Math 546 Problem Set 8 1. Prove: If R is a symmetric and transitive relation on X, and every element x of X is related to something in X, then R is also a reflexive relation. Proof: Suppose that x is any element of X. Then x is related to something in X, say to y. Hence, we have xRy, and so by symmetry, we must have yRx. But then by transitivity, xRy and yRx imply that xRx. Thus every element is related to itself and so the relation is reflexive. 2. Find a symmetric and transitive relation on X = {a,!b,!c} that is not reflexive. Example: R = {(a,b),!(b,a),!(a,a),!(b,!b)} on the set X = {a,!b,!c} . Recall, n m ! n divides m ! m = kn for some integer k (n > 0) . Define x ! y mod n " n (x # y). (We say that x is congruent to y modulo n) So, for example, 27 ! 2mod5, !!!!!39 ! 6mod11, !!!!!!20 !/ 6mod5 3. Show that for any n > 0, x ! y mod n is an equivalence relation on the set Z of integers. 4. Verify the following properties of congruence modulo n > 0. Let x ! y mod n and a ! b mod n and let k be an integer. (i). kx ! kymod n (ii). k + x ! k + y mod n (iii). ax ! by mod n (iv). If k ≥ 0, then xk ! yk mod n (Argue by induction.) (v). x ! y mod n if and only if x and y leave the same remainder upon division by n. Solution: See the hand out on congruence. 5. What is the remainder when 2102 is divided by 11? ( Hint: 25 ! "1mod11 ) Solution: Suppose that r is the remainder when 2102 is divided by 11. Then 2102 ! r mod11. So we must figure out what value of r, 0 ≤ r ≤ 10 satisfies 2102 ! r mod11. 20 20 But since 25 ! "1mod11 , then we get (25 ) ! ("1) mod11 # 2100 ! 1mod11. But now 4 ! 2100 " 4 !1mod11. Thus 2102 ! 4 mod11 and so the remainder is 4. 6. For the relation, x ! y mod 5, list four elements from the equivalence class of 3 and from the equivalence class of 7. Solution: [3] contains 3, 8, 13, and 8. [7] contains -3, 2, 7, 12, 7. The relation R on the set A = {a, b, c} is transitive but not reflexive. Three of the ordered pairs that belong to R are: (b,!a), (a,!b) and (c,!a) . From this information: (a). What other ordered pairs must belong to R? Solution: (b,!b),!(c,!b),!(a,!a) . (b). Is there some ordered pair that cannot belong to R? Solution: Yes, (c,!c) . (c). Is R symmetric? Explain. Solution: No, it cannot be symmetric. But why? .