Sec.%207%20Functions

Sec.%207%20Functions - Math 350 Advanced Calculus I Fall...

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Unformatted text preview: 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) ≥ , so f ( n ) 6 = − 1 = k , as required. Thus f is not surjective. 1 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....
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This note was uploaded on 10/20/2009 for the course MATH 350 taught by Professor Qian during the Spring '08 term at CSU Fullerton.

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Sec.%207%20Functions - Math 350 Advanced Calculus I Fall...

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