3.9N_Mathematical_Induction

3.9N_Mathematical_Induction - Math 280 Strategies of Proof...

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Unformatted text preview: Math 280 Strategies of Proof Fall 2007 Class Notes 3.9 Proof by Mathematical Induction Proof by mathematical induction is a special method of proof, used for proving universal statements of the form ∀ n p ( n ), where the domain for n is the set N of positive integers or a subset of N . The method is based upon the following principle, which may be considered an axiom of the set of positive integers. Note that as an axiom, the Principle of Mathematical Induction does not require proof. Principle of Mathematical Induction. Let S be a subset of the set N of positive integers. Suppose (1) 1 ∈ S , and (2) for all k ∈ N , if k ∈ S , then k + 1 ∈ S . Then S = N . To see why this should be considered an axiom, suppose S is a subset of N and suppose S satisfies the two properties in the Principle. According to the first property, 1 ∈ S . The second property says that for all positive integers k , whenever k ∈ S , then the next integer k + 1 ∈ S . Applying this property, since 1 ∈ S , then the next integer 1 + 1 = 2 ∈ S . Applying the second property again, since 2 ∈ S , then the next integer 2 + 1 = 3 ∈ S . Since 3 ∈ S , then 3 + 1 = 4 ∈ S . Since 4 ∈ S , then 4 + 1 = 5 ∈ S . Continuing, we will obtain that every positive integer is in S . Thus S = N , as concluded by the principle. We now want to apply the Principle of Mathematical Induction to proving statements. Con- sider a universal statement of the form ( ∀ n ∈ N ) p ( n ), where the domain for n is the set N of positive integers. To prove ( ∀ n ∈ N ) p ( n ) is true, we need to show that p ( n ) is true for all positive integers n . Let S be the truth set of p ( n ), the subset of positive integers n for which the proposi- tional function p ( n ) is true. To show that ( ∀ n ∈ N ) p ( n ) is true, we need to prove that the truth set S = N . The first condition in the Principle of Mathematical Induction, 1 ∈ S , is then equivalent to the condition that p (1) is true. The second condition ( ∀ k ∈ N ) , [( k ∈ S ) → ( k + 1 ∈ S )] , is equivalent to the condition ( ∀ k ∈ N ) [ p ( k ) → p ( k + 1)] . We can then restate the principle in terms of the propositional function as follows: Principle of Mathematical Induction. Suppose p ( n ) is a propositional function with domain the set N of positive integers. Suppose (1) p (1) is true, and (2) ( ∀ k ∈ N ) [ p ( k ) → p ( k + 1)] is true. Then ( ∀ n ∈ N ) p ( n ) is true. We can use the principle in this form to obtain the following procedure for proving a statement of the form ∀ n p ( n ). In order to prove ( ∀ n ∈ N ) p ( n ) is true, we need to prove the two conditions in the Principle hold. The proof of (1) is usually called the Basis Step of the proof and the proof of (2) is called the Induction Step. Note also that it’s often convenient in induction proofs to use the notation p ( n ) for the propositional function in the proof, instead of explicitly writing out the propositional function.propositional function....
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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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3.9N_Mathematical_Induction - Math 280 Strategies of Proof...

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