a7-sol

a7-sol - MATH 135 Assignment 7 Winter 2010 Due 8:30am on...

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MATH 135 Assignment 7, Winter 2010 Due 8:30am on Wednesday March 10, 2010 in the drop boxes opposite the Tutorial Centre, MC 4066. Question 1. Missing proofs in Section 5.1 [20 marks] Consider the set S = Z × ( Z -{ 0 } ) = { ( a, b ) | a, b Z , b = 0 } and a relation defined on S as ( a, b ) ( c, d ) iff ad = bc . ( i ) [10 marks] Show that is an equivalence relation on S . ( ii ) [10 marks] Let a b denote the equivalence class of containing ( a, b ). Addition and multiplication are defined on the equivalence classes as: a b + c d = ad + bc bd , a b · c d = ac bd . Show that they are well defined. In other words: if ( a, b ) ( a , b ) and ( c, d ) ( c , d ), then ( ad + bc, bd ) ( a d + b c , b d ) and ( ac, bd ) ( a c , b d ). Answer: ( i ) Since ( a, b ) S , it holds ab = ba so that ( a, b ) ( a, b ), therefore is reflexive. Suppose ( a, b ) ( c, d ). Then, ad = bc , and this can be rewritten as cb = da . So ( c, d ) ( a, b ). Since this holds for any ( a, b ), ( c, d ) in S , is symmetric. Suppose ( a, b ) ( c, d ) and ( c, d ) ( e, f ). Then, (i) ad = bc and (ii) cf = de . Multiplying (i) and (ii) gives adcf = bcde . Since d = 0, we have (iii) acf = bce . Now, we treat the cases c = 0 and c = 0 separately. If c = 0, then, by (i),(ii) and that d = 0, we conclude a = e = 0. Thus, af = be = 0. If c = 0, we conclude from (iii) that af = be . In either case, af = be so ( a, b ) ( e, f ). Since this holds for any ( a, b ), ( c, d ), and ( e, f ) in S , is transitive. Answer: ( ii ) Suppose ( a, b ) ( a , b ) and ( c, d ) ( c , d ). Therefore (i) ab = ba and (ii) cd = dc . First consider ( ad + bc ) b d : ( ad + bc ) b d = ab dd + cd bb (by expanding and rearranging) = ba dd + dc bb (by using (i) and (ii)) = ( a d + c b ) bd (by collecting the commom factor bd ) Therefore ( ad + bc, bd ) ( a d +

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