inductionkey1 - McCombs Math 381 Mathematical Induction...

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Mathematical Induction Main Idea: We will develop a technique for proving statements about the set of positive integers. Proof Template: To prove a statement of the form, “Every positive integer n satisfies the propositional function P n ( ) .” Basis Step: Show that P 1 ( ) is true. Induction Hypothesis: Assume that P k ( ) is true. Inductive Step: Show that P k ( ) ! P k + 1 ( ) is true. Examples Use mathematical induction to prove each of the following. 1. For every integer n > 0 , 10 2 n ! 1 is divisible by 11. Let P n ( ) :10 2 n ! 1 is divisible by 11. Basis Step: P 1 ( ) :10 2 ! 1 = 100 ! 1 = 99 Clearly, 11 divides 99. Induction Hypothesis: Assume P k ( ) is true. That is, assume 10 2 k ! 1 is divisible by 11. This means that 10 2 k ! 1 = 11 a , for some integer a . Inductive Step: Show P k ( ) ! P k + 1 ( ) . 10 2 k ! 1 = 11 a " 100 10 2 k ! 1 ( ) = 100 11 a ( ) 10 2 10 2 k ! 1 ( ) = 100 11 a ( ) 10 2 k + 2 ! 10 2 = 100 11 a ( ) 10 2 k + 2 ! 100 + 99 = 100 11 a ( ) + 99 10 2 k + 2 ! 1 = 100 11 a ( ) + 99 10 2 k + 1 ( ) ! 1 = 11 100 a + 9 ( ) Therefore, 10 2 k + 1 ( ) ! 1 is divisible by 11. Thus,
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This note was uploaded on 06/16/2011 for the course MATH 381 taught by Professor Na during the Summer '11 term at University of North Carolina School of the Arts.

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inductionkey1 - McCombs Math 381 Mathematical Induction...

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