Then there is some with . 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. Consider the relation on given by if . Assume is nonempty. Congruence is an example of an equivalence relation. For example, if S is a set of numbers one relation is ≤. This is equivalent to showing . We could have used a similar notation for equivalence classes, and this would have been perfectly acceptable. \end{array}\], The main results that we want to use now are Theorem 3.31 and Corollary 3.32 on page 150. If a 1 (mod 4), then a2121 (mod 4). This will be explored in Exercise (12). This proves that \(y \in [a]\) and, hence, that \([b] \subseteq [a]\). Each congruence class consists of those integers with the same remainder when divided by 3. Proof. Elements of the same class are said to be equivalent. We will illustrate this with congruence modulo 3. That is. 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. Consequently, the integer \(a\) must be congruent to 0, 1, or 2, and it cannot be congruent to two of these numbers. Hence 1 and 3 must be in different equivalence classes. That is, We read [\(a\)] as "the equivalence class of \(a\)" or as "bracket \(a\). Notice that the quotient of by an equivalence relation is a set of sets of elements of . This exhibits one of the main distinctions between equivalence relations and relations that are not equivalence relations. 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 \(a^2 + b^2 = c^2 + d^2\). 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. But as we have seen, there are really only three distinct equivalence classes. Definition: congruence class of \(a\) modulo \(n\). Specify each congruence class using the roster method. But notice that and not only overlap, but in fact are equal. Since \([a] \cap [b] \ne \emptyset\), there is an element \(x\) in \(A\) such that. 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. This corollary tells us that for any \(a \in \mathbb{Z}\), \(a\) is congruent to precisely one of the integers 0, 1, or 2. 8. \(a\ S\ b\) \(a\ S\ d\) \(b\ S\ c\) ()): Assume [a] = [b]. The third clause is trickier, mostly because we need to understand what it means. for the first problem 0 ∼ 4, 1 ∼ 3, 2 ∼ 2 so you have 3 equivalence classes (note that R is an equivalence realation). Therefore each element of an equivalence class has a direct path of length \(1\) to another element of the class. Theorem. To get the other set inclusion, suppose is an equivalence class. and it's easy to see that all other equivalence classes will be circles centered at the origin. Find the equivalence class [(1, 3)]. PREVIEW ACTIVITY \(\PageIndex{2}\): Congruence Modulo 3. is the set of all pairs of the form . Suppose . Let be an equivalence relation on . equivalence classes. We introduce the following formal definition. First, assume that \(x \in [a]\). Add texts here. This means that we can conclude that if \(a \sim b\), then \([a] = [b]\). A partition of a set \(A\) is a collection of subsets of \(A\) that “breaks up” the set \(A\) into disjoint subsets. . Example 5.1.1 Equality ($=$) is an equivalence relation. Consider the case of , . Given an equivalence relation on , the set of all equivalence classes is called the {\em quotient of by }. For the third part of the theorem, let \(a, b \in A\). We denote aEb as a ~ b. In a similar manner, if we use congruence modulo 2, we simply divide the integers into two classes. Which of the sets \(R[a]\), \(R[b]\), \(R[c]\), \(R[d]\) and \(R[e]\) are disjoint? Since , we have , so by definition of , we have . So, in Example 6.3.2, [S2] = [S3] = [S1] = {S1, S2, S3}. 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. Technically, each pair of distinct subsets in the collection must be disjoint. 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. We know that each integer has an equivalence class for the equivalence relation of congruence modulo 3. Proof. Let R be the equivalence relation on A × A defined by (a, b)R(c, d) iff a + d = b + c . distinct equivalence classes do not overlap. Let Rbe an equivalence relation on a nonempty set A. 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. We can also define subsets of the integers based on congruence modulo \(n\). A convenient way to represent them is , , , etc. We will now prove that the two sets \([a]\) and \([b]\) are equal. If a 0 (mod 4), then a2020 (mod 4). 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. We use the notation [\(a\)] when only one equivalence relation is being used. Using the first part of the theorem, we know that \(a \in [a]\) and since the two sets are equal, this tells us that \(a \in [b]\). Note that we have . For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. In Exercise (6) of Section 7.2, we proved that \(\sim\) is an equivalence relation on \(\mathbb{R}\). Then there is some . Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. The collection of subsets \(\mathcal{C}\) is a partition of \(A\) provided that. E.g. equivalence classes do not overlap. Congruence modulo \(n\) is an equivalence relation on \(\mathbb{Z}\). Any two equivalence classes are either equal or they are disjoint. 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