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Unformatted text preview: 1 COMP170 Discrete Mathematical Tools for Computer Science Discrete Math for Computer Science K. Bogart, C. Stein and R.L. Drysdale Section 4.1, pp. 127142 Intro to Induction Version 2.0: Last updated, May 13, 2007 Slides c 2005 by M. J. Golin and G. Trippen 2 4.1 Mathematical Induction • Smallest Counterexamples • The Principle of Mathematical Induction • Strong Induction • Induction in General 3 • We start by reviewing proof by smallest counterexample to try and understand what it is really doing • This will lead us to transform the indirect proof technique of proof by counterexample to direct proof technique. This direct proof technique will be induction • We conclude by distinguishing between the weak principle of mathematical induction and the strong principle of mathematical induction • Note that the strong principle can actually be de rived from the weak principle . The difference be tween them has less to do with the power of the techniques, than with proof format 4 Proof by smallest counterexample that statement P ( n ) is true for all n = 0 , 1 , 2 ... works by (i) Assuming that a nonzero counterexample exists, i.e., There is some n > for which P ( n ) is not true (ii) Letting m > be smallest value for which P ( m ) is not true (iii) Then use fact that P ( m ) is true for all ≤ m < m to show that P ( m ) is true , contradicting original choice of m . ⇒ P ( n ) true for all n = 0 , 1 , 2 ,... 2 3 4 5 m 1 1 m P ( m ) true; ≤ m < m P ( m ) not true P ( m ) true contradiction! 5 • Suppose (*) is not always true • Then there must be a smallest n ∈ N such that (*) does not hold for n • Then, for any nonnegative integer i < n , 1 + 2 + ... + i = i ( i +1) 2 . • Because 0 = 0 · 1 / 2 , (*) holds when n = 0 . • Therefore, the smallest counterexample n is larger than . Example 1: (*) 0 + 1 + 2 + 3 + 4 + ... + n = n ( n +1) 2 . Use proof by s.c. to show that, ∀ n ∈ N , Example 1: (nonnegative ints) 6 • So, (i) smallest counterexample n is greater than , and (ii) (*) holds for n 1 • Substituting n 1 for i gives 1 + 2 + ... + n 1 = ( n 1) n 2 . • Adding n to both sides gives 1 + 2 + ... + n 1 + n = ( n 1) n 2 + n = n ( n +1) 2 . • Thus, n is not a counterexample after all. • Therefore, there is no counterexample for (*) . • Hence, (*) holds for all positive integers n . 7 What implication did we have to prove? The crucial step was proving that p ( n 1) ⇒ p ( n ) where p ( n ) is the statement 1+2+ . . . + n = n ( n +1) 2 . 8 2 n +1 ≥ n 2 + 2 ....
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 Spring '10
 M.J.Golin
 Computer Science

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