final-sol - MATH 114 Final Exam - Solutions 1. Is the...

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Unformatted text preview: MATH 114 Final Exam - Solutions 1. Is the function f ( x ) = x 2 from N to N (a) one-to-one? Yes. Suppose f ( x 1 ) = f ( x 2 ) . For real numbers we have x 2 1 = x 2 2 → x 1 = ± x 2 . Since x 1 and x 2 are natural numbers, x 1 = x 2 . (b) onto? No. For example, 2 is not in the image because there is no x ∈ N such that x 2 = 2 . 2. Use Mathematical Induction to prove that 2 n < n ! for every positive integer n with n ≥ 4. Basis step. If n = 4 , 2 4 = 16 < 24 = 4! . Inductive step. Assume 2 k < k ! . Then 2 k +1 = 2 k · 2 < k ! · 2 < k !( k + 1) = ( k + 1)! . 3. Let P ( x, y ) denote the proposition y = x + 5 where x and y are positive integers. Determine the truth value of the following propositions. (a) ∀ x ∃ yP ( x, y ) True. For any positive integer x , x + 5 is a positive integer, so we can choose y = x + 5 . (b) ∀ y ∃ xP ( x, y ) False. Counterexample: if y = 1 , there is no positive integer x that satisfies 1 = x + 5 . (c) ∃ y ∀ xP ( x, y ) False. We will show that ¬∃ y ∀ xP ( x, y ) is true. ¬∃ y ∀ xP ( x, y ) ≡ ∀ y ¬∀ xP ( x, y ) ≡ ∀ y ∃ x ¬ P ( x, y ) ≡ ∀ y ∃ x y 6 = x + 5 . For any positive integer y we can choose x = y + 4 . Then y = x- 4 , so y 6 = x + 5 . 4. Consider the following graph. a b c d e f (a) How many vertices does this graph have? 6 (b) How many edges does this graph have?...
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This note was uploaded on 12/16/2011 for the course CS cs102 taught by Professor Rsharma during the Spring '11 term at IIT Kanpur.

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final-sol - MATH 114 Final Exam - Solutions 1. Is the...

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