Elementary Methods Y. İlker TOPCU, Ph.D. twitter.com/yitopcu.
Chapter 3: Elementary Number Theory and Methods of Proofs
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Transcript of Chapter 3: Elementary Number Theory and Methods of Proofs
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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
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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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3Introduction to Abstract Mathematics
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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4Introduction to Abstract Mathematics
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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5Introduction to Abstract Mathematics
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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6Introduction to Abstract Mathematics
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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7Introduction to Abstract Mathematics
├ (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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8Introduction to Abstract Mathematics
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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9Introduction to Abstract Mathematics
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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10Introduction to Abstract Mathematics
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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11Introduction to Abstract Mathematics
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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12Introduction to Abstract Mathematics
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
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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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14Introduction to Abstract Mathematics
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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15Introduction to Abstract Mathematics
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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16Introduction to Abstract Mathematics
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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17Introduction to Abstract Mathematics
Irrational Numbers
So far all the examples were of rational numbers
How about some irrationals? – e– sqrt(2)
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18Introduction to Abstract Mathematics
Simple Exercises
The sum of two even numbers is even.
The product of two odd numbers is odd.
direct proof.
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19Introduction to Abstract Mathematics
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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20Introduction to Abstract Mathematics
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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21Introduction to Abstract Mathematics
Divisibility by a Prime
Theorem. Any integer n > 1 is divisible by a prime number.
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22Introduction to Abstract Mathematics
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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23Introduction to Abstract Mathematics
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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24Introduction to Abstract Mathematics
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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25Introduction to Abstract Mathematics
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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26Introduction to Abstract Mathematics
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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27Introduction to Abstract Mathematics
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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28Introduction to Abstract Mathematics
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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29Introduction to Abstract Mathematics
• 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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30Introduction to Abstract Mathematics
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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31Introduction to Abstract Mathematics
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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32Introduction to Abstract Mathematics
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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33Introduction to Abstract Mathematics
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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34Introduction to Abstract Mathematics
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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35Introduction to Abstract Mathematics
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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36Introduction to Abstract Mathematics
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
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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
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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