nyuuzyou/wikis huggingface dataset:
Equivalence Class/Examples/Months that Start on the Same Day of the Week

== Examples of Equivalence Class ==
Let $M$ be the set of months of the (calendar) year according to the (usual) Gregorian calendar.
Let $\sim$ be the relation on $M$ defined as:
:$\forall x, y \in M: x \sim y \iff \text {$x$ and $y$ both start on the same day of the week}$
The set of equivalence classes under $\sim$ depends on whether the year is a leap year.
For a non-leap year, the set of equivalence classes is:
:$\set {\set {\text {January}, \text {October} }, \set {\text {February}, \text {March}, \text {November} }, \set {\text {April}, \text {July} }, \set {\text {May} }, \set {\text {June} }, \set {\text {August} }, \set {\text {September}, \text {December} } }$
For a leap year, the set of equivalence classes is:
:$\set {\set {\text {January}, \text {April}, \text {July} }, \set {\text {February}, \text {August} }, \set {\text {March}, \text {November} }, \set {\text {May} }, \set {\text {June} }, \set {\text {September}, \text {December} }, \set {\text {October} } }$
== Proof ==
We have that:
:The months with $30$ days are:
::$\text {April}, \text {June}, \text {September}, \text {November}$
:The months with $31$ days are:
::$\text {January}, \text {March}, \text {May}, \text {July}, \text {August}, \text {October}, \text {December}$
:In a non-leap year, $\text {February}$ has $28$ days
:In a leap year, $\text {February}$ has $29$ days.
Let month $m$ have $m_d$ days in it.
Let month $m$ start on day $d$, where $d$ is in the range $0$ to $6$ (which day of the week corresponds to which number is irrelevant at this stage).
Then month $m + 1$ starts on day $\paren {d + m_d} \pmod 7$.
For reference:
, let $\text {January}$ start on day $0$.
Then the sequence of the days which are the $1$st of the month is as follows:
For a non-leap year:
:$\tuple {0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5}$
For a leap year:
:$\tuple {0, 3, 4, 0, 2, 5, 0, 3, 6, 1, 4, 6}$
The result follows.
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Exercise $7$
Category:Examples of Equivalence Classes
Category:Calendars
Equivalence Class/Examples/People Born in Same Year

== Example of Equivalence Relation ==
Let $P$ be the set of people.
Let $\sim$ be the relation on $P$ defined as:
:$\forall \tuple {x, y} \in P \times P: x \sim y \iff \text { $x$ and $y$ were born in the same year}$
Then the elements of the equivalence class of $x \in P$ is:
:$\eqclass x \sim = \set {\text {All people born in the same year as $x$} }$
== Sources ==
* : Appendix $\text{A}.3$: Equivalence Relations: Problem Set $\text{A}.3$: $13$
Category:Examples of Equivalence Classes
Equivalence Class/Examples/People of Same Age

== Example of Equivalence Relation ==
Let $P$ be the set of people.
Let $\sim$ be the relation on $P$ defined as:
:$\forall \tuple {x, y} \in P \times P: x \sim y \iff \text { the age of $x$ and $y$ on their last birthdays was the same}$
Then the equivalence class of $x \in P$ is:
:$\eqclass x \sim = \set {\text {All people the same age as $x$ on their last birthday} }$
== Sources ==
* : $\S 2.4$. Equivalence classes
Category:Examples of Equivalence Classes
Equivalence Class/Examples/People with Same Height

== Example of Equivalence Relation ==
Let $A = \set {u, v, w, x, y, z}$ be a set of people.
Let $\sim$ be the relation on $A$ defined as:
:$\forall \tuple {a, b} \in A \times A: x \sim y \iff \text { $a$ and $b$ are the same height}$
Let:
:$u, v, w$ be the same height
:$x, y, z$ be the same height but different from the height of $u, v, w$.
Then the equivalence class of $x \in A$ are $\set {u, v, w}$ and $\set {x, y, z}$.
== Sources ==
* 
* 
Category:Examples of Equivalence Classes
Equivalence Class Equivalent Statements

== Theorem ==
Let $\RR$ be an equivalence relation on $S$.
Let $x, y \in S$.
:$(1): \quad x$ and $y$ are in the same $\RR$-class
:$(2): \quad \eqclass x \RR = \eqclass y \RR$
:$(3): \quad x \mathrel \RR y$
:$(4): \quad x \in \eqclass y \RR$
:$(5): \quad y \in \eqclass x \RR$
:$(6): \quad \eqclass x \RR \cap \eqclass y \RR \ne \O$
== Proof ==
=== $(1)$ Equivalent to $(2)$ ===
=== $(2)$ Equivalent to $(3)$ ===
=== $(3)$ Equivalent to $(4)$ ===
=== $(3)$ Equivalent to $(5)$ ===
=== $(3)$ Equivalent to $(6)$ ===
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Theorem $10.4$
Category:Equivalence Classes
Category:Equivalence Class Equivalent Statements
Equivalence Class Equivalent Statements/1 iff 2

== Theorem ==
Let $\RR$ be an equivalence relation on $S$.
Let $x, y \in S$.
:$(1): \quad x$ and $y$ are in the same $\RR$-class
:$(2): \quad \eqclass x \RR = \eqclass y \RR$
== Proof ==
By Equivalence Class is Unique:
:$\eqclass x \RR$ is the unique $\RR$-class to which $x$ belongs
and:
:$\eqclass y \RR$ is the unique $\RR$-class to which $y$ belongs.
As these are unique for each, they must be the same set.
== Sources ==
* : $\S 17.7$: Equivalence classes
Category:Equivalence Class Equivalent Statements
Equivalence Class Equivalent Statements/2 iff 3

== Theorem ==
Let $\RR$ be an equivalence relation on $S$.
Let $x, y \in S$.
:$\eqclass x \RR = \eqclass y \RR$
:$x \mathrel \RR y$
== Proof ==
By Equivalence Class holds Equivalent Elements:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Theorem $10.4$
* : $\S 17.7$: Equivalence classes
Category:Equivalence Class Equivalent Statements
Equivalence Class Equivalent Statements/3 iff 4

== Theorem ==
Let $\RR$ be an equivalence relation on $S$.
Let $x, y \in S$.
:$x \mathrel \RR y$
:$x \in \eqclass y \RR$
== Proof ==
This follows directly by the definition of equivalence class.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Theorem $10.4$
Category:Equivalence Class Equivalent Statements
Equivalence Class Equivalent Statements/3 iff 5

== Theorem ==
Let $\RR$ be an equivalence relation on $S$.
Let $x, y \in S$.
:$x \mathrel \RR y$
:$y \in \eqclass x \RR$
== Proof ==
This follows through dint of the symmetry of $\RR$ and the definition of Equivalence Class.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Theorem $10.4$
Category:Equivalence Class Equivalent Statements
Equivalence Class Equivalent Statements/3 iff 6

== Theorem ==
Let $\RR$ be an equivalence relation on $S$.
Let $x, y \in S$.
:$x \mathrel \RR y$
:$\eqclass x \RR \cap \eqclass y \RR \ne \O$
== Proof ==
Follows directly from Equivalence Classes are Disjoint.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Theorem $10.4$
Category:Equivalence Class Equivalent Statements
Equivalence Class holds Equivalent Elements

== Theorem ==
Let $\RR$ be an equivalence relation on a set $S$.
Then:
:$\tuple {x, y} \in \RR \iff \eqclass x \RR = \eqclass y \RR$
== Proof ==
=== Necessary Condition ===
First we prove that $\tuple {x, y} \in \RR \implies \eqclass x \RR = \eqclass y \RR$.
Suppose:
: $\tuple {x, y} \in \RR: x, y \in S$
Then:
So:
:$\eqclass x \RR \subseteq \eqclass y \RR$
Now:
... so we have shown that:
:$\tuple {x, y} \in \RR \implies \eqclass x \RR = \eqclass y \RR$
=== Sufficient Condition ===
Next we prove that $\eqclass x \RR = \eqclass y \RR \implies \tuple {x, y} \in \RR$.
By definition of set equality:
:$\eqclass x \RR = \eqclass y \RR$
means:
:$\paren {x \in \eqclass x \RR \iff x \in \eqclass y \RR}$
So by definition of equivalence class:
:$\tuple {y, x} \in \RR$
Hence by definition of equivalence relation: $\RR$ is symmetric
:$\tuple {x, y} \in \RR$
So we have shown that
:$\eqclass x \RR = \eqclass y \RR \implies \tuple {x, y} \in \RR$
Thus, we have:
So by equivalence:
:$\tuple {x, y} \in \RR \iff \eqclass x \RR = \eqclass y \RR$
== Sources ==
* : $\S 2.4$. Equivalence classes: Lemma $\text{(i)}$
* : Chapter $\text{I}$: Sets and Functions: Relations: Theorem $4$
* : Chapter $1$: Equivalence Relations: $\S 17$
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.3 \ (2)$
* : $\S 17.2, \ \S 17.3$: Equivalence classes
* : Appendix $\text{A}$: Set Theory: Equivalence Relations
Category:Equivalence Classes
Equivalence Class is Unique

== Theorem ==
Let $\RR$ be an equivalence relation on $S$.
For each $x \in S$, the one and only one $\RR$-class to which $x$ belongs is $\eqclass x \RR$.
== Proof ==
This follows directly from the Fundamental Theorem on Equivalence Relations: the set of $\RR$-classes forms a partition of $S$.
== Sources ==
* : $\S 17.6$: Equivalence classes
Category:Equivalence Classes
Equivalence Class is not Empty

== Theorem ==
Let $\RR$ be an equivalence relation on a set $S$.
Then no $\RR$-class is empty.
== Proof ==
== Also see ==
* Fundamental Theorem on Equivalence Relations
* Union of Equivalence Classes is Whole Set
* Equivalence Classes are Disjoint
== Sources ==
* : Appendix: Elementary set and number theory
* : Chapter $1$: Theory of Sets: $\S 7$: Relations
* : Appendix $\text{A}.3$: Equivalence Relations: Theorem $\text{A}.2$
* : $\S 17.5 \ \text{(iii)}$: Equivalence classes
* : Chapter $1$: Preliminaries: Exercises: $1.8$
Category:Equivalence Classes
Category:Empty Set
Category:Fundamental Theorem on Equivalence Relations
Equivalence Class of Equal Elements of Cross-Relation

== Theorem ==
Then:
:$\forall c, d \in S_1 \cap S_2: \tuple {c, c} \boxtimes \tuple {d, d}$
== Proof ==
Note that in order for $\tuple {c, c}$ and $\tuple {d, d}$ to be defined, $c$ and $d$ must be in both $S_1$ and $S_2$.
Hence the restriction given:
:$\forall c, d \in S_1 \cap S_2$
Then:
Hence the result.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Cross-Relations
Category:Examples of Equivalence Classes
Equivalence Class of Fixed Element

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\sigma \in S_n$.
Let $\RR_\sigma$ denote the equivalence defined in Permutation Induces Equivalence Relation.
Let $i \in \N^*_{\le n}$.
Then:
:$i \in \Fix \sigma$ $\eqclass i {\RR_\sigma} = \set i$
where:
:$\eqclass i {\RR_\sigma}$ denotes the equivalence class of $i$ under $\RR_\sigma$
:$\Fix \sigma$ denotes the set of fixed elements of $\sigma$.
 
=== Corollary ===
== Proof ==
By the definition of an equivalence relation it is easily seen that $\set i \subseteq \eqclass i {\RR_\sigma}$.
Suppose that $i \in \Fix \sigma$.
Let $j \in \eqclass i {\RR_\sigma}$.
Then by Condition for Membership of Equivalence Class and Permutation Induces Equivalence Relation:
:$j \in \eqclass i {\RR_\sigma} \iff i \mathrel {\RR_\sigma} j \implies \exists m \in \Z: \sigma^m \paren i = j$
And by Fixed Point of Permutation is Fixed Point of Power:
:$\sigma^m \paren i = i \implies i = j$
Therefore:
:$\eqclass i {\RR_\sigma} \subseteq \set i$
and so:
:$\eqclass i {\RR_\sigma} = \set i$
For the converse, suppose that:
:$\eqclass i {\RR_\sigma} = \set i$.
It is seen that:
:$i \mathrel {\RR_\sigma} \map \sigma i \iff \map \sigma i \in \eqclass i {\RR_\sigma}$
Therefore:
:$\map \sigma i = i \implies i \in \Fix \sigma$
Hence the result.
Category:Symmetric Groups
Category:Equivalence Classes
Equivalence Class of Fixed Element/Corollary

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\sigma \in S_n$.
Let $\RR_\sigma$ denote the equivalence defined in Permutation Induces Equivalence Relation.
Let $i \in \N^*_{\le n}$.
Then:
:$i \notin \Fix \sigma$ $\eqclass i {\RR_\sigma}$ contains more than one element
where:
:$\eqclass i {\RR_\sigma}$ denotes the equivalence class of $i$ under $\RR_\sigma$
:$\Fix \sigma$ denotes the set of fixed elements of $\sigma$.
== Proof ==
From Equivalence Class of Fixed Element and Biconditional Equivalent to Biconditional of Negations:
:$i \notin \Fix \sigma \iff \set i \ne \eqclass i {\RR_\sigma}$
Because the Biconditional is Transitive, it suffices to show that:
:$\set i \ne \eqclass i {\RR_\sigma}$
:
:$\eqclass i {\RR_\sigma}$ contains more than one element.
Suppose that $\set i \ne \eqclass i {\RR_\sigma}$.
From the definition of an equivalence relation it is seen that:
:$\set i \subseteq \eqclass i {\RR_\sigma}$
And from the hypothesis:
:$\set i \subset \eqclass i {\RR_\sigma}$
Therefore $\eqclass i {\RR_\sigma}$ contains more than one element.
Conversely, if $\eqclass i {\RR_\sigma}$ contains more than one element, then it is seen that:
:$\set i \ne \eqclass i {\RR_\sigma}$
Category:Symmetric Groups
Category:Equivalence Classes
Equivalence Class of Isometries is Subset of Equivalence Class of Homeomorphisms

== Theorem ==
Let $\CC_I$ be an equivalence class of isometries on the set of metric spaces.
Then $\CC_I$ is a subset of an equivalence class of homeomorphisms on the set of metric spaces.
== Proof ==
From Isometry of Metric Spaces is Equivalence Relation, all isometries can be partitioned into equivalence classes.
From Homeomorphism of Metric Spaces is Equivalence Relation, all homeomorphisms can be partitioned into equivalence classes.
Let $\CC_I$ be an equivalence class of isometries.
Let $f: M_1 \to M_2$ be an element of $\CC_I$ where $M_1$ and $M_2$ are some metric spaces.
From Isometry of Metric Spaces is Homeomorphism, $\CC_I$ is a homeomorphism.
Thus $\CC_I$ is an element of some equivalence class of homeomorphisms on the set of metric spaces.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces
Category:Homeomorphisms (Metric Spaces)
Category:Isometries (Metric Spaces)
Category:Equivalence Classes
Equivalence Classes are Disjoint

== Theorem ==
Let $\RR$ be an equivalence relation on a set $S$.
Then all $\RR$-classes are pairwise disjoint:
:$\tuple {x, y} \notin \RR \iff \eqclass x \RR \cap \eqclass y \RR = \O$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Fundamental Theorem on Equivalence Relations
* Union of Equivalence Classes is Whole Set
* Equivalence Class is not Empty
== Sources ==
* : Chapter $1$: Equivalence Relations: $\S 17$
* : Chapter $2$: 'And you do addition?': $\S 2.3$: Relations: Ponderable $2.3.3$
Category:Equivalence Classes
Category:Equivalence Classes are Disjoint
Category:Fundamental Theorem on Equivalence Relations
Equivalence Classes are Disjoint/Proof 1

== Theorem ==
== Proof ==
First we show that:
:$\tuple {x, y} \notin \RR \implies \eqclass x \RR \cap \eqclass y \RR = \O$
Suppose two $\RR$-classes are not disjoint:
So:
:$\eqclass x \RR \cap \eqclass y \RR \ne \O \implies \tuple {x, y} \in \RR$
By the Rule of Transposition:
:$\tuple {x, y} \notin \RR \implies \eqclass x \RR \cap \eqclass y \RR = \O$
Now we show that:
:$\eqclass x \RR \cap \eqclass y \RR = \O \implies \tuple {x, y} \notin \RR$
Suppose:
:$\tuple {x, y} \in \RR$.
Then:
So:
:$\tuple {x, y} \in \RR \implies \eqclass x \RR \cap \eqclass y \RR \ne \O$
By the Rule of Transposition:
:$\eqclass x \RR \cap \eqclass y \RR = \O \implies \paren {x, y} \notin \RR$
Using the rule of Biconditional Introduction on these results:
:$\eqclass x \RR \cap \eqclass y \RR = \O \iff \paren {x, y} \notin \RR$
and the proof is complete.
== Sources ==
* : $\S 2.4$. Equivalence classes: Lemma $\text{(ii)}$
* : Appendix $\text{A}.3$: Equivalence Relations: Theorem $\text{A}.2$
* : $\S 17.4, \ \S 17.5 \ \text{(ii)}$: Equivalence classes
Category:Equivalence Classes are Disjoint
Equivalence Classes are Disjoint/Proof 2

== Theorem ==
== Proof ==
Suppose that for $x, y \in S$:
:$\eqclass x \RR \cap \eqclass y \RR \ne \O$
Let:
:$z \in \eqclass x \RR$
:$z \in \eqclass y \RR$
Then by definition of equivalence class:
:$\tuple {x, z} \in \RR$
:$\tuple {y, z} \in \RR$
Let $c \in \eqclass x \RR$.
That is:
:$\tuple {x, c} \in \RR$
By definition of equivalence relation, $\RR$ is symmetric so:
:$\tuple {z, x} \in \RR$
By definition of equivalence relation, $\RR$ is transitive so:
:$\tuple {z, x} \in \RR \land \tuple {x, c} \in \RR \implies \tuple {z, c} \in \RR$
and
:$\tuple {y, z} \in \RR \land \tuple {z, c} \in \RR \implies \tuple {y, c} \in \RR$
So we have $c \in \eqclass y \RR$.
By definition of subset:
:$\eqclass x \RR \subseteq \eqclass y \RR$
Similarly, let $c \in \eqclass y \RR$.
That is:
:$\tuple {y, c} \in \RR$
By definition of equivalence relation, $\RR$ is symmetric so:
:$\tuple {z, y} \in \RR$
By definition of equivalence relation, $\RR$ is transitive so:
:$\tuple {z, y} \in \RR \land \tuple {y, c} \in \RR \implies \tuple {z, c} \in \RR$
and
:$\tuple {x, z} \in \RR \land \tuple {z, c} \in \RR \implies \tuple {x, c} \in \RR$
So we have $c \in \eqclass x \RR$.
By definition of subset:
:$\eqclass y \RR \subseteq \eqclass x \RR$
That is:
:$\eqclass x \RR \subseteq \eqclass y \RR$
and
:$\eqclass y \RR \subseteq \eqclass x \RR$
By definition of set equality:
:$\eqclass x \RR = \eqclass y \RR$
Thus:
:$\eqclass x \RR \cap \eqclass y \RR \ne \O \implies \eqclass x \RR = \eqclass y \RR$
and the result follows.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Relations: Theorem $4$
* : Appendix: Elementary set and number theory
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.3 \ (3)$
* : Chapter $1$: Theory of Sets: $\S 7$: Relations
* : Chapter $1$: Preliminaries: Exercises: $1.8$
* : $1$: Set Theory and Logic: $\S 3$: Relations: Lemma $3.1$
* : Appendix $\text{A}$: Set Theory: Equivalence Relations
Category:Equivalence Classes are Disjoint
Equivalence Classes induced by Derivative Function on Set of Functions

== Theorem ==
Let $X$ be the set of real functions $f: \R \to \R$ which possess continuous derivatives.
Let $\RR \subseteq X \times X$ be the equivalence relation on $X$ defined as:
:$\RR = \set {\tuple {f, g} \in X \times X: D f = D g}$
where $D f$ denotes the first derivative of $f$.
Then the equivalence classes of $\RR$ are defined as:
:$\map {\eqclass f \RR} x = \set {g \in X: \exists c \in \R: \forall x \in \R: \map g x = \map f x + c}$
That is, it consists of the set of all real functions $f \in X$ which differ by a real constant.
== Proof ==
Follows directly from Derivative Function on Set of Functions induces Equivalence Relation.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 7$: Relations: Exercise $4$
Category:Differential Calculus
Category:Examples of Equivalence Classes
Equivalence Classes of Cross-Relation on Natural Numbers

== Theorem ==
Let $\struct {\N, +}$ be the semigroup of natural numbers under addition.
Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$.
Let $\boxtimes$ be the cross-relation defined on $\N \times \N$ by:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
Then $\eqclass {\tuple {x, y} } \boxtimes$ is the equivalence class of $\tuple {x, y}$ under $\boxtimes$, where:
:$\eqclass {\tuple {x_1, y_1} } \boxtimes = \eqclass {\tuple {x_2, y_2} } \boxtimes \iff \tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2}$
The equivalence class $\eqclass {\tuple {x, y} } \boxtimes$ is more often denoted more simply as $\eqclass {x, y} {}$.
== Proof ==
We have that $\struct {\N, +}$ is a commutative semigroup from Natural Number Addition is Commutative.
We also have the result Cross-Relation on Natural Numbers is Equivalence Relation.
The result follows from the definition of equivalence class.
== Sources ==
* : $\S 2.5$: Theorem $2.21$
Category:Natural Numbers
Category:Cartesian Product
Category:Examples of Equivalence Classes
Equivalence Classes of Diagonal Relation

== Theorem ==
Let $S$ be a set.
Let $\Delta_S$ denote the diagonal relation on $S$.
The set $\EE_S$ of equivalence classes of $S$ can be expressed as:
:$\EE_S = \set {\set x: x \in S}$
That is, it is the set of all singletons of $S$.
== Proof ==
Let $x \in S$.
Then by definition of the diagonal relation:
:$y \mathrel {\Delta_S} x \iff y = x$
Hence:
:$y \in \eqclass x {\Delta_S} \iff y = x$
That is:
:$\eqclass x {\Delta_S} = \set x$
Hence the result.
Category:Diagonal Relation
Category:Examples of Equivalence Classes
Equivalence Classes of Dipper Relation

== Theorem ==
Let $m \in \N$ be a natural number.
Let $n \in \N_{>0}$ be a non-zero natural number.
Let $\RR_{m, n}$ be the dipper relation on $\N$:
:$\forall x, y \in \N: x \mathrel {\RR_{m, n} } y \iff \begin {cases} x = y \\ m \le x < y \text { and } n \divides \paren {y - x} \\ m \le y < x \text { and } n \divides \paren {x - y} \end {cases}$
The equivalence classes of $\RR_{m, n}$ are:
:$\eqclass x {\RR_{m, n} } = \begin {cases} \set x & : x < m \\ \set {y \in \N: \paren {m \le y} \land \paren {\exists k \in \Z: x = y + k n} } & : x \ge m \end {cases}$
Thus the classes are:
and it is seen that there are exactly $m + n$ such equivalence classes.
== Proof ==
From Dipper Relation is Equivalence Relation we have that $\RR_{m, n}$ is indeed an equivalence relation.
Let $x < m$.
Then by definition:
:$y \in \eqclass x {\RR_{m, n} } \iff x = y$
and so:
:$\eqclass x {\RR_{m, n} } = \set x$
Let $m \le x$.
Let $x \mathrel {\RR_{m, n} } y$.
First we note that $m \le y$.
Let $x = y$.
Then $y \in \eqclass x {\RR_{m, n} }$ immediately, and:
:$x = y + 0 \times n$
That is:
:$\exists k \in \Z: x = y + k n$
where in this instance $k = 0$.
Let $x > y$.
We have that:
Let $x < y$.
We have that:
Hence when $m \le x$, we have that:
:$\eqclass x {\RR_{m, n} } = \set {y \in \N: \paren {m \le y} \land \paren {\exists k \in \Z: x = y + k n} }$
Category:Dipper Relations
Category:Examples of Equivalence Classes
Equivalence Induced by Epimorphism is Congruence Relation

== Theorem ==
Let $\struct {S, \oplus}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \oplus} \to \struct {T, *}$ be an epimorphism.
Let $\RR_\phi$ be the equivalence induced by $\phi$.
Then the induced equivalence $\RR_\phi$ is a congruence relation for $\oplus$.
== Proof ==
Let $x, x', y, y' \in S$ such that:
:$x \mathrel {\RR_\phi} x' \land y \mathrel {\RR_\phi} y'$
By definition of induced equivalence:
Then:
Thus $\paren {x \oplus y} \mathrel {\RR_\phi} \paren {x' \oplus y'}$ by definition of induced equivalence.
So $\RR_\phi$ is a congruence relation for $\oplus$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.5$
Category:Epimorphisms (Abstract Algebra)
Category:Congruence Relations
Category:Examples of Equivalence Relations
Equivalence Principle/Examples/Spaceship

== Examples of Equivalence Principle ==
Consider a spaceship $\SS$ far out in space, under the influence of no gravitational fields.
Everything in $\SS$ would be weightless.
Let a uniform acceleration be applied to $\SS$, corresponding to Acceleration Due to Gravity on the surface of Earth.
Then everything within $\SS$ would behave as though $\SS$ were stationary on the surface of Earth.
== Sources ==
* 
* 
Category:Examples of Equivalence Principle
Equivalence Relation/Examples

== Examples of Equivalence Relations ==
=== Equality ===
=== Same Age Relation ===
=== Same Parents Relation ===
=== People with Same First Name ===
=== Books with Same Number of Pages ===
=== Even Sum Relation ===
=== Months that Start on the Same Day of the Week ===
=== Points on Same Horizontal Line ===
=== $z^4 = w^4$ on Complex Numbers ===
=== $\sin \dfrac {\pi x} 6 = \sin \dfrac {\pi y} 6$ on Integers ===
== Examples of Non-Equivalence Relations ==
Category:Examples of Equivalence Relations
Equivalence Relation/Examples/Books with Same Number of Pages

== Example of Equivalence Relation ==
Let $P$ be the set of books.
Let $\sim$ be the relation on $P$ defined as:
:$\forall \tuple {x, y} \in P \times P: x \sim y \iff \text {$x$ and $y$ have the same number of pages}$
Then $\sim$ is an equivalence relation.
== Sources ==
* : Appendix $\text{A}.3$: Equivalence Relations: $(6)$
Category:Examples of Equivalence Relations
Equivalence Relation/Examples/Equal Fourth Powers over Complex Numbers

== Example of Equivalence Relation ==
Let $\C$ denote the set of complex numbers.
Let $\RR$ denote the relation on $\C$ defined as:
:$\forall w, z \in \C: z \mathrel \RR w \iff z^4 = w^4$
Then $\RR$ is an equivalence relation.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.1 \ \text{(a)}$
Category:Examples of Equivalence Relations
Category:Equal Fourth Powers over Complex Numbers