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

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