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CSE 240
Logic and Discrete Math
Lecture notes 8
Weixiong Zhang
Washington University in St. Louis
http://www.cse.wustl.edu/~zhang/teaching
/cse240/Spring10/index.html
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Today
Chapter 3.3
Divisibility
Chapter 3.4
Quotient-remainder theorem
Chapter 3.5
Floor & ceiling
Chapter 3.7
Infinitude of primes
Irrationality of sqrt(p)
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Rational / Irrational
For more information see:
http://mathworld.wolfram.com/
http://mathworld.wolfram.com/IrrationalNumber.html
http://mathworld.wolfram.com/Pi.html
http://mathworld.wolfram.com/e.html
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Divisibility
Integers n, d; d≠0
d
d
d
n
n
d
(wholly) divides n
is a divisor of n
is a factor of n
is a multiple of d
is divisible by d
|n
iff:
∃k∈Z [ n=dk ]
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Positive Divisors
If a|b and a,b>0 then a≤b
Proof ?
Use definition b = a*k, k integer and k≥1
prime(n) iff (n>1 & n’s only positive
divisors are 1 and n)
Proof ?
Use definition b|n -> n = b*k.
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Properties
Transitivity:
a|b, b|c  a|c
Reflexivity:
a|a
Anti-symmetry (for naturals):
a|b & b|a  a=b
Proof ?
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Quotient-remainder Theorem
For any integer n
For any integer d>0
There exist unique integers q and r
Such that
n=dq+r
0≤r<d
q is the quotient : q=n div d
r is the remainder : r=n mod d
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Proof
Section 4.4 of the text
We will (hopefully) do it later in class
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Connection to odd/even
Suppose d=2, then 0 ≤ r < 2
Then for any integer n:
r=0
r=1


n=2q
n=2q+1
 n is even
 n is odd
So every number is odd or even integer
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Factors
Recall that if x|a and a>0,x>0 then x≤a
Consider set F(a)={x∈Z s.t. x|a}
∀x∈F(a) [ x≤a ]
So F(a) is upper bounded thus has a
maximum element (by the well-ordering
principle)
Call max F(a) highest factor
Example?
Give me a number ….
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Highest Common Factor
Consider integers a>0, b>0
Consider set F(a,b)={x∈Z s.t. x|a & x|b}
∀x∈F(a,b) [ x≤a & x≤b ]
So F(a,b) is upper bounded thus has a
maximum element (by the well-ordering
principle)
Call max F(a,b) highest common factor:
hcf(a,b)
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Example
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Euclid’s Algorithm Idea
Works fine for small numbers
What about hcf(4453,1314) ?
Here is an idea:
Lemma. a>b, a=qb+r then hcf(a,b)=hcf(b,r)
Proof.
Consider F(a,b)={x st x|a & x|b}
Consider F(b,r)={x st x|r & x|b}
If x1∈F(b,r) then x1|a, thus x1∈F(a,b)
If x2∈F(a,b) then x2|r, thus x2∈F(b,r)
What’s the implication?
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Euclid’s Algorithm
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Euclid’s Algorithm
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Flow Chart
Correctness
The algorithm is correct:
For any valid inputs
It terminates in a finite amount of time
And produces a correct output
Challenge:
Prove this at home
Must be your original proof
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Another Example
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Going Back
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Going Back
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Corollary
For any integer p,q (one of them is not 0)
If
h=hcf(p,q)
then
there exist integers x,y s.t. xp+qy=h
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Factoring
Finding divisors (factors) of a number is
called factoring or factorization
Consider 24:
24=1*24
24=(-1)*(-24)
24=1*2*12
24=(-1)*(-2)*12
24=(-1)*2*(-12)
24=(-1)*4*2*(-3)
…
Is there a “canonical” representation?
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Fundamental Theorem Of Arithmetic
Any integer z≠0 can be represented as:
z = (-1)k p1k1 p2k2 … pnkn
where:
p1<…<pn are prime numbers
k is 0 or 1
k1,…,kn are natural >0
This factorization is unique
Examples:
24=(-1)0 23 31
-7=(-1)1 71
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Proof (existence)
Let’s prove that for any n∈Z, n>0 such a
representation exists
Steps:
Lemma 0. There are n integers between 1 and n
Lemma 1. Non-trivial factors of a natural n are
strictly less than n
Lemma 2. Every integer n>1 is divisible by a
prime number
Corollary: can get exclusively prime factors for
n>1
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Proof (uniqueness)
Let’s prove that such representation is unique
Steps:
Lemma 3. If h=hcf(p,q) then there exist integers x,y s.t.
xp+qy=h
Lemma 4. If p|ab and prime(p) then p|a or p|b
Proof:
Suppose not. Then two different factorizations exist
Then arrive at a contradiction
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Further Information
Some background:
http://mathworld.wolfram.com/EuclidsTheorems.html
How to discover the proof:
http://www.dpmms.cam.ac.uk/~wtg10/FTA.html
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Next
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Floor & Ceiling
Definitions
Different from the textbook’s
floor(x) = max{n∈Z st n≤x}
ceiling(x) = min{n∈Z st n≥x}
Examples
floor(5.75)
5
floor(-5.75)
-6
ceiling(5.75)
6
ceiling(-5.75)
-5
floor(x)
x
ceiling(x)
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Equivalence to text’s defs and more
Theorem
For any x∈R\Z, floor(x) and ceiling(x) are defined,
unique, and ceiling(x)=floor(x)+1
Proof
Part 1: existence
Part 2: uniqueness
Part 3: relationship
Corollary
∀x∈R [floor(x) ≤ x < floor(x)+1]
∀x∈R [ceiling(x)-1 < x ≤ ceiling(x)]
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Lemma
∀m∈Z floor(m)=ceiling(m)=m
∃x,y∈R ceiling(x+y)≠ceiling(x)+ceiling(y)
Example?
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Lemma
∀x∈R ∀m∈Z floor(x+m)=floor(x)+m
Proof?
Take floor(x) ≤ x,
Then floor(x)+m ≤ x+m
∀n∈Z floor(n/2)=n/2 iff n is even and
floor(n/2)=(n-1)/2 iff n is odd
Proof?
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Infinitude of Primes
There is no greatest prime:
∀n ∃m [ prime(n)  m>n & prime(m) ]
Theorem 3.7.4 in the book
Will prove three lemmas first…
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Lemma 0
If p|a and p|a+1 then p=1 v p=-1
Proof
direct
p|a  a=pn
p|a+1  a+1=pm
p(m-n)=1
p=+1 v p=-1
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Lemma 1
For any integer a and a prime p if p|a
then ~(p|a+1)
Proof
indirect
Suppose such prime p exists
p|a and p|a+1
Then by Lemma 0: p=+1 or p=-1
p cannot be prime
contradiction
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Lemma 2
A natural n>1 is not prime iff there is a prime p<n
such that p|n
Proof (direct):

If n is not prime then it has non-trivial divisors (proved
before)
Then one of them has a prime factor p (proved before)

Know that p<n and p|n
Then p is a non-trivial factor of n
Thus n is not prime
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Proof: Infinitude of Primes
Indirect (i.e., by contradiction)
Suppose not
Then
∃n ∀m [ prime(n) & (m≤n v ~prime(m)) ]
(*)
Thus, denote the only primes as p1, …, pk=n
Then consider m=p1 * … pk + 1, m not prime (supposition)
m>pi
m>n=pk
Is prime(m)?
None of the primes pi divides it (by lemma 1)
But there are no other prime numbers (by supposition)
Thus, m is a prime (by lemma 2)
Contradiction with (*)
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Irrationality of sqrt(2) - Lemma
If q∈Q then there exist n,m∈Z such that
q=n/m and hcf(n,m)=1
Proof
Suppose we have n/m=q and hcf(n,m)>1
Then compute n’, m’ :
n’=n/hcf(n,m)
m’=m/hcf(n,m)
q=n’/m’ and n’,m’ are less than n and m
Either hcf(n’,m’)=1 : then done
Or hcf(n’,m’)>1 : then repeat the process again
Must terminate since n’ and m’ are decreasing
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Another proof ?
How about another constructive proof?
Hint:
Fundamental theorem of arithmetic
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Irrationality of sqrt(2)
Define sqrt(x)=y such that ∃y∈R, y*y=x
Let’s prove that sqrt(2) is irrational
Proof
Indirect (by contradiction)
Suppose not: sqrt(2)=n/m and hcf(n,m)=1
Then 2=n2/m2, 2m2=n2
n2 is even  n is even (proved earlier)
Then 2m2=4k2
Then m2 is even and so m is even
Thus, hcf(n,m) is at least 2
Contradiction
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