ==
Let $P$ be a regular $n$-gon.
Let $C$ be an incircle of $P$.
Let the radius of $C$ be $r$.
Then the perimeter $\PP$ of $P$ is given by:
:$\PP = 2 n r \tan \dfrac \pi n$
== Proof ==
:
From Regular Polygon is composed of Isosceles Triangles, let $\triangle OAB$ be one of the $n$ isosceles triangles that compose $P$.
Then $\AA$ is equal to $n$ times the area of $\triangle OAB$.
Also, $r$ is the length of the altitude of $\triangle OAB$.
Let $d$ be the length of one side of $P$.
Then $d$ is the length of the base of $\triangle OAB$.
The angle $\angle AOB$ is equal to $\dfrac {2 \pi} n$.
Then:
:$d = 2 r \tan \dfrac \pi n$
So:
== Sources ==
* : $\S 4$: Geometric Formulas: Regular Polygon of $n$ Sides Circumscribing a Circle of Radius $r$: $4.20$
* : $\S 7$: Geometric Formulas: Regular Polygon of $n$ Sides Circumscribing a Circle of Radius $r$: $7.20.$
Category:Regular Polygons
Category:Perimeter Formulas
Perimeter of Trapezium

== Theorem ==
:
Let $ABCD$ be a trapezium:
:whose parallel sides are of lengths $a$ and $b$
:whose height is $h$.
and
:whose non-parallel sides are at angles $\theta$ and $\phi$ with the parallels.
The perimeter $P$ of $ABCD$ is given by:
:$P = a + b + h \paren {\csc \theta + \csc \phi}$
where $\csc$ denotes cosecant.
== Proof ==
The perimeter $P$ of $ABCD$ is given by:
:$P = AB + BC + CD + AD$
where the lines are used to indicate their length.
Thus:
Hence:
Hence the result.
== Usage Differences ==
== Sources ==
* : $\S 4$: Geometric Formulas: Trapezoid of Altitude $h$ and Parallel Sides $a$ and $b$: $4.8$
* : $\S 7$: Geometric Formulas: Trapezoid of Altitude $h$ and Parallel Sides $a$ and $b$: $7.8.$
Category:Trapezia
Category:Perimeter Formulas
Perimeter of Triangle

== Theorem ==
Let $ABC$ be a triangle.
Then the perimeter $P$ of $ABC$ is given by:
:$P = a + b + c$
where $a, b, c$ are the lengths of the sides of $ABC$.
== Proof ==
The perimeter of a plane geometric figure is defined as the total length of the boundary.
By definition, the boundary of a triangle comprises the three sides of that triangle.
Hence the result.
== Sources ==
* : $\S 4$: Geometric Formulas: Triangle of Altitude $h$ and Base $b$: $4.6$
* : $\S 7$: Geometric Formulas: Triangle of Altitude $h$ and Base $b$: $7.6.$
Category:Triangles
Category:Perimeter Formulas
Period of Complex Exponential Function

== Theorem ==
Let $z \in \C$, and let $k \in \Z$.
Then:
:$\map \exp {z + 2 k \pi i} = \map \exp z$
== Proof ==
== Sources ==
* : $\S 2$. Geometrical Representations: $(2.21)$
* : $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$: $\text{(ii)}$
* : $\S 1.5$
Category:Exponential Function
Period of Harmonic Wave

== Theorem ==
Let $\phi$ be a harmonic wave expressed as:
:$\forall x, t \in \R: \map \phi {x, t} = a \map \cos {\omega \paren {x - c t} }$
The period $\tau$ of $\phi$ can be expressed as:
:$\tau = \dfrac \lambda c$
where $\lambda$ is the wavelength of $\phi$.
== Proof 1 ==
== Proof 2 ==
Category:Period of Harmonic Wave
Category:Harmonic Waves
Period of Harmonic Wave/Proof 2

== Theorem ==
== Proof ==
By definition, $\tau$ is the time taken for one complete wavelength of $\phi$ to pass an arbitrary point on the $x$-axis.
From Equation of Harmonic Wave, we have:
:$(1): \quad \map \phi {x, t} = a \map \cos {\omega \paren {x - c t} }$
From Wavelength of Harmonic Wave:
:$\lambda = \dfrac {2 \pi} \omega$
Hence:
:$\omega = \dfrac {2 \pi} \lambda$
and we can express $(1)$ in the form:
:$(2): \quad \map \phi {x, t} = a \map \cos {\dfrac {2 \pi} \lambda \paren {x - c t} }$
From $(2)$ it follows that $\dfrac {2 \pi} \lambda \paren {x - c t}$ must pass through a complete cycle of values as $t$ is increased by $\tau$.
Thus:
:$\dfrac {2 \pi c \tau} \lambda = 2 \pi$
and so:
:$\tau = \dfrac \lambda c$
Hence the result.
== Sources ==
* : Chapter $\text {I}$: The Equation of Wave Motion: $\S 3$: $(5)$
Category:Period of Harmonic Wave
Period of Oscillation of Underdamped Cart attached to Wall by Spring

== Theorem ==
=== Problem Definition ===
Let $C$ be underdamped.
Let $C$ be pulled aside to $x = x_0$ and released from stationary at time $t = 0$.
Then the period of oscillation of $C$ can be expressed as:
:$T = \dfrac {2 \pi} {\sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} } }$
== Proof ==
Let:
:$a^2 = \dfrac k m$
:$2 b = \dfrac c m$
From Position of Cart attached to Wall by Spring under Damping: Underdamped: $x = x_0$ at $t = 0$:
:$x = \dfrac {x_0} \alpha e^{-b t} \left({\alpha \cos \alpha t + b \sin \alpha t}\right)$
where $\alpha = \sqrt {a^2 - b^2}$.
Let $T$ be the period of oscillation of $C$.
Then:
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(21)$
Category:Mechanics
Period of Oscillation of Underdamped System is Regular

== Theorem ==
Consider a physical system $S$ whose behaviour can be described with the second order ODE in the form:
:$\dfrac {\d^2 x} {\d t^2} + 2 b \dfrac {\d x} {\d t} + a^2 x = 0$
for $a, b \in \R_{>0}$.
Let $b < a$, so as to make $S$ underdamped.
Then the period of its movement is well-defined, in the sense that its zeroes are regularly spaced, and given by:
:$T = \dfrac {2 \pi} {\sqrt {a^2 - b^2} }$
== Proof ==
Let the position of $S$ be described in the canonical form:
:$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \map \cos {\alpha t - \theta}$
where $\alpha = \sqrt {a^2 - b^2}$.
The zeroes of $(1)$ occur exactly where:
:$\map \cos {\alpha t - \theta} = 0$
Thus the period $T$ of $\map \cos {\alpha t - \theta}$ is given by:
:$\alpha T = 2 \pi$
and so:
:$T = \dfrac {2 \pi} {\sqrt {a^2 - b^2} }$
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(21)$
Category:Mechanics
Period of Periodic Wave

== Theorem ==
Let $\phi$ be a periodic wave expressed as:
:$\forall x, t \in \R: \map \phi {x, t} = \map f {x - c t}$
The period $\tau$ of $\phi$ can be expressed as:
:$\tau = \dfrac \lambda c$
where $\lambda$ is the wavelength of $\phi$.
== Proof ==
By definition, $\tau$ is the time taken for one complete wavelength of $\phi$ to pass an arbitrary point on the $x$-axis.
We have:
:$\map \phi {x, t} = \map f {x - c t} = \map f {x - c t + \lambda}$
It follows that $x - c t$ must pass through a complete cycle of values as $t$ is increased by $\tau$.
Thus:
:$\lambda = c \tau$
and so:
:$\tau = \dfrac \lambda c$
Hence the result.
== Examples ==
== Sources ==
* : Chapter $\text {I}$: The Equation of Wave Motion: $\S 3$
Category:Period of Periodic Wave
Category:Periodic Waves
Period of Real Cosine Function

== Theorem ==
The period of the real cosine function is $2 \pi$.
That is, $2 \pi$ is the smallest value $L \in \R_{>0}$ such that:
:$\forall x \in \R: \cos x = \map \cos {x + L}$
== Proof ==
From Sine and Cosine are Periodic on Reals, we have that $\cos$ is a periodic real function.
Let $L$ be that period.
From Cosine of Angle plus Full Angle:
:$\map \cos {x + 2 \pi} = \cos x$
So $L = 2 \pi$ satisfies:
:$\forall x \in \R: \cos x = \map \cos {x + L}$
It remains to be shown that $2 \pi$ is the smallest such $L \in \R_{>0}$ with this property.
We have that:
and for no other $x \in \closedint 0 {2 \pi}$ is $\cos x = 0$.
Hence if there is another smaller $L \in \R_{>0}$ with the property that classifies it as the period of the real cosine function, it can only be $\pi$.
But then we note:
Hence $\pi$ is not the period of the real cosine function.
Thus $2 \pi$ is the smallest value $L \in \R_{>0}$ such that:
:$\forall x \in \R: \cos x = \map \cos {x + L}$
and the result follows.
Category:Cosine Function
Category:Periodic Functions
Period of Real Sine Function

== Theorem ==
The period of the real sine function is $2 \pi$.
That is, $2 \pi$ is the smallest value $L \in \R_{>0}$ such that:
:$\forall x \in \R: \sin x = \map \sin {x + L}$
== Proof ==
From Sine and Cosine are Periodic on Reals, we have that $\sin$ is a periodic real function.
Let $L$ be that period.
From Sine of Angle plus Full Angle:
:$\map \sin {x + 2 \pi} = \sin x$
So $L = 2 \pi$ satisfies:
:$\forall x \in \R: \sin x = \map \sin {x + L}$
It remains to be shown that $2 \pi$ is the smallest such $L \in \R_{>0}$ with this property.
We have that:
and for no other $x \in \closedint 0 {2 \pi}$ is $\sin x = 0$.
Hence if there is another smaller $L \in \R_{>0}$ with the property that classifies it as the period of the real sine function, it can only be $\pi$.
But then we note:
Hence $\pi$ is not the period of the real sine function.
Thus $2 \pi$ is the smallest value $L \in \R_{>0}$ such that:
:$\forall x \in \R: \sin x = \map \sin {x + L}$
and the result follows.
== Sources ==
* : $1$. Functions: $1.5$ Trigonometric or Circular Functions: $1.5.2$ Sine Function
Category:Sine Function
Category:Periodic Functions
Period of Reciprocal of 43 is Odd

== Theorem ==
The decimal expansion of the reciprocal of $43$ has an odd period, that is, $21$:
:$\dfrac 1 {43} = 0 \cdotp \dot 02325 \, 58139 \, 53488 \, 37209 \, \dot 3$
== Proof ==
From Reciprocal of $43$:
Counting the digits, it is seen that this has a period of recurrence of $21$.
Hence the result.
Category:43
Category:Examples of Reciprocals
Period of Reciprocal of 53 is One Quarter of Maximal

== Theorem ==
The decimal expansion of the reciprocal of $53$ has $\dfrac 1 4$ the maximum period, that is: $13$:
:$\dfrac 1 {53} = 0 \cdotp \dot 01886 \, 79245 \, 28 \dot 3$
== Proof ==
From Reciprocal of $53$:
Counting the digits, it is seen that this has a period of recurrence of $13$.
From Maximum Period of Reciprocal of Prime, the maximum period of recurrence of $\dfrac 1 p$ is $p - 1$.
We have that:
:$13 = \dfrac {53 - 1} 4$
Hence the result.
== Sources ==
* : $53$
* : $53$
Category:53
Category:Examples of Reciprocals
Period of Reciprocal of 729 is 81

== Theorem ==
The decimal expansion of the reciprocal of $729$ has $\dfrac 1 9$ the maximum period, that is, $81$:
:$\dfrac 1 {729} = 0 \cdotp \dot 00137 \, 17421 \, 12482 \, 85322 \, 35939 \, 64334 \, 70507 \, 54458 \, 16186 \, 55692 \, 72976 \, 68038 \, 40877 \, 91495 \, 19890 \, 26063 \, \dot 1$
The recurring part can be arranged in groups of $9$ digits each, revealing an interesting pattern:
that is, each row (apart from the last) can be obtained from the previous one by adding $111 \, 111 \, 111$ to it.
== Proof ==
Performing the calculation using long division:
 0.00137174211248285322359396433470507544581618655692729766803840877914951989026063100
 ---------------------------------------------------------------------------------------
729)1.000000000000000000000000000000000000000000000000000000000000000000000000000000000000
 729 729 3645 2187 5103 5832 3645 5103 5832 3645 4374
 --- --- ---- ---- ---- ---- ---- ---- ---- ---- ----
 2710 910 2350 7030 3700 1180 5050 4870 5680 1450 2260
 2187 729 2187 6561 3645 729 4374 4374 5103 729 2187
 ---- --- ---- ---- ---- ---- ---- ---- ---- ---- ----
 5230 1810 1630 4690 5500 4510 6760 4960 5770 7210 730
 5103 1458 1458 4374 5103 4374 6561 4374 5103 6561 729
 ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---
 1270 3520 1720 3160 3970 1360 1990 5860 6670 6490 1000
 729 2916 1458 2916 3645 729 1458 5832 6561 5832 729
 ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ----
 5410 6040 2620 2440 3250 6310 5320 2800 1090 6580 ...
 5103 5832 2187 2187 2916 5832 5103 2187 729 6561
 ---- ---- ---- ---- ---- ---- ---- ---- ---- ----
 3070 2080 4330 2530 3340 4780 2170 6130 3610 1900
 2916 1458 3645 2187 2916 4374 1458 5832 2916 1458
 ---- ---- ---- ---- ---- ---- ---- ---- ---- ----
 1540 6220 6850 3430 4240 4060 7120 2980 6940 4420
 1458 5832 6561 2916 3645 3645 6561 2916 6561 4374
 ---- ---- ---- ---- ---- ---- ---- ---- ---- ----
 820 3880 2890 5140 5950 4150 5590 6400 3790 4600
 729 3645 2187 5103 5832 3645 5103 5832 3645 4374
== Historical Note ==
== Sources ==
* : $729$
* : $729$
Category:729
Category:Examples of Reciprocals
Period of Reciprocal of 729 is 81/Mistake

== Source Work ==
:
:The Dictionary
::$729$
:
:The Dictionary
::$729$
== Mistake ==
:$1 / 729$ has a decimal period of $81$ digits, which can be arranged in groups of $9$ digits, reading across each row, in this pattern:
The digit is wrong.
Instead of $6$ it should be $0$:
== Sources ==
* : $729$
* : $729$
Category:Mistakes/Curious and Interesting Numbers
Period of Reciprocal of 79 is One Sixth of Maximal

== Theorem ==
The decimal expansion of the reciprocal of $79$ has $\dfrac 1 6$ the maximum period, that is: $13$:
:$\dfrac 1 {79} = 0 \cdotp \dot 01265 \, 82278 \, 48 \dot 1$
== Proof ==
From Reciprocal of $79$:
Counting the digits, it is seen that this has a period of recurrence of $13$.
From Maximum Period of Reciprocal of Prime, the maximum period of recurrence of $\dfrac 1 p$ is $p - 1$.
We have that:
:$13 = \dfrac {79 - 1} 6$
Category:79
Category:Examples of Reciprocals
Period of Reciprocal of Prime

== Theorem ==
Consider the decimal expansion of the reciprocal $\dfrac 1 p$ of a prime $p$.
If $p \nmid a$, the decimal expansion of $\dfrac 1 p$ is periodic in base $a$ and its period of recurrence is the order of $a$ modulo $p$.
If $p \divides a$, the decimal expansion of $\dfrac 1 p$ in base $a$ terminates.
== Proof ==
=== Case $1$: $p \divides a$ ===
Let $q = \dfrac a p$.
Then $\dfrac 1 p = \dfrac q a$.
So the decimal expansion of $\dfrac 1 p$ in base $a$ is $0.q$ and terminates.
=== Case $2$: $p \nmid a$ ===
Let $d$ be the order of $a$ modulo $p$.
By definition, $d$ is the smallest integer such that:
:$a^d \equiv 1 \pmod p$
By definition of modulo arithmetic:
:$\exists x \in \Z: a^d - 1 = xp$
We can rearrange the terms to achieve the following expression:
so the decimal expansion of $\dfrac 1 p$ in base $a$ is $0.xxx...$ and is periodic of length $d$, which is the order of $a$ modulo $p$.
Category:Reciprocals
Category:Prime Numbers
Period of Reciprocal of Repunit 1031 is 1031

== Theorem ==
The decimal expansion of the reciprocal of the repunit prime $R_{1031}$ has a period of $1031$.
:$\dfrac 1 {R_{1031}} = 0 \cdotp \underbrace{\dot 000 \ldots 000}_{1030} \dot 9$
This is the only prime number to have a period of exactly $1031$.
== Proof ==
The reciprocal of a repunit $R_n$ is of the form:
:$\dfrac 1 {R_n} = 0 \cdotp \underbrace{\dot 000 \ldots 000}_{n - 1} \dot 9$
Thus $\dfrac 1 {R_{1031}}$ has a period of $1031$.
From Period of Reciprocal of Prime, for prime numbers such that:
:$p \nmid 10$
we have that the period of such a prime is the order of $10$ modulo $p$.
That is, the smallest integer $d$ such that:
:$10^d \equiv 1 \pmod p$
The only other possible primes $p$ whose reciprocals might have a period of $1031$ must also satisfy:
:$10^{1031} \equiv 1 \pmod p$
that is:
:$p \divides \paren {10^{1031} - 1} = 9 \times R_{1031}$
Therefore the only other possible prime whose reciprocal might have a period of $1031$ is $3$.
Trivially:
:$\dfrac 1 3 = 0 \cdotp \dot 3$
which has a period of $1$.
Hence the result.
== Sources ==
* : $1111 \ldots 111111$
Category:1031
Category:Examples of Reciprocals