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Sec.%207%20Functions

Sec.%207%20Functions - Math 350 Homework Solutions Section...

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Math 350 Advanced Calculus I Fall 2009 Homework Solutions Section 7 FUNCTIONS 7.7 Classify each function as injective, surjective, bijective, or none of these. (e) f : N Z defined by f ( n ) = n 2 n . Injectivity. This function is injective but not surjective. To show that f is injective, we need to show ( n 1 N ) ( n 2 N ) [ f ( n 1 ) = f ( n 2 ) n 1 = n 2 ] . Proof: Suppose n 1 , n 2 N are arbitrary. Suppose f ( n 1 ) = f ( n 2 ) . We need to show n 1 = n 2 . Since f ( n 1 ) = n 2 1 n 1 and f ( n 2 ) = n 2 2 n 2 , then f ( n 1 ) = f ( n 2 ) implies n 2 1 n 1 = n 2 2 n 2 ( n 2 1 n 2 2 ) ( n 1 n 2 ) = 0 ( n 1 n 2 )( n 1 + n 2 ) ( n 1 n 2 ) = 0 ( n 1 n 2 )( n 1 + n 2 1) = 0 This implies n 1 n 2 = 0 or n 1 + n 2 1 = 0 . Since n 1 and n 2 are natural numbers, then both are 1 so n 1 + n 2 2, which means n 1 + n 2 1 6 = 0. Therefore n 1 n 2 = 0 so n 1 = n 2 , as required. Therefore f is injective. Surjectivity. To show that f is not surjective, we need to show ( k Z ) ( n N ) [ f ( n ) 6 = k ] . That is, we need to show there exists k Z such that f ( n ) 6 = k for all n N . Proof: Find a k . For all n N , f ( n ) = n ( n 1) 0. This tells us we should choose k < 0. For instance, choose k = 1. Verify k satisfies: (i) Domain: k Z (ii) Prop. Function: ( n N ) [ f ( n ) 6 = k ] Clearly k = 1 Z . To prove (ii), suppose n is an arbitrary natural number. Then f ( n ) = n 2 n = n ( n 1) 0 , so f ( n ) 6 = 1 = k , as required. Thus f is not surjective. 1
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Section 7 Functions Math 350 Homework Solutions page 2 7.7 (continued) (f) f : [3 , ) [5 , ) defined by f ( x ) = ( x 3) 2 + 5 Injectivity. This function is injective. To show that f is injective, we need to show ° x 1 [3 , ) ¢ ° x 2 [3 , ) ¢ [ f ( x 1 ) = f ( x 2 ) x 1 = x 2 ] . Proof: Suppose x 1 , x 2 [3 , ) are arbitrary. Suppose f ( x 1 ) = f ( x 2 ) . We need to show x 1 = x 2 . Since f ( x 1 ) = f ( x 2 ), then ( x 1 3) 2 + 5 = ( x 2 3) 2 + 5 ( x 1 3) 2 = ( x 2 3) 2 Taking square roots of both sides of this equation, we obtain | x 1 3 | = | x 2 3 | . Since x 1 , x 2 3, then x 1 3 0 and x 2 3 0, so | x 1 3 | = x 1 3 and | x 2 3 | = x 2 3. The above equality then simplifies to x 1 3 = x 2 3 x 1 = x 2 Therefore f is injective. Surjectivity. To show that f is surjective, we need to show ° y [5 , ) ¢ ( x [3 , ) ¢ [ f ( x ) = y ] .
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