3.2N_Proving_Existential_Statements

# 3.2N_Proving_Existential_Statements - Math 280 Strategies...

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Unformatted text preview: Math 280 Strategies of Proof Fall 2007 Class Notes 3.2 Proving Existential Statements Many statements of both definitions and theorems in advanced mathematics involve the existential quantifier. In order to show that a particular mathematical object satisfies such a definition or to prove such a theorem, we need to be able to show that an existential statement is true. According to the definition, an existentially quantified statement ∃ x p ( x ) is true provided that there exists at least one element x in the domain, an example, for which p ( x ) is true. Numbers We’ll start with some definitions concerning numbers that involve the existential quantifier. Definition. An integer n is even provided that n = 2 k for some integer k . An integer n is odd provided that n = 2 k + 1 for some integer k . In symbolic form, n is even provided that ( ∃ k ∈ Z ) [ n = 2 k ] and n is odd provided that ( ∃ k ∈ Z ) [ n = 2 k + 1] . Example. Verify the following results concerning even and odd integers. (a) − 294 is even. Solution. According to the definition, to show − 294 is even, we need to show ( ∃ k ∈ Z ) [ − 294 = 2 k ] . That is, we need to show there exists an integer k such that − 294 = 2 k . Solving this equation for k , we see that we should choose k = − 147. Then k is an integer and − 294 = 2 k , as required. Thus − 294 is even. (b) 307 is odd. Solution. To show 307 is odd, we need to show ( ∃ k ∈ Z ) [307 = 2 k + 1] . That is, we need to show there exists an integer k such that 307 = 2 k +1. Solving this equation for k , we see that we should choose k = 153. Then k is an integer and 307 = 2 k +1, as required. Thus 307 is odd. (c) 0 is even. Solution. To show 0 is even, we need to show ( ∃ k ∈ Z ) [0 = 2 k ] . Choose k = 0. Then k is an integer and 0 = 2 k , as required. Thus 0 is even. Definition. A real number r is rational provided that there exist integers p and q , with q 6 = 0, such that r = p/q . In symbolic form, r is rational provided that ( ∃ p ) ( ∃ q 6 = 0) [ r = p/q ] , where the domain for p is the set of integers and the domain for q is the set of nonzero integers. A real number r is irrational provided that it is not rational. Example. Show that 17 is rational. Solution. To show that 17 is rational, we need to show ( ∃ p ) ( ∃ q 6 = 0) [17 = p/q ] . That is, we need to show there exist integers p and q , with q 6 = 0, such that 17 = p/q . There are various possible choices for p and q . For instance, choose p = 17 and q = 1, or choose p = − 51 and q = − 3. For both sets of choices, p and q are integers with q 6 = 0 and 17 = p/q . Thus 17 is rational. 67 68 Chapter 3 Proof Techniques Constructive Proofs of Existence There are two main methods of proving an existential statement. The first method is called a constructive proof of existence . In such a proof, we either explicitly find an example x in the domain D for which the propositional function p ( x ) is true, or we give a set of directions for finding...
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## This note was uploaded on 10/20/2009 for the course MATH 280 taught by Professor Solheid during the Spring '08 term at CSU Fullerton.

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3.2N_Proving_Existential_Statements - Math 280 Strategies...

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