7. The proof is found in your book, but I reproduce it here. Properties When R is an equivalence relation defined on a set A, it partitions A into non overlapping distinct equivalence classes. For example, one may distinguish fractions from rational numbers, the latter being equivalence classes of fractions: the fractions / and / are distinct as fractions (as different strings of symbols) but they "represent" the same rational number (the same point on a number line). 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. For each \(x \in A\), there exists a \(V \in \mathcal{C}\) such that \(x \in V\). We should note, however, that the sets \(S[y]\) were not equal and were not disjoint. ()): Assume [a] = [b]. Thus, the first two triangles are in the same equivalence class, while the third and fourth triangles are … Without using the terminology at that time, we actually determined the equivalence classes of the equivalence relation \(R\) in Preview Activity \(\PageIndex{1}\). Which of the sets \(R[a]\), \(R[b]\), \(R[c]\), \(R[d]\) and \(R[e]\) are equal? 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. Every element of \(A\) is in its own equivalence class. An equivalence class can be represented by any element in that equivalence class. E.g. 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 \(R\). So we have. In Exercise (15) of Section 7.2, we proved that \(\sim\) is an equivalence relation on \(\mathbb{R} \times \mathbb{R}\). Since this part of the theorem is a disjunction, we will consider two cases: Either. In this case, [\(a\)] is called the congruence class of \(a\) modulo \(n\). Theorem 7.14 gives the primary properties of equivalence classes. Then, by definition, \(x \sim a\). distinct equivalence classes do not overlap. We will prove it by proving two conditional statements. So let \(a, b \in A\) and assume that \(a \sim b\). EXAMPLE 29. We are asked to show set equality. The third clause is trickier, mostly because we need to understand what it means. Exercise. Missed the LibreFest? Theorem. For any a A we define the equivalence class of a, written [a], by [a] = { x A : x R a}. We'll show . This means that \(y \sim b\), and hence by the symmetric property, that \(b \sim y\). 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. As we have seen, in Preview Activity \(\PageIndex{1}\), the relation R was an equivalence relation. This means that we can conclude that if \(a \sim b\), then \([a] = [b]\). Claim. its class). From the de nition of an equivalence class, we then have a2[a]. We introduce the following formal definition. From our assumption, a2[b]. and it's easy to see that all other equivalence classes will be circles centered at the origin. Notice that the quotient of by an equivalence relation is a set of sets of elements of . That is, \(C[0] = \{a \in \mathbb{Z}\ |\ a \equiv 0\text{ (mod 3)}\}.\). For example, using \(y = b\), we see that \(S[b] = \{a, b\}\) since \((a, b) \in S\) and \((b, b) \in S\). That is, \(A = \{(a, b) \in \mathbb{Z} \times \mathbb{Z}\ |\ b \ne 0\}\). 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 \([x] = [b]\). For each \(y \in A\), define the subset \(S[y]\) of \(A\) as follows: In Progress Check 7.9 of Section 7.2, we showed that the relation \(\sim\) is an equivalence relation on \(\mathbb{Q}\). Hence, \([a] = [b]\), and we have proven that \([a] = [b]\) or \([a] \cap [b] = \emptyset\). for the second one a ∼ a, b ∼ d, c ∼ c. Let \(A = \mathbb{Z} \times (\mathbb{Z} - \{0\})\). and we are all together. This will be explored in Exercise (12). The following table restates the properties in Theorem 7.14 and gives a verbal description of each one. For each \(a, b \in A\), \([a] = [b]\) or \([a] \cap [b] = \emptyset\). That is, We read [\(a\)] as "the equivalence class of \(a\)" or as "bracket \(a\). Do not use fractions in your proof. For example, using \(y = a\), we see that \(a\ R\ a\), \(b\ R\ a\), and \(e\ R\ a\), and so \(R[a] = \{a, b, e\}\). 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