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Cappello Mathematical Induction Goals Explain amp illustrate construction of proofs of a variety of theorems using mathematical induction Copyright Peter Cappello Motivation Mathematics uses 2 kinds of arguments ID: 415416

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Slide1

Mathematical Induction

Goals

Explain & illustrate construction of

proofs of a variety of theorems using mathematical induction.Slide2

Motivation

Mathematics uses 2 kinds of arguments:

deductive

inductive

Proposition:

): 1 + 2 + … +

(

+ 1 )/2.

Observe that P(1), P(2), P(3), & P(4). Conjecture:





Mathematical induction

is a

finite

proof

pattern

for proving propositions of the form





).Slide3

The Principle of Mathematical Induction

Let P(

) be a predicate function:





) is a

proposition

To prove





), it

suffices

to prove:

P( 1 ) is true.





( P(

)



+ 1 ) ).

This is not magic.

It is a recipe for constructing a finite proof for

arbitrary



N.Slide4

Proving P( 3 )

Given P( 1 )

 

 1 ( P(

)  P(

+ 1) ).

Proof:

1. P( 1 ).

[premise 1]

2. P( 1 )  P( 2 ).

[U.S. of premise 2 for

= 1]

3. P( 2 ).

[step 1, 2, & modus ponens]

4. P( 2 )  P( 3 ).

[U.S. of premise 2 for

= 2]

5. P( 3 ).

[step 2, 3, & modus ponens]

Construct a finite proof for P(

1,999,765

).Slide5

Mathematical Induction as the Domino Principle

the

domino falls over

and

the

domino falls over

implies

that the

(

+ 1 )

domino falls over

then

domino

falls over for all



.Slide6

Mathematical Induction as the Domino Principle

+1Slide7

The 3-Step Method

The implication in step 2 typically is proved

directly

The proof pattern thus has

steps:

Prove P( 1 ).

[called the

basis

]

Assume P(

)

[called the

induction hypothesis]

Prove P( n + 1 ) [called the

inductive step]The last 2 steps are for arbitrary n

 Z+.Using P( n ) to prove P( n + 1 ) implies a recursive formulation of P( n ).Slide8

Induction as a Creative Process

Mathematical induction is similar to, but not identical to,

scientific induction

In both cases, a

“

theory

”

is created.

Look at specific cases; perceive a

pattern

Hypothesizing a pattern, a theory, is a

creative process

(only people who are bad at

mathematics

say otherwise).

With

mathematical

induction, a “theory” can be

proved.Slide9

Scientific theories

cannot be proved

They can be

disproved

A scientific theory can be based on a mathematical model.

Propositions can be

proved within the model

Like axioms, the

relationship

between:

the mathematical model

physical reality

cannot

proven

correct.Slide10

Example

1 = 1

3 = 1 + 2

6 = 1 + 2 + 3

10 = 1 + 2 + 3 + 4

What is a general formula, if any, for 1 + 2 + … +

Let F(

): 1 + 2 + . . . +

recursive

formulation: F(

) = F(

- 1 ) + n.Slide11

Put these blocks, which represent numbers, together to form sums:

1 + 2 =

1 + 2 + 3 =

A Geometric InterpretationSlide12

Area is

/2 +

/2 =

(

+ 1)/2Slide13

1 + 2 + … +

(

+ 1)/2

A Mathematical Induction Proof

F( 1 ) = 1( 1 + 1 )/2 = 1.

Assume F(

) =

(

+ 1 )/2

Show F(

+ 1 ) = (

n + 1 )( n

+ 2 )/2.F( n + 1 ) = 1 + 2 + . . . +

n + ( n + 1 ) [Definition] = F( n ) + n + 1 [Recursive formulation] = n( n + 1 ) / 2 + n + 1 [Induction hyp.] = n( n + 1 ) / 2 +

( n + 1 ) (2/2) = ( n + 1 ) ( n + 2 ) / 2.Slide14

finding

a recursive formulation,

we focused on the:

similarities

differences

for successive values of

Sometimes

, it is useful to:

Note the difference between F(

) & F(

– 1

Find a pattern in this sequence of differences.Slide15

Example: 13

+ 2

+ . . . +

= ?

Let F(

) = 1

+ 2

+ . . . +

What is a formula for F(n)?

1 = 13

9 = 13 + 23

36 = 13 + 23 + 3

3 100 = 13 + 23 + 33 + 43Do you see a pattern?Slide16

Prove that



) = [

(

+ 1 )/2 ]

1. F( 1 ) = 1

= 1 = [ 1( 2 ) / 2 ]

2. Assume F(

) = [ n

( n + 1 ) / 2 ]2 .  I.H.

3. Prove F( n + 1 ) = [ ( n

+ 1 )( n + 2 ) / 2 ]2.F ( n + 1 ) = 13 + 23 + . . . + n3 + ( n +1 )3  Defn. of F( n + 1 ) = F(

n ) + ( n + 1 )3  Recursive formulation = [ n( n + 1 )/ 2 ]2 + ( n + 1 )3  Use I. H. = ( n + 1 )2[ ( n / 2 )

2 + ( n + 1 ) ] = ( n + 1 )2[ n2 / 4 + ( 4 / 4 )( n + 1 ) ] = ( n +1 )2[ ( n2 + 4n + 4 ) / 4 ] = [ ( n + 1)

(n + 2) / 2 ]2.Slide17

Translating the starting point

we:

know P(

) is false for 1



think

) is true for

> 9.

Then define Q( n

) = P( n + 9 ).Use mathematical induction to show that n

N Q( n

).We thus can start the induction at any natural number.Slide18

Example: Stamps

Suppose the US Post Office prints only

5 & 9

cent stamps.

Prove



> 34

you can make postage for

cents, using only

5 & 9

cent stamps.

Let

)

denote the statement: You can make postage for n

cents using only 5-cent & 9-cent stamps.1. Basis: For n = 35: Use

7 5-cent stamps.2. I.H.: Assume S( n ) for

n > 34.Slide19

3. Prove S(

+ 1

Case: For S(

), # of 9-cent stamps used

= 0

Only 5-cent stamps are used for S(

# of 5-cent stamps

≥ 7

Replace

5-cent stamps with

9-cent stamps.

Case: For S( n ), # of 9-cent stamp used > 0

: Replace 1 9-cent stamp with 2 5-cent stamps.Slide20

Generalizing the Basis

To prove





), if suffices to show:

P( 1 )



P( 2 ).





(

[ P( n )

 P( n + 1 ) ]  P(

n + 2 ) )

If:We can push over the first 2 dominos;Pushing over any 2 adjacent dominos implies pushing over the next domino. then we can push over all the dominos.Slide21

The Fibonacci Formula

Define

the n

Fibonacci number, F( n )

, as:

F( 0 ) = 0, F( 1 ) = 1,

) = F(

– 1 ) + F(

– 2 ).

Prove

) = 5

-1/2 ( [ ( 1 + 51/2 ) / 2]n

- [ ( 1 - 51/2 ) / 2 ]n ).

Basis: F( 0 ) = 5-1/2 ( [ ( 1 + 51/2 ) /2 ]0 - [ (1 - 51/2) / 2 ]0 ) = 0. F( 1 ) = 5-1/2 ( [ ( 1 + 51/2 ) / 2 ]1 - [ ( 1 - 5

1/2 ) / 2 ]1 ) = ( 5-1/2 / 2 ) ( 1 + 51/2 - 1 + 51/2 ) = 1.Slide22

The Fibonacci Formula

) = 5

-1/2

( [ ( 1 + 5

1/2

) / 2]

- [ ( 1 - 5

1/2

) / 2 ]

)

Let

= ( 1 + 51/2 ) / 2

b = ( 1 - 51/2 ) / 2

. Note: a + 1 = a2 & b + 1 = b2.Induction hypotheses:F( n ) = 5-1/2 ( an –

bn ) F( n +1 ) = 5-1/2 ( an + 1 – bn + 1 ).

Induction step: Show F( n + 2 ) = 5-1/2 ( an + 2 - bn + 2 ).Slide23

Cappello

) = 5

-1/2

( [ ( 1 + 5

1/2

) / 2]

- [ ( 1 - 5

1/2

) / 2 ]

)

Proof of Induction Step

+ 2 ) = F (

+ 1 ) + F ( n ) [

Definition] = 5-1/2 ( a

n + 1 – bn + 1 ) + 5-1/2 ( an – bn ) [I.H.] = 5-1/2 ( a

n + 1 + an – bn + 1 – bn ) = 5-1/2 ( an ( a + 1 ) – bn ( b + 1 ) ). = 5

-1/2 ( an + 2 – bn + 2 )Slide24

Generalizing this ...

( 1 )

 P( 2 ) 

. . . P(

)



{

[ P(

+ 1 )

 P(

+ 2 ) 

. . . P(

+ k ) ]  P( n

+ k + 1 ) }then

n  N P( n ). Slide25

EndSlide26

Strong Mathematical Induction

P( 1 )

 P( 2 ) 

. . .



P( k )

and

for n

 k

[ P( 1 )

 P( 2 ) 

. . .



P( n ) ]



P( n + 1 )

then,

n  N

P(n).Slide27

Example: Fundamental Theorem of Arithmetic

Prove that all natural numbers

 2 can be represented as a product of primes.

Basis

: 2: 2 is a prime.

Assume

that 1, 2, . . . , n can be represented as a product of primes.Slide28

Show

that n + 1can be represented as a product of primes.

Case

n + 1 is a prime

: It can be represented as a product of 1 prime, itself.

Case

n + 1 is composite

: n = ab, for some a,b < n.

Therefore, a = p

. . . p

& b = q

. . . q

, where the pis & qis are primes.Represent n = p

1p2 . . . pkq1

q2 . . . ql.

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