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1 Section 7.5 Equivalence Relations. 2 Definition A relation on a set is called an equivalence relation if it is: –Reflexive, –Symmetric and –Transitive.

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Presentation on theme: "1 Section 7.5 Equivalence Relations. 2 Definition A relation on a set is called an equivalence relation if it is: –Reflexive, –Symmetric and –Transitive."— Presentation transcript:

1 1 Section 7.5 Equivalence Relations

2 2 Definition A relation on a set is called an equivalence relation if it is: –Reflexive, –Symmetric and –Transitive Two elements related by an equivalence relation are said to be equivalent

3 3 Example 1 Congruence modulo m is a widely used equivalence relation: –Let m be a positive integer (m > 1); show that R = {(a,b) | a  b (mod m)} is an equivalence relation on the set of integers –a  b (mod m) means a is congruent to b mod m; this is true if and only if m divides (is a factor of) a-b

4 4 Example 1 To show that R is an equivalence relation, we must prove it is: –Reflexive: a-a is divisible by m, since 0=0 * m;  a  a (mod m), and the relation is reflexive –Symmetric: suppose a  b (mod m); then a-b is divisible by m, so a-b = km; it follows that b- a = (-k)m, so b  a (mod m), and the relation is symmetric –Transitive: see next slide

5 5 Example 1 –Transitive: suppose a  b (mod m) and b  c (mod m); so m divides both a-b and b-c; this means there exist integers k and n with a-b = km and b-c = nm; adding these equations, we get a-c = (a-b)+(b-c) = km+nm = (k+n)m; thus, a  c (mod m) and the relation is transitive  a  b (mod m) is an equivalence relation

6 6 Example 2 Suppose A is a nonempty set, and f is a function that has A as its domain; let R be a relation on A consisting of all ordered pairs (x,y) where f(x) = f(y) Is R an equivalence relation on A?

7 7 Example 2 R is reflexive: f(x) = f(x) for all x  A R is symmetric: if f(x) = f(y), then f(y) = f(x) (by the fundamental property of equality) R is transitive: if f(x) = f(y) and f(y) = f(z), then f(x) = f(z) (also by the fundamental property of equality)  R is an equivalence relation on A

8 8 Equivalence classes Let R be an equivalence relation on set A; the set of all elements that are related to an element e  A is called the equivalence class of e The equivalence class of e with respect to R is denoted [e] R If R is an equivalence relation on A, the equivalence class of element e  A is [e] R = {s | (e,s)  R} If b  [e] R b is a representative of this equivalence class

9 9 Example 3 What are equivalence classes of 0 and 1 for congruence modulo 4? –For 0, the equivalence class contains all integers e such that e  0 (mod 4); in other words, a-0 is divisible by 4, so [0] = {…, -8, -4, 0, 4, 8, …} –For 1, the equivalence class is all integers e such that e  1 (mod 4); so [1] = {…, -7, -3, 1, 5, 9, …}

10 10 Generalizing example 3 We can replace 4 in the previous example with any integer m The equivalence classes of the relation congruence modulo m are called congruence classes modulo m; such a congruence class of integer a is denoted [a] m and [a] m = {…, a-2m, a-m, a, a+m, a+2m, …}

11 11 Example 4 Suppose R is the equivalence relation on non-empty set A such that R={(x,y)|f(x)=f(y)}; what are the equivalence classes on R? The equivalence class of x is the set of all y  A such that f(y)=f(x) By definition, this is the inverse image of f(x);  the equivalence classes are those sets f -1 (b) for every b in the range of f

12 12 Equivalence classes of 2 elements of a set are either identical or disjoint Let R be an equivalence relation on set A. The following statements are equivalent: –(i) aRb –(ii) [a] R = [b] R –(iii) [a] R  [b] R ≠ 

13 13 Proof Assuming aRb, prove [a]=[b] by showing [a]≤[b] and [b]≤[a]: –Suppose c  [a] – so aRc; –Since aRb and R is symmetric, we know that bRa; –Since R is transitive, bRa, and aRc, so bRc; –Hence c  [b]; –This shows that [a]≤[b] –We can use the same logic to show b]≤[a]

14 14 Proof Assume [a]=[b] It follows that [a]  [b] ≠  because [a] is nonempty (since a  [a] because R is reflexive) Assume [a] R  [b] R ≠  ; then there is some element c with c  [a] and c  [b]: –  aRc and bRc –Since R is symmetric, cRb –Since R is transitive, aRc and cRb, so aRb So aRb  [a] R = [b] R, [a] R = [b] R  [a] R  [b] R ≠ , and [a] R  [b] R ≠   aRb; thus the statements are equivalent

15 15 Partitions Let R be an equivalence relation on set A –the union of the equivalence classes of R is all of A, since each a  A is its own equivalence class, [a] R –in other words,  [a] R = A a  A From the theorem previously proved, it follows that equivalence classes are either equal or disjoint, so [a] R  [b] R =  only when [a] R  [b] R

16 16 Partitions Equivalence classes split set A into disjoint subsets, called partitions: collections of disjoint non-empty subsets of a set which have the set as their union In other words, the collection of subsets A i forms a partition of S if and only if: –Ai  –Ai   –A i  A j =  when i  j and –  A i = S

17 17 Example 5 Suppose S = {a,b,c,d,e,f} The collection of sets A 1 = {a,b,c}, A 2 = {d} and A 3 = {e,f} form partitions of S, since: –they are disjoint and –their union is S

18 18 Equivalence classes & partitions Equivalence classes of an equivalence relation on a set form a partition of the set; subsets in this partition are the equivalence classes Conversely, every partition of a set can be used to form an equivalence relation: –2 elements are equivalent with respect to this relation if and only if they are in the same subset of the partition

19 19 Example 6 Assume {A i } is a partition on S Let R be a relation on S consisting of the pair (x,y) where x,y  A i To prove R is an equivalence relation, we must show that it is reflexive, symmetric, and transitive

20 20 Example 6 Reflexive: (a,a)  R for every a  S, since a is in the same subset as itself; Symmetric: if (a,b)  R then b and a are in the same subset, so (b,a)  R Transitive: if (a,b)  R and (b,c)  R then a and b are both in the same subset (x) and b and c are both in the same subset (y) (continued next slide)

21 21 Example 6 Since by definition the subsets are disjoint, and b is in both x and y, x and y must be equal So a and c are in the same subset and R is transitive Therefore R is an equivalence relation

22 22 One more theorem Let R be an equivalence relation on a set S Then the equivalence classes of R form a partition of S Conversely, given a partition {A i } or set S, there is an equivalence relation R that has the sets A i, where  A i = S, as its equivalence classes

23 23 Example 7 What are the ordered pairs in the equivalence relation of {0,1,2,3,4,5} in the following partitions? A 1 = {0}, A 2 = {1,2}, A 3 = {3,4,5}: (0,0) belongs to R since A 1 is an equivalence class; (1,1), (1,2), (2,1) and (2,2) belong to R because A 2 is an equivalence class; (3,3), (3,4), (3,5), (4,4), (4,3), (4,5), (5,5), (5,3,), (5,4) belong to R because A 3 is an equivalence class


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