nyuuzyou/wikis huggingface dataset:
Sum of Internal Angles of Polygon/Proof 2

== Theorem ==
== Proof ==
This proof assumes that the polygon is convex.
Name a vertex as $A_1$, go clockwise and name the vertices as $A_2, A_3, \ldots, A_n$.
By joining $A_1$ to every vertex except $A_2$ and $A_n$, one can form $\paren {n - 2}$ triangles in a fan triangulation of the convex polygon.
From Sum of Angles of Triangle equals Two Right Angles, the sum of the internal angles of a triangle is $180 \degrees$.
Therefore, the sum of internal angles of a polygon with $n$ sides is $\paren {n - 2} 180 \degrees$.
Category:Sum of Internal Angles of Polygon
Sum of Larger Ideals is Larger

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $\mathfrak a_1, \mathfrak a_2, \mathfrak b_1, \mathfrak b_2$ be ideals of $R$ such that:
:$\mathfrak a_1 \subseteq \mathfrak a_2$
and:
:$\mathfrak b_1 \subseteq \mathfrak b_2$
Then their sums satisfy:
:$\mathfrak a_1 + \mathfrak b_1 \subseteq \mathfrak a_2 + \mathfrak b_2$
== Proof ==
Let $x \in \mathfrak a_1 + \mathfrak b_1$.
By , there are $a \in \mathfrak a_1$ and $a \in \mathfrak b_1$ such that:
:$x = a + b$
Then:
:$a \in \mathfrak a_1 \subseteq \mathfrak a_2$
and:
:$b \in \mathfrak b_1 \subseteq \mathfrak b_2$
Thus, by :
:$x \in \mathfrak a_2 + \mathfrak b_2$
Category:Radical of Ideals
Sum of Little-O Estimates/Sequences

== Theorem ==
Let $\sequence {a_n}, \sequence {b_n}, \sequence {c_n}, \sequence {d_n}$ be sequences of real or complex numbers.
Let:
:$a_n = \map \oo {b_n}$
:$c_n = \map \oo {d_n}$
where $\oo$ denotes little-$\oo$ notation.
Then:
:$a_n + c_n = \map \oo {\size {b_n} + \size {d_n} }$
== Proof ==
Let $\epsilon > 0$.
Then by definition of little-$\oo$ notation:
:$\exists n_1 \in \N: \paren {n \ge n_1 \implies \size {a_n} \le \epsilon \cdot \size {b_n}}$
:$\exists n_2 \in \N: \paren {n \ge n_2 \implies \size {c_n} \le \epsilon \cdot \size {d_n}}$
For $n \ge \max \set {n_1, n_2}$:
Hence by definition of little-$\oo$ notation:
:$a_n + c_n = \map \oo {\size {b_n} + \size {d_n} }$
Category:Little-O Notation
Sum of Logarithms/General Logarithm

== Theorem ==
Let $x, y, b \in \R$ be strictly positive real numbers such that $b > 1$.
Then:
:$\log_b x + \log_b y = \map {\log_b} {x y}$
where $\log_b$ denotes the logarithm to base $b$.
== Proof 1 ==
== Proof 2 ==
== Notes ==
If one presupposes Exponent Combination Laws, the proofs for logarithms to the general base can be used to directly prove the laws for $\log_e$.
Whether this would be circular is ultimately dependent on which definition of $e^x$ one chooses.
== Sources ==
* : $\S 7$: Laws of Logarithms: $7.10$
* : Chapter $1$: Computations Using Logarithms
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: $(11)$
Category:Sum of Logarithms
Sum of Logarithms/Natural Logarithm

== Theorem ==
Let $x, y \in \R$ be strictly positive real numbers.
Then:
:$\ln x + \ln y = \map \ln {x y}$
where $\ln$ denotes the natural logarithm.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Sources ==
* : Chapter $\text{II}$: Groups: Morphisms
Category:Sum of Logarithms
Sum of Logarithms/Natural Logarithm/Proof 1

== Theorem ==
== Proof ==
Let $y \in \R_{>0}$ be fixed.
Consider the function:
:$\map f x = \ln x y - \ln x$
From the definition of the natural logarithm, the Fundamental Theorem of Calculus and the Chain Rule for Derivatives:
:$\forall x > 0: \map {f'} x = \dfrac 1 {x y} y - \dfrac 1 x = \dfrac 1 x - \dfrac 1 x = 0$
Thus from Zero Derivative implies Constant Function, $f$ is constant:
:$\forall x > 0: \ln x y - \ln x = c$
To determine the value of $c$, put $x = 1$.
From Logarithm of 1 is 0:
:$\ln 1 = 0$
Thus:
:$c = \ln y - \ln 1 = \ln y$
and hence the result.
== Sources ==
* : $\S 14.2 \ \text{(i)}$
Category:Sum of Logarithms
Sum of Logarithms/Natural Logarithm/Proof 3

== Theorem ==
== Proof ==
Let $\sequence {f_n}$ be the sequence of mappings $f_n : \R_{>0} \to \R$ defined as:
:$\map {f_n} x = n \paren {\sqrt [n] x - 1}$
Let $\map M t = \max \set {\size {t - 1}, \size {\dfrac {t - 1} t} }$
From Bounds of Natural Logarithm:
:$\forall t \in \R_{>0}: \size {\map {f_n} t} \le \map M t$
Fix $x, y \in \R_{>0}$.
Then:
Thus:
:$\ds \lim_{n \mathop \to \infty} \map {f_n} {x y} = \lim_{n \mathop \to \infty} \paren {\map {f_n} x + \map {f_n} y}$
 
Hence the result, from the definition of $\ln$.
== Sources ==
Category:Sum of Logarithms
Sum of Logarithms/Natural Logarithm/Proof 4

== Theorem ==
== Proof ==
Recall the definition of the natural logarithm as the definite integral of the reciprocal function:
:$\ds \ln x := \int_1^x \frac {\d t} t$
:
Consider the diagram above.
The value of $\ln x$ is represented by the area between the points:
:$\tuple {1, 0}, \tuple {1, 1}, \tuple {x, \dfrac 1 x}, \tuple {x, 0}$
which is represented by the yellow region above.
Similarly, the value of $\ln y$ is represented by the area between the points:
:$\tuple {1, 0}, \tuple {1, 1}, \tuple {y, \dfrac 1 y}, \tuple {y, 0}$
Let the second of these areas be transformed by dividing its height by $x$ and multiplying its length by $x$.
This will preserve its area, while making it into the area between the points:
:$\tuple {x, 0}, \tuple {x, 1 / x}, \tuple {x y, \dfrac 1 {x y} }, \tuple {x y, 0}$
which is exactly the green area.
The total of the green and yellow areas represents the value of $\ln x y$.
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $23$
Category:Sum of Logarithms
Sum of Maximum and Minimum

== Theorem ==
For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
:$a + b = \max \set {a, b} + \min \set {a, b}$
== Proof ==
From the definitions of max and min:
:$\max \set {a, b} = \begin{cases}
b: & a \le b \\
a: & b \le a
\end{cases}$
and
:$\min \set {a, b} = \begin{cases}
a: & a \le b \\
b: & b \le a
\end{cases}$
Let $a < b$.
Then:
:$\max \set {a, b} + \min \set {a, b} = b + a$
Let $a > b$.
Then:
:$\max \set {a, b} + \min \set {a, b} = a + b$
Finally, let $a = b$.
Then:
:$\max \set {a, b} = \min \set {a, b} = a = b$
Hence:
:$\max \set {a, b} + \min \set {a, b} = 2a = 2b = a + b$
Note that this result does not apply to $a, b \in \C$ as there is no concept of ordering on the complex numbers $\C$.
Category:Algebra
Sum of Möbius Function over Divisors

== Theorem ==
Let $n \in \Z_{>0}$, i.e. let $n$ be a strictly positive integer.
Let $\ds \sum_{d \mathop \divides n}$ denote the sum over all of the divisors of $n$.
Let $\map \mu d$ be the Möbius function.
Then:
:$\ds \sum_{d \mathop \divides n} \map \mu d = \floor {\frac 1 n}$
where $\floor {\dfrac 1 n}$ is the floor of $\dfrac 1 n$.
== Proof ==
The theorem is clearly true if $n = 1$.
Assume, then, that $n > 1$ and write, by the Fundamental Theorem of Arithmetic:
:$n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}$
In the sum $\ds \sum_{d \mathop \divides n} \map \mu d$ the only non-zero terms come from $d = 1$ and the divisors of $n$ which are products of distinct primes.
From the definition of the Möbius function, the Möbius function of a product of $m$ distinct primes is $\paren {-1}^m$.
From the definition of Binomial coefficient, there are $\dbinom k m$ ways to choose $m$ primes from $p_1, p_2, \dots, p_k$ to multiply together.
So among the divisors of $p_1 p_2 \dotsm p_k$, there are exactly $\dbinom k m$ numbers that are the product of $m$ distinct primes.
Thus:
Hence, the sum is $1$ for $n = 1$, and $0$ for $n > 1$, which are precisely the values of $\floor {\dfrac 1 n}$.
== Examples ==
== Sources ==
* : $2.2$: The Möbius function $\map \mu n$
Category:Sum of Möbius Function over Divisors
Category:Möbius Function
Sum of Non-Consecutive Fibonacci Numbers

== Theorem ==
Let $S$ be a non-empty set of distinct non-consecutive Fibonacci numbers not containing $F_0$ or $F_1$.
Let the largest element of $S$ be $F_j$.
Then:
:$\ds \sum_{F_i \mathop \in S} F_i < F_{j + 1}$
That is, the sum of all the elements of $S$ is strictly less than the next largest Fibonacci number.
That is, given some increasing sequence $\sequence {c_i}$ satisfying $c_i \ge 2$ and $c_{i + 1} \ge c_i + 1$:
:$\ds F_{c_k + 1} > \sum_{i \mathop = 0}^k F_{c_i}$
== Proof ==
The proof proceeds by induction on $j$ for $j \ge 2$.
For all $j \in \N_{>0}$, let $\map P j$ be the proposition:
:$\ds \sum_{F_i \mathop \in S} F_i < F_{j + 1}$
Let the term allowable set be used to mean a non-empty set of distinct non-consecutive Fibonacci numbers not containing $F_0$ or $F_1$.
=== Basis for the Induction ===
The only possible allowable set whose largest member is $F_2 = 1$ is the set:
:$\set {F_2} = \set 1$
This has sum $1$, which is strictly less than $F_3 = 2$.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \sum_{F_i \mathop \in S} F_i < F_{k + 1}$
where $S$ is an allowable set whose largest element is $F_k$.
Then we need to show:
:$\ds \sum_{F_i \mathop \in S} F_i < F_{k + 2}$
where $S$ is an allowable set whose largest element is $F_{k + 1}$.
=== Induction Step ===
This is our induction step:
Let $S$ be an allowable set whose greatest element is $F_{k + 1}$.
Let $S' := S \setminus \set {F_{k + 1} }$.
That is, $S'$ is $S$ without its largest element $F_{k + 1}$.
We have that $S$ cannot contain consecutive Fibonacci numbers.
Hence the largest element of $S'$ is at most $F_{k - 1}$.
So by the induction hypothesis:
:$\ds \sum S' < F_k$
Then:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall j \in \N_{> 0}: \sum_{F_i \mathop \in S} F_i < F_{j + 1}$
Category:Fibonacci Numbers
Sum of Odd Digits equal to Sum of Even Digits

== Puzzle ==
:Place the $9$ digits,
:so that the sum of the odd digits may be equal to the sum of the even digits.
== Solution ==
:$2 + 4 + 6 + .8 = 12.8 = 1 + 3 + 7 + \dfrac 9 5$
== Sources ==
* 
* : Rational Amusements for Winter Evenings: $148$
Category:Trick Puzzles
Sum of Odd Index Binomial Coefficients

== Theorem ==
:$\ds \sum_{i \mathop \ge 0} \binom n {2 i + 1} = 2^{n - 1}$
where $\dbinom n i$ is a binomial coefficient.
That is:
:$\dbinom n 1 + \dbinom n 3 + \dbinom n 5 + \dotsb = 2^{n - 1}$
== Proof ==
From Sum of Binomial Coefficients over Lower Index we have:
:$\ds \sum_{i \mathop \in \Z} \binom n i = 2^n$
That is:
:$\dbinom n 0 + \dbinom n 1 + \dbinom n 2 + \dbinom n 3 + \cdots + \dbinom n n = 2^n$
as $\dbinom n i = 0$ for $i < 0$ and $i > n$.
This can be written more conveniently as:
:$(1): \quad \dbinom n 0 + \dbinom n 1 + \dbinom n 2 + \dbinom n 3 + \dbinom n 4 + \cdots = 2^n$
Similarly, from Alternating Sum and Difference of Binomial Coefficients for Given n we have:
:$\ds \sum_{i \mathop \in \Z} \paren {-1}^i \binom n i = 0$
That is:
:$(2): \quad \dbinom n 0 - \dbinom n 1 + \dbinom n 2 - \dbinom n 3 + \dbinom n 4 - \cdots = 0$
Subtracting $(2)$ from $(1)$, we get:
:$2 \dbinom n 1 + 2 \dbinom n 3 + 2 \dbinom n 5 + \cdots = 2^n$
as the even index coefficients cancel out.
Dividing by $2$ throughout gives us the result.
== Also see ==
* Sum of Even Index Binomial Coefficients
== Sources ==
* : $\text I$. Algebra: The Binomial Theorem: Relations between coeffficients
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.11$
* : $\text {3-1}$ Permutations and Combinations: Exercise $7$
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem: Problems $1.2$: $3 \ \text{(e)}$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.11.$
Category:Binomial Coefficients
Sum of Odd Number of Odd Numbers is Odd

== Theorem ==
== Proof ==
Let $S = \set {r_1, r_2, \ldots, r_n}$ be a set of $n$ odd numbers, where $n = 2 m + 1$.
By definition of odd number, this can be expressed as:
:$S = \set {2 s_1 + 1, 2 s_2 + 1, \ldots, 2 s_n + 1}$
where:
:$\forall k \in \closedint 1 n: r_k = 2 s_k + 1$
Then:
Thus, by definition, $\ds \sum_{k \mathop = 1}^n r_k$ is odd.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Sum of Odd Positive Powers

== Theorem ==
Let $n \in \N$ be an odd positive integer.
Let $x, y \in \Z_{>0}$ be (strictly) positive integers.
Then $x + y$ is a divisor of $x^n + y^n$.
== Proof ==
Given that $n \in \N$ be odd, it can be expressed in the form:
:$n = 2 m + 1$
where $m \in \N$.
The proof proceeds by strong induction.
For all $m \in \N$, let $\map P m$ be the proposition:
: $x^{2 m + 1} + y^{2 m + 1} = \paren {x + y} \paren {x^{2 m} + \cdots + y^{2 m} }$
$\map P 0$ is the case:
: $x + y = x + y$
which is trivially an identity.
=== Basis for the Induction ===
$\map P 1$ is the case:
:$x^3 + y^3 = \paren {x + y} \paren {x^2 - x y + y^2}$
which follows by Sum of Two Cubes.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\exists k \in \N: \forall j: 1 \le j \le k: x^{2 j + 1} + y^{2 j + 1} = \paren {x + y} \map {P_{2 j} } {x, y}$
where $\map {P_{2 j} } {x, y}$ is a polynomial of degree $2 j$ in $x$ and $y$.
from which it is to be shown that:
:$x^{2 k + 3} + y^{2 k + 3} = \paren {x + y} \map {P_{2 k + 2} } {x, y}$
where $\map {P_{2 k + 2} } {x, y}$ is a polynomial of degree $2 k + 2$ in $x$ and $y$.
=== Induction Step ===
This is the induction step:
We have that:
:$\paren {x^{2 k + 1} + y^{2 k + 1} } \paren {x^2 + y^2} = x^{2 k + 3} + y^{2 k + 3} + x^2 y^{2 k + 1} + x^{2 k + 1} y^2$
So:
But $\paren {x^{2 k - 1} + y^{2 k - 1} }$ itself is of the form:
:$\paren {x + y} \map {P_{2 k - 2} } {x, y}$
So:
:$x^{2 k + 3} + y^{2 k + 3} = \paren {x + y} \paren {\paren {x^2 + y^2} \map {P_{2 k} } {x, y} - x^2 y^2 \map {P_{2 k - 2} } {x, y} }$
Hence the result by the Second Principle of Mathematical Induction.
== Sources ==
* : $\text {1-1}$ Principle of Mathematical Induction: Exercise $18$
Category:Number Theory
Sum of Odd Sequence of Products of Consecutive Fibonacci Numbers

== Theorem ==
Let $F_k$ be the $k$'th Fibonacci number.
Then:
:$\ds \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} = {F_{2 n} }^2$
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} = {F_{2 n} }^2$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says $F_1 F_2 = 1 \times 1 = 1 = {F_2}^2$.
This is our basis for the induction.
=== 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:
:$\ds \sum_{j \mathop = 1}^{2 k - 1} F_j F_{j + 1} = {F_{2 k} }^2$
Then we need to show:
:$\ds \sum_{j \mathop = 1}^{2 k + 1} F_j F_{j + 1} = {F_{2 \paren {k + 1} } }^2$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \ge 1: \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} = {F_{2 n} }^2$
== Sources ==
* : $\text {1-1}$ Principle of Mathematical Induction: Exercise $11$
Category:Sums of Sequences
Category:Fibonacci Numbers
Category:Proofs by Induction
Sum of Pair of Elements of Geometric Sequence with Three Elements in Lowest Terms is Coprime to other Element

== Theorem ==
Let $P = \tuple {a, b, c}$ be a geometric sequence of integers in its lowest terms.
Then $\paren {a + b}$, $\paren {b + c}$ and $\paren {a + c}$ are all coprime to each of $a$, $b$ and $c$.
== Proof ==
Let the common ratio of $P$ in canonical form be $\dfrac q p$.
By Form of Geometric Sequence of Integers in Lowest Terms:
:$P = \tuple {p^2, p q, q^2}$
Then:
Similarly:
Then:
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Geometric Sequences
Sum of Pandigital Triplet of 3-Digit Primes

== Theorem ==
The smallest integer which is the sum of a set of $3$ three-digit primes using all $9$ digits from $1$ to $9$ once each is $999$:
:$149 + 263 + 587 = 999$
== Proof ==
All three-digit primes end in $1, 3, 7, 9$.
Suppose $1$ is used as the units digit of a prime.
Since the digit $1$ cannot be used again, the sum of the primes is at least:
:$221 + 333 + 447 = 1001$
so $1$ cannot be used as a units digit .
The units digits of the primes are $3, 7, 9$.
To minimise the sum, the hundreds digits must be $1, 2, 4$.
This leaves $5, 6, 8$ be the tens digits.
The primes satisfying these conditions are:
:$157, 163, 167$
:$257, 263, 269, 283$
:$457, 463, 467, 487$
Only $269$ contain $9$, so we must choose it.
Only $157$ does not contain $6$, so must choose it next.
But then all primes beginning with $4$ have some digit coinciding with the above primes.
Hence the next minimal sum of the primes (if they exist) is:
:$10^2 \paren {1 + 2 + 5} + 10 \paren {4 + 6 + 8} + \paren {3 + 7 + 9} = 999$
and we have shown that these primes do exist.
== Sources ==
* : $999$
* : $999$
Category:Prime Numbers
Category:999
Sum of Positive and Negative Parts

== Theorem ==
Let $X$ be a set, and let $f: X \to \overline \R$ be an extended real-valued function.
Let $f^+, f^-: X \to \overline \R$ be the positive and negative parts of $f$, respectively.
Then $\size {f} = f^+ + f^-$, where $\size {f}$ is the absolute value of $f$.
== Proof ==
Let $x \in X$.
Suppose that $\map f x \ge 0$, where $\ge$ signifies the extended real ordering.
Then $\size {\map f x} = \map f x$, and:
:$\map {f^+} x = \map \max {\map f x, 0} = \map f x$
:$\map {f^-} x = - \map \min {\map f x, 0} = 0$
Hence $\map {f^+} x + \map {f^-} x = \map f x = \size {\map f x}$.
Next, suppose that $\map f x < 0$, again in the extended real ordering.
Then $\size {\map f x} = - \map f x$, and:
:$\map {f^+} x = \map \max {\map f x, 0} = 0$
:$\map {f^-} x = - \map \min {\map f x, 0} = - \map f x$
Hence $\map {f^+} x + \map {f^-} x = - \map f x = \size {\map f x}$.
Thus, for all $x \in X$:
:$\map {f^+} x + \map {f^-} x = \size {\map f x}$
That is, $f^+ + f^- = \size {f}$.
== Sources ==
* : $8.7 \ \text{(v)}$, $\S 8$: Problem $6$
Category:Mapping Theory
Category:Positive Parts
Category:Negative Parts
Sum of Powers of 2/Proof 2

== Theorem ==
== Proof ==
Let $S \subseteq \N_{>0}$ denote the set of (strictly positive) natural numbers for which $(1)$ holds.
=== Basis for the Induction ===
We have:
So $1 \in S$.
This is our basis for the induction.
=== Induction Hypothesis ===
We now show that, if $k \in S$ is true, where $k \ge 1$, then it logically follows that $k + 1 \in S$.
So this is our induction hypothesis:
:$\ds \sum_{j \mathop = 0}^{k - 1} 2^j = 2^k - 1$
Then we need to show:
:$\ds \sum_{j \mathop = 0}^k 2^j = 2^{k + 1} - 1$
=== Induction Step ===
This is our induction step:
So $k \in S \implies k + 1 \in S$.
It follows by the Principle of Finite Induction that $S = \N_{>0}$.
That is:
:$\ds \forall n \in \N_{> 0}: \sum_{j \mathop = 0}^{n - 1} 2^j = 2^n - 1$
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction
Category:Sum of Powers of 2
Category:Proofs by Induction
Sum of Powers of Primitive Complex Roots of Unity

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $U_n$ denote the complex $n$th roots of unity:
:$U_n = \set {z \in \C: z^n = 1}$
Let $\alpha = \exp \paren {\dfrac {2 k \pi i} n}$ denote a primitive complex $n$th root of unity.
Let $s \in \Z_{>0}$ be a (strictly) positive integer.
Then:
where:
:$n \divides s$ denotes that $n$ is a divisor of $s$
:$n \nmid s$ denotes that $n$ is not a divisor of $s$.
== Proof ==
First we address the case where $n \divides s$.
Then:
Hence:
Now let $n \nmid s$.
Then:
Hence:
== Sources ==
* : $\S 3$. Roots of Unity: Example $3$.
Category:Complex Roots of Unity
Sum of Products of nth Roots of Unity taken up to n-1 at a Time is Zero

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $U_n = \set {z \in \C: z^n = 1}$ be the complex $n$th roots of unity.
Then the sum of the products of the elements of $U_n$ taken $2, 3, 4, \dotsc n - 1$ at a time is zero.
== Proof ==
The elements of $U_n = \set {z \in \C: z^n = 1}$ are the solutions to the equation:
:$z^n - 1 = 0$
Thus by definition the coefficients of the powers of $z$:
:$z^2, z^3, \ldots, z^{n - 1}$
are all zero.
The result follows directly from Viète's Formulas.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The $n$th Roots of Unity: $108$
Category:Viète's Formulas
Category:Complex Roots of Unity
Sum of Projections/Binary Case

== Theorem ==
Let $H$ be a Hilbert space.
Let $P, Q$ be projections.
Then $P + Q$ is a projection $\Rng P \perp \Rng Q$.
Here, $\Rng P$ denotes range, and $\perp$ denotes orthogonality.
== Proof ==
=== Necessary Condition ===
Suppose $P + Q$ is a projection. Then:
Now suppose that $h \in \Rng Q$; say $h = Q q$ for $q \in H$.
Then it follows that $Q h = Q Q q = Q q = h$ as $Q$ is idempotent.
It follows that:
:$0 = P Q h + Q P h = P h + Q P h$
From Characterization of Projections, statement $(6)$, $\innerprod {Q P h} {P h}_H \ge 0$.
Next, observe:
:$0 = \innerprod {P h + Q p h} {P h}_H = \innerprod {P h} {P h}_H + \innerprod {Q P h} {P h}_H$
As the second term is non-negative, the first is non-positive; it follows that $P h = \mathbf 0_H$ from the definition of the inner product.
Hence:
:$h \in \ker P = \paren {\Rng P}^\perp$
It follows that $\Rng Q \perp \Rng P$, as asserted.
=== Sufficient Condition ===
Suppose that $\Rng P \perp \Rng Q$.
Then as $P, Q$ are projections, have:
:$\Rng P \subseteq \ker Q$
:$\Rng Q \subseteq \ker P$
That is, for all $h \in H$:
:$Q P h = P Q h = \mathbf 0_H$
Hence:
That is, $P + Q$ is an idempotent.
Furthermore, by Adjoining is Linear, have:
:$\paren {P + Q}^* = P^* + Q^* = P + Q$
where the latter follows from Characterization of Projections, statement $(4)$.
This same statement implies that $P + Q$ is also a projection.
== Also see ==
* Product of Projections
* Difference of Projections
== Sources ==
* $\text {II}.3$: Exercise $4$
Category:Linear Transformations on Hilbert Spaces
Sum of Projections/General Case

== Theorem ==
Let $H$ be a Hilbert space.
Let $\family {M_i}_{i \mathop \in I}$ be an $I$-indexed set of closed linear subspaces of $H$.
Let $M_i$ and $M_j$ be orthogonal whenever $i \ne j$.
Denote, for each $i \in I$, by $P_i$ the orthogonal projection onto $M_i$.
Denote by $P$ the orthogonal projection onto the closed linear span $\vee \set {M_i: i \in I}$ of the $M_i$.
Then for all $h \in H$:
:$\ds \sum \set {P_i h: i \in I} = P h$
where $\ds \sum$ denotes a generalized sum.
== Proof ==
== Also see ==
* Product of Projections
* Difference of Projections
== Sources ==
* $\text {II}.3$ Exercise $5$
* $\text {II}.7.1$
Category:Linear Transformations on Hilbert Spaces
Sum of Quaternion Conjugates

== Theorem ==
Let $\mathbf x, \mathbf y \in \mathbb H$ be quaternions.
Let $\overline {\mathbf x}$ be the conjugate of $\mathbf x$.
Then:
:$\overline {\mathbf x + \mathbf y} = \overline {\mathbf x} + \overline {\mathbf y}$
== Proof ==
Let:
:$\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$
:$\mathbf y = e \mathbf 1 + f \mathbf i + g \mathbf j + h \mathbf k$
Then:
Category:Quaternions
Sum of Quotients of Real Numbers

== Theorem ==
:$\forall x, w \in \R, y, z \in \R_{\ne 0}: \dfrac x y + \dfrac w z = \dfrac {\paren {x \times z} + \paren {y \times w} } {y \times z}$
== Proof ==
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(o)}$
Category:Real Addition
Category:Real Division