CS 23022Discrete Mathematical Structures

Mehdi GhayoumiMSB rm 132mghayoum@kent.edu Ofc hr: Thur, 9:30-11:30a

Announcements

Midterm…!!!

Do HW5 again for Thursday

This session is from chapter 5 Rosen

and some other articles

Mathematical InductionOne 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)

Mathematical InductionSuppose we want to prove everyone likes Fruit

LoopsNeed 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))

Mathematical InductionSuppose we want to prove everyone likes Fruit

LoopsNeed 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.

• 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

Mathematical InductionIf 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 nN. (inductive step)

Then P(n) must be true for any nN. (conclusion)

• Use induction to prove that the • 1 + 2 + 22 + … + 2n = 2n+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 HypothesisAssume P(k) = 1 + 2 + 22 + … + 2k = 2 k+1 – 1

Inductive hypothesi

s

n = 0 10 = 21-1.

P(n) = 1 + 2 + 22 + … + 2n = 2 n+1 – 1for all non-negative integers n.

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 By inductive

hypothesisp(k) = 2 2k+1 - 1

P(k+1) = 2k+2 - 1

= 2(k+1)+1 - 1

Mathematical Induction

Mathematical InductionUse induction to prove that the sum of the first n

odd integers is n2.Prove a base case (n=1)

Base case (n=1): the sum of the first 1 odd integer is 12. Yup, 1

= 12.

Prove P(k)P(k+1)

Assume P(k): the sum of the first k odd ints is k2.

1 + 3 + … + (2k - 1) = k2Prove that 1 + 3 + … + (2k - 1) + (2k + 1) = (k+1)2

Inductive hypothesi

s

1 + 3 + … + (2k-1) + (2k+1) =

k2 + (2k + 1)

By inductive hypothesis

= (k+1)2By arithmetic

Mathematical InductionProve that 11! + 22! + … + nn! = (n+1)!

- 1, nBase case (n=1): 11! = (1+1)! - 1?Yup, 11! = 1, 2! - 1 = 1

Assume P(k): 11! + 22! + … + kk! = (k+1)! - 1Prove that 11! + … + kk! + (k+1)(k+1)! = (k+2)! - 1

Inductive hypothesi

s

11! + … + kk! + (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

Mathematical InductionProve that if a set S has |S| = n, then |P(S)| =

2nBase case (n=0): S=ø, P(S) = {ø} and |P(S)| =

1 = 20Assume P(k): If |S| = k, then |P(S)| = 2k

Prove that if |S’| = k+1, then |P(S’)| = 2k+1

Inductive hypothesi

s

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.

Mathematical InductionAssume 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) 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 InductionAssume 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) 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)|

= 22k

= 2k+1

Mathematical Induction - a cool example

Deficient TilingA 2n x 2n sized grid is deficient if all but one cell is

tiled.

2n

2n

Mathematical Induction - a cool example

We want to show that all 2n x 2n sized deficient grids can be tiled with tiles shaped like:

Mathematical Induction - a cool example

Is it true for 20 x 20 grids?Yup!

Is it true for 21 x 21 grids?

Yup!

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.

Mathematical Induction - a cool example

2k

2k 2k

2k

2k+1

OK!! (by IH)

???

Mathematical Induction - a cool example

2k

2k 2k

2k

2k+1

OK!! (by IH)

OK!! (by IH)

OK!! (by IH)

OK!! (by IH)

Mathematical Induction - a cool example

Mathematical Induction - a cool example