nyuuzyou/wikis huggingface dataset:
Sum of Consecutive Triangular Numbers is Square

== Theorem ==
The sum of two consecutive triangular numbers is a square number.
== Proof ==
Let $T_{n - 1}$ and $T_n$ be two consecutive triangular numbers.
From Closed Form for Triangular Numbers‎, we have:
:$T_{n - 1} = \dfrac {\paren {n - 1} n} 2$
:$T_n = \dfrac {n \paren {n + 1} } 2$
So:
 
== Visual Demonstration ==
== Historical Note ==
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $1 \ \text {(c)}$
* : $15$
* : $25$
* : $15$
* : $25$
Category:Triangular Numbers
Category:Square Numbers
Category:Sum of Consecutive Triangular Numbers is Square
Sum of Consecutive Triangular Numbers is Square/Historical Note

== Historical Note on Sum of Consecutive Triangular Numbers is Square ==
According to , in his of $1980$, this result has been attributed to , circa $100$ C.E.
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $1 \ \text {(c)}$
Category:Historical Notes
Sum of Continuous Functions on Topological Ring is Continuous

== Theorem ==
Let $X$ be a topological space.
Let $R$ be a topological ring.
Let $f, g : X \to R$ be continuous mappings.
Then $f + g : X \to R$ is continuous.
== Proof ==
Equip the Cartesian product $R \times R$ with its product topology.
Define $h : X \to R \times R$ by: 
:$\map h x = \tuple {\map f x, \map g x}$ 
for each $x \in X$ and $s : R \times R \to R$ by: 
:$\map s {x, y} = x + y$ 
for each $\tuple {x, y} \in R \times R$. 
From Continuous Mapping to Product Space, $h$ is continuous.
From the definition of a topological ring, $s$ is continuous.
From Composite of Continuous Mappings is Continuous, $s \circ h : X \to R$ is continuous.
Then: 
for $x \in X$.
Category:Continuous Mappings (Topology)
Category:Topological Rings
Sum of Convergent Nets in Topological Vector Space is Convergent

== Theorem ==
Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $x, y \in X$.
Let $\family {x_\lambda}_{\lambda \in \Lambda}$ and $\family {y_\lambda}_{\lambda \in \Lambda}$ be nets converging to $x$ and $y$ respectively.
Then the net $\family {x_\lambda + y_\lambda}_{\lambda \in \Lambda}$ converges to $x + y$. 
== Proof ==
For ease of reading, let $\succeq$ be the inverse relation of $\preceq$.
Let $W$ be an open neighborhood of $x + y$.
From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods there exists an open neighborhood $U$ of $x$ and an open neighborhood $V$ of $y$ such that:
:$U + V \subseteq W$
Since $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$, there exists $\lambda_1 \in \Lambda$ such that:
:$x_\lambda \in U$ for $\lambda \succeq \lambda_1$.
Since $\family {y_\lambda}_{\lambda \in \Lambda}$ converges to $y$, there exists $\lambda_2 \in \Lambda$ such that:
:$y_\lambda \in V$ for $\lambda \succeq \lambda_2$.
Since $\struct {\Lambda, \preceq}$ is directed, there exists $\lambda_\ast \in \Lambda$ such that: 
:$\lambda_\ast \succeq \lambda_1$ and $\lambda_\ast \succeq \lambda_2$.
Since $\preceq$ is transitive, if $\lambda \in \Lambda$ has $\lambda \succeq \lambda_\ast$, then $\lambda \succeq \lambda_1$ and $\lambda \succeq \lambda_2$.
Hence, for $\lambda \succeq \lambda_\ast$ we have $x_\lambda \in U$ and $y_\lambda \in V$.
Hence:
:$x_\lambda + y_\lambda \in U + V \subseteq W$ for $\lambda \succeq \lambda_\ast$. 
Since $W$ was an arbitrary open neighborhood of $x + y$, the net $\family {x_\lambda + y_\lambda}_{\lambda \in \Lambda}$ converges to $x + y$. 
Category:Nets (Set Theory)
Category:Topological Vector Spaces
Sum of Convergent Sequences in Topological Vector Space is Convergent

== Theorem ==
Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $x, y \in X$.
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be convergent sequences with:
:$x_n \to x$
and:
:$y_n \to y$
Then $\sequence {x_n + y_n}_{n \in \N}$ converges with:
:$x_n + y_n \to x + y$
== Proof ==
Let $W$ be an open neighborhood of $x + y$.
From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods there exists an open neighborhood $U$ of $x$ and an open neighborhood $V$ of $y$ such that:
:$U + V \subseteq W$
Since $\sequence {x_n}_{n \in \N}$ converges to $x$, there exists $N_1 \in \N$ such that:
:$x_n \in U$ for $n \ge N_1$.
Since $\sequence {y_n}_{n \in \N}$ converges to $y$, there exists $N_2 \in \N$ such that:
:$y_n \in V$ for $n \ge N_2$.
Let $N = \max \set {N_1, N_2}$. 
Then for $n \ge N$ we have: 
:$x_n + y_n \in U + V \subseteq W$
Since $W$ was an arbitrary open neighborhood of $x + y$, we have that $\sequence {x_n + y_n}$ converges to $x + y$.
Category:Topological Vector Spaces
Sum of Convex Sets in Vector Space is Convex

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $A$ and $B$ be convex subsets of $X$.
Then: 
:$A + B$ is convex.
=== Corollary ===
== Proof ==
Let $x, y \in A + B$ and $t \in \closedint 0 1$. 
Then there exists $a, a' \in A$ and $b, b' \in B$ such that: 
:$x = a + b$
and:
:$y = a' + b'$
Then:
Since $A$ and $B$ are convex, we have:
:$t a + \paren {1 - t} a' \in A$
and:
:$t b + \paren {1 - t} b' \in B$
so that:
:$t x + \paren {1 - t} y \in A + B$
So $A + B$ is convex.
Category:Convex Sets (Vector Spaces)
Category:Sum of Convex Sets in Vector Space is Convex
Sum of Convex Sets in Vector Space is Convex/Corollary

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $A$ and $B$ be convex subsets of $X$.
Let $\lambda, \mu \in \Bbb F$.
Then:
:$\lambda A + \mu B$ is convex.
== Proof ==
From Dilation of Convex Set in Vector Space is Convex, we have: 
:$\lambda A$ and $\mu B$ are convex.
From Sum of Convex Sets in Vector Space is Convex, we have:
:$\lambda A + \mu B$ is convex.
Category:Sum of Convex Sets in Vector Space is Convex
Sum of Cosets of Ideals is Sum in Quotient Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\powerset R$ be the power set of $R$.
Let $J$ be an ideal of $R$.
Let $X$ and $Y$ be cosets of $J$.
Let $X +_\PP Y$ be the sum of $X$ and $Y$, where $+_\PP$ is the operation induced on $\powerset R$ by $+$.
The sum $X +_\PP Y$ in $\powerset R$ is also their sum in the quotient ring $R / J$.
== Proof ==
As $\struct {R, +, \circ}$ is a ring, it follows that $\struct {R, +}$ is an abelian group.
Thus by Subgroup of Abelian Group is Normal, all subgroups of $\struct {R, +, \circ}$ are normal.
So from the definition of quotient group, it follows directly that $X +_\PP Y$ in $\powerset R$ is also the sum in the quotient ring $R / J$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old
Category:Ring Operations on Coset Space of Ideal
Sum of Cosines of 2 k pi over 5

== Theorem ==
== Proof ==
Let $z_1, z_2, \dotsc, z_5$ be the complex $5$th roots of unity.
From Sum of Powers of Primitive Complex Roots of Unity, setting $s = 1$:
:$z_1 + z_2 + z_3 + z_4 + z_5 = 0$
Thus from Complex 5th Roots of Unity:
:$1 + \cis 72 \degrees + \cis 144 \degrees + \cis 216 \degrees + \cis 288 \degrees = 0$
Equating the real parts:
:$1 + \cos 72 \degrees + \cos 144 \degrees + \cos 216 \degrees + \cos 288 \degrees = 0$
:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The $n$th Roots of Unity: $106$
Category:Complex 5th Roots of Unity
Sum of Cosines of Angles in Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle.
Then:
:$\cos A + \cos B + \cos C = 1 - 4 \sin \dfrac A 2 \sin \dfrac B 2 \sin \dfrac C 2$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(33)$
Category:Sum of Cosines of Angles in Triangle
Category:Triangles
Category:Cosine Function
Sum of Cosines of Arithmetic Sequence of Angles

== Theorem ==
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
=== Formulation 1 ===
=== Formulation 2 ===
== Also see ==
* Sum of Sines of Arithmetic Sequence of Angles
Category:Cosine Function
Sum of Cosines of Arithmetic Sequence of Angles/Formulation 1

== Theorem ==
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
== Proof ==
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles: Formulation 1:
:$\ds \sum_{k \mathop = 0}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} }$
It is noted that, from Sine of Multiple of Pi, when $\alpha = 2 \pi k$ for $k \in \Z$, $\map \sin {\alpha / 2} = 0$ and the is not defined.
From Euler's Formula, this can be expressed as:
:$\ds \sum_{k \mathop = 0}^n \paren {\map \cos {\theta + k \alpha} + i \map \sin {\theta + k \alpha} } = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} }$
Equating real parts:
:$\ds \sum_{k \mathop = 0}^n \map \cos {\theta + k \alpha} = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \map \cos {\theta + \frac {n \alpha} 2}$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $143 \ \text{(a)}$
Category:Cosine Function
Sum of Cosines of Arithmetic Sequence of Angles/Formulation 2

== Theorem ==
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
== Proof ==
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles: Formulation 2:
:$\ds \sum_{k \mathop = 1}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$
It is noted that, from Sine of Multiple of Pi, when $\alpha = 2 \pi k$ for $k \in \Z$, $\map \sin {\alpha / 2} = 0$ and the is not defined.
From Euler's Formula, this can be expressed as:
:$\ds \sum_{k \mathop = 1}^n \paren {\map \cos {\theta + k \alpha} + i \map \sin {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$
Equating real parts:
:$\ds \sum_{k \mathop = 1}^n \map \cos {\theta + k \alpha} = \map \cos {\theta + \frac {n + 1} 2 \alpha} \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$
== Sources ==
* : $\S 2$. Geometrical Representations: Example $1$: $(2.23)$
Category:Cosine Function
Sum of Cosines of Four Times Angles of Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle.
Then:
:$\cos 4 A + \cos 4 B + \cos 4 C = 4 \cos 2 A \cos 2 B \cos 2 C - 1$
== Proof ==
First we note that:
That is, $2 C$ is the conjugate of $2 A + 2 B$.
Then:
== Also presented as ==
This result can also be presented as:
:$\cos 4 A + \cos 4 B + \cos 4 C + 1 = 4 \cos 2 A \cos 2 B \cos 2 C$
== Sources ==
* : $\text V$. Trigonometry: Exercises $\text {XXXII}$: $20$.
Category:Triangles
Category:Cosine Function
Sum of Cosines of Fractions of Pi

== Theorem ==
Let $n \in \Z$ such that $n > 1$.
Then:
:$\ds \sum_{k \mathop = 1}^{n - 1} \cos \frac {2 k \pi} n = -1$
== Proof ==
Consider the equation:
:$z^n - 1 = 0$
whose solutions are the complex roots of unity:
:$1, e^{2 \pi i / n}, e^{4 \pi i / n}, e^{6 \pi i / n}, \ldots, e^{2 \paren {n - 1} \pi i / n}$
By Sum of Roots of Polynomial:
:$1 + e^{2 \pi i / n} + e^{4 \pi i / n} + e^{6 \pi i / n} + \cdots + e^{2 \paren {n - 1} \pi i / n} = 0$
From Euler's Formula:
:$e^{i \theta} = \cos \theta + i \sin \theta$
from which comes:
:$\paren {1 + \cos \dfrac {2 \pi} n + \cos \dfrac {4 \pi} n + \cdots + \cos \dfrac {2 \paren {n - 1} \pi} n} + i \paren {\sin \dfrac {2 \pi} n + \sin \dfrac {4 \pi} n + \cdots + \sin \dfrac {2 \paren {n - 1} \pi} n} = 0$
Equating real parts:
:$1 + \cos \dfrac {2 \pi} n + \cos \dfrac {4 \pi} n + \cdots + \cos \dfrac {2 \paren {n - 1} \pi} n = 0$
whence the result.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: The $n$th Roots of Unity: $38 \ \text{(a)}$
Category:Cosine Function
Sum of Cosines of Twice 2 Angles minus Cosine of Twice Third Angle of Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle.
Then:
:$\cos 2 A + \cos 2 B - \cos 2 C = 1 - 4 \sin A \sin B \cos C$
== Proof ==
First we note that:
That is, $C$ is the supplement of $A + B$.
Then:
== Sources ==
* : $\text V$. Trigonometry: Exercises $\text {XXXII}$: $15$.
Category:Triangles
Category:Cosine Function
Sum of Cosines of Twice Angles of Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle.
Then:
:$\cos 2 A + \cos 2 B + \cos 2 C = -1 - 4 \cos A \cos B \cos C$
== Proof ==
First we note that:
That is, $C$ is the supplement of $A + B$.
Then:
== Also presented as ==
This result can also be presented as:
:$\cos 2 A + \cos 2 B + \cos 2 C + 1 = -4 \cos A \cos B \cos C$
== Sources ==
* : $\text V$. Trigonometry: Exercises $\text {XXXII}$: $17$.
Category:Triangles
Category:Cosine Function
Sum of Cosines of k pi over 5

== Theorem ==
:$\cos 36 \degrees + \cos 72 \degrees + \cos 108 \degrees + \cos 144 \degrees = 0$
== Proof ==
We have:
and:
Thus:
:$\cos 36 \degrees + \cos 72 \degrees + \cos 108 \degrees + \cos 144 \degrees = 0$
:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The $n$th Roots of Unity: $107$
Category:Complex 5th Roots of Unity
Sum of Cotangents of Half Angles in Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle.
Then:
:$\cot \dfrac A 2 + \cot \dfrac B 2 + \cot \dfrac C 2 = \cot \dfrac A 2 \cot \dfrac B 2 \cot \dfrac C 2$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(23)$
Category:Sum of Cotangents of Half Angles in Triangle
Category:Triangles
Category:Cotangent Function
Sum of Cube Roots of Unity

== Theorem ==
Let $U_3 = \set {1, \omega, \omega^2}$ denote the Cube Roots of Unity.
Then:
:$1 + \omega + \omega^2 = 0$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Cube Roots of Unity
Category:Sum of Cube Roots of Unity
Sum of Cube Roots of Unity/Proof 2

== Theorem ==
== Proof ==
From Sum of Powers of Primitive Complex Roots of Unity:
:$\ds \sum_{j \mathop = 0}^{n - 1} \alpha^{j s} = \begin {cases} n & : n \divides s \\ 0 & : n \nmid s \end {cases}$
Here we have that $n = 3$ and $s = 1$.
Thus $n$ is not a divisor of $s$.
Hence the result.
Category:Sum of Cube Roots of Unity
Sum of Cubes of 3 Consecutive Integers which is Square

== Theorem ==
The following sequences of $3$ consecutive (strictly) positive integers have cubes that sum to a square:
:$1, 2, 3$
:$23, 24, 25$
No other such sequence of $3$ consecutive positive integers has the same property.
However, if we allow sequences containing zero and negative integers, we also have:
:$-1, 0, 1$
:$0, 1, 2$
== Proof ==
Any sequence of $3$ consecutive integers that have cubes that sum to a square would satisfy:
:$m^2 = \paren {n - 1}^3 + n^3 + \paren {n + 1}^3$
where $n$ is the middle number of the sequence, with $m, n \in \Z$.
Expanding the :
Substituting $y = 3 m$ and $x = 3 n$:
which is an elliptic curve.
According to LMFDB, this elliptic curve has exactly $7$ lattice points:
:$\tuple {0, 0}, \tuple {3, \pm 9}, \tuple {6, \pm 18}, \tuple {72, \pm 612}$
which correspond to these values of $n$:
:$0, 1, 2, 24$
Hence there are no more solutions.
== Also see ==
* Sum of Cubes of 5 Consecutive Integers which is Square
== Sources ==
* : $204$
* : $204$
* 
Category:Sums of Cubes
Sum of Cubes of 5 Consecutive Integers which is Square

== Theorem ==
The following sequences of $5$ consecutive (strictly) positive integers have cubes that sum to squares:
:$1, 2, 3, 4, 5$
:$25, 26, 27, 28, 29$
:$96, 97, 98, 99, 100$
:$118, 119, 120, 121, 122$
No other such sequence of $5$ consecutive positive integers has the same property.
However, if we allow sequences containing zero and negative integers, we also have:
:$0, 1, 2, 3, 4$
:$-2, -1, 0, 1, 2$
== Proof ==
Then we also have:
and finally the degenerate case:
Any sequence of $5$ consecutive integers that have cubes that sum to a square would satisfy:
:$m^2 = \paren {n - 2}^3 + \paren {n - 1}^3 + n^3 + \paren {n + 1}^3 + \paren {n + 2}^3$
where $n$ is the middle number of the sequence, with $m, n \in \Z$.
Expanding the :
Substituting $y = 5 m$ and $x = 5 n$:
which is an elliptic curve.
According to LMFDB, this elliptic curve has exactly $13$ lattice points:
:$\tuple {0, 0}, \tuple {10, \pm 50}, \tuple {15, \pm 75}, \tuple {24, \pm 132}, \tuple {135, \pm 1575}, \tuple {490, \pm 10 \, 850}, \tuple {600, \pm 14 \, 700}$
which correspond to these values of $n$:
:$0, 2, 3, \dfrac {24} 5, 27, 98, 120$
Note that $\dfrac {24} 5$ is not an integer.
Hence there are no more solutions.
== Also see ==
* Sum of Cubes of 3 Consecutive Integers which is Square
== Historical Note ==
== Sources ==
* 
* : Chapter $21$
* : $118$
* 
Category:Sums of Cubes
Sum of Cubes of 5 Consecutive Integers which is Square/Historical Note

== Historical Note on Sum of Cubes of 5 Consecutive Integers which is Square ==
This result was originally published by in $1873$, where he wrote:
:The sum of the cubes of $5$ consecutive numbers is never equal to a square, except for the solutions of which the middle numbers are $2$, $3$, $27$, $98$ or $120$.
Subsequently it was reported by , who made a mistake by omitting the $27$.
The erroneous sequence is still mistakenly published on occasion.
== Sources ==
* 
* : Chapter $21$
* : $118$
Category:Sums of Cubes
Sum of Cubes of 5 Consecutive Integers which is Square/Mistake

== Source Work ==
:
:The Dictionary
::$118$
== Mistake ==
:The sum of the cubes of the $5$ consecutive positive integers $118$ to $122$ is a square.
:The only other similar sequences with this property start with $1$ and $96$.
== Correction ==
$25$ also has that property.
 published this result in $1873$.
However, when reported on it in his of $1920$, he appears to have erroneously omitted the sequence starting with $25$.
The correct sequence appears on .
The erroneous sequence is documented on .
== Sources ==
* 
* : Chapter $21$
* : $118$
Category:Mistakes/Curious and Interesting Numbers
Sum of Cubes of Sine and Cosine

== Theorem ==
:$\cos^3 x + \sin^3 x = \paren {\cos x + \sin x} \paren {1 - \cos x \sin x}$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Exercise $\text {XXXI}$: $5.$
Category:Trigonometric Identities
Category:Sine Function
Category:Cosine Function
Sum of Cubes of Three Indeterminates Minus 3 Times their Product

== Theorem ==
For indeterminates $x, y, z$:
:$x^3 + y^3 + z^3 - 3 x y z = \paren {x + y + z} \paren {x + \omega y + \omega^2 z} \paren {x + \omega^2 y + \omega z}$
where $\omega = -\dfrac 1 2 + \dfrac {\sqrt 3} 2$
== Proof 1 ==
== Proof 2 ==
Category:Algebra
Category:Cube Roots of Unity
Category:Sum of Cubes of Three Indeterminates Minus 3 Times their Product
Sum of Cubes of Three Indeterminates Minus 3 Times their Product/Proof 2

== Theorem ==
== Proof ==
Consider the determinant:
:$\Delta = \begin {vmatrix} x & z & y \\ y & x & z \\ z & y & x \end {vmatrix}$
We have:
Then we note that adding rows $2$ and $3$ to rows $1$ gives:
Let $\omega$ denote the complex cube root of unity:
:$\omega = -\dfrac 1 2 + \dfrac {\sqrt 3} 2$
Hence adding $\omega$ times row $2$ and $\omega^2$ times row $3$ to rows $1$:
and adding $\omega^2$ times row $2$ and $\omega$ times row $3$ to rows $1$:
Thus we have $3$ divisors of $x^3 + y^3 + z^3 - 3 x y z$, which is a polynomial of degree $3$.
There can be no other divisors except for a constant.
By examining the coefficient of $x^3$, for example, the constant is seen to be $1$.
Hence the result.
== Sources ==
* : $\S 3$. Roots of Unity: Example $1$.
Category:Sum of Cubes of Three Indeterminates Minus 3 Times their Product
Sum of Cuts is Cut

== Theorem ==
Let $\alpha$ and $\beta$ be cuts.
Let $\gamma$ be the set of all rational numbers $r$ such that:
:$\exists p \in \alpha, q \in \beta: r = p + q$
Then $\gamma$ is also a cut.
Thus the operation of addition on the set of cuts is closed.
== Proof ==
By definition of cut, neither $\alpha$ nor $\beta$ are empty.
Hence there exist $p \in \alpha$ and $q \in \beta$.
Hence there exists $r = p + q$ and so $\gamma$ is likewise not empty.
Let $s, t \in \Q$ such that $s \notin \alpha$ and $t \notin \beta$, where $\Q$ denotes the set of rational numbers.
Such $s$ and $t$ are bound to exist because by definition of cut, neither $\alpha$ nor $\beta$ equal $\Q$.
We have:
Thus it is demonstrated that $\gamma$ does not contain every rational number.
Thus condition $(1)$ of the definition of a cut is fulfilled.
Let $r \in \gamma$.
Let $s \in \Q$ such that $s < r$.
Then $r = p + q$ for some $p \in \alpha, q \in \beta$.
Let $t \in \Q$ such that $s = t + q$.
Then $t < p$.
Hence $t \in \alpha$.
Hence by definition of $\gamma$, $t + q = s \in \gamma$.
Thus we have that $r \in \gamma$ and $s < r$ implies that $s \in \gamma$.
Thus condition $(2)$ of the definition of a cut is fulfilled.
 $r \in \gamma$ is the greatest element of $\gamma$.
Then $r = p + q$ for some $p \in \alpha, q \in \beta$.
By definition of a cut, $\alpha$ has no greatest element:
Hence:
:$\exists s \in \Q: s > p: s \in \alpha$
But then $s + q \in \gamma$ while $s + q > r$.
This contradicts the supposition that $r$ is the greatest element of $\gamma$.
Hence $\gamma$ itself can have no greatest element.
Thus condition $(3)$ of the definition of a cut is fulfilled.
Thus it is seen that all the conditions are fulfilled for $\gamma$ to be a cut.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.12$. Theorem
Category:Cuts
Category:Addition
Sum of Degenerate Linear Transformation is Degenerate

== Theorem ==
Let $U, V$ be vector spaces over a field $K$.
Let $S: U \to V$ be a degenerate linear transformation.
Let $T: U \to V$ be a degenerate linear transformation.
Then $S + T$ is a degenerate linear transformation.
== Proof ==
Let $\set {s_1, \ldots, s_m}$ be a generator of $\Img S$.
Let $\set {t_1, \ldots, t_n}$ be a generator of $\Img T$.
Then $\set {s_1, \ldots, s_m, t_1, \ldots, t_n}$ is a generator of $\Img {S + T}$.
By Cardinality of Generator of Vector Space is not Less than Dimension:
:$\map \dim {\Img {S + T}} \le m + n$
== Sources ==
* : $2.2$: Index of a Linear Map
Category:Linear Algebra
Sum of Deviations from Mean

== Theorem ==
Let $S = \set {x_1, x_2, \ldots, x_n}$ be a set of real numbers.
Let $\overline x$ denote the arithmetic mean of $S$.
Then: 
:$\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x} = 0$
== Proof ==
For brevity, let us write $\ds \sum$ for $\ds \sum_{i \mathop = 1}^n$.
Then:
== Also see ==
* Sum of Squared Deviations from Mean
Category:Arithmetic Mean
Sum of Digamma of n plus 2 over n plus 2 squared

== Theorem ==
where:
:$\map \psi n$ is the digamma function of $n$
:$\map \zeta 2$ is the Riemann $\zeta$ function of $2$
:$\map \zeta 3$ is Apéry's constant: the Riemann $\zeta$ function of $3$.
Note the use of the $\cdot$ notation for product, or you wonder what the $\gamma$ function is.
== Proof ==
== Sources ==
* https://www.instagram.com/p/C03-XIwBZ34/
Category:Riemann Zeta Function
Category:Digamma Function
Category:Apéry's Constant