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Unformatted text preview: MATH 109 Assignment 4 Due: 4/25/11 #9.1 (i) We claim that f ( x ) = 2 x + 5 is both an injection and a surjection, and hence a bijection. To prove that f is an injection we have to show that f ( x ) = f ( y ) implies that x = y . Hence, suppose that 2 x +5 = f ( x ) = f ( y ) = 2 y +5. By subtracting 5 from both sides of the equation, and then dividing by 2, we see that x = y as desired, so f is an injection. To prove that f is a surjection, we must show that for any y ∈ R , there exists x ∈ R such that f ( x ) = y . Given y ∈ R , let x = ( y 5) / 2. We then have f ( x ) = f y 5 2 = 2 y 5 2 + 5 = ( y 5) + 5 = y, as desired. Hence, f is both an injection and a surjection, and hence a bijection. (ii) We claim that f is neither an injection nor a surjection, and hence not a bijection. To see that f is not an injection, observe that f ( 2) = f (0) = 1. To see that f is not a surjection, observe that f ( x ) = ( x 1) 2 ≥ 0; hence there does not exist x ∈ R for which f ( x ) =...
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 Spring '06
 Knutson
 Math

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