FP1 Chapter 5 - Series
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Transcript FP1 Chapter 5 - Series
FP1 Chapter 6 β 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).
Example
Show that
π
π=1 π
1
= 2 π π + 1 for all π β β.
Showing itβs true for a few values of π is not a proof.
We need to show itβs true for ALL values of π.
If itβs true for π = 1, it must be true for π = 2.
π=π
π=π
Suppose we could prove the
above is true for the first
value of π, i.e. π = 1.
π=π
If itβs true for π = 3, it must be true for π = 4.
π=π
π=π
β¦
If itβs true for π = 2, it must be true for π = 3.
And so on. Hence the statement must be true for all π.
Suppose that if we assumed itβs true
for π = π, then we could prove itβs
true for π = π + 1. Then:
Chapter Overview
We will use Proof of Induction for 4 different types of proof:
π
π=1 π
1
1
Summation Proofs
Show that
2
Divisibility Proofs
Prove that π3 β 7π + 9 is divisible by 3 for all π β β€+ .
3
Recurrence Relation
Proofs
Given that π’π+1 = 3π’π + 4,
prove that π’π = 3π β 2
Matrix Proofs
1
Prove that
0
4
= 2 π π + 1 for all π β β.
β1
3
π
1
=
0
π’1 = 1,
1 β 2π
for all π β β€+ .
π
2
Bro Tip: Recall that β€ is the set of all integers, and β€+ is the set of all
positive integers. Thus β = β€+ .
The Four Steps of Induction
For all these types the process of proof is the same:
! Proof by induction:
Step 1: Basis:
Prove the general statement is true for π = 1.
Step 2: Assumption: Assume the general statement is true for π = π.
Step 3: Inductive: Show that the general statement is then true for π = π + 1.
Step 4: Conclusion: The general statement is then true for all positive integers π.
(Step 1 is commonly known as the βbase caseβ and Step 3 as the βinductive caseβ)
Type 1: Summation Proofs
Prove the general statement is true for π = 1.
Assume the general statement is true for π = π.
Show that the general statement is then true for π = π + 1.
The general statement is then true for all positive integers π.
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Show that
π
π=1(2π
β 1) = π2 for all π β β.
1 Basis Step
When π = 1, πΏπ»π = 1π=1 2π β 1 = 2 1 β 1 = 1
π
π»π = 12 = 1. πΏπ»π = π
π»π so?summation true for π = 1.
2 Assumption
Assume summation true for π = π.
?
So ππ=1(2π β 1) = π 2
3 Inductive
When π = π + 1:
π+1
π
2π β 1 =
π=1
2π β 1
π=1
= π 2 + 2π + 1
= π+1 2
Hence true when π = π + 1.
4 Conclusion
?
+ 2 π+1 β1
Since true for π = 1 and if true for π = π, true for π = π + 1,
?
β΄ true for all π.
More on the βConclusion Stepβ
βAs result is true for π = 1, this implies true for all positive integers and hence
true by induction.β
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:
β’ βtrue for π = πβ
β’ βassume true for π = πβ
β’ βtrue for π = π + πβ
β’ βtrue for all π/positive integersβ
Test Your Understanding
Edexcel FP1 Jan 2010
?
Exercise 6A
All questions except Q2.
Type 2: Divisibility Proofs
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Prove the general statement is true for π = 1.
Assume the general statement is true for π = π.
Show that the general statement is then true for π = π + 1.
The general statement is then true for all positive integers π.
Prove by induction that 32π + 11 is divisible by 4 for all positive integers π.
1 Basis Step
Let π π = 32π + 11, where π β β€+ .
π 1 = 32 + 11 = 20 which is divisible
? by 4.
2 Assumption
Assume that for π = π, π π = 32π + 11 is divisible by 4 for π β β€+ .
?
β¦
Type 2: Divisibility Proofs
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Prove the general statement is true for π = 1.
Assume the general statement is true for π = π.
Show that the general statement is then true for π = π + 1.
The general statement is then true for all positive integers π.
Prove by induction that 32π + 11 is divisible by 4 for all positive integers π.
3 Inductive
Method 1 (the textbookβs method, which I donβt like)
Find the difference between successive terms and show this is
divisible by the number of interest.
π π + 1 = 32 π+1 + 11
Bro Tip: Putting in terms of 32π allows us to
better compare with the assumption case
= 32π β
32 + 11
since the power is now the same.
= 9 32π + 11
?
β΄ π π + 1 β π π = 9 32π + 11 β 32π + 11
= 8 32π = 4 2 32π
Method: You need to explicitly factor out 4 to show divisibility by 4.
32π
Method: Then write π π + 1 back in term of π(π) and the
difference. We know both the terms on the RHS are divisible by 4.
β΄π π+1 =π π +4 2
Therefore π(π) is divisible by 4 when π = π + 1.
4 Conclusion
Since true for π = 1 and if true for π = π, true for π = π + 1,
?
β΄ true for all π.
Type 2: Divisibility Proofs
Prove by induction that 32π + 11 is divisible by 4 for all positive integers π.
3 Inductive
Method 2 (presented as βAlternative Methodβ in mark schemes)
Directly write π(π + 1) in form π π + π(β¦ ) by βsplitting offβ π(π)
or some multiple of it, where π is the divisibility number.
π π + 1 = 32 π+1 + 11
= 32π β
32 + 11
= 9 32π + 11
= 32π + 11 + 8 32π
?
Method: We have separated π πππ into
32π + 8 32π because it allowed us to then
replace 32π + 11 with π(π)
= π π + 4 2 32π
Therefore π(π) is divisible by 4 when π = π + 1.
This method is far superior for two reasons:
a) The idea of βseparating out π(π)β is much more conceptually similar to both summation
proofs and (when we encounter it) matrix proofs, whereas using π π + 1 β π(π) is
specific to divisibility proofs. Thus we neednβt see divisibility proofs as a different
method from the other types.
b) Weβll encounter a question next where we need to get in the form π π + 1 =
π π π + π(β¦ ), i.e. we in fact separate out a multiple of π(π). Thus causes absolute
havoc for Method 1 as our difference π π + 1 β π(π) has to refer to π(π), which is
mathematically inelegant.
Type 2: Divisibility Proofs
Prove by induction that 8π β 3π is divisible by 5.
2 Assumption
Assume that for π = π, π π = 8π β
? 3π is divisible by 5 for π β β€+.
3 Inductive
When π = π + 1:
π π + 1 = 8π+1 β 3π+1
= 8 8π β 3 3π
?
= 3 8π β 3π + 5 8π
Using Method 2
= 3π π + 5(8π )
therefore true for π = π + 1.
Method: We have 3 lots of 8π and
3π we can split off.
Test Your Understanding
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Prove the general statement is true for π = 1.
Assume the general statement is true for π = π.
Show that the general statement is then true for π = π + 1.
The general statement is then true for all positive integers π.
Prove by induction that π3 β 7π + 9 is divisible by 3 for all positive integers π.
1 Basis Step
Let π π = π3 β 7π + 9, where π β β€+ .
π 1 = 3 which is divisible by 3. ?
2 Assumption
Assume that for π = π, π π = π 3 β 7π + 9 is divisible by 3 for π β
?
β€+ .
3 Inductive
Method 1:
π π+1 = π+1 3β7 π+1 +9
=β―
= π 3 + 3π 2 β 4π + 3
β΄ π π + 1 β π π = β― = 3 π2 + π β 2
β΄ π π + 1 = π π + 3 π2 + π β 2
Therefore π(π) is divisible by 3 when π = π + 1.
?
4 Conclusion
Method 2:
π π+1 = π+1 3β7 π+1 +9
=β―
= π 3 + 3π 2 β 4π + 3
= π 3 β 7π + 9 + 3π 2 + 3π β 6
= π π + 3 π2 + π β 2
Therefore π(π) is divisible by 3 when π = π + 1.
Since true for π = 1 and if true for π = π, true for π = π + 1,
?
β΄ true for all π.
Exercise 6B
Page 130.
All questions.
Type 3: Recurrence Relation Proofs
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Prove the general statement is true for π = 1.
Assume the general statement is true for π = π.
Show that the general statement is then true for π = π + 1.
The general statement is then true for all positive integers π.
Notice that weβre trying
to prove a position-toterm formula from a
term-to-term one.
Given that π’π+1 = 3π’π + 4 and π’1 = 1, prove by induction that π’π = 3π β 2
1 Basis Step
When π = 1, π’1 = 31 β 2 = 1 as given.
?
IMPORTANT NOTE: The textbook gets this wrong (bottom of Page 131),
saying that ππ is required. I checked a mark scheme β it isnβt.
Bro Tip: You need to carefully reflect on what statement you are trying to prove, i.e. π’π = 3π β 2, and what
statement is GIVEN (and thus already know to be true), i.e.π’π+1 = 3π’π + 4
2 Assumption
Assume that for π = π, π’π = 3π β 2 is true for π β β€+ .
3 Inductive
Then π’π+1 = 3π’π + 4
?
3π
=3
β2 +4
?
= 3π+1 β 6 + 4
= 3π+1 β 2
Method: Sub in your
assumption expression
into the recurrence.
Bro Tip: For harder questions, write out what youβre trying to prove in advance, i.e.
that we want to get 3π+1 β 2, , to help guide your manipulation.
4 Conclusion
Since true for π = 1 and if true for π = π, true for π = π + 1, β΄ true for all π.
?
Test Your Understanding
Edexcel FP1 Jan 2011
?
Recurrence Relations based on two previous terms
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Prove the general statement is true for π = 1.
Assume the general statement is true for π = π.
Show that the general statement is then true for π = π + 1.
The general statement is then true for all positive integers π.
Given that π’π+2 = 5π’π+1 β 6π’π and π’1 = 13, π’2 = 35, prove by induction that
π’π = 2π+1 + 3π+1
1 Basis Step
When π = 1, π’1 = 22 + 32 = 13 as given.
When π = 2, π’2 = 23 + 33 = 35 as
? given.
We have two base cases so had to check 2 terms!
2 Assumption
3 Inductive
Assume that for π = π and π = π + 1, π’π = 2π+1 + 3π+1 and
π’π+1 = 2π+2 + 3π+2 is true for π β?β€+ . We need two previous instances of
Let π = π + 1
π in our assumption case now.
(Brominder: itβs worth noting in advance weβre aiming for π’π+1 = 2
π+1 +1
Then π’π+2 = 5π’π+1 β 6π’π
= 5 2π+1 + 3π+1 ?
β 6 2π+1 + 3π+1
=β―
= 2π+3 + 3π+3
= 2π+2+1 + 3π+2+1
4 Conclusion
+3
π+1 +1
)
Since true for π = 1 and π = 2 and if true for π = π and π = π + 1,
?
true for π = π + 2,
β΄ true for all π. (Warning: slightly different to before)
Exercise 6C
Page 132.
Q1, 3, 5, 7
Type 4: Matrix Proofs
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Prove the general statement is true for π = 1.
Assume the general statement is true for π = π.
Show that the general statement is then true for π = π + 1.
The general statement is then true for all positive integers π.
1
Prove by induction that
0
1 Basis Step
2 Assumption
3 Inductive
4 Conclusion
β1
2
π
1
=
0
1 β 2π
for all π β β€+ .
π
2
1 β1 1
1 β1
When π = 1, πΏπ»π =
=
0 2
0 2
1
1 β1 ?
π
π»π = 1 1 β12 =
. As πΏπ»π = π
π»π, true for π = 1.
0
2
0
2
π
1 β1 π
1
1
β
2
Assume true π = π, i.e.
=
0 3 ?
0
2π
When π = π + 1,
π
1 β1 π+1
1 β1 π 1 β1
1 β1
1
1
β
2
=
=
0 2
0 2
0 2
0 2
0
2π
π+1
2
? Method: As with summation and divisibility
= β― = 1 1 βπ+1
0
2
proofs, itβs just a case of βseparating offβ the
Therefore true when π = π + 1.
assumption expression.
Since true for π = 1 and if true for π = π, true for π = π + 1,
?
β΄ true for all π.
Test Your Understanding
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Prove the general statement is true for π = 1.
Assume the general statement is true for π = π.
Show that the general statement is then true for π = π + 1.
The general statement is then true for all positive integers π.
β2 9
Prove by induction that
β1 4
1 Basis Step
2 Assumption
3 Inductive
4 Conclusion
π
=
β3π + 1
9π
for all π β β€+ .
βπ
3π + 1
β2 9 1
β2 9
When π = 1, πΏπ»π =
=
β1 4
β1 4
β3 + 1
9
β2 ? 9
π
π»π =
=
. As πΏπ»π = π
π»π, true for π = 1.
β1
3+1
β1 4
β2 9 π
β3π + 1
9π
Assume true π = π, i.e.
=
βπ
3π + 1
β1 4 ?
When π = π + 1,
β2 9 π+1
β2 9 π β2 9
β3π + 1
9π
β2 9
=
=
βπ
3π + 1 β1 4
β1 4
β1 4
β1 4
β3
? π + 1 + 1 9(π + 1)
β3π β 2 9π + 9
=β―=
=
β(π + 1)
3 π+1 +1
βπ β 1 3π + 4
Therefore true when π = π + 1.
Since true for π = 1 and if true for π = π, true for π = π + 1,
?
β΄ true for all π.
Exercise 6D
Page 134.
All questions.