9/27/2011 Lecture Mathematical Induction1 Lecture 3.1: Mathematical Induction* CS 250, Discrete Structures, Fall 2011 Nitesh Saxena *Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag

9/27/2011 Lecture Mathematical Induction2 Course Admin Mid-Term 1 Hope it went well! Thanks cooperating with the TA as a proctor We are grading them now, and should have the results by next weekend Solution will be posted today

9/27/2011 Lecture Mathematical Induction3 Course Admin HW1 graded Scores have been posted To be distributed at the end of lecture Thanks for your patience waiting for the results Any questions after taking a careful look, please contact TA If that doesn’t help, please contact me HW2 due Sep 30 (this Friday)

9/27/2011 Lecture Mathematical Induction4 Outline Mathematical Induction Principle Examples Why it all works

9/27/2011 Lecture Mathematical Induction5 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)

9/27/2011 Lecture Mathematical Induction6 Mathematical Induction So sequence of propositions is a sequence of dominos. … P (n+1)P (n) P (2)P (1)P (0) …

9/27/2011 Lecture Mathematical Induction7 Mathematical Induction When the domino falls, the corresponding proposition is considered true: P (n)

9/27/2011 Lecture Mathematical Induction8 Mathematical Induction When the domino falls (to right), the corresponding proposition is considered true: P (n) true

9/27/2011 Lecture Mathematical Induction9 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)

9/27/2011 Lecture Mathematical Induction10 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)

9/27/2011 Lecture Mathematical Induction11 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

9/27/2011 Lecture Mathematical Induction12 Mathematical Induction Then can conclude that all the dominos fall! … P (n+1)P (n) P (2)P (1)P (0)

9/27/2011 Lecture Mathematical Induction13 Mathematical Induction Then can conclude that all the dominos fall! … P (n+1)P (n) P (2)P (1)P (0)

9/27/2011 Lecture Mathematical Induction14 Mathematical Induction Then can conclude that all the dominos fall! …P (0) true P (n+1)P (n) P (2)P (1)

9/27/2011 Lecture Mathematical Induction15 Mathematical Induction Then can conclude that all the dominos fall! …P (0) true P (1) true P (n+1)P (n) P (2)

9/27/2011 Lecture Mathematical Induction16 Mathematical Induction Then can conclude that all the dominos fall! P (2) true …P (0) true P (1) true P (n+1)P (n)

9/27/2011 Lecture Mathematical Induction17 Mathematical Induction Then can conclude that all the dominos fall! P (2) true …P (0) true P (1) true P (n+1)P (n)

9/27/2011 Lecture Mathematical Induction18 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)

9/27/2011 Lecture Mathematical Induction19 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

9/27/2011 Lecture Mathematical Induction20 Mathematical Induction Principle of Mathematical Induction: If: 1) [basis] P (0) is true 2) [induction]  n P(n)  P(n+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

9/27/2011 Lecture Mathematical Induction21 Example 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

9/27/2011 Lecture Mathematical Induction22 Example 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

9/27/2011 Lecture Mathematical Induction23 Example 3 Prove that if a set S has |S| = n, then |P(S)| = 2 n Base case (n=0): S=ø, P(S) = {ø} and |P(S)| = 1 = 2 0 Assume P(k): If |S| = k, then |P(S)| = 2 k Prove that if |S’| = k+1, then |P(S’)| = 2 k+1 S’ = S U {a} for some S  S’ with |S| = k, and a  S’. Partition the power set of S’ into the sets containing a and those not. We count these sets separately.

9/27/2011 Lecture Mathematical Induction24 Example 3 (contd) Assume P(k): If |S| = k, then |P(S)| = 2 k Prove that if |S’| = k+1, then |P(S’)| = 2 k+1 S’ = S U {a} for some S  S’ with |S| = k, and a  S’. Partition the power set of S’ into the sets containing a and those not. P(S’) = {X : a  X} U {X : a  X} P(S’) = {X : a  X} U P(S) Since these are all the subsets of elements in S. Subsets containing a are made by taking any set from P(S), and inserting an a.

9/27/2011 Lecture Mathematical Induction25 Example 3 (contd) Assume P(k): If |S| = k, then |P(S)| = 2 k Prove that if |S’| = k+1, then |P(S’)| = 2 k+1 S’ = S U {a} for some S  S’ with |S| = k, and a  S’. P(S’) = {X : a  X} U {X : a  X} P(S’) = {X : a  X} U P(S) Subsets containing a are made by taking any set from P(S), and inserting an a. So |{X : a  X}| = |P(S)| |P(S’)| = |{X : a  X}| + |P(S)| = 2 |P(S)| = 2  2 k = 2 k+1

9/27/2011 Lecture Mathematical Induction26 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)

9/27/2011 Lecture Mathematical Induction27 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.

9/27/2011 Lecture Mathematical Induction28 More examples – prove by induction 1. Recall sum of arithmetic sequence: 2. Recall sum of geometric sequence:

9/27/2011 Lecture Mathematical Induction29 Today’s Reading Rosen 5.1