Unformatted text preview: Solution. Let g : Y → X and h : Y → X both be inverses of f . Then h = h 1 Y = h ( f g ) = ( h f ) g = 1 X g = g . 2.10 Show that f : R → R , de f ned by f ( x ) = 3 x + 5, is a bijection, and f nd its inverse. Solution. The function g , de f ned by g ( x ) = 1 3 ( x − 5 ) , is the inverse of f , and so f is a bijection. (Alternatively, one could prove that f is a bijection by showing directly that it is injective and surjective.) 2.11 Determine whether f : Q × Q → Q , given by f ( a / b , c / d ) = ( a + c )/( b + d ) is a function. Solution. f is not a function: 1 2 = 2 4 and 2 6 = 1 3 , but f ( 1 2 , 2 6 ) = 3 8 ±= 3 7 = f ( 2 4 , 1 3 )....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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