CS2800-Induction_v.6

CS2800-Induction_v.6 - Discrete Math CS 2800 Prof. Bart...

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1 Discrete Math CS 2800 Prof. Bart Selman [email protected] Module Induction Rosen, Chapter 4
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2 What’s is Induction About? Many statements assert that a property is an universal true – i.e., all the elements of the universe exhibit that property; Examples: 1. For every positive integer n: n! ≤ n n 1. For every set with n elements, the cardinality of its power set is 2 n . Induction is one of the most important techniques for proving statements about universal properties.
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We know that: 1. We can reach the first rung of this ladder; 2. If we can reach a particular rung of the ladder, then we can reach the next rung of the ladder. Can we reach every step of this infinite ladder? Yes, using Mathematical Induction which is a rule of inference that tells us: P(1) 2200 k (P(k) P(k+1)) -------------------------- 2200 n (P(n)
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4 Principle of Mathematical Induction Hypothesis: P(n) is true for all integers n b To prove that P(n) is true for all integers n b (*), where P(n) is a propositional function, follow the steps: Basic Step or Base Case : Verify that P(b) is true; Inductive Hypothesis : assume P(k) is true for some k b; Inductive Step : Show that the conditional statement P(k) P(k+1) is true for all integers k b. This can be done by showing that under the inductive hypothesis that P(k) is true, P(k+1) must also be true. (*) quite often b=1, but b can be any integer number.
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5 Writing a Proof by Induction 1. State the hypothesis very clearly: P(n) is true for all integers n b – state the property P in English 1. Identify the the base case P(b) holds because … 1. Inductive Hypothesis Assume P(k) 1. Inductive Step - Assuming the inductive hypothesis P(k), prove that P(k+1) holds; i.e., P(k) P(k+1) Conclusion By induction we have shown that P(k) holds for all k b (b is what was used for the base case).
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6 Mathematical Induction What’s the hypothesis? Prove a base case (n=1) 2 - Base case (n=1): the sum of the first 1 odd integer is 1 2 . Since 1 = 1 2 Prove P(k) P(k+1) 3 - Assume P(k): the sum of the first k odd ints is k 2 . 1 + 3 + … + (2k - 1) = k 2 4 – Inductive Step: show that 2200 (k) P(k) P(k+1), assuming P(k). How? Inductive hypothesis P(k+1)= 1 + 3 + … + (2k-1) + (2k+1) = k 2 + (2k + 1) By inductive = (k+1) 2 p(k) QED 1 – Hypothesis: P(n) – sum of first n odd integers = n 2 .
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7 Mathematical Induction Use induction to prove that the 1 + 2 + 2 2 + … + 2 = 2 - 1 for all non-negative integers n. 1 – Hypothesis? Prove a base case (n=?) 2 - Base case? Prove P(k) P(k+1) 3 – Inductive Hypothesis Assume P(k) = 1 + 2 + 2 2 + … + 2 k = 2 k+1 – 1 Inductive hypothesis n = 0 1 0 = 2 1 -1. P(n) = 1 + 2 + 2 2 + … + 2 n = 2 n+1 – 1 for all non-negative integers n. not n=1! The base case can be negative, zero, or positive
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8 Mathematical Induction 4 – Inductive Step: show that 2200 (k) P(k) P(k+1), assuming P(k). How? P(k+1)= 1 + 2 + 2 2 + … + 2 k + 2 k+1 = (2 k+1 – 1) + 2 k+1 By inductive hypothesis p(k) QED = 2 2 k+1 - 1 P(k+1) = 2 k+2 - 1 = 2 (k+1)+1 - 1
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Mathematical Induction Prove that 1 1! + 2 2! + … + n n! = (n+1)! - 1, 2200 positive integers 2 - Base case (n=1): 1 1! = (1+1)! - 1?
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This note was uploaded on 01/16/2010 for the course CS 2800 at Cornell.

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CS2800-Induction_v.6 - Discrete Math CS 2800 Prof. Bart...

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