FP1 Chapter 6 – Proof By Induction

FP1 Chapter 6 – Proof By Induction

Dr J Frost ([email protected])

Last modified: 31st March 2015

What is Proof by Induction?We tend to use proof by induction whenever we want to show some property holds for all integers (usually positive).

Show that for all .

Example

𝒏=𝟏 𝒏=𝟐 𝒏=𝟑 𝒏=𝟒 𝒏=𝟓 …

Showing it’s true for a few values of is not a proof. We need to show it’s true for ALL values of .

Suppose we could prove the above is true for the first value of , i.e. . Suppose that if we assumed it’s true

for , then we could prove it’s true for . Then:

If it’s true for , it must be true for .



And so on. Hence the statement must be true for all .

Chapter Overview

We will use Proof of Induction for 4 different types of proof:

Summation Proofs

Divisibility Proofs

Recurrence Relation Proofs

Matrix Proofs

1

2

3

4

Given that , prove that

Prove that for all .

Show that for all .

Prove that is divisible by 3 for all .

Bro Tip: Recall that is the set of all integers, and is the set of all positive integers. Thus .

The Four Steps of Induction

! Proof by induction:

Step 1: Basis: Prove the general statement is true for .Step 2: Assumption: Assume the general statement is true for .Step 3: Inductive: Show that the general statement is then true for .Step 4: Conclusion: The general statement is then true for all positive integers .

For all these types the process of proof is the same:

(Step 1 is commonly known as the ‘base case’ and Step 3 as the ‘inductive case’)

Type 1: Summation ProofsStep 1: Basis: Prove the general statement is true for .Step 2: Assumption: Assume the general statement is true for .Step 3: Inductive: Show that the general statement is then true for .Step 4: Conclusion: The general statement is then true for all positive integers .

Show that for all .

1 Basis Step When , . so summation true for .

2 Assumption Assume summation true for .So

3 Inductive When :

Hence true when .

4 Conclusion Since true for and if true for , true for , true for all .

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More on the ‘Conclusion Step’

“As result is true for , this implies true for all positive integers and hence true by induction.”

I lifted this straight from a mark scheme, hence use this exact wording!The mark scheme specifically says:

(For method mark)Any 3 of these seen anywhere in the proof:• “true for ”• “assume true for ”• “true for ”• “true for all /positive integers”

Test Your Understanding

Edexcel FP1 Jan 2010

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Exercise 6AAll questions except Q2.

Type 2: Divisibility ProofsStep 1: Basis: Prove the general statement is true for .Step 2: Assumption: Assume the general statement is true for .Step 3: Inductive: Show that the general statement is then true for .Step 4: Conclusion: The general statement is then true for all positive integers .

Prove by induction that is divisible by 4 for all positive integers .

1 Basis Step Let , where . which is divisible by 4.

2 Assumption Assume that for , is divisible by 4 for .?

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…

Since true for and if true for , true for , true for all .

Type 2: Divisibility ProofsStep 1: Basis: Prove the general statement is true for .Step 2: Assumption: Assume the general statement is true for .Step 3: Inductive: Show that the general statement is then true for .Step 4: Conclusion: The general statement is then true for all positive integers .


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3 Inductive

4 Conclusion

Therefore is divisible by 4 when .

Bro Tip: Putting in terms of allows us to better compare with the assumption case since the power is now the same.

Method 1 (the textbook’s method, which I don’t like)Find the difference between successive terms and show this is divisible by the number of interest.

Method: Then write back in term of and the difference. We know both the terms on the RHS are divisible by 4.

Method: You need to explicitly factor out 4 to show divisibility by 4.

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Type 2: Divisibility ProofsProve by induction that is divisible by 4 for all positive integers .

3 Inductive


Method 2 (presented as ‘Alternative Method’ in mark schemes)Directly write in form by ‘splitting off’ or some multiple of it, where is the divisibility number.

Method: We have separated into because it allowed us to then replace with

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This method is far superior for two reasons:a) The idea of ‘separating out ’ is much more conceptually similar to both summation

proofs and (when we encounter it) matrix proofs, whereas using is specific to divisibility proofs. Thus we needn’t see divisibility proofs as a different method from the other types.

b) We’ll encounter a question next where we need to get in the form , i.e. we in fact separate out a multiple of . Thus causes absolute havoc for Method 1 as our difference has to refer to , which is mathematically inelegant.

Type 2: Divisibility ProofsProve by induction that is divisible by 5.

2 Assumption Assume that for , is divisible by 5 for .

3 Inductive When :

therefore true for .Method: We have 3 lots of and we can split off.

?Using Method 2

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Test Your UnderstandingStep 1: Basis: Prove the general statement is true for .Step 2: Assumption: Assume the general statement is true for .Step 3: Inductive: Show that the general statement is then true for .Step 4: Conclusion: The general statement is then true for all positive integers .


1 Basis Step Let , where . which is divisible by 3.

2 Assumption Assume that for , is divisible by 3 for .

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3 Inductive

4 Conclusion

Method 1:


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Method 2:

Therefore is divisible by 3 when .?

Exercise 6BPage 130.All questions.

Method: Sub in your assumption expression into the recurrence.


Type 3: Recurrence Relation ProofsStep 1: Basis: Prove the general statement is true for .Step 2: Assumption: Assume the general statement is true for .Step 3: Inductive: Show that the general statement is then true for .Step 4: Conclusion: The general statement is then true for all positive integers .

Given that and , prove by induction that

1 Basis Step When , as given.IMPORTANT NOTE: The textbook gets this wrong (bottom of Page 131), saying that is required. I checked a mark scheme – it isn’t.

2 Assumption Assume that for , is true for .

3 Inductive

4 Conclusion

Then

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Notice that we’re trying to prove a position-to-term formula from a term-to-term one.

Bro Tip: You need to carefully reflect on what statement you are trying to prove, i.e. , and what statement is GIVEN (and thus already know to be true), i.e.

Bro Tip: For harder questions, write out what you’re trying to prove in advance, i.e. that we want to get , , to help guide your manipulation.

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Test Your UnderstandingEdexcel FP1 Jan 2011

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Since true for and and if true for and , true for , true for all .

Recurrence Relations based on two previous termsStep 1: Basis: Prove the general statement is true for .Step 2: Assumption: Assume the general statement is true for .Step 3: Inductive: Show that the general statement is then true for .Step 4: Conclusion: The general statement is then true for all positive integers .

Given that and , , prove by induction that

1 Basis Step When , as given.When , as given.We have two base cases so had to check 2 terms!

2 Assumption Assume that for and , and is true for .

? (Warning: slightly different to before)

3 Inductive

4 Conclusion

Let (Brominder: it’s worth noting in advance we’re aiming for )Then

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We need two previous instances of in our assumption case now.

Exercise 6CPage 132.Q1, 3, 5, 7

Type 4: Matrix Proofs



Prove by induction that for all .

1 Basis Step When , . As , true for .

2 Assumption Assume true , i.e.

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3 Inductive

4 Conclusion

When ,

Therefore true when .

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Method: As with summation and divisibility proofs, it’s just a case of ‘separating off’ the assumption expression.

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Test Your Understanding



Prove by induction that for all .

1 Basis Step When , . As , true for .

2 Assumption Assume true , i.e.

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3 Inductive

4 Conclusion

When ,

Therefore true when . ?

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Exercise 6DPage 134.All questions.

FP1 Chapter 6 – Proof By Induction

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