MATHEMATICAL INDUCTION notes
Transcript of MATHEMATICAL INDUCTION notes
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FURTHER MATHEMATICAL INDUCTION
TOPIC OUTLINE
• Proof by mathematical induction
• Summation and Product
• Divisibility
• Inequalities
• Geometry, Trigonometry, Combinatorics, Calculus, …
• Recursive
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MATHEMATICAL INDUCTION
The process of mathematical induction is as follows:
• Show the statement is true for 𝑛 1 (an initial value)
• Assume it is true some 𝑘 ℤ
• Prove that it is true for 𝑛 𝑘 1
• Since it is true for 𝑛 1 and proven true for 𝑛 1 1 2 and so on, therefore it is true for all 𝑛
SUMMATION TYPE
Prove by induction that, for integers 𝑛 1,
1 212
312
⋯ 𝑛12
4𝑛 22
Step 1: Show true for 𝑛 1.
𝐿𝐻𝑆 1
𝑅𝐻𝑆 43
21
Therefore, true for 𝑛 1.
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SUMMATION TYPE
Prove by induction that, for integers 𝑛 1,
1 212
312
⋯ 𝑛12
4𝑛 22
Step 2: Assume true for some 𝑘 ∈ ℤ .
1 212
312
⋯ 𝑘12
4𝑘 22
SUMMATION TYPE
Prove by induction that, for integers 𝑛 1,
1 212
312
⋯ 𝑛12
4𝑛 22
Step 3: Prove true for 𝑛 𝑘 1.
𝑅𝑇𝑃: 1 212
312
⋯ 𝑘 112
4𝑘 3
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SUMMATION TYPE
𝑅𝑇𝑃: 1 212
312
⋯ 𝑘 112
4𝑘 3
2
𝑆 𝑆 𝑇
4𝑘 22
𝑘 112
𝑏𝑦 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛
42𝑘 4
2𝑘
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SUMMATIONTYPE
𝑅𝑇𝑃: 1 212
312
⋯ 𝑘 112
4𝑘 3
2
𝑆 42𝑘 4
2𝑘
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42𝑘 4 𝑘 1
2
4𝑘 3
2Therefore, true for 𝑛 𝑘 1. Therefore, true for all 𝑛.
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PRODUCT TYPE
By the process of induction, prove that for all integers 𝑛 2:
11𝑟
𝑛 12𝑛
Step 1: Show true for 𝑛 2.
𝐿𝐻𝑆 114
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𝑅𝐻𝑆2 1
434
Therefore, true for 𝑛 2.
PRODUCT TYPE
By the process of induction, prove that for all integers 𝑛 2:
11𝑟
𝑛 12𝑛
Step 2: Assume true for some 𝑘 ∈ ℤ .
Let 𝑃 be the product of 𝑘 terms.
𝑃𝑘 1
2𝑘
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PRODUCT TYPE
By the process of induction, prove that for all integers 𝑛 2:
11𝑟
𝑛 12𝑛
Step 3: Prove true for 𝑛 𝑘 1.
𝑅𝑇𝑃: 𝑃𝑘 2
2 𝑘 1
PRODUCT TYPE
𝑅𝑇𝑃: 𝑃𝑘 2
2 𝑘 1
𝑃 𝑃 𝑇
𝑘 12𝑘
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𝑘 1 𝑏𝑦 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛
𝑘 12𝑘
𝑘 1 1𝑘 1
𝑘 2𝑘 1 12𝑘 𝑘 1
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PRODUCT TYPE
𝑅𝑇𝑃: 𝑃𝑘 2
2 𝑘 1
𝑃𝑘 2𝑘 1 1
2𝑘 𝑘 1
𝑘 𝑘 22𝑘 𝑘 1
𝑘 22 𝑘 1
Therefore, true for 𝑛 𝑘 1. Therefore, true for all 𝑛.
DIVISIBILITY PROOFS
Prove that 1 𝑥 is divisible by 1 𝑥 for all 𝑛 ∈ ℤ .
Step 1: Show true for 𝑛 1.
1 𝑥 1 𝑥 which is divisible by 1 𝑥.
Step 2: Assume true for some 𝑘 ∈ ℤ .
1 𝑥 1 𝑥 𝑀 where 𝑀 is an integer.
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DIVISIBILITY PROOFS
Prove that 1 𝑥 is divisible by 1 𝑥 for all 𝑛.
Step 3: Prove true for 𝑛 𝑘 1.
1 𝑥 1 𝑥 . 𝑥
1 𝑥 1 1 𝑥 𝑀 𝑏𝑦 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛
1 𝑥 𝑥 1 𝑥 𝑀
1 𝑥 1 𝑀𝑥 which is divisible by 1 𝑥
Therefore, true for 𝑛 𝑘 1. Therefore, true for all 𝑛.
INEQUALITY PROOFS
With induction proofs involving inequalities, we often make use of the
following:
If 𝑎 𝑐 then 𝑎 𝑏 𝑐 where 𝑏 ∈ ℝ
If 𝑎 𝑏 𝑐 then 𝑎 𝑏 where 𝑐 ∈ ℝ
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INEQUALITY PROOFS
Use the principle of mathematical induction to show that
4 7 7𝑛 0 for all integers 𝑛 2.
Step 1: Show true for 𝑛 3.
4 7 21 36 0
Step 2: Assume true for some 𝑘 ∈ ℤ .
4 7 7𝑘 0
INEQUALITY PROOFS
Use the principle of mathematical induction to show that
4 7 7𝑛 0 for all integers 𝑛 2.Step 3: Prove true for 𝑛 𝑘 1.
4 7 7 𝑘 1 4 . 4 7 7𝑘 7
4 4 7 7𝑘 14 21𝑘
0 since 4 7 7𝑘 0 (by assumption)
and 14 2𝑘 0Therefore, true for 𝑛 𝑘 1. Therefore, true for all 𝑛.
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INEQUALITY PROOFS
Use mathematical induction to prove that 2𝑛 ! 2 𝑛! for all
positive integers 𝑛.
Step 1: Show true for 𝑛 1.
𝐿𝐻𝑆 2! 2
𝑅𝐻𝑆 2 1! 2
Step 2: Assume true for some 𝑘 ∈ ℤ .
2𝑘 ! 2 𝑘!
2𝑘 ! 2 𝑘! 0
INEQUALITY PROOFS
Use mathematical induction to prove that 2𝑛 ! 2 𝑛! for all
positive integers 𝑛.
Step 3: Prove true for 𝑛 𝑘 1.
𝑅𝑇𝑃: 2 𝑘 1 ! 2 𝑘 1 !
2 𝑘 1 ! 2 𝑘 1 ! 0
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INEQUALITY PROOFS
2𝑘 2 ! 2 𝑘 1 !
2𝑘 2 2𝑘 1 2𝑘 ! 2. 2 𝑘 1 𝑘!
4𝑘 6𝑘 2 2𝑘 ! 2 𝑘 2𝑘 1 2 𝑘!
4𝑘 6𝑘 2 2𝑘 ! 2𝑘 4𝑘 2 2 𝑘!
4𝑘 6𝑘 2 2𝑘 ! 2 𝑘! 2𝑘 2𝑘 2 𝑘!
0 by assumption and since 4𝑘 6𝑘 2 0 and
2𝑘 2𝑘 2 𝑘! 0
GEOMETRICAL PROOFS
Prove that the angle sum of an 𝑛‐sided polygon is 𝑛 2 180 for all integers 𝑛 3.
Step 1: Show true for 𝑛 3.
For 𝑛 3 we have a triangle.
The angle sum is 3 2 180 180 which is true.
Step 2: Assume true for some 𝑘 ∈ ℤ .
A 𝑘‐sided polygon has an angle sum of 𝑘 2 180 .
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GEOMETRICAL PROOFS
Prove that the angle sum of an 𝑛‐sided polygon is 𝑛 2 180 for all integers 𝑛 3.
Step 3: Prove true for 𝑛 𝑘 1.
Taking the 𝑘 1 ‐sided polygon, split it into two: a 𝑘‐sided polygon and a triangle.
The 𝑘‐sided polygon has an angle sum of 𝑘 2 180 and the triangle has an angle sum of
180 . Total is 𝑘 1 180 .
TRIGONOMETRIC PROOFS
It is given that 2cos𝐴sin𝐵 sin 𝐴 𝐵 sin 𝐴 𝐵 . Prove by induction that, for integers 𝑛 1,
cos𝜃 cos3𝜃 ⋯ cos 2𝑛 1 𝜃sin 2𝑛𝜃
2sin𝜃Step 1: Show true for 𝑛 1.
𝐿𝐻𝑆 cos𝜃 𝑅𝐻𝑆sin2𝜃2sin𝜃
2sin𝜃cos𝜃2sin𝜃
cos𝜃
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TRIGONOMETRIC PROOFS
It is given that 2cos𝐴sin𝐵 sin 𝐴 𝐵 sin 𝐴 𝐵 . Prove by induction that, for integers 𝑛 1,
cos𝜃 cos3𝜃 ⋯ cos 2𝑛 1 𝜃sin 2𝑛𝜃
2sin𝜃Step 2: Assume true for some 𝑘 ∈ ℤ .
cos𝜃 cos3𝜃 ⋯ cos 2𝑘 1 𝜃sin 2𝑘𝜃
2sin𝜃
TRIGONOMETRIC PROOFS
It is given that 2cos𝐴sin𝐵 sin 𝐴 𝐵 sin 𝐴 𝐵 . Prove by induction that, for integers 𝑛 1,
cos𝜃 cos3𝜃 ⋯ cos 2𝑛 1 𝜃sin 2𝑛𝜃
2sin𝜃Step 3: Prove true for 𝑛 𝑘 1.
RTP: cos𝜃 cos3𝜃 ⋯ cos 2𝑘 1 𝜃sin2 𝑘 1 𝜃
2sin𝜃
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TRIGONOMETRIC PROOFS
RTP: cos𝜃 cos3𝜃 ⋯ cos 2𝑘 1 𝜃sin2 𝑘 1 𝜃
2sin𝜃𝑆 𝑆 𝑇
sin 2𝑘𝜃2sin𝜃
cos 2𝑘 1 𝜃 𝑏𝑦 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛
sin 2𝑘𝜃 2sinθcos 2𝑘 1 𝜃2sin𝜃
sin 2𝑘𝜃 sin 2𝑘 1 𝜃 𝜃 sin 2𝑘 1 𝜃 𝜃2sin𝜃
TRIGONOMETRIC PROOFS
RTP: cos𝜃 cos3𝜃 ⋯ cos 2𝑘 1 𝜃sin2 𝑘 1 𝜃
2sin𝜃
𝑆sin 2𝑘𝜃 sin 2𝑘 1 𝜃 𝜃 sin 2𝑘 1 𝜃 𝜃
2sin𝜃
sin 2𝑘𝜃 sin 2 𝑘 1 𝜃 sin 2𝑘𝜃2sin𝜃
sin2 𝑘 1 𝜃2sin𝜃
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COMBINATORICS
A 2 by 𝑛 grid is made up of two rows of 𝑛 square tiles, as shown.
The tiles of the 2 by 𝑛 grid are to be painted so that tiles sharing an
edge are painted using different colours. There are 𝑥 different colours
available, where 𝑥 2.
It is NOT necessary to use all the colours.
COMBINATORICS
Consider the case of the 2 by 2 grid with tiles labelled A, B, C and D, as
shown:
There are 𝑥 𝑥 1 ways to choose colours for the first column
containing tiles A and B. Do NOT prove this.
i) Assume the colours for tiles A and B have been chosen. There are two
cases to consider when choosing colours for the second column. Either
C is the same colour as tile B or tile C is a different colour from B.
By considering these two cases, show that the number of ways of
choosing colours for the second column is 𝑥 3𝑥 3.
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COMBINATORICS
There are 𝑥 𝑥 1 ways to choose colours for the first column
containing tiles A and B.
Either C is the same colour as tile B:
There is one way to choose C and 𝑥 1 ways to choose the colour for D
Or C is a different colour to B:
There are 𝑥 2 ways to choose C and then 𝑥 2 ways to choose D
Altogether there are 𝑥 1 𝑥 2 𝑥 3𝑥 3
COMBINATORICS
ii) Prove by mathematical induction that the number of ways in which the
2 by 𝑛 grid can be painted is 𝑥 𝑥 1 𝑥 3𝑥 3 , for 𝑛 1
Step 1: Show true for 𝑛 1
𝑥 𝑥 1 𝑥 3 3 𝑥 𝑥 1 true for 𝑛 1
Step 2: Assume true for some 𝑘 ∈ ℤ
Let 𝑁 be the number of ways to paint a 2 by 𝑘 grid
𝑁 𝑥 𝑥 1 𝑥 3𝑥 3
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COMBINATORICS
ii) Prove by mathematical induction that the number of ways in which the
2 by 𝑛 grid can be painted is 𝑥 𝑥 1 𝑥 3𝑥 3 , for 𝑛 1
Step 3: Prove true for 𝑛 𝑘 1
𝑅𝑇𝑃:𝑁 𝑥 𝑥 1 𝑥 3𝑥 3
𝑁 𝑁 𝑥 3𝑥 3
𝑥 𝑥 1 𝑥 3𝑥 3 𝑥 3𝑥 3
𝑥 𝑥 1 𝑥 3𝑥 3
Therefore, true for 𝑛 𝑘 1. Therefore, true for all 𝑛.
COMBINATORICS
iii) In how many ways can a 2 by 5 grid be painted if 3 colours are
available and each colour must now be used at least once?
𝑁 𝑥 𝑥 1 𝑥 3𝑥 3
𝑁 3 2 9 9 3
486
but this includes the cases where only two colours are used.
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COMBINATORICS
iii) In how many ways can a 2 by 5 grid be painted if 3 colours are
available and each colour must now be used at least once?
We need to subtract these cases.
There are six arrangements where only two colours are used:
A B C A B C
B A A C C B
Therefore, total arrangements = 486 – 6 = 480
A B A B A
B A B A B
CALCULUS PROOF
Use mathematical induction to prove that the derivative of 𝑦 𝑥 is
𝑛𝑥 for all 𝑛 1.
Step 1: Show true for 𝑛 1.
That is, for 𝑦 𝑥, 𝑦′ 1 𝑥 1.
We have to prove this by first principles.
𝑓′ 𝑥 lim→
𝑓 𝑥 ℎ 𝑓 𝑥ℎ
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CALCULUS PROOF
Use mathematical induction to prove that the derivative of 𝑦 𝑥 is
𝑛𝑥 for all 𝑛 1.
Step 1: Show true for 𝑛 1
𝑓′ 𝑥 lim→
𝑓 𝑥 ℎ 𝑓 𝑥ℎ
lim→
𝑥 ℎ 𝑥ℎ
lim→
1 1
CALCULUS PROOF
Use mathematical induction to prove that the derivative of 𝑦 𝑥 is
𝑛𝑥 for all 𝑛 1.
Step 2: Assume true for some 𝑘 ∈ ℤ
For 𝑦 𝑥 ,𝑑𝑦𝑑𝑥
𝑘𝑥
Step 3: Prove true for 𝑛 𝑘 1
For 𝑦 𝑥 ,𝑑𝑦𝑑𝑥
𝑘 1 𝑥
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CALCULUS PROOF
𝑅𝑇𝑃: For 𝑦 𝑥 , 𝑦′ 𝑘 1 𝑥
How is 𝑦 𝑥 related to 𝑦 𝑥 ?
𝑥 𝑥 𝑥
𝑑𝑑𝑥
𝑥𝑑𝑑𝑥
𝑥 𝑥
𝑘𝑥 𝑥 𝑥 1 (using assumption, initial case)
𝑘𝑥 𝑥
𝑘 1 𝑥
REFLECTION QUESTIONS
Why did we have to use first principles
to show true for 𝑛 1?
Why were we allowed to use the
product rule?
Do we have to use induction to prove
this or is there an easier proof?
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RECURSION
A recursive formula is one where a
term is defined by referring to one
or more previous terms.
eg. A sequence 𝑎 is defined by
𝑎 2𝑎 𝑎 for 𝑛 2 with
𝑎 𝑎 2.
RECURSIVE PROOFS
A sequence 𝑎 is defined by 𝑎 2𝑎 𝑎 for 𝑛 2 with
𝑎 𝑎 2. Use mathematical induction to prove that:
𝑎 1 2 1 2 for all 𝑛 0.
Step 1: Show true for 𝑛 0 and 1.
𝑎 1 2 1 2 1 1 2
𝑎 1 2 1 2 1 1 2
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RECURSIVE PROOFS
A sequence 𝑎 is defined by 𝑎 2𝑎 𝑎 for 𝑛 2 with
𝑎 𝑎 2. Use mathematical induction to prove that:
𝑎 1 2 1 2 for all 𝑛 0.
Step 2: Assume true for some 𝑘 and 𝑘 1 ∈ ℤ
𝑎 1 2 1 2
𝑎 1 2 1 2
RECURSIVE PROOFS
A sequence 𝑎 is defined by 𝑎 2𝑎 𝑎 for 𝑛 2 with
𝑎 𝑎 2. Use mathematical induction to prove that:
𝑎 1 2 1 2 for all 𝑛 0.
Step 3: Prove true for 𝑛 𝑘 2.
𝑅𝑇𝑃: 𝑎 1 2 1 2
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RECURSIVE PROOFS
𝑅𝑇𝑃: 𝑎 1 2 1 2
We were given that 𝑎 2𝑎 𝑎 .
𝑎 2𝑎 𝑎
2 1 2 1 2 1 2 1 2
1 2 2 1 2 1 1 2 2 1 2 1
1 2 3 2 2 1 2 3 2 2
RECURSIVE PROOFS
1 2 3 2 2 1 2 3 2 2
Now: 1 2 1 2 2 2 3 2 2
Similarly 1 2 1 2 2 2 3 2 2
Therefore 1 2 3 2 2 1 2 3 2 2
1 2 1 2 1 2 1 2
1 2 1 2
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RECURSIVE PROOFS
Let 𝑧 1 𝑖 and for 𝑛 2 let 𝑧 𝑧 1𝑖
𝑧.
Use mathematical induction to prove that 𝑧 𝑛 for all integers 𝑛 2.
Step 1: Show true for 𝑛 2.
𝑧 1 1 2
RECURSIVE PROOFS
Let 𝑧 1 𝑖 and for 𝑛 2 let 𝑧 𝑧 1𝑖
𝑧.
Use mathematical induction to prove that 𝑧 𝑛 for all integers 𝑛 2.
Step 2: Assume true for some 𝑘 ∈ ℤ .
𝑧 𝑘
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RECURSIVE PROOFS
Let 𝑧 1 𝑖 and for 𝑛 2 let 𝑧 𝑧 1𝑖
𝑧.
Use mathematical induction to prove that 𝑧 𝑛 for all integers 𝑛 2.
Step 3: Prove true for 𝑛 𝑘 1.
𝑅𝑇𝑃: 𝑧 𝑘 1
RECURSIVE PROOFS
We are given 𝑧 𝑧 1𝑖
𝑧.
𝑧 𝑧 1𝑖𝑧
𝑧 1𝑖
𝑘 by assumption
𝑧 𝑧 1𝑖
𝑘
𝑧 1𝑖
𝑘
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RECURSIVE PROOFS
𝑧 𝑧 1𝑖
𝑘
𝑘 11𝑘
𝑘1 𝑘𝑘
𝑘 1
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