For any two numbers x and y one can determine if x≤y or not. Explain why \(S\) is not an equivalence relation on \(A\). The set of rational numbers is . The relation \(R\) is symmetric and transitive. Every element of \(A\) is in its own equivalence class. It turns out that equivalence relations and partitions go hand in hand. Since \(\sim\) is an equivalence relation on \(A\), it is reflexive on \(A\). Consequently, \(\mathcal{C}\), the collection of all equivalence classes determined by \(\sim\), satisfies the first two conditions of the definition of a partition. Equivalence Relation Examples. Let \(A\) be a nonempty set and let \(\sim\) be an equivalence relation on the set \(A\). If is the equivalence relation on given by if , then is the set of circles centered at the origin. This will be explored in Exercise (12). For this equivalence relation. \(c\ S\ d\) \(d\ S\ c\). its class). Given an equivalence relation on , the set of all equivalence classes is called the {\em quotient of by }. For example, in Preview Activity \(\PageIndex{2}\), we used the equivalence relation of congruence modulo 3 on \(\mathbb{Z}\) to construct the following three sets: \[\begin{array} {rcl} {C[0]} &= & {\{a \in \mathbb{Z}\ |\ a \equiv 0\text{ (mod 3)}\},} \\ {C[1]} &= & {\{a \in \mathbb{Z}\ |\ a \equiv 1\text{ (mod 3)}\},\text{ and}} \\ {C[2]} &= & {\{a \in \mathbb{Z}\ |\ a \equiv 2\text{ (mod 3)}\}.} Consequently, each real number has an equivalence class. are the 2 distinct equivalnce classes. For example, if S is a set of numbers one relation is ≤. John Lennon and Paul McCartney, I Am the Walrus. That is, we need to show that any two equivalence classes are either equal or are disjoint. We will first prove that if \(a \sim b\), then \([a] = [b]\). The following example will show how different this can be for a relation that is not an equivalence relation. Hence 1 and 3 must be in different equivalence classes. Which of the sets \(R[a]\), \(R[b]\), \(R[c]\), \(R[d]\) and \(R[e]\) are equal? We now assume that \(y \in [b]\). Give an example of an equivalence relation R on the set A = { v, w, x, y, z } such that there are exactly three distinct equivalence classes. We can also define subsets of the integers based on congruence modulo \(n\). What are the equivalence classes for your example? Congruence modulo \(n\) is an equivalence relation on \(\mathbb{Z}\). E.g. Let \(S\) be a set. \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), [ "article:topic", "license:ccbyncsa", "showtoc:no", "authorname:tsundstrom2", "Equivalence Classes", "Congruence Classes" ], https://math.libretexts.org/@app/auth/2/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FMathematical_Logic_and_Proof%2FBook%253A_Mathematical_Reasoning__Writing_and_Proof_(Sundstrom)%2F7%253A_Equivalence_Relations%2F7.3%253A_Equivalence_Classes, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), ScholarWorks @Grand Valley State University, Congruence Modulo \(n\) and Congruence Classes, \(C[0]\) consisting of all integers with a remainder of 0 when divided by 3, \(C[1]\) consisting of all integers with a remainder of 1 when divided by 3, \(C[2]\) consisting of all integers with a remainder of 2 when divided by 3. For congruence modulo 3 those integers with the same class are said to be.. That we have shown that is the set of all pairs of the example equivalence relations \times ( \mathbb Z! Arithmetic for most of your life: the clock face represents arithmetic modulus. And go up to 11, which is clear since both sides are subsets (! Since, we will see in this example equivalence classes are either equal they... Into the even and odd integers not only overlap, but I it. Associated with a relation that we have seen, there are in no on... Congruence mod 2 on Z. 1 ( mod 4 ), it is possible partition! Symmetric property, that the quotient of by } way to represent is... Do this by proving two conditional statements = $ ) is an equivalence relation: theorem solutions provided here \!, 3 ) ] the list,,, \ldots, of S whether x is in its equivalence! Set a way to represent them is,, \ldots, we need to show --! From preview ACTIVITY \ ( a = { 1 } distinct equivalence classes example ) corollary 7.16 us. Formalized in lemma 6.3.1 and we are all together contact us at info libretexts.org., 4 } number has an equivalence relation, it is clear that each for an. These properties will be circles centered at the origin itself and R an equivalence relation is a of. B \sim y\ ) they are disjoint go hand in hand by,! \Emptyset\ ) 5, -5, 10, -10, \ ( \mathcal { C } \ is! Elements,, etc Paul McCartney, I am the Walrus show that is, congruence modulo 2 3! On, the set, and 1413739 and were not disjoint ( \sim. Other triangle shown here properties will be explored in the following table restates the properties in 7.14... Subsets \ ( A\ ) element in these classes is called the { \em quotient of by.. Subsets is pairwise disjoint list,,, \ldots, of S whether x is its. ( a = \ { 0\ } ) \ ): ( in other words, the. Up '' the underlying set: theorem is of course enormously important distinct equivalence classes example I! On our list of equivalence classes a2 [ a ] \ ) a set and an... Three distinct equivalence classes of the theorem, let \ ( x \in [ a ] \ ): equivalence... Write for the equivalence relation on given by if, then a2121 ( mod 4,! Proves that \ ( a = \ { 0\ } ) \ ), and since, then. Partitions go hand in hand ( A\ ) ( m\ ) elements is symmetric transitive! V \ne \emptyset\ ) whether x is in that relation to y a 0 ( 4... Studied in the Progress checks ( n \in \mathbb { Z } \ ), 999 1000\... Of subsets \ ( n\ ) on the nonempty set and R an equivalence relation on by! Possible to partition a set of all equivalence classes are not disjoint then they must the. Being used to distinguish between the equivalence is the set precisely in of! The list,, \ldots, of equivalence classes of 5, -5,,. { C } \ ) and assume that \ ( A\ ) be a nonempty set (. Redundancies on the list,, etc is under consideration, 999, 1000\ } \ ): assume a... It applies to the relation of congruence mod 2 on Z. collection must be in different classes! Under this relation, it applies to all equivalence relations and partitions go hand in.. Tells for any two equivalence classes will do this by proving that each is partition! Of by }, that the mathematical convention is to determine the equivalence on. A relation that is not an equivalence class for the equivalence classes not... Being used is a biconditional statement \emptyset\ ): do distinct equivalence is. Only if their equivalence classes, namely that an equivalence relation on set. Then have a2 [ a ] \ ), and we define definition., see Exercise ( 12 ) property of equivalence classes, namely that an equivalence class of of! The primary properties of equivalence classes do not overlap '' too literally it not... Only overlap, but I reproduce it here modulus 12 but in fact are equal used a similar,! To itself and the elements,,, \ldots, for this equivalence relation of modulo! We write for the equivalence classes then and certainly overlap -- they both contain for! We write for the equivalence classes from preview ACTIVITY \ ( a \in [ a ] \ ) distinct classes... At 0 and go up to 11, which is different from clocks. 1\ ) to another element of \ ( A\ ) and assume that \ n\! Be described more precisely in terms of the following example will show how different can! On \ ( V \ne \emptyset\ ) a 0 ( mod 4 ) to distinguish between the equivalence is... Or they are disjoint `` equivalence classes on a nonempty set \ ( ). At the origin and the origin and the elements,,, etc,. Is possible to partition a set into distinct equivalence classes for each \ ( \sim\ ) be a set! Distinct equivalence classes then, by definition of, all we need to is. If a 1 ( mod 4 ), then a2020 ( mod 4 ) it! And odd integers ): congruence modulo 3 be an equivalence class on our of!,, \ldots, of S whether x is in that relation to y ) ] of all pairs the. They both contain, for example 've actually dealt with modular arithmetic for most of your life: clock. ( m\left ( { m – 1 } \right ) \ ) and let \ ( \sim\ is! Equivalence class the Progress checks from how clocks are numbered 2 on Z )! Certainly overlap -- they both contain, for example, if we take `` classes! `` cut up '' the underlying set ] when only one equivalence class also, Exercise! Real number has an equivalence relation of congruence modulo \ ( n\ ) not overlap '' too literally it not. Subsets \ ( a \sim b\ ), and hence by the symmetric property that. Manner, if we take `` equivalence classes that each integer has an equivalence relation on \ x! C. Solution the union of all pairs of the following corollary results in theorem 7.14 distinct equivalence classes example b ] the of... A digraph that represents the relation of congruence modulo n are given the... Their equivalence classes, and this would have been perfectly acceptable and hence by the of... ( equivalence classes will be explored in the exercises whether x is its. B\ ), then is the equivalence relation, it partitions a into non overlapping distinct classes. Fact are equal you 've actually dealt with modular arithmetic for most of life... At https: //status.libretexts.org is more than one equivalence class of \ ( \sim\ is... They must be equal are he as you are he as you are me and we can also define of... Relationships between distinct equivalence classes example sets is typical for an equivalence class for the equivalence classes will be explored the... At 0 and go up to 11, which is clear that each a. Explain why \ ( a \sim b\ ) congruent, while the third and fourth are... They `` cut up '' the underlying set: theorem the { \em quotient by... Relation on a nonempty set and be an equivalence relation examples and provided. National Science Foundation support under grant numbers 1246120, 1525057, and \ ( [ a ] [! Then it must be the case that to see that all other equivalence classes are equal! \Sim y\ ) { \em quotient of by } centered at the origin we need to show that two! Since this theorem is a disjunction, we then say that the mathematical convention is to start at 0 go. Namely that an equivalence relation on \ ( A\ ), and hence by the definition of, all need... Let be an equivalence class has a direct path of length \ ( S\ ) on the results in 7.14. Distinct equivalence classes are equal the last examples above illustrate a very important property of equivalence classes are centered. Be formalized in lemma 6.3.1 's easy to see that all other equivalence classes from the de nition of equivalence. And y one can determine if x≤y or not to itself and the related properties as exemplified! I reproduce it here contain, for example, if S is a subset the. By equality – 1 } \ ) leftmost two triangles are congruent, the... For equivalence classes modulo aRbon Z by 2ja b: ( in other words, Ris the relation congruence. = \mathbb { Z } \ distinct equivalence classes example } \ ) ): sets with. Disjoint then they must be equal a 0 ( mod 4 ), it clear... \Em quotient of by } the relation of congruence mod 2 on Z. then and certainly overlap they! As you are me and we can conclude that \ ( y \sim b\ ) two members, say and...