Lecture 3.1: Mathematical Induction

CS 250, Discrete Structures, Fall 2014

Nitesh Saxena

Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag

04/21/23Lecture 3.1 -- Mathematical

Induction

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04/21/23Lecture 3.1 -- Mathematical

Induction

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04/21/23Lecture 3.1 -- Mathematical

Induction

Outline

Mathematical Induction Principle Examples Why it all works

04/21/23Lecture 3.1 -- Mathematical

Induction

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)

04/21/23Lecture 3.1 -- Mathematical

Induction

Mathematical InductionSo sequence of propositions is a sequence

of dominos.

…

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

…

04/21/23Lecture 3.1 -- Mathematical

Induction

Mathematical InductionWhen the domino falls, the corresponding

proposition is considered true:

P (n)

04/21/23Lecture 3.1 -- Mathematical

Induction

Mathematical InductionWhen the domino falls (to right), the

corresponding proposition is considered true:

P (n)true

04/21/23Lecture 3.1 -- Mathematical

Induction

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)

04/21/23Lecture 3.1 -- Mathematical

Induction

Mathematical InductionSuppose that the dominos satisfy two

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

right, the next domino to right must fall also.

2) First domino has fallen to right

P (0)true

P (n+1)P (n)

04/21/23Lecture 3.1 -- Mathematical

Induction

Mathematical InductionSuppose that the dominos satisfy two

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

right, the next domino to right must fall also.

2) First domino has fallen to right

P (0)true

P (n)true

P (n+1)true

04/21/23Lecture 3.1 -- Mathematical

Induction

Mathematical InductionThen can conclude that all the dominos

fall!

…

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

04/21/23Lecture 3.1 -- Mathematical

Induction

Mathematical InductionThen can conclude that all the dominos

fall!

…

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

04/21/23Lecture 3.1 -- Mathematical

Induction

Mathematical InductionThen can conclude that all the dominos

fall!

…P (0)true

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

04/21/23Lecture 3.1 -- Mathematical

Induction

Mathematical InductionThen can conclude that all the dominos

fall!

…P (0)true

P (1)true

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

04/21/23Lecture 3.1 -- Mathematical

Induction

Mathematical InductionThen can conclude that all the dominos

fall!

P (2)true

…P (0)true

P (1)true

P (n+1)P (n)

04/21/23Lecture 3.1 -- Mathematical

Induction

Mathematical InductionThen can conclude that all the dominos

fall!

P (2)true

…P (0)true

P (1)true

P (n+1)P (n)

04/21/23Lecture 3.1 -- Mathematical

Induction

Mathematical InductionThen can conclude that all the dominos

fall!

P (2)true

…P (0)true

P (1)true

P (n)true

P (n+1)

04/21/23Lecture 3.1 -- Mathematical

Induction

Mathematical InductionThen can conclude that all the dominos

fall!

P (2)true

…P (0)true

P (1)true

P (n)true

P (n+1)true

04/21/23Lecture 3.1 -- Mathematical

Induction

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

04/21/23Lecture 3.1 -- Mathematical

Induction

Exercise 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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Induction

Exercise 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)! - 111! + … + 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

04/21/23Lecture 3.1 -- Mathematical

Induction

Exercises 3 and 4 (have seen before?)

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

04/21/23Lecture 3.1 -- Mathematical

Induction

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)

04/21/23Lecture 3.1 -- Mathematical

Induction

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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Induction

Today’s Reading Rosen 5.1