avewright/proofwiki-math huggingface dataset:
Theorem: Taylor Series reaches closest Singularity

Let $F$ be a complex function.
Let $F$ be analytic everywhere except at a finite number of singularities.
Let a singularity of $F$ be one of the following:
:a pole
:an essential singularity
:a branch point
In the latter case $F$ is a restriction of a multifunction to one of its branches.
Let $x_0$ be a real number.
Let $F$ be analytic at the complex number $\tuple {x_0, 0}$.
Let $R \in \R_{>0}$ be the distance from the complex number $\tuple {x_0, 0}$ to the closest singularity of $F$.
Let the restriction of $F$ to $\R \to \C$ be a real function $f$.
This means:
:$\forall x \in \R: \map f x = \map \Re {\map F {x, 0} }, 0 = \map \Im {\map F {x, 0} }$
where $\tuple {x, 0}$ denotes the complex number with real part $x$ and imaginary part $0$.
Then:
:the Taylor series of $f$ about $x_0$ converges to $f$ at every point $x \in \R$ satisfying $\size {x - x_0} < R$

Proof:
We have that $F$ is analytic everywhere except at its singularities.
Also, the distance from the complex number $\tuple {x_0, 0}$ to the closest singularity of $F$ is $R$.
Therefore:
:$F$ is analytic at every point $z \in \C$ satisfying $\size {z - \tuple {x_0, 0} } < R$
where $\tuple {x_0 , 0}$ denotes the complex number with real part $x_0$ and imaginary part $0$.
The result follows by Convergence of Taylor Series of Function Analytic on Disk.
{{qed}}
Theorem: Number of Binary Digits in Power of 10

Let $n$ be a positive integer.
Expressed in binary notation, the number of digits in the $n$th power of $10$:
:$10^n$
is equal to:
:$\ceiling {n \log_2 10}$
where $\ceiling x$ denotes the ceiling of $x$.

Proof:
Let $10^n$ have $m$ digits when expressed in binary notation.
By the Basis Representation Theorem and its implications, a positive integer $x$ has $m$ digits {{iff}}:
:$2^{m - 1} \le x < 2^m$
Thus:
{{begin-eqn}}
{{eqn | l = 2^{m - 1}
      | o = \le
      | m = 10^n
      | mo= <
      | r = 2^m
      | c = 
}}
{{eqn | ll= \leadsto
      | l = m - 1
      | o = \le
      | m = \map {\log_2} {10^n}
      | mo= <
      | r = m
      | c = 
}}
{{eqn | ll= \leadsto
      | l = m - 1
      | o = \le
      | m = n \log_2 10
      | mo= <
      | r = m
      | c = 
}}
{{end-eqn}}
Because a power of $10$ cannot equal a power of $2$, it will always be the case that:
:$m - 1 < n \log_2 10 < m$
and so:
:$m - 1 < n \log_2 10 \le m$
Hence from Integer equals Ceiling iff Number between Integer and One Less:
:$m = \ceiling {n \log_2 10}$
{{qed}}
Theorem: Number of Binary Digits in Power of 10/Example/1000

When expressed in binary notation, the number of digits in $1000$ is $10$.

Proof:
Let $m$ be the number of digits in $1000$.
From Number of Binary Digits in Power of 10:
:$m = \ceiling {3 \log_2 10}$
From Logarithm Base 2 of 10:
:$\log_2 10 \approx 3 \cdotp 32192 \, 8 \ldots$
and so:
:$m \approx 9 \cdotp 96$
Hence the result.
The actual number is:
:$1000_{10} = 1 \, 111 \, 101 \, 100_2$
{{qed}}
Theorem: Bound for Analytic Function and Derivatives

Let $f$ be a complex function.
Let $z_0$ be a point in $\C$.
Let $r$ be a real number in $\R_{>0}$.
Let $\Gamma$ be a circle in $\C$ with center at $z_0$ and radius $r$.
Let $f$ be analytic on $\Gamma$ and its interior.
Let $t \in \C$ be such that $\cmod {t - z_0} < r$.
Then a real number $M$ exists such that, for every $n \in \N$:
:$\cmod {\map {f^{\paren n} } t} \le \dfrac {M r \, n!} {\paren {r - \cmod {t - z_0} }^{n + 1} }$

Proof:
=== Lemma (Analytic Function Bounded on Circle) ===
{{:Bound for Analytic Function and Derivatives/Analytic Function Bounded on Circle}}{{qed|lemma}}
We have that $f$ is bounded on $\Gamma$ by Lemma (Analytic Function Bounded on Circle).
Therefore, there is a positive real number $M$ that satisfies:
:$(1): \quad \forall z \in \Gamma: \cmod {\map f z} \le M$
We have $\cmod {t - z_0} < r$.
Therefore:
:$0 < r - \cmod {t - z_0}$
We observe that $r - \cmod {t - z_0}$ is the minimum distance between $t$ and $\Gamma$.
Therefore:
:$(2): \quad \forall z \in \Gamma: r - \cmod {t - z_0} \le \cmod {z - t}$
We have:
:$f$ is analytic on $\Gamma$ and its interior
:$t$ is in the interior of $\Gamma$
We get:
{{begin-eqn}}
{{eqn | l = \cmod {\map {f^{\paren n} } t}
      | r = \cmod {\frac {n!} {2 \pi i} \oint_\Gamma \frac {\map f z} {\paren {z - t}^{n + 1} } \rd z}
      | c = Cauchy's Integral Formula for Derivatives
}}
{{eqn | o = \le
      | r = \frac {n!} {2 \pi} \oint_\Gamma \frac {\cmod {\map f z} } {\cmod {z - t}^{n + 1} } \cmod {\d z}
      | c = {{Defof|Absolute Value of Complex Differential}}
}}
{{eqn | o = \le
      | r = \frac {n!} {2 \pi} \oint_\Gamma \frac M {\cmod {z - t}^{n + 1} } \cmod {\d z}
      | c = from $(1)$
}}
{{eqn | o = \le
      | r = \frac {n!} {2 \pi} \oint_\Gamma \frac M {\paren {r - \cmod {t - z_0} }^{n + 1} } \cmod {\d z}
      | c = from $(2)$
}}
{{eqn | r = \frac {n!} {2 \pi} \frac M {\paren {r - \cmod {t - z_0} }^{n + 1} } \oint_\Gamma \cmod {\d z}
      | c = Linear Combination of Contour Integrals
}}
{{eqn | r = \frac {n!} {2 \pi} \frac M {\paren {r - \cmod {t - z_0} }^{n + 1} } 2 \pi r
}}
{{eqn | r = \frac {M r \, n!} {\paren {r - \cmod {t - z_0} }^{n + 1} }
}}
{{end-eqn}}
{{qed}}
Category:Analytic Complex Functions
tobzqv5ui6hbezpep8v30eu2onkmw1c
Theorem: Regular Octahedron is Dual of Cube

The regular octahedron is the dual of the cube.

Proof:
:300px
{{ProofWanted}}
Theorem: Plane Figure with Bilateral Symmetry about Two Lines has 4 Congruent Parts

Let $F$ be a plane figure.
Let $F$ have two different axes of bilateral symmetry.
Then those two axes divide $F$ into $4$ congruent parts.
:500px

Proof:
{{ProofWanted|More background needed on the geometry of symmetry.}}
Theorem: Divisibility of n-1 Factorial by Composite n

Let $n \in \Z$ be composite.
Then:
:$n \divides \paren {n - 1}! \iff n \ne 4$
where:
:$\divides$ denotes divisibility
:$n!$ denotes the factorial of $n$.

Proof:
=== Necessary Condition ===
We have that $3! = 6$ and that $4$ does not divide $6$.
So in order for $n$ to divide $\paren {n - 1}!$ it is necessary that $n \ne 4$.
{{qed|lemma}}
Theorem: Divisibility by 4

An integer $N$ expressed in decimal notation is divisible by $4$ {{iff}} the $2$ {{LSD}}s of $N$ form a $2$-digit integer divisible by $4$.
That is:
:$N = \sqbrk {a_n \ldots a_2 a_1 a_0}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $4$ 
{{iff}}:
:$10 a_1 + a_0$ is divisible by $4$.

Proof:
Let $N$ be divisible by $4$.
Then:
{{begin-eqn}}
{{eqn | l = N
      | o = \equiv
      | r = 0 \pmod 4
}}
{{eqn | ll= \leadstoandfrom
      | l = \sum_{k \mathop = 0}^n a_k 10^k
      | o = \equiv
      | r = 0 \pmod 4
}}
{{eqn | ll= \leadstoandfrom
      | l = 10 a_1 + a_0 + 10^2 \sum_{k \mathop = 2}^n a_k 10^{k - 2}
      | o = \equiv
      | r = 0 \pmod 4
}}
{{eqn | ll= \leadstoandfrom
      | l = 10 a_1 + a_0
      | o = \equiv
      | r = 0 \pmod 4
      | c = as $10^2 \equiv 0 \pmod 4$
}}
{{end-eqn}}
{{qed}}
Theorem: Ratio of Number to Reversal which is Multiple

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 ===
$8712 = 4 \times 2178$
$9801 = 9 \times 1089$
Theorem: Smallest Pythagorean Triangle is 3-4-5

The smallest Pythagorean triangle has sides of length $3$, $4$ and $5$.
:300px

Proof:
From Solutions of Pythagorean Equation, all Pythagorean triangles, the set of all primitive Pythagorean triples is generated by:
:$\tuple {2 m n, m^2 - n^2, m^2 + n^2}$
where:
:$m, n \in \Z_{>0}$ are (strictly) positive integers
:$m \perp n$, that is, $m$ and $n$ are coprime
:$m$ and $n$ are of opposite parity
:$m > n$.
The smallest two (strictly) positive integers which satisfy the above criteria are:
:$n = 1$
:$m = 2$
Hence:
:$2 m n = 2 \times 2 \times 1 = 4$
:$m^2 - n^2 = 2^2 - 1^2 = 3$
:$m^2 + n^2 = 2^2 + 1^2 = 5$
and to confirm:
:$3^2 + 4^2 = 9 + 16 = 25 = 5^2$
{{qed}}
Theorem: Pythagorean Triangle with Sides in Arithmetic Sequence

The $3-4-5$ triangle is the only Pythagorean triangle such that:
:the lengths of whose sides are in arithmetic sequence
and:
:the lengths of whose sides form a primitive Pythagorean triple.

Proof:
Let $a, b, c$ be the lengths of the sides of a Pythagorean triangle such that $a < b < c$.
Let $a, b, c$ be in arithmetic sequence:
:$b - a = c - b$
Let $a, b, c$ form a primitive Pythagorean triple:
:$a \perp b$
By definition of primitive Pythagorean triple, $a, b, c$ are in the form:
:$2 m n, m^2 - n^2, m^2 + n^2$
We have that $m^2 + n^2$ is always the hypotenuse.
There are two cases:
:$(1): \quad 2 m n > m^2 - n^2$, as in, for example, $3, 4, 5$, where $m = 2, n = 1$.
:$(2): \quad 2 m n < m^2 - n^2$, as in, for example, $8-15-17$, where $m = 4, n = 1$.
First, let $2 m n > m^2 - n^2$:
:$a = m^2 - n^2$
:$b = 2 m n$
:$c = m^2 + n^2$
Then:
{{begin-eqn}}
{{eqn | l = 2 m n - \paren {m^2 - n^2}
      | r = m^2 + n^2 - 2 m n
      | c = {{Defof|Arithmetic Sequence}}
}}
{{eqn | ll= \leadsto
      | l = 4 m n
      | r = 2 m^2
      | c = adding $2 m n + m^2 - n^2$ to both sides
}}
{{eqn | ll= \leadsto
      | l = 2 n
      | r = m
      | c = dividing both sides by $2 m$
}}
{{end-eqn}}
From Solutions of Pythagorean Equation: Primitive, $m$ and $n$ must be coprime.
Hence $n = 1$ and $m = 2$ are the only $m$ and $n$ which fulfil the requirements.
This leads to the $3-4-5$ triangle.
Now let $2 m n < m^2 - n^2$:
:$a = 2 m n$
:$b = m^2 - n^2$
:$c = m^2 + n^2$
Then:
{{begin-eqn}}
{{eqn | l = m^2 - n^2 - 2 m n
      | r = m^2 + n^2 - \paren {m^2 - n^2}
      | c = {{Defof|Arithmetic Sequence}}
}}
{{eqn | ll= \leadsto
      | l = m^2 - n^2 - 2 m n
      | r = 2 n^2
      | c = simplifying {{RHS}}
}}
{{eqn | ll= \leadsto
      | l = m^2 - 2 m n - 3 n^2
      | r = 0
      | c = subtracting $2 n^2$ from both sides
}}
{{end-eqn}}
In order for $a, b, c$ to form a primitive Pythagorean triple, then $m$ and $n$ must be of opposite parity.
If $m$ is even, then $m^2 - 2 m n$ is even.
But then $3 n^2$ is even, which makes $n$ even.
Otherwise, if $m$ is odd, then $m^2 - 2 m n$ is odd.
But then $3 n^2$ is odd, which makes $n$ odd.
So when $2 m n < m^2 - n^2$, $a, b, c$ cannot be both in arithmetic sequence and be a primitive Pythagorean triple.
Hence follows the result.
{{qed}}
Theorem: Injectivity of Laplace Transform

Let $f$, $g$ be functions from $\R_{\ge 0} \to K$ of a real variable $t$, where $K \in \set {\R, \C}$.
Further let $f$ and $g$ be continuous everywhere on their domains.
Let $f$ and $g$ both admit Laplace transforms.
Suppose that the Laplace transforms $\laptrans f$ and $\laptrans g$ satisfy:
:$\forall t \in \R_{\ge 0}: \laptrans {\map f t} = \laptrans {\map g t}$
Then $f = g$ everywhere on $\R_{\ge 0}$.

Proof:
{{ProofWanted}}
Category:Injectivity of Laplace Transform
Category:Laplace Transforms
31tsnqyswld6dmy5qw1bm23lx3pdbnd
Theorem: Pythagorean Triangle whose Area is Half Perimeter

The $3-4-5$ triangle is the only Pythagorean triangle whose area is half its perimeter.

Proof:
Let $a, b, c$ be the lengths of the sides of a Pythagorean triangle $T$.
Thus $a, b, c$ form a Pythagorean triple.
By definition of Pythagorean triple, $a, b, c$ are in the form:
:$2 m n, m^2 - n^2, m^2 + n^2$
We have that $m^2 + n^2$ is always the hypotenuse.
Thus the area of $T$ is given by:
:$\AA = m n \paren {m^2 - n^2}$
The perimeter of $T$ is given by:
:$\PP = m^2 - n^2 + 2 m n + m^2 + n^2 = 2 m^2 + 2 m n$
We need to find all $m$ and $n$ such that $\PP = 2 \AA$. 
Thus:
{{begin-eqn}}
{{eqn | l = 2 m^2 + 2 m n
      | r = 2 m n \paren {m^2 - n^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = m + n
      | r = n \paren {m + n} \paren {m - n}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n \paren {m - n}
      | r = 1
      | c = 
}}
{{end-eqn}}
As $m$ and $n$ are both (strictly) positive integers, it follows immediately that:
:$n = 1$
:$m - n = 1$
and so:
:$m = 2, n = 1$
and the result follows.
{{qed}}
Theorem: Square Modulo 5/Corollary

When written in conventional base $10$ notation, no square number ends in one of $2, 3, 7, 8$.

Proof:
The absence of $2$ and $3$ from the digit that can end a square follows directly from Square Modulo 5.
As $7 \equiv 2 \pmod 5$ and $8 \equiv 3 \pmod 5$, the result for $7$ and $8$ follows directly.
{{qed}}
Theorem: Prime equals Plus or Minus One modulo 6

Let $p$ be a prime number greater than $3$.
Then $p$ is either of the form:
:$p = 6 n + 1$
or:
:$p = 6 n - 1$
That is:
:$p = \pm 1 \pmod 6$

Proof:
To demonstrate that there are prime numbers of either form, note:
:$5 = 6 \times 1 - 1$
:$7 = 6 \times 1 + 1$
The only other possibilities for $p$ are:
:$p = 6 n$, in which case $6 \divides p$ and so $p$ is not prime
:$p = 6 n + 2$, in which case $2 \divides p$ and so $p$ is not prime
:$p = 6 n + 3$, in which case $3 \divides p$ and so $p$ is not prime
:$p = 6 n + 4$, in which case $2 \divides p$ and so $p$ is not prime
:$p = 6 n + 5$, which is the same as $p = 6 \paren {n + 1} - 1$, which is covered above.
Hence the result.
{{qed}}
Theorem: Number is Sum of Five Cubes

Let $n \in \Z$ be an integer.
Then $n$ can be expressed as the sum of $5$ cubes (either positive or negative) in an infinite number of ways.

Proof:
We have for any $m, n \in \Z$:
{{begin-eqn}}
{{eqn | l = \paren {6 m + n}^3
      | o = \equiv
      | r = n^3
      | rr = \pmod 6
      | c = Congruence of Powers
}}
{{eqn | o = \equiv
      | r = n
      | rr = \pmod 6
      | c = Euler's Theorem (Number Theory): {{EulerPhiLink|6}} $= 2$
}}
{{end-eqn}}
By definition of modulo arithmetic:
:$\exists k \in \Z: \paren {6 m + n}^3 = n + 6 k$
We also have:
{{begin-eqn}}
{{eqn | l = \paren {k + 1}^3 + \paren {k - 1}^3 - k^3 - k^3 
      | r = \paren {k^3 + 3 k^2 + 3 k + 1} + \paren {k^3 - 3 k^2 + 3 k - 1} - k^3 - k^3
      | c = Cube of Sum
}}
{{eqn | r = 6 k
}}
{{end-eqn}}
Thus $n = \paren {6 m + n}^3 + k^3 + k^3 - \paren {k + 1}^3 - \paren {k - 1}^3$ is an expression of $n$ as a sum of $5$ cubes.
In the equation above, $m$ is arbitrary and $k$ depends on both $m$ and $n$.
As there is an infinite number of choices for $m$, there is an infinite number of such expressions.
{{qed}}
Theorem: Lamé's Theorem

Let $a, b \in \Z_{>0}$ be (strictly) positive integers.
Let $c$ and $d$ be the number of digits in $a$ and $b$ respectively when expressed in decimal notation.
Let the Euclidean Algorithm be employed to find the GCD of $a$ and $b$.
Then it will take fewer than $5 \times \min \set {c, d}$ integer divisions to find $\gcd \set {a, b}$.

Proof:
=== Lemma===
{{:Lamé's Theorem/Lemma}}{{qed|lemma}}
{{WLOG}} suppose $a \ge b$. 
Then $\min \set {c, d}$ is the number of digits in $b$.
By Number of Digits in Number, we have:
:$\min \set {c, d} = \floor {\log b} + 1$
{{AimForCont}} it takes at least $5 \paren {\floor {\log b} + 1}$ cycles around the Euclidean Algorithm to find $\gcd \set {a, b}$.
Then we have:
{{begin-eqn}}
{{eqn | l = b
      | o = \ge
      | r = F_{5 \paren {\floor {\log b} + 1} + 2}
      | c = Lemma
}}
{{eqn | o = \ge
      | r = \phi^{5 \paren {\floor {\log b} + 1} }
      | c = Fibonacci Number greater than Golden Section to Power less Two
}}
{{eqn | o = >
      | r = \phi^{5 \log b}
      | c = {{Defof|Floor Function|subdef = Definition 3}}
}}
{{end-eqn}}
For $b = 1$, both sides are equal to $1$, giving $1 > 1$, which is a contradiction.
Hence we consider $b > 1$ and take $\log$ on both sides:
{{begin-eqn}}
{{eqn | ll= \leadsto
      | l = \log b
      | o = >
      | r = \paren {5 \log b} \log \phi
      | c = Logarithm of Power
}}
{{eqn | ll= \leadsto
      | l = \frac 1 {\log \phi}
      | o = >
      | r = 5
}}
{{end-eqn}}
However, $\dfrac 1 {\log \phi} \approx 4.785 < 5$.
This is a contradiction.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Volume of Unit Hypersphere

The volume of the unit sphere in $n$-dimensional space increases as $n$ goes up to $5$, but decreases thereafter.

Proof:
{{ProofWanted|Needs a considerable amount of background work to be completed first.}}
Theorem: Finite Subset Bounds Element of Finite Infima Set and Upper Closure

Let $L = \struct {S, \wedge, \preceq}$ be meet semilattice.
Let $F$ be filter in $L$.
Let $X$ be non empty finite subset of $S$.
Let $x \in S$ such that
:$x \in \paren {\map {\operatorname{fininfs} } {F \cup X} }^\succeq$
where
:$\operatorname{fininfs}$ denotes the finite infima set
:$X^\succeq$ denotes the upper closure of $X$.
Then there exists $a \in S$: $a \in F \land x \succeq a \wedge \inf X$

Proof:
By definition of upper closure of subset:
:$\exists u \in \map {\operatorname{fininfs} } {F \cup X}: u \preceq x$
By definition of finite infima set:
:there exists finite subset $Y$ of $F \cup X$:
:$Y$ admits an infimum and $u = \inf Y$
We will prove that
:$Y \setminus X \subseteq F$
Let $a \in Y \setminus X$.
By definition of difference:
:$a \in Y$ and $a \notin X$
By definition of subset:
:$a \in F \cup X$
Thus by definition of union:
:$a \in F$
{{qed|lemma}}
Define $Z := Y \setminus X$.
;Case $Z = \O$:
By $F$ is non-empty:
:$\exists b: b \in F$
By Meet Precedes Operands:
:$b \wedge \inf X \preceq \inf X$
By Set Difference with Superset is Empty Set:
:$Y \subseteq X$
By Existence of Non-Empty Finite Infima in Meet Semilattice:
:$X$ admits an infimum.
By Infimum of Subset:
:$\inf X \preceq u$
Thus by definition of trantisivity:
:$b \wedge \inf X \preceq x$
{{qed|lemma}}
;Case $Z \ne \O$:
By Set Difference is Subset:
:$Z \subseteq Y$
By Subset of Finite Set is Finite:
:$Z$ is finite.
By Union of Finite Sets is Finite:
:$Z \cup X$ is finite.
By definitions of non-empty set and union:
:$Z \cup X \ne \O$
By Existence of Non-Empty Finite Infima in Meet Semilattice:
:$X$ admits an infimum
and
:$Z \cup X$ admits an infimum.
By Set is Subset of Union:
:$Y \subseteq Y \cup X$
By Set Difference Union Second Set is Union:
:$Y \subseteq Z \cup X$
By Infimum of Subset:
:$\map \inf {Z \cup X} \preceq u$
By Existence of Non-Empty Finite Infima in Meet Semilattice:
:$Z$ admits an infimum
By Infimum of Infima:
:$\map \inf {Z \cup X} = \paren {\inf Z} \wedge \inf X$
By Filtered in Meet Semilattice with Finite Infima:
:$\inf Z \in F$
By definition of transitivity:
:$\paren {\inf Z} \wedge \inf X \preceq x$
Thus:
:$\exists a \in S: a \in F \land x \succeq a \wedge \inf X$
{{qed}}
Theorem: Length of Reciprocal of Product of Powers of 2 and 5

Let $n \in \Z$ be an integer.
Let $\dfrac 1 n$, when expressed as a decimal expansion, terminate after $m$ digits.
Then $n$ is of the form $2^p 5^q$, where $m$ is the greater of $p$ and $q$.

Proof:
Since $\dfrac 1 n$ terminates after $m$ digits:
:$\dfrac {10^m} n$ is an integer
:$\dfrac {10^{m - 1}} n$ is not an integer
From the first condition, we have $n = 2^p 5^q$ for some positive integers $p, q \le m$.
This gives $m \ge \max \set {p, q}$.
From the second condition, we cannot have both $p, q \le m - 1$.
Therefore at least one of $p, q$ is equal to $m$.
This gives $m \le \max \set {p, q}$.
These results give $m = \max \set {p, q}$.
{{qed}}
Theorem: Structure of Recurring Decimal

Let $\dfrac 1 m$, when expressed as a decimal expansion, recur with a period of $p$ digits with no nonperiodic part.
Let $\dfrac 1 n$, when expressed as a decimal expansion, terminate after $q$ digits.
Then $\dfrac 1 {m n}$ has a nonperiodic part of $q$ digits, and a recurring part of $p$ digits.

Proof:
Let $b \in \N_{>1}$ be the base we are working on.
Note that $b^p \times \dfrac 1 m$ is the result of shifting the decimal point of $\dfrac 1 m$ by $p$ digits.
Hence $b^p \times \dfrac 1 m - \dfrac 1 m$ is an integer, and $\paren {b^i - 1} \dfrac 1 m$ is not an integer for integers $0 < i < p$.
Therefore $m \divides b^p - 1$.
Also note that $b^q \times \dfrac 1 n$ is the result of shifting the decimal point of $\dfrac 1 n$ by $q$ digits.
Hence $b^q \times \dfrac 1 n$ is an integer, and $b^j \times \dfrac 1 n$ is not an integer for integers $0 < j < q$.
Therefore $n \divides b^q$ and $n \nmid b^{q - 1}$.
Write $m x = b^p - 1$ and $n y = b^q$.
Then $\dfrac 1 {m n} = \dfrac {x y} {\paren {b^p - 1} b^q}$.
By Division Theorem:
:$\exists! r, s \in \Z: x y = s \paren {b^p - 1} + r, 0 \le r < b^p - 1$
Then we would have:
:$\dfrac 1 {m n} = \dfrac {s \paren {b^p - 1} + r} {\paren {b^p - 1} b^q} = \dfrac s {b^q} + \dfrac r {\paren {b^p - 1} b^q}$
which is a fraction with a nonperiodic part of $q$ digits equal to $s$, followed by a recurring part of $p$ digits equal to $r$.
To show that the nonperiodic part and recurring part of $\dfrac 1 {m n}$ cannot be shorter, we must show:
:$r \not \equiv s \pmod b$: or else the nonperiodic part could be shortened by at least $1$ digit
:$\dfrac r {b^p - 1} \paren {b^i - 1}$ is not an integer for integers $0 < i < p$: or else the recurring part could be shortened to $i$ digits
To show the first condition, suppose the contrary.
Then:
{{begin-eqn}}
{{eqn | l = x y
      | r = s \paren {b^p - 1} + r
}}
{{eqn | o = \equiv
      | r = s \paren {b^p - 1} + s
      | rr= \pmod b
}}
{{eqn | o = \equiv
      | r = s b^p
      | rr= \pmod b
}}
{{eqn | o = \equiv
      | r = 0
      | rr= \pmod b
}}
{{end-eqn}}
From GCD with Remainder:
:$\gcd \set {b, b^p - 1} = 1$
Since $x \divides b^p - 1$, by Divisor of One of Coprime Numbers is Coprime to Other:
:$\gcd \set {b, x} = 1$
By Euclid's Lemma:
:$b \divides y$
From $n y = b^q$:
:$n \times \dfrac y b = b^{q - 1}$
so $n \divides b^{q - 1}$, which contradicts the properties of $n$.
To show the second condition, suppose the contrary.
Then:
{{begin-eqn}}
{{eqn | l = \paren {b^i - 1} \dfrac 1 m
      | r = \dfrac {n \paren {b^i - 1} } {m n}
}}
{{eqn | r = n \paren {b^i - 1} \paren {\dfrac s {b^q} + \dfrac r {\paren {b^p - 1} b^q} }
}}
{{eqn | r = \dfrac n {b^q} \paren {s \paren {b^i - 1} + \dfrac {r \paren {b^i - 1} } {b^p - 1} }
}}
{{end-eqn}}
Both fractions are integers, so $\paren {b^i - 1} \dfrac 1 m$ is also an integer, which contradicts the properties of $\dfrac 1 m$.
Hence the result.
{{qed}}
Theorem: Characteristics of Pentatope

A pentatope has $5$ cells, $10$ faces, $10$ edges and $5$ vertices.

Proof:
{{ProofWanted|... also worth mentioning that an astute reader would correlate this result with the similar characteristics of a tetrahedron and triangle, and wonder whether Pascal's triangle comes into this anywhere.}}
Theorem: Pentatope is Self-Dual

A pentatope is self-dual.

Proof:
{{ProofWanted|Plenty of background work on hyperdimensional geometry to be done before we can even start.}}
Theorem: Exponential is of Exponential Order Real Part of Index

Let $\map f t = e^{\psi t}$ be the complex exponential function, where $t \in \R, \psi \in \C$.
Let $a = \map \Re \psi$.
Then $e^{\psi t}$ is of exponential order $a$.

Proof:
{{begin-eqn}}
{{eqn | q = \forall t \ge 1
      | l = \size {e^{\psi t} }
      | r = e^{a t}
      | c = Modulus of Exponential is Exponential of Real Part
}}
{{eqn | o = <
      | r = 2 e^{a t}
      | c = Exponential of Real Number is Strictly Positive
}}
{{end-eqn}}
The result follows from the definition of exponential order with $M = 1$, $K = 2$, and $a = a$.
{{qed}}
Category:Exponential Order
d1jpdpl98duorghnm3z0rnqd5v4s48r
Theorem: Cosine is of Exponential Order Zero

Let $\cos t$ be the cosine of $t$, where $t \in \R$.
Then $\cos t$ is of exponential order $0$.

Proof:
{{begin-eqn}}
{{eqn | q = \forall t \ge 1
      | l = \size {\cos t}
      | o = \le
      | r = 1
      | c = Real Cosine Function is Bounded
}}
{{eqn | ll= \leadsto
      | l = \size {\cos t}
      | o = <
      | r = 2
}}
{{eqn | r = 2 e^{0 t}
      | c = Exponential of Zero
}}
{{end-eqn}}
The result follows from the definition of exponential order with $M = 1$, $K = 2$, and $a = 0$.
{{qed}}
Theorem: Cosine is of Exponential Order Zero

Let $\cos t$ be the cosine of $t$, where $t \in \R$.
Then $\cos t$ is of exponential order $0$.

Proof:
The result follows from Real Cosine Function is Bounded and Bounded Function is of Exponential Order Zero.
{{qed}}
Theorem: Sine is of Exponential Order Zero

Let $\sin t$ be the sine of $t$, where $t \in \R$.
Then $\sin t$ is of exponential order $0$.

Proof:
{{begin-eqn}}
{{eqn | l = \size {\sin t}
      | o = \le
      | r = 1
      | c = Real Sine Function is Bounded
}}
{{eqn | ll =\leadsto
      | l = \size {\sin t}
      | o = <
      | r = 2
}}
{{eqn | r = 2 e^{0 t}
      | c = Exponential of Zero
}}
{{end-eqn}}
{{qed}}