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Unformatted text preview: Homework #1 Solutions p 241, #2 The identity element is easily seen to be 6. Indeed, in Z 10 we have 2 Â· 6 = 12 = 2 4 Â· 6 = 24 = 4 6 Â· 6 = 36 = 6 8 Â· 6 = 48 = 8 . p 241, #4 There are many possible examples. Probably the simplest occurs in Z 4 , where both 1 and 3 are solutions to 2 x = 2. We know that such a situation cannot happen in a group, for in that case the equation ax = b has the unique solution x = a 1 b . p 241, #14 We prove the result for all nonnegative m first. If m = 0 the result is obvious. Now assume that m â‰¥ 1. Then m Â· ( ab ) = ab + ab + Â·Â·Â· + ab  {z } m times = ( a + a + Â·Â·Â· + a )  {z } m times b = ( m Â· a ) b. If we had instead factored a out on the right side, we would have obtained instead m Â· ( ab ) = a ( m Â· b ). Thus, m Â· ( ab ) = ( m Â· a ) b = a ( m Â· b ) for all m âˆˆ Z + . If m < 0 then m = n for some n > 0. We then have, using part 2 of Theorem 12.1 and the preceding result m Â· ( ab ) = ( n ) Â· ( ab ) = n Â· ( ( ab )) = n Â· (( a...
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This note was uploaded on 02/29/2012 for the course MATH 4363 taught by Professor Ryandaileda during the Spring '07 term at Trinity University.
 Spring '07
 RyanDaileda

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