This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ) =...
View Full
Document
 Spring '06
 Knutson
 Math

Click to edit the document details