In the above example, for instance, the class of 0, , may is the set of all pairs of the form . We denote aEb as a ~ b. Equivalence Classes Given an equivalence relation R over a set A, for any x ∈ A, the equivalence class of x is the set [x] R = { y ∈ A | xRy} [x] R is the set of all elements of A that are related to x. Theorem: If R is an equivalence relation over A, then every a ∈ A belongs to exactly one equivalence class. PREVIEW ACTIVITY $$\PageIndex{1}$$: Sets Associated with a Relation. In any case, always remember that when we are working with any equivalence relation on a set A if $$a \in A$$, then the equivalence class [$$a$$] is a subset of $$A$$. So, in Example 6.3.2, [S2] = [S3] = [S1] = {S1, S2, S3}. Which of the sets $$R[a]$$, $$R[b]$$, $$R[c]$$, $$R[d]$$ and $$R[e]$$ are disjoint? First we show that every . That is, congruence modulo 2 simply divides the integers into the even and odd integers. iff . So we'll amend, distinct equivalence classes do not overlap. For each $$a \in A$$, the equivalence class of $$a$$ determined by $$\sim$$ is the subset of $$A$$, denoted by [$$a$$], consisting of all the elements of $$A$$ that are equivalent to $$a$$. Legal. $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, [ "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, $$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, ScholarWorks @Grand Valley State University, Congruence Modulo $$n$$ and Congruence Classes, $$C$$ consisting of all integers with a remainder of 0 when divided by 3, $$C$$ consisting of all integers with a remainder of 1 when divided by 3, $$C$$ consisting of all integers with a remainder of 2 when divided by 3. For the equivalence relation of congruence modulo $$n$$, Theorem 3.31 and Corollary 3.32 tell us that each integer is congruent to its remainder when divided by $$n$$, and that each integer is congruent modulo $$n$$ to precisely one of one of the integers $$0, 1, 2, ..., n - 1$$. $$C \cap C = \emptyset$$, $$C \cap C = \emptyset$$, and $$C \cap C = \emptyset$$. The last examples above illustrate a very important property of equivalence classes, namely that an equivalence class may have many di erent names. What are the equivalence classes under the relation ? Question 1: Let assume that F is a relation on the set R real numbers defined by xFy if and only if x-y is an integer. As was indicated in Section 7.2, an equivalence relation on a set $$A$$ is a relation with a certain combination of properties (reflexive, symmetric, and transitive) that allow us to sort the elements of the set into certain classes. Then , , etc. Show that is the set of all pairs of the form . Without using the terminology at that time, we actually determined the equivalence classes of the equivalence relation $$R$$ in Preview Activity $$\PageIndex{1}$$. Also, see Exercise (9) in Section 7.2. Let A be a nonempty set and assume that $$\sim$$ is an equivalence relation on $$A$$. Which of the sets $$S[a]$$, $$S[b]$$, $$S[c]$$, $$S[d]$$, and $$S[e]$$ are equal? equivalence classes. Every element of $$A$$ is in its own equivalence class. The proof is found in your book, but I reproduce it here. Since is transitive, we have . If [x][[y] = X, we are done (there are two equivalence classes); if not, choose z 2Xn([x][[y]), compute its equivalence classes and keep going until the union of the equivalence classes we explicitly computed is the entire set X. Now, to gure out the equivalence classes, let’s think about the four possibilities for an integer: it can be congruent to 0, 1, 2, or 3 modulo 4. $$\mathbb{Z} =  \cup  \cup  \cup \cdot\cdot\cdot \cup [n - 1]$$. Then. Find the equivalence class [(1, 3)]. Since is symmetric, this means , i.e. For these examples: Do distinct equivalence classes have a non-empty intersection? From our assumption, a2[b]. Thus, in this example equivalence classes are circles centered at the origin and the origin itself. The following definition makes this idea precise. So we have. For example, we can define $$C$$ to be the set of all integers a that are congruent to 0 modulo 3. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. We'll prove the contrapositive: if , then . Note that we have . We know that each integer has an equivalence class for the equivalence relation of congruence modulo 3. So if we take equivalence classes do not overlap" too literally it cannot be true. $$[a] \cap [b] = \emptyset$$ or $$[a] \cap [b] \ne \emptyset$$. Example 5.1.1 Equality ($=$) is an equivalence relation. Lemma Let A be a set and R an equivalence relation on A. If is the equivalence relation on given by if , then is the set of circles centered at the origin. Consider an equivalence class consisting of $$m$$ elements. For every $$V, W \in \mathcal{C}$$, $$V = W$$ or $$V \cap W = \emptyset$$. Have questions or comments? Using the notation from the definition, they are: For each $$a, b \in A$$, $$a \sim b$$ if and only if $$[a] = [b]$$. Consequently, $$x \in a$$ and $$x \in b$$, and so we can use the first part of the theorem to conclude that $$[x] = [a]$$ and $$[x] = [b]$$. Since Ris re exive, we know aRa. equivalence class); if not, we can choose some y 2X n[x] and compute its equivalence class [y]. Watch the recordings here on Youtube! If Ris an equivalence relation on a set A, and a2A, then the set [a] = fx2Ajx˘ag is called the equivalence class of a. We will now prove that the two sets $$[a]$$ and $$[b]$$ are equal. Since we have assumed that $$a \sim b$$, we can use the transitive property of $$\sim$$ to conclude that $$x \sim b$$, and this means that $$x \in [b]$$. Do not use fractions in your proof. The equivalence class of under the equivalence is the set. Since this theorem applies to all equivalence relations, it applies to the relation of congruence modulo $$n$$ on the integers. Assume is nonempty. Consequently, $$\mathcal{C}$$, the collection of all equivalence classes determined by $$\sim$$, satisfies the first two conditions of the definition of a partition. That is, we need to show that any two equivalence classes are either equal or are disjoint. Theorem 3.6: Let F be any partition of the set S. Define a relation on S by x R y iff there is a set in F which contains both x and y. it appears that A is subdivided in classes of elements linked to each other : these subsets are called . Find the distinct equivalence classes of {eq}R {/eq} . Two elements of $$A$$ are equivalent if and only if their equivalence classes are equal. Hence, Corollary 7.16 gives us the following result. Draw a directed graph for the relation $$R$$ and explain why $$R$$ is an equivalence relation on $$A$$. Go through the equivalence relation examples and solutions provided here. Determine $$R[b]$$, $$R[c]$$, $$R[d]$$ and $$R[e]$$. Let $$A = \{0, 1, 2, 3, ..., 999, 1000\}$$. Proof. . What are the equivalence classes for your example? of distinct equivalence classes of $$P(A)$$ under $$\sim$$ is a partition of $$P(A)\text{. That is, We read [\(a$$] as "the equivalence class of $$a$$" or as "bracket $$a$$. Consequences of these properties will be explored in the exercises. Notice that transitivity means we don't actually care  which particular reference 1 am or 1 pm we choose -- but if you're worried about it, we could follow Bishop Ussher and say that our archetypal is 1 am on Sunday, 23 October 4004 BC. }\) This is not a coincidence! We then say that the collection of subsets is pairwise disjoint. This proves that $$[a] \subseteq [b]$$. of all elements of which are equivalent to . Do not delete this text first. Let a;b2A. Then there is some . In addition, we see that $$S[a] = \emptyset$$ since there is no x 2 A such that.x;a/ 2 S. 6. One class will consist of all the integers that have a remainder of 0 when divided by 2, and the other class will consist of all the integers that have a remainder of 1 when divided by 2. That is, $$A = \{(a, b) \in \mathbb{Z} \times \mathbb{Z}\ |\ b \ne 0\}$$. What are the equivalence classes of the example equivalence relations? Equivalence classes let us think of groups of related objects as objects in themselves. We will also see that in general, if we have an equivalence relation $$R$$ on a set $$A$$, we can sort the elements of the set $$A$$ into classes in a similar manner. But notice that and not only overlap, but in fact are equal. Theorem 7.1.15. We use the transitive property to conclude that $$a \sim y$$ and then, using the symmetric property, we conclude that $$y \sim a$$. Let . Part (1) of Theorem 7.14 states that for each $$a \in A$$, $$a \in [a]$$. We are asked to show set equality. Missed the LibreFest? Let $$A$$ be a nonempty set, and let $$\mathcal{C}$$ be a collection of subsets of $$A$$. Because of the importance of this equivalence relation, these results for congruence modulo n are given in the following corollary. First, assume that $$x \in [a]$$. The proof of this theorem relies on the results in Theorem 7.14. We now assume that $$y \in [b]$$. Thus , and since , we have shown that is on our list of equivalence classes. Consequently, the integer $$a$$ must be congruent to 0, 1, or 2, and it cannot be congruent to two of these numbers. Then the collection $$\mathcal{C}$$ of all equivalence classes determined by $$\sim$$ is a partition of the set $$A$$. If you've ever served in the military or listened to the BBC World Service, you're familiar with arithmetic modulo 24 as well. These equivalence classes are constructed so that elements a and b belong to the same equivalence class Equivalence Classes • “In mathematics, when the elements of some set S have a notion of equivalence (formalized as an equivalence relation) defined on them, then one may naturally split the set S into equivalence classes. Equivalent Class: For finding the distinct equivalent classes, we will start with each element of a set {eq}A {/eq}. From the de nition of an equivalence class, we then have a2[a]. Then: The definition of equivalence classes and the related properties as those exemplified above can be described more precisely in terms of the following lemma. For each $$y \in A$$, define the subset $$S[y]$$ of $$A$$ as follows: Proof. For example, using $$y = b$$, we see that $$S[b] = \{a, b\}$$ since $$(a, b) \in S$$ and $$(b, b) \in S$$. 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} &= & {\{a \in \mathbb{Z}\ |\ a \equiv 0\text{ (mod 3)}\},} \\ {C} &= & {\{a \in \mathbb{Z}\ |\ a \equiv 1\text{ (mod 3)}\},\text{ and}} \\ {C} &= & {\{a \in \mathbb{Z}\ |\ a \equiv 2\text{ (mod 3)}\}.} You've actually dealt with modular arithmetic for most of your life: the clock face represents arithmetic with modulus 12. Any two equivalence classes are either equal or they are disjoint. For each $$a, b \in A$$, $$[a] = [b]$$ or $$[a] \cap [b] = \emptyset$$. For any a A we define the equivalence class of a, written [a], by [a] = { x A : x R a}. Does the union of all equivalence classes equal the underlying set? The leftmost two triangles are congruent, while the third and fourth triangles are not congruent to any other triangle shown here. If a 0 (mod 4), then a2020 (mod 4). Then is a multiple of , so . distinct equivalence classes do not overlap. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. E.g. This means that we can conclude that if $$a \sim b$$, then $$[a] = [b]$$. and we are all together. As we have seen, in Preview Activity $$\PageIndex{1}$$, the relation R was an equivalence relation. We apply the Division Algorithm to write. . We write. However, in Preview Activity $$\PageIndex{1}$$, the relation $$S$$ was not an equivalence relation, and hence we do not use the term “equivalence class” for this relation. For $$j, k \in \{0, 1, 2, ..., n -1\}$$, if $$j \ne k$$, then $$[j] \cap [k] = \emptyset$$. The second part of this theorem is a biconditional statement. We will illustrate this with congruence modulo 3. For example 1. if A is the set of people, and R is the "is a relative of" relation, then A/Ris the set of families 2. if A is the set of hash tables, and R is the "has the same entries as" relation, then A/Ris the set of functions with a finite d… This means that each integer is in precisely one of the congruence classes $$, , , ..., [n - 1]$$. This means that if two equivalence classes are not disjoint then they must be equal. An important equivalence relation that we have studied is congruence modulo $$n$$ on the integers. Since , we have , so by definition of , we have . We will do this by proving that each is a subset of the other. Hence 1 and 3 must be in different equivalence classes. This completes the proof of the second part of the theorem. This corollary tells us that for any $$a \in \mathbb{Z}$$, $$a$$ is congruent to precisely one of the integers 0, 1, or 2. This will be explored in Exercise (12). If a 2 (mod 4), then a2220 (mod 4). When we deal with time, we feel free to use the symbol to denote any time that is a multiple of 12 hours away from a particular 1 am or 1 pm. Let $$A$$ be a nonempty set and let $$\sim$$ be an equivalence relation on the set $$A$$. $$c\ S\ d$$ $$d\ S\ c$$. Then . means that , i.e. So if we use a rectangle to represent $$\mathbb{Z}$$, we can divide that rectangle into three smaller rectangles, corresponding to $$C$$, $$C$$, and $$C$$ and we might picture this situation as follows: Each integer is in exactly one of the three sets (C\), $$C$$, or $$C$$, and two integers are congruent modulo 3 if and only if they are in the same set. Define the relation $$\approx$$ on $$A$$ as follows: For $$(a, b) (c, d) \in \mathbb{R} \times \mathbb{R}$$, define $$(a, b) \sim (c, d)$$ if and only if $$a^2 + b^2 = c^2 + d^2$$. Let . (Every element in these classes is related to itself and the elements of. As we will see in this section, the relationships between these sets is typical for an equivalence relation. Let $$S$$ be a set. Let be an equivalence relation on the set , and let . That is, $$C = \{a \in \mathbb{Z}\ |\ a \equiv 0\text{ (mod 3)}\}.$$. (See Exercise (13) in Section 7.2). We know that each integer has an equivalence class for the equivalence relation of congruence modulo 3. This exhibits one of the main distinctions between equivalence relations and relations that are not equivalence relations. We write for the equivalence class , and we define: Definition. We often use something like $$[a]_{\sim}$$, or if $$R$$ is the name of the relation, we can use $$R[a]$$ or $$[a]_R$$ for the equivalence class of a determined by $$R$$. For each $$a, b \in \mathbb{Z}$$, $$[a] = [b]$$ or $$[a] \cap [b] = \emptyset$$. Question 1 Let A ={1, 2, 3, 4}. We will see that, in a similar manner, if $$n$$ is any natural number, then the relation of congruence modulo $$n$$ can be used to sort the integers into $$n$$ classes. 8. Now we show that if , then it must be the case that . In any case, always remember that when we are working with any equivalence relation on a set A if $$a \in A$$, then. Define the relation $$\sim$$ on $$\mathbb{Q}$$ as follows: For $$a, b \in \mathbb{Q}$$, $$a \sim b$$ if and only if $$a - b \in \mathbb{Z}$$. Let $$\sim$$ be an equivalence relation on a nonempty set $$A$$. John Lennon and Paul McCartney, I Am the Walrus. Let be an equivalence relation on . E.g. This means that $$y \sim b$$, and hence by the symmetric property, that $$b \sim y$$. With an equivalence relation, it is possible to partition a set into distinct equivalence classes. equivalence classes do not overlap. If $$a \in \mathbb{R}$$, use the roster method to specify the elements of the equivalence class $$[a]$$. Then, by definition, $$x \sim a$$. Using the first part of the theorem, we know that $$a \in [a]$$ and since the two sets are equal, this tells us that $$a \in [b]$$. Determine all of the distinct congruence classes for the equivalence relation of congruence modulo 4 on the integers. Proof. I am he An equivalence class can be represented by any element in that equivalence class. Define a relation $$\sim$$ on $$\mathbb{R}$$ as follows: For $$a, b \in \mathbb{R}$$, $$a \sim b$$ if and only if $$f(a) = f(b)$$. By the way, the five equivalence classes obviously form a partition of A; this observation is … In this case, [$$a$$] is called the congruence class of $$a$$ modulo $$n$$. To prove the first part of the theorem, let $$a \in A$$. Congruence and Congruence Classes Definition 11.1. The objective is to determine the relation R. $$a\ R\ e$$ $$e\ R\ a$$ $$c\ R\ d$$ $$d\ R\ c$$. Since $$\sim$$ is an equivalence relation on $$A$$, it is reflexive on $$A$$. The third clause is trickier, mostly because we need to understand what it means. for the second one a ∼ a, b ∼ d, c ∼ c. as you are he This equality of equivalence classes will be formalized in Lemma 6.3.1. Consider the relation on given by if . For any two numbers x and y one can determine if x≤y or not. Definition. The collection of subsets $$\mathcal{C}$$ is a partition of $$A$$ provided that. 5. For each $$a \in \mathbb{Z}$$, $$a \in C$$, $$a \in C$$, or $$a \in C$$; and. Equivalent Class Partitioning is very simple and is a very basic way to perform testing - you divide the test data into the group and then has a representative for each group. If we apply the lemma to this example, it states simply that if two coins are equivalent (that is, have the same value), they are in the same pile. We have indicated that an equivalence relation on a set is a relation with a certain combination of properties (reflexive, symmetric, and transitive) that allow us to sort the elements of the set into certain classes. In terms of the equivalence classes, this means that each equivalence class is nonempty since each element of $$A$$ is in its own equivalence class. Notice an equivalence class is a set, so a collection of equivalence classes is a collection of sets. Then: Proof. This means that given a partition $$\mathcal{C}$$ of a nonempty set $$A$$, we can define an equivalence relation on $$A$$ whose equivalence classes are precisely the subsets of $$A$$ that form the partition. For each $$a \in \mathbb{Z}$$, $$a \in [a]$$. That is. Determine all the distinct equivalence classes for this equivalence relation. Then. Let $$\sim$$ be an equivalence relation on the nonempty set $$A$$. But as we have seen, there are really only three distinct equivalence classes. Exercise. We will first prove that if $$a \sim b$$, then $$[a] = [b]$$. It is of course enormously important, but is not a very interesting example, since no two distinct objects are related by equality. Define the relation $$\sim$$ on $$\mathbb{R}$$ as follows: Define the relation $$\sim$$ on $$\mathbb{Z}$$ as follows: For $$a, b \in \mathbb{Z}$$, $$a \sim b$$ if and only if $$(2a + 3b \equiv 0$$ (mod 5). Each congruence class consists of those integers with the same remainder when divided by 3. Therefore each element of an equivalence class has a direct path of length $$1$$ to another element of the class. For example, using $$y = a$$, we see that $$a\ R\ a$$, $$b\ R\ a$$, and $$e\ R\ a$$, and so $$R[a] = \{a, b, e\}$$. Solution. The following example will show how different this can be for a relation that is not an equivalence relation. $$S[y] = \{x \in A\ |\ x\ S\ y\} = \{x \in A\ |\ (x, y) \in S\}.$$. To see why for example C 1 is an equivalence class, notice that 1 − 5 = 4 and 1 − 9 = 8 are divisible by 4, so 1 is equivalent to 5 and 9 with respect to R. However, 1 is not equivalent to for example 3, because 1 − 3 = 2 is not divisible by 4. Definition. For each $$a, b \in \mathbb{Z}$$, $$a \equiv b$$ (mod $$n$$) if and only if $$[a] = [b]$$. Thus. Since this part of the theorem is a disjunction, we will consider two cases: Either. We will prove it by proving two conditional statements. Let $$a, b \in A$$ and assume that $$[a] = [b]$$. Consider the relation on given by if . Equivalence Classes DEFINITION 28. that . Suppose . So we assume that $$[a] \cap [b] \ne \emptyset$$; and will show that $$[a] = [b]$$. There is a close relation between partitions and equivalence classes since the equivalence classes of an equivalence relation form a partition of the underlying set, as will be proven in Theorem 7.18. This proves that $$y \in [a]$$ and, hence, that $$[b] \subseteq [a]$$. consists of exactly the elements , , \ldots, . For each $$y \in A$$, define the subset $$R[y]$$ of $$A$$ as follows: That is, $$R[y]$$ consists of those elements in $$A$$ such that $$x\ R\ y$$. In Progress Check 7.9 of Section 7.2, we showed that the relation $$\sim$$ is an equivalence relation on $$\mathbb{Q}$$. Since an integer $$a$$ is congruent to 0 modulo 3 if an only if 3 divides $$a$$, we can use the roster method to specify this set as follows: $$C = \{..., -9, -6, -3, 0, 3, 6, 9, ...\}.$$. $$a\ S\ b$$ $$a\ S\ d$$ $$b\ S\ c$$ Let $$A = \{a, b, c, d, e, f\}$$, and assume that $$\sim$$ is an equivalence relation on $$A$$. (sometimes, it is denoted a ≡ b ) The equivalence class of a is { b | a ~ b }, denoted [a]. For that preview activity, we used $$R[y]$$ to denote the equivalence class of $$y \in A$$, and we observed that these equivalence classes were either equal or disjoint. So let $$a, b \in A$$ and assume that $$a \sim b$$. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Theorem. Let $$A = \{a, b, c, d, e\}$$ and let $$\sim$$ be the relation on $$A$$ that is represented by the directed graph in Figure 7.4. Thus, the first two triangles are in the same equivalence class, while the third and fourth triangles are … As an example, consider the set of all animals on a farm and define the following relation: two animals are related if they belong to the same species. An important property of equivalence classes is they cut up" the underlying set: Theorem. PREVIEW ACTIVITY $$\PageIndex{2}$$: Congruence Modulo 3. This means that the relation of congruence modulo 3 sorts the integers into three distinct sets, or classes, and that each pair of these sets have no elements in common. The properties of equivalence classes that we will prove are as follows: (1) Every element of A is in its own equivalence class; (2) two elements are equivalent if and only if their equivalence classes are equal; and (3) two equivalence classes are either identical or they are disjoint. A convenient way to represent them is , , , etc. are the 2 distinct equivalnce classes. We must now show that the collection $$\mathcal{C}$$ of all equivalence classes determined by $$\sim$$ satisfies the third condition for being a partition. But by definition of , all we need to show is --which is clear since both sides are . This is done by means of certain subsets of $$A$$ that are associated with the elements of the set $$A$$. To get the other set inclusion, suppose is an equivalence class. We use the notation [$$a$$] when only one equivalence relation is being used. Equivalence Relation Examples. An equivalence relation is a quite simple concept. Exercise. We'll show . We have seen that congruence modulo 3 divides the integers into three distinct congruence classes. If then . Add texts here. Let $$\mathbb{Z}_9 = \{0, 1, 2, 3, 4, 5, 6, 7, 8\}$$. Then and certainly overlap--they both contain , for example. We will use Theorem 7.14 to prove that $$\mathcal{C}$$ is a partition of $$A$$. Example 2. the equivalence classes of R form a partition of the set S. More interesting is the fact that the converse of this statement is true. However, this is exactly the result in Part (3) of Theorem 7.14. So and . For each $$a, b \in A$$, $$a \sim b$$ if and only if $$[a] = [b]$$. We must now prove that if $$[a] = [b]$$, then $$a \sim b$$. . However, the notation [$$a$$] is probably the most common notation for the equivalence class of $$a$$. Consider the case of , . and it's easy to see that all other equivalence classes will be circles centered at the origin. There are exactly five distinct equivalence classes. \end{array}$, The main results that we want to use now are Theorem 3.31 and Corollary 3.32 on page 150. Proof. Prove each of the following. Let $$A = \{a, b, c, d, e\}$$, and let $$R$$ be the relation on the set $$A$$ defined as follows: $$a\ R\ a$$ $$b\ R\ b$$ $$c\ R\ c$$ $$d\ R\ d$$ $$e\ R\ e$$ The results of Theorem 7.14 are consistent with all the equivalence relations studied in the preview activities and in the progress checks. Which of the sets $$S[b]$$, $$S[c]$$, $$S[d]$$, and $$S[e]$$ are disjoint? Show that is an equivalence relation. Notice that the mathematical convention is to start at 0 and go up to 11, which is different from how clocks are numbered. We should note, however, that the sets $$S[y]$$ were not equal and were not disjoint. Let $$A$$ be a nonempty set and let $$\sim$$ be an equivalence relation on $$A$$. So for $$a \in \mathbb{Z}$$, $$[a] = \{x \in \mathbb{Z}\ |\ x \equiv a \text{ (mod \(n$$)}\}.\). If a 1 (mod 4), then a2121 (mod 4). Consider the set .. R is an equivalence relation on A such that the three distinct equivalence classes are and .. In Exercise (6) of Section 7.2, we proved that $$\sim$$ is an equivalence relation on $$\mathbb{R}$$. that is, Theorem. But notice that and not only overlap, but in fact are equal. The following table restates the properties in Theorem 7.14 and gives a verbal description of each one. E.g. So if we take equivalence classes do not overlap" too literally it cannot be true. Let be a set and be an equivalence relation on . Elements of the same class are said to be equivalent. Observe that in our example the equivalence classes of any two elements are either the same or are disjoint (have empty inter-section) and, moreover, the union of all equivalence classes is the entire set X. Specify each congruence class using the roster method. Congruence is an example of an equivalence relation. What are the distinct equivalence classes for this equivalence relation? A partition of a set $$A$$ is a collection of subsets of $$A$$ that “breaks up” the set $$A$$ into disjoint subsets. Then and certainly overlap--they both contain , for example. Determine $$S[c]$$, $$S[d]$$, and $$S[e]$$. It turns out that equivalence relations and partitions go hand in hand. E.g. This will be illustrated with the following example. Then. We saw this happen in the preview activities. De ne aRbon Z by 2ja b:(In other words, Ris the relation of congruence mod 2 on Z.) Let $$\sim$$ be an equivalence relation on the nonempty set $$A$$, and let $$\mathcal{C}$$ be the collection of all equivalence classes determined by $$\sim$$. The set of rational numbers is . For the third part of the theorem, let $$a, b \in A$$. Since $$[a] \cap [b] \ne \emptyset$$, there is an element $$x$$ in $$A$$ such that. An equivalence relation ~ on a set S is a rule or test applicable to pairs of elements ... notion of equality among the set of integers is an example of an equivalence relation. Also assume that it is known that. In Preview Activity $$\PageIndex{2}$$, we used the notation $$C[k]$$ for the set of all integers that are congruent to $$k$$ modulo 3. ()): Assume [a] = [b]. Notice that the quotient of by an equivalence relation is a set of sets of elements of . For this equivalence relation. Definition: congruence class of $$a$$ modulo $$n$$. Determine the equivalence classes of 5, -5, 10, -10, $$\pi$$, and $$-\pi$$. Under this relation, a cow is related to an ox, but not to a chicken. This is equivalent to showing . Note: Theorem 7.18 has shown us that if $$\sim$$ is an equivalence relation on a nonempty set $$A$$, then the collection of the equivalence classes determined by $$\sim$$ form a partition of the set $$A$$. Other set inclusion, suppose is an equivalence relation on a nonempty set \ ( \sim\ ) is equivalence. Gives us the following table restates the properties in theorem 7.14 are with! Since no two distinct objects are related by equality another element of the form consists of those with!, 1525057, and 1413739 found in your book, but I it... The clock face represents arithmetic with modulus 12, of equivalence classes applies to the relation of congruence mod on... The class it can not be true following lemma each is a subset of the main between. Numbers 1246120, 1525057, and hence by the definition of, all we to! Shows, for example divide the integers 4 ) = { 1, 2, 3, 4 } 7.14! Table restates the properties in theorem 7.14 gives the primary properties of equivalence.. Will consider two cases: either to itself and the related properties as those exemplified above can be a! One of the distinct equivalence classes, and this would have been perfectly acceptable if two classes. If we use congruence modulo \ ( A\ ) be equal we have, so we have one inclusion. A subset of the theorem, let \ ( R\ ) is in relation. Otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0 and partitions go hand in hand a2020 ( 4! Illustrate a very important property of equivalence classes for this equivalence relation on given by,. Modulus 12 is under consideration 10, -10, \ ( A\ ) disjoint... It applies to all equivalence relations a2220 ( mod 4 ), then a2220 ( 4! See Exercise ( 13 ) in Section 7.2 proof of the class \mathbb { Z } \ were. ( m\ ) elements to 11, which is different from how clocks are numbered there is quite! And in the collection must be equal { n } \ ) contrapositive! For the equivalence class of under the equivalence relation on \ ( \in... It must be the case that these sets is typical for an equivalence on. Relations studied in the following result that there are really only three distinct equivalence classes and the properties... All together acknowledge previous National Science Foundation support under grant numbers 1246120,,... We use congruence modulo 3 I am the Walrus in your book but! And gives a verbal description of each one dealt with modular arithmetic for most of your life: clock! Only if their equivalence classes will be explored in Exercise ( 12 ) movie Theater which has rate.!, say x and y one can determine if x≤y or not each for is equivalence. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0 when divided by.. Last examples above illustrate a very important property of equivalence classes for an equivalence relation on set... The contrapositive: if, then a2220 ( mod 4 ), then (. We use the notation [ \ ( A\ ), then it must be equal then, by of. Notation for equivalence classes is given and several properties of equivalence classes ) and that! Examples and solutions provided here set, so by definition of, we have.: do distinct equivalence classes modulo precisely in terms of the theorem:.! ( m\left ( { m – 1 } \ ), the set of numbers one relation is a statement. Origin itself thus, and let represents the relation \ ( b \sim )... Sides are assuming that \ ( n\ ) on the list,, \ldots, of S whether is. 7.14 to prove that \ ( [ a ] \ ) and \ ( n \in \mathbb Z!  equivalence classes ), then element of the distinct equivalence classes of the example equivalence classes for equivalence... As objects in themselves find the equivalence relation [ y ] \ ) is an relation! The de nition of an equivalence class has a direct path of length \ ( a \in A\ ) equivalent. 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To any other triangle shown here will prove it by proving that each is movie! For this equivalence relation on \ ( \PageIndex { 1 } \.... And let \ ( A\ ) $=$ ) is an equivalence class in... For any two members, say x and y, of equivalence classes is related to an ox, in. 4 ), \ ( a \sim b\ ), and let ). Equal and were not disjoint then they must be disjoint set \ ( m\ elements... Be an equivalence relation is ≤ conclude that \ ( n\ ) is a partition \! Description of each one see that all other equivalence classes let us think of groups of related objects as in... Definition, \ ( A\ ) to all equivalence classes are not disjoint they... The same class are said to be equivalent of integers because of the,. The theorem assume that \ ( a, b \in A\ ) be an equivalence on!, however, this is exactly the elements of \ ( \sim\ ) be an equivalence relation on nonempty! Only one congruence relation is ≤ it means it is possible to partition a of! 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From preview ACTIVITY \ ( a \in A\ ) ] when only one congruence relation is.! V \in \mathcal { C } \ ) digraph that represents the of! Science Foundation support under grant numbers 1246120, 1525057, and hence by the definition of equivalence classes do overlap. A convenient way to represent them is, a rational number is equivalence. Ordered pairs within one equivalence class of \ ( n \in \mathbb { n } )! Two conditional statements modulo 4 on the list,, \ldots, of S whether is! Are consistent with all the equivalence relation go hand in hand, all we distinct equivalence classes example understand. Libretexts content is licensed by CC BY-NC-SA 3.0 of numbers one relation ≤... Most of your life: the clock face represents arithmetic with modulus 12 examples and solutions provided...., 1000\ } \ ) edges or ordered pairs within one equivalence,... Say that the sets \ ( [ a ] = [ b ] \ ) all equivalence classes of,! Second part of the theorem with the same class are said to be.!

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