Download Introduction to Discrete Mathematics

Elementary Number Theory
and
Methods of Proof
Basic Definitions
An integer n is an even number
if there exists an integer k such that n = 2k.
An integer n is an odd number
if there exists an integer k such that n = 2k+1.
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.
Simple Exercises
The sum of two even numbers is even.
The product of two odd numbers is odd.
direct proof.
Rational Number
R is rational  there are integers a and b such that
numerator
and b ≠ 0.
denominator
Is 0.281 a rational number?
Is 0 a rational number?
If m and n are non-zero integers, is (m+n)/mn a rational number?
Is the sum of two rational numbers a rational number?
Is 0.12121212…… a rational number?
Divisibility
a “divides” b
(a|b):
b = ak for some integer k
5|15 because 15 = 35
n|0 because 0 = n0
1|n because n = 1n
n|n because n = n1
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.
Simple Divisibility Facts
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.
Proof of (1)
direct proof.
Simple Divisibility Facts
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.
Proof of (2)
direct proof.
Divisibility by a Prime
Theorem. Any integer n > 1 is divisible by a prime number.
Idea of induction.
Fundamental Theorem of Arithmetic
Every integer, n>1, has a unique factorization into primes:
p0 ≤ p1 ≤ ··· ≤ pk
p0 p1 ··· pk = n
Example:
61394323221 = 3·3·3·7·11·11·37·37·37·53
Prime Products
Claim: Every integer > 1 is a product of primes.
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
Claim: Every integer > 1 is a product of primes.
…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
Idea of induction (or smallest counterexample).
(The proof of the fundamental theorem will be given later.)
The Quotient-Reminder Theorem
For b > 0 and any a, there are unique numbers
q ::= quotient(a,b), r ::= remainder(a,b), such that
a = qb + r
We also say
and
q = a div b
0  r < b.
and
r = a mod b.
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.
The Division Theorem
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.
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
-b
0
b
2b
r = the offset in this block
kb
a
(k+1)b
Clearly, given a and b, q and r are uniquely defined.
The Square of an Odd Integer
Idea 0: find counterexample.
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
The Square of an Odd Integer
Idea 3: Use quotient-remainder theorem.
Proof by cases.
Trial and Error Won’t Work!
Fermat (1637): If an integer n is greater than 2,
then the equation an + bn = cn has no solutions in non-zero integers a, b, and c.
has no solutions in non-zero integers a, b, and c.
Claim:
False. But smallest counterexample has more than 1000 digits.
Euler conjecture:
has no solution for a,b,c,d positive integers.
Open for 218 years, until Noam Elkies found
958004  2175194  4145604  4224814
The Square Root of an Even Square
Statement: If m2 is even, then m is even
Contrapositive: If m is odd, then m2 is odd.
Proof (the contrapositive):
Since m is an odd number, m = 2l+1 for some natural number l.
So m2 = (2l+1)2
= (2l)2 + 2(2l) + 1
So m2 is an odd number.
Proof by contrapositive.
Irrational Number
Theorem:
2
is irrational.
Proof (by contradiction):
• Suppose
2
was rational.
• Choose m, n integers without common prime factors (always possible)
such that
m
2
n
• Show that m and n are both even, thus having a common factor 2,
a contradiction!
Irrational Number
Theorem:
2
is irrational.
Proof (by contradiction):
m
2
n
Want to prove both m and n are even.
so can assume
m  2l
m 2  4l 2
2n  m
2n 2  4l 2
2n 2  m 2
n 2  2l 2
so m is even.
so n is even.
Proof by contradiction.
Infinitude of the Primes
Theorem. There are infinitely many prime numbers.
Let p1, p2, …, pN be all the primes.
Consider p1p2…pN + 1.
Claim: if p divides a, then p does not divide a+1.
Proof by contradiction.
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.
a=12, b=8 => 12 = 8 + 4
gcd(12,8) = 4
gcd(8,4) = 4
a=21, b=9 => 21 = 2x9 + 3
gcd(21,9) = 3
gcd(9,3) = 3
a=99, b=27 => 99 = 3x27 + 18
gcd(99,27) = 9
gcd(27,18) = 9
Euclid: gcd(a,b) = gcd(b,r)!
Euclid’s GCD Algorithm
a = qb + r
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)
Example 1
gcd(a,b)
if b = 0, then answer = a.
else
write a = qb + r
answer = gcd(b,r)
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(a,b)
if b = 0, then answer = a.
else
write a = qb + r
answer = gcd(b,r)
GCD(252, 189)
= GCD(189, 63)
252 = 1x189 + 63
189 = 3x63 + 0
= GCD(63, 0)
Return value: 63.
Example 3
gcd(a,b)
if b = 0, then answer = a.
else
write a = qb + r
answer = gcd(b,r)
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.
Correctness of Euclid’s GCD Algorithm
a = qb + r
Euclid: gcd(a,b) = gcd(b,r)
When r = 0:
Correctness of Euclid’s GCD Algorithm
a = qb + r
Euclid: gcd(a,b) = gcd(b,r)
When r > 0:
Let d be a common divisor of b, r
Correctness of Euclid’s GCD Algorithm
a = qb + r
Euclid: gcd(a,b) = gcd(b,r)
When r > 0:
Let d be a common divisor of a, b.