avewright/proofwiki-math huggingface dataset:
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:
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:The sum of all the entries in the $n$th row of Pascal's triangle is equal to $2^n$.
=== Basis for the Induction ===
$\map P 0$ is the case:
:The sum of all the entries in the row $0$ of Pascal's triangle is equal to $2^0 = 1$.
This is true, as the only non-zero entry in row $0$ is $\dbinom 0 0$ which equals $1$.
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:The sum of all the entries in the row $k$ of Pascal's triangle is equal to $2^k$.
from which it is to be shown that:
:The sum of all the entries in the row $k + 1$ of Pascal's triangle is equal to $2^{k + 1}$.
=== Induction Step ===
This is the induction step:
In row $k + 1$ there are $k + 2$ entries:
:$\dbinom {k + 1} 0, \dbinom {k + 1} 1, \dbinom {k + 1} 2, \ldots, \dbinom {k + 1} k, \dbinom {k + 1} {k + 1}$
From Pascal's Rule:
:$\dbinom {k + 1} j = \dbinom k {j - 1} + \dbinom k j$
Thus for each of the $k + 1$ entries in row $k$:
:$\dbinom k 0, \dbinom k 1, \ldots, \dbinom k k$
Element $\dbinom k j$ contributes to exactly $2$ elements of row $k + 1$
:$\dbinom {k + 1} j$ and $\dbinom {k + 1} {j + 1}$
So the sum of all the elements of row $k + 1$ is equal to $2$ times the sum of all the elements of row $k$.
By the induction hypothesis, that means the sum of all the elements of row $k + 1$ is equal to $2 \times 2^k$.
That is, the sum of all the entries in the row $k + 1$ of Pascal's triangle is equal to $2^{k + 1}$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:For all $n \in \Z_{\ge 0}$, the sum of all the entries in the $n$th row of Pascal's triangle is equal to $2^n$.
{{qed}}
Theorem: Columns of Pascal's Triangle contain Simplicial Polytopic Numbers

The columns of Pascal's triangle contain the simplicial polytopic numbers:
: Column $0$: repeated instances of number $1$
: Column $1$: the (strictly) positive integers
: Column $2$: the triangular numbers
: Column $3$: the tetrahedral numbers
: Column $4$: the pentatope numbers
and so on.

Proof:
{{ProofWanted|Need to define the simplicial polytopic numbers and demonstrate that the $n$th simplicial polytopic number of dimension $m$ is $\dbinom m n$ or whatever the formula is.}}
Theorem: Equivalence Class of Fixed Element/Corollary

:$i \notin \Fix \sigma$ {{iff}} $\eqclass i {\RR_\sigma}$ contains more than one element

Proof:
From Equivalence Class of Fixed Element and Biconditional Equivalent to Biconditional of Negations:
:$i \notin \Fix \sigma \iff \set i \ne \eqclass i {\RR_\sigma}$
Because the Biconditional is Transitive, it suffices to show that:
:$\set i \ne \eqclass i {\RR_\sigma}$
{{iff}}:
:$\eqclass i {\RR_\sigma}$ contains more than one element.
Suppose that $\set i \ne \eqclass i {\RR_\sigma}$.
From the definition of an equivalence relation it is seen that:
:$\set i \subseteq \eqclass i {\RR_\sigma}$
And from the hypothesis:
:$\set i \subset \eqclass i {\RR_\sigma}$
Therefore $\eqclass i {\RR_\sigma}$ contains more than one element.
{{qed|lemma}}
Conversely, if $\eqclass i {\RR_\sigma}$ contains more than one element, then it is seen that:
:$\set i \ne \eqclass i {\RR_\sigma}$
{{qed}}
Category:Symmetric Groups
Category:Equivalence Classes
p80trof9qxw8s3pzafsqody2pmfv8er
Theorem: Sum of Entries in Lesser Diagonal of Pascal's Triangle equal Fibonacci Number

The sum of the entries in the $n$th lesser diagonal of Pascal's triangle equals the $n + 1$th Fibonacci number.

Proof:
By definition, the entries in the $n$th lesser diagonal of Pascal's triangle are:
:$\dbinom n 0, \dbinom {n - 1} 1, \dbinom {n - 2} 2, \dbinom {n - 3} 3, \ldots$
and so the statement can be written:
:$F_{n + 1} = \ds \sum_{k \mathop \ge 0} \dbinom {n - k} k$
The proof proceeds by strong induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$F_{n + 1} = \ds \sum_{k \mathop \ge 0} \dbinom {n - k} k$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \dbinom {0 - k} k
      | r = \dbinom 0 0
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | r = F_1
      | c = 
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
Theorem: Extension of Infima Preserving Mapping to Complete Lattice Preserves Infima

Let $L_1 = \struct {S_1, \preceq_1}$, $L_2 = \struct {S_2, \preceq_2}$ be ordered sets.
Let $L_3 = \struct {S_3, \preceq_3}$ be a complete lattice.
Suppose that.
:$L_2$ is an infima inheriting ordered subset of $L_3$.
Let $f: S_1 \to S_2$ be a mapping such that:
:$f$ preserves infima.
Then
$f: S_1 \to S_3$ preserves infima.

Proof:
By definition of ordered subset:
:$S_2 \subseteq S_3$
Then define $g = f:S_1 \to S_3$
Let $X$ be a subset of $S_1$ such that
:$X$ admits a infimum in $L_1$.
Thus by definition of complete lattice:
:$g \sqbrk X$ admits a infimum in $L_3$.
By definition of mapping preserves infima:
:$f \sqbrk X$ admits a infimum in $L_2$ and $\map {\inf_{L_2} } {f \sqbrk X} = \map f {\map {\inf_{L_1} } X}$
Thus by definition of infima inheriting:
:$\map {\inf_{L_3} } {g \sqbrk X} = \map g {\inf_{L_1} } X$
{{qed}}
Theorem: Extension of Directed Suprema Preserving Mapping to Complete Lattice Preserves Directed Suprema

Let $L_1 = \struct {S_1, \preceq_1}$ and $L_2 = \struct {S_2, \preceq_2}$ be ordered sets.
Let $L_3 = \struct {S_3, \preceq_3}$ be a complete lattice.
Suppose that
:$L_2$ is directed suprema inheriting ordered subset of $L_3$.
Let $f:S_1 \to S_2$ be a mapping such that
:$f$ preserves directed suprema.
Then
$f:S_1 \to S_3$ preserves directed suprema.

Proof:
By definition of ordered subset:
:$S_2 \subseteq S_3$
Then define $g = f: S_1 \to S_3$
Let $X$ be a directed subset of $S_1$ such that
:$X$ admits a supremum in $L_1$.
Thus by definition of complete lattice:
:$g \sqbrk X$ admits a supremum in $L_3$.
By definition of mapping preserves directed suprema:
:$f \sqbrk X$ admits a supremum in $L_2$ and $\map {\sup_{L_2} } {f \sqbrk X} = \map f {\sup_{L_1} X}$
By Directed Suprema Preserving Mapping is Increasing:
:$f$ is an increasing mapping.
By Image of Directed Subset under Increasing Mapping is Directed:
:$f \sqbrk X$ is directed.
Thus by definition of directed suprema inheriting:
:$\map {\inf_{L_3} } {g \sqbrk X} = \map g {\inf_{L_1} X}$
{{qed}}
Theorem: Rows in Pascal's Triangle containing Numbers in Arithmetic Sequence

There are an infinite number of rows of Pascal's triangle which contain $3$ integers in arithmetic sequence.

Proof:
Suppose $\dbinom n k$, $\dbinom n {k + 1}$ and $\dbinom n {k + 2}$ are in an arithmetic sequence.
Then:
{{begin-eqn}}
{{eqn | l = \dbinom n {k + 2} - \dbinom n {k + 1}
      | r = \dbinom n {k + 1} - \dbinom n k
      | c = {{Defof|Arithmetic Sequence}}
}}
{{eqn | l = \frac {n!} {\paren {n - k - 2}! \paren {k + 2}!} - \frac {n!} {\paren {n - k - 1}! \paren {k + 1}!}
      | r = \frac {n!} {\paren {n - k - 1}! \paren {k + 1}!} - \frac {n!} {\paren {n - k}! \paren k!}
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | l = \paren {n - k - 1} \paren {n - k} - \paren {n - k} \paren {k + 2}
      | r = \paren {n - k} \paren {k + 2} - \paren {k + 1} \paren {k + 2}
      | c = Multiply both sides by $\dfrac {\paren {n - k}! \paren {k + 2}!} {n!}$
}}
{{eqn | l = n^2 - k n - k n + k^2 - n + k - n k - 2 n + k^2 + 2 k
      | r = n k + 2 n - k^2 - 2 k - k^2 - 2 k - k - 2
      | c = 
}}
{{eqn | l = n^2 - \paren {4 k + 5} n + \paren {4 k^2 + 8 k + 2}
      | r = 0
      | c = 
}}
{{eqn | l = n
      | r = \frac {4 k + 5 \pm \sqrt {\paren {4 k + 5}^2 - 4 \paren {4 k^2 + 8 k + 2} } } 2
      | c = Quadratic Formula
}}
{{eqn | r = \frac {4 k + 5 \pm \sqrt {16 k^2 + 40 k + 25 - 16 k^2 - 32 k - 8} } 2
      | c = 
}}
{{eqn | r = \frac {4 k + 5 \pm \sqrt {8 k + 17} } 2
      | c = 
}}
{{end-eqn}}
Since $n$ is rational, we require $8 k + 17$ to be a square.
Since $8 k + 17$ is odd, if $8 k + 17$ is square, then $\sqrt {8 k + 17}$ is odd.
Write $\sqrt {8 k + 17} = 2 x + 1$.
Notice that:
{{begin-eqn}}
{{eqn | l = \paren {2 x + 1}^2
      | r = 4 x^2 + 4 x + 1
}}
{{eqn | r = 8 \paren {\frac {x^2 + x - 4} 2} + 17
}}
{{end-eqn}}
Using the substitution $k = \dfrac {x^2 + x - 4} 2$:
{{begin-eqn}}
{{eqn | l = n
      | r = \frac {2 x^2 + 2 x - 8 + 5 \pm \paren {2 x + 1} } 2
}}
{{eqn | r = x^2 - 2 \text { or } x^2 + 2 x - 1
}}
{{eqn | r = x^2 - 2 \text { or } \paren {x + 1}^2 - 2
}}
{{end-eqn}}
Each $x$ with $k = \dfrac {x^2 + x - 4} 2 > 0$ give a value for $n$.
Therefore there are an infinite number of rows of Pascal's triangle which contain $3$ integers in arithmetic sequence.
{{qed}}
Theorem: Rows in Pascal's Triangle containing Numbers in Geometric Sequence

There exist no rows of Pascal's triangle which contain $3$ integers in geometric sequence.

Proof:
Suppose $\dbinom n k$, $\dbinom n {k + 1}$ and $\dbinom n {k + 2}$ are in a geometric sequence.
Then:
{{begin-eqn}}
{{eqn | l = \dbinom n {k + 2} / \dbinom n {k + 1}
      | r = \dbinom n {k + 1} / \dbinom n k
      | c = {{Defof|Geometric Sequence}}
}}
{{eqn | l = \paren {\frac {n!} {\paren {n - k - 2}! \paren {k + 2}!} } \paren {\frac {\paren {n - k - 1}! \paren {k + 1}!} {n!} }
      | r = \paren {\frac {n!} {\paren {n - k - 1}! \paren {k + 1}!} } \paren {\frac {\paren {n - k}! \paren k!} {n!} }
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | l = \frac {n - k - 1} {k + 2}
      | r = \frac {n - k} {k + 1}
      | c = 
}}
{{eqn | l = \paren {n - k - 1} \paren {k + 1}
      | r = \paren {n - k} \paren {k + 2}
      | c = 
}}
{{eqn | l = n k - k^2 - k + n -k - 1
      | r = n k + 2 n - k^2 - 2 k
      | c = 
}}
{{eqn | l = n
      | r = -1
      | c = 
}}
{{end-eqn}}
Since $n \ge 0$, no row of Pascal's triangle contains $3$ integers in geometric sequence.
However, suppose one extends the definition of binomial coefficients to allow $n < 0$.
Then by Negated Upper Index of Binomial Coefficient, we have: 
:$\dbinom {-1} k = \paren {-1}^k$
which indeed forms a geometric sequence.
{{qed}}
Theorem: Rows in Pascal's Triangle containing Numbers in Harmonic Sequence

There exist no rows of Pascal's triangle which contain $3$ integers in harmonic sequence.

Proof:
Suppose $\dbinom n k$, $\dbinom n {k + 1}$ and $\dbinom n {k + 2}$ are in a harmonic sequence.
Then:
{{begin-eqn}}
{{eqn | l = \dbinom n {k + 2}^{-1} - \dbinom n {k + 1}^{-1}
      | r = \dbinom n {k + 1}^{-1} - \dbinom n k^{-1}
      | c = {{Defof|Harmonic Sequence}}
}}
{{eqn | l = \frac {\paren {n - k - 2}! \paren {k + 2}!} {n!} - \frac {\paren {n - k - 1}! \paren {k + 1}!} {n!}
      | r = \frac {\paren {n - k - 1}! \paren {k + 1}!} {n!} - \frac {\paren {n - k}! \paren k!} {n!}
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | l = \paren {k + 2} \paren {k + 1} - \paren {n - k - 1} \paren {k + 1}
      | r = \paren {n - k - 1} \paren {k + 1} - \paren {n - k} \paren {n - k - 1}
      | c = Multiply both sides by $\dfrac {n!} {\paren {n - k - 2}! \paren k!}$
}}
{{eqn | l = k^2 + 3 k + 2 - n k + k^2 + k - n + k + 1
      | r = n k - k^2 - k + n - k - 1 - n^2 + n k + n + k n - k^2 - k
      | c = 
}}
{{eqn | l = n^2 - \paren {4 k + 3} n + \paren {4 k^2 + 8 k + 4}
      | r = 0
      | c = 
}}
{{end-eqn}}
This is a quadratic equation in $n$, so we can calculate its discriminant.
Notice that for each $k \ge 0$:
:$\paren {4 k + 3}^2 - 4 \paren {4 k^2 + 8 k + 4} = - \paren {8 k + 7} < 0$
By Solution to Quadratic Equation with Real Coefficients, there is no real solution for $n$.
Therefore there is no row of Pascal's triangle which contain $3$ integers in harmonic sequence.
{{qed}}
Theorem: Reciprocal as Summation of Binomial Coefficients of Reciprocals

:$\forall n \in \Z_{>0}: \dfrac 1 n = \ds \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \dbinom {n - 1} k \dfrac 1 {k + 1}$
where $\dbinom {n - 1} k$ denotes a binomial coefficient.
That is, for example:
{{begin-eqn}}
{{eqn | l = \dfrac 1 1
      | r = 1
}}
{{eqn | l = \dfrac 1 2
      | r = 1 - \dfrac 1 2
}}
{{eqn | l = \dfrac 1 3
      | r = 1 - 2 \times \dfrac 1 2 + \dfrac 1 3
}}
{{eqn | l = \dfrac 1 4
      | r = 1 - 3 \times \dfrac 1 2 + 3 \times \dfrac 1 3 - \dfrac 1 4
}}
{{eqn | l = \dfrac 1 5
      | r = 1 - 4 \times \dfrac 1 2 + 6 \times \dfrac 1 3 - 4 \times \dfrac 1 4 + \dfrac 1 5
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \dbinom {n - 1} k \dfrac 1 {k + 1}
      | r = \frac 1 n \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \binom n {k + 1}
      | c = Factors of Binomial Coefficient
}}
{{eqn | r = \frac 1 n \sum_{k \mathop = 1}^n \paren {-1}^{k + 1} \binom n k
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = \frac 1 n \paren {\sum_{k \mathop = 0}^n \paren {-1}^{k + 1} \binom n k + 1}
      | c = $\paren {-1}^1 \dbinom n 0 = -1$
}}
{{eqn | r = \frac 1 n \paren {-\paren {1 - 1}^n + 1}
      | c = Binomial Theorem
}}
{{eqn | r = \frac 1 n
}}
{{end-eqn}}
{{qed}}
Theorem: Element of Leibniz Harmonic Triangle is Sum of Numbers Below

The elements in the Leibniz harmonic triangle are the sum of the elements immediately below them.
{{refactor|Rework this as another definition of LHT, establishing that column and diagonal $0$ are defined as the reciprocals.|level = medium}}

Proof:
By definition of Leibniz harmonic triangle, element $\tuple {n, m}$ is:
:$\dfrac 1 {\paren {n + 1} \binom n m}$
Thus we have:
{{begin-eqn}}
{{eqn | r = \dfrac 1 {\paren {n + 2} \binom {n + 1} m} + \dfrac 1 {\paren {n + 2} \binom {n + 1} {m + 1} }
      | o = 
      | c = Elements of Leibniz Harmonic Triangle immediately below
}}
{{eqn | r = \dfrac {\binom {n + 1} {m + 1} + \binom {n + 1} m} {\paren {n + 2} \binom {n + 1} m \binom {n + 1} {m + 1} }
      | c = 
}}
{{eqn | r = \dfrac {\binom {n + 2} {m + 1} } {\paren {n + 2} \binom {n + 1} m \binom {n + 1} {m + 1} }
      | c = Pascal's Rule
}}
{{eqn | r = \dfrac {\frac {\paren {n + 2}!} {\paren {m + 1}! \paren {n - m + 1}!} } {\paren {n + 2} \frac {\paren {n + 1}!} {m! \paren {n - m + 1}!} \frac {\paren {n + 1}!} {\paren {m + 1}! \paren {n - m}!} }
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \dfrac {\paren {n + 1}!} {\frac {\paren {n + 1}!} {m!} \frac {\paren {n + 1}!} {\paren {n - m}!} }
      | c = some simplification
}}
{{eqn | r = \dfrac 1 {\frac {\paren {n + 1}!} {m! \paren {n - m}!} }
      | c = further simplification
}}
{{eqn | r = \dfrac 1 {\paren {n + 1} \frac {n!} {m! \paren {n - m}!} }
      | c = further simplification
}}
{{eqn | r = \dfrac 1 {\paren {n + 1} \binom n m}
      | c = {{Defof|Binomial Coefficient}}
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Complement of Irreducible Topological Subset is Prime Element

Let $T = \struct {S, \tau}$ be a topological space.
Let $X$ be an irreducible subset of $S$ such that:
:$\relcomp S X \in \tau$
Let $L = \struct {\tau, \preceq}$ be an inclusion ordered set of topology $\tau$.
Then $\relcomp S X$ is prime element in $L$.

Proof:
{{tidy|This proof lends itself to be rendered neatly by means of the {{TL|eqn}} template}}
Let $Y, Z \in \tau$ such that
:$Y \wedge Z \preceq \relcomp S X$
By definition of topological space:
:$Y \cap Z \in \tau$
By Meet in Inclusion Ordered Set:
:$Y \cap Z = Y \wedge Z$
By definition of inclusion ordered set:
:$Y \cap Z \subseteq \relcomp S X$
By Relative Complement inverts Subsets and Relative Complement of Relative Complement:
:$X \subseteq \relcomp S {Y \cap Z}$
By De Morgan's Laws: Relative Complement of Intersection:
:$X \subseteq \relcomp S Y \cup \relcomp S Z$
By Intersection with Subset is Subset:
:$X = \paren {\relcomp S Y \cup \relcomp S Z} \cap X$
By Intersection Distributes over Union:
:$X = \paren {\relcomp S Y \cap X} \cup \paren {\relcomp S Z \cap X}$
By definition of closed set and Relative Complement of Relative Complement:
:$X$, $\relcomp S Y$, and $\relcomp S Z$ are closed sets.
By Intersection of Closed Sets is Closed in Topological Space:
:$\relcomp S Y \cap X$, $\relcomp S Z \cap X$ are closed sets.
By definition of irreducible:
:$\relcomp S Y \cap X = X$ or $\relcomp S Z \cap X = X$
By Intersection with Subset is Subset:
:$X \subseteq \relcomp S Y$ or $X \subseteq \relcomp S Z$
By Relative Complement inverts Subsets and Relative Complement of Relative Complement:
:$Y \subseteq \relcomp S X$ or $Z \subseteq \relcomp S X$
Thus by definition of inclusion ordered set:
:$Y \preceq \relcomp S X$ or $Z \preceq \relcomp S X$
{{qed}}
Theorem: Element of Leibniz Harmonic Triangle as Sum of Elements on Diagonal from Below

Consider the Leibniz harmonic triangle:
{{:Definition:Leibniz Harmonic Triangle}}
Let $\tuple {n, m}$ be the element in the $n$th row and $m$th column.
Then:
:$\tuple {n, m} = \ds \sum_{k \mathop \ge 0} \tuple {n + 1 + k, m + k}$

Proof:
Taking $r \to \infty$ in Lemma 2:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \tuple {n + 1 + k, m + k}
      | r = \sum_{k \mathop \ge 1} \tuple {n + k, m + k - 1}
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = \lim_{r \to \infty} \paren {\tuple {n, m} - \tuple {n + r, m + r} }
      | c = Lemma 2
}}
{{eqn | r = \tuple {n, m} - \lim_{r \to \infty} \frac 1 {\paren {n + r + 1} \binom {n + r}{m + r} }
}}
{{eqn | r = \tuple {n, m}
}}
{{end-eqn}}
{{qed}}
Theorem: Element of Pascal's Triangle is Sum of Diagonal or Column starting above it going Upwards

Consider Pascal's triangle:
:{{Definition:Pascal's Triangle}}
Let $\tuple {n, m}$ be the element in the $n$th row and $m$th column.
Then:
:$\tuple {n, m} = \ds \sum_{k \mathop \ge 0} \tuple {n - k - 1, m - 1}$
and:
:$\tuple {n, m} = \ds \sum_{k \mathop \ge 0} \tuple {n - k - 1, m - k - 1}$

Proof:
We have $\tuple {n, m} = \dbinom n m$, and we have:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom {n - k - 1} {m - 1}
      | r = \sum_{k \mathop = 0}^{n - m} \binom {m - 1 + k} {m - 1}
}}
{{eqn | r = \binom {m - 1 + n - m + 1} {m - 1 + 1}
      | c = Rising Sum of Binomial Coefficients
}}
{{eqn | r = \binom n m
}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom {n - k - 1} {m - k - 1}
      | r = \sum_{k \mathop \ge 0} \binom {n - k - 1} {n - k - 1 - \paren {m - k - 1} }
      | c = Symmetry Rule for Binomial Coefficients
}}
{{eqn | r = \sum_{k \mathop \ge 0} \binom {n - k - 1} {n - m}
}}
{{eqn | r = \sum_{k \mathop = 0}^{m - 1} \binom {n - m + k} {n - m}
}}
{{eqn | r = \binom {n - m + m - 1 + 1} {m - 1 + 1}
      | c = Rising Sum of Binomial Coefficients
}}
{{eqn | r = \binom n m
}}
{{end-eqn}}
{{qed}}
Theorem: Power of Moved Element is Moved

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

Proof:
{{AimForCont}} that there exists some $i \notin \Fix \sigma$ and some $m \in \Z$ such that $\map {\sigma^m} i \in \Fix \sigma$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {\sigma \circ \sigma^m} } i
      | r = \map {\sigma^{m + 1} } i
}}
{{eqn | r = \map {\sigma^m} i
      | c = {{Defof|Fixed Element under Permutation}}
}}
{{eqn | ll= \leadsto
      | l = \map {\sigma^{-m} \sigma^{m + 1} } i
      | r = \map {\sigma^{-m} \sigma^m} i
}}
{{eqn | ll= \leadsto
      | l = \map \sigma i
      | r = \map e i
}}
{{eqn | r = i
}}
{{end-eqn}}
But it was previously established that $i$ was moved by $\sigma$.
This is a contradiction.
Therefore $\map {\sigma^m} i$ is moved by $\sigma$.
{{qed}}
Category:Symmetric Groups
9yl6gvwr11b2vjqgzfuniq91gml0lys
Theorem: Triangular Numbers in Geometric Sequence

The numbers:
:$1, 6, 36$
are the smallest triangular numbers in geometric sequence.

Proof:
{{begin-eqn}}
{{eqn | l = 6 \div 1
      | r = 6
      | c = 
}}
{{eqn | l = 36 \div 6
      | r = 6
      | c = 
}}
{{end-eqn}}
Hence the common ratio is $6$.
{{qed}}
Theorem: Square Numbers which are Divisor Sum values

The sequence of square numbers which are the divisor sum value of a (strictly) positive integer begins:
:$1, 4, 36, 121, 144, 256, 324, 400, 576, 784, 900, 961, \ldots$
{{OEIS|A038688}}

Proof:
{{begin-eqn}}
{{eqn | l = 1
      | r = \map {\sigma_1} 1
      | c = {{DSFLink|1}}
}}
{{eqn | l = 4
      | r = \map {\sigma_1} 3
      | c = {{DSFLink|3}}
}}
{{eqn | l = 36
      | r = \map {\sigma_1} {22}
      | c = {{DSFLink|22}}
}}
{{eqn | l = 121
      | r = \map {\sigma_1} {81}
      | c = {{DSFLink|81}}
}}
{{eqn | l = 144
      | r = \map {\sigma_1} {66}
      | c = {{DSFLink|66}}
}}
{{eqn | r = \map {\sigma_1} {70}
      | c = {{DSFLink|70}}
}}
{{eqn | r = \map {\sigma_1} {94}
      | c = {{DSFLink|94}}
}}
{{eqn | r = \map {\sigma_1} {115}
      | c = {{DSFLink|115}}
}}
{{eqn | r = \map {\sigma_1} {119}
      | c = {{DSFLink|119}}
}}
{{eqn | l = 256
      | r = \map {\sigma_1} {217}
      | c = {{DSFLink|217}}
}}
{{eqn | l = 324
      | r = \map {\sigma_1} {170}
      | c = {{DSFLink|170}}
}}
{{eqn | r = \map {\sigma_1} {214}
      | c = {{DSFLink|214}}
}}
{{eqn | r = \map {\sigma_1} {265}
      | c = {{DSFLink|265}}
}}
{{eqn | l = 400
      | r = \map {\sigma_1} {343}
      | c = {{DSFLink|343}}
}}
{{eqn | l = 576
      | r = \map {\sigma_1} {210}
      | c = {{DSFLink|210}}
}}
{{eqn | r = \map {\sigma_1} {282}
      | c = {{DSFLink|282}}
}}
{{eqn | r = \map {\sigma_1} {310}
      | c = {{DSFLink|310}}
}}
{{eqn | r = \map {\sigma_1} {345}
      | c = {{DSFLink|345}}
}}
{{eqn | r = \map {\sigma_1} {357}
      | c = {{DSFLink|357}}
}}
{{eqn | r = \map {\sigma_1} {382}
      | c = {{DSFLink|382}}
}}
{{eqn | r = \map {\sigma_1} {385}
      | c = {{DSFLink|385}}
}}
{{eqn | l = 784
      | r = \map {\sigma_1} {364}
      | c = {{DSFLink|364}}
}}
{{eqn | l = 900
      | r = \map {\sigma_1} {472}
      | c = {{DSFLink|472}}
}}
{{eqn | l = 961
      | r = \map {\sigma_1} {400}
      | c = {{DSFLink|400}}
}}
{{end-eqn}}
{{qed}}
Theorem: Join and Meet in Inclusion Ordered Set of Topology

Let $T = \left({S, \tau}\right)$ be a topological space.
Let $L = \left({\tau, \preceq}\right)$ be an inclusion ordered set of $\tau$.
Let $X, Y \in \tau$.
Then $X \vee Y = X \cup Y$ and $X \wedge Y = X \cap Y$

Proof:
By definition of topological space:
:$X \cup Y, X \cap Y \in \tau$
Thus by Join in Inclusion Ordered Set ans Meet in Inclusion Ordered Set:
:the result holds.
{{qed}}
Theorem: Join in Inclusion Ordered Set

Let $P = \left({X, \subseteq}\right)$ be an inclusion ordered set.
Let $A, B \in X$ such that
:$A \cup B \in X$
Then $A \vee B = A \cup B$

Proof:
By Set is Subset of Union:
:$A \subseteq A \cup B$ and $B \subseteq A \cup B$
By definition:
:$A \cup B$ is upper bound for $\left\{ {A, B}\right\}$
We will prove that
:$\forall C \in X: C$ is upper bound for $\left\{ {A, B}\right\} \implies A \cup B \subseteq C$
Let $C \in X$ such that
:$C$ is upper bound for $\left\{ {A, B}\right\}$.
By definition of upper bound:
:$A \subseteq C$ and $B \subseteq C$
Thus by Union is Smallest Superset:
:$A \cup B \subseteq C$
{{qed|lemma}}
By definition of supremum:
:$\sup \left\{ {A, B}\right\} = A \cup B$
Thus by definition of join:
:$A \vee B = A \cup B$
{{qed}}
Theorem: Characterization of Prime Element in Inclusion Ordered Set of Topology

Let $T = \struct {S, \tau}$ be a topological space.
Let $L = \struct {\tau, \preceq}$  be an inclusion ordered set of $\tau$.
Let $Z \in \tau$.
Then $Z$ is prime element in $L$ {{iff}}:
:$\forall X, Y \in \tau: X \cap Y \subseteq Z \implies X \subseteq Z \lor Y \subseteq Z$

Proof:
=== Sufficient Condition ===
Assume that
:$Z$ is prime element in $L$.
Let $X, T \in \tau$ such that:
:$X \cap Y \subseteq Z$
By Join and Meet in Inclusion Ordered Set of Topology and definition of inclusion ordered set:
:$X \wedge Y \preceq Z$
By definition of prime element:
:$X \preceq Z$ or $Y \preceq Z$
Thus by definition of inclusion ordered set:
:$X \subseteq Z$ or $Y \subseteq Z$
{{qed|lemma}}