FP1 Chapter 6 – Proof By Induction
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Transcript of FP1 Chapter 6 – Proof By Induction
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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 .
If it’s true for , it must be true for .
If it’s true for , it must be true for .
And so on. Hence the statement must be true for all .
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Chapter Overview
We will use Proof of Induction for 4 different types of proof:
Summation Proofs
Divisibility Proofs
Recurrence Relation Proofs
Matrix Proofs
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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 .
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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’)
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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”
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Test Your Understanding
Edexcel FP1 Jan 2010
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Exercise 6AAll questions except Q2.
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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 .
Prove by induction that is divisible by 4 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
Therefore is divisible by 4 when .
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.
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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.
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Since true for and if true for , true for , true for all .
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 .
Prove by induction that is divisible by 3 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:
Therefore is divisible by 3 when .
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Method 2:
Therefore is divisible by 3 when .?
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Exercise 6BPage 130.All questions.
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Method: Sub in your assumption expression into the recurrence.
Since true for and if true for , true for , true for all .
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 .
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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.
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Exercise 6CPage 132.Q1, 3, 5, 7
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Type 4: Matrix Proofs
Since true for and if true for , true for , true for all .
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 .
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
Since true for and if true for , true for , true for all .
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 .
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.