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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 denition 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 quantied 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 plugandchug. 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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 Fall '10
 DannyHeap
 Computer Science

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