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Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 1
CMPUT 272CMPUT 272Formal Systems Logic in Formal Systems Logic in
CSCS
CMPUT 272CMPUT 272Formal Systems Logic in Formal Systems Logic in
CSCS
I. E. LeonardUniversity of Alberta
http://www.cs.ualberta.ca/~isaac/cmput272/f03
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 2
TodayTodayTodayToday
Chapter 3.4Chapter 3.4
Chapter 3.5Chapter 3.5
Chapter 3.6Chapter 3.6
Other things…Other things…
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are
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to t
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resp
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d h
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aca
dem
ic p
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s u
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the c
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“fair
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”.
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 3
AnnouncementsAnnouncementsAnnouncementsAnnouncements
Assignment #2 is due Thursday Oct 9.
Quiz #1 is on Thursday Oct 9.
3 Sample Midterms from Winter 2003 are on the course webpage (with solutions)
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 4
Theorem(Division Theorem(Division Algorithm)Algorithm)
Theorem(Division Theorem(Division Algorithm)Algorithm)
For any integer aFor any integer b>0There exist unique integers q and rSuch that
a=qb+r0r<b
q is the quotient when a divided by br is the remainder when a divided by b
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 5
QuestionsQuestionsQuestionsQuestions
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 6
Greatest Common Greatest Common DivisorDivisor
(Highest Common (Highest Common Factor)Factor)
Greatest Common Greatest Common DivisorDivisor
(Highest Common (Highest Common Factor)Factor)Consider integers a>0, b>0
Consider set F(a,b)={xZ s.t. x|a & x|b}xF(a,b) [ xa & xb ]So F(a,b) is nonempty and bounded above thus has a largest element (by the well-ordering principle)Call max F(a,b) greatest common divisor:
gcd(a,b)
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 7
ExampleExampleExampleExample
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 8
Euclid’s Algorithm IdeaEuclid’s Algorithm IdeaEuclid’s Algorithm IdeaEuclid’s Algorithm Idea
Works fine for small numbersWhat about gcd(4453,1314) ?Here is an idea:Lemma. a>b, a=qb+r then gcd(a,b)=gcd(b,r)Proof.
Consider F(a,b)={x st x|a & x|b}Consider F(b,r)={x st x|r & x|b}If x1F(b,r) then x1|a, thus x1F(a,b)
If x2F(a,b) then x2|r, thus x2F(b,r)
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 9
Euclid’s AlgorithmEuclid’s AlgorithmEuclid’s AlgorithmEuclid’s Algorithm
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 10
Euclid’s AlgorithmEuclid’s AlgorithmEuclid’s AlgorithmEuclid’s Algorithm
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 11
QuestionsQuestionsQuestionsQuestions
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 12
Another ExampleAnother ExampleAnother ExampleAnother Example
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 13
Going BackGoing BackGoing BackGoing Back
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 14
Going BackGoing BackGoing BackGoing Back
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 15
CorollaryCorollaryCorollaryCorollary
Corollary For any integers p,q not both 0, if d=gcd(p,q) then there exist integers x,y s.t.
I xp+qy=d
Proof: Done in class.
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 16
QuestionsQuestionsQuestionsQuestions
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 17
Fundamental Theorem Of Fundamental Theorem Of ArithmeticArithmetic
Fundamental Theorem Of Fundamental Theorem Of ArithmeticArithmetic
Any integer z0 can be represented as:z = (-1)k p1
k1 p2
k2
… pnkn
where: p1<…<pn are prime numbersk is 0 or 1k1,…,kn are >0
This factorization is unique
Examples:24=(-1)0 23 31
-7=(-1)1 71
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 18
Proof (existence)Proof (existence)Proof (existence)Proof (existence)
Let’s prove that for any nZ, n>0 such a representation exists
Steps:Lemma 0. There are n integers between 1 and nLemma 1. Non-trivial factors of a natural n are strictly less than nLemma 2. Every integer n>1 is divisible by a prime numberCorollary: can get exclusively prime factors for n>1
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 19
Proof (uniqueness)Proof (uniqueness)Proof (uniqueness)Proof (uniqueness)
Let’s prove that such representation is unique
Steps:Lemma 3. If d=gcd(p,q) then there exist integers x,y s.t. xp+qy=dLemma 4 (Euclid’s 1st Theorem).
If p|ab and prime(p) then p|a or p|b Main Proof:
Suppose not. Then two different factorizations existThen arrive at a contradiction
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 20
Further InformationFurther InformationFurther InformationFurther Information
Lecture notes (we reproduced some parts from):
http://www.mat.bham.ac.uk/P.J.Flavell/teaching/Foundation/LectureNotes/
Some background:http://mathworld.wolfram.com/EuclidsTheorems.html
How to discover the proof:http://www.dpmms.cam.ac.uk/~wtg10/FTA.html
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 21
QuestionsQuestionsQuestionsQuestions
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 22
DefinitionDefinitionDefinitionDefinitionA binary relation R from a set A to a set B is a subset of the Cartesian Product AB, that is,
a set of ordered pairs (a,b) with aA and b BA binary relation R on a set A is a subset of the Cartesian product AA, that is, ordered pairs of the form (a,b) where a A and b A
Instead of (x,y) R AAWe use the shorthand notation: x R y
Examples:3 < 4Angela likes Belinda
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 23
ExamplesExamplesExamplesExamples
Binary relations on numbers:Domain : R x R<, , , etc.
Binary relations on sets:Domain : 2U x 2U
, etc.
Binary relations on people:Domain : people x peopleFather of, son of, in love with, took C272 from prof.
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 24
Arrow DiagramsArrow DiagramsArrow DiagramsArrow Diagrams
A convenient way of illustrating relations:a R a, b R c, c R d
aa bb cc dd
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 25
ReflexivityReflexivityReflexivityReflexivity
A binary relation R is reflexive iff for any x holds x R x
Examples:=
Counter-examples:<
xx
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 26
IrreflexivityIrreflexivityIrreflexivityIrreflexivity
A binary relation R is irreflexive iff for any x holds NOT x R x
Counter-Examples:=
Examples:<
xx
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 27
QuestionQuestionQuestionQuestion
Is every binary relation either reflexive or irreflexive?
No
Challenge : find a natural counter example
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 28
SymmetrySymmetrySymmetrySymmetry
A binary relation R is symmetric iff for any x,y [x R y implies y R x]
Examples:=
Counter-examples:<
xx yy
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 29
AsymmetryAsymmetryAsymmetryAsymmetry
A binary relation R is asymmetric iff for any x,y [x R y implies NOT y R x]Counter-examples:
=
Examples:<
xx yy
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 30
QuestionQuestionQuestionQuestion
Is every binary relation either symmetric or asymmetric?
No
Challenge : find an original natural counter-example
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 31
Equivalence RelationEquivalence RelationEquivalence RelationEquivalence Relation
A relation is an equivalence relation iffit is reflexive, transitive, and symmetric
Examples:= on numbers, sets, etc.
mod z on integers
“logic equivalence” on formulae
Counter examples:<
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 32
RemaindersRemaindersRemaindersRemainders
Remainder: suppose n,k,z are integersDivide n by k get z as the remainder:
n=ak+z : remainder(n,k)=z
Examples:remainder(14,7)=0remainder(15,7)=1remainder(22,7)=1
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 33
Congruence RelationCongruence RelationCongruence RelationCongruence RelationAny two integers n,m are congruent modulo z iff remainder(n,z) = remainder(m,z)Alternatively: n,m are congruent modulo z iff z divides (n-m)
Notation n m (mod z)Examples:
14 21 (mod 7)
15 22 (mod 7)
NOT 14 15 (mod7)
6 0 (mod 2)
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 34
More CongruenceMore CongruenceMore CongruenceMore Congruence
(mod 2) defines the concept ofParity : even and odd numbers
(mod 1) defines the entire set
of integers Z, since for any integers
m and n, the difference m-n is always
divisible by 1
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 35
Congruence & Congruence & EquivalenceEquivalence
Congruence & Congruence & EquivalenceEquivalence
Theorem. For any integer n > 0 (mod n) is an equivalence relation
Properties to prove:Reflexive: for any a [a a (mod n)]
Symmetric: for any a,b [a b (mod n) b a mod(n)]
Transitive: for any a,b,c [a b (mod n) & b c (mod n)
a c (mod n)]
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 36
Equivalence ClassesEquivalence ClassesEquivalence ClassesEquivalence ClassesSuppose R is an equivalence relation over A x AThen we can define subsets of A in this way:
[x] = {aA | a R x} , where xA (a representative of the class)
Example:A=N
R is (mod 2)
[0] = {nN | n 0 (mod 2)} even numbers
[1] = {nN | n 1 (mod 2)} odd numbers
Question: For a positive integer k, how many equivalence classes does (mod k) have?
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 37
QuestionsQuestionsQuestionsQuestions
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 38
PartitioningPartitioningPartitioningPartitioningA Set A is partitioned into sets {Ai} iffThe union of all Ai equals to A, that is,
A = Ai
and the sets Ai are pairwise disjoint, that is,
if i k then Ai Ak = Examples:
{a,b,c} = {a,b} {c}N={0,1,…,9} {10,…,19}…N=[0] [1]
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 39
Equivalence Classes & Equivalence Classes & PartitioningPartitioning
Equivalence Classes & Equivalence Classes & PartitioningPartitioning
The last example demonstrated that (mod 2) with its two equivalence
classes [0] and [1] partitions N into the disjoint subsets of even and odd integersWill this always be the case?In other words: do equivalence classes always partition of a set into disjoint
subsets?
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 40
Theorem 1Theorem 1Theorem 1Theorem 1
For any set A and any equivalence relation R defined on it, the equivalence classes induced by R form a partition of A
Example:N= [0] [1] … [n-1] with respect to (mod n)
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 41
Theorem 2: ConverseTheorem 2: ConverseTheorem 2: ConverseTheorem 2: Converse
For any set A and any partition {Ai} of A the binary relation R defined by a R b iff a,bA & i such that a,bAi}is an equivalence relation on A
Proof structure:Show that R is reflexiveShow that R is symmetricShow that R is transitive
Oct 2, 2003 © Vadim Bulitko : CMPUT 272, Fall 2003, UofA 42
QuestionsQuestionsQuestionsQuestions