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Lecture 3.1: Mathematical Induction*

Lecture 3.1: Mathematical Induction*. CS 250, Discrete Structures, Fall 2011 Nitesh Saxena. * Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag. Course Admin. Mid-Term 1 Hope it went well! Thanks cooperating with the TA as a proctor

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Lecture 3.1: Mathematical Induction*

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

  2. 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 Lecture 3.1 -- Mathematical Induction

  3. 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) Lecture 3.1 -- Mathematical Induction

  4. Outline • Mathematical Induction Principle • Examples • Why it all works Lecture 3.1 -- Mathematical Induction

  5. 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: P (n) Lecture 3.1 -- Mathematical Induction

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

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

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

  9. 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 Lecture 3.1 -- Mathematical Induction

  10. 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 Lecture 3.1 -- Mathematical Induction

  11. 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 Lecture 3.1 -- Mathematical Induction

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

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

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

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

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

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

  18. 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 Lecture 3.1 -- Mathematical Induction

  19. 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 Lecture 3.1 -- Mathematical Induction

  20. Mathematical Induction Principle of Mathematical Induction: If: • [basis] P (0) is true • [induction] nP(n)P(n+1) is true Then: nP(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 Lecture 3.1 -- Mathematical Induction

  21. Prove a base case (n=1) Prove P(k)P(k+1) By arithmetic Example 1 Use induction to prove that the sum of the first n odd integers is n2. 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 Lecture 3.1 -- Mathematical Induction

  22. 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 Lecture 3.1 -- Mathematical Induction

  23. Example 3 Prove that if a set S has |S| = n, then |P(S)| = 2n Base case (n=0): S=ø, P(S) = {ø} and |P(S)| = 1 = 20 Assume P(k): If |S| = k, then |P(S)| = 2k Prove that if |S’| = k+1, then |P(S’)| = 2k+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. Lecture 3.1 -- Mathematical Induction

  24. 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. Example 3 (contd) Assume P(k): If |S| = k, then |P(S)| = 2k Prove that if |S’| = k+1, then |P(S’)| = 2k+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) Lecture 3.1 -- Mathematical Induction

  25. Subsets containing a are made by taking any set from P(S), and inserting an a. So |{X : a  X}| = |P(S)| Example 3 (contd) Assume P(k): If |S| = k, then |P(S)| = 2k Prove that if |S’| = k+1, then |P(S’)| = 2k+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) |P(S’)| = |{X : a  X}| + |P(S)| = 2 |P(S)| = 22k = 2k+1 Lecture 3.1 -- Mathematical Induction

  26. Proof by contradiction. n P(n) 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) Lecture 3.1 -- Mathematical Induction

  27. Since N is well ordered, S has a least element. Call it k. But by (2), P(k-1)  P(k). Contradicts P(k-1) true, P(k) false. Done. n P(n) Mathematical Induction - why does it work? Assume • P(0) • k P(k)  P(k+1) • n P(n) 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. Lecture 3.1 -- Mathematical Induction

  28. More examples – prove by induction • Recall sum of arithmetic sequence: • Recall sum of geometric sequence: Lecture 3.1 -- Mathematical Induction

  29. Today’s Reading • Rosen 5.1 Lecture 3.1 -- Mathematical Induction

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