Understanding Mathematical Induction in Discrete Mathematics

1. CS 23022Discrete Mathematical Structures Mehdi Ghayoumi MSB rm 132 mghayoum@kent.edu Ofc hr: Thur, 9:30-11:30a

2. Announcements Midterm…!!! Do HW5 again for Thursday This session is from chapter 5 Rosen and some other articles

3. Everyone likes Ice cream! Suggests a proof technique n FL(n) Mathematical Induction One rule: if the person “before” you likes Ice cream, then you like Ice cream. Person 1 likes Ice cream. What can we conclude?

4. Mathematical Induction Suppose we want to prove everyone likes FruitLoops Need to show two things: Person 1 likes Fruit Loops (FL(1)) If person k likes Fruit Loops, Then person k+1 does too. (FL(k)  FL(k+1))

5. First part is a simple proposition we call the base case. Second part is a conditional. Start by assuming FL(k), and show that FL(k+1) follows. Mathematical Induction Suppose we want to prove everyone likes Fruit Loops Need to show two things: Person 1 likes Fruit Loops (FL(1)) If person k likes Fruit Loops, then person k+1 does too. (FL(k)  FL(k+1))

6. Mathematical Induction • We know that: We can reach the first rung of this ladder; If we can reach a particular rung of the ladder, then we can reach the next rung of the ladder. Can we reach every step of this infinite ladder? Yes, using Mathematical Induction which is a rule of inference that tells us: P(1) k (P(k)  P(k+1)) --------------------------  n (P(n)

7. Mathematical Induction If we have a propositional function P(n), and we want to prove that P(n) is true for any natural number n, we do the following: Show that P(0) is true. (basis step) Show that if P(n) then P(n + 1) for any nN.(inductive step) Then P(n) must be true for any nN. (conclusion)

8. Prove a base case (n=?) Prove P(k)P(k+1) Inductive hypothesis Mathematical Induction • Use induction to prove that the • 1 + 2 + 22 + … + 2n = 2n+1 - 1 • for all non-negative integers n. • 1 – Hypothesis? P(n) = 1 + 2 + 22 + … + 2n = 2 n+1 – 1 for all non-negative integers n. • 2 - Base case? n = 0 10 = 21-1. 3 – Inductive Hypothesis Assume P(k) = 1 + 2 + 22 + … + 2k = 2 k+1 – 1

9. By inductive hypothesis p(k) Mathematical Induction • 4 – Inductive Step: show that (k) P(k)  P(k+1), assuming P(k). • How? P(k+1)= 1 + 2 + 22 + … + 2k+ 2k+1 = (2k+1 – 1) + 2k+1 = 22k+1 - 1 P(k+1) = 2k+2 - 1 = 2(k+1)+1 - 1

10. Prove a base case (n=1) Prove P(k)P(k+1) Inductive hypothesis By inductive hypothesis By arithmetic Mathematical Induction 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

11. Inductive hypothesis Mathematical Induction 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

12. Inductive hypothesis Mathematical Induction 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.

13. 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. Mathematical Induction 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)

14. Subsets containing a are made by taking any set from P(S), and inserting an a. So |{X : a  X}| = |P(S)| Mathematical Induction 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

15. Mathematical Induction - a cool example Deficient Tiling A 2n x 2n sized grid is deficient if all but one cell is tiled. 2n 2n

16. Mathematical Induction - a cool example We want to show that all 2n x 2n sized deficient grids can be tiled with tiles shaped like:

17. Yup! Yup! Mathematical Induction - a cool example Is it true for 20 x 20 grids? Is it true for 21 x 21 grids?

18. Mathematical Induction - a cool example Inductive Hypothesis: We can tile a 2k x 2k deficient board using our fancy designer tiles. Use this to prove: We can tile a 2k+1 x 2k+1 deficient board using our fancy designer tiles.

19. 2k 2k ? ? 2k ? 2k Mathematical Induction - a cool example 2k+1 OK!! (by IH)

20. 2k 2k 2k 2k Mathematical Induction - a cool example OK!! (by IH) OK!! (by IH) 2k+1 OK!! (by IH) OK!! (by IH)

21. Mathematical Induction - a cool example

22. Mathematical Induction - a cool example