Lecture 3.1: Mathematical Induction*

9/27/2011 Lecture 3.1 -- Mathematical Induction

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Lecture 3.1: Mathematical Induction*

CS 250, Discrete Structures, Fall 2011

Nitesh Saxena *Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag


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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


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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)


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Outline

Mathematical Induction Principle Examples Why it all works


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Mathematical InductionSuppose 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)


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Mathematical InductionSo sequence of propositions is a sequence

of dominos.

…

P (n+1)P (n)P (2)P (1)P (0)

…


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Mathematical InductionWhen the domino falls, the corresponding

proposition is considered true:

P (n)


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Mathematical InductionWhen the domino falls (to right), the

corresponding proposition is considered true:

P (n)true


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Mathematical InductionSuppose 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)


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constraints.1) Well-positioned: If any domino falls to

right, the next domino to right must fall also.


P (0)true

P (n+1)P (n)


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constraints.1) Well-positioned: If any domino falls to

right, the next domino to right must fall also.


P (0)true

P (n)true

P (n+1)true


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Mathematical InductionThen can conclude that all the dominos

fall!

…

P (n+1)P (n)P (2)P (1)P (0)


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fall!

…

P (n+1)P (n)P (2)P (1)P (0)


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fall!

…P (0)true

P (n+1)P (n)P (2)P (1)


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fall!

…P (0)true

P (1)true

P (n+1)P (n)P (2)


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fall!

P (2)true

…P (0)true

P (1)true

P (n+1)P (n)


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fall!

P (2)true

…P (0)true

P (1)true

P (n+1)P (n)


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fall!

P (2)true

…P (0)true

P (1)true

P (n)true

P (n+1)


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fall!

P (2)true

…P (0)true

P (1)true

P (n)true

P (n+1)true


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Mathematical InductionPrinciple of Mathematical Induction: If:1) [basis] P (0) is true2) [induction] n P(n)P(n+1) is true

Then: n P(n) is trueThis formalizes what occurred to dominos.

P (2)true

…P (0)true

P (1)true

P (n)true

P (n+1)true


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Example 1Use 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) = k2

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

1 + 3 + … + (2k-1) + (2k+1) =k2 + (2k + 1)= (k+1)2 By arithmetic


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Example 2

Prove 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

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


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Example 3

Prove that if a set S has |S| = n, then |P(S)| = 2nBase 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.


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Example 3 (contd)Assume P(k): If |S| = k, then |P(S)| = 2k


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.


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Example 3 (contd)Assume P(k): If |S| = k, then |P(S)| = 2k


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


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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.

Assume1. P(0)2. k P(k) P(k+1)3. n P(n) n P(n)


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Mathematical Induction - why does it work?

Assume1. 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.


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More examples – prove by induction

1. Recall sum of arithmetic sequence:

2. Recall sum of geometric sequence:

2

)1(

1

nni

n

i

1

)1(...

12

0

r

raarararaar

nn

n

i

i


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Today’s Reading Rosen 5.1

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Lecture 3.1: Mathematical Induction*