nyuuzyou/wikis huggingface dataset:
Sum from -m to m of 1 minus Cosine of n + alpha of theta over n + alpha

== Theorem ==
For $0 < \theta < 2 \pi$:
:$\ds \sum_{n \mathop = -m}^m \dfrac {1 - \cos \paren {n + \alpha} \theta} {n + \alpha} = \int_0^\theta \map \sin {\alpha u} \dfrac {\sin \paren {m + \frac 1 2} u \rd u} {\sin \frac 1 2 u}$
== Proof ==
We have:
Note that the at $(1)$ is not defined when $e^{i u} = 1$.
This happens when $u = 2 k \pi$ for $k \in \Z$.
For the given range of $0 < \theta < 2 \pi$ it is therefore seen that $(1)$ does indeed hold.
Then:
Hence the result.
== Sources ==
* : Exercises on Chapter $\text {II}$: $2$.
Category:Cosine Function
Sum from -m to m of Sine of n + alpha of theta over n + alpha

== Theorem ==
For $0 < \theta < 2 \pi$:
:$\ds \sum_{n \mathop = -m}^m \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} = \int_0^\theta \map \cos {\alpha \theta} \dfrac {\sin \paren {m + \frac 1 2} \theta \rd \theta} {\sin \frac 1 2 \theta}$
== Proof ==
We have:
Note that the at $(1)$ is not defined when $e^{i \theta} = 1$.
This happens when $\theta = 2 k \pi$ for $k \in \Z$.
For the given range of $0 < \theta < 2 \pi$ it is therefore seen that $(1)$ does indeed hold.
Then:
Hence the result.
== Sources ==
* : Exercises on Chapter $\text {II}$: $1$.
Category:Sine Function
Sum of 1 + sin pi by 5 plus i cos pi by 5 to Fifth Power plus i times its Conjugate

== Theorem ==
:$\paren {1 + \sin \dfrac \pi 5 + i \cos \dfrac \pi 5}^5 + i \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5 = 0$
== Proof ==
== Sources ==
* : $\S 2$. Geometrical Representations: Exercise $11$
Category:Examples of Complex Powers
Sum of 2 Lucky Numbers in 4 Ways

== Theorem ==
The number $34$ is the smallest positive integer to be the sum of $2$ lucky numbers in $4$ different ways.
== Proof ==
The sequence of lucky numbers begins:
:$1, 3, 7, 9, 13, 15, 21, 25, 31, 33, \ldots$
Thus we have:
== Sources ==
* : $34$
Category:Lucky Numbers
Category:34
Sum of 2 Squares in 2 Distinct Ways

== Theorem ==
Let $m, n \in \Z_{>0}$ be distinct positive integers that can be expressed as the sum of two distinct square numbers.
Then $m n$ can be expressed as the sum of two square numbers in at least two distinct ways.
=== Sequence ===
== Proof ==
Let:
:$m = a^2 + b^2$
:$n = c^2 + d^2$
Then:
It remains to be shown that if $a \ne b$ and $c \ne d$, then the four numbers:
:$a c + b d, a d - b c, a c - b d, a d + b c$
are distinct.
Because $a, b, c, d > 0$, we have:
:$a c + b d \ne a c - b d$
:$a d + b c \ne a d - b c$
We also have:
Thus $a \ne b$ and $c \ne d$ implies $a c \pm b d \ne a d \pm b c$.
The case for $a c \pm b d \ne a d \mp b c$ is similar.
== Examples ==
== Sources ==
* : $50$
* : $50$
Category:Square Numbers
Category:Sums of Squares
Category:Brahmagupta-Fibonacci Identity
Category:Sum of 2 Squares in 2 Distinct Ways
Sum of 2 Squares in 2 Distinct Ways/Examples

== Examples of Sum of 2 Squares in 2 Distinct Ways ==
=== $50$ as the Sum of 2 Squares ===
=== $65$ as the Sum of 2 Squares ===
=== $145$ as the Sum of 2 Squares ===
Category:Sum of 2 Squares in 2 Distinct Ways
Sum of 2 Squares in 2 Distinct Ways/Examples/145

== Theorem ==
$145$ can be expressed as the sum of two square numbers in two distinct ways:
== Proof ==
We have that:
:$145 = 5 \times 29$
Both $5$ and $29$ can be expressed as the sum of two distinct square numbers:
Thus:
and:
== Sources ==
* : $145$
* : $145$
Category:Sum of 2 Squares in 2 Distinct Ways
Category:145
Sum of 2 Squares in 2 Distinct Ways/Examples/145/Mistake/First Edition

== Source Work ==
:
:The Dictionary
::$145$
== Mistake ==
:The $4$th number to be the sum of $2$ squares in two different ways:
::$145 = 12^2 + 1^2 = 8^2 + 9^2$
It is in fact the $6$th after $50$, $65$, $85$, $125$, $130$.
 has it as the $5$th such number, which is also wrong.
== Sources ==
* : $145$
Category:Mistakes/Curious and Interesting Numbers
Sum of 2 Squares in 2 Distinct Ways/Examples/145/Mistake/Second Edition

== Source Work ==
:
:The Dictionary
::$145$
== Mistake ==
:The $5$th number to be the sum of $2$ squares in $2$ different ways:
::$145 = 12^2 + 1^2 = 8^2 + 9^2$
It is in fact the $6$th after $50$, $65$, $85$, $125$, $130$.
 has it as the $4$th such number. This is clearly an incorrect correction.
== Sources ==
* : $145$
Category:Mistakes/Curious and Interesting Numbers
Sum of 2 Squares in 2 Distinct Ways/Examples/50

== Theorem ==
$50$ is the smallest positive integer which can be expressed as the sum of two square numbers in two distinct ways:
== Proof ==
The smallest two positive integers which can be expressed as the sum of two distinct square numbers are:
We have that:
:$50 = 5 \times 10$
Thus:
and:
== Sources ==
* : $50$
* : $50$
Category:Sum of 2 Squares in 2 Distinct Ways
Category:50
Sum of 2 Squares in 2 Distinct Ways/Examples/65

== Theorem ==
$65$ can be expressed as the sum of two square numbers in two distinct ways:
== Proof ==
We have that:
:$65 = 5 \times 13$
Both $5$ and $13$ can be expressed as the sum of two distinct square numbers:
Thus:
and:
== Sources ==
* : $50$
* : $65$
* : $50$
* : $65$
Category:Sum of 2 Squares in 2 Distinct Ways
Category:65
Sum of 2 Squares in 2 Distinct Ways which is also Sum of Cubes

== Theorem ==
The smallest positive integer which is both the sum of $2$ square numbers in two distinct ways and also the sum of $2$ cube numbers is $65$:
=== Sequence ===
== Proof ==
From Sum of 2 Squares in 2 Distinct Ways, the smallest $2$ positive integer which are the sum of $2$ square numbers in two distinct ways are $50$ and $65$.
But $50$ cannot be expressed as the sum of $2$ cube numbers:
Hence $65$ is that smallest number.
== Historical Note ==
== Sources ==
* : $65$
Category:Sums of Squares
Category:Sums of Cubes
Category:65
Category:Sum of 2 Squares in 2 Distinct Ways which is also Sum of Cubes
Sum of 2 Squares in 2 Distinct Ways which is also Sum of Cubes/Historical Note

== Historical Note on Sum of 2 Squares in 2 Distinct Ways which is also Sum of Cubes ==
This result is attributed by in his of $1997$ to a source going by the name of Thayer, but it has not been possible to find out any further information.
== Sources ==
* : $250$
Category:Historical Notes
Sum of 2 Squares in 2 Distinct Ways which is also Sum of Cubes/Sequence

== Theorem ==
The sequence of positive integers which are both the sum of $2$ square numbers in two distinct ways and also the sum of $2$ cube numbers begins:
:$65, 250, \ldots$
== Proof ==
== Historical Note ==
== Sources ==
* : $250$
Category:Sum of 2 Squares in 2 Distinct Ways which is also Sum of Cubes
Sum of 2 Squares in 3 Distinct Ways

== Theorem ==
The following positive integers can be expressed as the sum of $2$ square numbers in $3$ distinct ways:
:$325, 425, 650, 725, 845, 850, 925, 1025, 1105, 1250, \dots$
== Examples ==
Category:Square Numbers
Category:Sums of Squares
Category:Sum of 2 Squares in 3 Distinct Ways
Sum of 2 Squares in 3 Distinct Ways/Examples/325

== Example of Sum of 2 Squares in 3 Distinct Ways ==
$325$ is the smallest positive integer which can be expressed as the sum of two square numbers in three distinct ways:
== Proof ==
== Sources ==
* : $325$
* : $325$
Category:Sum of 2 Squares in 3 Distinct Ways
Category:325
Sum of 3 Unit Fractions that equals 1

== Theorem ==
There are $3$ ways to represent $1$ as the sum of exactly $3$ unit fractions.
== Proof ==
Let:
:$1 = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c$
where:
:$0 < a \le b \le c$
and:
 $a = 1$.
Then:
:$1 = \dfrac 1 1 + \dfrac 1 b + \dfrac 1 c$
and so:
:$\dfrac 1 b + \dfrac 1 c = 0$
which contradicts the stipulation that $b, c > 0$.
So there is no solution possible when $a = 1$.
Therefore $a \ge 2$.
=== $a = 2$ ===
Let $a = 2$.
==== $b = 2$ ====
Let $b = 2$.
Then:
:$\dfrac 1 a + \dfrac 1 b = 1$
leaving no room for $c$.
Hence there are no solutions where $a = 2$ and $b = 2$.
==== $b = 3$ ====
Let $b = 3$.
Thus we have:
:$(1): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 6$
==== $b = 4$ ====
Let $b = 4$.
Thus we have:
:$(2): \quad 1 = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 4$
==== $b > 4$ ====
Let $b > 4$.
Then:
Hence there are no solutions such that $a = 2, b > 4$.
=== $a = 3$ ===
Let $a = 3$.
==== $b = 3$ ====
Let $b = 3$.
Thus we have:
:$(3): \quad 1 = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 3$
=== $a > 3$ ===
Let $a > 3$.
Then:
Hence there are no solutions for $a>3$.
=== Summary ===
Hence our $3$ solutions:
== Sources ==
* 
Category:Unit Fractions
Category:Recreational Mathematics
Sum of 4 Consecutive Binomial Coefficients forming Square

== Theorem ==
Consider the Diophantine equation:
:$\dbinom n 0 + \dbinom n 1 + \dbinom n 2 + \dbinom n 3 = m^2$
where:
:$\dbinom a b$ denotes a binomial coefficient
:$n$ is an integer
:$m$ is a non-negative integer.
Then $n$ has one of the following values:
:$-1, 0, 2, 7, 15, 74, 767$
The corresponding values of $m$ are:
:$0, 1, 2, 8, 24, 260, 8672$
== Proof ==
== Sources ==
* 
* : $767$
Category:Binomial Coefficients
Category:Square Numbers
Sum of 4 Consecutive Binomial Coefficients forming Square/Mistake

== Source Work ==
:
:The Dictionary
::$767$
== Mistake ==
Mistake $1$:
:$\dbinom {767} 1 + \dbinom {767} 2 + \dbinom {767} 3 + \dbinom {767} 4$ is a perfect square, $8672^2$.
Mistake $2$:
:The smaller solutions are $7$, $15$ and $74$.
== Correction ==
The first mistake is that it is $\dbinom {767} 0 + \dbinom {767} 1 + \dbinom {767} 2 + \dbinom {767} 3 = 8672^2$.
The second mistake is that the numbers $-1$, $0$ and $2$ have the same property.
 is perhaps excused the first two, as they may not have been in his purview, but omitting $2$ is a less forgivable oversight, considering the complete set is given in , which he himself references.
== Sources ==
* : $767$
Category:Mistakes/Curious and Interesting Numbers
Sum of 714 and 715

== Theorem ==
The sum of $714$ and $715$ is a $4$-digit integer which has $6$ anagrams which are prime.
== Proof ==
We have that:
:$714 + 715 = 1429$
Hence we investigate its anagrams.
We bother only to check those which do not end in either $2$ or $4$, as those are even.
Of the above, $6$ are seen to be prime.
== Sources ==
* : $714$
* : $714$
Category:Prime Numbers
Category:714
Category:715
Sum of Absolute Values on Ordered Integral Domain

== Theorem ==
Let $\struct {D, +, \times, \le}$ be an ordered integral domain.
For all $a \in D$, let $\size a$ denote the absolute value of $a$.
Then:
:$\size {a + b} \le \size a + \size b$
== Proof ==
Let $P$ be the (strict) positivity property on $D$.
Let $<$ be the (strict) total ordering defined on $D$ as:
:$a < b \iff a \le b \land a \ne b$
Let $N$ be the (strict) negativity property on $D$.
Let $a \in D$.
If $\map P a$ or $a = 0$ then $a \le \size a$.
If $\map N a$ then by Properties of Strict Negativity: $(1)$ and definition of absolute value:
:$a < 0 < \size a$
and hence by transitivity $<$ we have:
:$a < \size a$
By similar reasoning:
:$-a < \size a$
Thus for all $a, b \in D$ we have:
:$a \le \size a, b \le \size b$
As $<$ is compatible with $+$, we have:
:$a + b \le \size a + \size b$
and:
:$-\paren {a + b} = \paren {-a} + \paren {-b} \le \size a + \size b$
But either:
:$\size {a + b} = a + b$
or:
:$\size {a + b} = -\paren {a + b}$
Hence the result:
:$\size {a + b} \le \size a + \size b$
== Also see ==
* Triangle Inequality
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $11 \ \text{(ii)}$
Category:Absolute Value Function
Sum of Absolutely Continuous Functions is Absolutely Continuous

== Theorem ==
Let $I \subseteq \R$ be a real interval.
Let $f, g : I \to \R$ be absolutely continuous real functions.
Then $f + g$ is absolutely continuous.
== Proof ==
Let $\epsilon$ be a positive real number.
Since $f$ is absolutely continuous, there exists real $\delta_1 > 0$ such that for all sets of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_1$
we have: 
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon 2$
Similarly, since $g$ is absolutely continuous, there exists real $\delta_2 > 0$ such that whenever: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_2$
we have: 
:$\ds \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon 2$
Let: 
:$\delta = \map \min {\delta_1, \delta_2}$
Then, for all sets of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
we have: 
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon 2$
and:
:$\ds \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon 2$
We then have:
whenever: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
Since $\epsilon$ was arbitrary:
:$f + g$ is absolutely continuous.
Category:Absolutely Continuous Real Functions
Sum of Absolutely Convergent Series

== Theorem ==
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be two real or complex series that are absolutely convergent.
Then the series $\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}$ is absolutely convergent, and:
:$\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n} = \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n$
== Proof ==
Let $\epsilon \in \R_{>0}$.
From Tail of Convergent Series tends to Zero, it follows that there exists $M \in \N$ such that:
:$\ds \sum_{n \mathop = M + 1}^\infty \cmod {a_n} < \dfrac \epsilon 2$
and:
:$\ds\sum_{n \mathop = M + 1}^\infty \cmod {b_n} < \dfrac \epsilon 2$
For all $m \ge M$, it follows that:
By definition of convergent series, it follows that:
To show that $\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}$ is absolutely convergent, note that:
== Sources ==
* : $\S 4.4$. Power Series
* : $\S 1.2.3$: Operations with series
Category:Absolute Convergence
Sum of Absorbing Sets is Absorbing

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$. 
Let $A, B \subseteq X$ be absorbing sets.
Then $A + B$ is absorbing.
== Proof ==
Let $u \in A + B$. 
Then there exists $a \in A$, $b \in B$ such that $u = a + b$.
Now, there exists $t_1 \in \R_{> 0}$ such that: 
:$a \in \alpha A$ for $\alpha \in \C$ with $\cmod \alpha \ge t_1$
There also exists $t_2 \in \R_{> 0}$ such that:
:$b \in \alpha B$ for $\alpha \in \C$ with $\cmod \alpha \ge t_2$
Let $t = \max \set {t_1, t_2}$. 
Then for $\alpha \in \C$ with $\cmod \alpha \ge t$, we have:
:$u = a + b \in \alpha A + \alpha B = \alpha \paren {A + B}$
from Dilation of Subset of Vector Space Distributes over Sum.
So $A + B$ is absorbing.
Category:Absorbing Sets
Sum of Additive Function Values is Well-Defined

== Theorem ==
Let $\SS$ be an algebra of sets.
Let $f: \SS \to \overline \R$ be an additive function on $\SS$.
Let $A, B \in \SS$.
Then the sum
:$\map f A + \map f B$
is well-defined in the extended real numbers $\overline \R$.
== Proof ==
Suppose the sum $\map f A + \map f B$ is void.
By , this happens when the sum is $\paren{ +\infty } + \paren{ -\infty }$, or $\paren{ -\infty } + \paren{ +\infty }$.
, assume that $\map f A = +\infty$, and $\map f B = -\infty$.
Set Difference and Intersection are Disjoint shows that $A \setminus B$ and $A \cap B$ are disjoint.
Set Difference and Intersection are Disjoint shows that $B \setminus A$ and $A \cap B$ are disjoint.
By definition of additive function and Set Difference Union Intersection, it follows that:
:$ \map f { A \setminus B } + \map f { A \cap B } = \map f A = +\infty$
:$ \map f { B \setminus A } + \map f { A \cap B } = \map f B = -\infty$
Suppose $\map f { A \cap B } = +\infty$.
It follows that:
:$\map f { B \setminus A } + \paren{ +\infty } = -\infty$
which is a contradiction.
Suppose $\map f { A \cap B } = -\infty$.
It follows that:
:$\map f { A \setminus B } + \paren{ -\infty } = +\infty$
which is a contradiction.
Finally, suppose $\map f { A \cap B } \in \R$.
As $\map f { A \setminus B } + \map f { A \cap B } = +\infty$, it follows that $\map f { A \setminus B } = +\infty$.
As $\map f { B \setminus A } + \map f { A \cap B } = -\infty$, it follows that $\map f { B \setminus A } = -\infty$.
By definition of additive function and Set Difference is Disjoint with Reverse, it follows that the sum:
:$\map f { A \setminus B } + \map f { B \setminus A }$
is well-defined.
As this sum is equal to $\paren{ +\infty } + \paren{ -\infty}$, this is a contradiction.
Category:Set Systems
Category:Additive Functions
Sum of Adjacent Sequences of Triangular Numbers

== Theorem ==
and so on.
The $n$th line of the pattern can be written as:
:$\ds \sum_{k \mathop = n^2 + n - 1}^{n^2 + 2 n} T_n = \sum_{k \mathop = n^2 + 2 n + 1}^{n^2 + 3 n} T_n$
== Proof ==
== Historical Note ==
== Sources ==
* : $15$
* : $15$
Category:Sums of Sequences
Category:Triangular Numbers
Sum of All Ring Products is Additive Subgroup

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +}$ and $\struct {T, +}$ be additive subgroups of $\struct {R, +, \circ}$.
Let $S + T$ be defined as subset product.
Let $S T$ be defined as:
:$\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$
Then both $S + T$ and $S T$ are additive subgroups of $\struct {R, +, \circ}$.
== Proof ==
As $\struct {R, +}$ is abelian (from the definition of a ring), we have:
:$S + T = T + S$
from Subset Product of Commutative is Commutative.
So from Subset Product of Subgroups it follows that $S + T$ is an additive subgroup of $\struct {R, +, \circ}$.
Let $x, y \in S T$.
We have that $\struct {S T, +}$ is closed.
So $x + y \in S T$.
So, if $\ds y = \sum s_i \circ t_i \in S T$, it follows that:
:$\ds -y = \sum \paren {-s_i} \circ t_i \in S T$
By the Two-Step Subgroup Test, we have that $S T$ is an additive subgroup of $\struct {R, +, \circ}$.
== Sources ==
* : $\S 2.1$: Subrings: Lemma $2.3 \ \text{(ii)}$
Category:Ring Theory
Category:Subset Products
Sum of All Ring Products is Associative

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +}, \struct {T, +}, \struct {U, +}$ be additive subgroups of $\struct {R, +, \circ}$.
Let $S T$ be defined as:
:$\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$
Then:
:$\paren {S T} U = S \paren {T U}$
== Proof ==
We have by definition that $S T$ is made up of all finite sums of elements of the form $s \circ t$ where $s \in S, t \in T$.
From Sum of All Ring Products is Closed under Addition, this set is closed under ring addition.
Therefore, so are $\paren {S T} U$ and $S \paren {T U}$.
Let $z \in \paren {S T} U$.
Then $z$ is a finite sum of elements in the form $x \circ u$ where $x \in ST$ and $u \in U$.
So $x$ is a finite sum of elements in the form $s \circ t$ where $s \in S, t \in T$.
Therefore $z$ is a finite sum of elements in the form $\paren {s \circ t} \circ u$ where $s \in S, t \in T, u \in U$.
As $\struct {R, +, \circ}$ is a ring, $\circ$ is associative.
So $z$ is a finite sum of elements in the form $s \circ \paren {t \circ u}$ where $s \in S, t \in T, u \in U$.
So these elements all belong to $S \paren {T U}$.
Since $S \paren {T U}$ is closed under addition, $z \in S \paren {T U}$.
So:
:$\paren {S T} U \subseteq S \paren {T U}$
By a similar argument in the other direction:
:$S \paren {T U} \subseteq \paren {S T} U $
and so by definition of set equality:
:$\paren {S T} U = S \paren {T U}$
== Sources ==
* : $\S 2.1$: Subrings: Lemma $2.3 \ \text{(i)}$
Category:Ring Theory
Sum of All Ring Products is Closed under Addition

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +}$ and $\struct {T, +}$ be additive subgroups of $\struct {R, +, \circ}$.
Let $S T$ be defined as:
:$\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$
Then $\struct {S T, +}$ is a closed subset of $\struct {R, +}$.
== Proof ==
Let $x_1, x_2 \in S T$.
Then:
:$\ds x_1 = \sum_{i \mathop = 1}^j s_i \circ t_i, x_2 = \sum_{i \mathop = 1}^k s_i \circ t_i$
for some $s_i, t_i, j, k$, etc.
By renaming the indices, we can express $x_2$ as:
:$\ds x_2 = \sum_{i \mathop = j + 1}^{j + k} s_i \circ t_i$
and hence:
:$\ds x_1 + x_2 = \sum_{i \mathop = 1}^j s_i \circ t_i + \sum_{i \mathop = j + 1}^{j + k} s_i \circ t_i = \sum_{i \mathop = 1}^k s_i \circ t_i$
So $x_1 + x_2 \in S T$ and $\struct {S T, +}$ is shown to be closed.
== Sources ==
* : $\S 2.1$: Subrings: Lemma $2.3 \ \text{(i)}$
Category:Ring Theory
Sum of Alternating Sign Reciprocals of Sequence of Pairs of Consecutive Fibonacci Numbers is Reciprocal of Golden Mean Squared

== Theorem ==
where:
:$F_k$ denotes the $k$th Fibonacci number
:$\phi$ denotes the golden mean.
== Proof ==
== Historical Note ==
== Sources ==
* : $5$
* : $5$
Category:Fibonacci Numbers
Category:Golden Mean
Sum of Alternating Sign Reciprocals of Sequence of Pairs of Consecutive Fibonacci Numbers is Reciprocal of Golden Mean Squared/Historical Note

== Historical Note on Sum of Alternating Sign Reciprocals of Sequence of Pairs of Consecutive Fibonacci Numbers is Reciprocal of Golden Mean Squared ==
This result is attributed to .
== Sources ==
* : $5$
:::(although erroneously given as Schue)
* : $5$
Category:Historical Notes
Sum of Angles between Straight Lines at Point form Four Right Angles

== Corollary to Two Angles on Straight Line make Two Right Angles ==
If any number of straight lines are drawn from a given point, the sum of the consecutive angles so formed is $4$ right angles.
== Proof ==
Let $OA_1, OA_2, \ldots, OA_n$ be straight lines drawn from a point $O$ to points $A_1, A_2, \ldots, A_n$.
Let $OA_1$ be produced past $O$ to $B$.
Then $OB$ either coincides with $OA_j$ for some $j$ between $1$ and $n$, or $OB$ divides angle $A_j O A_k$ for some $j, k$ between $1$ and $n$.
First suppose $OB$ coincides with $OA_j$.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.1$: Corollary $1$
Category:Angles
Sum of Angles of Triangle equals Two Right Angles

== Theorem ==
In a triangle, the sum of the three interior angles equals two right angles.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $180$
* 
* : $180$
* : $(1)$
* : $(1)$
* : Chapter $2$: The Logic of Shape: Euclid
* 
* : $\S 1$: What Is Curvature? The Euclidean Plane
* 
Category:Sum of Angles of Triangle equals Two Right Angles
Category:Triangles
Sum of Angles of Triangle equals Two Right Angles/Proof 1

== Theorem ==
== Proof ==
:
Let $\triangle ABC$ be a triangle.
Let $BC$ be extended to a point $D$.
From External Angle of Triangle equals Sum of other Internal Angles:
: $\angle ACD = \angle ABC + \angle BAC$
Bby by Euclid's Second Common Notion:
: $\angle ACB + \angle ACD = \angle ABC + \angle BAC + \angle ACB$
But from Two Angles on Straight Line make Two Right Angles, $\angle ACB + \angle ACD$ equals two right angles.
So by Euclid's First Common Notion, $\angle ABC + \angle BAC + \angle ACB$ equals two right angles.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Sum of Angles of Triangle equals Two Right Angles
Sum of Angles of Triangle equals Two Right Angles/Proof 2

== Theorem ==
== Proof ==
:
Let $\Delta ABC$ be a triangle.
Let $DAE$ be a line such that $DE \parallel BC$.
By Parallelism implies Equal Alternate Angles:
:$\angle DAB = \angle ABC$
and:
:$\angle EAC = \angle ACB$
Therefore, the sum of the three angles is:
:$\angle ABC + \angle BCA + \angle CAB = \angle DAB + \angle BAC + \angle CAE = 180 \degrees$
Category:Sum of Angles of Triangle equals Two Right Angles
Sum of Antecedent and Consequent of Proportion

== Theorem ==
That is, if $a : b = c : d$ and $a$ is the greatest and $d$ is the least, then:
:$a + d > b + c$
== Proof ==
Let the four magnitudes $AB, CD, E, F$ be proportional, so that $AB : CD = E : F$.
Let $AB$ be the greatest and $F$ the least.
We need to show that $AB + F > CD + E$.
:
Let $AG = E, CH = F$.
We have that $AB : CD = E : F$, $AG = E, F = CH$.
So $AB : CD = AG : CH$.
So from Proportional Magnitudes have Proportional Remainders $GB : HD = AB : CD$.
But $AB > CD$ and so $GB > HD$.
Since $AG = E$ and $CH = F$, it follows that $AG + F = CH + E$.
We have that $GB > HD$.
So add $AG + F$ to $GB$ and $CH + E$ to $HD$.
It follows that $AB + F > CD + E$.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios