nyuuzyou/wikis huggingface dataset:
Sum of Two Cubes in Complex Domain

== Theorem ==
:$a^3 + b^3 = \paren {a + b} \paren {a \omega + b \omega^2} \paren {a \omega^2 + b \omega}$
where:
: $\omega = -\dfrac 1 2 + \dfrac {\sqrt 3} 2$
== Proof ==
From Sum of Cubes of Three Indeterminates Minus 3 Times their Product:
:$x^3 + y^3 + z^3 - 3 x y z = \paren {x + y + z} \paren {x + \omega y + \omega^2 z} \paren {x + \omega^2 y + \omega z}$
Setting $x \gets 0, y \gets a, z \gets b$:
:$0^3 + a^3 + b^3 - 3 \times 0 \times a b = \paren {0 + a + b} \paren {0 + \omega a + \omega^2 b} \paren {0 + \omega^2 a + \omega b}$
The result follows.
== Sources ==
* : $\S 3$. Roots of Unity: Exercise $5 \ \text {(i)}$
Category:Cube Roots of Unity
Category:Algebra
Sum of Two Fifth Powers

== Theorem ==
:$x^5 + y^5 = \paren {x + y} \paren {x^4 - x^3 y + x^2 y^2 - x y^3 + y^4}$
== Proof ==
From Sum of Two Odd Powers:
:$a^{2 n + 1} + b^{2 n + 1} = \paren {a + b} \paren {a^{2 n} - a^{2 n - 1} b + a^{2 n - 2} b^2 - \dotsb + a b^{2 n - 1} + b^{2 n} }$
We have that $5 = 2 \times 2 + 1$.
The result follows directly by setting $n = 2$.
== Sources ==
* : $\S 2$: Special Products and Factors: $2.16$
* : $\S 2$: Special Products and Factors: $2.16.$
Category:Fifth Powers
Category:Examples of Use of Sum of Two Odd Powers
Sum of Two Odd Powers

== Theorem ==
Let $\F$ be one of the standard number systems, that is $\Z, \Q, \R$ and so on.
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
== Proof ==
== Examples ==
== Also see ==
* Factors of Sum of Two Odd Powers
== Sources ==
* : $\S 2$: Special Products and Factors: $2.21$
* : $\S 2$: Special Products and Factors: $2.21.$
Category:Sum of Two Odd Powers
Category:Sum of Two Powers
Category:Polynomial Theory
Category:Algebra
Sum of Two Odd Powers/Examples/Sum of Two Cubes

== Theorem ==
:$x^3 + y^3 = \paren {x + y} \paren {x^2 - x y + y^2}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 2$: Special Products and Factors: $2.13$
* : $\S 2$: Special Products and Factors: $2.13.$
Category:Sum of Two Cubes
Category:Examples of Use of Sum of Two Odd Powers
Sum of Two Odd Powers/Examples/Sum of Two Cubes/Proof 1

== Theorem ==
== Proof ==
From Difference of Two Powers:
:$\ds a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
Let $x = a$ and $y = -b$.
Then:
Category:Sum of Two Cubes
Sum of Two Odd Powers/Examples/Sum of Two Cubes/Proof 2

== Theorem ==
== Proof ==
From Sum of Two Odd Powers:
:$a^{2 n + 1} + b^{2 n + 1} = \paren {a + b} \paren {a^{2 n} - a^{2 n - 1} b + a^{2 n - 2} b^2 - \dotsb + a b^{2 n - 1} + b^{2 n} }$
We have that $3 = 2 \times 1 + 1$.
Hence setting $n = 1$ gives the result.
Category:Sum of Two Cubes
Sum of Two Positive Elements of C*-Algebra is Positive

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $a, b$ be positive.
Then $a + b$ is positive.
=== Corollary ===
== Proof ==
From $(\text C^\ast 2)$ in the definition of an involution, $a + b$ is Hermitian.
First suppose that $A$ is unital.
From Characterization of Positive Element of Unital C*-Algebra we have:
:$\norm {a - \norm a {\mathbf 1}_A} \le \norm a$
and:
:$\norm {b - \norm b {\mathbf 1}_A} \le \norm b$ 
Hence we have:
From , we have:
:$\norm {a + b} \le \norm a + \norm b$
Hence by Characterization of Positive Element of Unital C*-Algebra, we have that $a + b$ is positive.
Now suppose that $A$ is not unital.
Let $A_+$ be the unitization of $A$. 
Define:
:$\tuple {u, \lambda}^\ast = \tuple {u^\ast, \overline \lambda}$
for each $\tuple {u, \lambda} \in A_+$.
From Existence of Unitization of C*-Algebra, there exists a unique algebra norm $\norm {\, \cdot \,}_\ast$ such that $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ is a unital $\text C^\ast$-algebra.
With this definition $\tuple {a, 0}$, $\tuple {b, 0}$ and $\tuple {a + b, 0}$ are all Hermitian in $A_+$. 
Further:
:$\map {\sigma_{A_+} } {\tuple {a, 0} } = \map {\sigma_A} a \subseteq \hointr 0 \infty$
:$\map {\sigma_{A_+} } {\tuple {b, 0} } = \map {\sigma_A} b \subseteq \hointr 0 \infty$
:$\map {\sigma_{A_+} } {\tuple {a + b, 0} } = \map {\sigma_A} {a + b}$
From the unital case, we have that $\tuple {a + b, 0}$ is positive in $A_+$. 
Hence $\map {\sigma_{A_+} } {\tuple {a + b, 0} } \subseteq \hointr 0 \infty$.
Hence $\map {\sigma_A} {a + b} \subseteq \hointr 0 \infty$. 
Hence $a + b$ is positive.
== Sources ==
* : $2.2$: Positive Elements of $C^\ast$-Algebras
Category:Positive Elements of C*-Algebras
Category:Sum of Two Positive Elements of C*-Algebra is Positive
Sum of Two Positive Elements of C*-Algebra is Positive/Corollary

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $a_1, a_2, \ldots, a_n \in A$ be positive.
Then:
:$\ds \sum_{j \mathop = 1}^n a_j$ is positive.
== Proof ==
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:for all $a_1, a_2, \ldots, a_n \in A$ positive, $\ds \sum_{j \mathop = 1}^n a_j$ is positive.
=== Basis for the Induction ===
The case $n = 1$ is immediate.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:for all $a_1, a_2, \ldots, a_k \in A$ positive, $\ds \sum_{j \mathop = 1}^k a_j$ is positive.
Then we need to show:
:for all $a_1, a_2, \ldots, a_{k + 1} \in A$ positive, $\ds \sum_{j \mathop = 1}^{k + 1} a_j$ is positive.
=== Induction Step ===
This is our induction step:
From the induction hypothesis, we have:
:$\ds \sum_{j \mathop = 1}^k a_j$ is positive.
From Sum of Two Positive Elements of C*-Algebra is Positive, we have:
:$\ds \sum_{j \mathop = 1}^{k + 1} a_j = a_{k + 1} + \sum_{j \mathop = 1}^k a_j$ is positive.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Category:Sum of Two Positive Elements of C*-Algebra is Positive
Sum of Two Rational 4th Powers but not Two Integer 4th Powers

== Theorem ==
$5906$ is the smallest integer which can be expressed as the sum of two rational $4$th powers, but not two integer $4$th powers.
== Proof ==
:$5906 = \paren {\dfrac {149} {17} }^4 + \paren {\dfrac {25} {17} }^4$
Suppose $5906$ is a sum of two integer $4$th powers.
We have:
:$9^4 = 6561 > 5906$
which shows that no $4$th power greater than $8^4$ is in the sum.
:$7^4 + 7^4 = 4802 < 5906$
which shows that some $4$th power greater than $7^4$ is in the sum.
So the sum must contain $8^4$.
We have:
:$5906 - 8^4 = 1810$
but $1810$ is not an integer $4$th power.
Therefore $5906$ is not a sum of two integer $4$th powers.
== Sources ==
* 
* 
* : $5906$
Category:Fourth Powers
Category:5906
Sum of Two Sides of Triangle Greater than Third Side

== Theorem ==
Given a triangle $ABC$, the sum of the lengths of any two sides of the triangle is greater than the length of the third side.
== Proof ==
:
Let $ABC$ be a triangle.
By , we can produce $BA$ past $A$ in a straight line.
By Construction of Equal Straight Lines from Unequal, there exists a point $D$ such that $DA = CA$.
Therefore, from Isosceles Triangle has Two Equal Angles:
:$\angle ADC = \angle ACD$
Thus by :
:$\angle BCD > \angle BDC$
Thus $\triangle DCB$ is a triangle having $\angle BCD$ greater than $\angle BDC$
Hence from Greater Angle of Triangle Subtended by Greater Side:
:$BD > BC$
But:
:$BD = BA + AD$
and:
:$AD = AC$
Thus:
:$BA + AC > BC$
A similar argument shows that $AC + BC > BA$ and $BA + BC > AC$.
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.18$
* 
* 
* 
* 
* 
Category:Triangles
Category:Triangle Inequality
Sum of Two Squares not Congruent to 3 modulo 4

== Theorem ==
Let $n \in \Z$ such that $n = a^2 + b^2$ where $a, b \in \Z$.
Then $n$ is not congruent modulo $4$ to $3$.
== Proof ==
Let $n \equiv 3 \pmod 4$.
 $n$ can be expressed as the sum of two squares:
:$n = a^2 + b^2$.
From Square Modulo 4, either $a^2 \equiv 0$ or $a^2 \equiv 1 \pmod 4$.
Similarly for $b^2$.
So $a^2 + b^2 \not \equiv 3 \pmod 4$ whatever $a$ and $b$ are.
Thus $n$ cannot be the sum of two squares.
The result follows by Proof by Contradiction.
Category:Sums of Squares
Sum of Union of Subsets of Vector Space and Subset

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $\sequence {E_\alpha}_{\alpha \mathop \in I}$ be a $I$-indexed family of subsets of $X$. 
Let $C \subseteq X$. 
Then: 
:$\ds C + \bigcup_{\alpha \mathop \in I} E_\alpha = \bigcup_{\alpha \mathop \in I} \paren {C + E_\alpha}$
== Proof ==
We have: 
:$\ds v \in \bigcup_{\alpha \mathop \in I} \paren {C + E_\alpha}$
 there exists $c \in C$, $\alpha \in I$, $x \in E_\alpha$ such that: 
:$v = c + x$
That is, $v = c + x$ for $c \in C$ and:
:$\ds x \in \bigcup_{\alpha \mathop \in I} E_\alpha$ 
This is equivalent to: 
:$\ds v \in C + \bigcup_{\alpha \mathop \in I} E_\alpha$
so, we get:
:$\ds C + \bigcup_{\alpha \mathop \in I} E_\alpha = \bigcup_{\alpha \mathop \in I} \paren {C + E_\alpha}$
by the definition of set equality.
Category:Vector Spaces
Category:Set Union
Sum of Unitary Divisors is Multiplicative

== Theorem ==
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 $a b$ $a_i, b_j$ 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:
:$d = a_i b_j \perp \dfrac a {a_i} \dfrac b {b_j} = \dfrac {a b} d$
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:
Summing over all $a_i$:
Category:Sum of Unitary Divisors
Category:Multiplicative Functions
Sum of Unitary Divisors of Integer

== Theorem ==
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.
Category:Sum of Unitary Divisors
Sum of Unitary Divisors of Power of Prime

== Theorem ==
Let $n = p^k$ be the power of a prime number $p$.
Then the sum of all positive unitary divisors of $n$ is $1 + n$.
== Proof ==
Let $d \divides n$.
By Divisors of Power of Prime, $d = p^a$ for some positive integer $a \le k$.
We have $\dfrac n d = p^{k - a}$.
Suppose $d$ is a unitary divisor of $n$.
Then $d$ and $\dfrac n d$ are coprime.
If both $a, k - a \ne 0$, $p^a$ and $p^{k - a}$ have a common divisor: $p$.
Hence either $a = 0$ or $k - a = 0$.
This leads to $d = 1$ or $p^k$.
Hence the sum of all positive unitary divisors of $n$ is:
:$1 + p^k = 1 + n$
Category:Prime Numbers
Category:Sum of Unitary Divisors
Sum of Variances of Independent Trials

== Theorem ==
Let $\EE_1, \EE_2, \ldots, \EE_n$ be a sequence of experiments whose outcomes are independent of each other.
Let $X_1, X_2, \ldots, X_n$ be discrete random variables on $\EE_1, \EE_2, \ldots, \EE_n$ respectively.
Let $\var {X_j}$ be the variance of $X_j$ for $j \in \set {1, 2, \ldots, n}$.
Then:
:$\ds \var {\sum_{j \mathop = 1}^n X_j} = \sum_{j \mathop = 1}^n \var {X_j}$
That is, the sum of the variances equals the variance of the sum.
== Proof ==
Category:Variance
Sum of Wholly Real Numbers is Wholly Real

== Theorem ==
Let $x = \tuple {a, 0}$ and $y = \tuple {b, 0}$ be wholly real complex numbers.
Then $x + y$ is also wholly real.
== Proof ==
We have:
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 1$. Complex Numbers: The abbreviated notation
Category:Complex Addition
Sum of nth Fibonacci Number over nth Power of 2

== Theorem ==
:$\ds \sum_{n \mathop = 0}^\infty \frac {F_n} {2^n} = 2$
where $F_n$ is the $n$th Fibonacci number.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
Category:Sum of nth Fibonacci Number over nth Power of 2
Category:Fibonacci Numbers
Sum of nth Fibonacci Number over nth Power of 2/Proof 1

== Theorem ==
== Proof ==
Let us define a sample space which satisfies the Kolmogorov Axioms such that it is the set of all combinations of flipping a fair coin until you receive two heads in a row.
Let $X_n$ be the event of some outcome from flipping $n$ fair coins in a row, then $\map \Pr {X_n} = \dfrac 1 {2^n}$.
In the sample space defined above, we now demonstrate that for a given number of flips $n$, there are exactly $F_{n - 1}$ outcomes contained in the sample space.
=== Illustration ===
:$\begin{array}{c|c|cc}
n & \map f n & \text {Sample Space}: \Omega \\
\hline
1 & 0 & \text {impossible} \\
2 & 1 & HH \\
3 & 1 & THH \\
4 & 2 & (HTHH), (TTHH) \\
5 & 3 & (THTHH), (HTTHH), (TTTHH) \\
6 & 5 & (HTHTHH), (TTHTHH), (THTTHH), (HTTTHH), (TTTTHH) \\
\hline
\cdots & \cdots & \cdots \\
\hline
n & F_{n - 1} & \cdots \\
\hline
\end{array}$
Reviewing the illustration above, for any given value of $n$:
For ALL combinations displayed in row $n$ (that is $\map f n$) , we can place a $T$ in front and that new combination would exist in the sample space for $\paren {n + 1}$.
For example:
:$\paren {HTHH}, \paren {TTHH} \to \paren {THTHH}, \paren {TTTHH}$
However, we also see that for only those combinations starting with a $T$ (that is $\map f {n - 1}$), can we place an $H$ in front and that new combination will also exist in the sample space for $\paren {n + 1}$.
For example:
:$\paren {TTHH} \to \paren {HTTHH}$
Therefore, we have:
The sum of the probabilities of outcomes in a sample space is one by the second Kolmogorov Axiom. 
Hence:
Category:Sum of nth Fibonacci Number over nth Power of 2
Sum of nth Fibonacci Number over nth Power of 2/Proof 4

== Theorem ==
== Proof ==
First, let:
:$S = \ds \sum_{n \mathop = 0}^\infty \frac {F_n} {2^n}$
Some sundry results:
and:
We have:
We have:
Hence we can show that $S$ is absolutely convergent by the Ratio Test:
For $S$ we get $a_n = \dfrac {F_n} {2^n}$ therefore
Therefore the double summation is absolutely convergent and we can exchange the order of the sums.
We note that:
:$0 \le k \le n < \infty$
Therefore by Fubini's Theorem for Infinite Sums:
:$\ds \sum_{n \mathop = 0}^\infty \sum_{k \mathop = 0}^n \frac {F_k} {2^n} = \sum_{k \mathop = 0}^\infty \sum_{n \mathop = k}^\infty \frac{F_k} {2^n}$
and it follows that:
Hence the result.
Category:Sum of nth Fibonacci Number over nth Power of 2
Sum of r+k Choose k up to n

== Theorem ==
Let $r \in \R$ be a real number.
Then:
:$\ds \forall n \in \Z: n \ge 0: \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$
where $\dbinom r k$ is a binomial coefficient.
== Proof ==
Proof by induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition
:$\ds \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$
=== Basis for the Induction ===
$\map P 0$ is true, as $\dbinom r 0 = 1 = \dbinom {r + 1} 0$ by definition.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P m$ is true, where $m \ge 0$, then it logically follows that $\map P {m + 1}$ is true.
So this is our induction hypothesis:
:$\ds \sum_{k \mathop = 0}^m \binom {r + k} k = \binom {r + m + 1} m$
Then we need to show:
:$\ds \sum_{k \mathop = 0}^{m + 1} \binom {r + k} k = \binom {r + m + 2} {m + 1}$
=== Induction Step ===
This is our induction step:
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$
== Also see ==
* Rising Sum of Binomial Coefficients, where the result is proved for integer $r$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $\text{E} \ (10)$
* : $\S 1.2.6$: Binomial Coefficients: Exercise $13$
Category:Binomial Coefficients
Sum of r Powers is between Power of Maximum and r times Power of Maximum

== Theorem ==
Let $a_1, a_2, \ldots, a_r$ be non-negative real numbers.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $a = \max \set {a_1, a_2, \ldots, a_r}$.
Then:
:$a^n \le a_1^n + a_2^n + \cdots + a_r^n \le r a^n$
== Proof ==
This proof is divided into $2$ parts:
=== Power of Maximum is not Greater than Sum of Powers ===
=== Sum of $r$ Powers is not Greater than $r$ times Power of Maximum ===
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 11$
Category:Inequalities
Sum of r Powers is not Greater than r times Power of Maximum

== Theorem ==
Let $a_1, a_2, \ldots, a_r$ be non-negative real numbers.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $a = \max \set {a_1, a_2, \ldots, a_r}$.
Then:
:$a_1^n + a_2^n + \cdots + a_r^n \le r a^n$
== Proof ==
By definition of the $\max$ operation:
:$\exists k \in \set {1, 2, \ldots, r}: a_k = a$
Then:
:$\forall i \in \set {1, 2, \ldots, r}: a_i \le a_k$
Hence:
Hence the result.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 11$
Category:Inequalities
Sum of two Fourth Powers cannot be Fourth Power

== Theorem ==
$\forall a, b, c \in \Z_{>0}$, the equation $a^4 + b^4 = c^4$ has no solutions.
== Proof ==
This is a direct consequence of Fermat's Right Triangle Theorem.
== Historical Note ==
== Sources ==
* : Chapter $\text{IV}$: The Prince of Amateurs
Category:Number Theory
Sum of two Fourth Powers cannot be Fourth Power/Historical Note

== Historical Note on Sum of two Fourth Powers cannot be Fourth Power ==
 gave an outline of the proof of Sum of two Fourth Powers cannot be Fourth Power using the Method of Infinite Descent.
It was proved in full by .
It can of course be seen to be a special case of Fermat's Last Theorem.
== Sources ==
* : Chapter $\text{IV}$: The Prince of Amateurs
Category:Historical Notes
Sum of two Incommensurable Medial Areas give rise to two Irrational Straight Lines

== Theorem ==
== Proof ==
:
Let $AB$ and $CD$ be medial areas.
It is to be demonstrated that the "side" of their combined area $AD$ is either:
:$(1): \quad$ a second bimedial
or:
:$(2): \quad$ a side of the sum of two medial areas.
WLOG let $AB > CD$.
Let a rational straight line $EF$ be set out.
Let the rectangle $EG = AB$ be applied to $EF$, producing $EH$ as breadth.
Let the rectangle $HI = DC$ be applied to $EF$, producing $HK$ as breadth.
We have that $AB$ and $CD$ are medial areas and equals $EG$ and $HI$ respectively.
Therefore $EG$ and $HI$ are medial areas.
By :
:$EH$ and $HK$ are rational straight lines which are both incommensurable in length with $EF$.
We have that:
:$AB$ is incommensurable with $CD$
and:
:$AB = EG, CD = HI$
Therefore:
:$EG$ is incommensurable with $HI$.
But from :
:$EG : HI = EH : HK$
Therefore from :
:$EH$ is incommensurable in Length with $HK$.
Therefore $EH$ and $HK$ are rational straight lines which are commensurable in square only.
Therefore, by definition, $EK$ is a binomial straight line which is divided at $H$.
We have that:
:$AB > CD$
and:
:$AB = EG, CD = HI$
Therefore:
:$EH > HK$
Thus $EH^2$ is greater than $HK^2$ by either:
:the square on a straight line which is commensurable in length with $EH$
or:
:the square on a straight line which is incommensurable in length with $EH$.
Let $EH^2 = HK^2 + \lambda^2$.
First, let $\lambda$ be commensurable in length with $EH$.
Neither of $EH$ or $HK$ is commensurable in length with the rational straight line $EF$.
Therefore $EK$ is a third binomial.
But $EF$ is rational.
From :
:the "side" of the square equal to the rectangle contained by a rational straight line and a first binomial is second bimedial.
Therefore the "side" of $EI$ is second bimedial.
That is, the "side" of $AD$ is second bimedial.
Next, let $\lambda$ be incommensurable in length with $EH$.
Neither of $EH$ or $HK$ is commensurable in length with the rational straight line $EF$.
Therefore $EK$ is a sixth binomial.
But $EF$ is rational.
From :
:the "side" of the square equal to the rectangle contained by a rational straight line and a first binomial is the side of the sum of two medial areas.
Therefore the "side" of $EI$ is the side of the sum of two medial areas.
That is, the "side" of $AD$ is the side of the sum of two medial areas.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Sum over Complement of Finite Set

== Theorem ==
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f: S \to \mathbb A$ be a mapping.
Let $T \subseteq S$ be a subset.
Let $S \setminus T$ be its relative complement.
Then we have the equality of summations over finite sets:
:$\ds \sum_{s \mathop \in S \setminus T} \map f s = \sum_{s \mathop \in S} \map f s - \sum_{t \mathop \in T} \map f t$
== Proof ==
Note that by Subset of Finite Set is Finite, $T$ is indeed finite.
By Set is Disjoint Union of Subset and Relative Complement, $S$ is the disjoint union of $S \setminus T$ and $T$.
The result now follows from Sum over Disjoint Union of Finite Sets.
== Also see ==
* Sum over Union of Finite Sets
Category:Summations
Sum over Disjoint Union of Finite Sets

== Theorem ==
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ and $T$ be finite disjoint sets.
Let $S \cup T$ be their union.
Let $f: S \cup T \to \mathbb A$ be a mapping.
Then we have the equality of summations over finite sets:
:$\ds \sum_{u \mathop \in S \mathop \cup T} \map f u = \sum_{s \mathop \in S} \map f s + \sum_{t \mathop \in T} \map f t$
== Outline of Proof ==
We reduce to the case of indexed summations by choosing appropriate bijections in the definition of summation over finite set.
We thereby reduce the problem to Indexed Summation over Adjacent Intervals.
== Proof ==
Note that by Union of Finite Sets is Finite, the union $S \cup T$ is finite.
Let $m$ be the cardinality of $S$ and $n$ be the cardinality of $T$.
Let $\N_{< m}$ denote an initial segment of the natural numbers.
Let $\sigma: \N_{<m} \to S$ and $\tau: \N_{<n} \to T$ be bijections.
Let $\alpha: \N_{< n} \to \closedint m {m + n - 1}$ be the mapping defined as:
:$\map \alpha k = k + m$
By Translation of Integer Interval is Bijection, $\alpha$ is a bijection.
By Composite of Bijections is Bijection and Disjoint Union of Bijections is Bijection, the union:
:$\sigma \cup \paren {\tau \circ \alpha} : \N_{<m} \cup \closedint m {m + n - 1} \to S \cup T$ is a bijection.
By Union of Integer Intervals, $\N_{<m} \cup \closedint m {m + n - 1} = \N_{< m + n}$.
We have:
== Also see ==
* Sum over Union of Finite Sets
* Sum over Complement of Finite Set
Category:Summations