avewright/proofwiki-math huggingface dataset:
Theorem: Vector Cross Product Distributes over Addition

The vector cross product is distributive over addition.
That is, in general:
:$\mathbf a \times \paren {\mathbf b + \mathbf c} = \paren {\mathbf a \times \mathbf b} + \paren {\mathbf a \times \mathbf c}$
for $\mathbf a, \mathbf b, \mathbf c \in \R^3$.

Proof:
We draw a triangular prism whose parallel edges are in the direction of $\mathbf a$ and with its end faces as triangles with sides $\mathbf b$, $\mathbf c$ and $\mathbf b + \mathbf c$.
:600px
From Magnitude of Vector Cross Product equals Area of Parallelogram Contained by Vectors, the vector areas of these triangular end faces are $\dfrac {\mathbf b \times \mathbf c} 2$ and $\dfrac {\mathbf c \times \mathbf b} 2$.
The remaining vector areas are $\mathbf b \times \mathbf a$, $\mathbf c \times \mathbf a$ and $\mathbf a \times \paren {\mathbf b + \mathbf c}$.
From Total Vector Area of Polyhedron is Zero:
:$\paren {\mathbf b \times \mathbf a} + \paren {\mathbf c \times \mathbf a} + \paren {\mathbf a \times \paren {\mathbf b + \mathbf c} } + \dfrac {\mathbf b \times \mathbf c} 2 + \dfrac {\mathbf c \times \mathbf b} 2 = 0$
from which we get:
:$\paren {-\mathbf b \times \mathbf a} + \paren {-\mathbf c \times \mathbf a} = \paren {\mathbf a \times \paren {\mathbf b + \mathbf c} }$
The result follows from Vector Cross Product is Anticommutative:
:$\mathbf a \times \mathbf b + \mathbf a \times \mathbf c = \mathbf a \times \paren {\mathbf b + \mathbf c}$
{{qed}}
Theorem: Vector Cross Product Distributes over Addition

The vector cross product is distributive over addition.
That is, in general:
:$\mathbf a \times \paren {\mathbf b + \mathbf c} = \paren {\mathbf a \times \mathbf b} + \paren {\mathbf a \times \mathbf c}$
for $\mathbf a, \mathbf b, \mathbf c \in \R^3$.

Proof:
Let $\mathbf b'$ and $\mathbf c'$ be the projections of $\mathbf b$ and $\mathbf c$ onto the plane perpendicular to $\mathbf a$.
Then $\mathbf b' + \mathbf c'$ is the projection of $\mathbf b + \mathbf c$ onto that plane.
We have:
{{begin-eqn}}
{{eqn | l = \mathbf a \times \mathbf b'
      | r = \mathbf a \times \mathbf b
      | c = 
}}
{{eqn | l = \mathbf a \times \mathbf c'
      | r = \mathbf a \times \mathbf c
      | c = 
}}
{{eqn | l = \mathbf a \times \paren {\mathbf b' + \mathbf c'}
      | r = \mathbf a \times \paren {\mathbf b + \mathbf c}
      | c = 
}}
{{end-eqn}}
Because:
:$\mathbf a$ is perpendicular to $\mathbf b'$ and $\mathbf c'$
:$\mathbf a \times \mathbf b'$ lies in the plane perpendicular to $\mathbf a$
it follows that:
:$\mathbf a \times \mathbf b'$ is $\norm {\mathbf a}$ times the length of $\mathbf b'$ and is perpendicular to $\mathbf b'$.
Similarly:
:$\mathbf a \times \mathbf c'$ is $\norm {\mathbf a}$ times the length of $\mathbf c'$ and is perpendicular to $\mathbf c'$.
Hence:
:$\mathbf a \times \mathbf b' + \mathbf a \times \mathbf c'$ also lies in the plane perpendicular to $\mathbf a$.
:$\mathbf a \times \mathbf b' + \mathbf a \times \mathbf c'$ is $\norm {\mathbf a}$ times the length of $\mathbf b' + \mathbf c'$ and is perpendicular to $\mathbf b' + \mathbf c'$.
In other words:
:$\mathbf a \times \mathbf b + \mathbf a \times \mathbf c = \mathbf a \times \paren {\mathbf b + \mathbf c}$
{{qed}}
Theorem: Jacobi's Equation is Variational Equation of Euler's Equation

The Variational equation of Euler's equation is Jacobi's equation.

Proof:
Let Euler's equation be
:$\map {F_y} {x, \hat y, \hat y'} - \dfrac \d {\d x} \map {F_{y'} } {x, \hat y, \hat y'} = 0$
which is derived from:
:$\ds \int_a^b \paren {\map {F_y} {x, \hat y, \hat y'} - \frac \d {\d x} \map {F_{y'} } {x, \hat y, \hat y'} } \rd x = 0$
Let $\map {\hat y} x = \map y x$ and $\map {\hat y} x = \map y x + \map h x$ be solutions of Euler's equation.
By Taylor's Theorem:
{{begin-eqn}}
{{eqn | r = \map {F_y} {x, y + h, y' + h'} - \frac \d {\d x} \map {F_{y'} } {x, y + h, y' + h'}
      | o = 
}}
{{eqn | r = F_y + F_{yy} h + F_{yy'} h' + \map \OO {h^2, h'^2} - \map {\frac \d {\d x} } {F_{y'} + F_{y'y} h + F_{y'y'} h' + \map \OO {h^2, h'^2} }
}}
{{eqn | r = \paren {F_y - \frac \d {\d x} F_{y'} } + F_{yy} h + \map \OO {h^2, h h', h'^2} - F_{y'y}' h - F_{y'y'}' h' - F_{y'y'} h'' - \frac \d {\d x} \map \OO {h^2, h h', h'^2}
}}
{{eqn | r = \paren {F_{yy} - \frac \d {\d x} F_{y'y} } h - \map {\frac \d {\d x} } {F_{y'y'} h'} + \map \OO {h^2, h h', h'^2}
}}
{{end-eqn}}
where the omitted ordered tuple of variables is $\tuple {x, y, y'}$, and $\map {\hat y} x = \map y x$ has been used as a solution to $F_y - \dfrac \d {\d x} F_{y'} = 0$.
Therefore, Euler's equation is to be derived from
:$\ds \int_a^b \paren {\paren {F_{yy} - \frac \d {\d x} F_{y'y} } h - \map {\frac \d {\d x} } {F_{y'y'} h'} + \map \OO {h^2, h h', h'^2} } \rd x = 0$
By integration by parts, 
:$\ds \int_a^b \map \OO {h^2, h h', h'^2} \rd x = \int_a^b \map \OO {h^2} \rd x$
Thus, the equivalent differential equation is:
:$\paren {F_{yy} - \dfrac \d {\d x} F_{y'y} } h - \map {\dfrac \d {\d x} } {F_{y'y'} h'} + \map \OO {h^2} = 0$
Omission of $\map \OO {h^2}$ and multiplication of equation by $\frac 1 2$ yields Jacobi's equation.
{{qed}}
Theorem: Palindromes in Base 10 and Base 2

The following $n \in \Z$ are palindromic in both decimal and binary:
:$0, 1, 3, 5, 7, 9, 33, 99, 313, 585, 717, 7447, 9009, 15 \, 351, 32 \, 223, 39 \, 993, \ldots$
{{OEIS|A007632}}

Proof:
:{| border="1"
|-
! align="center" style = "padding: 2px 10px" | $n_{10}$ 
! align="center" style = "padding: 2px 10px" | $n_2$ 
|-
| align="right" style = "padding: 2px 10px" | $0$
| align="right" style = "padding: 2px 10px" | $0$
|-
| align="right" style = "padding: 2px 10px" | $1$
| align="right" style = "padding: 2px 10px" | $1$
|-
| align="right" style = "padding: 2px 10px" | $3$
| align="right" style = "padding: 2px 10px" | $11$
|-
| align="right" style = "padding: 2px 10px" | $5$
| align="right" style = "padding: 2px 10px" | $101$
|-
| align="right" style = "padding: 2px 10px" | $7$
| align="right" style = "padding: 2px 10px" | $111$
|-
| align="right" style = "padding: 2px 10px" | $9$
| align="right" style = "padding: 2px 10px" | $1001$
|-
| align="right" style = "padding: 2px 10px" | $33$
| align="right" style = "padding: 2px 10px" | $100 \, 001$
|-
| align="right" style = "padding: 2px 10px" | $99$
| align="right" style = "padding: 2px 10px" | $1 \, 100 \, 011$
|-
| align="right" style = "padding: 2px 10px" | $313$
| align="right" style = "padding: 2px 10px" | $100 \, 111 \, 001$
|-
| align="right" style = "padding: 2px 10px" | $585$
| align="right" style = "padding: 2px 10px" | $1 \, 001 \, 001 \, 001$
|-
| align="right" style = "padding: 2px 10px" | $717$
| align="right" style = "padding: 2px 10px" | $1 \, 011 \, 001 \, 101$
|-
| align="right" style = "padding: 2px 10px" | $7447$
| align="right" style = "padding: 2px 10px" | $1 \, 110 \, 100 \, 010 \, 111$
|-
| align="right" style = "padding: 2px 10px" | $15 \, 351$
| align="right" style = "padding: 2px 10px" | $11 \, 101 \, 111 \, 110 \, 111$
|-
| align="right" style = "padding: 2px 10px" | $32 \, 223$
| align="right" style = "padding: 2px 10px" | $111 \, 110 \, 111 \, 011 \, 111$
|-
| align="right" style = "padding: 2px 10px" | $39 \, 993$
| align="right" style = "padding: 2px 10px" | $1 \, 001 \, 110 \, 000 \, 111 \, 001$
|}
{{qed}}
Theorem: Bottom in Compact Closure

Let $L = \struct {S, \preceq}$ be a bounded below ordered set.
Let $x \in S$.
Then:
:$\bot \in x^{\mathrm{compact} }$
where:
:$\bot$ denotes the smallest element in $L$
:$ x^{\mathrm{compact} }$ denotes the compact closure of $x$.

Proof:
By Bottom is Compact:
:$\bot$ is a compact element.
By definition of the smallest element:
:$\bot \preceq x$
Thus by definition of compact closure:
:$\bot \in x^{\mathrm{compact} }$
{{qed}}
Category:Way Below Relation
2z4a1pcn8t5hdg1ggxblyw896p9vkg1
Theorem: Triplets of Products of Two Distinct Primes

The following triplets of consecutive positive integers are the smallest in which each number is the product of $2$ distinct prime numbers:
:$33, 34, 35$
:$85, 86, 87$
:$93, 94, 95$
:$141, 142, 143$
:$201, 202, 203$
:$213, 214, 215$
:$217, 218, 219$

Proof:
Taking each triplet in turn:
{{begin-eqn}}
{{eqn | l = 33
      | r = 3 \times 11
}}
{{eqn | l = 34
      | r = 2 \times 17
}}
{{eqn | l = 35
      | r = 5 \times 7
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 85
      | r = 5 \times 17
}}
{{eqn | l = 86
      | r = 2 \times 43
}}
{{eqn | l = 87
      | r = 3 \times 29
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 93
      | r = 3 \times 31
}}
{{eqn | l = 94
      | r = 2 \times 47
}}
{{eqn | l = 95
      | r = 5 \times 19
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 141
      | r = 3 \times 47
}}
{{eqn | l = 142
      | r = 2 \times 71
}}
{{eqn | l = 143
      | r = 11 \times 13
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 201
      | r = 3 \times 67
}}
{{eqn | l = 202
      | r = 2 \times 101
}}
{{eqn | l = 203
      | r = 7 \times 29
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 213
      | r = 3 \times 71
}}
{{eqn | l = 214
      | r = 2 \times 107
}}
{{eqn | l = 215
      | r = 5 \times 43
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 217
      | r = 7 \times 31
}}
{{eqn | l = 218
      | r = 2 \times 109
}}
{{eqn | l = 219
      | r = 3 \times 73
}}
{{end-eqn}}
It is noted that the triplet:
:$121, 122, 123$
while consisting of semiprimes, not all of these are the product of $2$ distinct prime numbers, as $121 = 11^2$.
{{qed}}
Theorem: Product of Two Distinct Primes has 4 Positive Divisors

Let $n \in \Z_{>0}$ be a positive integer which is the product of $2$ distinct primes.
Then $n$ has exactly $4$ positive divisors.

Proof:
Let $n = p \times q$ where $p$ and $q$ are primes.
We have by definition of divisor:
{{begin-eqn}}
{{eqn | l = 1
      | o = \divides
      | r = n
      | c = One Divides all Integers
}}
{{eqn | l = p
      | o = \divides
      | r = n
      | c = {{Defof|Divisor of Integer}}
}}
{{eqn | l = q
      | o = \divides
      | r = n
      | c = {{Defof|Divisor of Integer}}
}}
{{eqn | l = p \times q
      | o = \divides
      | r = n
      | c = Integer Divides Itself
}}
{{end-eqn}}
where $\divides$ denotes divisibility.
Suppose $a \divides n$ such that $1 \le a < n$.
Suppose $a \ne p$.
By definition of prime number:
:$a \perp p$
where $\perp$ denotes coprimality.
From Euclid's Lemma:
:$a \divides q$
and so by definition of prime number:
:$a = q$
or:
:$a = 1$
Similarly, suppose $a \ne q$.
By the same reasoning:
:$a = p$
or:
:$a = 1$
Thus the only positive divisors are as above.
{{qed}}
{{questionable|The above is of course rubbish.}}
Theorem: Product of Two Distinct Primes is Multiplicatively Perfect

Let $n \in \Z_{>0}$ be a positive integer which is the product of $2$ distinct primes.
Then $n$ is multiplicatively perfect.

Proof:
Let $n = p \times q$ where $p$ and $q$ are primes.
From Product of Two Distinct Primes has 4 Positive Divisors, the positive divisors of $n$ are:
:$1, p, q, pq$
Thus the product of all the divisors of $n$ is:
:$1 \times p \times q \times p q = p^2 q^2 = n^2$
Hence the result, by definition of multiplicatively perfect.
{{qed}}
Theorem: Cube of Prime has 4 Positive Divisors

Let $n \in \Z_{>0}$ be a positive integer which is the cube of a prime number.
Then $n$ has exactly $4$ positive divisors.

Proof:
Let $n = p^3$ where $p$ is prime.
The positive divisors of $n$ are:
:$1, p, p^2, p^3$
This result follows from Divisors of Power of Prime.
{{qed}}
Theorem: Cube of Prime is Multiplicatively Perfect

Let $n \in \Z_{>0}$ be a positive integer which is the cube of a prime number.
Then $n$ is multiplicatively perfect.

Proof:
Let $n = p^3$ where $p$ is prime.
From Cube of Prime has 4 Positive Divisors, the positive divisors of $n$ are:
:$1, p, p^2, p^3$
Thus the product of all the divisors of $n$ is:
:$1 \times p \times p^2 \times p^3 = p^6 = n^2$
Hence the result, by definition of multiplicatively perfect.
{{qed}}
Theorem: Compact Closure is Directed

Let $L = \left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.
Let $x \in S$.
Then $x^{\mathrm{compact} }$ is directed
where $x^{\mathrm{compact} }$ denotes the compact closure of $x$.

Proof:
By Bottom in Compact Closure:
:$\bot \in x^{\mathrm{compact} }$
where $\bot$ denotes the smallest element in $L$.
Thus by definition:
:$x^{\mathrm{compact} }$ is non-empty.
Let $y, z \in x^{\mathrm{compact} }$
By definition of compact closure:
:$y$ and $z$ are compact elements and $y \preceq x$ and $z \preceq x$
By definitions of supremum and upper bound:
:$y \vee z \preceq x$
By definition of compact element:
:$y \ll y$ and $z \ll z$
where $\ll$ denotes the way below relation.
By Join Succeeds Operands:
:$y \preceq y \vee z$ and $z \preceq y \vee z$
By Preceding and Way Below implies Way Below and definition of reflexivity:
:$y \ll y \vee z$ and $z \ll y \vee z$
By Join is Way Below if Operands are Way Below:
:$y \vee z \ll y \vee z$
By definition:
:$y \vee z$ is compact element.
By definition of compact closure:
:$y \vee z \in x^{\mathrm{compact} }$
Thus
:$\exists v \in x^{\mathrm{compact} }: y \preceq v \land z \preceq v$
{{qed}}
Theorem: Sum of 2 Lucky Numbers in 4 Ways

The number $34$ is the smallest positive integer to be the sum of $2$ lucky numbers in $4$ different ways.

Proof:
The sequence of lucky numbers begins:
:$1, 3, 7, 9, 13, 15, 21, 25, 31, 33, \ldots$
Thus we have:
{{begin-eqn}}
{{eqn | l = 34
      | r = 1 + 33
      | c = 
}}
{{eqn | r = 3 + 31
      | c = 
}}
{{eqn | r = 9 + 25
      | c = 
}}
{{eqn | r = 13 + 21
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: 35 Hexominoes

There exist $35$ distinct free hexominoes:
:600px

Proof:
{{ProofWanted|Work to be done yet to establish method of creation}}
Theorem: Fixed Point of Permutation is Fixed Point of Power

Let $S_n$ denote the symmetric group on $n$ letters.
Let $\sigma \in S_n$.
Let $i \in \Fix \sigma$, where $\Fix \sigma$ denotes the set of fixed elements of $\sigma$.
Then for all $m \in \Z$:
:$i \in \Fix {\sigma^m}$

Proof:
It follows from Integers form Ordered Integral Domain that for any integer $m$ either:
:$m = 0$
or
:$m > 0$
or:
:$m < 0$
Theorem: Real Symmetric Matrix is Hermitian

Every real symmetric matrix is Hermitian.

Proof:
Let $\mathbf A$ be a real symmetric matrix.
Then we have:
{{begin-eqn}}
{{eqn | l = \sqbrk {\mathbf A}^\dagger_{i j}
      | r = \overline {\sqbrk {\mathbf A}_{ji} }
      | c = {{Defof|Hermitian Conjugate}}
}}
{{eqn | r = \sqbrk {\mathbf A}_{ji}
      | c = Complex Number equals Conjugate iff Wholly Real: $\mathbf A$ is Real
}}
{{eqn | r = \sqbrk {\mathbf A}_{ij}
      | c = $\mathbf A$ is Symmetric
}}
{{end-eqn}}
So:
:$\mathbf A = \mathbf A^\dagger$
Thus, by definition, $\mathbf A$ is Hermitian.
{{qed}}
Theorem: Hexominoes cannot form Rectangle

While there are a total of $210$ squares in a complete set of hexominoes, it is impossible to build them into a rectangle of side lengths $a$ and $b$ where $a \times b = 210$.

Proof:
{{ProofWanted|I'll have to check this, but I think the proof is along the lines that if you coloured them in a checkerboard pattern, you can't get an equal number of black and white squares.}}
Theorem: Number of Heptominoes

There exist $108$ distinct free heptominoes, one of which has a hole:
:600px

Proof:
{{ProofWanted|Work to be done yet to establish method of creation}}
Theorem: 369 Octominoes

There exist $369$ distinct free octominoes, $6$ of which have a hole:
:600px

Proof:
{{ProofWanted|Work to be done yet to establish method of creation}}
Theorem: 1285 9-ominoes

There exist $1285$ distinct free $9$-ominoes, $37$ of which have a hole:
:600px

Proof:
{{ProofWanted|Work to be done yet to establish method of creation}}
Theorem: Mapping Assigning to Element Its Compact Closure is Order Isomorphism

Let $L = \struct {S, \vee, \preceq}$ be a bounded below algebraic join semilattice.
Let $C = \struct {\map K L, \preceq'}$ be an ordered subset of $L$
where $\map K L$ denotes the compact subset of $L$.
Let $I = \struct {\map {\mathit {Ids} } C, \precsim}$ be an inclusion ordered set
where $\map {\mathit {Ids} } C$ denotes the set of all ideals in $C$.
Let $f: S \to \map {\mathit {Ids} } C$ be a mapping such that
:$\forall x \in S: \map f x = x^{\mathrm {compact} }$
where $x^{\mathrm {compact} }$ denotes the compact closure of $x$.
Thus $f$ is order isomorphism between $L$ and $I$.

Proof:
We will prove that
:$f$ is an order embedding.
Let $x, y \in S$.
'''Sufficient condition'''
Assume that
:$x \preceq y$
By Compact Closure is Increasing:
:$x^{\mathrm {compact} } \subseteq y^{\mathrm {compact} }$
By definition of $f$:
:$\map f x \subseteq \map f y$
Thus by definition of inclusion ordered set:
:$\map f x \precsim \map f y$
{{qed|lemma}}
'''Necessary condition'''
Assume that:
:$\map f x \precsim \map f y$
By definition of inclusion ordered set:
:$\map f x \subseteq \map f y$
By definition of $f$:
:$x^{\mathrm {compact} } \subseteq y^{\mathrm{compact} }$
By Supremum of Subset:
:$\map \sup {x^{\mathrm {compact} } } \preceq \map \sup {y^{\mathrm {compact} } }$
By definition of algebraic:
:$L$ satisfies the axiom of $K$-approximation.
Thus by the axiom of $K$-approximation:
:$x \preceq y$
{{qed|lemma}}
We will prove that
:$f$ is a surjection.
Let $y \in \map {\mathit {Ids} } C$
Define $x = \sup_L y$
Thus $x \in S$.
We will prove that:
:$x^{\mathrm {compact} } \subseteq y$
Let $d \in x^{\mathrm {compact} }$
By definition of compact closure:
:$d$ is a compact element and $d \preceq x$
By definition of compact subset:
:$d \in \map K L$
By definition of compact element:
:$d \ll d$
where $\ll$ denotes the way below relation.
By definition of way below relation:
:$\exists z \in y: d \preceq z$
By definition of ordered subset:
:$d \preceq' z$
Thus by definition of lower section:
:$d \in y$
{{qed|lemma}}
We will prove that
:$y \subseteq x^{\mathrm {compact} }$
Let $d \in y$.
By definition of subset:
:$d \in \map K L$
By definition of compact subset:
:$d$ is a compact element.
By definition of supremum:
:$x$ is upper bound for $y$.
By definition of upper bound:
:$d \preceq x$
Thus by definition of compact closure:
:$d \in x^{\mathrm {compact} }$
{{qed|lemma}}
By definition of set equality:
:$y = x^{\mathrm {compact} }$
Hence:
:$y = \map f x$
{{qed|lemma}}
Hence $f$ is order isomorphism.
{{qed}}
Theorem: Maximum Length of Non-Crossing Knight's Tour

The maximum length of a non-crossing knight's tour on a standard chessboard is $35$ moves.

Proof:
{{ProofWanted|Lots of background material needed first.}}
Theorem: Prime Factors of 35, 36, 4734 and 4735

The integers:
:$35, 4374$
have the same prime factors between them as the integers:
:$36, 4375$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = 35
      | r = 5 \times 7
      | c = 
}}
{{eqn | l = 4374
      | r = 2 \times 3^7
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 36
      | r = 2^2 \times 3^2
      | c = 
}}
{{eqn | l = 4375
      | r = 5^4 \times 7
      | c = 
}}
{{end-eqn}}
Thus both pairs of integers can be seen to have the same prime factors:
:$2, 3, 5, 7$
{{qed}}
Theorem: Equivalence Class of Fixed Element

Let $S_n$ denote the symmetric group on $n$ letters.
Let $\sigma \in S_n$.
Let $\RR_\sigma$ denote the equivalence defined in Permutation Induces Equivalence Relation.
Let $i \in \N^*_{\le n}$.
Then:
:$i \in \Fix \sigma$ {{iff}} $\eqclass i {\RR_\sigma} = \set i$
where:
:$\eqclass i {\RR_\sigma}$ denotes the equivalence class of $i$ under $\RR_\sigma$
:$\Fix \sigma$ denotes the set of fixed elements of $\sigma$.

Proof:
By the definition of an equivalence relation it is easily seen that $\set i \subseteq \eqclass i {\RR_\sigma}$.
Suppose that $i \in \Fix \sigma$.
Let $j \in \eqclass i {\RR_\sigma}$.
Then by Condition for Membership of Equivalence Class and Permutation Induces Equivalence Relation:
:$j \in \eqclass i {\RR_\sigma} \iff i \mathrel {\RR_\sigma} j \implies \exists m \in \Z: \sigma^m \paren i = j$
And by Fixed Point of Permutation is Fixed Point of Power:
:$\sigma^m \paren i = i \implies i = j$
Therefore:
:$\eqclass i {\RR_\sigma} \subseteq \set i$
and so:
:$\eqclass i {\RR_\sigma} = \set i$
{{qed|lemma}}
For the converse, suppose that:
:$\eqclass i {\RR_\sigma} = \set i$.
It is seen that:
:$i \mathrel {\RR_\sigma} \map \sigma i \iff \map \sigma i \in \eqclass i {\RR_\sigma}$
Therefore:
:$\map \sigma i = i \implies i \in \Fix \sigma$
Hence the result.
{{qed}}
Category:Symmetric Groups
Category:Equivalence Classes
pnk9h7r4rvcg60113ff3gaafm77d7y6
Theorem: Number of Odd Entries in Row of Pascal's Triangle is Power of 2

The number of odd entries in a row of Pascal's triangle is a power of $2$.

Proof:
Let $n, k \in \Z$.
Let the representations of $n$ and $k$ to the base $p$ be given by:
:$n = 2^r a_r + \cdots + 2 a_1 + a_0$
:$k = 2^r b_r + \cdots + 2 b_1 + b_0$
By {{Corollary|Lucas' Theorem}}:
:$\ds \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod 2$
By definition, $a_j$ and $b_j$ are either $0$ or $1$.
We have:
:$\dbinom 0 1 = 0$
:$\dbinom 0 0 = \dbinom 1 0 = \dbinom 1 1 = 1$
Thus $\dbinom n k$ is odd {{iff}} none of $\dbinom {a_j} {b_j} = 0$.
We are to determine how many values of $k$ there are such that $\dbinom n k$ is odd.
We construct a $k$ such that $\dbinom n k$ is odd.
For each $a_j = 1$, we can have $b_j = 0$ or $1$.
For each $a_j = 0$, we must have $b_j = 0$.
Hence for each $a_j$, there are $2^{a_j}$ ways for the corresponding $b_j$ to be such that $\dbinom {a_j} {b_j} \ne 0$.
By Product Rule for Counting, the number of choices for $a_0, \dots, a_r$ for which $k$ is such that $\dbinom n k$ is odd is given by: 
:$\ds \prod_{j \mathop = 0}^r 2^{a_j} = 2^{\sum_{j = 0}^r {a_j} }$
which is a power of $2$.
Thus the number of odd entries in a row of Pascal's triangle is a power of $2$.
{{qed}}
Theorem: Compact Closure is Increasing

Let $\left({S, \preceq}\right)$ be an ordered set.
Let $x, y \in S$ such that
:$x \preceq y$
Then $x^{\mathrm{compact} } \subseteq y^{\mathrm{compact} }$
where $x^{\mathrm{compact} }$ denotes the compact closure of $x$.

Proof:
Let $z \in x^{\mathrm{compact} }$
By definition of compact closure:
:$z$ is a compact element and $z \preceq x$
By definition of transitivity:
:$z \preceq y$
Thus by definition of compact closure:
:$z \in y^{\mathrm{compact} }$
{{qed}}
Theorem: Sum of Entries in Row of Pascal's Triangle

The sum of all the entries in the $n$th row of Pascal's triangle is equal to $2^n$.

Proof:
By definition, the entries in $n$th row of Pascal's triangle are exactly the binomial coefficients:
:$\dbinom n 0, \dbinom n 1, \ldots, \dbinom n n$
The result follows from Sum of Binomial Coefficients over Lower Index.
{{qed}}