mat300q2ans

mat300q2ans - n 1 whenever it holds for n the result is...

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MAT 300 , Mathematical Structures Name: Answers Quiz 2 March 25, 2010 1. (10 points) Complete the following definitions: (a) For sets A and B , the Cartesian product A × B is the set of all ordered pairs ( a, b ) with a A and b B . (b) A function f : A B is one-to-one if for all x 1 , x 2 A , ( f ( x 1 ) = f ( x 2 )) ( x 1 = x 2 ) . (c) A function f : A B is onto if for all y B , there exists x A such that f ( x ) = y . 2. (10 points) Prove by induction that, for all n N + , 1 2 + 2 2 + 3 2 + ··· + n 2 = n ( n + 1)(2 n + 1) 6 . Proof. By induction. First, for n = 1, note that 1 2 = 1 = 6 6 = (1)(2)(3) 6 , as required. Next, assume that there is some integer n for which 1 2 + 2 2 + ··· + n 2 = n ( n + 1)(2 n + 1) 6 . We wish to prove that the result holds for n +1. Now, using the induction assumption, 1 2 + 2 2 + ··· + n 2 + ( n + 1) 2 = n ( n + 1)(2 n + 1) 6 + ( n + 1) 2 = n ( n + 1)(2 n + 1) + 6( n + 1) 2 6 = ( n + 1)[ n (2 n + 1) + 6( n + 1)] 6 = ( n + 1)[2 n 2 + 7 n + 6] 6 = ( n + 1)( n + 2)(2 n + 3) 6 = ( n + 1)([ n + 1] + 1)(2[ n + 1] + 1) 6 , as required. Since the result holds for
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Unformatted text preview: n + 1 whenever it holds for n , the result is true for all n ∈ N + , by induction. 1 3. (10 points) Prove that the function f : R + → R + given by f ( x ) = 1 x is one-to-one and onto. Proof. First, we will prove that f is one-to-one. Let x 1 , x 2 ∈ R + , and assume that f ( x 1 ) = f ( x 2 ). Then 1 x 1 = 1 x 2 , and therefore, by cross-multiplying, we conclude that x 1 = x 2 . This shows that f is one-to-one. Now, we will prove that f is onto. Let y ∈ R + . Then define x = 1 /y . Since y ∈ R + , y 6 = 0, so x ∈ R , and since y > 0, x > 0. Therefore, x ∈ R + . Finally, f ( x ) = f ± 1 y ! = 1 ± 1 y ! = y, as required. This shows that f : R + → R + is onto....
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This note was uploaded on 04/04/2010 for the course MAT 300 taught by Professor Thieme during the Spring '07 term at ASU.

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mat300q2ans - n 1 whenever it holds for n the result is...

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