nyuuzyou/wikis huggingface dataset:
Even Functions in 2-Lebesgue Space form Closed Subspace of 2-Lebesgue Space

== Theorem ==
Let $\map {\LL^2} \R$ be the Lebesgue $2$-space.
For all $f \in \map {\LL^2} \R$ denote $\map {\check f} x = \map f {-x}$.
Let $Y = \set {f \in \map {\LL^2} \R : f = \check f}$ be the set of all even functions in $\map {\LL^2} \R$.
Then $Y$ is a closed subspace of $\map {\LL^2} \R$.
== Proof ==
Let $R : \map {\LL^2} \R \to \map {\LL^2} \R$ be the reflection mapping such that:
:$\map R f = \check f$
Then:
Furthermore:
Thus, $R$ is a linear transformation:
:$R \in \map L {\map {\LL^2} \R}$.
Moreover, because the reflection $y$-axis preserves the total area under the curve $\map f x$:
:$\forall f \in \map {\LL^2} \R : \norm {f}_2 = \norm {\check f}_2$
By Continuity of Linear Transformation between Normed Vector Spaces, $R$ is continuous.
Let $I : \map {\LL^2} \R \to \map {\LL^2} \R$ be the identity map:
:$\map I f = f$
By Identity Mapping on Normed Vector Space is Bounded Linear Operator, $I$ is a linear transformation:
:$I \in \map L {\map {\LL^2} \R}$
Moreover:
:$\forall f \in \map {\LL^2} \R : \norm {f}_2 = \norm {f}_2$
By Continuity of Linear Transformation between Normed Vector Spaces, $I$ is continuous.
By Linear Mappings between Vector Spaces form Vector Space, $I - R$ is a linear transformation:
:$I - R \in \map L {\map {\LL^2} \R}$
By definition of pointwise addition of linear operators:
:$\map {\paren {I - R} } f = f - \check f$
Then:
By Continuity of Linear Transformation between Normed Vector Spaces, $I - R$ is continuous.
Suppose $f$ is even.
Then:
:$\check f = f$.
It follows that:
:$\map {\paren {I - R} } f = 0$.
So the set of even functions is the kernel of $I - R$:
:$Y = \map \ker {I - R}$
We have that $I - R$ is continuous.
By Singleton in Normed Vector Space is Closed, $\set 0$ is closed in $\map {\LL^2} \R$.
$Y$ is the inverse image of $\set 0$ under $I - R$.
By Mapping is Continuous iff Inverse Images of Closed Sets are Closed, $Y$ is closed.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$
Category:Even Functions
Category:Lebesgue Spaces
Category:Examples of Vector Subspaces
Even Impulse Pair is Fourier Transform of Cosine Function

== Theorem ==
Consider the (real) cosine function $\map \cos x: \R \to \R$.
:$\map f x = \map \cos {\pi x}$
Then:
where:
:$\map {\hat f} s$ is the Fourier transform of $\map f x$.
:$\operatorname {II}$ denotes the even impulse pair function.
== Proof ==
By the definition of a Fourier transform:
== Sources ==
* : Chapter $1$: Introduction
* 
Category:Cosine Function
Category:Even Impulse Pair Function
Category:Examples of Fourier Transforms
Even Integer Plus 5 is Odd/Direct Proof

== Theorem ==
== Direct Proof ==
Let $x$ be an even integer.
Then by definition:
:$x = 2 n$
for some integer $n$.
Let $y = 2 n + 5$.
Then:
Hence $y = 2 n + 5$ is an odd integer by definition.
== Historical Note ==
== Sources ==
* : Appendix $\text{A}.5$: Theorems and Proofs: Example $\text A.3$
Category:Even Integer Plus 5 is Odd
Even Integer Plus 5 is Odd/Historical Note

== Historical Note on Even Integer Plus 5 is Odd ==
There is nothing profound about this result.
 used it as a simple demonstration of the construction of various kinds of proof in his of $1977$.
It is questionable whether the indirect proof and the Proof by Contradiction actually constitute different proofs of this result, but both are included on anyway, in case they are found to be instructional.
He sets a similar theorem as an exercise:
:Prove the implication "If $x$ is an odd integer, then $y = x - 3$ is an even integer" using the three proof techniques: ...
but it has been considered not sufficiently different from this one to be actually included on as a separate result to be proved.
For similar reasons, several other of the trivial exercises in applied logic that he sets have also been omitted from this site.
== Sources ==
* : Appendix $\text{A}.5$: Theorems and Proofs: Problem Set $\text A.5$: $27$
Category:Historical Notes
Even Integer Plus 5 is Odd/Indirect Proof

== Theorem ==
== Proof ==
Let $x$ be an even integer.
Let $y = 2 n + 5$.
Assume $y = x + 5$ is not an odd integer.
Then:
:$y = x + 5 = 2 n$
where $n \in \Z$.
Then:
Hence $x$ is odd.
That is, it is false that $x$ is even.
It follows by the Rule of Transposition that if $x$ is even, then $y$ is odd.
== Historical Note ==
== Sources ==
* : Appendix $\text{A}.5$: Theorems and Proofs: Example $\text A.3$
Category:Even Integer Plus 5 is Odd
Even Integer Plus 5 is Odd/Proof by Contradiction

== Theorem ==
== Proof ==
Let $x$ be an even integer.
Then by definition:
:$x = 2 n$
for some integer $n$.
 $y = x + 5 = 2 m$ for some integer $m$.
Then:
Hence $x$ is odd.
But this contradicts our premise that $x$ is even.
Hence the result by Proof by Contradiction.
== Historical Note ==
== Sources ==
* : Appendix $\text{A}.5$: Theorems and Proofs: Example $\text A.3$
Category:Even Integer Plus 5 is Odd
Even Integer with Abundancy Index greater than 9

== Theorem ==
Let $n \in \Z_{>0}$ have an abundancy index greater than $9$.
Then $n$ has at least $35$ distinct prime factors.
== Proof ==
As Divisor Sum Function is Multiplicative, it follows easily that abundancy index is multiplicative as well.
We have for any prime $p$ and positive integer $k$:
In fact this is the limit of the abundancy index of a prime power.
The greater the prime, the smaller this value is.
Therefore finding the least number of prime factors a number needs to have its abundancy index exceed $9$ is to find the least $m$ such that:
:$\ds \prod_{i \mathop = 1}^m \frac {p_i} {p_i - 1} > 9$
where $p_i$ is the $i$th prime ordered by size.
The result can be verified by direct computation.
== Sources ==
* 
* : $55$
Category:Abundancy
Even Integer with Abundancy Index greater than 9/Mistake

== Source Work ==
:
:The Dictionary
::$55$
== Mistake ==
:Every even number $n$ whose abundancy index, defined to equal $\map \sigma n / n$, is greater than $9$, has at least $55$ distinct prime factors.
== Correction ==
The author appears to have confused the numbers of distinct prime factors for abundancy indices over $9$ and $10$.
The correct statements are:
:Every even number $n$ whose abundancy index ... is greater than $9$, has at least $35$ distinct prime factors
and:
:Every even number $n$ whose abundancy index ... is greater than $10$, has at least $55$ distinct prime factors.
== Sources ==
* : $55$
Category:Mistakes/Curious and Interesting Numbers
Even Integers do not form Integral Domain

== Theorem ==
Let $2 \Z$ be the set of even integers.
Then $\struct {2 \Z, +, \times}$ is not an integral domain.
== Proof ==
From Integer Multiples form Commutative Ring, $\struct {2 \Z, +, \times}$ is a commutative ring.
As $2 \ne 1$, we also have from Integer Multiples form Commutative Ring that $\struct {2 \Z, +, \times}$ has no unity.
Hence by definition it is not an integral domain.
== Sources ==
* : Exercises: Chapter $1$: Exercise $1 \ \text{(i)}$
Category:Integers
Category:Integral Domains
Even Integers form Commutative Ring

== Theorem ==
Let $2 \Z$ be the set of even integers.
Then $\struct {2 \Z, +, \times}$ is a commutative ring.
However, $\struct {2 \Z, +, \times}$ is not an integral domain.
== Proof ==
From Integer Multiples form Commutative Ring, $\struct {2 \Z, +, \times}$ is a commutative ring.
As $2 \ne 1$, we also have from Integer Multiples form Commutative Ring that $\struct {2 \Z, +, \times}$ has no unity.
Hence by definition it is not an integral domain.
== Sources ==
* : Chapter $5$: Rings: $\S 18$. Definition of a Ring: Example $28$
Category:Integers
Category:Commutative Rings
Even Integers not Expressible as Sum of 3, 5 or 7 with Prime

== Theorem ==
The even integers that cannot be expressed as the sum of $2$ prime numbers where one of those primes is $3$, $5$ or $7$ begins:
:$98, 122, 124, 126, 128, 148, 150, \ldots$
== Proof ==
These are the primes which coincide with the upper end of a prime gap greater than $6$.
These can be found at:
:$89$ to $97$: prime gap of $8$
:$113$ to $127$: prime gap of $14$
:$139$ to $149$: prime gap of $10$
and so on.
We have that:
and so on.
== Sources ==
* : $98$
Category:98
Category:Prime Numbers
Even Integers not Sum of 2 Twin Primes

== Theorem ==
The following even integers cannot be expressed as the sum of $2$ prime numbers which are each one of a pair of twin primes:
:$2, 4, 94, 96, 98, 400, 402, 404, \ldots$
== Proof ==
== Sources ==
* : $94$
* : $94$
Category:Twin Primes
Category:94
Even Integers not Sum of Two Abundant Numbers

== Theorem ==
The even integers which are not the sum of $2$ abundant numbers are:
:All even integers less than $24$;
:$26, 28, 34, 46$
== Proof ==
From Sequence of Abundant Numbers, the first few abundant numbers are:
:$12, 18, 20, 24, 30, 36, 40, 42, 48$
Immediately we see that any number less than $2 \times 12 = 24$ cannot be expressed as a sum of $2$ abundant numbers.
The sum of the first $2$ abundant numbers is $12 + 18 = 30$, so $26$ and $28$ are not sums of $2$ abundant numbers.
 
Since none of the differences above are abundant numbers, $34$ and $46$ are not sums of $2$ abundant numbers.
We demonstrate that $32$ and all even numbers from $36$ to $66$ except $46$ are sums of $2$ abundant numbers:
The numbers $36, 42, 48, 54, 60$ and $66$ are multiples of $6$.
By Multiple of Perfect Number is Abundant, any multiple of $6$ greater than $6$ is abundant.
Hence these numbers can be expressed as:
:$12 + \paren {n - 12}$
which are sums of $2$ multiples of $6$ greater than $6$.
Now that we show that all even numbers greater than $66$ are sums of $2$ abundant numbers.
By Multiple of Abundant Number is Abundant, any multiple of $20$ is abundant.
By Largest Number not Expressible as Sum of Multiples of Coprime Integers/Generalization, any even number greater than:
:$\dfrac {6 \times 20} {\gcd \set {6, 20}} - 6 - 20 = 34$
is a sum of (possibly zero) multiples of $6$ and $20$.
Hence any even number greater than:
:$34 + 6 \times 2 + 20 = 66$
is a sum of a multiple of $6$ greater than $6$ and a non-zero multiple of $20$, which by above is a sum of $2$ abundant numbers.
This shows that the list above is complete.
== Also see ==
* Largest Integer not Sum of Two Abundant Numbers
== Sources ==
* 
* 
Category:Abundant Numbers
Even Multiple Angle Formula for Cosine

== Theorem ==
:$\ds \cos 2 n \theta = \paren {-1}^n \prod_{k \mathop = 1}^n \paren {1 - \frac {\cos^2 \theta} {\map {\cos^2} {\frac {\paren {2 k - 1} \pi} {4 n} } } }$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $162$
Category:Cosine Function
Even Natural Numbers are Infinite

== Theorem ==
The set of even natural numbers is infinite.
== Proof ==
Let $E$ denote the set of even natural numbers.
 $E$ is finite.
Then there exists $n \in \N$ such that $E$ has $n$ elements.
Let $m$ be the greatest element of $E$.
But then $m + 2$ is an even natural number.
But $m + 2 > m$, and $m$ is the greatest element of $E$.
Therefore $m + 2$ is an even natural number that is not an element of $E$.
So $E$ does not contain all the even natural numbers.
From that contradiction it follows by Proof by Contradiction that $E$ is not finite.
== Sources ==
* : Chapter $1$: General Background: $\S 1$ What is infinity?
Category:Even Integers
Category:Infinite Sets
Even Number minus Even Number is Even

== Theorem ==
== Proof ==
Let $a$ and $b$ be even numbers.
Then by definition of even number:
:$\exists c \in \Z: a = 2 c$
:$\exists d \in \Z: b = 2 d$
So:
Hence the result by definition of even number.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Even Number minus Odd Number is Odd

== Theorem ==
== Proof ==
Let $a$ be even and $b$ be odd.
Then by definition of even number:
:$\exists c \in \Z: a = 2 c$
and by definition of odd number:
:$\exists d \in \Z: b = 2 d + 1$
So:
Hence the result by definition of odd number.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Even Order Derivative of Odd Function Vanishes at Zero

== Theorem ==
Let $X$ be a symmetric subset of $\R$ containing $0$. 
Let $n$ be a positive integer. 
Let $f:X \to \R$ be an odd function. 
Let $f$ be at least $\paren {2 n}$-times differentiable.
Then: 
:$\map {f^{\paren {2 n} } } 0 = 0$
== Proof ==
From the definition of an odd function, for all $x \in X$ we have:
:$\map f x = -\map f {-x}$
Differentiating $2 n$ times, we have, by the Chain Rule for Derivatives:
:$\map {f^{\paren {2 n} } } x = -\paren {-1}^{2 n} \map {f^{\paren {2 n} } } {-x} = -\map {f^{\paren {2 n} } } {-x}$
Setting $x = 0$ gives: 
:$\map {f^{\paren {2 n} } } 0 = -\map {f^{\paren {2 n} } } {- 0}$ 
That is:
:$2 \map {f^{\paren {2 n} } } 0 = 0$
The result follows on division by $2$. 
Category:Differential Calculus
Category:Odd Functions
Even Order Group has Odd Number of Order 2 Elements

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $G$ be of even order.
Then $G$ has an odd number of elements of order $2$.
== Proof ==
Let $S = \set {x \in G: \order x > 2}$ be the set of all elements of $G$ whose order is strictly greater than $2$.
Let $h \in S$.
Then from Order of Group Element equals Order of Inverse:
:$h^{-1} \in S$.
Because $\order h > 2$ it follows that $h^{-1} \ne h$ by definition of self-inverse.
Thus every element in $S$ can be paired off with its inverse which is also in $S$.
It follows that $S$ has an even number of elements.
Thus $G \setminus S$ also contains an even number of elements.
From Identity is Only Group Element of Order 1, there is precisely $1$ element of $G$ whose order is less than $2$.
This leaves $G$ with an odd number of elements of order $2$.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: Examples: $(10)$
Category:Finite Groups
Category:Order of Group Elements
Even Order Group has Order 2 Element

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $G$ be of even order.
Then:
:$\exists x \in G: \order x = 2$
That is:
:$\exists x \in G: x \ne e: x^2 = e$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.12 \ \text{(a)}$
* : $\S 1.9$: Exercise $5.7$
* : Chapter $2$: Subgroups and Cosets: $\S 38 \delta$
* : $\S 1.1$: Exercise $14$
* : Chapter $6$: An Introduction to Groups: Exercise $10$
* : $\S 3.4$: Cyclic groups: Exercise $8$
Category:Finite Groups
Category:Order of Group Elements
Category:Even Order Group has Order 2 Element
Even Order Group has Order 2 Element/Proof 1

== Theorem ==
== Proof ==
In any group $G$, the identity element $e$ is self-inverse with Identity is Only Group Element of Order 1, and is the only such.
That leaves an odd number of elements.
Each element in $x \in G: \order x > 2$ can be paired off with its inverse, as $\order {x^{-1} } = \order x > 2$ from Order of Group Element equals Order of Inverse.
Hence there must be at least one element which has not been paired off with any of the others which is therefore self-inverse.
The result follows from Group Element is Self-Inverse iff Order 2.
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.13$
Category:Even Order Group has Order 2 Element
Even Perfect Number except 6 is Congruent to 1 Modulo 9

== Theorem ==
Let $n$ be an even perfect number, but not $6$.
Then:
:$n \equiv 1 \pmod 9$
== Proof ==
From Theorem of Even Perfect Numbers:
:$n = 2^{p - 1} \paren {2^p - 1} = \dfrac {2^p \paren {2^p - 1} } 2$
where $p$ is prime.
From Odd Power of 2 is Congruent to 2 Modulo 3:
:$2^p \equiv 2 \pmod 3$
for odd $p$.
Thus:
So $n$ is $1$ more than $9$ times the $k$th triangular number for some $k$.
That is:
:$n \equiv 1 \pmod 9$
When $n = 6$ the situation is different.
We have:
and so the result does not hold.
== Sources ==
* : $28$
* : $28$
Category:Perfect Numbers
Even Perfect Number is Hexagonal

== Theorem ==
All perfect numbers which are even are hexagonal.
== Proof ==
Let $a$ be an even perfect number.
From the Theorem of Even Perfect Numbers, $a$ is in the form $2^{p - 1} \paren {2^p - 1}$ where $2^p - 1$ is prime.
Thus:
The result follows from Closed Form for Hexagonal Numbers.
== Sources ==
* : $28$
* : $45$
* : $28$
* : $45$
Category:Hexagonal Numbers
Category:Perfect Numbers
Even Perfect Number is Triangular/Proof 1

== Theorem ==
== Proof ==
Let $a$ be an even perfect number.
From the Theorem of Even Perfect Numbers, $a$ is in the form $2^{p - 1} \left({2^p - 1}\right)$ where $2^p - 1$ is prime.
Thus:
The result follows from Closed Form for Triangular Numbers.
Category:Even Perfect Number is Triangular
Even Permutation/Examples/231

== Example of Even Permutation ==
:$\tuple {2, 3, 1}$ is an even permutation of $\tuple {1, 2, 3}$.
== Proof ==
The permutations $\tuple {1 2}$ and then $\tuple {2 3}$ transform $\tuple {1, 2, 3}$ to $\tuple {2, 3, 1}$.
That is two transpositions, which is an even integer.
Hence the result.
== Sources ==
* 
* 
Category:Examples of Even Permutations
Even Permutation/Examples/312

== Example of Even Permutation ==
:$\tuple {3, 1, 2}$ is an even permutation of $\tuple {1, 2, 3}$.
== Proof ==
The permutations $\tuple {1 3}$ and then $\tuple {1 2}$ transform $\tuple {1, 2, 3}$ to $\tuple {3, 1, 2}$.
That is two transpositions, which is an even integer.
Hence the result.
== Sources ==
* 
* 
Category:Examples of Even Permutations
Even Permutation is Product of 3-Cycles

== Theorem ==
Let $n \in \N$ such that $n \ge 3$.
Let $\pi: \N_n \to \N_n$ be a permutation on $n$ letters.
Let $\pi$ be an even permutation.
Then $\pi$ can be expressed as the composition of cyclic permutations of length $3$.
== Proof ==
Let $A_n$ denote the alternating group on $n$ letters.
By Alternating Group is Set of Even Permutations, $A_n$ is the set of all even permutations of $S_n$.
$\pi$ can therefore be analysed in the context of elements of $A_n$.
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\pi \in A_n$ can be expressed as the composition of cyclic permutations of length $3$
From Identity Mapping on Symmetric Group is Even Permutation, the identity permutation $e$ on $S_n$ is even.
But there are no cyclic permutations of length $3$ in $S_1$ or $S_2$.
So $\map P 1$ and $\map P 2$ are false.
=== Basis for the Induction ===
$\map P 3$ is the case:
:$\pi \in A_3$ can be expressed as the composition of cyclic permutations of length $3$.
The even permutations of $A_3$ are $e$, $\tuple {1, 2, 3}$ and $\tuple {1, 3, 2}$.
Thus $\map P 3$ is seen to hold trivially.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P {k - 1}$ is true, where $k > 3$, then it logically follows that $\map P k$ is true.
So this is the induction hypothesis:
:$\pi \in A_{k - 1}$ can be expressed as the composition of cyclic permutations of length $3$.
from which it is to be shown that:
:$\pi \in A_k$ can be expressed as the composition of cyclic permutations of length $3$.
=== Induction Step ===
This is the induction step:
Let $\pi \in A_n$.
Consider the permutation:
:$\sigma := \tuple {\pi_n, n, i} \circ \pi$ 
where $\pi_i = n$.
We have:
and is even.
Because $\sigma$ fixes $n$ and is even, $\sigma$ is an even permutation of $\set {1, 2, \ldots, n - 1}$.
By the |inductive hypothesis, $\sigma$ is the composition of $3$-cycles, say:
:$\sigma = \alpha_1 \alpha_2 \dotsm \alpha s$
Setting $\alpha_0 := \tuple {\pi_n, i, n}$, we have:
and we have expressed $\pi$ as the composition of $3$-cycles
So $\map P {k - 1} \implies \map P k$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \pi \in A_n$ can be expressed as the composition of cyclic permutations of length $3$.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 82$
Category:Even Permutations
Category:Cyclic Permutations‎
Even Power is Non-Negative

== Theorem ==
Let $x \in \R$ be a real number.
Let $n \in \Z$ be an even integer.
Then $x^n \ge 0$.
That is, all even powers are positive.
== Proof ==
Let $n \in \Z$ be an even integer.
Then $n = 2 k$ for some $k \in \Z$.
Thus:
:$\forall x \in \R: x^n = x^{2 k} = \paren {x^k}^2$
But from Square of Real Number is Non-Negative:
:$\forall x \in \R: \paren {x^k}^2 \ge 0$
and so there is no real number whose square is negative.
The result follows from Solution to Quadratic Equation.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.12 \ (1)$
Category:Powers
Even Power of 3 as Sum of Consecutive Positive Integers

== Theorem ==
Take the positive integers and group them in sets such that the $n$th set contains the next $3^n$ positive integers:
:$\set 1, \set {2, 3, 4}, \set {5, 6, \ldots, 13}, \set {14, 15, \cdots, 40}, \ldots$
Let the $n$th such set be denoted $S_{n - 1}$, that is, letting $S_0 := \set 1$ be considered as the zeroth.
Then the sum of all the elements of $S_n$ is $3^{2 n}$.
== Proof ==
The total number of elements in $S_0, S_1, \ldots, S_r$ is:
Thus for any given $S_n$:
:$\ds \sum S_n = \sum k \sqbrk {\dfrac {3^n - 1} 2 < k \le \dfrac {3^{n + 1} - 1} 2}$
using Iverson's convention.
So $\ds \sum S_n$ can be evaluated as the difference between two triangular numbers:
Hence the result.
== Examples ==
== Historical Note ==
== Sources ==
* : $81$
* : $81$
Category:Powers of 3
Category:Even Power of 3 as Sum of Consecutive Positive Integers
Even Power of Negative Real Number

== Theorem ==
Let $x \in \R$ be a real number.
Let $n \in \Z$ be an even integer.
Then:
:$\paren {-x}^n = x^n$
== Proof ==
From Real Numbers form Totally Ordered Field, $\R$ is a field .
By definition, $\R$ is therefore a ring.
The result follows from Power of Ring Negative.
Category:Real Numbers
Even Vertex of Graph/Examples/Graph with All Even Vertices

== Examples of Even Vertices ==
Examples of simple graphs whose vertices are all even include the cycle graphs.
For example, the cycle graph of order $4$:
:
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $7 \ \text{(a)}$
Category:Examples of Even Vertices
Even and Odd Integers form Partition of Integers

== Theorem ==
The odd integers and even integers form a partition of the integers.
== Proof ==
Let $n \in \Z$ be an integer.
Let $\Bbb O$ be the set of odd integers and $\Bbb E$ be the set of even integers.
By the Division Theorem:
:$\forall n \in \Z: \exists! q, r \in \Z: n = 2 q + r, 0 \le r < 2$
from which it follows that either:
:$n = 2 q \in \Bbb E$
or:
:$n = 2 q + 1 \in \Bbb O$
Thus:
:$(1): \quad$ each element of $\Z$ is in no more than one of $\Bbb E$ and $\Bbb O$
:$(2): \quad$ each element of $\Z$ is in at least one of $\Bbb E$ and $\Bbb O$
and:
:$(3): \quad$ setting $q = 0$ it is seen that $0 \in \Bbb E$ and $1 \in \Bbb O$ and so neither $\Bbb E$ or $\Bbb O$ is empty.
Thus $\set {\Bbb E \mid \Bbb O}$ is a partition of $\Z$ by definition.
== Sources ==
* : $2$
* : $2$
* : Chapter $1$: Elementary, my dear Watson: $\S 1.2$: Elements, my dear Watson
Category:Set Partitions
Category:Odd Integers
Category:Even Integers
Category:Integers
Category:2
Event/Examples/Arbitrary Space of Cardinality 3

== Examples of Events ==
Let $\EE$ be an experiment whose sample space is defined as $\Sigma = \set {e_1, e_2, e_3}$.
The complete set of events of $\EE$ is:
:$\set {\set {e_1}, \set {e_2}, \set {e_3}, \set {e_1, e_2}, \set {e_1, e_3}, \set {e_2, e_3}, \set {e_1, e_2, e_3}, \O}$
The simple events of $\EE$ are:
:$E_1 = \set {e_1}, E_2 = \set {e_2}, E_3 = \set {e_3}$
while $\O$ is the trivial event.
== Sources ==
* : Chapter $2$: Probability and Discrete Sample Spaces: $2.2$ Sample spaces and events
Category:Examples of Events
Event/Examples/Even Number on 6-Sided Die

== Example of Event ==
Let $\EE$ be the experiment of throwing a standard $6$-sided die.
The sample space of $\EE$ is $\Omega = \set {1, 2, 3, 4, 5, 6}$.
Consider the subset $E \subseteq \Omega$ defined as:
:$E = \set {2, 4, 6}$
Then $E$ is the event that the result of $\EE$ is even.
== Sources ==
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Category:Examples of Events