Lecture 21(Wedneday) - induction Course Notes pp. 56{58...

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Unformatted text preview: induction Course Notes pp. 56{58 Danny Heap heap@cs.toronto.edu http://www.cdf.toronto.edu/ f heap/165/F10/ CSC 165 slide 1 implication, over and over Earlier we saw that implication is transitive. If P (n) is a predicate of the natural numbers, and we have a chain: P (0) A P (1) P (1) A P (2) P (12) A P (13) . . . then P (0) A P (13). A property of the natural numbers (often this property is built in to the de nition of the natural numbers) is that any property of zero that can be passed from any natural number to its successor will \spread" to all natural numbers. In symbols: (P (0) (Vn P N; P (n) A P (n + 1))) A (Vn P N; P (n)) Of course, the starting point doesn't have to be zero, since the key idea is passing a property from some natural number to its successors: (P (k) (Vn P N f0; : : : ; k 1g; P (n) A P (n + 1))) A (Vn P N f0; : : : ; k 1g; P (n)) slide 2 simple induction decomposed The two pieces of mathematical induction are: 1. Establish that the property is true at the starting point. Simply check that P (k) is true | usually not too hard. 2. Establish that the property spreads from each natural number to its successor: V P N f0 n ;:::;k 1g ;P n ( ) A P (n + 1) That second step should seem familiar | it's universally quanti ed implication. We've got a proof structure for that. slide 3 comparing growth Exponental functions grow faster than polynomials. It's almost true that: V PN 3 n ; n > n 3 Explain the almost. Fix up the claim, and prove it (at least ll in the proof structure). slide 4 scratch slide 5 bonacci Sometimes induction to prove an equality becomes pure plug-and-chug. Consider the Fibonacci sequence: 8 >n; < if n < 2 Fn = > :Fn 2 + Fn 1 if n ! 2 Make a conjecture about the sum of the rst k Fibonacci numbers: k X i=0 Fi Set up a proof of your conjecture by simple induction. slide 6 ...
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Lecture 21(Wedneday) - induction Course Notes pp. 56{58...

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