1. 9/27/2011 Lecture 3.1 -- Mathematical Induction1 Lecture 3.1: Mathematical Induction* CS 250, Discrete Structures, Fall 2011 Nitesh Saxena *Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag

2. 9/27/2011 Lecture 3.1 -- 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

3. 9/27/2011 Lecture 3.1 -- 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)

4. 9/27/2011 Lecture 3.1 -- Mathematical Induction4 Outline Mathematical Induction Principle Examples Why it all works

5. 9/27/2011 Lecture 3.1 -- 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)

6. 9/27/2011 Lecture 3.1 -- Mathematical Induction6 Mathematical Induction So sequence of propositions is a sequence of dominos. … P (n+1)P (n) P (2)P (1)P (0) …

7. 9/27/2011 Lecture 3.1 -- Mathematical Induction7 Mathematical Induction When the domino falls, the corresponding proposition is considered true: P (n)

8. 9/27/2011 Lecture 3.1 -- Mathematical Induction8 Mathematical Induction When the domino falls (to right), the corresponding proposition is considered true: P (n) true

9. 9/27/2011 Lecture 3.1 -- 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)

10. 9/27/2011 Lecture 3.1 -- 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)

11. 9/27/2011 Lecture 3.1 -- 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

12. 9/27/2011 Lecture 3.1 -- Mathematical Induction12 Mathematical Induction Then can conclude that all the dominos fall! … P (n+1)P (n) P (2)P (1)P (0)

13. 9/27/2011 Lecture 3.1 -- Mathematical Induction13 Mathematical Induction Then can conclude that all the dominos fall! … P (n+1)P (n) P (2)P (1)P (0)

14. 9/27/2011 Lecture 3.1 -- Mathematical Induction14 Mathematical Induction Then can conclude that all the dominos fall! …P (0) true P (n+1)P (n) P (2)P (1)

15. 9/27/2011 Lecture 3.1 -- 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)

16. 9/27/2011 Lecture 3.1 -- 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)

17. 9/27/2011 Lecture 3.1 -- 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)

18. 9/27/2011 Lecture 3.1 -- 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)

19. 9/27/2011 Lecture 3.1 -- 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

20. 9/27/2011 Lecture 3.1 -- 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

21. 9/27/2011 Lecture 3.1 -- 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 2. 1 + 3 + … + (2k - 1) = k 2 Prove that 1 + 3 + … + (2k - 1) + (2k + 1) = (k+1) 2 1 + 3 + … + (2k-1) + (2k+1) =k 2 + (2k + 1) = (k+1) 2 By arithmetic

22. 9/27/2011 Lecture 3.1 -- 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 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

23. 9/27/2011 Lecture 3.1 -- 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.

24. 9/27/2011 Lecture 3.1 -- 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.

25. 9/27/2011 Lecture 3.1 -- 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

26. 9/27/2011 Lecture 3.1 -- 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)

27. 9/27/2011 Lecture 3.1 -- 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.

28. 9/27/2011 Lecture 3.1 -- Mathematical Induction28 More examples – prove by induction 1. Recall sum of arithmetic sequence: 2. Recall sum of geometric sequence:

29. 9/27/2011 Lecture 3.1 -- Mathematical Induction29 Today’s Reading Rosen 5.1