hw5sol

# hw5sol - MATH 55 Prof Bernoff Fall 2007 Harvey Mudd College...

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Unformatted text preview: MATH 55 Prof. Bernoff Fall 2007 Harvey Mudd College Homework 5 – Solutions E1: Scheinerman, Section 23 (p. 204) # 2: Let A = { 1 , 2 , 3 } and B = { 4 , 5 } . Write down all functions f : A → B . Indicate which are one-to-one and which are onto B . Solution: Our eight functions are: { (1 , 4) , (2 , 4) , (3 , 4) } , { (1 , 4) , (2 , 4) , (3 , 5) } , { (1 , 4) , (2 , 5) , (3 , 4) } , { (1 , 5) , (2 , 4) , (3 , 4) } , { (1 , 4) , (2 , 5) , (3 , 5) } , { (1 , 5) , (2 , 4) , (3 , 5) } , { (1 , 5) , (2 , 5) , (3 , 4) } , and { (1 , 5) , (2 , 5) , (3 , 5) } . None of them are one-to-one, and all but the first and last are onto B . E2: Scheinerman, Section 23 (p. 204) # 9 (a),(b),(c), (d): For each of the following, determine if the function is one-to-one, onto, or both. Prove your assertions. (a) f : Z → Z defined by f ( x ) = 2 x . Solution: One-to-one only. f is not onto, since there is no integer x such that f ( x ) = 1. For x 6 = y , we have f ( x ) = 2 x 6 = 2 y = f ( y ), so f is one-to-one only. (b) f : Z → Z defined by f ( x ) = 10 + x . Solution: Both one-to-one and onto. For x,y distinct integers, f ( x ) = 10 + x 6 = 10 + y = f ( y ), so f is one-to-one. For x ∈ Z , f ( x- 10) = x , so f is onto. (c) f : N → N defined by f ( x ) = 10 + x . Solution: One-to-one only. Since N ⊂ Z and f is one-to-one on the integers, f is one-to-one on the naturals. But there is no natural number x such that f ( x ) = 1, so f is not onto N . (d) f : Z → Z defined by f ( x ) = x/ 2 if x is even ( x- 1) / 2 if x is odd. Solution: Onto only. f is not one-to-one as f (0) = f (1) = 0. For x ∈ Z , f (2 x ) = x , so f is onto....
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hw5sol - MATH 55 Prof Bernoff Fall 2007 Harvey Mudd College...

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