Elementary Number Theory and Methods of Proof

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Elementary Number Theory andMethods of Proof
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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 2k1.
An integer n is a prime number if and only if ngt1
and if nrs for some positive integers r and s
then r1 or s1.
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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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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
(mn)/mn a rational number? Is the sum of two
rational numbers a rational number? Is
0.12121212 a rational number?
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Divisibility
a divides b (ab) b ak for
some integer k
515 because 15 3?5 n0 because 0 n?0 1n
because n 1?n nn because n n?1
A number p gt 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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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.
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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.
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Divisibility by a Prime
Theorem. Any integer n gt 1 is divisible by a
prime number.
Idea of induction.
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Fundamental Theorem of Arithmetic
Every integer, ngt1, has a unique factorization
into primes p0 p1 pk p0 p1 pk n
Example 61394323221 3337111137373753
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Prime Products
Claim Every integer gt 1 is a product of primes.
Proof (by contradiction) Suppose not. Then set
of non-products is nonempty. There is a smallest
integer n gt 1 that is not a product of primes. In
particular, n is not prime.
So n km for integers k, m where n gt k,m
gt1. 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
Claim Every integer gt 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 gt 1 must be empty.
  QED

Idea of induction (or smallest counterexample).
(The proof of the fundamental theorem will be
given later.)
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The Quotient-Reminder Theorem
For b gt 0 and any a, there are unique numbers q
quotient(a,b), r remainder(a,b), such
that a qb r and 0 ? r lt b.
We also say q a div b and r a
mod b.
When b2, this says that for any a, there is a
unique q such that a2q or a2q1.
When b3, this says that for any a, there is a
unique q such that a3q or a3q1 or a3q2.
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The Division Theorem
For b gt 0 and any a, there are unique numbers q
quotient(a,b), r remainder(a,b), such
that a qb r and 0 ? r lt 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
r the offset in this block
a
(k1)b
kb
2b
b
-b
0
Clearly, given a and b, q and r are uniquely
defined.
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The Square of an Odd Integer
Idea 0 find counterexample.
32 9 81, 52 25 3x81 1312
17161 2145x8 1,
Idea 1 prove that n2 1 is divisible by 8.
Idea 2 consider (2k1)2
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The Square of an Odd Integer
Idea 3 Use quotient-remainder theorem.
Proof by cases.
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Trial and Error Wont 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
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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 2l1 for some
natural number l.
So m2 (2l1)2
(2l)2 2(2l) 1
So m2 is an odd number.
Proof by contrapositive.
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Irrational Number
Theorem is irrational.
Proof (by contradiction)

• 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!

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Irrational Number
Theorem is irrational.
Proof (by contradiction)
Want to prove both m and n are even.
so can assume
so n is even.
Proof by contradiction.
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Infinitude of the Primes
Theorem. There are infinitely many prime numbers.
Let p1, p2, , pN be all the primes.
Consider p1p2pN 1.
Claim if p divides a, then p does not divide a1.
Proof by contradiction.
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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?

• Lets say agtb.
• If akb, then gcd(a,b)b, and we are done.
• Otherwise, by the Division Theorem, a qb r
  for rgt0.

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Greatest Common Divisors

• Lets say agtb.
• If akb, then gcd(a,b)b, and we are done.
• Otherwise, by the Division Theorem, a qb r
  for rgt0.

gcd(8,4) 4
gcd(12,8) 4
a12, b8 gt 12 8 4
gcd(9,3) 3
a21, b9 gt 21 2x9 3
gcd(21,9) 3
gcd(99,27) 9
a99, b27 gt 99 3x27 18
gcd(27,18) 9
Euclid gcd(a,b) gcd(b,r)!
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Euclids 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)
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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.
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Example 2
gcd(a,b) if b 0, then answer a. else write
a qb r answer gcd(b,r)
GCD(252, 189) 252 1x189
63 GCD(189, 63) 189 3x63
0 GCD(63, 0) Return value 63.
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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.
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Correctness of Euclids GCD Algorithm
a qb r
Euclid gcd(a,b) gcd(b,r)
When r 0
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Correctness of Euclids GCD Algorithm
a qb r
Euclid gcd(a,b) gcd(b,r)
When r 0 Then gcd(b, r) gcd(b, 0) b since
every number divides 0. But a qb so gcd(a, b)
b gcd(b, r), and we are done.
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Correctness of Euclids GCD Algorithm
Euclid gcd(a,b) gcd(b,r)
a qb r
When r gt 0
Let d be a common divisor of b, r
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Correctness of Euclids GCD Algorithm
Euclid gcd(a,b) gcd(b,r)
a qb r
When r gt 0
Let d be a common divisor of a, b.
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Correctness of Euclids GCD Algorithm
Euclid gcd(a,b) gcd(b,r)
a qb r
When r gt 0

• Let d be a common divisor of b, r
• b k1d and r k2d for some k1, k2.
• a qb r qk1d k2d (qk1 k2)d gt d
  is a divisor of a
• Let d be a common divisor of a, b
• a k3d and b k1d for some k1, k3.
• r a qb k3d qk1d (k3 qk1)d gt d
  is a divisor of r
• So d is a common factor of a, b iff d is a common
  factor of b, r
• d gcd(a, b) iff d gcd(b, r)