avewright/proofwiki-math huggingface dataset:
Theorem: Two-Person Zero-Sum Game with Multiple Solutions

There exists a two-person zero-sum game with more than one solution.

Proof:
Consider the game defined by the following payoff table:
{{PayoffTable|table = <nowiki>$\begin{array} {r {{|}} c {{|}} c {{|}} c {{|}} }
& B_1 & B_2 & B_3 \\
\hline A_1 & 1 & 2 & 4 \\
\hline A_2 & 4 & 2 & 1 \\
\hline \end{array}$</nowiki>}}
This has two solutions:
:$(1): \quad A: \tuple {1/3, 2/3}, B: \tuple {0, 1, 0}$
:$(2): \quad A: \tuple {2/3, 1/3}, B: \tuple {0, 1, 0}$
{{explain|The reason why the above are solutions is not indicated.}}
It follows that the strategies:
:$\tuple {\dfrac t 3, \dfrac {2 \paren {1 - t} } 3}, B: \tuple {0, 1, 0}$
for all $0 \le t \le 1$ are also solutions.
{{explain|Again, the explanation of this needs to wait till more learning has been done.}}
Theorem: Two-Person Zero-Sum Game with Finite Strategies has Solution

Let $G$ be a two-person zero-sum game.
Let each player of $G$ have a finite set of strategies available.
Then $G$ has at least one solution.

Proof:
{{ProofWanted|Proof given later in book. Chapter III gets technical.}}
Theorem: Card Game with Bluffing is Completely Mixed Game

The game of cards with bluffing is a completely mixed game.

Proof:
From Solution to Card Game with Bluffing, the optimum strategy for the [[Definition:Card Game with Bluffing|card game with bluffing, is:
:For $A$, to use both pure strategies but favour $A_1$ by $2/3$ to $1/3$.
:For $B$, to use both pure strategies but favour $B_1$ by $2/3$ to $1/3$.
Hence the result by definition of completely mixed game.
Theorem: Matching Pennies is Completely Mixed Game

The game of matching pennies is a completely mixed game.

Proof:
The optimum strategy for matching pennies, for both players, is to randomise both their pure strategies.
Hence the result by definition of completely mixed game.
Theorem: Dominated Strategy may be Optimal

A dominated strategy of a game may be the optimal strategy for a player of that game.

Proof:
Consider the game defined by the following payoff table:
{{PayoffTable|table = <nowiki>$\begin{array} {r {{|}} c {{|}} }
& B_1 & B_2 \\
\hline A_1 & 1 & 2 \\
\hline A_2 & 1 & 3  \\
\hline \end{array}$</nowiki>}}
This has two solutions:
:$(1): \quad A: \tuple {1, 0}, B: \tuple {1, 0}$
:$(2): \quad A: \tuple {0, 1}, B: \tuple {1, 0}$
{{explain|The reason why the above are solutions is not indicated.}}
Thus both pure strategies for $A$ are optimal, but $A_1$ is dominated by $A_1$.
{{qed}}
Theorem: Value of Skew-Symmetric Game is Zero

Let $G$ be a two-person zero-sum game.
Let $G$ be represented by a payoff table that is skew-symmetric.
Then the value of $G$ is zero.

Proof:
{{proof wanted|Once we have a definition for skew-symmetric we can start thinking about how to present this.}}
Theorem: Prime Element iff Complement of Lower Closure is Filter

Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded above lattice.
Let $p \in S$ such that
:$p \ne \top$
where $\top$ denotes the top of $L$.
Then:
:$p$ is a prime element
{{iff}}
:$\relcomp S {p^\preceq}$ is filter in $L$
where
:$p^\preceq$ denotes the lower closure of $p$
:$\relcomp S {p^\preceq}$ denotes the relative complement of $p^\preceq$ relative to $S$.

Proof:
=== Sufficient Condition ===
Assume that $p$ is a prime element.
By definition of the greatest element:
:$p \preceq \top$
By definition of $\prec$:
:$p \prec \top$
By definition of antisymmetry:
:$\top \npreceq p$
By definition of lower closure of element:
:$\top \notin p^\preceq$
By definition of relative complement:
:$\top \in \relcomp S {p^\preceq}$
By definition
:$X := \relcomp S {p^\preceq}$ is a non-empty set
We will prove that
:$X$ is filtered.
Let $x, y \in X$.
By definition of relative complement:
:$x \notin p^\preceq$ and $y \notin p^\preceq$
By definition of lower closure of element:
:$x \npreceq p$ and $y \npreceq p$
Then by definition of prime element:
:$x \wedge y \npreceq p$
By definition of lower closure of element:
:$x \wedge y \notin p^\preceq$
Thus by definition of relative complement:
:$x \wedge y \in X$
Thus by Meet Precedes Operands:
:$x \wedge y \preceq x$ and $x \wedge y \preceq y$
Thus
:$\exists z \in X: z \preceq x \land z \preceq y$
{{qed|lemma}}
We will prove that $X$ is an upper section.
Let $x \in X, y \in S$ such that:
:$x \preceq y$
By definition of relative complement:
:$x \notin p^\preceq$
By definition of lower closure of element:
:$x \npreceq p$
By definition of transitivity:
:$y \npreceq p$
By definition of lower closure of element:
:$y \notin p^\preceq$
Thud by definition of relative complement:
:$y \in X$
{{qed|lemma}}
Hence, by definition, $X$ is filter in $L$.
{{qed|lemma}}
Theorem: Oscillation at Point (Infimum) equals Oscillation at Point (Epsilon-Neighborhood)

Let $f: D \to \R$ be a real function where $D \subseteq \R$.
Let $x$ be a point in $D$.
Let $N_x$ be the set of neighborhoods of $x$.
Let $E_x$ be the set of $\epsilon$-neighborhoods of $x$.
Let $\map {\omega_f} x$ be the oscillation of $f$ at $x$ based on $N_x$:
:$\map {\omega_f} x = \inf \set {\map {\omega_f} {I \cap D}: I \in N_x}$
where $\map {\omega_f} {I \cap D}$ is the oscillation of $f$ on $I \cap D$:
:$\map {\omega_f} {I \cap D} = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
Let $\map {\omega^E_f} x$ be the oscillation of $f$ at $x$ based on $E_x$:
:$\map {\omega^E_f} x = \inf \set {\map {\omega_f} {I \cap D}: I \in E_x}$
Then:
:$\map {\omega_f} x \in \R$ {{iff}} $\map {\omega^E_f} x \in \R$
and, if $\map {\omega_f} x$ and $\map {\omega^E_f} x$ exist as real numbers:
:$\map {\omega_f} x = \map {\omega^E_f} x$

Proof:
=== Lemma ===
Let $f: D \to \R$ be a real function where $D \subseteq \R$.
Let $x$ be a point in $D$.
Let $N_x$ be the set of neighborhoods of $x$.
Let $E_x$ be the set of $\epsilon$-neighborhoods of $x$.
Let $\map {\omega_f} x$ be the oscillation of $f$ at $x$ based on $N_x$:
:$\map {\omega_f} x = \inf \set {\map {\omega_f} {I \cap D}: I \in N_x}$
where $\map {\omega_f} {I \cap D}$ is the oscillation of $f$ on $I \cap D$:
:$\map {\omega_f} {I \cap D} = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
Let $\map {\omega^E_f} x$ be the oscillation of $f$ at $x$ based on $E_x$:
:$\map {\omega^E_f} x = \inf \set {\map {\omega_f} {I \cap D}: I \in E_x}$
Let $\map {\omega_f} x \in \R$.
Let $\map {\omega^E_f} x \in \R$.
Then $\map {\omega_f} x = \map {\omega^E_f} x$.
Theorem: Prime Element iff Element Greater is Top

Let $L = \struct {S, \vee, \wedge, \preceq}$ be a Boolean lattice.
Let $p \in S$ such that:
:$p \ne \top$
Then:
:$p$ is prime element
{{iff}}:
:$\forall x \in S: \paren {p \prec x \implies x = \top}$

Proof:
=== Sufficient Condition ===
Suppose that:
:$p$ is prime element
Let $x \in S$ such that:
:$p \prec x$
By definition of Boolean lattice:
:$L$ is complemented distributive lattice.
By definition of complemented:
:$\exists y \in S: y$ is complement of $x$.
By definition of complement:
:$x \wedge y = \bot$
By definition of smallest element:
:$x \wedge y \preceq p$
By definition of prime element:
:$x \preceq p$ or $y \preceq p$
Then by definition of antisymmetry:
:$y \prec x$
By definition of $\prec$:
:$y \preceq x$
By definition of complement:
:$x \vee y = \top$
Thus by Preceding iff Join equals Larger Operand:
:$x = \top$
{{qed|lemma}}
Theorem: Prime Element iff There Exists Way Below Open Filter which Complement has Maximum

Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous distributive lattice.
Let $p \in S$ such that:
:$p \ne \top$
where $\top$ denotes the top of $L$.
Then
:$p$ is a prime element
{{iff}}:
:there exists a way below open filter $F$ in $L$: $p = \map \max {\relcomp S F}$

Proof:
=== Sufficient Condition ===
Let $p$ be a prime element.
By definition of continuous:
:$\forall x \in S: x^\ll$ is directed
We will prove that:
:$\forall x \in S: \paren {x \in \relcomp S {p^\preceq} \implies \exists y \in S: y \in \relcomp S {p^\preceq} \land y \ll x}$
Let $x \in S$ such that:
:$x \in \relcomp S {p^\preceq}$
By definition of relative complement:
:$x \notin p^\preceq$
By definition of lower closure of element:
:$x \npreceq p$
By Axiom of Approximation in Up-Complete Semilattice:
:$\exists y \in S: y \ll x \land y \npreceq p$
By definition of lower closure of element:
:$y \notin p^\preceq$
By definition of relative complement:
:$y \in \relcomp S {p^\preceq}$
Thus:
:$\exists y \in S: y \in \relcomp S {p^\preceq} \land y \ll x$
{{qed|lemma}}
By Prime Element iff Complement of Lower Closure is Filter:
:$F := \relcomp S {p^\preceq}$ is way below open filter in $L$.
By definitions of antisymmetry and lower closure of element:
:$\lnot \exists y \in S: y \in F \land y \prec p$
By Relative Complement of Relative Complement::
:$\relcomp S F = p^\preceq$
By definitions of reflexivity and lower closure of element:
:$p \in \relcomp S F$
Thus by definition of greatest element:
:$p = \map \max {\relcomp S F}$
{{qed|lemma}}
Theorem: Characterization of Prime Ideal

Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $I$ be an proper ideal in $L$.
Then
:$I$ is a prime ideal
{{iff}}:
:$\forall x, y \in S: \paren {x \wedge y \in I \implies x \in I \lor y \in I}$

Proof:
=== Sufficient Condition ===
Assume that
:$I$ is a prime ideal.
Let $x, y \in S$ such that
:$x \wedge y \in I$
By definition of relative complement:
:$x \wedge y \notin \relcomp S I$
By definition of prime ideal:
:$\relcomp S I$ is filter in $L$.
By Filtered in Meet Semilattice:
:$x \notin \relcomp S I$ or $y \notin \relcomp S I$
Thus by definition of relative complement:
:$x \in I$ or $y \in I$
{{qed|lemma}}
Theorem: Matching Pennies is Strictly Competitive

The game of Matching Pennies is strictly competitive.

Proof:
By definition, a strictly competitive game is a game in which the interests of each player are diametrically opposed.
Recall the payoff table of Matching Pennies:
{{:Definition:Matching Pennies/Payoff Table}}
It can be seen by inspection that exchanging $\text A$ and $\text B$ results in exactly the same entries in the payoff table.
From the nature of a payoff table it follows that the fortunes of each player are opposite and equal.
Hence the result.
{{qed}}
Theorem: Intersection Condition for Direct Sum of Subspaces

Let $U$ and $W$ be subspaces of a vector space $V$.
Then $U + W$ is a direct sum {{iff}} $U \cap W = 0$.

Proof:
We must first prove that if $U + W$ is a direct sum, then:
:$U \cap W = 0$
Let $U + W$ be a direct sum.
Let $\mathbf v \in U \cap W$ be an arbitrary vector in $U \cap W$.
Then:
:$\mathbf 0 = \mathbf v + \paren {-\mathbf v}$
where:
:$\mathbf v \in U$
:$-\mathbf v \in W$
{{ProofWanted}}
Theorem: Second Price Auction has Inefficient Equilibria

A second price auction has Nash equilibria which are inefficient.

Proof:
{{ProofWanted|Awaiting a reliable definition of inefficient Nash equilibrium}}
Theorem: Epic Equalizer is Isomorphism

Let $\mathbf C$ be a metacategory.
Let $e: E \to C$ be the equalizer of two morphisms $f, g: C \to D$.
Let $e$ be an epimorphism.
Then $e$ is an isomorphism.

Proof:
As $e$ equalises $f$ and $g$, $f \circ e = g \circ e$.
Since $e$ is epic, it follows that $f = g$.
Then in the equaliser diagram: 
:::<nowiki>$\begin{xy}\xymatrix{
 E 
  \ar[r]^*{e}
&
 A
  \ar[r]<2pt>^*{f}
  \ar[r]<-2pt>_*{g} 
&
 B
\\
 A
  \ar@{.>}[u]^*{k}
  \ar[ur]_*{\operatorname{id}_A}
}\end{xy}</nowiki>$
We have that:
:$f \circ 1_A = g \circ 1_A$
so there is a unique $k$ with $e \circ k = \operatorname{id}_A$.
Then:
{{begin-eqn}}
{{eqn | l = e \circ k \circ e
      | r = {\operatorname {id}_A} \circ e
      | c = 
}}
{{eqn | r = e
      | c = 
}}
{{eqn | r = e \circ \operatorname{id}_E
      | c = 
}}
{{end-eqn}}
By Equalizer is Monomorphism, it follows that $k \circ e = \operatorname{id}_E$.
This gives $k$ as an inverse to $e$.
Thus $e$ is an isomorphism.
{{qed}}
Theorem: Infimum of Subset of Extended Real Numbers is Arbitrarily Close

Let $A \subseteq \overline \R$ be a subset of the extended real numbers.
Let $b$ be an infimum (in $\R$) of $A$.
Let $\epsilon \in \R_{>0}$.
Then:
:$\exists x \in A \cap \R: x - b < \epsilon$

Proof:
We have that:
:$A$ is a a set of extended real numbers
:$A$ is bounded below (in $\R$) as the real number $b$ is a lower bound for $A$.
From this follows by Infimum of Real Subset:
:$\map \inf {A \cap \R} \in \R$ as $\inf A \in \R$
Let $\epsilon \in \R_{>0}$.
Then:
{{begin-eqn}}
{{eqn | l = \map \inf {A \cap \R}
      | o = \in
      | r = \R
}}
{{eqn | ll= \leadsto
      | q = \exists x \in A \cap \R
      | l = x - b
      | o = <
      | r = \epsilon
      | c = Infimum of Subset of Real Numbers is Arbitrarily Close
}}
{{end-eqn}}
{{qed}}
Category:Extended Real Numbers
nwt407r04hv70yx4q3qh3nnb5f7zvtb
Theorem: Generating Fraction for Lucas Numbers

The fraction:
:$\dfrac {199} {9899}$
has a decimal expansion which contains within it the start of the Lucas sequence:
:$0 \cdotp 02010 \, 30407 \, 11 \ldots$

Proof:
By long division:
<pre>
         0.0201030407111829
     ----------------------
9899 ) 199.0000000000000000
       197 98
       --- --
         1 0200
           9899
         - ----
            30100
            29697
            -----
              40300
              39596
              -----
                70400
                69293
                -----
                 11070
                  9899
                 -----
                  11710
                   9899
                  -----
                   18110
                    9899
                   -----
                    82110
                    79192
                    -----
                     29180
                     19798
                     -----
                      93820
                      89091
                      -----
                       4729
</pre>
By Generating Function for Lucas Numbers:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} L_k 10^{-2 k - 2}
      | r = \frac {2 - 10^{-2} } {1 - 10^{-2} - 10^{-4} } \times 10^{-2}
}}
{{eqn | r = \frac {2 \times 10^2 - 1} {10^4 - 10^2 - 1}
}}
{{eqn | r = \frac {199} {9899}
}}
{{end-eqn}}
The first few terms are contained in the decimal expansion, as long as $L_{k + 1} < 100$, where there is no carry.
{{qed}}
Theorem: Transcendental Numbers are Uncountable

The set of transcendental real numbers is uncountable.

Proof:
By definition, a transcendental number (in this context) is a real number which is not an algebraic number.
Recall that the Real Numbers are Uncountable.
Also recall that the Algebraic Numbers are Countable.
The result follows from Uncountable Set less Countable Set is Uncountable.
{{qed}}
Theorem: Almost All Real Numbers are Transcendental

Almost all real numbers are transcendental.

Proof:
By definition, a transcendental number (in this context) is a real number which is not an algebraic number.
Recall that the Real Numbers are Uncountable.
Also recall that the Algebraic Numbers are Countable.
Thus the subset of all real numbers which are not transcendental is countable.
The result follows by the definition of '''almost all'''.
{{qed}}
Theorem: Anomalous Cancellation/Variants/37 + 13 over 37 + 24

:$\dfrac {37 + 13} {37 + 24} = \dfrac {37^3 + 13^3} {37^3 + 24^3}$

Proof:
{{begin-eqn}}
{{eqn | l = 37^3 + 13^3
      | r = \paren {37 + 13}^3 - \paren {3 \times 37^2 \times 13 + 3 \times 37 \times 13^2}
      | c = 
}}
{{eqn | r = \paren {37 + 13}^3 - 3 \times 13 \times 37 \paren {37 + 13}
      | c = 
}}
{{eqn | r = \paren {37 + 13} \paren {\paren {37 + 13}^2 - 3 \times 13 \times 37}
      | c = 
}}
{{eqn | r = \paren {37 + 13} \paren {2500 - 1443}
      | c = 
}}
{{eqn | r = \paren {37 + 13} \times 1057
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 37^3 + 24^3
      | r = \paren {37 + 24}^3 - \paren {3 \times 37^2 \times 24 + 3 \times 37 \times 24^2}
      | c = 
}}
{{eqn | r = \paren {37 + 24}^3 - 3 \times 24 \times 37 \paren {37 + 24}
      | c = 
}}
{{eqn | r = \paren {37 + 24} \paren {\paren {37 + 24}^2 - 3 \times 24 \times 37}
      | c = 
}}
{{eqn | r = \paren {37 + 24} \paren {3721 - 2664}
      | c = 
}}
{{eqn | r = \paren {37 + 24} \times 1057
      | c = 
}}
{{end-eqn}}
Hence:
:$\dfrac {37^3 + 13^3} {37^3 + 24^3} = \dfrac {1057 \paren {37 + 13} } {1057 \paren {37 + 24} } = \dfrac {37 + 13} {37 + 24}$
{{qed}}
Theorem: Anomalous Cancellation/Variants/3 + 25 + 38 over 7 + 20 + 39

:$\dfrac {3 + 25 + 38} {7 + 20 + 39} = \dfrac {3^4 + 25^4 + 38^4} {7^4 + 20^4 + 39^4}$

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac {3^4 + 25^4 + 38^4} {7^4 + 20^4 + 39^4}
      | r = \dfrac {81 + 390625 + 2085136} {2401 + 160000 + 2313441}
      | c = 
}}
{{eqn | r = \dfrac {2475842} {2475842}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \dfrac {3 + 25 + 38} {7 + 20 + 39}
      | r = \dfrac {66} {66}
      | c = 
}}
{{eqn | r = \dfrac {2475842} {2475842}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Anomalous Cancellation/Examples/3544 over 7531

:$\dfrac {344} {731} = \dfrac {3544} {7531} = \dfrac {35544} {75531} = \cdots$

Proof:
{{begin-eqn}}
{{eqn | l = \frac {355 \cdots 544} {755 \cdots 531}
      | r = \paren {3 \times 10^n + \paren {\sum_{i \mathop = 2}^{n - 1} 5 \times 10^i} + 44} \Big / \paren {7 \times 10^n + \paren {\sum_{i \mathop = 2}^{n - 1} 5 \times 10^i} + 31}
      | c = 
}}
{{eqn | r = \paren {3 \times 10^n + \paren {500 \times \sum_{i \mathop = 0}^{n - 3} \times 10^i} + 44} \Big / \paren {7 \times 10^n + \paren {500 \times \sum_{i \mathop = 0}^{n - 3} \times 10^i} + 31}
      | c = 
}}
{{eqn | r = \paren {3 \times 10^n + \paren {500 \times \frac {10^{n - 2} - 1} {10 - 1} } + 44} \Big / \paren {7 \times 10^n + \paren {500 \times \frac {10^{n - 2} - 1} {10 - 1} } + 31}
      | c = Sum of Geometric Sequence
}}
{{eqn | r = \dfrac {3 \times 10^n \times \paren {10 - 1} + 500 \times \paren {10^{n - 2} - 1} + 44 \times \paren {10 - 1} } {7 \times 10^n \times \paren {10 - 1} + 500 \times \paren {10^{n - 2} - 1} + 31 \times \paren {10 - 1} }
      | c = simplifying
}}
{{eqn | r = \dfrac {27 \times 10^n + 500 \times 10^{n - 2} - 104} {63 \times 10^n + 500 \times 10^{n - 2} - 221}
      | c = simplifying
}}
{{eqn | r = \dfrac {2700 \times 10^{n - 2} + 500 \times 10^{n - 2} - 104} {6300 \times 10^{n - 2} + 500 \times 10^{n - 2} - 221}
      | c = simplifying
}}
{{eqn | r = \dfrac {3200 \times 10^{n - 2} - 104} {6800 \times 10^{n - 2} - 221}
      | c = simplifying
}}
{{eqn | r = \dfrac {8 \times \paren {400 \times 10^{n - 2} - 13} } {17 \times \paren {400 \times 10^{n - 2} - 13} }
      | c = factoring
}}
{{eqn | r = \dfrac 8 {17}
      | c = factoring
}}
{{eqn | r = \dfrac {8 \times 43} {17 \times 43}
      | c = 
}}
{{eqn | r = \dfrac {344} {731}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Number of Digits in Power of 2

Let $n$ be a positive integer.
Expressed in conventional decimal notation, the number of digits in the $n$th power of $2$:
:$2^n$
is equal to:
:$\ceiling {n \log_{10} 2}$
where $\ceiling x$ denotes the ceiling of $x$.

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

:$\dfrac 1 3 = \dfrac {1 + 3} {5 + 7} = \dfrac {1 + 3 + 5} {7 + 9 + 11} = \dfrac {1 + 3 + 5 + 7} {9 + 11 + 13 + 15} = \cdots$

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = n + 1}^{2 n} \paren {2 k - 1}
      | r = \sum_{k \mathop = 1}^{2 n} \paren {2 k - 1} - \sum_{k \mathop = 1}^n \paren {2 k - 1}
      | c = 
}}
{{eqn | r = \paren {2 n}^2 - n^2
      | c = Odd Number Theorem
}}
{{eqn | r = 3 n^2
      | c = 
}}
{{eqn | r = 3 \sum_{k \mathop = 1}^n \paren {2 k - 1}
      | c = Odd Number Theorem
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Euler's Number as Sum of Egyptian Fractions

The reciprocal of Euler's number $e$ can be approximated by the following sequence of Egyptian fractions:
:$\dfrac 1 e = \dfrac 1 3 + \dfrac 1 {29} + \dfrac 1 {15 \, 786} + \dfrac 1 {513 \, 429 \, 610} + \cdots$
{{OEIS|A006526}}

Proof:
We have by definition of the reciprocal of $e$ that:
:$\dfrac 1 e \approx 0 \cdotp 36787 \, 94411 \, 71442 \, 32159 \, 55237 \, 70161 \, 46086 \, 74458 \, 11131 \, 031 \ldots$
By inspection:
:$\dfrac 1 3 < \dfrac 1 e < \dfrac 1 2$
Thus:
:$\dfrac 1 e - \dfrac 1 3 \approx 0 \cdotp 03454 \, 61078 \, 38109 \, 08826 \, 21904 \, 36828 \, 12753 \, 41124 \, 77797 \, 70 \ldots$
Then:
:$\dfrac 1 {29} = 0 \cdotp 03448 \, 27586 \, 20689 \, 65517 \, 24137 \, 931 \ldots$ repeating
and:
:$\dfrac 1 {28} = 0 \cdotp 03571 \, 42857 \, 14285 \ldots$ repeating
and so:
:$\dfrac 1 e - \dfrac 1 3 - \dfrac 1 {29} \approx 0 \cdotp 00006 \, 33482 \, 17499 \, 43308 \, 97766 \, 33724 \, 67925 \, 82504 \, 08832 \, 19 \cdots$
Thus one can generate this sequence of denominators $\sequence {D_n}$ by:
:$D_n = \ceiling {\paren {\dfrac 1 e - \ds \sum_{i \mathop = 0}^{n - 1} \frac 1 {D_i} }^{-1} }$
Theorem: One Half as Pandigital Fraction

There are $12$ ways $\dfrac 1 2$ can be expressed as a pandigital fraction:
{{begin-eqn}}
{{eqn | l = \dfrac 1 2
      | r = \dfrac {6729} {13 \, 458}
}}
{{eqn | l = \dfrac 1 2
      | r = \dfrac {6792} {13 \, 584}
}}
{{eqn | l = \dfrac 1 2
      | r = \dfrac {6927} {13 \, 854}
}}
{{eqn | l = \dfrac 1 2
      | r = \dfrac {7269} {14 \, 538}
}}
{{eqn | l = \dfrac 1 2
      | r = \dfrac {7293} {14 \, 586}
}}
{{eqn | l = \dfrac 1 2
      | r = \dfrac {7329} {14 \, 658}
}}
{{eqn | l = \dfrac 1 2
      | r = \dfrac {7692} {15 \, 384}
}}
{{eqn | l = \dfrac 1 2
      | r = \dfrac {7923} {15 \, 846}
}}
{{eqn | l = \dfrac 1 2
      | r = \dfrac {7932} {15 \, 864}
}}
{{eqn | l = \dfrac 1 2
      | r = \dfrac {9267} {18 \, 534}
}}
{{eqn | l = \dfrac 1 2
      | r = \dfrac {9273} {18 \, 546}
}}
{{eqn | l = \dfrac 1 2
      | r = \dfrac {9327} {18 \, 654}
}}
{{end-eqn}}

Proof:
Can be verified by brute force.
Theorem: One Seventh as Pandigital Fraction

There are $7$ ways $\dfrac 1 7$ can be expressed as a pandigital fraction:
:$\dfrac 1 7 = \dfrac {2394} {16 \, 758}$
:$\dfrac 1 7 = \dfrac {2637} {18 \, 459}$
:$\dfrac 1 7 = \dfrac {4527} {31 \, 689}$
:$\dfrac 1 7 = \dfrac {5274} {36 \, 918}$
:$\dfrac 1 7 = \dfrac {5418} {37 \, 926}$
:$\dfrac 1 7 = \dfrac {5976} {41 \, 832}$
:$\dfrac 1 7 = \dfrac {7614} {53 \, 298}$

Proof:
Can be verified by brute force.
Theorem: One Third as Pandigital Fraction

There are $2$ ways $\dfrac 1 3$ can be expressed as a pandigital fraction:
{{begin-eqn}}
{{eqn | l = \dfrac 1 3 
      | r = \dfrac {5823} {17469}
}}
{{eqn | l = \dfrac 1 3 
      | r = \dfrac {5832} {17496}
}}
{{end-eqn}}

Proof:
Can be verified by brute force.
Theorem: One Quarter as Pandigital Fraction

There are $4$ ways $\dfrac 1 4$ can be expressed as a pandigital fraction:
:$\dfrac 1 4 = \dfrac {3942} {15768}$
:$\dfrac 1 4 = \dfrac {4392} {17568}$
:$\dfrac 1 4 = \dfrac {5796} {23184}$
:$\dfrac 1 4 = \dfrac {7956} {31824}$

Proof:
Can be verified by brute force.
Theorem: One Fifth as Pandigital Fraction

There are $12$ ways $\dfrac 1 5$ can be expressed as a pandigital fraction:
:$\dfrac 1 5 = \dfrac {2697} {13485}$
:$\dfrac 1 5 = \dfrac {2769} {13845}$
:$\dfrac 1 5 = \dfrac {2937} {14685}$
:$\dfrac 1 5 = \dfrac {2967} {14835}$
:$\dfrac 1 5 = \dfrac {2973} {14865}$
:$\dfrac 1 5 = \dfrac {3297} {16485}$
:$\dfrac 1 5 = \dfrac {3729} {18645}$
:$\dfrac 1 5 = \dfrac {6297} {31485}$
:$\dfrac 1 5 = \dfrac {7629} {38145}$
:$\dfrac 1 5 = \dfrac {9237} {46185}$
:$\dfrac 1 5 = \dfrac {9627} {48135}$
:$\dfrac 1 5 = \dfrac {9723} {48615}$

Proof:
Can be verified by brute force.