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Unformatted text preview: MATH 74 HOMEWORK 5 SOLUTIONS 1. Just so people can see how these go I am including proofs. They were not a required part of the homework. 1(a). Injective but not surjective. Proof. To see that f is not surjective, notice that 2 is not in the range of f . Indeed, 1 2 = 1 is less than 2, and if x ≥ 2, then x 2 ≥ 4 is bigger than 2, so there is no n ∈ N for which n 2 = 2. To see that f is injective, fix x,y ∈ N and suppose f ( x ) = f ( y ), ie, that x 2 = y 2 . We cannot have x < y because then x 2 = x · x < x · y < y · y = y 2 , and similarly (interchanging x and y in the proof just given) we cannot have x > y . So x = y . There is a general fact at work here. The point is that squaring, on the given domain, is strictly increasing : if x < y , then f ( x ) < f ( y ). Any strictly increasing function is injective. 1(b). Injective but not surjective. Proof. To see that it is injective, consider the function g : R → R given by g ( x ) = x 1 / 3 . Then g ( f ( x )) = ( x 3 ) 1 / 3 = x, x ∈ R . so f is injective by “Theorem 18” from class. To see that f is not surjective, notice that every number in the range of f , being a cube of a nonnegative number, must be nonnegative. So any negative number (eg- 1) is not in the range of f . 1(c). Injective and surjective. Proof. Without a precise definition of the sine function (or the number π ) we are not really in a position to “prove” this, but it is clear from calculus. On the givennot really in a position to “prove” this, but it is clear from calculus....
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This homework help was uploaded on 04/02/2008 for the course MATH 74 taught by Professor Courtney during the Fall '07 term at Berkeley.
- Fall '07