oct 7, 2003© vadim bulitko : cmput 272, fall 2003, uofa1 cmput 272 formal systems logic in cs i. e....

Oct 7, 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


Quiz I (October 9Quiz I (October 9thth))Quiz I (October 9Quiz I (October 9thth))

Coverage:Everything up to and including October 7th

lecture:textbook lectures seminars

Focus on the overlapping part:textbook lectures

Format:Problems similar in format to the assignments50 minutes in-classOpen book, notes, etc.Calculators are allowedNo team-work of ANY kind


ABOUT QUIZ 1ABOUT QUIZ 1ABOUT QUIZ 1ABOUT QUIZ 1

Exams were around the corner. During a lecture in

mathematical analysis the students questioned the professor about the contents

of the forthcoming paper.

“It will contain some interesting problems,” he said. “Right now faculty members are busy working on one of them. If we solve it,

it will be included in the examination paper.”

Mathematics and Informatics Quarterly, Volume 6, No. 1, 1996


TodayTodayTodayToday

Binary relationsEquivalence classesFloor & ceilingProof by contradictionand contraposition

Infinitude of primes

Irrationality of sqrt(p)


Binary RelationsBinary RelationsBinary RelationsBinary Relations A 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


PropertiesPropertiesPropertiesProperties

Reflexivity

Irreflexivity

Symmetry

Asymmetry

Anti-symmetry

Transitivity


QuestionsQuestionsQuestionsQuestions


Equivalence RelationEquivalence RelationEquivalence RelationEquivalence Relation

A relation is an equivalence relation iffit is reflexive, symmetric, and transitive

Examples:= on numbers, sets, etc.

mod n on integers

“logic equivalence” on formulae

Counter examples:<


Congruence RelationCongruence RelationCongruence RelationCongruence Relation

Any two integers a,b are congruent modulo n iff remainder(a,n) = remainder(b,n)Alternatively: a,b are congruent modulo n iff n divides a - b

Notation a b (mod n)Examples:

14 21 (mod 7)

15 22 (mod 7)

NOT 14 15 (mod 7)

6 0 (mod 2)


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)]


Equivalence ClassesEquivalence ClassesEquivalence ClassesEquivalence ClassesSuppose Ris an equivalence relation on AThen we can define subsets of A in this way:

[x] = {aA | a R x}, 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: how many equivalence classes does (mod 7) have?


PartitioningPartitioningPartitioningPartitioning

A set A is partitioned into sets {Ai} iff

The union of all Ai equals to A

Sets Ai are disjoint (i.e., don’t intersect)

Examples:{a,b,c} = {a,b} {c}N={0,1,…,9} {10,…,19}…N= [0] [1]


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] partitioned N

Will it always be the case?

In other words: are equivalence classes always going to form a partition of the set?


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)


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 st a,bAi}is an equivalence relation on A

Proof structure:Show that R is reflexiveShow that R is symmetricShow that R is transitive


Floor & CeilingFloor & CeilingFloor & CeilingFloor & Ceiling

DefinitionsDifferent from the textbook’sfloor(x) = max{nZ st nx}, that is

the greatest integer less than or equal to xceiling(x) = min{nZ st nx}, that is

the smallest integer greater than or equal to x

Examplesfloor(5.75) = 5floor(-5.75) = -6ceiling(5.75) = 6ceiling(-5.75) = -5

xx ceiling(x)ceiling(x)floor(x)floor(x)


Equivalence to text’s defs and Equivalence to text’s defs and moremore

Equivalence to text’s defs and Equivalence to text’s defs and moremore

TheoremFor any xR\Z, floor(x) and ceiling(x) are defined, unique, and ceiling(x)=floor(x)+1

ProofPart 1: existencePart 2: uniquenessPart 3: relationship

CorollaryxR [floor(x) x < floor(x)+1]xR [ceiling(x)-1 < x ceiling(x)]


Lemmata Galore…Lemmata Galore…Lemmata Galore…Lemmata Galore…

mZ floor(m)=ceiling(m)=m

x,yR ceiling(x+y)ceiling(x)+ceiling(y)

xR mZ floor(x+m)=floor(x)+m

nZ floor(n/2)=n/2 iff n is even and floor(n/2)=(n-1)/2 iff n is odd


Types of ProofsTypes of ProofsTypes of ProofsTypes of Proofs

Many interesting statements are of the type:

n S(n)

Two primary proof methods:Direct

Take an arbitrary n, prove S(n), generalizeIf S(n) P(n)Q(n)

Then can prove ~Q(n)~P(n) instead

IndirectShow that n ~S(n) would lead to a contradiction


Contraposition & Contraposition & ContradictionContradiction

Contraposition & Contraposition & ContradictionContradiction

Suppose the statement to prove is:n [ P(n)Q(n) ]

Direct proof by contraposition:Take an arbitrary nShow that if ~Q(n) holds for that n then ~P(n) holds

Indirect proof (by contradiction):Assume P(n) and ~Q(n) hold for some nShow ~P(n)

Contradiction : cannot have P(n) and ~P(n)


IllustrationIllustrationIllustrationIllustration

If n2 is even then n is evenn [ P(n)Q(n) ]

Direct proof by contraposition:Assume ~Q(n) : n is not evenn is oddThen n=2k+1n2=4k2+4k+1n2 is odd : n2 is not even : ~P(n)


IllustrationIllustrationIllustrationIllustration

If n2 is even then n is evenn [ P(n)Q(n) ]

Indirect proof (by contradiction):Assume P(n) and ~Q(n)n2 is evenn is not even : n is oddThen n=2k+1n2=4k2+4k+1n2 is odd : n2 is even : contradiction


The third proofThe third proofThe third proofThe third proof

Theorem: n2 is even n is evenHow about a direct proof without contraposition?

ProofAssume n2 is even2 | n2

p|ab p|a v p|b (Euclid’s 1st theorem)2|n v 2|nThen n is even


Infinitude of PrimesInfinitude of PrimesInfinitude of PrimesInfinitude 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…


Lemma 0Lemma 0Lemma 0Lemma 0

If a|n and a|n+1 then a=1 v a=-1

Proofdirecta|n n=kaa|n+1 n+1=jaa(j-k)=1a=+1 v a=-1 (proved before)



For any integer a and any prime p if p | a then ~(p | a+1)

ProofindirectSuppose such prime p existsp|a and p|a+1Then by Lemma 0: p=+1 or p=-1p cannot be primecontradiction



A natural number 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|nThen p is a non-trivial factor of nThus n is not prime


Proof: Infinitude of Proof: Infinitude of PrimesPrimes

Proof: Infinitude of Proof: Infinitude of PrimesPrimes

Indirect (i.e., by contradiction)Suppose notThen

n m [ prime(n) & (mn v ~prime(m)) ] (*)

Thus, denote the only primes as p1, …, pk=n

Then consider m=p1 * …* pk + 1m>pi

m>n=pk

Is prime(m)?None of the primes pi divides it (by lemma 2)But there are no other prime numbers (by supposition)Thus, m is a prime (by lemma 1)This contradicts (*)


LemmaLemmaLemmaLemma

If qQ then there exist n,mZ such that q=n/m and gcd(n,m)=1Proof

Suppose we have n/m=q and gcd(n,m)>1Then compute n’, m’ :

n’=n/gcd(n,m)m’=m/gcd(n,m)

q=n’/m’ and n’,m’ are less than n and mEither gcd(n’,m’)=1 : then doneOr gcd(n’,m’)>1 : then repeat the process againMust terminate since n’ and m’ are decreasing


Another proof ?Another proof ?Another proof ?Another proof ?

How about another constructive proof?

Hint:Fundamental theorem of arithmetic


Irrationality of Irrationality of sqrt(2)sqrt(2)Irrationality of Irrationality of sqrt(2)sqrt(2)Define sqrt(x)=y such that yR, y*y=xLet’s prove that sqrt(2) is irrationalProof

Indirect (by contradiction)Suppose not: sqrt(2)=n/m and gcd(n,m)=1Then 2=n2/m2, 2m2=n2

n2 is even n is even (proved earlier)Then 2m2=4k2

Then m2 is even and so m is evenThus, gcd(n,m) is at least 2Contradiction


Irrationality of Irrationality of sqrt(p)sqrt(p)Irrationality of Irrationality of sqrt(p)sqrt(p)

Claim: for any prime p, sqrt(p) is irrational

How do we generalize the previous proof for this?

oct 7, 2003© vadim bulitko : cmput 272, fall 2003, uofa1 cmput 272 formal systems logic in cs i. e....

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