avewright/proofwiki-math huggingface dataset:
Theorem: Group of Order 15 is Cyclic Group

Let $G$ be a group whose order is $15$.
Then $G$ is cyclic.

Proof:
{{AimForCont}} $G$ is non-abelian.
Let $n_3$ denote the number of elements of $G$ of order $3$.
From Number of Elements of Order p in Group of Order pq is Multiple of q, $n_3$ is a multiple of $5$.
From Number of Order p Elements in Group with m Order p Subgroups, $n_3$ is a multiple of $2$.
Therefore $n_3$ is a multiple of $10$.
Let $n_5$ denote the number of elements of $G$ of order $5$.
From Number of Elements of Order p in Group of Order pq is Multiple of q, $n_5$ is a multiple of $3$.
From Number of Order p Elements in Group with m Order p Subgroups, $n_5$ is a multiple of $4$.
Therefore $n_5$ is a multiple of $12$.
Together with the identity element which has order $1$, that makes $1 + 12 a + 10 b = 15$ for some positive integers $a$ and $b$.
This is impossible.
Hence by Proof by Contradiction it follows that $G$ must be abelian.
Since $15$ is a product of $2$ distinct primes, by Abelian Group of Semiprime Order is Cyclic, $G$ is cyclic.
{{qed}}
Theorem: Number of Abelian Groups

Let $n \in \Z_{\ge 1}$ be a (strictly) positive integer.
Let:
:$n = \ds \prod_{i \mathop = 1}^s p_i^{m_i}$
where the $p_i$ are distinct primes.
Let $\map {\nu_a} n$ denote the number of abelian groups of order $n$.
Then:
:$\map {\nu_a} n = \ds \prod_{i \mathop = 1}^s \map {\nu_a} {p_i^{m_i} }$
where:
:$\map {\nu_a} {p_i^{m_i} }$ is the number of integer partitions of $m_i$.

Proof:
{{ProofWanted|long and heavy proof which needs plenty work}}
Theorem: Order of Quotient Group

Let $G$ be a finite group.
Let $N$ be a normal subgroup of $G$.
Let $G / N$ be the quotient group of $G$ by $N$.
Then:
:$\dfrac {\order G} {\order N} = \order {G / N}$
where $\order G$ denotes the order of $G$.

Proof:
From Lagrange's Theorem:
:$\dfrac {\order G} {\order N} = \index G N$
where $\index G N$ is the index of $N$ in $G$.
By definition of index:
:$\index G N = \order {G / N}$
{{qed}}
Theorem: Non-Abelian Simple Finite Groups are Infinitely Many

There exist infinitely many types of group which are non-abelian and finite.

Proof:
We have that Alternating Group is Simple except on 4 Letters.
So for all $n \in \N$ such that $n \ne 4$, the alternating group $A_n$ is a simple group.
We also have that $A_n$ is non-abelian for all $n > 3$.
Hence the result.
{{qed}}
Theorem: Centralizer of Self-Inverse Element of Non-Abelian Finite Simple Group is not That Group

Let $G$ be a non-abelian finite simple group.
Let $t \in G$ be a self-inverse element of $G$.
Then:
:$\map {C_G} t \ne G$
where $\map {C_G} t$ denotes the centralizer of $t$ in $G$.

Proof:
Let $G$ be a non-abelian finite simple group.
Let $t \in G$ which is not the identity.
By definition of a simple group and Center of Group is Normal Subgroup:
:either $\map Z G = G$ or $\map Z G$ is the trivial group.
By definition of an abelian group:
:$\map Z G = G$ {{iff}} $G$ is abelian
Hence we must have $\map Z G$ is the trivial group.
Thus $t \notin \map Z G$.
From definition of center:
:$\exists x \in G: t x \ne x t$
For this $x$, $x \notin \map {C_G} t$.
Hence $\map {C_G} t \ne G$.
{{Questionable|So trivial it's scary: something stronger was proved </br> Remark: the identity is a self-inverse element with $\map {C_G} e {{=}} G$ trivially}}
Theorem: Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 1

:$\norm {\, \cdot \,}_1$ is well-defined.
That is, 
:$(1): \quad \ds \forall \eqclass {x_n}{}: \lim_{n \mathop \to \infty} \norm{x_n}$ exists
:$(2): \quad \ds \forall \eqclass {x_n}{}, \eqclass {y_n}{} \in \CC \,\big / \NN: \eqclass {x_n}{} = \eqclass {y_n}{} \implies \lim_{n \mathop \to \infty} \norm{x_n} = \lim_{n \mathop \to \infty} \norm{y_n}$

Proof:
By Norm Sequence of Cauchy Sequence has Limit then:
:for each $\eqclass {x_n}{}$ the $\ds \lim_{n \mathop \to \infty} \norm{x_n}$ exists.
Suppose $\eqclass {x_n}{} = \eqclass {y_n}{}$.
By Left Cosets are Equal iff Difference in Subgroup then:
:$\sequence {x_n} - \sequence {y_n} =  \sequence {x_n - y_n} \in \NN$
By Equivalent Cauchy Sequences have Equal Limits of Norm Sequences then:
:$\ds \lim_{n \mathop \to \infty} \norm{x_n} = \lim_{n \mathop \to \infty} \norm{y_n}$
Hence:
:$\ds \norm {\eqclass {x_n}{} }_1 = \lim_{n \mathop \to \infty} \norm{x_n} = \lim_{n \mathop \to \infty} \norm{y_n} =  \norm {\eqclass {x_n}{} }_1$
The result follows.
{{qed}}
Theorem: Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 2

:$\norm {\, \cdot \,}_1$ satisfies {{Norm-axiom-mult|1}}
That is:
:$\forall \eqclass {x_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n} {} }_1 = 0 \iff \eqclass {x_n} {} = \eqclass {0_R} {} $

Proof:
By Quotient Ring of Cauchy Sequences is Division Ring the zero of $\CC \,\big / \NN$ is $\eqclass {0_R} {}$.
{{begin-eqn}}
{{eqn | l = \norm {\eqclass {0_R} {} }_1 = 0
      | o = \leadstoandfrom
      | r = \lim_{n \mathop \to \infty} \norm {x_n} = 0
      | c = Definition of $\norm {\,\cdot\,}_1$
}}
{{eqn | o = \leadstoandfrom
      | r = \sequence {x_n} \in \NN
      | c = Definition of $\NN$
}}
{{eqn | o = \leadstoandfrom
      | r = \eqclass {x_n} {} = \eqclass {0_R} {}
      | c = Left Cosets are Equal iff Product with Inverse in Subgroup
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 3

:$\norm {\, \cdot \,}_1$ satisfies the {{Norm-axiom-mult|2}}.
That is:
:$\forall \eqclass {x_n} {}, \eqclass {y_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n} {} \eqclass {y_n} {} }_1 = \norm {\eqclass {x_n} {} }_1 \times \norm {\eqclass {y_n} {} }_1$

Proof:
Let $\eqclass {x_n} {}, \eqclass {y_n} {} \in \CC \,\big / \NN$
{{begin-eqn}}
{{eqn | l = \norm {\eqclass {x_n} {} \eqclass {y_n} {} }_1
      | r = \norm {\eqclass {x_n y_n} {} }_1
      | c = Multiplication on quotient ring
}}
{{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n y_n}
      | c = Definition of $\norm {\,\cdot\,}_1$
}}
{{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n} \norm {y_n}
      | c = {{Norm-axiom-mult|2}}
}}
{{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n} \times \lim_{n \mathop \to \infty} \norm {y_n}
      | c = Product Rule for Real Sequences
}}
{{eqn | r = \norm {\eqclass {x_n} {} }_1 \times \norm {\eqclass {y_n} {} }_1
      | c = Definition of $\norm {\,\cdot\,}_1$
}}
{{end-eqn}}
{{qed}}
Theorem: Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 4

:$\norm {\, \cdot \,}_1$ satisfies the {{Norm-axiom-mult|3}}.
That is:
:$\forall \eqclass {x_n} {}, \eqclass {y_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n} {} +  \eqclass {y_n} {} }_1 \le \norm {\eqclass {x_n} {} }_1 + \norm {\eqclass {y_n} {} }_1$

Proof:
Let $\eqclass {x_n} {}, \eqclass {y_n} {} \in \CC \,\big / \NN$
{{begin-eqn}}
{{eqn | l = \norm {\eqclass {x_n} {} + \eqclass {y_n} {} } _1
      | r = \norm {\eqclass {x_n + y_n} {} }_1
      | c = Addition on quotient ring
}}
{{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n + y_n}
      | c = Definition of $\norm {\,\cdot\,}_1$
}}
{{end-eqn}}
By {{Norm-axiom-mult|3}}:
:$\forall n: \norm {x_n + y_n} \le \norm {x_n} + \norm {y_n}$
So:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \norm {x_n + y_n}
      | o = \le
      | r = \lim_{n \mathop \to \infty} \norm {x_n} + \norm {y_n}
      | c = Inequality Rule for Real Sequences 
}}
{{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n} + \lim_{n \mathop \to \infty} \norm {y_n}
      | c = Sum Rule for Real Sequences
}}
{{eqn | r = \norm {\eqclass {x_n} {} }_1 + \norm {\eqclass {y_n} {} } _1
      | c = Definition of $\norm {\,\cdot\,}_1$
}}
{{end-eqn}}
{{qed}}
Theorem: Invertible Elements of Monoid form Subgroup

Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.
Let $U \subseteq S$ be the subset of $S$ consisting of the invertible elements of $S$.
Then $\struct {U, \circ}$ forms a subgroup of $S$.

Proof:
We have from Inverse of Identity Element is Itself that $e$ is invertible.
Hence $e \in U$ and so $U \ne \O$.
Let $x, y \in U$.
As $x$ and $y$ are invertible, it follows that $x^{-1}$ and $y^{-1}$ both exist in $S$.
Both $x^{-1}$ and $y^{-1}$ also have inverses $x$ and $y$ respectively, and so themselves are invertible.
Hence both $x^{-1} \in U$ and $y^{-1} \in U$.
Then we have:
{{begin-eqn}}
{{eqn | l = \paren {x \circ y} \circ \paren {y^{-1} \circ x^{-1} }
      | r = x \circ \paren {y \circ y^{-1} } \circ x^{-1}
      | c = $\circ$ is associative in $S$
}}
{{eqn | r = x \circ e \circ x^{-1}
      | c = {{Defof|Inverse Element}}
}}
{{eqn | r = x \circ x^{-1}
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = e
      | c = {{Defof|Inverse Element}}
}}
{{end-eqn}}
and similarly for $\paren {y^{-1} \circ x^{-1} } \circ \paren {x \circ y}$
Hence $x \circ y$ is invertible.
The result follows by the Two-Step Subgroup Test.
{{qed}}
Theorem: Combination Theorem for Sequences/Normed Division Ring/Inverse Rule/Lemma

:$\ds \lim_{n \mathop \to \infty} {y_n}^{-1} = l^{-1}$

Proof:
By Limit of Subsequence equals Limit of Sequence then $\sequence {y_n}$ is convergent with:
:$\ds \lim_{n \mathop \to \infty} y_n = l$
Let $\epsilon > 0$ be given.
Let $\epsilon' = \dfrac {\epsilon  {\norm l}^2 } {2}$.
Then:
:$ \epsilon' > 0$
As $\sequence {y_n} \to l$, as $n \to \infty$, we can find $N_1$ such that:
:$\forall n > N_1: \norm {y_n - l} < \epsilon'$
As $\sequence {y_n}$ converges to $l \ne 0$, by Sequence Converges to Within Half Limit:
:$\exists N_2 \in \N: \forall n > \N_2: \dfrac {\norm l} 2 < \norm {y_n}$
or equivalently:
:$\exists N_2 \in \N: \forall n > \N_2: 1 < \dfrac {2 \norm {y_n} } {\norm l}$
Let $N = \max \set {N_1, N_2}$. 
Then $\forall n > N$:
:$(1): \quad \norm {y_n - l} < \epsilon'$
:$(2): \quad 1 < \dfrac {2 \norm {y_n} } {\norm l}$
Hence:
{{begin-eqn}}
{{eqn | l = \norm { {y_n }^{-1} - l^{-1} }
      | o = <
      | r = \dfrac {2 \norm {y_n} } {\norm l} \norm { {y_n}^{-1} - l^{-1} }
      | c = from $(1)$
}}
{{eqn | r = \dfrac 2 {\norm l^2} \paren {\norm {y_n} \norm { {y_n}^{-1} - l^{-1} } \norm l}
      | c = multiplying and dividing by $\norm l$
}}
{{eqn | r = \dfrac 2 {\norm l^2} \norm {y_n \paren { {y_n}^{-1} - l^{-1} } l}
      | c = {{Norm-axiom-mult|2}}
}}
{{eqn | r = \dfrac 2 {\norm l^2} \norm {y_n y_n^{-1} l - y_n l^{-1} l}
      | c = {{Ring-axiom|D}}
}}
{{eqn | r = \dfrac 2 {\norm l^2} \norm {l - y_n}
      | c = Inverse Property of Division Ring
}}
{{eqn | o = <
      | r = \dfrac 2 {\norm l^2} \epsilon'
      | c = from $(2)$
}}
{{eqn | r = \dfrac 2 {\norm l^2} \paren {\dfrac {\epsilon \norm l^2 } 2}
      | c = Definition of $\epsilon'$
}}
{{eqn | r = \epsilon
      | c = cancelling terms
}}
{{end-eqn}}
Hence:
:$\sequence { {y_n}^{-1} }$ is convergent with $\ds \lim_{n \mathop \to \infty} {y_n}^{-1} = l^{-1}$.
{{qed}}
Category:Combination Theorem for Sequences in Normed Division Rings
s3qt4k5v764dmrfiezl3tedvqn6h603
Theorem: Division Ring Norm is Continuous on Induced Metric Space

Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
The mapping $\norm {\,\cdot\,} : \struct {R, d} \to \R$ is continuous.

Proof:
Let $x_0 \in R$.
Let $\epsilon \in \R_{>0}$. 
Let $x \in R: \norm {x - x_0} < \epsilon$.
Then:
{{begin-eqn}}
{{eqn | l = \size {\norm x - \norm {x_0} }
      | o = \le
      | r = \norm {x - x_0}
      | c = Reverse Triangle Inequality on Normed Division Ring
}}
{{eqn | o = <
      | r = \epsilon
}}
{{end-eqn}}
By the definition of metric induced by a norm and the definition of a continuous mapping, $\norm {\,\cdot\,}$ is continuous.
{{qed}}
Category:Normed Division Rings
Category:Norm Theory
7th26w1unw3wb35jl5blce3f62jh7cs
Theorem: Sum of Sequence of Squares of Fibonacci Numbers

:$\forall n \ge 1: \ds \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
That is:
:${F_1}^2 + {F_2}^2 + {F_3}^2 + \cdots + {F_n}^2 = F_n F_{n + 1}$

Proof:
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
Theorem: Properties of Norm on Division Ring/Norm of Negative

:$\norm {-x} = \norm x$

Proof:
By Norm of Negative of Unity:
:$\norm {-1_R} = 1$
Then:
{{begin-eqn}}
{{eqn | l = \norm {-x}
      | r = \norm {-1_R \circ x}
      | c = Product with Ring Negative
}}
{{eqn | r = \norm {-1_R} \norm x
      | c = {{Norm-axiom-mult|2}}
}}
{{eqn | r = \norm x
      | c = Norm of Negative of Unity
}}
{{end-eqn}}
as desired.
{{qed}}
Theorem: Properties of Norm on Division Ring/Norm of Unity

:$\norm {1_R} = 1$.

Proof:
By {{Norm-axiom-mult|2}}:
:$\forall x, y \in R: \norm {x \circ y} = \norm x \norm y$
In particular:
:$\norm {1_R} = \norm {1_R \circ 1_R} = \norm {1_R} \norm {1_R}$
By {{Norm-axiom-mult|1}}:
:$\norm {1_R} \ne 0$
So $\norm {1_R}$ has an inverse in $R$.
Multiplying by this inverse:
:$\norm {1_R} \norm {1_R} = \norm {1_R} \iff \norm {1_R} = 1$
as desired.
{{qed}}
Theorem: Properties of Norm on Division Ring/Norm of Negative of Unity

:$\norm {-1_R} = 1$

Proof:
By Product of Ring Negatives:
:$-1_R \circ -1_R = 1_R \circ 1_R = 1_R$
So:
{{begin-eqn}}
{{eqn | l = \norm {-1_R}^2
      | r = \norm {-1_R} \norm {-1_R}
}}
{{eqn | r = \norm {-1_R \circ -1_R}
      | c = {{Norm-axiom-mult|2}}
}}
{{eqn | r = \norm {1_R}
      | c = Product of Ring Negatives
}}
{{eqn | r = 1
      | c = Norm of Unity of Division Ring
}}
{{end-eqn}}
Thus:
:$\norm {-1_R} = \pm 1$
By {{Norm-axiom-mult|1}}: 
:$\norm {-1_R} \ge 0$
Hence:
:$\norm {-1_R} = 1$
{{qed}}
Theorem: Repunit is Zuckerman Number

Let $n$ be a repunit.
Then $n$ is also a Zuckerman number.

Proof:
The digits of a repunit are by definition all $1$.
Thus the product of the digits of a repunit is $1$.
By One Divides all Integers, $1$ is a divisor of $n$.
Hence the result, by definition of Zuckerman number.
Theorem: Properties of Norm on Division Ring/Norm of Difference

:$\norm {x - y} \le \norm x + \norm y$

Proof:
Then:
{{begin-eqn}}
{{eqn | l = \norm {x - y}
      | r = \norm {x + \paren {-y} }
}}
{{eqn | o = \le
      | r = \norm x + \norm {-y}
      | c = {{Norm-axiom-mult|3}}
}}
{{eqn | r = \norm x + \norm y
      | c = Norm of Ring Negative
}}
{{end-eqn}}
as desired.
{{qed}}
Theorem: Properties of Norm on Division Ring/Norm of Inverse

:$x \ne 0_R \implies \norm {x^{-1} } = \dfrac 1 {\norm x}$

Proof:
Let $x \ne 0_R$.
By {{Norm-axiom-mult|1}}:
:$\norm x \ne 0$
So:
{{begin-eqn}}
{{eqn| l = \norm x \norm {x^{-1} }
     | r = \norm {x \circ x^{-1} }
     | c = {{Norm-axiom-mult|2}}
}}
{{eqn| r = \norm {1_R}
     | c = {{Defof|Product Inverse}}
}}
{{eqn| r = 1
     | c = Norm of Unity of Division Ring
}}
{{eqn| ll= \leadsto
     | l = \norm {x^{-1} }
     | r = \dfrac 1 {\norm x}
     | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Properties of Norm on Division Ring/Norm of Quotient

:$y \ne 0_R \implies \norm {x y^{-1} } = \norm {y^{-1} x} = \dfrac {\norm x} {\norm y}$

Proof:
Let $y \ne 0_R$.
By {{Norm-axiom-mult|1}} then:
:$\norm y \ne 0$
So:
{{begin-eqn}}
{{eqn| l = \norm {x \circ y^{-1} }
     | r = \norm x \norm {y^{-1} }
     | c = {{Norm-axiom-mult|2}}
}}
{{eqn| r = \dfrac {\norm x} {\norm y}
     | c = Norm of Inverse
}}
{{end-eqn}}
Similarly:
:$\norm {y^{-1} x} = \dfrac {\norm x} {\norm y}$
{{qed}}
Theorem: Properties of Norm on Division Ring/Norm of Power Equals Unity

:$\forall n \in \N_{>0}: \norm {x^n} = 1 \implies \norm x = 1$

Proof:
Let $n \in \N_{>0}$.
Let $\norm {x^n} = 1$.
By {{Norm-axiom-mult|2}}:
:$\norm x^n = 1$
Since $\norm x \ge 0$, by Positive Real Complex Root of Unity:
:$\norm x = 1$
as desired.
{{qed}}
Theorem: Properties of Norm on Division Ring/Norm of Integer

For all $n \in \N_{>0}$, let $n \cdot 1_R$ denote the sum of $1_R$ with itself $n$-times. That is:
:$n \cdot 1_R = \underbrace {1_R + 1_R + \dots + 1_R}_{\text {$n$ times} }$
Then:
:$\norm {n \cdot 1_R} \le n$.

Proof:
Let $n \in \N_{>0}$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {n \cdot 1_R}
      | r = \norm {1_R + 1_R + \dots + 1_R}
}}
{{eqn | o = \le
      | r = \underbrace {\norm {1_R} + \norm {1_R} + \dots + \norm {1_R} }_{\text {$n$ times} }
      | c = {{Norm-axiom-mult|3}}
}}
{{eqn | r = \underbrace {1 + 1 + \dots + 1 }_{\text {$n$ times} }
      | c = Norm of Unity
}}
{{eqn | r = n
}}
{{end-eqn}}
{{qed}}
Theorem: Conversion from Hexadecimal to Binary

Let $n$ be a (positive) integer expressed in hexadecimal notation as:
:$n = \sqbrk {a_r a_{r - 1} \dotso a_1 a_0}_H$
Then $n$ can be expressed in binary notation as:
:$n = \sqbrk {b_{r 3} b_{r 2} b_{r 1} b_{r 0} b_{\paren {r - 1} 3} b_{\paren {r - 1} 2} b_{\paren {r - 1} 1} b_{\paren {r - 1} 0} \dotso b_{1 3} b_{1 2} b_{1 1} b_{1 0} b_{0 3} b_{0 2} b_{0 1} b_{0 0} }_2$
where $\sqbrk {b_{j 3} b_{j 2} b_{j 1} b_{j 0} }_2$ is the expression of the hexadecimal digit $a_j$ in binary notation.
That is, you take the binary expression of each hexadecimal digit, padding them out with zeroes to make them $4$ bits long, and simply concatenate them.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = n
      | r = \sqbrk {a_r a_{r - 1} \dotso a_1 a_0}_H
      | c = 
}}
{{eqn | r = \sum_{j \mathop = 0}^r a_j 16^j
      | c = {{Defof|Hexadecimal Notation}}
}}
{{end-eqn}}
We have that:
:$0 \le a_j < 16$
and so:
{{begin-eqn}}
{{eqn | l = a_j
      | r = \sqbrk {b_{j 3} b_{j 2} b_{j 1} b_{j 0} }_2
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^3 b_{j k} 2^k
      | c = {{Defof|Binary Notation}}
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = n
      | r = \sqbrk {a_r a_{r - 1} \dotso a_1 a_0}_H
      | c = 
}}
{{eqn | r = \sum_{j \mathop = 0}^r a_j 16^j
      | c = {{Defof|Hexadecimal Notation}}
}}
{{eqn | r = \sum_{j \mathop = 0}^r \paren {\sum_{k \mathop = 0}^3 b_{j k} 2^k} 16^j
      | c = {{Defof|Binary Notation}}
}}
{{eqn | r = \sum_{j \mathop = 0}^r \paren {\sum_{k \mathop = 0}^3 b_{j k} 2^k} 2^{4 j}
      | c = 
}}
{{eqn | r = \sum_{j \mathop = 0}^r \paren {\sqbrk {b_{j 3} b_{j 2} b_{j 1} b_{j 0} }_2} 2^{4 j}
      | c = {{Defof|Binary Notation}}
}}
{{eqn | r = \sqbrk {b_{r 3} b_{r 2} b_{r 1} b_{r 0} }_2 2^{4 r} + \sqbrk {b_{\paren {r - 1} 3} b_{\paren {r - 1} 2} b_{\paren {r - 1} 1} b_{\paren {r - 1} 0} }_2 2^{4 {r - 1} } + \cdots + \sqbrk {b_{1 3} b_{1 2} b_{1 1} b_{1 0} }_2 2^4 + \sqbrk {b_{0 3} b_{0 2} b_{0 1} b_{0 0} }_2
      | c = 
}}
{{eqn | r = \sqbrk {b_{r 3} b_{r 2} b_{r 1} b_{r 0} b_{\paren {r - 1} 3} b_{\paren {r - 1} 2} b_{\paren {r - 1} 1} b_{\paren {r - 1} 0} \dotso b_{1 3} b_{1 2} b_{1 1} b_{1 0} b_{0 3} b_{0 2} b_{0 1} b_{0 0} }_2
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Birthday Paradox/General/3

Let $n$ be a set of people.
Let the probability that at least $3$ of them have the same birthday be greater than $50 \%$.
Then $n \ge 88$.

Proof:
Let $\map F {r, n}$ be the number of ways to distribute $r$ objects into $n$ cells such that there are no more than $2$ objects in each cell.
Let there be $d$ cells which are each occupied by $2$ objects.
These can be chosen in $\dbinom n d$ ways.
There remain $s = r - 2 d$ objects which can then be distributed among $n - d$ cells in $\dbinom {n - d} s$ ways.
In each such arrangement, the $r$ objects may be permuted in:
:$\dbinom r 2 \dbinom {r - 2} 2 \cdots \dbinom {r - 2 d + 2} 2 \paren {r - 2 d}! = \dfrac {r!} {2^d}$
different ways.
Hence:
:$\map F {r, n} = \dbinom n d \dbinom {n - d} s \dfrac {r!} {2^d}$
So the probability of exactly $d$ pairs and $s$ singletons, where $d - s \le n$, is given by:
:$\dfrac {\map F {r, n} } {n^r}$
If we assume a $365$-day year, we have that the probability that at least $3$ of them have the same birthday is given by:
:$\map \Pr r = 1 - \ds \sum_{d \mathop = 0}^{\floor {r / 2} } \dfrac {n! \, r!} {n^r 2^d d! \paren {r - 2 d}! \paren {n + d - r}!}$
where $n = 365$.
We require the smallest $r$ for which $\map \Pr r > \dfrac 1 2$.
The result yields to calculation.
{{qed}}
Theorem: Convergent Sequence is Cauchy Sequence/Normed Division Ring

Let $\struct {R, \norm {\,\cdot\,}} $ be a normed division ring.
Every convergent sequence in $R$ is a Cauchy sequence.

Proof:
Let $\sequence {x_n}$ be a sequence in $R$ that converges to the limit $l \in R$.
Let $\epsilon > 0$. 
Then also $\dfrac \epsilon 2 > 0$.
Because $\sequence {x_n}$ converges to $l$, we have:
:$\exists N: \forall n > N: \norm {x_n - l} < \dfrac \epsilon 2$
So if $m > N$ and $n > N$, then:
{{begin-eqn}}
{{eqn | l = \norm {x_n - x_m}
      | r = \norm {x_n - l + l - x_m}
}}
{{eqn | o = \le
      | r = \norm {x_n - l} + \norm {l - x_m}
      | c = {{Norm-axiom-mult|3}}
}}
{{eqn | o = <
      | r = \frac \epsilon 2 + \frac \epsilon 2
      | c = by choice of $N$
}}
{{eqn | r = \epsilon
      | c = 
}}
{{end-eqn}}
Thus $\sequence {x_n}$ is a Cauchy sequence.
{{qed}}
Theorem: Convergent Sequence is Cauchy Sequence/Normed Division Ring

Let $\struct {R, \norm {\,\cdot\,}} $ be a normed division ring.
Every convergent sequence in $R$ is a Cauchy sequence.

Proof:
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.
Let $\sequence {x_n}$ be a sequence in $R$ that converges to the limit $l$ in $\struct {R, \norm {\,\cdot\,}}$.
Thus, by definition, $\sequence {x_n} $ converges to the limit $l$ in $\struct {R, d}$.
By Convergent Sequence is Cauchy Sequence in metric space, $\sequence {x_n} $ is a Cauchy sequence in $\struct {R, d}$.
Thus, by definition, $\sequence {x_n} $ is a Cauchy sequence in $\struct {R, \norm {\,\cdot\,}}$.
{{qed}}
Theorem: Sum of 3 Squares in 2 Distinct Ways

$27$ is the smallest positive integer which can be expressed as the sum of $3$ square numbers in $2$ distinct ways:
{{begin-eqn}}
{{eqn | l = 27
      | r = 3^2 + 3^2 + 3^2
}}
{{eqn | r = 5^2 + 1^2 + 1^2
}}
{{end-eqn}}

Proof:
Can be performed by brute-force investigation.
Theorem: Triangular Numbers which are Sum of Two Cubes

The sequence of triangular numbers which are the sum of $2$ cubes begins:
:$28, 91, 351, 2926, 8001, 46971, 58653, 93528, 97461, \dots$
{{OEIS|A113958}}

Proof:
Can be demonstrated by brute force.
For example:
{{begin-eqn}}
{{eqn | l = 28
      | r = 1 + 27
      | c = 
}}
{{eqn | r = 1^3 + 3^3
      | c = 
}}
{{eqn | r = \dfrac {7 \paren {7 + 1} } 2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 91
      | r = 27 + 64
      | c = 
}}
{{eqn | r = 3^3 + 4^3
      | c = 
}}
{{eqn | r = \dfrac {13 \paren {13 + 1} } 2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 351
      | r = 125 + 216
      | c = 
}}
{{eqn | r = 5^3 + 6^3
      | c = 
}}
{{eqn | r = \dfrac {26 \paren {26 + 1} } 2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 2976
      | r = 729 + 2197
      | c = 
}}
{{eqn | r = 5^3 + 6^3
      | c = 
}}
{{eqn | r = \dfrac {76 \paren {76 + 1} } 2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 8001
      | r = 1 + 8000
      | c = 
}}
{{eqn | r = 1^3 + 20^3
      | c = 
}}
{{eqn | r = \dfrac {126 \paren {126 + 1} } 2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 46 \, 971
      | r = 4096 + 42 \, 875
      | c = 
}}
{{eqn | r = 16^3 + 35^3
      | c = 
}}
{{eqn | r = \dfrac {306 \paren {306 + 1} } 2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 58 \, 653
      | r = 8000 + 50 \, 653
      | c = 
}}
{{eqn | r = 20^3 + 37^3
      | c = 
}}
{{eqn | r = \dfrac {342 \paren {343 + 1} } 2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 93 \, 528
      | r = 42 \, 875 + 50 \, 653
      | c = 
}}
{{eqn | r = 35^3 + 37^3
      | c = 
}}
{{eqn | r = \dfrac {432 \paren {433 + 1} } 2
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 97 \, 461
      | r = 125 + 97 \, 336
      | c = 
}}
{{eqn | r = 5^3 + 46^3
      | c = 
}}
{{eqn | r = \dfrac {441 \paren {442 + 1} } 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Product of Factors of Even Perfect Number

Let $P$ be the perfect number $2^{n - 1} \paren {2^n - 1}$.
Then:
:$\ds \prod_{d \mathop \divides P} d = P^n$

Proof:
The factors of $P$ are:
:$1, 2, 4, \dots, 2^{n - 1}, 2^n - 1, 2 \paren {2^n - 1}, \dots, 2^{n - 1} \paren {2^n - 1}$
Therefore their product is:
{{begin-eqn}}
{{eqn | l = \prod_{d \mathop \divides P} d
      | r = \paren {\prod_{i \mathop = 0}^{n - 1} 2^i} \paren {\prod_{i \mathop = 0}^{n - 1} 2^i \paren {2^n - 1} }
}}
{{eqn | r = \paren {\prod_{i \mathop = 0}^{n - 1} 2^i}^2 \paren {2^n - 1}^n
}}
{{eqn | r = \paren {2^\frac {n \paren {n - 1} } 2}^2 \paren {2^n - 1}^n
}}
{{eqn | r = \paren {2^{n - 1} }^n \paren {2^n - 1}^n
}}
{{eqn | r = P^n
}}
{{end-eqn}}
{{qed}}