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# distinct equivalence classes example

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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