Copyright © Peter Cappello Mathematical Induction Goals Explain & illustrate construction of proofs of a variety of theorems using mathematical induction

Copyright © Peter Cappello

Mathematical Induction

Goals

Explain & illustrate construction of proofs of a variety of

theorems using mathematical induction.


Motivation

• Mathematics uses 2 kinds of arguments:

• deductive

• inductive

• Proposition: P( n ): 1 + 2 + … + n = n( n + 1 )/2.

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

• Mathematical induction is a finite proof pattern for proving

propositions of the form nN P( n ).


The Principle of Mathematical Induction

• Let P( n ) be a predicate function: nN P( n ) is a

proposition.

• To prove nN P( n ), it suffices to prove:

1. P( 1 ) is true.

2. nN ( P( n ) P( n + 1 ) ).

• This is not magic.

• It is a recipe for constructing a finite proof for arbitrary nN.


Proving P( 3 )• Given P( 1 ) n 1 ( P( n ) P( n + 1) ).

• Proof:

1. P( 1 ). [premise 1]

2. P( 1 ) P( 2 ). [U.S. of premise 2 for n = 1]

3. P( 2 ). [step 1, 2, & modus ponens]

4. P( 2 ) P( 3 ). [U.S. of premise 2 for n = 2]

5. P( 3 ). [step 2, 3, & modus ponens]

• Construct a finite proof for P( 1,999,765 ).


Mathematical Induction as the Domino Principle

If

the 1st domino falls over

and

the nth domino falls over implies that the ( n + 1 )st domino

falls over

then

domino n falls over for all n N.


Mathematical Induction as the Domino Principle

1 2 3 n n +1


The 3-Step Method

• The implication in step 2 typically is proved directly.

• The proof pattern thus has 3 steps:

1. Prove P( 1 ). [called the basis]

2. Assume P( n ) [called the induction hypothesis]

3. Prove P( n + 1 ) [called the inductive step]

• The last 2 steps are for arbitrary n N.

• Using P( n ) to prove P( n + 1 ) implies a recursive formulation of P( n ).


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.


• 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 be proven correct.


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 + … + n?

• Let F( n ): 1 + 2 + . . . + n.

• A recursive formulation: F( n ) = F( n - 1 ) + n.


• 1:

• 2:

• 3:

• Put these blocks, which represent numbers, together to

form sums:

• 1 + 2 =

• 1 + 2 + 3 =

A Geometric Interpretation


n

n

Area is n2/2 + n/2 = n(n + 1)/2


1 + 2 + … + n = n(n + 1)/2 A Mathematical Induction Proof

• F( 1 ) = 1( 1 + 1 )/2 = 1.

• Assume F( n ) = n( n + 1 )/2

• Show F( n + 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.


• In finding a recursive formulation, we focused on the:

• similarities

• differences

for successive values of n.

• Sometimes, it is useful to:

• Note the difference between F( n ) & F( n – 1 ).

• Find a pattern in this sequence of differences.


Example: 13 + 23 + . . . + n3 = ?

• Let F( n ) = 13 + 23 + . . . + n3.

• What is a formula for F(n)?

• 1 = 13

• 9 = 13 + 23

• 36 = 13 + 23 + 33

• 100 = 13 + 23 + 33 + 43

• Do you see a pattern?


Prove that n F( n ) = [ n( n + 1 )/2 ]2

1. F( 1 ) = 13 = 1 = [ 1( 2 ) / 2 ]2.

2. Assume F( n ) = [ 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.


Translating the starting point

• If we:

• know P( n ) is false for 1 n 9

• think P( n ) is true for n > 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.


Example: Stamps

• Suppose the US Post Office prints only 5 & 9 cent stamps.

• Prove n > 34, you can make postage for n cents, using

only 5 & 9 cent stamps.

Let S( n ) 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 ).


3. Prove S( n + 1 ).

Case: For S( n ), # of 9-cent stamps used = 0:

Only 5-cent stamps are used for S( n ).

# of 5-cent stamps ≥ 7.

Replace 7 5-cent stamps with 4 9-cent

stamps.

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

Replace 1 9-cent stamp with 2 5-cent

stamps.


Generalizing the Basis

• To prove nN P( n ), if suffices to show:• P( 1 ) P( 2 ).• nN( [ 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.


The Fibonacci Formula

• Define the nth Fibonacci number, F( n ), as:• F( 0 ) = 0, F( 1 ) = 1, • F( n ) = F( n – 1 ) + F( n – 2 ).

• Prove F( n ) = 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 - 51/2 ) / 2 ]1 ) = ( 5-1/2 / 2 ) ( 1 + 51/2 - 1 + 51/2 ) = 1.


The Fibonacci Formula F( n ) = 5-1/2 ( [ ( 1 + 51/2 ) / 2]n - [ ( 1 - 51/2 ) / 2 ]n )

• Let a = ( 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 ).


F( n ) = 5-1/2 ( [ ( 1 + 51/2 ) / 2]n - [ ( 1 - 51/2 ) / 2 ]n ) Proof of Induction Step

F( n + 2 ) = F ( n + 1 ) + F ( n ) [Definition]

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

= 5-1/2 ( an + 1 + an – bn + 1 – bn )

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

= 5-1/2 ( an + 2 – bn + 2 )


Generalizing this ...

• If

• P( 1 ) P( 2 ) . . . P( k )

• n { [ P( n + 1 ) P( n + 2 ) . . . P( n + k ) ] P( n + k + 1 ) }

• then n N P( n ).


End


Strong Mathematical Induction

• If

• P( 1 ) P( 2 ) . . . P( k ) and

• for n k,

[ P( 1 ) P( 2 ) . . . P( n ) ] P( n + 1 )

• then, n N P(n).


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.


• 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 = p1p2 . . . pk & b = q1q2 . . . ql, where the

pis & qis are primes.

• Represent n = p1p2 . . . pkq1q2 . . . ql.

Documents

Copyright © Peter Cappello Mathematical Induction Goals Explain & illustrate construction of proofs of a variety of theorems using mathematical induction