1. CS 23022 Discrete 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. 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? Everyone likes Ice cream!Suggests a proof technique  n FL(n)

4. 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))

5. 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)) 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.

6. 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) Mathematical Induction

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

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

10. 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 Inductive hypothesis 1 + 3 + … + (2k-1) + (2k+1) =k 2 + (2k + 1) By inductive hypothesis = (k+1) 2 By arithmetic

11. 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 Inductive hypothesis 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. Mathematical Induction 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 Inductive hypothesis 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. Mathematical Induction 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.

14. Mathematical Induction 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

15. Mathematical Induction - a cool example Deficient Tiling A 2 n x 2 n sized grid is deficient if all but one cell is tiled. 2n2n 2n2n

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

17. Mathematical Induction - a cool example Is it true for 2 0 x 2 0 grids? Yup! Is it true for 2 1 x 2 1 grids? Yup!

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

19. Mathematical Induction - a cool example 2k2k 2k2k 2k2k 2k2k 2 k+1 OK!! (by IH) ? ? ?

20. Mathematical Induction - a cool example 2k2k 2k2k 2k2k 2k2k 2 k+1 OK!! (by IH)

21. Mathematical Induction - a cool example