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COMP 250 Lecture 10 mathematical induction Sept. 29, 2017 1 For all 1 , 1 + 2 + 3 + . + 1 + = ( + 1) 2 How to prove such a statement ? By proof, we mean a formal logical argument that convincely

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COMP 250

Lecture 10

mathematical induction

  • Sept. 29, 2017
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For all π‘œ β‰₯ 1, How to prove such a statement ? By β€œproof”, we mean a formal logical argument that convincely shows the statement is true. Note that β€œconvincely” is itself not well defined.

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1 + 2 + 3 + … . + π‘œ βˆ’ 1 + π‘œ = π‘œ(π‘œ + 1) 2

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1 + 2 + … + π‘œ βˆ’ 1 + π‘œ Rewrite by considering π‘œ/2 pairs : 1 + 2 + β‹― + π‘œ 2 + ( π‘œ 2 + 1) … + π‘œ βˆ’ 1 + π‘œ If π‘œ is even, then adding up the π‘œ/2 pairs gives π‘œ/2 *(π‘œ+ 1). What if π‘œ is odd?

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What if π‘œ is odd? Then, π‘œ-1 is even. So, 1 + 2 + … + π‘œ βˆ’ 1 + π‘œ = (

π‘œβˆ’1 2 βˆ— π‘œ ) + π‘œ

= (

π‘œβˆ’1 2 + 1) βˆ— π‘œ

= π‘œ+1

2 βˆ— π‘œ which is the same formula as before.

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Mathematical Induction

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Consider a statement of the form: β€œFor all π‘œ β‰₯ π‘œ0, 𝑄 π‘œ is true” where π‘œ0 is some constant and proposition 𝑄(π‘œ) has value true or false for each π‘œ. Mathematical induction is a general technique for proving such a statement.

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β€œFor all π‘œ β‰₯ π‘œ0, 𝑄 π‘œ is true”

For all π‘œ β‰₯ 1, 1 + 2 + … + π‘œ βˆ’ 1 + π‘œ =

π‘œ(π‘œ+1) 2

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Mathematical induction requires proving two things: Base case: β€œP(π‘œ0) is true.” Induction step: β€œFor any 𝑙 β‰₯ π‘œ0, if P(𝑙) is true, then P(𝑙 + 1) is also true.”

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Base case:

β€œP(π‘œ0) is true.”

Induction step:

β€œFor any k β‰₯ π‘œ0, if P(k) is true, then P(k+1) is also true.” The statement β€œP(k) is true” is called the β€œinduction hypothesis”. 1 2 π‘œ0 k k+1

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For any k >= π‘œ0, if P(k) is true then P(k+1) is true. P(π‘œ0) is true.

Thus we have proved:

For any n >= π‘œ0, P(n) is true. Base case: Induction step: 1 2 n0 k k+1 1 2 n0 k k+1

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Statement: For all π‘œ β‰₯ 1, 1 + 2 + 3 + … . + π‘œ βˆ’ 1 + π‘œ = π‘œ(π‘œ + 1) 2 Proof (base case, n=1):

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Statement: For all π‘œ β‰₯ 1, 1 + 2 + 3 + … . + π‘œ βˆ’ 1 + π‘œ = π‘œ(π‘œ + 1) 2 Proof (base case, n=1): 1 =

1(1+1) 2

(true)

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For any k >= π‘œ0, if P(k) is true then P(k+1) is true. P(1) is true. Base case: Induction step: 1 2 k k+1

induction hypothesis is that P(𝑙) is true:

1 + 2 + … + 𝑙 = 𝑙(𝑙 + 1) 2

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Proof of Induction Step: ( 1 + 2 + 3 + … . + 𝑙) + 𝑙 + 1 =

𝑙(𝑙+1) 2

+ 𝑙 + 1 =

by induction hypothesis

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Proof of Induction Step: ( 1 + 2 + 3 + … . + 𝑙) + 𝑙 + 1 =

𝑙(𝑙+1) 2

+ 𝑙 + 1 = (

𝑙 2 + 1) (𝑙 + 1)

=

1 2 ( 𝑙 + 2 ) (𝑙 + 1)

Thus, 𝑄 𝑙 𝑗𝑑 𝑒𝑠𝑣𝑓 π‘—π‘›π‘žπ‘šπ‘—π‘“π‘‘ 𝑄 𝑙 + 1 𝑗𝑑 𝑒𝑠𝑣𝑓.

by induction hypothesis

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Possible confusion

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𝑄 𝑙 has value true or false (Boolean). So, 𝑄 𝑙 𝑗𝑑 𝑒𝑠𝑣𝑓 means what?

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Examples

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β€œ3 = 2 + 1” is true. β€œ3 = 2 + 2” is false.

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Examples

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β€œ3 = 2 + 1” is true. β€œ3 = 2 + 2” is false. β€œIf 3 = 2 + 2 then 5 > 7” is true.

If this is a mystery to you, then I strongly advise you to take MATH 240 or MATH 318 (logic).

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Mathematical Induction: Example 2

Prove the following statement: For all π‘œ β‰₯ 3, 2π‘œ + 1 < 2π‘œ. 2π‘œ + 1 2π‘œ π‘œ

0 1 2 3 4

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Statement: For all π‘œ β‰₯ 3, 2π‘œ + 1 < 2π‘œ. Note: P( π‘œ ) is false for n= 1, 2. But that has nothing to do with what we need to prove. Proof (base case, π‘œ =3):

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Statement: For all π‘œ β‰₯ 3, 2π‘œ + 1 < 2π‘œ. Note: P( π‘œ ) is false for n= 1, 2. But that has nothing to do with what we need to prove. Proof (base case, π‘œ =3): 2*3 + 1 < 8 (true)

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Proof of Induction Step: We want to show that P(k) implies P(k+1).

Statement: For all π‘œ β‰₯ 3, 2π‘œ + 1 < 2π‘œ.

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Proof of Induction Step: We want to show that P(k) implies P(k+1). 2 𝑙 + 1 + 1 = ? 2𝑙 + 2 + 1 (This is 2π‘œ + 1 where π‘œ = 𝑙 + 1.)

Statement: For all π‘œ β‰₯ 3, 2π‘œ + 1 < 2π‘œ.

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Proof of Induction Step: We want to show that P(k) implies P(k+1). 2 𝑙 + 1 + 1 = 2𝑙 + 2 + 1 < 2𝑙 + 2

by induction hypothesis Statement: For all π‘œ β‰₯ 3, 2π‘œ + 1 < 2π‘œ.

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Proof of Induction Step: We want to show that P(k) implies P(k+1). 2 𝑙 + 1 + 1 = 2𝑙 + 2 + 1 < 2𝑙 + 2 < 2𝑙 + 2𝑙, for k β‰₯ 3 = 2𝑙+1

by induction hypothesis Statement: For all π‘œ β‰₯ 3, 2π‘œ + 1 < 2π‘œ. This inequality is also true for k >=2 but we don’t care because we are trying to prove for k>=3.

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For any k >= π‘œ0, if P(k) then P(k+1). P(π‘œ0) is true.

Thus,

For any n >= π‘œ0, P(n) is true. Base case: Induction step: 1 2 n0 k k+1 1 2 n0 k k+1

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Example 3

Statement: For all π‘œ β‰₯ 5, π‘œ2 < 2π‘œ. 2π‘œ π‘œ2 π‘œ

0 1 2 3 4 5

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Statement: For all π‘œ β‰₯ 5, π‘œ2 < 2π‘œ. Base case (n = 5): Induction step:

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Statement: For all π‘œ β‰₯ 5, π‘œ2 < 2π‘œ. Base case (n = 5): 25 < 32 Induction step: What do we assume ? What do we want to prove ?

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Statement: For all π‘œ β‰₯ 5, π‘œ2 < 2π‘œ. Base case (n = 5): 25 < 32 Induction step: What do we assume ? 𝑙2 < 2𝑙, 𝑙 β‰₯ 5 What do we want to show ? (𝑙 + 1)2 < 2𝑙+1

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Statement: For all π‘œ β‰₯ 5, π‘œ2 < 2π‘œ. Base case (n = 5): 25 < 32 Induction step: (𝑙 + 1) 2 = 𝑙2 + 2𝑙 + 1

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Statement: For all π‘œ β‰₯ 5, π‘œ2 < 2π‘œ. Base case (n = 5): 25 < 32 Induction step: (𝑙 + 1) 2 = 𝑙2 + 2𝑙 + 1 < 2𝑙 + 2𝑙 + 1

by induction hypothesis

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Statement: For all π‘œ β‰₯ 5, π‘œ2 < 2π‘œ. Base case (n = 5): 25 < 32 Induction step: (𝑙 + 1) 2 = 𝑙2 + 2𝑙 + 1 < 2𝑙 + 2𝑙 + 1 < 2𝑙 + 2𝑙 = 2𝑙+1

by induction hypothesis by Example 2

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Example 4 : Fibonacci Sequence

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, …. 𝐺 0 = 0 𝐺 1 = 1 𝐺 π‘œ + 2 = 𝐺 π‘œ + 1 + 𝐺(π‘œ) , for π‘œ β‰₯ 0. Statement: For all π‘œ β‰₯ 0, 𝐺 π‘œ < 2π‘œ

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𝐺 0 = 0 𝐺 1 = 1 𝐺 π‘œ + 2 = 𝐺 π‘œ + 1 + 𝐺(π‘œ) , for π‘œ β‰₯ 0. Base case(s): π‘œ = 0: 0 < 20 is true. π‘œ = 1: 1 < 21 is true.

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𝐺 0 = 0 𝐺 1 = 1 𝐺 π‘œ + 2 = 𝐺 π‘œ + 1 + 𝐺(π‘œ) , for π‘œ β‰₯ 0. Induction step: 𝐺 𝑙 + 1 = 𝐺 𝑙 + 𝐺 𝑙 βˆ’ 1 < 2𝑙 + 2π‘™βˆ’1

by induction hypothesis

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𝐺 0 = 0 𝐺 1 = 1 𝐺 π‘œ + 2 = 𝐺 π‘œ + 1 + 𝐺(π‘œ) , for π‘œ β‰₯ 0. Induction step: 𝐺 𝑙 + 1 = 𝐺 𝑙 + 𝐺 𝑙 βˆ’ 1 < 2𝑙 + 2π‘™βˆ’1 < 2𝑙 + 2𝑙 = 2𝑙+1

by induction hypothesis