Lecture 3.1: Mathematical Induction CS 250, Discrete Structures, Fall 2014 Nitesh Saxena Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag

10/2/2015 Lecture Mathematical Induction Course Admin Mid-Term 1 Hope it went well! Thanks for your cooperation running it smoothly We are grading them now, and should have the results no less than two weeks Solution will be provided soon Questions?

10/2/2015 Lecture Mathematical Induction Course Admin HW2 (don’t forget) Due Oct 14 (Tues)

10/2/2015 Lecture Mathematical Induction Outline Mathematical Induction Principle Examples Why it all works

10/2/2015 Lecture Mathematical Induction Mathematical Induction Suppose we have a sequence of propositions which we would like to prove: P (0), P (1), P (2), P (3), P (4), … P (n), … EG: P (n) = “The sum of the first n positive odd numbers is equal to n 2 ” We can picture each proposition as a domino: P (n)

10/2/2015 Lecture Mathematical Induction Mathematical Induction So sequence of propositions is a sequence of dominos. … P (n+1)P (n) P (2)P (1)P (0) …

10/2/2015 Lecture Mathematical Induction Mathematical Induction When the domino falls, the corresponding proposition is considered true: P (n)

10/2/2015 Lecture Mathematical Induction Mathematical Induction When the domino falls (to right), the corresponding proposition is considered true: P (n) true

10/2/2015 Lecture Mathematical Induction Mathematical Induction Suppose that the dominos satisfy two constraints. 1) Well-positioned: If any domino falls (to right), next domino (to right) must fall also. 2) First domino has fallen to right P (0) true P (n+1)P (n)

10/2/2015 Lecture Mathematical Induction Mathematical Induction Suppose that the dominos satisfy two constraints. 1) Well-positioned: If any domino falls to right, the next domino to right must fall also. 2) First domino has fallen to right P (0) true P (n+1)P (n)

10/2/2015 Lecture Mathematical Induction Mathematical Induction Suppose that the dominos satisfy two constraints. 1) Well-positioned: If any domino falls to right, the next domino to right must fall also. 2) First domino has fallen to right P (0) true P (n) true P (n+1) true

10/2/2015 Lecture Mathematical Induction Mathematical Induction Then can conclude that all the dominos fall! … P (n+1)P (n) P (2)P (1)P (0)

10/2/2015 Lecture Mathematical Induction Mathematical Induction Then can conclude that all the dominos fall! … P (n+1)P (n) P (2)P (1)P (0)

10/2/2015 Lecture Mathematical Induction Mathematical Induction Then can conclude that all the dominos fall! …P (0) true P (n+1)P (n) P (2)P (1)

10/2/2015 Lecture Mathematical Induction Mathematical Induction Then can conclude that all the dominos fall! …P (0) true P (1) true P (n+1)P (n) P (2)

10/2/2015 Lecture Mathematical Induction Mathematical Induction Then can conclude that all the dominos fall! P (2) true …P (0) true P (1) true P (n+1)P (n)

10/2/2015 Lecture Mathematical Induction Mathematical Induction Then can conclude that all the dominos fall! P (2) true …P (0) true P (1) true P (n+1)P (n)

10/2/2015 Lecture Mathematical Induction Mathematical Induction Then can conclude that all the dominos fall! P (2) true …P (0) true P (1) true P (n) true P (n+1)

10/2/2015 Lecture Mathematical Induction Mathematical Induction Then can conclude that all the dominos fall! P (2) true …P (0) true P (1) true P (n) true P (n+1) true

10/2/2015 Lecture Mathematical Induction Mathematical Induction Principle of Mathematical Induction: If: 1) [basis] P (0) is true 2) [induction]  k P(k)  P(k+1) is true Then:  n P(n) is true This formalizes what occurred to dominos. P (2) true …P (0) true P (1) true P (n) true P (n+1) true

10/2/2015 Lecture Mathematical Induction Exercise 1 Use induction to prove that the sum of the first n odd integers is n 2. Prove a base case (n=1) Base case (n=1): the sum of the first 1 odd integer is 1 2. Yup, 1 = 1 2. Prove P(k)  P(k+1) Assume P(k): the sum of the first k odd ints is k … + (2k - 1) = k 2 Prove that … + (2k - 1) + (2k + 1) = (k+1) … + (2k-1) + (2k+1) =k 2 + (2k + 1) = (k+1) 2 By arithmetic

10/2/2015 Lecture Mathematical Induction Exercise 2 Prove that 1  1! + 2  2! + … + n  n! = (n+1)! - 1,  n Base case (n=1): 1  1! = (1+1)! - 1? Yup, 1  1! = 1, 2! - 1 = 1 Assume P(k): 1  1! + 2  2! + … + k  k! = (k+1)! - 1 Prove that 1  1! + … + k  k! + (k+1)(k+1)! = (k+2)!  1! + … + k  k! + (k+1)(k+1)! =(k+1)! (k+1)(k+1)! = (1 + (k+1))(k+1)! - 1 = (k+2)(k+1)! - 1 = (k+2)! - 1

10/2/2015 Lecture Mathematical Induction Exercises 3 and 4 (have seen before?) 1. Recall sum of arithmetic sequence: 2. Recall sum of geometric sequence:

10/2/2015 Lecture Mathematical Induction Mathematical Induction - why does it work? Proof of Mathematical Induction: We prove that (P(0)  (  k P(k)  P(k+1)))  (  n P(n)) Proof by contradiction. Assume 1. P(0) 2.  k P(k)  P(k+1) 3.  n P(n)  n  P(n)

10/2/2015 Lecture Mathematical Induction Mathematical Induction - why does it work? Assume 1. P(0) 2.  k P(k)  P(k+1) 3.  n P(n)  n  P(n) Let S = { n :  P(n) } Since N is well ordered, S has a least element. Call it k. What do we know? P(k) is false because it’s in S. k  0 because P(0) is true. P(k-1) is true because P(k) is the least element in S. But by (2), P(k-1)  P(k). Contradicts P(k-1) true, P(k) false. Done.

10/2/2015 Lecture Mathematical Induction Today’s Reading Rosen 5.1