Proof by Induction


















Suppose you want to prove (for n >= 1):

	P(n): 1 + 3 + 5 + ... + (2n-1) = n*n
	The sum of the first n positive odd integers is n-squared.

Can you prove a Specific Case (Base Case) such as when n is 1?

	when n is 1
	   the sum of the first n positive odd integers is 1
	   n-squared is 1
	P(n) is true when n is 1

	when n is 2
	   the sum of the first n positive odd integers is 1+3=4
	   n-squared is 4
	P(n) is true when n is 2

	when n is 3
	   the sum of the first n positive odd integers is 1+3+5=9
	   n-squared is 9
	P(n) is true when n is 3


















Suppose you could prove a General Step (Inductive Step).

Suppose you could prove: P(k) -> P(k+1)
  If the statement is true when n is k,
  then the statement is true when n is k+1.

What would this allow you to conclude when combined with the base case?

	1. the base case says P(1) is true

	2. the inductive step says P(1) -> P(2)
	   conclude P(2) is true

	3. the inductive step says P(2) -> P(3)
	   conclude P(3) is true

	4. ...


















How do you prove the General Step?

To prove the General Step (Inductive Step)
you need to prove that for all k >= 1, P(k) -> P(k+1).

How do you prove an implication such as this?

	1. assume P(k) is true (inductive hypothesis)

	2. use the assumption to prove P(k+1) is true


What do you get to assume?

What do you need to prove?

How do you prove it using the assumption?


How do you conclude the proof?

	state the conclusion like this:

	Since P(1) is true
	and P(k) -> P(k+1) for all k >= 1,
	we conclude that P(n) is true for all n >= 1.


















What's the Principle of Mathematical Induction?

	P(1), P(k) -> P(k+1) => P(n), for n >= 1


What are the steps needed for a Proof by Induction?

	1. prove the base case.
	2. assume P(k).
	3. prove P(k+1).


















Classwork
You may work with a partner.

Use induction to prove (for n >= 1):

	P(n): 1 + 2 + 3 + ... + n = n(n+1)/2
	The sum of the first n positive integers is n(n+1)/2.


















Classwork
You may work with a partner.

Use induction to prove (for n >= 0):

	P(n): 2^0 + 2^1 + 2^2 + ... + 2^n = 2^(n+1) - 1
	The sum of the powers of 2 from 0 to n is 2^(n+1) - 1.