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


1 212

312

⋯ 𝑛12

4𝑛 22

Step 2: Assume true for some 𝑘 ∈ ℤ .

1 212

312

⋯ 𝑘12

4𝑘 22

SUMMATION TYPE


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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2

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


𝐿𝐻𝑆 114

34

𝑅𝐻𝑆2 1

434

Therefore, true for 𝑛 2.

PRODUCT TYPE


11𝑟

𝑛 12𝑛


Let 𝑃 be the product of 𝑘 terms.

𝑃𝑘 1

2𝑘

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PRODUCT TYPE


11𝑟

𝑛 12𝑛


𝑅𝑇𝑃: 𝑃𝑘 2

2 𝑘 1

PRODUCT TYPE


2 𝑘 1

𝑃 𝑃 𝑇

𝑘 12𝑘

11

𝑘 1 𝑏𝑦 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛

𝑘 12𝑘

𝑘 1 1𝑘 1

𝑘 2𝑘 1 12𝑘 𝑘 1

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PRODUCT TYPE


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 .


1 𝑥 1 𝑥 which is divisible by 1 𝑥.


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.


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


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.


4 7 21 36 0


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.


𝐿𝐻𝑆 2! 2

𝑅𝐻𝑆 2 1! 2


2𝑘 ! 2 𝑘!

2𝑘 ! 2 𝑘! 0

INEQUALITY PROOFS

Use mathematical induction to prove that 2𝑛 ! 2 𝑛! for all

positive integers 𝑛.


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.


For 𝑛 3 we have a triangle.

The angle sum is 3 2 180 180 which is true.


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.


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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2sin𝜃Step 2: Assume true for some 𝑘 ∈ ℤ .

cos𝜃 cos3𝜃 ⋯ cos 2𝑘 1 𝜃sin 2𝑘𝜃

2sin𝜃




2sin𝜃Step 3: Prove true for 𝑛 𝑘 1.

RTP: cos𝜃 cos3𝜃 ⋯ cos 2𝑘 1 𝜃sin2 𝑘 1 𝜃

2sin𝜃

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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



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


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.


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



Step 1: Show true for 𝑛 1

𝑓′ 𝑥 lim→

𝑓 𝑥 ℎ 𝑓 𝑥ℎ

lim→

𝑥 ℎ 𝑥ℎ

lim→

1 1

CALCULUS PROOF



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 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 1 2 1 2 for all 𝑛 0.


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.


𝑧 1 1 2

RECURSIVE PROOFS


𝑧.



𝑧 𝑘

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RECURSIVE PROOFS


𝑧.



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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MATHEMATICAL INDUCTION notes