Chapter 3: Elementary Number Theory and Methods of Proofs

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Chapter 3: Elementary Number Theory and Methods of Proofs. 3.1-.3.4 Direct Methods and Counterexamples Introduction Rational Numbers Divisibility Division Algorithm. Instructor: Hayk Melikya melikyan@nccu.edu. Basic Definitions. - PowerPoint PPT Presentation

Transcript of Chapter 3: Elementary Number Theory and Methods of Proofs

1Introduction to Abstract Mathematics

Chapter 3: Elementary Number Theory and Methods of Proofs

Instructor: Hayk Melikya melikyan@nccu.edu

3.1-.3.4 Direct Methods and Counterexamples• Introduction• Rational Numbers• Divisibility• Division Algorithm

2Introduction to Abstract Mathematics

Basic Definitions

Definition: An integer n is an even number if there exists an integer k such that n = 2k.

Def: An integer n is an odd number if there exists an integer k such that n = 2k+1.

Def: An integer n is a prime number if and only if n>1

and if n=rs for some positive integers r and s then r=1 or s=1.

Symbolically: Let Even(n) := “an integer n is even”: E(n) = ( k Z)( n = 2k) .

Symbolically: Let O(n) := “an integer is odd”: Odd(n) = ( k Z)( n = 2k +1) .

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Primes and CompositesDef: An integer n is a prime number if and only if n>1

and if n=rs for some positive integers r and s then r=1 or s=1.

Symbolically: Prime(n):= n is prime

positive integers r and s, if n = rs then r =1 or s =1

Def: A positive integer n is a composite if and only if n=rs for some

positive integers r and s then r ≠ 1 and s ≠ 1.

Symbolically: Cpmposite(n):= n is compoeite positive integers r and s,

such that n = rs and r ≠ 1 and s ≠ 1

The RSA Challenge (up to US$200,000)http://www.rsasecurity.com/rsalabs/challenges/factoring/numbers.html

Examples: Find the truth value of the following prpopositionsE(6), P(12), C(17)

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Existential Statements

x P(x)Proofs:

– Constructive Construct an example of such a such that P(a) is true

– Non-constructive By contradiction

– Show that if such x does NOT exist than a contradiction can be derived

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Example

Let G(n):= a b ((a+b=n) Prime(a) Prime(b))

Prove that (nN)G(n) Proof:

– n=210 – a=113– b=97

Piece of cake…

What about (nN) G(n) ( many Million $ baby)

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Universal Statements

x P(x) x [Q(x) R(x)]

Proof techniques:– Exhaustion– By contradiction

Assume the statement is not true Arrive at a contradiction

– Direct Generalizing from an arbitrary particular member

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├ (xU)P(x)

To prove a theorem of the form (xU)P(x) (same as ├ (xU)P(x))

which states “for all elements x in a given universe U, the

proposition P(x) is true” we select an arbitrary aU from the

universe, and then prove the assertion P(a).

Then by Universal generalization we conclude P(a)├ (xU)P(x)

For arguments of the form├ x [Q(x) R(x)]

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Example 1

Exhaustion:– Any even number between 4 and 30 can be written as a

sum of two primes:– 4=2+2– 6=3+3– 8=3+5– …– 30=11+19

Works for finite domains only

What if I want to prove that for any integer n the product of n and n+1 is even?

Can I exhaust all integer values of n?

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Example 2

Theorem: (nZ)( even(n*(n+1)) )

Proof:– Consider a particular but arbitrary chosen integer n – n is odd or even– Case 1: n is odd

Then n=2k+1, n+1=2k+2 n(n+1) = (2k+1)(2k+2) = 2(2k+1)(k+1) = 2p for some

integer p So n(n+1) is even

Case 2: n is evenThen n=2k, n+1=2k+1 n(n+1) = 2k(2k+1) = 2p for some integer pSo n(n+1) is even

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Fallacy

Generalizing from a particular but NOT arbitrarily chosen example

I.e., using some additional properties of nExample:

– “all odd numbers are prime”– “Proof”:

Consider odd number 3 It is prime Thus for any odd n prime(n) holds

Such “proofs” can be given for correct statements as well!

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Prevention

Try to stay away from specific instances (e.g., 3) Make sure that you are not using any additional properties of n

considered Challenge your proof

– Try to play the devil’s advocate and find holes in it…

• Using the same letter to mean different things• Jumping to a conclusion• Insufficient justification• Begging the question assuming the claim first

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Rational Numbers

A real number is rational iff it can be representedas a ratio/quotient/fraction of integers a and b(b0)

rR [rQ a,bZ [r=a/b & b0]]

Notes:– a is numerator– b is denominator– Any rational number can be represented in infinitely many

ways– The fractional part of any rational number written in any

natural radix has a period in it

13Introduction to Abstract Mathematics

Rational or not? -12

– -12/1 3.1459

– 3+1459/10000 0.56895689568956895689…

– 5689/9999 1+1/2+1/4+1/8+…

– 2 0

– 0/1

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Theorem 1

Any number with a periodic fractional part in a natural

radix representation is rational

Proof:– Constructive:

– x=0.n1…nmn1…nm…

– x=0.(n1…nm)

– x*10m-x=n1…nm

– x=n1…nm/(10m-1)

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Theorem 2

Any geometric series:– S=q0+q1+q2+q3+…– where -1<q<1– evaluates to S=1/(1-q)

Proof– Proof idea – More formal proof– Definitions of limits and partial sums

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Z Q

Every integer is a rational numberProof : set the denominator to 1

Book : page 127

The set of rational numbers is closed with respect to arithmetic operations +, -, *, /

Partial proofs : textbook pages 121-131Formal proof

Q

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Irrational Numbers

So far all the examples were of rational numbers

How about some irrationals? – e– sqrt(2)

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Simple Exercises

The sum of two even numbers is even.

The product of two odd numbers is odd.

direct proof.

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a “divides” b or is b divisible by a (a|b ):

b = ak for some integer k

Also we say that

b is multiple of a

a is a factor of b

b is divisor for a

Divisibility

5|15 because 15 = 35

n|0 because 0 = n0

1|n because n = 1n

n|n because n = n1

A number p > 1 with no positive integer divisors other than 1 and itself

is called a prime. Every other number greater than 1 is called

composite. 2, 3, 5, 7, 11, and 13 are prime,

4, 6, 8, and 9 are composite.

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1. If a | b, then a | bc for all c.

2. If a | b and b | c, then a | c.

3. If a | b and a | c, then a | sb + tc for all s and t.

4. For all c ≠ 0, a | b if and only if ca | cb.

Simple Divisibility Facts

Proof of (??)

direct proof.

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Divisibility by a Prime

Theorem. Any integer n > 1 is divisible by a prime number.

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Every integer, n>1, has a unique factorization into primes:

p0 ≤ p1 ≤ ··· ≤ pk

p0 p1 ··· pk = n

Fundamental Theorem of Arithmetic

Example:

61394323221 = 3·3·3·7·11·11·37·37·37·53

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Claim: Every integer > 1 is a product of primes.

Prime Products

Proof: (by contradiction)

Suppose not. Then set of non-products is nonempty.

There is a smallest integer n > 1 that is not a product of

primes.

In particular, n is not prime.

So n = k·m for integers k, m where n > k,m >1.

Since k,m smaller than the least nonproduct, both are prime

products, eg.,

k = p1 p2 p94

m = q1 q2 q214

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Prime Products

…So

n = k m = p1 p2 p94 q1 q2 q214

is a prime product, a contradiction.

The set of nonproducts > 1 must be empty. QED

Claim: Every integer > 1 is a product of primes.

(The proof of the fundamental theorem will be given later.)

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For b > 0 and any a, there are unique numbers

q : quotient(a,b), r : remainder(a,b), such that

a = qb + r and 0 r < b.

The Quotient-Reminder Theorem

When b=2, this says that for any a,

there is a unique q such that a=2q or a=2q+1.

When b=3, this says that for any a,

there is a unique q such that a=3q or a=3q+1 or a=3q+2.

We also say q = a div b r = a mod b.

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For b > 0 and any a, there are unique numbers

q : quotient(a,b), r : remainder(a,b), such that

a = qb + r and 0 r < b.

The Division Theorem

0 b 2b kb (k+1)b

Given any b, we can divide the integers into many blocks of b numbers.

For any a, there is a unique “position” for a in this line.

q = the block where a is in

a

r = the offset in this block

Clearly, given a and b, q and r are uniquely defined.

-b

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The Square of an Odd Integer

32 = 9 = 8+1, 52 = 25 = 3x8+1 …… 1312 = 17161 = 2145x8 + 1, ………

Idea 1: prove that n2 – 1 is divisible by 8.

Idea 2: consider (2k+1)2

Idea 0: find counterexample.

Idea 3: Use quotient-remainder theorem.

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Since m is an odd number, m = 2l+1 for some natural number l.

So m2 is an odd number.

Contrapositive Proof

Statement: If m2 is even, then m is evenContrapositive: If m is odd, then m2 is odd.

So m2 = (2l+1)2= (2l)2 + 2(2l) + 1

Proof (the contrapositive):

Proof by contrapositive.

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• Suppose was rational.

• Choose m, n integers without common prime factors (always

possible) such that

• Show that m and n are both even, thus having a common

factor 2,

a contradiction!

n

m2

Theorem: is irrational.2

Proof (by contradiction):

Irrational Number

2

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lm 2so can assume

2 24m l

22 2ln

so n is even.

n

m2

mn2

222 mn

so m is even.

2 22 4n l

Theorem: is irrational.2

Proof (by contradiction): Want to prove both m and n are even.

Proof by contradiction.

Irrational Number

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Infinitude of the Primes

Theorem. There are infinitely many prime numbers.

Claim: if p divides a, then p does not divide a+1.

Let p1, p2, …, pN be all the primes.

Proof by contradiction.

Consider p1p2…pN + 1.

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Floor and Ceiling

If k is an integer, what are x and x + 1/2 ?

Is x + y = x + y? ( what if x = 0.6 and y = 0.7)

For all real numbers x and all integers m, x + m = x + m

For any integer n, n/2 is n/2 for even n and (n–1)/2 for odd n

Def: For any real number x, the floor of x, written x, is the unique integer n such that n x < n + 1.

It is the largest integer not exceeding x ( x).

Def: For any real number x, the ceiling of x, written x, is the unique integer n such that n – 1 < x n. What is n?

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Exercises

Is it true that for all real numbers x and y:

x – y = x - y

x – 1 = x - 1

x + y = x + y

x + 1 = x + 1

For positive integers n and d, n = d * q + r, where d = n / d and r = n – d * n / d with 0 r < d

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Greatest Common Divisors

Given a and b, how to compute gcd(a,b)?

Can try every number, but can we do it more efficiently?

Let’s say a>b.

1. If a=kb, then gcd(a,b)=b, and we are done.

2. Otherwise, by the Division Theorem, a = qb + r for r>0.

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Greatest Common Divisors

Let’s say a>b.

1. If a=kb, then gcd(a,b)=b, and we are done.

2. Otherwise, by the Division Theorem, a = qb + r for r>0.

Euclid: gcd(a,b) = gcd(b,r)!

a=12, b=8 => 12 = 8 + 4 gcd(12,8) = 4

a=21, b=9 => 21 = 2x9 + 3 gcd(21,9) = 3

a=99, b=27 => 99 = 3x27 + 18 gcd(99,27) = 9

gcd(8,4) = 4

gcd(9,3) = 3

gcd(27,18) = 9

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Euclid’s GCD Algorithm

Euclid: gcd(a,b) = gcd(b,r)

gcd(a,b)

if b = 0, then answer = a.

else

write a = qb + r

answer = gcd(b,r)

a = qb + r

37Introduction to Abstract Mathematics

gcd(a,b)

if b = 0, then answer = a.

else

write a = qb + r

answer = gcd(b,r)

Example 1

GCD(102, 70) 102 = 70 + 32

= GCD(70, 32) 70 = 2x32 +

6

= GCD(32, 6) 32 = 5x6 +

2

= GCD(6, 2) 6 = 3x2 + 0

= GCD(2, 0)

Return value: 2.Example 2

GCD(252, 189) 252 = 1x189 + 63

= GCD(189, 63) 189 = 3x63 +

0

= GCD(63, 0)

Return value: 63.

GCD(662, 414) 662 = 1x414 + 248

= GCD(414, 248) 414 = 1x248 +

166

= GCD(248, 166) 248 = 1x166 +

82

= GCD(166, 82) 166 = 2x82 +

2

= GCD(82, 2) 82 = 41x2 + 0

= GCD(2, 0)

Return value: 2.

Example 3

39Introduction to Abstract Mathematics

Practice problems

1. Study the Sections 3.1- 3.4 from your textbook.

2. Be sure that you understand all the examples discussed in class and in textbook.

3. Do the following problems from the textbook:

Exercise 3.1 # 13, 16, 32, 36, 45 Exercise 3.2 # 15, 19, 21, 32, Exercise 3.3 # 13, 16, 25, 26, Exercise 3.4 # 4, 6, 8, 10, 18, 33