Solutions to HW 1

Solutions to HW 1 - Math 350 Homework Solutions 1 Textbook...

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Advanced Calculus I Spring 2008 Homework Solutions 1 Textbook Chapter 7 7.7 Classify each function as injective, surjective, bijective, or none of these. (a) f : N Z defined by f ( n ) = n 2 - n Solution This function is injective but not surjective, hence is not bijective. Injectivity. 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. We show that there exists k Z such that f ( n ) 6 = k for all n N. Choose k = - 1. Clearly k = - 1 Z. 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. (d) f : [3 , ) [5 , ) defined by f ( x ) = ( x - 3) 2 + 5. Solution This function is both injective and surjective, hence is bijective. Injectivity. 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 | x 1 - 3 | = | x 2 - 3 | upon taking square roots of both sides of the equality. 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 and so x 1 = x 2 . Therefore f is injective. Surjectivity.
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Solutions to HW 1 - Math 350 Homework Solutions 1 Textbook...

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