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Unformatted text preview: SOLUTIONS FOR HOMEWORK 2 1.51. (a) Yes , I f ( X Y ) = I f ( X ) I f ( Y ). To establish this equality, note that a I f ( X Y ) iff f ( a ) X Y . The last equality holds when either f ( a ) X , or f ( a ) Y . By the definition of I f , f ( a ) X is equivalent to a I f ( X ), and similarly, f ( a ) Y is equivalent to a I f ( Y ). Thus, f ( a ) X Y iff a I f ( X ) I f ( Y ). (b) Yes , I f ( X Y ) = I f ( X ) I f ( Y ). Indeed, a I f ( X Y ) iff f ( a ) X Y , that is, f ( a ) X , and f ( a ) Y . As noted in (a), f ( a ) X iff a I f ( X ), and f ( a ) Y iff a I f ( Y ). Thus, f ( a ) X Y iff a I f ( X ) I f ( Y ). 1.55. Multiplication. Show first that x x = 0, or equivalently, x = x 1 . By Definition 1.39, x has the opposite element x 1 = y F s.t. x y = 0. We have to show that y = x , or in other words, y negationslash = 0, and y negationslash = 1. However, these cases are easy to rule out: x 0 = 0 negationslash = 1, and x 1 = x negationslash = 1. The only remaining possibility is= 1....
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This note was uploaded on 01/27/2010 for the course MATH 347 taught by Professor ? during the Fall '09 term at University of Illinois at Urbana–Champaign.
 Fall '09
 ?
 Math

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