1 Introduction to Abstract Mathematics 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 [email protected]

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


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) .


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)


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


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)


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


├ (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)]


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?


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


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!


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


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


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


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)


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


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


Irrational Numbers

So far all the examples were of rational numbers

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


Simple Exercises

The sum of two even numbers is even.

The product of two odd numbers is odd.

direct proof.


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.


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.


Divisibility by a Prime

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


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


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


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.)


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.


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


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.


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.


• 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


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


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.


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?


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


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.


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


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


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


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

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1 Introduction to Abstract Mathematics Chapter 3: Elementary Number Theory and Methods of Proofs...

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