Lecture Week 5

# Discrete Mathematics and Its Applications with MathZone

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• davidvictor
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UCI ICS/Math 6A, Summer 2007 5-Recursion -1 after [email protected] Strong Induction Normal” Induction Normal” Induction: If we prove that 1) P(n 0 ) is true for some n 0 (typically 0 or 1), and 2) If P(k) is true for any k≥n 0 , then P(k+1) is also true. Then P(n) is true for all n≥n 0 . Strong” Induction Strong” Induction : If we prove that 1) Q(n 0 ) is true for some n 0 (typically 0 or 1), and 2) If Q(j) is true for all j from n 0 to k (for any k≥n 0 ), then Q(k+1) is also true. Then Q(k) is true for all k≥n 0 . These 2 forms are equivalent: Let P(n) = “ These 2 forms are equivalent: Let P(n) = “n 0 ≤j≤n Q(j)” Ref:

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UCI ICS/Math 6A, Summer 2007 5-Recursion -2 after [email protected] Proof by Strong Induction: Jigsaw Puzzle Each “step” in assembling a jigsaw puzzle consists of putting together 2 already assembled blocks of pieces where each single piece is considered a block itself. P(n) = “It takes exactly n-1 steps to assemble a jigsaw puzzle of n pieces.” Basis Step: P(1) is (trivially) true. Inductive Step: We assume P(k) true for k≤n and we’ll argue P(n+1): The last step in assembling a puzzle with n+1 steps is to put together 2 blocks: one of size j>0 and one of size n+1-j. By since 0<j,n+1-j≤n, P(j) and P(n+1-j) are both assumed true. And so, the total number of steps to assemble a puzzle with n+1 pieces is 1 + (j-1) + ((n+1-j)-1) = n = (n+1)-1. (this implies P(n+1), and hence ends the inductive part, and thus also the whole proof)
UCI ICS/Math 6A, Summer 2007 5-Recursion -3 after [email protected] More Examples of Theorems with easy Proofs using Strong Induction Thm1: The second player always wins the following game: Starting with 2 piles each containing the same number of matches, players alternately remove any non-zero number of matches from one of the piles. The winner is the person who removes the last match. Thm2: Every n>1 can be written as the product of primes. Thm3: Every postage amount of at least 18 cents can be formed using just 4-cent and 7-cent stamps.

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UCI ICS/Math 6A, Summer 2007 5-Recursion -4 after [email protected] Basis for Induction: Integers are well ordered Induction is based on the fact that the integers are well ordered ” in the following sense: Any non-empty set of integers which is bounded below i.e. there exists b (not necessarily in S) s.t. b ≤ x for all x in S contains a least element i.e. element e in S s.t. for all x in S we have e ≤ x.
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