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Unformatted text preview: MODEL ANSWERS TO THE FIRST HOMEWORK 1. Chapter 1, 1: 1. Suppose that a and b are elements of S . By rule (1) a b = a. But by rule (2), a b = b a. Applying rule (1) we get a b = b a = b . Thus a = a b = b . As a and b are arbitrary, S can have at most one element. 1. Chapter 1 1: 2. (a) Suppose that a and b are two integers and that a b = b a . Now a b = a b and b a = b a so that then a b = b a . Applying the standard rules of arithmetic, we get 2 a = 2 b and so a = b . (b) Suppose that a , b and c are integers. Then a ( b c ) = a ( b c ) = a ( b c ) = a + c b. On the other hand ( a b ) c = ( a b ) c = ( a b ) c = a ( b + c ) . Thus equality holds iff a + c b = a ( b + c ), that is cancelling c = c so that c = 0. Thus is not associative. For example, (0 1) = 1 but (0 0) 1 = 1 . (c) Let a be an integer. Then a 0 = a 0 = a. (d) Let a be an integer. Then a a = a a = 0 ....
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This note was uploaded on 07/15/2008 for the course MATH 111 taught by Professor Long during the Spring '08 term at UCSB.
 Spring '08
 LONG
 Math, Algebra

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