Mathematical Thinking: Problem-Solving and Proofs (2nd Edition)

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4.26 Suppose f ( x ) = f ( y ). Then | f ( x ) - f ( y ) | = 0, so c | x - y | α 0. Since c > 0 and α > 0, this implies that | x - y | = 0 and hence x = y . Thus f is injective. 4.32 We have f ( - x ) = - ( - x ) = x and g ( y - 1 ) = ( y - 1 ) - 1 = y for every x F and y F - { 0 } , so f and g are surjective. If - x 1 = f ( x 1 ) = f ( x 2 ) = - x 2 and if y - 1 1 = g ( y 1 ) = g ( y 2 ) = y - 1 2 , then addition of x 1 + x 2 to both sides of the first equation and multiplication of both sides of the second equation by y 1 y 2 yields x 2 = x 1 and y 2 = y 1 . Therefore
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This document was uploaded on 03/25/2011.

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