Presentation is loading. Please wait.

Presentation is loading. Please wait.

Induction Chapter 4.1-4.2.

Similar presentations


Presentation on theme: "Induction Chapter 4.1-4.2."— Presentation transcript:

1 Induction Chapter

2 Odd Powers Are Odd If m is odd and n is odd, then nm is odd
Proposition: for an odd number m, mk is odd for all non-negative integer k. Let P(i) be the proposition that mi is odd. Proof by induction

3 Proof by induction P(1) is true by definition.
P(2) is true by P(1) and the fact. P(3) is true by P(2) and the fact. P(i+1) is true by P(i) and the fact. So P(i) is true for all i.

4 Mathematical Induction
1 and (from n to n +1) are valid, then, proves 1, 2, 3,…. Expressed as a rule of inference To prove that P(n) is true for all positive integers, where P(n) is a propositional function Two steps: Basis: verify P(1) is true Show that conditional statement if P(k) then P(k+1) is true for all positive integers k

5 Induction Basis step: Very easy to prove
Induction step: Much easier to prove with P(n) as an assumption Domino effects.

6 Proof by Induction Example: summation of integers Proof: P(1) is true
Assume: Then

7 Example Prove: Basis: Induction: if Then add to both side rn+1

8 Example Prove an inequity Basis: Induction:

9 Strong Induction To prove that P(n) is true for all positive integers n, where P(n) is a propositional function Tow steps: Basis step: verify that P(1) is true Inductive step: show that the conditional statement is true for ALL positive integers k Conclusion:

10 Equivalent to Ordinary Induction
From ordinary induction: 0 to 1, 1 to 2, 2 to 3, …, n-1 to n So by the time we got to n+1, already know all of P(0), P(1), …, P(n) So strong induction is equivalent to ordinary one

11 Example: Prime Products
Claim: Every integer > 1 is a product of primes. Proof:(by strong induction) Base case is easy, 2 is a prime itself; Suppose the claim is true for all 2 <= i < n. Consider an integer n; In particular, n is not prime since if it is we do not need to prove. So n = k·m for integers k, m where n > k,m>1. Since k,m smaller than n, by the induction hypothesis, both k and m are product of prime Therefore, n is a prime product

12 Example: Postage Available stamps:
5¢ ¢ What amount can we get from these stamps Theorem: Can form any amount larger or equal to 8¢ Prove by strong induction on all n larger or equal to 8 P(n) ::= can form (n+8)¢.

13 Postage: Strong induction proof
Basis step: n=0 (0+8)¢=8¢. Inductive Step: assume (m +8)¢ for 0 <= m < n then prove ((n +1) + 8)¢ Case 1: n +1=1, 9¢ Case 2: n +1=2, 10¢

14 Postage: Strong induction proof
Case 3: n +1 >= 3: let m=n-2 now n >= m >= 0, so by induction hypothesis + = (n+1)+8 can be formed (n-2)+8 is formed Add a 3¢


Download ppt "Induction Chapter 4.1-4.2."

Similar presentations


Ads by Google