Mathematical Induction

F(1) = 1; F(n+1) = F(n) + (2n+1) for n≥1

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F(n)

n

F(n) = n2 for all n ≥ 1

Prove it!

Prove statements of the form: "p(n) for n ≥ 1"

Verify p(1). "Base case"

Assume p(n) for some n ≥ 1 "Induction hypothesis"

Show p(n+1) must be true for this value of n. "Induction step"

Conclude p(n) is true for all n ≥ 1

Prove: F(n) = n2 for n ≥ 1.

RecallF(1) = 1; F(n+1) = F(n) + (2n+1) for n>1

Proof:F(1) = 1 = 12

Assume F(n) = n2 for some n ≥ 1. Then F(n+1) = F(n) + (2n+1), since n+1 > 1.

= n2 + 2n + 1 by hypothesis = (n+1)2

Therefore F(n) = n2 for all n ≥ 1.

Induction hypothesisBase Case

Algebra

Not just for some!

First principle of mathematical induction

Let S be a set of integers containing a. Suppose S has the property that whenever some integer n ≥ a belongs to S, then the integer n + 1 also belongs to S.

Then S contains every integer greater than or equal to a.

What does our proof have to do with S?

Let S = { n ≥ 1 | F(n) = n2 }F(n) = 1 = 12

Base case shows 1 is in SF(n) = n2 for some n ≥ 1 => F(n+1) = (n+1)2

Induction step shows that if n is in S, then n+1 is in S.

By the first principle of induction, S contains all integers ≥ 1.

That is, F(n) = n2 for all n ≥ 1.

Why does the first principle of induction work?

Proof: Given S = { n ≥ a | p(n)} has the property that a is in S and n in S implies n+1 is in S.

It remains to show that S contains all integers ≥ a. Let T = { n ≥ a | n is not in S} Assume, towards a contradiction, T is not empty. By the WOP, T has a smallest element, t Now t > a, and n = t – 1 belongs to S. But then n+1 = (t–1)+1 = t belongs to S. This contradiction shows that T is empty.

Second Principle of Mathematical Induction

Let S be the set of integers containing a. Suppose S has the property that n belongs to S whenever every integer less than n and greater than or equal to a belongs to S.

Then S contains all integers greater than or equal to a.

Proof of the 2nd Principle of Mathematical Induction

Almost identical to the proof of the first. Try it!

Proof of the Fundamental Theorem of Arithmetic.

Let S be the set of integers greater than 1 which are primes or products of primes.

2 belongs to S Assume that for some integer n ≥ 2, S contains all

integers k with 2 ≤ k < n. We must show that n is in S. If n is prime, then n is in S. If n is composite, n = ab where a,b are in S.

By hypothesis, a and b are primes or products of primes. So n is a product of primes.