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**Unformatted text preview: **1 Mathematical Induction. Induction is the most important proof method in computer science. Suppose you want to prove that the proposition P ( n ) is true for all n N , where N = {0, 1, 2, } (an infinite set). ( i. e. every natural number n has some property P ) You might be able to prove it for 0, 1, 2, But you cant check one-by-one that all natural numbers have property P. 2 The key idea of mathematical induction is start with 0 and repeatedly add 1. Suppose you can show that i) 0 has property P and ii) whenever you add 1 to a number that has property P the resulting number also has property P. This guarantees that as you go through the list of all natural numbers, every number you encounter must have property P . 3 The proof method of mathematical induction says that you just need to do two things: i) prove it for n = 0 (basis case) ii) prove if its true for n=k, then its true for n =k+ 1 Induction Hypothesis: fix some k 0 assume P ( k ) Induction Step Using assumption P ( k ) prove that P ( k +1) In this way we prove that for any n [ P ( n ) P ( n +1)] 4 Why does this prove it for all n N ? Actually it relies upon the property of integers N= {0, 1, 2, } The pattern is: To prove 2200 n 0 ( P ( n )) its enough to show: 1) (Base) P (0) 2) 2200 n 0 ( P ( n ) P ( n +1)) So, in the second step we say: take some k 0 and assume that property P holds for n = k (Induction Hypothesis). Then, (based upon this assumption) we need to show that the property P can be implied for n = k +1. How can we prove 2200 n 0 ( P ( n ) P ( n +1)) ? Pick arbitrary k 0 and prove it for n = k. 5 Suppose we want to prove the formula for the sum of geometric progression q q q q q q n n-- = + + + + + + 1 1 ... 1 1 3 2 where 1 < < q , and n is any integer, n How can we prove that the formula is correct for any n ? Show it for arbitrary fixed n Use induction on n Notation: = = + + + + = n i i n n q q q q S 2 ... 1 q q S n n-- = + 1 1 1 6 Show it for some n 1 2 1 2 1 ) ... 1 ( ......

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