Equivalence Relations 4/10/2017 Section 7.5 Equivalence Relations ch7.5

4/10/2017 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 ch7.5

Example 1 Congruence modulo m is a widely used equivalence relation: 4/10/2017 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 ch7.5

4/10/2017 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 ch7.5

Example 1  a  b (mod m) is an equivalence relation 4/10/2017 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 ch7.5

4/10/2017 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? ch7.5

Example 2 R is reflexive: f(x) = f(x) for all x  A 4/10/2017 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 ch7.5

4/10/2017 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 ch7.5

4/10/2017 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, …} ch7.5

4/10/2017 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, …} ch7.5

4/10/2017 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 ch7.5

4/10/2017 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 ≠  ch7.5

Proof Assuming aRb, prove [a]=[b] by showing [a]≤[b] and [b]≤[a]: 4/10/2017 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] ch7.5

4/10/2017 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 ch7.5

Partitions Let R be an equivalence relation on set A 4/10/2017 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 ch7.5

4/10/2017 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 Ai forms a partition of S if and only if: Ai   Ai  Aj =  when i  j and Ai = S ch7.5

Example 5 Suppose S = {a,b,c,d,e,f} 4/10/2017 Example 5 Suppose S = {a,b,c,d,e,f} The collection of sets A1 = {a,b,c}, A2 = {d} and A3 = {e,f} form partitions of S, since: they are disjoint and their union is S ch7.5

Equivalence classes & partitions 4/10/2017 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 ch7.5

Example 6 Assume {Ai} is a partition on S 4/10/2017 Example 6 Assume {Ai} is a partition on S Let R be a relation on S consisting of the pair (x,y) where x,y  Ai To prove R is an equivalence relation, we must show that it is reflexive, symmetric, and transitive ch7.5

4/10/2017 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) ch7.5

4/10/2017 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 ch7.5

One more theorem Let R be an equivalence relation on a set S 4/10/2017 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 {Ai} or set S, there is an equivalence relation R that has the sets Ai, where Ai = S, as its equivalence classes ch7.5

4/10/2017 Example 7 What are the ordered pairs in the equivalence relation of {0,1,2,3,4,5} in the following partitions? A1= {0}, A2 = {1,2}, A3 = {3,4,5}: (0,0) belongs to R since A1 is an equivalence class; (1,1), (1,2), (2,1) and (2,2) belong to R because A2 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 A3 is an equivalence class ch7.5