Lesson 11 – Proof by induction Further Pure 1 Lesson 11 – Proof by induction

Mathematical Statements In this lesson we are going to look at proving mathematical statements. For example: 3n-1 is always an even number The problem is though, how can we prove this to be true as there are infinitely many values of n to try? We can see that if n = 1, then 31-1 = 2 = Even and if n = 2, then 32-1 = 8 = Even and if n = 3, then 33-1 = 26 = Even But it is impossible to test every single value of n to see if the statement is true. We are going to now look at the idea of proof by induction.

Proof by induction Here is an example to think about. An old woman is asked how old she is, and replies “I am 100 years old”. No record has been kept of her birthday so how do we now she is 100? Answer: Because she was 99 last year. With proof by induction we assume that a statement is true and then try to prove that the next result will also be true. If this principle holds then as long as the first value works then automatically so must the second, then the third and so on and so on. It’s a bit like pushing over a domino line.

Proof by induction Now we can go back to our original problem from the first slide. Prove that 3n-1 is always an even number Lets assume that 3k-1 is always even, for n = k. Now 3k+1-1 = 3×3k-1 = 3×3k-3 + 2 = 3×(3k-1) + 2 = 3×(Even) + Even = Even So we have shown that if 3n-1 is even for n = k then 3n-1 will be even for n = k+1. Now if k = 1, 3k-1 = Even The domino effect kicks in and it must be true for n = 2,3,…………∞

Questions 1) Prove that, for all positive integers n, 1 + 2 + 3 + …… + n = (n/2)(n+1) 2) Prove that, for all positive integers n, 12 + 22 + 32 + …… + n2 = (n/6)(n+1)(2n+1) 3) Prove that, for all positive integers n, 13 + 23 + 33 + …… + n3 = (n2/4) (n+1)2 These are important results and we shall learn some more about them in lesson 12

Solution 1 Prove that, for all positive integers n, 1 + 2 + 3 + …… + n = (n/2)(n+1) First we start by assuming that it is true for n = k and try to show that it will be true for n = k+1 This shows that if n = k works then n = k+1 will also work. So if n = 1, then the sum of 1 is 1 & 0.5×1×2 = 1 which is the sum of 1. Therefore by induction the original statement is true for all n.

Solution 2 Prove that, for all positive integers n, 12 + 22 + 32 + …… + n2 = (n/6)(n+1)(2n+1) First we start by assuming that it is true for n = k and try to show that it will be true for n = k+1 This shows that if n = k works then n = k+1 will also work. So if n = 1, then the sum of 1 is 1 & (1/6)×1×2×3= 1 which is the sum of 1. Therefore by induction the original statement is true for all n

Solution 3 Prove that, for all positive integers n, 13 + 23 + 33 + …… + n3 = (n/4)2(n+1)2 First we start by assuming that it is true for n = k and try to show that it will be true for n = k+1 This shows that if n = k works then n = k+1 will also work. So if n = 1, then the sum of 1 is 1 & 0.25×4 = 1 which is the sum of 1. Therefore by induction the original statement is true for all n

More Proofs by induction 1 A sequence is defined by un+1 = 4un – 3, u1 = 2 Prove that un = 4n-1 + 1 We start by assuming that n = k will work and investigate what happens with n = k+1 uk+1 = 4uk – 3 = 4(4k-1 + 1) – 3 = 4×4k-1 + 4 – 3 = 4k + 4 – 3 = 4k + 1

More Proofs by induction 2 Prove that un = 4n + 6n – 1 is divisible by 9 for all n > 0. uk+1 = 4k+1 + 6(k+1) – 1 = 4k+1 + 6k + 6 – 1 = 4k+1 + 6k + 5 = 4×4k + 24k – 18k – 4 + 4 + 5 With this line I have just added in values so that I can factorise to make un = (4×4k + 24k – 4) – 18k + 4 + 5 = 4(4k + 6k -1) – 18k + 9 = 4(uk) – 9(2k – 1) This proves that uk+1 is divisible by 9 because uk is divisible by 9 and the second part 9(2k – 1) must also be divisible by 9. Now if n = 1, then u1 = 41 + 6×1 – 1 = 9 which is divisible by 9. This proves that un must always be divisible by 9 for all n > 0.

Enrichment problem Let P(n) be the statement that is divisible by 44. I shall show this is true if n is any positive integer! If n=1 then is equal to 44… so is divisible by 44! So the statement is clearly true when n=1.

Assume statement is true for n=k We suppose that the statement is true for n=k where k is some positive integer. So we assume that is divisible by 44. We want to then show that the statement must also be true for n=k+1.

Considering n=k+1 We therefore want to consider (this is with n=k+1). We want to show this is divisible by 44. All we are allowed to assume is that Is divisible by 44 (the case n=k). So let’s see if we can express in terms of .

The trick! So if is divisible by 44 with n=k then it is also divisible by 44 with n=k+1.

The conclusion So we have shown that is divisible by 44 with n=1. If it is divisible by 44 with n=k, for any positive integer k, then it follows that it is divisible by 44 with n=k+1. Therefore it follows that must be divisible by 44 for all positive integers n!