Lecture 3.1: Mathematical Induction CS 250, Discrete Structures, Fall 2015 Nitesh Saxena Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag

Lecture 3.1 -- 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 by next class Solution was provided (emailed) Questions? 7/24/2019 Lecture 3.1 -- Mathematical Induction

Lecture 3.1 -- Mathematical Induction Course Admin HW2 Was due just now Will provide solution soon Will start grading soon Any questions? 7/24/2019 Lecture 3.1 -- Mathematical Induction

Lecture 3.1 -- Mathematical Induction Outline Mathematical Induction Principle Examples Why it all works 7/24/2019 Lecture 3.1 -- 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 n2” We can picture each proposition as a domino: Book typically starts with n=1 instead of n=0. Starting point is arbitrary. P (n) 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction So sequence of propositions is a sequence of dominos. P (0) P (1) P (2) P (n) P (n+1) … … 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction When the domino falls, the corresponding proposition is considered true: P (n) 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction When the domino falls (to right), the corresponding proposition is considered true: P (n) true 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction Suppose that the dominos satisfy two constraints. Well-positioned: If any domino falls (to right), next domino (to right) must fall also. First domino has fallen to right P (n) P (n+1) P (0) true 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction Suppose that the dominos satisfy two constraints. Well-positioned: If any domino falls to right, the next domino to right must fall also. First domino has fallen to right P (n) P (n+1) P (0) true 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction Suppose that the dominos satisfy two constraints. Well-positioned: If any domino falls to right, the next domino to right must fall also. First domino has fallen to right P (n) true P (n+1) true P (0) true 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction Then can conclude that all the dominos fall! P (0) P (1) P (2) P (n) P (n+1) … 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction Then can conclude that all the dominos fall! P (0) P (1) P (2) P (n) P (n+1) … 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction Then can conclude that all the dominos fall! P (1) P (2) P (n) P (n+1) P (0) true … 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction Then can conclude that all the dominos fall! P (2) P (n) P (n+1) P (0) true P (1) true … 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction Then can conclude that all the dominos fall! P (n) P (n+1) P (0) true P (1) true P (2) true … 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction Then can conclude that all the dominos fall! P (n) P (n+1) P (0) true P (1) true P (2) true … 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction Then can conclude that all the dominos fall! P (n+1) P (0) true P (1) true P (2) true … P (n) true 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction Then can conclude that all the dominos fall! P (0) true P (1) true P (2) true … P (n) true P (n+1) true 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction Principle of Mathematical Induction: If: [basis] P (0) is true [induction] k P(k)P(k+1) is true Then: n P(n) is true This formalizes what occurred to dominos. P (0) true P (1) true P (2) true … P (n) true P (n+1) true 7/24/2019 Lecture 3.1 -- Mathematical Induction

Lecture 3.1 -- Mathematical Induction Exercise 1 Use induction to prove that the sum of the first n odd integers is n2. Prove a base case (n=1) Prove P(k)P(k+1) Base case (n=1): the sum of the first 1 odd integer is 12. Yup, 1 = 12. Assume P(k): the sum of the first k odd ints is k2. 1 + 3 + … + (2k - 1) = k2 Prove that 1 + 3 + … + (2k - 1) + (2k + 1) = (k+1)2 1 + 3 + … + (2k-1) + (2k+1) = k2 + (2k + 1) = (k+1)2 By arithmetic 7/24/2019 Lecture 3.1 -- Mathematical Induction

Lecture 3.1 -- 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 11! + … + kk! + (k+1)(k+1)! = (k+1)! - 1 + (k+1)(k+1)! = (1 + (k+1))(k+1)! - 1 = (k+2)(k+1)! - 1 = (k+2)! - 1 7/24/2019 Lecture 3.1 -- Mathematical Induction

Exercises 3 and 4 (have seen before?) Recall sum of arithmetic sequence: Recall sum of geometric sequence: 7/24/2019 Lecture 3.1 -- 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)) Assume P(0) k P(k)  P(k+1) n P(n) Proof by contradiction. n P(n) 7/24/2019 Lecture 3.1 -- Mathematical Induction

Mathematical Induction - why does it work? Assume P(0) k P(k)  P(k+1) n P(n) n P(n) Since N is well ordered, S has a least element. Call it k. Let S = { n : P(n) } 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. 7/24/2019 Lecture 3.1 -- Mathematical Induction

Lecture 3.1 -- Mathematical Induction Today’s Reading Rosen 5.1 7/24/2019 Lecture 3.1 -- Mathematical Induction