# 77 classify each function as injective surjective

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7.7 Classify each function as injective, surjective, bijective, or none of these. (a) f : N N by f ( n ) = n + 3 Injective: f ( a ) = f ( b ) implies that a + 3 = b + 3 and so a = b . Not Surjective: There is no n N with f ( n ) = 1. (c) f : R R by f ( x ) = x 3 - x Not Injective: f (0) = 0 = f (1), 0 = 1. Surjective: The graph goes from -∞ to lim x →-∞ f ( x ) = -∞ and lim x →∞ f ( x ) = . (e) f : N Z by f ( n ) = n 2 - n Injective or not depending on whether 0 is in N : f (0) = 0 = f (1), 0 = 1. Not Surjective: f ( n ) = 3 for any n in N . (g) f : N Q by f ( n ) = 1 /n Injective: f ( a ) = f ( b ) implies that 1 /a = 1 /b and so a = b . Not Surjective: f ( n ) = 3 for any n n N . 7.11a Suppose f : S S . Prove that if f f is injective, then f is injective.

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2 (Show that f ( a ) = f ( b ) implies that a = b .) Suppose that f ( a ) = f ( b ). Apply f to both sides. Then f ( f ( a )) = f ( f ( b )). Since f f is injective, f ( f ( a )) = f ( f ( b )) implies that a = b . Therefore a = b and so f is injective. 7.22 Let A = { 1 , 2 , 3 } and B = { a, b, c } . Define the functions f = { (1 , a ) , (2 , b ) , (3 , a ) } and g = { ( a, 1) , ( b, 3) , ( c, 2) } . Describe each of the following functions by listing its ordered pairs. Then state its range. (a) g - 1 g - 1 = { (1 , a ) , (3 , b ) , (2 , c ) } , the range is { a, b, c } . (d) f g f First do f , then g , then f again: 1 a 1 a , so f ( g ( f (1))) = a , 2 b 3 a , so f ( g ( f (2))) = a , 3 a 1 a , so f ( g ( f (3))) = a . f g f = { (1 , a ) , (2 , a ) , (3 , a ) } , the range is { a } .
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