Number Theory and Techniques of Proof - PowerPoint Presentation

Slide1

Number Theory

and Techniques of ProofSlide2

Basic

definitions:Parity

An

integer

n is called

even

if, and only if

,

there exists

an integer k such that

n = 2*k

.

An integer n is called

odd

if, and only if,

it is not even.

Corollary: An integer n is called odd if, and only if, there exists an integer k such that

n = 2*k + 1

The property of an integer as being either odd or even is known as its

parity

.Slide3

Arguing the positive: Universal Statements

Let’s consider the following statement:

“The sum of an odd and an even integer is odd.”Slide4

Arguing the positive: Universal Statements

Let’s consider the following statement:

“The sum of an odd and an even integer is odd.”

Do you believe this statement?

Yes

NoSlide5

Arguing the positive: Universal Statements

Let’s consider the following statement:

“The sum of an odd and an even integer is odd.”

Do you believe this statement?

If you believe it,

you have to try to prove

that it’s

true (argue the positive/affirmative)

Yes

NoSlide6

Proof, take 1

Claim to be proven

true

(we argue its

affirmative):

“The sum of an odd and an even integer is odd.”Proof:Let

be

any odd integer.

Then,

Let

be

any even integer.

Then, (2)By (1) and (2), we have that (3). We set . Clearly, . (4)Substituting (4) into (3) yields: + 1, which means that is odd.End of proof.

 Slide7

Proof, take 1

Claim to be proven

true

(we argue its

affirmative):

“The sum of an odd and an even integer is odd.”Proof:Let

be

any odd integer.

Then,

Let

be

any even integer.

Then, (2)By (1) and (2), we have that (3). We set . Clearly, . (4)Substituting (4) into (3) yields: + 1, which means that is odd.End of proof.

WHOOPS

!

What does this proof actually prove?Slide8

Proof, take 1

Claim to be proven

true

(we argue its

affirmative):

“The sum of an odd and an even integer is odd.”Proof:Let

be

any odd integer.

Then,

Let

be

any even integer.

Then, (2)By (1) and (2), we have that (3). We set . Clearly, . (4)Substituting (4) into (3) yields: + 1, which means that is odd.End of proof.

WHOOPS

!

What does this proof actually prove?

It proves that two

consecutive

integers sum to an odd number!Slide9

Proof, take 2

Claim to be proven

true

(we argue its

affirmative):

“The sum of an odd and an even integer is odd.”Proof:Let

be

any odd integer.

Then,

Let

be

any even integer.

Then, (2)By (1) and (2), we have that (3). We set . Clearly, is an integer. (4)Substituting (4) into (3) yields:

+ 1, which means that

is odd.

End of proof.

 Slide10

Mathematical claims and theorems can be stated in various different ways!

Statements of claims / theorems

“The sum of an odd and an even integer is odd.”

“

Any two

integers of

opposite parity

sum to an

odd

number”

“

Every pair of

integers of

opposite parity

sum

s to an odd number”

 Slide11

Statements of claims / theorems

Mathematical claims and theorems can be stated in various different ways!

Other ideas?

“The sum of an odd and an even integer is odd.”

“

Any two

integers of

opposite parity

sum to an

odd

number”

“

Every pair of

integers of

opposite parity

sums to an odd number” Slide12

Your turn, class!

Let’s split into teams and prove the following claims

true

:

The square of an odd integer is also odd.

If

is an integer, then

is even.

 Slide13

Arguing the affirmative of

existential

statements

Two methods:

Constructive

Non-Constructive

In “constructive” proofs we either

explicitly show or construct an element of the domain that answers our query.In

non-constructive

proofs (very rare in this class) we prove that

it is a logical necessity

for such an element to exist!

But we neither explicitly, nor implicitly, show or construct such an element!Slide14

Our first constructive proof

Claim

: There exists a natural number that you

cannot

write as a sum of three squares of natural numbers.Slide15

Constructive proofs in Number Theory (and one non-constructive one)Slide16

Our first constructive proof

Claim

: There exists a natural number that you

cannot

write as a sum of three squares of natural numbers.

Examples of numbers you

can write as a sum of three squares:

Try to find a number that

cannot

be

written as such.

 Slide17

Proof

The natural number 7 cannot be written as the sum of three squares.

This we can prove by case analysis:

Can’t use 3, since

Can’t use

more than once, since

So, we can use 2, one or zero times.

If we use 2 once, we have

If we use 2 zero times, the maximum value is

 Slide18

Your turn, class!

Let’s split in teams and prove the following theorems:

There exists an integer

that can be written in two ways as a sum of two prime numbers.

There is a

perfect square

that can be written as a sum of two other

perfect squares.

Suppose

. Then, (

There exists an integer

that can be written in two ways as a sum of two cubed integers.

(Hard)

 Slide19

Your turn, class!

Let’s split in teams and prove the following theorems:

There exists an integer

that can be written in two ways as a sum of two prime numbers.

There is a

perfect square

that can be written as a sum of two other

perfect squares.

Suppose

. Then,

(

There exists an integer

that can be written in two ways as a sum of two cubed integers.

(Hard)

How is the 3rd proof different from the others?Slide20

Our first (and last?) non-constructive proof

Theorem

: There exists a pair of

irrational

numbers

and

such that

is a

rational

number.

Proof

: Let

. Since

is irrational,

and

are both irrational. Is rational? Two cases:If is rational, then we have proven the result. Done.If is irrational, then we will name it . Then, observe that is rational, since . Since both and are irrationals, but is rational, we are done.  Slide21

Divisibility

Let

and

. Then, we say or denote any one of the following:

d

divides

n

n

is divided by

d

d

|

n

d is a

divisor (or factor) of nn is a multiple of dif, and only if, We sometimes call k the quotient of the division of n by d.If d does not divide n, we denote that by (note the small strikethrough) Slide22

Pop Quizzes

3 | 6

Yes

NoSlide23

Pop Quizzes

3 | 6

Y

6 | 3

Yes

NoSlide24

Pop Quizzes

3 | 6

Y

6 | 3

N

10 | 10

Yes

NoSlide25

Pop Quizzes

3 | 6

Y

6 | 3

N

10 | 10

Y

-10

10

Yes

NoSlide26

Pop Quizzes

3 | 6

Y

6 | 3

N

10 | 10

Y

-10

10

N

5 | 0

Yes

NoSlide27

Pop Quizzes

3 | 6

Y

6 | 3

N

10 | 10

Y

-10

10

N

5 | 0

Y

0 | 5

Yes

NoSlide28

Pop Quizzes

3 | 6

Y

6 | 3

N

10 | 10

Y

-10

10

N

5 | 0

Y

0 | 5 N

Yes

NoSlide29

Pop Quizzes

3 | 6

Y

6 | 3

N

10 | 10

Y

-10

10

N

5 | 0

Y

0 | 5 N

N (

 YesNoSlide30

Pop Quizzes

3 | 6

Y

6 | 3

N

10 | 10

Y

-10

10

N

5 | 0

Y

0 | 5 N

N (

N (any non-zero integer divides 0) YesNoSlide31

Universal claims with divisibility

Let’s all try to prove the

affirmative

of this claim:

 Slide32

Universal claims with divisibility

Let’s all try to prove the

affirmative

of this claim:

Proof:

From (1) and (2), we have that

So

Done.

 Slide33

Proof by contradiction

Sometimes, proving a fact

directly

is tough.

In such cases, we can attempt an

indirect

proofThe most common type of indirect proof is

proof by contradictionBriefly: We want to prove a fact , so we assume

and hope that we reach a contradiction (a

falsehood

).

Example: We will prove that if a prime number divides an integer

, it cannot possible divide

.

 Slide34

Proofs by contradiction in Number TheorySlide35

First proof by contradiction

Claim: Let

Then, if

, then

 Slide36

First proof by contradiction

Claim: Let

Then, if

, then

Proof:

Assume that

. Then, this means that

(I)

We already know that

(II)

Substituting (II) into (I) yields:

which is a

contradiction

. Therefore,

 Slide37

Infinitude of primes

Assume that the primes are finite. Then, we can list them in ascending order:

 Slide38

Infinitude of primes

Assume that the primes are finite. Then, we can list them in ascending order:

Let’s create the number

 Slide39

Infinitude of primes

Clearly,

is bigger than any

. We have two cases:

N

is prime. Contradiction, since

is bigger than any prime.

N

is composite. This means that N has at least one factor

. Let’s take the smallest factor of

N,

and call it

.

Then, this number is prime (why?)

Since

is prime, it divides . By the previous theorem, this means that it cannot possibly divide . Contradiction, since we assumed that is a factor of N.Therefore, the primes are not finite. Slide40

Modular ArithmeticSlide41

Modular Arithmetic

We say that

(read

“a is congruent to b mod m”)

means that

.

Examples:

Convention:

THINK: Take large number

divide by

, remainder is

Terminology: “Reducing

 Slide42

vs

In Logic,

mean that

and

have the same truth table

(are logically equivalent)

In Number Theory,

, read

“a is

congruent

to b mod m”) means ! Slide43

vs

In Logic,

mean that

and

have the same truth table

(are logically equivalent)

In Number Theory,

, read

“a is

congruent

to b mod m”) means !THESE TWO ARE VERY DIFFERENT!!!! THEY HAVE NOTHING TO DO WITH EACH OTHER! Slide44

Properties of equivalence

If

and

, then:

 Slide45

Properties of equivalence

If

and

, then:

Proof:

(I)

Similarly,

(II)

Therefore, by (I) and (II) we have:

 Slide46

Properties of equivalence

If

and

, then

 Slide47

Properties of equivalence

If

and

, then

Proof: For you to figure out. Might be in:

Homework

Quiz

Midterm 1

Midterm 2

Final

Any combination of the above

How many possibilities are there?

 Slide48

First proof revisited

Recall that we proved that the sum of an even and an odd integer is odd.

Note that:

If

is even (so 2 divides it), then

If

is

odd, then

So now we can re-do the proof with modular arithmetic!

 Slide49

Proof with modular arithmetic

Claim: Any two integers of opposite parity sum to an odd number.

Proof:

Since

,

are opposite parity,

without loss of generality

, assume that

and

Using the

properties of modular arithmetic

, we obtain:

Done.

 Slide50

More proofs

Similarly, you can show that

 Slide51

More proofs

Similarly, you can show that

Proof:

is

throughout to save space.

We have two cases:

Then,

. Done.

. Then,

. Done.

 Slide52

Advantages of this notation

Theorem (clumsy): If

is such that when you divide x by 4 you get a remainder of 2, and

is such that when you divide y by 4 you get a remainder of 3, then when you divide

by 4 you get a remainder of 2.

 Slide53

Advantages of this notation

Theorem (clumsy): If

is such that when you divide x by 4 you get a remainder of 2, and

is such that when you divide y by 4 you get a remainder of 3, then when you divide

by 4 you get a remainder of 2.

THIS SOUNDS AWFUL!

 Slide54

Advantages of this notation

Theorem (clumsy): If

is such that when you divide x by 4 you get a remainder of 2, and

is such that when you divide y by 4 you get a remainder of 3, then when you divide

by 4 you get a remainder of 2.

THIS SOUNDS AWFUL!

Theorem

(elegant): If

and

, then

4).

 Slide55

Advantages of this notation

Theorem (clumsy): If

is such that when you divide x by 4 you get a remainder of 2, and

is such that when you divide y by 4 you get a remainder of 3, then when you divide

by 4 you get a remainder of 2.

THIS SOUNDS AWFUL!

Theorem

(elegant): If

and

, then

4).

Proof:

All

are mod 4. Then:

 Slide56

Proofs by contrapositive in Number TheorySlide57

Proof by contraposition

Applicable to all kinds of statements of type:

 Slide58

Proof by contraposition

Applicable to all kinds of statements of type:

Sometimes, proving the implication in this way is

hard

.

On the other hand, proving its

contrapositive

might be easier:

 Slide59

Examples

Proving this

directly

is somewhat

hard

On the other hand, the contrapositive is child’s (or 250 student’s) play:

]

 Slide60

Examples

Proving this

directly

is somewhat

hard

On the other hand, the contrapositive is child’s (or 250 student’s) play:

Proof: Since

, we have that

So,

 Slide61

Another example

If

, then

 Slide62

Another example

If

, then

Proof (contrapositive):

Cases (all

are mod 5):

0

Done.

 Slide63

A historical proof by contradictionSlide64

Proof that

is irrational

Let’s assume BY WAY OF CONTRADICTION that

is rational.

So

and

do not have common factors.

So

so

(1)

By the previous theorem,

this means that

S

o

for some integer . (2)Substituting (2) into (1) yields: So both and are both even, have common factor of 2.Contradiction. Slide65

Proof that

is irrational

Let’s assume BY WAY OF CONTRADICTION that

is rational.

So

and

do not have common factors.

So

so

(1)

By the previous theorem,

this means that

S

o

for some integer . (2)Substituting (2) into (1) yields: So both and are both even, have common factor of 5.Contradiction. Slide66

Proof that

is irrational (???)

Why can we

not

use this machinery to prove that

is irrational (which is wrong anyway)?

 Slide67

Using the Unique Factorization TheoremSlide68

Unique Factorization: examples

There is no other way to factor 91

into a product of primes.

Once again, no other way to factor 18 into a product of primes.

Since

is prime, there is trivially no other way to factor it into primes.

prime or not?

 Slide69

Unique Factorization: examples

There is no other way to factor 91

into a product of primes.

Once again, no other way to factor 18 into a product of primes.

Since

is prime, there is trivially no other way to factor it into primes.

prime or not?

Nope!

(1049 is prime)

 Slide70

Statement of Theorem

Every number

can be

uniquely

factored into a product of prime numbers

like so:

Proving

existence

is

easy (Jason)

Proving

uniqueness

is

hard (Bill)

 Slide71

What is “uniqueness”?

By “uniqueness” we mean that the product is unique

up to reordering of the factors

.

Examples:

 Slide72

Proof of

with PFT

Proof (once again by contradiction): Assume that

, so

 Slide73

Proof of

with PFT

Proof (once again by contradiction): Assume that

, so

Let

be the

largest

integer such that

(By UPFT)

Similarly, let

be the

largest

integer such that

(By UPFT) Slide74

Proof of

with UFT

Proof (once again by contradiction):

Assume that

, so

Since

 Slide75

Proof of

with UFT

Since

Let

be the

largest

integer such that

Let

be the

largest

integer such that

Since

are largest

ints, and are odd, so odd (we proved this)

=

Even number of 2s on left side, odd number of 2s on right

Contradiction.

 Slide76

Proof of

with UFT

Proof (by contradiction)

Assume that

Let

be

the largest integers

such that

.

Clearly,

, so

(make sure you’re convinced)

Even number of 5s on the left, odd on the right.

Contradiction.

 Slide77

Proof that

(???) with UFT

Why can we

not

use this machinery to prove that

is irrational (which is wrong anyway)?

 Slide78

Speed of Computations

in Number TheorySlide79

Basic assumptions

and

have

unit cost

This is not true if

are

too large

Jason

:

Do you mean

bits or something?

Bill:

Nobody cares, just say “large”.

 Slide80

First problem

How fast can we compute

Obviously, we can compute

and

mod that large number by

.

 Slide81

First problem

How fast can we compute

Obviously, we can compute

and

mod that large number by

.

Is this algorithm ?

Good

Bad

Ugly

 Slide82

First problem

How fast can we compute

Obviously, we can compute

and

mod that large number by

.

Is this algorithm ?

Good

Bad

Ugly

Because:

Jason:

Numbers can get above 32 bits, and that’s a storage and computation problem.

Bill: Numbers get “too freaking large”.Slide83

First

problem, second approach

We could start computing

until the product becomes larger than

, reduce and repeat until we’re done.

 Slide84

First

problem, second approach

We could start computing

until the product becomes larger than

, reduce and repeat until we’re done.

Is this better?

Yes

No

Something ElseSlide85

First

problem, second approach

We could start computing

until the product becomes larger than

, reduce and repeat until we’re done.

Is this better?

Yes

No

Something Else

We no longer produce huge numbers!

However, we still need

multiplications.

 Slide86

First problem

How fast can we compute

We always need

steps

We can do it in roughly

steps

We can do it in roughly

steps

Something ElseSlide87

First problem

How fast can we compute

We always need

steps

We can do it in roughly

steps

We can do it in roughly

steps

Something ElseSlide88

Example

Computing

in

steps.

All

are

(mod 99).

 Slide89

Example

Computing

in

steps.

All

are

(mod 99).

Aha!

 Slide90

Good news, bad news

Good news: By using

repeated squaring

, can compute

quickly (roughly

steps)

Bad news: What if our

exponent

is

not

a power of 2?

 Slide91

Example

Computing

with the same method

All

are

(mod 99).

 Slide92

Example (contd.)

To avoid large numbers, reduce product as you go:

 Slide93

Algorithm to compute

in

log

steps

Step 1: Write

,

Step 2: Note that

Step 3: Use

repeated squaring

to compute:

using

Step 4: Compute

mod m reducing when necessary to avoid large numbers

 Slide94

The key step

The key step is Step #3:

Use repeated squaring to compute:

using

When computing

mod m,

already have

computed

Note that all numbers are below

because we reduce mod m every step of the way

So

is

unit cost

and

anything mod m

is also unit cost! Slide95

Second problem: Greatest Common Divisor (GCD)

If

, then the GCD of

is the

largest

non-zero integer

such that

and

 Slide96

Second problem: Greatest Common Divisor (GCD)

If

, then the GCD of

is the

largest

non-zero integer

such that

and

What is the GCD of

…

and 15?

 Slide97

Second problem: Greatest Common Divisor (GCD)

If

, then the GCD of

is the

largest

non-zero integer

such that

and

What is the GCD of

…

and 15?

5

12 and 90?

 Slide98

Second problem: Greatest Common Divisor (GCD)

If

, then the GCD of

is the

largest

non-zero integer

such that

and

What is the GCD of

…

and 15?

5

12 and 90?

620 and 29?  Slide99

Second problem: Greatest Common Divisor (GCD)

If

, then the GCD of

is the

largest

non-zero integer

such that

and

What is the GCD of

…

and 15?

5

12 and 90?

620 and 29? 1 (20 and 29 are called co-prime or relatively prime)153 and 181 Slide100

Second problem: Greatest Common Divisor (GCD)

If

, then the GCD of

is the

largest

non-zero integer

such that

and

What is the GCD of

…

and 15?

5

12 and 90?

620 and 29? 1 (20 and 29 are called co-prime or relatively prime)153 and 181 17 Slide101

Euclid’s GCD algorithm

Recall: If

and

then

The GCD algorithm finds the

greatest

common divisor by executing this recursion (assume a > b):

Until its arguments are the same.

 Slide102

Greatest Common Divisor (GCD)

Recall: If

and

then

The GCD algorithm finds the

greatest

common divisor by executing this recursion (assume

a > b

):

Until its arguments are the same.

Question: If we implement this in a programming language,

it can only be done recursively

Yes

(why)No (Why)Something Else(What)Slide103

Greatest Common Divisor (GCD)

Recall: If

and

then

The GCD algorithm finds the

greatest

common divisor by executing this recursion:

Until its arguments are the same.

Question: If we implement this in a programming language,

it can only be done recursively

Yes

(why)

No

(Why)Something Else(What)left = a;right = b;while(left != right){ if(left > right) left = left – right; else right = right - left;}print "GCD is: " left; // Or rightTail recursionSlide104

GCD example

GCD(18, 100) =

GCD(18, 100

–

18) = GCD(18, 82)=

GCD(18, 82 – 18 = GCD(18, 64) =

GCD(18, 64 – 18) = GCD(18, 46) = GCD(18, 46 – 18) = GCD(18, 28) =GCD(18, 28

–

18) = GCD(18, 10) =

GCD(18 - 10, 10) = GCD(8, 10)=

GCD(8, 10 - 8)= GCD(8, 2) =

GCD(8 - 2, 2) = GCD(6, 2) =GCD(6 - 2, 2) = GCD(4, 2) = GCD(4- 2, 2) = GCD(2, 2) = 2Slide105

GCD example

GCD(18

, 100) =

GCD(18, 100

–

18) = GCD(18, 82)=GCD(18, 82 – 18 = GCD(18, 64) =

GCD(18, 64 – 18) = GCD(18, 46) = GCD(18, 46 – 18) = GCD(18, 28) =GCD(18, 28 –

18) = GCD(18, 10) =

GCD(18 - 10, 10) = GCD(8, 10)=

GCD(8, 10 - 8)= GCD(8, 2) =

GCD(8 - 2, 2) = GCD(6, 2) =

GCD(6 - 2, 2) = GCD(4, 2) = GCD(4- 2, 2) = GCD(2, 2) = 2

Given integers

with

(without loss of generality), approximately how many steps does this algorithm take? a stepsb stepsSomething Elsea-b stepsSlide106

GCD example

GCD(18

, 100) =

GCD(18, 100

–

18) = GCD(18, 82)=GCD(18, 82 – 18 = GCD(18, 64) =

GCD(18, 64 – 18) = GCD(18, 46) = GCD(18, 46 – 18) = GCD(18, 28) =GCD(18, 28 –

18) = GCD(18, 10) =

GCD(18 - 10, 10) = GCD(8, 10)=

GCD(8, 10 - 8)= GCD(8, 2) =

GCD(8 - 2, 2) = GCD(6, 2) =

GCD(6 - 2, 2) = GCD(4, 2) = GCD(4- 2, 2) = GCD(2, 2) = 2

Given integers

with

(without loss of generality), approximately how many steps does this algorithm take? a stepsb stepsSomething Elsea-b stepsRoughly  Slide107

Can we do better?

GCD(18

, 100) =

GCD(18, 100

–

18) = GCD(18, 82)=GCD(18, 82 – 18 = GCD(18, 64) =

GCD(18, 64 – 18) = GCD(18, 46) = GCD(18, 46 – 18) = GCD(18, 28) =GCD(18, 28 –

18) = GCD(18, 10) =

GCD(18 - 10, 10) = GCD(8, 10)=

GCD(8, 10 - 8)= GCD(8, 2) =

GCD(8 - 2, 2) = GCD(6, 2) =

GCD(6 - 2, 2) = GCD(4, 2) = GCD(4- 2, 2) = GCD(2, 2) = 2

Yes

No

Something ElseSlide108

Can we do better?

GCD(18

, 100) =

GCD(18, 100

–

18) = GCD(18, 82)=GCD(18, 82 – 18 = GCD(18, 64) =

GCD(18, 64 – 18) = GCD(18, 46) = GCD(18, 46 – 18) = GCD(18, 28) =GCD(18, 28 –

18) = GCD(18, 10) =

GCD(18 - 10, 10) = GCD(8, 10)=

GCD(8, 10 - 8)= GCD(8, 2) =

GCD(8 - 2, 2) = GCD(6, 2) =

GCD(6 - 2, 2) = GCD(4, 2) = GCD(4- 2, 2) = GCD(2, 2) = 2

Yes

No

Something ElseGCD(18, 100 – 5 x 18)GCD(8 – 3 x 2, 2)GCD(18, 100) = GCD(18, 100 – 5 x 18) = GCD(18, 10) = GCD(18 – 10, 10) = GCD(8, 10) = GCD(8, 10 - 8) = GCD(8, 2) = GCD(8 – 3 x 2, 2) = GCD(2, 2) = 2From 10 to 4 steps!Slide109

How fast is this new algorithm?

Given non-zero integers

with

,

roughly how many steps does this new algorithm take to compute GCD(a, b

)?

loga

Something ElseSlide110

How fast is this new algorithm?

Given non-zero integers

with

,

roughly how many steps does this new algorithm take to compute GCD(a, b

)?

In fact, it takes

, where

is the

golden ratio.

Proof by Gabriel Lam

é

in 1844,

considered

to

be the first ever result in Algorithmic Complexity theory.   logaSomething Else