avewright/proofwiki-math huggingface dataset:
Theorem: Sum of Unitary Divisors is Multiplicative

Let $\map {\sigma^*} n$ denote the sum of unitary divisors of $n$.
Then the function:
:$\ds \sigma^*: \Z_{>0} \to \Z_{>0}: \map {\sigma^*} n = \sum_{\substack d \mathop \divides n \\ d \mathop \perp \frac n d} d$
is multiplicative.

Proof:
Let $a, b$ be coprime integers.
Because $a$ and $b$ have no common divisor, the divisors of $a b$ are integers of the form $a_i b_j$, where $a_i$ is a divisor of $a$ and $b_j$ is a divisor of $b$.
That is, any divisor $d$ of $a b$ is in the form:
:$d = a_i b_j$
in a unique way, where $a_i \divides a$ and $b_j \divides b$.
First we show that:
:$d$ is an unitary divisor of $a b$ {{iff}} $a_i, b_j$ are unitary divisors of $a, b$ respectively
In the forward implication we are given $d \perp \dfrac {a b} d$.
By Divisor of One of Coprime Numbers is Coprime to Other:
:$a_i, b_j \perp \dfrac {a b} d$
By Divisor of One of Coprime Numbers is Coprime to Other again:
:$a_i \perp \dfrac a {a_i} \land b_j \perp \dfrac b {b_j}$
In the backward implication we are given $a_i \perp \dfrac a {a_i} \land b_j \perp \dfrac b {b_j}$.
By Divisor of One of Coprime Numbers is Coprime to Other:
:$a \perp b \implies \paren {a_i \perp b_j \land \dfrac a {a_i} \perp \dfrac b {b_j} }$
By Product of Coprime Pairs is Coprime:
:$d = a_i b_j \perp \dfrac a {a_i} \dfrac b {b_j} = \dfrac {a b} d$
{{qed|lemma}}
We can list the unitary divisors of $a$ and $b$ as 
:$1, a_1, a_2, \ldots, a$
and:
:$1, b_1, b_2, \ldots, b$
and thus the sum of their unitary divisors are:
:$\ds \map {\sigma^*} a = \sum_{i \mathop = 1}^r a_i$
:$\ds \map {\sigma^*} b = \sum_{j \mathop = 1}^s b_j$
Consider all unitary divisors of $a b$ with the same $a_i$.
Their sum is:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^s a_i b_j
      | r = a_i \sum_{j \mathop = 1}^s b_j
      | c = 
}}
{{eqn | r = a_i \map {\sigma^*} b
      | c = 
}}
{{end-eqn}}
Summing over all $a_i$:
{{begin-eqn}}
{{eqn | l = \map {\sigma^*} {a b}
      | r = \sum_{i \mathop = 1}^r \paren {a_i \map {\sigma^*} b}
      | c = 
}}
{{eqn | r = \paren {\sum_{i \mathop = 1}^r a_i} \map {\sigma^*} b
      | c = 
}}
{{eqn | r = \map {\sigma^*} a \map {\sigma^*} b
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Sum of Unitary Divisors
Category:Multiplicative Functions
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Theorem: Sum of Unitary Divisors of Integer

Let $n$ be an integer such that $n \ge 2$.
Let $\map {\sigma^*} n$ be the sum of all positive unitary divisors of $n$.
Let the prime decomposition of $n$ be:
:$\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} =  p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
Then:
:$\ds \map {\sigma^*} n = \prod_{1 \mathop \le i \mathop \le r} \paren {1 + p_i^{k_i} }$

Proof:
We have that the Sum of Unitary Divisors is Multiplicative.
From Value of Multiplicative Function is Product of Values of Prime Power Factors, we have:
:$\map {\sigma^*} n = \map {\sigma^*} {p_1^{k_1} } \map {\sigma^*} {p_2^{k_2} } \ldots \map {\sigma^*} {p_r^{k_r} }$
From Sum of Unitary Divisors of Power of Prime, we have:
:$\ds \map {\sigma^*} {p_i^{k_i} } = \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
Hence the result.
{{qed}}
Category:Sum of Unitary Divisors
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Theorem: Range of Infinite Sequence may be Finite

Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence.
Then it is possible for the range of $\sequence {x_n}$ to be finite.

Proof:
Consider the infinite sequence $\sequence {x_n}_{n \mathop \in \N}$ defined as:
:$\forall n \in \N: x_n = \dfrac {1 + \paren {-1}^n} 2$
Thus:
:$\sequence {x_n}_{n \mathop \in \N} = 1, 0, 1, 0, \dotsc$
Hence the range of $\sequence {x_n}$ is $\set {0, 1}$, which is finite.
{{qed}}
Theorem: Squares Ending in n Occurrences of m-Digit Pattern

Suppose there exists some integer $x$ such that $x^2$ ends in some $m$-digit pattern ending in an odd number not equal to $5$ and is preceded by another odd number, i.e.:
:$\exists x \in \Z: x^2 \equiv \sqbrk {1 a_1 a_2 \cdots a_m} \pmod {2 \times 10^m}$
where $a_m$ is odd, $a_m \ne 5$ and $m \ge 1$.
Then for any $n \ge 1$, there exists some integer with not more than $m n$-digits such that its square ends in $n$ occurrences of the $m$-digit pattern.

Proof:
We prove that there exists a sequence $\sequence {b_n}$ with the properties:
:$b_n < 10^{m n}$
:$b_n^2 \equiv \underbrace {\sqbrk {1 \paren {a_1 \cdots a_m} \cdots \paren {a_1 \cdots a_m}}}_{n \text { occurrences}} \pmod {2 \times 10^{m n}}$
by induction:
Theorem: Largest Number not Expressible as Sum of Multiples of Coprime Integers

Let $a, b$ be coprime integers, each greater than $1$.
Then the largest number not expressible as a sum of multiples of $a$ and $b$ (possibly zero) is the number:
:$a b - a - b = \paren {a - 1} \paren {b - 1} - 1$

Proof:
First we show that $a b - a - b$ is not expressible as a sum of multiples of $a$ and $b$.
{{AimForCont}} $a b - a - b = s a + t b$ for some $s, t \in \N$.
Note that $t b \le s a + t b < a b - b = \paren {a - 1} b$.
This gives $t < a - 1$.
We also have $\paren {a - t - 1} b = \paren {s + 1} a$.
Hence $a \divides \paren {a - t - 1} b$.
Since $a$ and $b$ are coprime, by Euclid's Lemma:
:$a \divides a - t - 1$
As $a - t - 1 > 0$, by Absolute Value of Integer is not less than Divisors:
:$a \le a - t - 1$
which is a contradiction.
Hence $a b - a - b$ is not expressible as a sum of multiples of $a$ and $b$.
{{qed|lemma}}
Next we need to show that all numbers greater than $a b - a - b$ can be so expressed.
{{WLOG}} assume that $a > b$ and we split the numbers into two cases:
Theorem: Fermat Quotient of 2 wrt p is Square iff p is 3 or 7/Generalization

The Fermat quotient of $2$ with respect to $p$:
:$\map {q_p} 2 = \dfrac {2^{p - 1} - 1} p$
is a perfect power {{iff}} $p = 3$ or $p = 7$.

Proof:
To show that these are the only ones, we observe that since $p$ is an odd prime, write:
:$p = 2 n + 1$ for $n \ge 1$.
Let $\map {q_p} 2$ be a perfect power.
Then $2^{p - 1} - 1 = p x^y$ for some integers $x, y$.
Note that:
:$2^{p - 1} - 1 = 2^{2 n} - 1 = \paren {2^n - 1} \paren {2^n + 1}$
and we have:
:$\gcd \set {2^n - 1, 2^n + 1} = \gcd \set {2^n - 1, 2} = 1$
so $2^n - 1$ and $2^n + 1$ are coprime.
Hence there are $2$ cases:
Theorem: Taxicab Norm is Norm

The taxicab norm is a norm on the real and complex numbers.

Proof:
By P-Norm is Norm, $\norm {\, \cdot \,}_p$ is a norm.
By definition, the taxicab norm is $\norm {\, \cdot \,}_1$.
Therefore, the taxicab norm is a norm.
{{qed}}
Theorem: Multiplication by 2 over 3 in Egyptian Fractions

Let $\dfrac 1 n$ be an Egyptian fraction not equal to $\dfrac 2 3$.
In order to multiply $\dfrac 1 n$ by $\dfrac 2 3$ and have it that $\dfrac 1 n \times \dfrac 2 3$ is also expressed in Egyptian form, we have:
:$\dfrac 1 n \times \dfrac 2 3 = \dfrac 1 {2 n} + \dfrac 1 {6 n}$

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {2 n} + \dfrac 1 {6 n}
      | r = \dfrac 3 {6 n} + \dfrac 1 {6 n}
      | c = 
}}
{{eqn | r = \dfrac {3 + 1} {6 n}
      | c = 
}}
{{eqn | r = \dfrac 2 {3 n}
      | c = 
}}
{{eqn | r = \dfrac 1 n \times \dfrac 2 3
      | c = 
}}
{{end-eqn}}
Note the case where we multiply $\dfrac 2 3$ by $\dfrac 2 3$ itself:
{{begin-eqn}}
{{eqn | l = \dfrac 2 3 \times \dfrac 2 3
      | r = \dfrac 4 9
      | c = 
}}
{{eqn | r = \dfrac 3 9 + \dfrac 1 9
      | c = 
}}
{{eqn | r = \dfrac 1 3 + \dfrac 1 9
      | c = which is in Egyptian form
}}
{{end-eqn}}
{{qed}}
Theorem: Upper Limit of Number of Unit Fractions to express Proper Fraction from Greedy Algorithm

Let $\dfrac p q$ denote a proper fraction expressed in canonical form.
Let $\dfrac p q$ be expressed as the sum of a finite number of distinct unit fractions using Fibonacci's Greedy Algorithm.
Then $\dfrac p q$ is expressed using no more than $p$ unit fractions.

Proof:
Let $\dfrac {x_k} {y_k}$ and $\dfrac {x_{k + 1} } {y_{k + 1} }$ be consecutive stages of the calculation of the unit fractions accordingly:
:$\dfrac {x_k} {y_k} - \dfrac 1 {\ceiling {y_n / x_n} } = \dfrac {x_{k + 1} } {y_{k + 1} }$
By definition of Fibonacci's Greedy Algorithm:
:$\dfrac {x_{k + 1} } {y_{k + 1} } = \dfrac {\paren {-y_k} \bmod {x_k} } {y_k \ceiling {y_k / x_k} }$
It is established during the processing of Fibonacci's Greedy Algorithm that:
:$\paren {-y_k} \bmod {x_k} < x_k$
Hence successive numerators decrease by at least $1$.
Hence there can be no more unit fractions than there are natural numbers between $1$ and $p$.
Hence the result.
{{Qed}}
Theorem: Smallest n for which 3 over n produces 3 Egyptian Fractions using Greedy Algorithm when 2 Sufficient

Consider proper fractions of the form $\dfrac 3 n$ expressed in canonical form.
Let Fibonacci's Greedy Algorithm be used to generate a sequence $S$ of Egyptian fractions for $\dfrac 3 n$.
The smallest $n$ for which $S$ consists of $3$ terms, where $2$ would be sufficient, is $25$.

Proof:
We have that:
{{begin-eqn}}
{{eqn | l = \frac 3 {25}
      | r = \frac 1 9 + \frac 2 {225}
      | c = as $\ceiling {25 / 3} = \ceiling {8.333\ldots} = 9$
}}
{{eqn | r = \frac 1 9 + \frac 1 {113} + \frac 1 {25 \, 425}
      | c = as $\ceiling {225 / 2} = \ceiling {112.5} = 113$
}}
{{end-eqn}}
But then we have:
{{begin-eqn}}
{{eqn | l = \frac 3 {25}
      | r = \frac 6 {50}
      | c = 
}}
{{eqn | r = \frac 5 {50} + \frac 1 {50}
      | c = 
}}
{{eqn | r = \frac 1 {10} + \frac 1 {50}
      | c = 
}}
{{end-eqn}}
By Condition for 3 over n producing 3 Egyptian Fractions using Greedy Algorithm when 2 Sufficient, we are to find the smallest $n$ such that:
:$n \equiv 1 \pmod 6$
:$\exists d: d \divides n$ and $d \equiv 2 \pmod 3$
The first few $n \ge 4$ which satisfies $n \equiv 1 \pmod 6$ are:
:$7, 13, 19, 25$
of which $7, 13, 19$ are primes, so they do not have a divisor of the form $d \equiv 2 \pmod 3$.
We see that $5 \divides 25$ and $5 \equiv 2 \pmod 3$.
Hence the result.
{{qed}}
Theorem: Supremum Norm is Norm/Continuous on Closed Interval Real-Valued Function

Let $I = \closedint a b$ be a closed interval.
Let $\struct {\map C I, +, \, \cdot \,}_\R$ be the vector space of real-valued functions, continuous on $I$.
Let $\map x t \in \map C I$ be a continuous real function.
Let $\size {\, \cdot \,}$ be the absolute value.
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on real-valued functions, continuous on $I$.
Then $\norm {\, \cdot \,}_\infty$ is a norm over $\struct {\map C I, +, \, \cdot \,}_\R$.

Proof:
=== Positive definiteness ===
{{begin-eqn}}
{{eqn | l = \norm x_\infty
      | r = \sup_{t \mathop \in I} \size {\map x t}
      | c = {{Defof|Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions}}
}}
{{eqn | r = \max_{t \mathop \in I} \size {\map x t}
      | c = Weierstrass Extreme Value Theorem
}}
{{eqn | o = \ge
      | r = \size {\map x t}
      | c = {{Defof|Max Operation}}
}}
{{eqn | o = \ge
      | r = 0
      | c = Complex Modulus is Non-Negative
}}
{{end-eqn}}
Suppose $\norm x_\infty = 0$.
Then:
{{begin-eqn}}
{{eqn | l = 0
      | r = \norm x_\infty
}}
{{eqn | r = \sup_{t \mathop \in I} \size {\map x t}
      | c = {{Defof|Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions}}
}}
{{eqn | r = \max_{t \mathop \in I} \size {\map x t}
      | c = Weierstrass Extreme Value Theorem
}}
{{eqn | o = \ge
      | r = \size {\map x t}
      | c = {{Defof|Max Operation}}
}}
{{eqn | o = \ge 
      | r = 0
      | c = Complex Modulus is Non-Negative
}}
{{eqn | ll = \leadsto
      | l = \size {\map x t}
      | r = 0
}}
{{eqn | ll = \leadsto
      | l = \map x t
      | r = 0
      | c = Complex Modulus equals Zero iff Zero
}}
{{end-eqn}}
Therefore:
:$\forall t \in I : \map x t = 0$
{{qed|lemma}}
Theorem: Union with Disjoint Singleton is Dependent if Element Depends on Subset

Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $A \subseteq S$.
Let $x \in S : x \notin A$.
Let $x$ depend on $A$.
Then $A \cup \set x$ is a dependent subset of $M$.

Proof:
We proceed by Proof by Contraposition.
Let $A \cup \set x$ be independent.
By matroid axiom $(\text I 2)$:
:$A$ is independent
We have:
{{begin-eqn}}
{{eqn | l = \map \rho {A \cup \set x} 
      | r = \size {A \cup \set x}
      | c = Rank of Independent Subset Equals Cardinality
}}
{{eqn | r = \size A + \size {\set x}
      | c = {{Corollary|Cardinality of Set Union}}
}}
{{eqn | r = \size A + 1
      | c = Cardinality of Singleton
}}
{{eqn | o = >
      | r = \size A
}}
{{eqn | r = \map \rho A
      | c = Rank of Independent Subset Equals Cardinality
}}
{{end-eqn}}
Then $x$ does not depend on $A$ by definition.
The theorem holds by the Rule of Transposition.
{{qed}}
Category:Matroid Dependence
Category:Matroid Dependent Subsets
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Theorem: Element Depends on Independent Set iff Union with Singleton is Dependent

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $X \in \mathscr I$.
Let $x \in S : x \notin X$.
Then:
:$x \in \map \sigma X$ {{iff}} $X \cup \set x$ is dependent.

Proof:
=== Necessary Condition ===
Let $x \in \map \sigma X$.
By definition of the closure:
:$x$ depends on $X$.
From Union with Disjoint Singleton is Dependent if Element Depends on Subset:
:$X \cup \set x$ is dependent.
{{qed|lemma}}
Theorem: Element Depends on Independent Set iff Union with Singleton is Dependent/Lemma

Let $A \in \mathscr I$ such that $A \subseteq X \cup \set x$.
Then:
:$\size A \le \size X$

Proof:
==== Case 1: $x \in A$ ====
Let $x \in A$.
We have:
{{begin-eqn}}
{{eqn | l = A \setminus \set x
      | o = \subseteq
      | r = \paren {X \cup \set x} \setminus \set x
      | c = Set Difference over Subset
}}
{{eqn | r = \paren {X \setminus \set x} \cup \paren {\set x \setminus \set x}
      | c = Set Difference is Right Distributive over Union
}}
{{eqn | r = X \cup \paren {\set x \setminus \set x}
      | c = Set Difference with Disjoint Set
}}
{{eqn | r = X \cup \O
      | c = Set Difference with Superset is Empty Set
}}
{{eqn | r = X
      | c = Union with Empty Set
}}
{{end-eqn}}
{{AimForCont}}:
:$A \setminus \set x = X$
Then:
{{begin-eqn}}
{{eqn | l = X \cup \set x
      | r = \paren {A \setminus \set x} \cup \set x
      | c = 
}}
{{eqn | r = A
      | c = Set Difference Union Second Set is Union
}}
{{end-eqn}}
So:
:$X \cup \set x$ is independent.
This contradicts:
:$X \cup \set x$ is dependent.
So:
:$A \setminus \set x \subsetneq X$
Then:
{{begin-eqn}}
{{eqn | l = \size X
      | o = >
      | r = \size {A \setminus \set x}
      | c = Cardinality of Proper Subset of Finite Set
}}
{{eqn | r = \size A - \size {\set x}
      | c = Cardinality of Set Difference with Subset
}}
{{eqn | r = \size A - 1
      | c = Cardinality of Singleton
}}
{{end-eqn}}
So:
:$\size A \le \size X$
{{qed|lemma}}
Theorem: Rank of Independent Subset Equals Cardinality

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\rho : \powerset S \to \Z$ be the rank function of $M$.
Let $X \in \mathscr I$
Then:
:$\map \rho X = \size X$

Proof:
{{begin-eqn}}
{{eqn | l = \map \rho X
      | r = \max \set {\size Y : Y \subseteq X \land X \in \mathscr I}
      | c = {{Defof|Rank Function (Matroid)}}
}}
{{eqn | r = \max \set {\size Y : Y \in \powerset X \land X \in \mathscr I}
      | c = {{Defof|Power Set}} of $X$
}}
{{eqn | r = \max \set {\size Y : Y \in \powerset X \cap \mathscr I}
      | c = {{Defof|Intersection}}
}}
{{eqn | r = \max \set {\size Y : Y \in \powerset X}
      | c = Matroid axiom $(\text I 2)$
}}
{{eqn | r = \size X
      | c = Cardinality of Proper Subset of Finite Set
}}
{{end-eqn}}
{{qed}}
Category:Matroid Rank Functions
Category:Matroid Independent Subsets
1vpd8q2jj7018gllwjv40k8jtab9uh3
Theorem: Generating Function for Lucas Numbers

Let $\map G z$ be the function defined as:
:$\map G z = \dfrac {2 - z} {1 - z - z^2}$
Then $\map G z$ is a generating function for the Lucas numbers.

Proof:
Let the form of $\map G z$ be assumed as:
{{begin-eqn}}
{{eqn | l = \map G z
      | r = \sum_{k \mathop \ge 0} L_k z^k
      | c = 
}}
{{eqn | r = L_0 + L_1 z + L_2 z^2 + L_3 z^3 + L_4 z^4 + \cdots
      | c = 
}}
{{eqn | r = 2 + z + 3 z^2 + 4 z^3 + 7 z^4 + \cdots
      | c = 
}}
{{end-eqn}}
where $L_n$ denotes the $n$th Lucas number.
Then:
{{begin-eqn}}
{{eqn | l = z \map G z
      | r = L_0 z + L_1 z^2 + L_2 z^3 + L_3 z^4 + L_4 z^5 + \cdots
      | c = 
}}
{{eqn | l = z^2 \map G z
      | r = L_0 z^2 + L_1 z^3 + L_2 z^4 + L_3 z^5 + L_4 z^6 + \cdots
      | c = 
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = \paren {1 - z - z^2} \map G z
      | r = L_0 + \paren {L_1 - L_0} z + \paren {L_2 - L_1 - L_0} z^2 + \paren {L_3 - L_2 - L_1} z^3 + \cdots
      | c = 
}}
{{eqn | r = L_0 + \paren {L_1 - L_0} z
      | c = {{Defof|Lucas Number}}: $L_n = L_{n - 1} + L_{n - 2}$
}}
{{eqn | r = 2 - z
      | c = {{Defof|Lucas Number}}: $L_0 = 2, L_1 = 1$
}}
{{end-eqn}}
Hence the result:
:$\map G z = \dfrac {2 - z} {1 - z - z^2}$
{{qed}}
Category:Lucas Numbers
Category:Examples of Generating Functions
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Theorem: 492 Cubed is Sum of 3 Positive Cubes in 13 Ways

The cube of $492$ can be expressed as the sum of $3$ positive cubes in $13$ different ways:
{{begin-eqn}}
{{eqn | l = 492^3
      | r = 24^3 + 204^3 + 480^3
}}
{{eqn | r = 48^3 + 85^3 + 491^3
}}
{{eqn | r = 72^3 + 384^3 + 396^3
}}
{{eqn | r = 113^3 + 264^3 + 463^3
}}
{{eqn | r = 114^3 + 360^3 + 414^3
}}
{{eqn | r = 149^3 + 336^3 + 427^3
}}
{{eqn | r = 176^3 + 204^3 + 472^3
}}
{{eqn | r = 190^3 + 279^3 + 449^3
}}
{{eqn | r = 207^3 + 297^3 + 438^3
}}
{{eqn | r = 226^3 + 332^3 + 414^3
}}
{{eqn | r = 243^3 + 358^3 + 389^3
}}
{{eqn | r = 246^3 + 328^3 + 410^3
}}
{{eqn | r = 281^3 + 322^3 + 399^3
}}
{{end-eqn}}

Proof:
Brute force.
Theorem: Largest Number not Expressible as Sum of Multiples of Coprime Integers/Generalization

Let $a, b$ be integers greater than $1$.
Let $d = \gcd \set {a, b}$.
Then the largest multiple of $d$ not expressible as a sum of multiples of $a$ and $b$ (possibly zero) is the number:
:$\dfrac {a b} d - a - b$

Proof:
By Integers Divided by GCD are Coprime:
:$\dfrac a d \perp \dfrac b d$
By Largest Number not Expressible as Sum of Multiples of Coprime Integers, the largest number not expressible as a sum of multiples of $\dfrac a d$ and $\dfrac b d$ is the number:
:$\dfrac {a b} {d^2} - \dfrac a d - \dfrac b d$
Let $k d$ be a multiple of $d$ expressible as a sum of multiples of $a$ and $b$:
:$\exists s, t \in \N: s a + t b = k d$
Then:
:$s \dfrac a d + t \dfrac b d = k$
showing that $k$ is a sum of multiples of $\dfrac a d$ and $\dfrac b d$.
This argument reverses.
Hence the largest multiple of $d$ not expressible as a sum of multiples of $a$ and $b$ is the number:
:$d \paren {\dfrac {a b} {d^2} - \dfrac a d - \dfrac b d} = \dfrac {a b} d - a - b$
{{qed}}
Category:Largest Number not Expressible as Sum of Multiples of Coprime Integers
Category:Integer Combinations
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Theorem: Maximum Area of Isosceles Triangle

Consider two line segments $A$ and $B$ of equal length $a$ which are required to be the legs of an isosceles triangle $T$.
Then the area of $T$ is greatest when the apex of $T$ is a right angle.
The area of $T$ in this situation is equal to $\dfrac {a^2} 2$.

Proof:
:500px
Let $\triangle OAB$ be the isosceles triangle $T$ formed by the legs $OA$ and $OB$.
Thus the apex of $T$ is at $O$.
Let $\theta$ be the angle $\angle AOB$.
We see that by keeping $OA$ fixed, $B$ can range over the semicircle $AOB$.
Thus $\theta$ can range from $0$ to $180 \degrees$, that is, $2$ right angles.
From Area of Triangle in Terms of Two Sides and Angle, the area $\AA$ of $T$ is:
:$\AA = \dfrac 1 2 a^2 \sin \theta$
This is a maximum when $\sin \theta = 1$, that is, when $\theta$ is a right angle.
The result follows.
{{qed}}
Theorem: Construction of Perpendicular using Rusty Compass

Let $AB$ be a line segment.
Using a straightedge and rusty compass, it is possible to construct a straight line at right angles to $AB$ from the endpoint $A$, without extending $AB$ past $A$.

Proof:
As $DE = CD = DA$, the points $A$, $C$ and $E$ all lie on a circle of radius $AC$.
$CE$ is a straight line through the centers of circle $ACE$ and so is a diameter of circle $ACE$.
Hence by Thales' Theorem, $\angle CAE$ is a right angle
{{qed}}
Theorem: Sum to Infinity of 2x^2n over n by 2n Choose n

For $\cmod x < 1$:
:$\ds \frac {2 x \arcsin x} {\sqrt {1 - x^2} } = \sum_{n \mathop = 1}^\infty \frac {\paren {2 x}^{2 n} } {n \dbinom {2 n} n}$

Proof:
By Gregory Series:
:$\ds \arctan t = \sum_{m \mathop = 0}^\infty \frac {\paren {-1}^m t^{2 m + 1} } {2 m + 1}$
Let $t = \dfrac x {\sqrt {1 - x^2} }$.
Let $y = \arcsin x$.
Then:
{{begin-eqn}}
{{eqn | l = t
      | r = \frac {\sin y} {\sqrt {1 - \sin^2 y} }
}}
{{eqn | r = \frac {\sin y} {\cos y}
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \tan y
}}
{{end-eqn}}
Hence $\arctan t = \arcsin x$.
We have:
{{begin-eqn}}
{{eqn | l = \frac {2 x \arcsin x} {\sqrt {1 - x^2} }
      | r = 2 t \arctan t
}}
{{eqn | r = 2 t \sum_{m \mathop = 0}^\infty \frac {\paren {-1}^m t^{2 m + 1} } {2 m + 1}
      | c = Gregory Series
}}
{{eqn | r = 2 t \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} t^{2 m - 1} } {2 m - 1}
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} t^{2 m} } {2 m - 1}
}}
{{eqn | r = 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} x^{2 m} } {\paren {2 m - 1} \paren {1 - x^2}^m}
}}
{{eqn | r = 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} x^{2 m} } {2 m - 1} \sum_{k \mathop = 0}^\infty \dbinom {m + k - 1} {m - 1} x^{2 k}
      | c = Binomial Theorem for Negative Index and Negative Parameter
}}
{{end-eqn}}
It remains to show the the coefficient of $x^{2 n}$ on the {{RHS}} is equal to $\dfrac {2^{2 n} } {n \dbinom {2 n} n}$, that is:
:$\ds 2 \sum_{r \mathop = 1}^n \frac {\paren {-1}^{r - 1} } {2 r - 1} \dbinom {r + n - r - 1} {r - 1} = \frac {2^{2 n} } {n \dbinom {2 n} n}$
The {{LHS}} above is generated by picking, for each $m > 0$, the corresponding $k = n - m$ from the right sum $\ds \sum_{k \mathop = 0}^\infty \dbinom {m + k - 1} {m - 1} x^{2 k}$.
We have:
{{begin-eqn}}
{{eqn | r = 2 n \dbinom {2 n} n \sum_{r \mathop = 1}^n \frac {\paren {-1}^{r - 1} } {2 r - 1} \dbinom {n - 1} {r - 1}
      | o =
}}
{{eqn | r = 2 n \dbinom {2 n} n \sum_{r \mathop = 0}^{n - 1} \frac {\paren {-1}^r} {2 r + 1} \dbinom {n - 1} r
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = 2 n \dbinom {2 n} n \int_0^1 \sum_{r \mathop = 0}^{n - 1} \paren {-1}^r \dbinom {n - 1} r y^{2 r} \d y
}}
{{eqn | r = 2 n \dbinom {2 n} n \int_0^1 \paren {1 - y^2}^{n - 1} \d y
      | c = Binomial Theorem
}}
{{eqn | r = 2 n \dbinom {2 n} n \int_{\frac \pi 2}^0 \sin^{2 n - 2} \theta \, \frac {\d y} {\d \theta} \d \theta
      | c = by substitution of $y = \cos \theta$
}}
{{eqn | r = 2 n \dbinom {2 n} n \int_0^{\frac \pi 2} \sin^{2 n - 1} \theta \, \d \theta
}}
{{eqn | r = 2 n \dbinom {2 n} n \frac {\paren {2^{n - 1} \paren {n - 1}!}^2} {\paren {2 n - 1}!}
      | c = Definite Integral from 0 to Half Pi of Odd Power of Sine x
}}
{{eqn | r = 2 n \paren {\frac {\paren {2 n}!} {n! \, n!} } \paren {\frac {\paren {2^{n - 1} \paren {n - 1}!}^2} {\paren {2 n - 1}!} }
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = 2 n \paren {\frac {2 n} {n^2} } \paren {2^{2 n - 2} }
}}
{{eqn | r = 2^{2 n}
}}
{{end-eqn}}
Hence the result.
{{qed}}