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IKI10600I: Discrete Mathematics IPropositional Logic

Adila A. Krisnadhi

Faculty of Computer Science, University of Indonesia

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General Introduction Propositions Examples

Propositional LogicOutline

1 General Introduction

2 Propositions

The BasicsCompound propositions

3 ExamplesTranslating English sentencesSystem specicationsBoolean searchesLogic puzzlesLogic and bit operations

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What is Discrete Mathematics?

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What is Discrete Mathematics?

discrete : consisting of distinct or unconnected elements;opposite of continuousdiscrete mathematics : mathematics of discrete objectstypical topics : logic, arithmetics, algorithms, numbertheory, counting, set theory, graph theory, etc.

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Why study Discrete Mathematics?

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Why study Discrete Mathematics?

Because it is needed for answering the following questions:

How many ways are there to choose a valid password on acomputer system?

Is there a link between two computers in a network?How can I identify spam e-mail messages?How can I encyrpt a message so that no unintendedrecipient can read it?What is the shortest path between two cities using atransportation system?How can a circuit that adds two integers be designed?How many valid Internet addresses are there?...

l d l

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List of Topics for Discrete Mathematics I

Logic: propositional, rst orderSets, functions, sequences and summationsNumber theoryMathematical induction

G l I d i P i i E l

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List of Topics for Discrete Mathematics I

Logic: propositional, rst orderSets, functions, sequences and summationsNumber theoryMathematical induction

Important emphasis

problem solving skills, especially on reasoning andconstructing mathematical proofs .


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The Basics

Propositions

Denitionproposition : a declarative sentence that has a denite truthvalue (either true or false, but not both)


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The Basics

Propositions


Propositions are typically represented using letters:p , q , r , s , p 1 , p 2 . . . , q 1 , q 2 , . . .


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The Basics

Propositions


Propositions are typically represented using letters:p , q , r , s , p 1 , p 2 . . . , q 1 , q 2 , . . .Possible truth values for propositions: true and false (youcan use T and F ; or 1 and 0, etc.)


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The Basics

Propositions


Propositions are typically represented using letters:p , q , r , s , p 1 , p 2 . . . , q 1 , q 2 , . . .Possible truth values for propositions: true and false (youcan use T and F ; or 1 and 0, etc.)An interpretation (truth assignment) : mapping that mapspropositional variables to their truth values.


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p p

The Basics

Propositions


Propositions are typically represented using letters:p , q , r , s , p 1 , p 2 . . . , q 1 , q 2 , . . .Possible truth values for propositions: true and false (youcan use T and F ; or 1 and 0, etc.)An interpretation (truth assignment) : mapping that mapspropositional variables to their truth values.

Example notation: p I = T means p is interpreted as true.Similarly, p I = F means p is false.


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p p

The Basics

Propositions: examples

ExampleJakarta is the capital of Indonesia.There are 19 new students of this class.The students are on time for today’s class.


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The Basics

Propositions: examples

ExampleJakarta is the capital of Indonesia.There are 19 new students of this class.The students are on time for today’s class.

Are the following propositions? Why?2 + 2 = 4.Study this subject well.x < 2.Blablabla.Do you know more examples of propositions?

I know more examples of propositions


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Compound propositions

Logical connectives

Propositions that you have encountered up to now is calledatomic proposition .If you have one or two propositions, you can form a new

proposition using a logical connective or logical operator .Such a proposition is called compound proposition .


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Logical connectives

Propositions that you have encountered up to now is calledatomic proposition .If you have one or two propositions, you can form a new

proposition using a logical connective or logical operator .Such a proposition is called compound proposition .There are two kinds of logical operator:

unary operator (needs only one operand): negation ( ¬)binary operator (needs two operands):

conjunction ( ∧)disjunction ( ∨)exclusive-or ( ⊕)implication/conditional ( →)biconditional ( ↔).


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Negation

DenitionLet p be a proposition. Then ¬p is also a proposition.

¬p is called the negation of p and read as “ not p ”.


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Negation

DenitionLet p be a proposition. Then ¬p is also a proposition.

¬p is called the negation of p and read as “ not p ”.

Important!

¬p has always the opposite meaning of p .

This can be described using the following truth table :

p ¬p T FF T


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Negation: examples

Examplep : “Jakarta is the capital of Indonesia”

¬p : “Jakarta is not the capital of Indonesia”;

or “It is not the case that Jakarta is the capital of Indonesia”q : “Today is Friday”

¬q : “Today is not Friday”or “It is not the case that today is Friday”


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Negation: examples

Examplep : “Jakarta is the capital of Indonesia”

¬p : “Jakarta is not the capital of Indonesia”;

or “It is not the case that Jakarta is the capital of Indonesia”q : “Today is Friday”

¬q : “Today is not Friday”or “It is not the case that today is Friday”

What is the negation of each of the following?Jeff never sleeps in class.Jakarta has more than 10 million inhabitants.


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Conjunction

DenitionLet p and q be propositions. Then p ∧q is also a proposition.p ∧q is called the conjunction of p and q , read as “ p and q ”.


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Conjunction

DenitionLet p and q be propositions. Then p ∧q is also a proposition.p ∧q is called the conjunction of p and q , read as “ p and q ”.

Important!p ∧q is true whenever both p and q are true .Otherwise, it is false.

p q p

∧

q

T T TT F FF T FF F F


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Conjunction: examples

Examplep : Today is Friday.q : It is raining today.p ∧q : Today is Friday and it is raining today.p : This class is interesting.q : Class participation is lacking.

p ∧

q : This class is interesting, but class participation islacking.


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Disjunction

DenitionLet p and q be propositions. Then p ∨q is also a proposition.p ∨q is called the disjunction of p and q , read as “ p or q ”.


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Disjunction

DenitionLet p and q be propositions. Then p ∨q is also a proposition.p ∨q is called the disjunction of p and q , read as “ p or q ”.

Important!p ∨q is true whenever at least one among p and q is true ,otherwise it is false.

p q p

∨

q T T TT F TF T TF F F


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Disjunction: examples

Examplep : Today is Friday.q : It is raining today.p

∨

q : Today is Friday or it is raining today.p : This job requires experience in web design.q : This job requires experience in web programming.p ∨q : This job requires experience in web design or webprogramming.


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Disjunction: examples

Examplep : Today is Friday.q : It is raining today.p

∨

q : Today is Friday or it is raining today.p : This job requires experience in web design.q : This job requires experience in web programming.p ∨q : This job requires experience in web design or webprogramming.

Important!The use of or in disjunction is inclusive , i.e., p ∨q is still truewhenever both p and q are true.For the exclusive or, see next slide.


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Exclusive Or

DenitionLet p and q be propositions. Then p ⊕q is also a proposition.p ⊕q is called the exclusive or of p and q , read as “ p xor q ”.


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Exclusive Or

DenitionLet p and q be propositions. Then p ⊕q is also a proposition.p ⊕q is called the exclusive or of p and q , read as “ p xor q ”.

Important!p ⊕q is true whenever exactly one among p and q is true ,otherwise it is false.

p q p

⊕

q T T FT F TF T TF F F


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Exclusive Or: examples

Example

p : I pass this course. q : I fail this course.p ⊕q : (Either ) I pass this course or I fail it.p : You can choose computer science.q : You can choose information systems.p

⊕

q : You can choose ( either ) computer science orinformation systems, ( but not both ).


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Implication/Conditional

DenitionLet p and q be propositions. Then p →q is also a proposition,called implication or conditional statement .p is called the hypothesis/antecedent/premise ;

and q is called the conclusion/consequence .


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DenitionLet p and q be propositions. Then p →q is also a proposition,called implication or conditional statement .p is called the hypothesis/antecedent/premise ;

and q is called the conclusion/consequence .

p →q is read as:if p , (then) q

p is sufcient for q q if p

q when p

a necessary condition for p is q

q unless ¬p

p implies q

a sufcient condition for q is p p only if q

q whenever p

q is necessary for p

q follows from p


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Important!p →q is true whenever both p and q are true; or when p isfalse.

Thus, p →q is only false when p is true and q is false.

p q p →q

T T T

T F FF T TF F T


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Implication: examples

Examplep : Jeff gets seasick. q : Jeff is on a boat.p →q : If Jeff gets seasick, then Jeff is on a boat.p : Tuesday is a holiday. q : 2 + 3 = 5.p →q : Tuesday is a holiday implies 2 + 3 = 5.

Try restate these examples using patterns in the previous slide.When do these implications become true?



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Implication: contrapositive, converse and inverse

DenitionContrapositive of p →q is the proposition ¬q → ¬p .Converse of p →q is the proposition q →p .

Inverse of p →q is the proposition ¬p → ¬q .



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What are the contrapositive, converse and inverse of theexamples in the previous slide?



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Given a conditional statement, which one is equivalent(always has the same truth value) to it? Its contrapositive,converse or inverse?



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Given a conditional statement, which one is equivalent(always has the same truth value) to it? Its contrapositive,converse or inverse?Prove it.



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p p p

Contrapositive, converse, inverse: truth table

contrapositive converse inverse

p q ¬p ¬q p →q ¬q → ¬p q →p ¬p → ¬q T T F F T T T TT F F T F F T TF T T F T T F F

F F T T T T T T



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p p p

Biconditional/bi-implication/equivalence

DenitionLet p and q be propositions. Then p ↔q is also a proposition,

called bi-implication or biconditional statement .



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Biconditional/bi-implication/equivalence

DenitionLet p and q be propositions. Then p ↔q is also a proposition,

called bi-implication or biconditional statement .

p ↔q is read as:p if and only if q

if p then q , and conversely.p is equivalent with q

p iff q

p is necessary and sufcient for q .p and q are equivalent



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Biconditional/Bi-implication: truth value

Important!p ↔q is true exactly when p and q have the same truthvalue.p

↔q is true when both p

→q and q

→p is true. See

truth table below.

p q p →q q →p (p →q )∧(q →p ) p ↔q T T T T T TT F F T F FF T T F F FF F T T T T



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Biconditional: examples

Examplep : You can take the ight. q : You buy a ticket.p ↔q : You can take the ight if and only if you buy a ticket.p : For you to win the contest.q : You have the only winning ticket.p

↔q : For you to win the contest, it is necessary and

sufcient that you have the only winning ticket.



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Truth tables of all logical connectives

negation conjunction disjunction exclusive or implication bi-implication

NOT AND OR XOR IMPLIES IFF

p q ¬p p ∧q p ∨q p ⊕q p →q p ↔q

T T F T T F T TT F F F T T F FF T T F T T T FF F T F F F T T



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Precedence of logical connectives

Operator Precedence

¬ 1

∧ 2

∨ 3

→ 4↔ 5

Use parantheses whenever necessary!

Examplep ∧q ∨r means (p ∧q )∨r

¬p ∧q means (¬p )∧q p ∨q →r means (p ∨q ) →r


Translating English sentences

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Translate the following sentence into logical expression:“You cannot ride the roller coaster if you are under 130 cm tallunless you are older than 16 years old.”



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Translate the following sentence into logical expression:“You cannot ride the roller coaster if you are under 130 cm tallunless you are older than 16 years old.”

AnswerLet p : “You can ride the roller coaster”,q : “You are under 130 cm tall”, andr : “You are older than 16 years old”.Then the sentence can be translated into (q ∧¬r ) → ¬p


System specications

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System specications are expected to be consistent , i.e., they should notcontain conicting requirements that can be used to derive a contradiction.


System specications

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Are the following specications consistent?The system is in multiuser state if and only if it is operating normally. If thesystem is operating normally, the kernel is functioning. The kernel is notfunctioning or the system is in interrupt mode. The system is not in interruptmode.


System specications

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Are the following specications consistent?The system is in multiuser state if and only if it is operating normally. If thesystem is operating normally, the kernel is functioning. The kernel is notfunctioning or the system is in interrupt mode. The system is not in interruptmode.

AnswerSuppose p : “The system is in multiuser state”, q : “The system is operatingnormally”, r : “The kernel is functioning” and s : “The system is in interruptmode”. Thus the specications can be written as: p ↔ q , q → r , ¬ r ∨ s , ¬s .

We want to nd a truth assignment for p , q , r , s that makes all statementstrue. First, s must be false to make ¬s true. Thus, to make ¬ r ∨ s true, r mustbe false. This means, q must also be false in order to make q → r true.Finally, p should be false to make p ↔ q true. Hence, the specications areconsistent because they are all true when p , q , r , s are all false. (Check thisusing truth table).


Boolean searches

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Boolean searches

Logical connectives are used for searching in large collections ofinformation. E.g., in search engine.

AND: match records containing both two search terms

OR: match records containing at least one of the two search terms

NOT: exclude records containing the search term.

ExampleWe can search pages that contains the term “universitas”, either the

term “Indonesia” or “Bogor”, but does not contain the term “Semarang”.search term: UNIVERSITAS AND (INDONESIA OR BOGOR) ANDNOT SEMARANG

in Google: universitas indonesia OR bogor -semarang


Logic puzzles

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Inhabitants of an island can be divided into the knights and theknaves. Knights always tell the truth and knaves always lie. Suppose

you meet two inhabitants of this island, say A and B. A says “At leastone of us is a knave” and B says nothing. Can you tell who’s theknight and/or who’s the knave?


Logic puzzles

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Inhabitants of an island can be divided into the knights and theknaves. Knights always tell the truth and knaves always lie. Suppose

you meet two inhabitants of this island, say A and B. A says “At leastone of us is a knave” and B says nothing. Can you tell who’s theknight and/or who’s the knave?

AnswerLet p : “A is a knight” and q : “B is a knight”, so that

¬p means “A is a

knave” and ¬q means “B is a knave”. We can thus express A’sstatement as ¬p ∨¬q .


Logic puzzles

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Inhabitants of an island can be divided into the knights and theknaves. Knights always tell the truth and knaves always lie. Supposeyou meet two inhabitants of this island, say A and B. A says “At leastone of us is a knave” and B says nothing. Can you tell who’s theknight and/or who’s the knave?


¬p means “A is a

knave” and ¬q means “B is a knave”. We can thus express A’sstatement as ¬p ∨¬q .First suppose p is true, i.e., A is a knight. Then A’s statement, i.e.,

¬p ∨¬q is true. It follows that ¬q must be true, i.e., B is a knave.


Logic puzzles

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¬p means “A is a


¬p ∨¬q is true. It follows that ¬q must be true, i.e., B is a knave.To ensure that this is the only answer, let’s take a look what happen if

we assume p as false. If p is false, then ¬p is true, i.e., A must be aknave. This means A’s statement, ¬p ∨¬q must be false. But tomake ¬p ∨¬q false, we require ¬p to be false too which is impossiblesince we already have that ¬p is true.


Logic puzzles

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¬p means “A is a


¬p ∨¬q is true. It follows that ¬q must be true, i.e., B is a knave.To ensure that this is the only answer, let’s take a look what happen if

we assume p as false. If p is false, then ¬p is true, i.e., A must be aknave. This means A’s statement, ¬p ∨¬q must be false. But tomake ¬p ∨¬q false, we require ¬p to be false too which is impossiblesince we already have that ¬p is true.Hence, we conclude that A is a knight and B is a knave.


Logic and bit operations

Bit ti

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Bit operation

Bit (Binary digit): a symbol with two possible values (0 and1).Computers represent information using bits.

Bit can be used to represent truth value: 1 represents trueand 0 represents false.Bit operations : analogous to operations using logicalconnectives (AND, OR, XOR).

Bit string : sequence of zero or more bits. Its length is thenumber of bits in the string.Example: 101010011 is a bit string of length nine.



Bitwise AND bitwise OR bitwise XOR

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Bitwise AND, bitwise OR, bitwise XOR

Find the bitwise AND, bitwise OR and bitwise XOR of the bitstrings 1011 0011 and 1110 0001.



Bitwise AND bitwise OR bitwise XOR

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Bitwise AND, bitwise OR, bitwise XOR

Find the bitwise AND, bitwise OR and bitwise XOR of the bitstrings 1011 0011 and 1110 0001.

Answer

1011 00111110 0001

−−−−−−−−−1010 0001 b i t w i s e AND

1111 0011 bi tw is e OR0101 0010 b i t w i s e XOR

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IKI10600I: Discrete Mathematics ILogical Equivalence

Adila A. Krisnadhi


Interpretation Propositional Equivalence

Logical Equivalence

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Logical EquivalenceOutline

4 Interpretation

5 Propositional Equivalence


Interpretation

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Interpretation

DenitionInterpretation is a truth assignment to a (possibly compound)proposition.

Interpretation can be described simply by truthassignments of all propositional variables of theproposition.Truth assignment of a compound proposition can then be

computed from truth assignments of its propositionalvariables.Each row of the truth table of a proposition corresponds toa particular interpretation of the proposition.


Truth tables of compound proposition

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Construct the truth table for (p ∨¬q ) →(p ∧q )

Interpretations

¬q p

∨¬q p

∧

q (p

∨¬q )

→(p

∧

q )

I 1 : p I 1 = T , q I 1 = T F T T T

I 2 : p I 2 = T , q I 2 = F T T F F

I 3 : p I 3 = F , q I 3 = T F F F T

I 4: p I 4 = F , q I 4 = F T T F F



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Construct the truth table for (p ∨¬q ) →(p ∧q )

p q ¬q p ∨¬q p ∧q (p ∨¬q ) →(p ∧q )T T F T T TT F T T F FF T F F F TF F T T F F

The compound proposition above used 2 variables. Its truthtable contains 4 rows. How many rows are there in the truthtable of a compound proposition that uses 3 variables? 4variables? n variables?


Validity, satisability and contradiction

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Validity, satisability and contradiction

DenitionLet G be a proposition.

G is valid iff G is true for every interpretation of G . In thiscase, G is also called a tautology .

G is satisable iff there exists at least one interpretation ofG for which G is true.G is falsiable iff there exists at least one interpretation ofG for which G is false.

G is unsatisable/contradictory iff G is false for everyinterpretation of G . In this case, G is also called acontradiction .

Note: A proposition that is both not valid and non contradictoryis also called contingency .


Equivalence

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Equivalence

DenitionLet p and q be two propositions. p is equivalent with q iff theproposition p ↔q is a tautology.

In other words, p and q are equivalent iff p and q have thesame truth value on every row of their truth tables.

Instead of using ↔, we often use ≡to denote equivalence.


Proving equivalence

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Proving equivalence

If you are given two propositions p and q , how can you

show that they are equivalent?There are two ways of proving two propositions equivalent:

using truth tables;using laws of equivalence.


Proving equivalence using truth table

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g q g

Show that p ∨q and ¬(¬p ∧¬q ) are equivalent (i.e.,p ∨q ↔ ¬(¬p ∧¬q ) is a tautology).

p q p

∨

q

¬p

¬q

¬p

∧¬q

¬(

¬p

∧¬q ) G

T T T F F F T TT F T F T F T TF T T T F F T T

F F F T T T F TNote: we dene G := p ∨q ↔ ¬(¬p ∧¬q )


Proving equivalence using truth table

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g q g

Show that p →q and ¬p ∨q are equivalent.

p q p →q ¬p ¬p ∨q

T T T F TT F F F FF T T T TF F T T T

Since p →q and ¬p ∨q have the same truth values on eachrow of the truth table, both of them are equivalent.


Some logical equivalences (1)

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g q ( )

p ∧T ≡p Identity lawsp ∨F ≡p p

∨

T

≡T Domination laws

p ∧

F ≡Fp ∨p ≡p Idempotent lawsp ∧p ≡p

¬(

¬p ) Double negation laws

p ∨

q ≡q ∨

p Commutative lawsp ∧q ≡q ∧p



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g q ( )

(p ∨q )∨r ≡p ∨(q ∨r ) Associative laws(p ∧q )∧r ≡p ∧(q ∧r )

p ∨(q ∧r ) ≡(p ∨q )∧(p ∨r ) Distributive laws

p ∧(q ∨

r ) ≡(p ∧

q )∨(p ∧

r )¬(p ∧q ) ≡ ¬p ∨¬q De Morgan’s laws

¬(p ∨q ) ≡ ¬p ∧¬q p ∨(p ∧q ) ≡p Absorption laws

p ∧(p ∨

q ) ≡p p ∨¬p ≡T Negation lawsp ∧¬p ≡F



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p →q ≡ ¬p ∨q p →q ≡ ¬q → ¬p

¬(p →q ) ≡p ∧¬q

(p →q )∧(p →r ) ≡p →(q ∧

r )(p →r )∧(q →r ) ≡(p ∨q ) →r (p →q )∨(p →r ) ≡p →(q ∨r )(p →r )∨(q →r ) ≡(p ∧q ) →r

p ↔q ≡(p →q )∧(q →p )p ↔q ≡ ¬p ↔ ¬q p ↔q ≡(p ∧q )∨(¬p ∧¬q )

(p q ) p q


Proving equivalence without truth table

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Show that ¬(p ∨(¬p ∧

q )) and ¬p ∧¬q are equivalent

Answer

¬(p ∨(¬p ∧q )) ≡ ¬p ∧¬(¬p ∧q ) by De Morgan laws

≡ ¬p

∧

(

¬(

¬p )

∨¬q ) by De Morgan laws

≡ ¬p ∧(p ∨¬q ) by double negationlaw

≡(¬p ∧p )∨(¬p ∧¬q ) by distributivity

≡F

∨

(

¬p

∧¬q ) because

¬p

∧

p

≡F

≡(¬p ∧¬q )∨F by commutativity ofdisjunction

≡ ¬p ∧¬q by identity law for F


Proving tautology without truth table

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Show that (p ∧

q ) →(p ∨

q ) is a tautology.

AnswerWe show that the statement is equivalent to T.

(p ∧

q ) →(p ∨

q ) ≡ ¬(p ∧

q )∨(p ∨

q ) since p →q ≡ ¬p ∨

q ≡(¬p ∨¬q )∨(p ∨q ) by De Morgan laws

≡(¬p ∨p )∨(¬q ∨q ) by associativity and

commutativity of ∨

≡T∨

T by commutativity of∨and since p ∨¬p ≡T

≡T by domination law

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IKI10600I: Discrete Mathematics ILogical Equivalence: continued

Adila A. KrisnadhiFaculty of Computer Science, University of Indonesia

Logical Equivalence: continuedOutline

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Outline

Proving tautology without truth table

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Show that (¬p ∨ ¬q ) → ¬(p ∧ q ) is a tautology.

AnswerTo show that the statement is valid, we assume that there exists aninterpretation making it false, and derive a contradiction.

Assume that (¬p ∨ ¬q ) → ¬(p ∧ q ) is false.

It must be the case that (¬p ∨ ¬q ) is true and ¬(p ∧ q ) is false.

Since ¬(p ∧ q ) is false, p ∧ q must be true.

Since p ∧ q is true, we have both p and q are true.

p that p and q are true implies that both ¬p and ¬q are false.

Since ¬p and ¬q are false, (¬p ∨ ¬q ) must be false.

Since we already have that (¬p ∨ ¬q ) is true, we obtain a contradiction.

Therefore, our assumption that (¬p ∨ ¬q ) → ¬(p ∧ q ) is false isincorrect, i.e., (¬p ∨ ¬q ) → ¬(p ∧ q ) is always true.

More examples

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Determine whether the following statements are valid, orcontradictory or neither.

(p →q ) →(¬p → ¬q )¬(p ↔q ) ↔(p ↔ ¬q )p ∧(p →q ) →q ((p →q )∧(q →r )) →(p →r )

(p ∨

q ) →(p ∧

q )

Determining validity of a proposition: summary

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Use a truth table:valid if the proposition is true on each row of the truth table;straightforward; difcult if there are too many variables (2 n

rows for n variables)

Use logical equivalence lawsUse a contradiction:

Assume that the proposition is falseDerive a contradictory truth assignment for the

variables/subpropositionsDue to contradiction, the initial assumption is incorrect, i.e.,the proposition is always true.

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IKI10600I: Discrete Mathematics IPredicates and Quantiers

Adila A. KrisnadhiFaculty of Computer Science, University of Indonesia

Predicates Quantiers

Predicates and QuantiersOutline

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6 Predicates

7 QuantiersUniversal quantierExistential quantier

More examples and remarks


Predicates

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The statement “ x > 3” or “x is greater than 3” consists oftwo parts:variable x whose possible values are taken from a certainset D predicate “is greater than 3”

The set D is called the domain or the universe of discourse .We can denote such statement as P (x ) where P denotesthe predicate and x is the variable .P (x ) has no truth value until x is assigned to a certain

element from D .A predicate is always associated to a certain arity : thenumber of variables associated to P An n -ary predicate is a predicate with arity n .


Let P (x ) be the statement “ x > 3 ” What are the truth

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Let P (x ) be the statement “ x > 3.” What are the truthvalues of P (5) and P (1)?

P (5) is obtained by setting x = 5 in the statement. Hence,P (5) is true since 5 > 3 is true.P (1) is false since 1 > 3 is false.P is a unary predicate, i.e., a predicate with arity 1.

Let A(c , n ) be “Computer c is connected to network n ”where c represents a computer and n represents anetwork.

A(c , n ) is true whenever c is a computer that is connectedto the network n , otherwise it is false.A is a binary predicate, i.e., its arity is 2.

Let R (x , y , z ) be “x + y = z ”.R is a ternary predicate, i.e., its arity is 3.R (1, 2, 3) is true whereas R (0, 0, 1) is false.


Variable Binding

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Variable x in P (x ) is bound if x is replaced with an element

of the domain D or bound by a quantier.Otherwise it is a free variable.Two mainly used quantiers are universal quantiers andexistential quantier.


Universal quantier

Universal quantier

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DenitionThe universal quantication of P (x ) (∀xP (x )) is the proposition“P (x ) is true for all (every) values of x in the domain D ”

∀is the universal quantier and read “for all”, “for every”,

“for each”, “for any”, “for arbitrary”, “all of”, “given any”, etc.An element for which P (x ) is false is calledcounterexample of∀xP (x )

∀xP (x ) is true when P (x ) is true for every x ∀xP (x ) is false when there is an x for which P (x ) is false.In∀xP (x , y ), x is a bound variable whereas y is a freevariable.


Universal quantier

Universal Quantier: examples

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Let P (x ) be “x + 1 > x ”. What is the truth value of ∀xP (x )with R as the domain ?Let Q (x ) be “x < 2”. What is the truth value of

∀

xQ (x ) with

R as the domain?What is the truth value of ∀xP (x ), where P (x ) denotes“x 2 < 10” and the domain consists of positive integers notexceeding 4?

Note: for a nite domain where x 1 , . . . ,x n are its elements,∀xP (x ) is equivalent with P (x 1)∧P (x 2)∧ · · ·∧P (x n )


Universal quantier


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“Every student in this class attends the course of Foundationsof Programming”


Universal quantier


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First alternative:Take D := {x | x is a student in this class }P

(x

) :=x attends the course of Foundations of

ProgrammingThen the statement can be written as ∀xP (x )


Universal quantier


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First alternative:Take D := {x | x is a student in this class }P (x ) := x attends the course of Foundations ofProgrammingThen the statement can be written as ∀xP (x )

Second alternative:Take as the domain: D := {x | x is a student }Q (x ) := x is in this class.P (x ) := x attends the course of Foundations ofProgrammingThen the statement can be written as ∀x (Q (x ) →P (x ))


Existential quantier


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DenitionThe existential quantication of P (x ) (∃xP (x )) is the proposition“P (x ) is true for some values of x in the domain D ”

∃is the existential quantier and read: “for some”, “there

exists”, “there is”, “at least one”, etc.

∃xP (x ) is true when there is at least one element x of D forwhich P (x ) is true.

∃xP (x ) is false when P (x ) is false for every x , i.e., noelement x of D for which P (x ) is true.

In∃xP (x , y ), x is a bound variable whereas y is a freevariable.



Existential Quantier: examples

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Let P (x ) be “x > 34”. What is the truth value of ∃xP (x )with R as the domain?Let Q (x ) be “x = x + 1”. What is the truth value of

∃

xQ (x )

with R as the domain?What is the truth value of ∃xP (x ), where P (x ) denotes“x 2 < 10” and the domain consists of positive integers notexceeding 4?

Note: for a nite domain where x 1 , . . . ,x n are its elements,∃xP (x ) is equivalent with P (x 1)∨P (x 2)∨ · · ·∨P (x n )



More examples

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“Everyone has a close friend”



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“Everyone has a close friend”Because friendship is a relation between two person, weneed two variables: x and y



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“Everyone has a close friend”Because friendship is a relation between two person, weneed two variables: x and y Take D :=

{x

|x is a person

}as the domain of both

variables x and y .



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{x

|x is a person


variables x and y .Take a predicate A(x , y ) := y is a close friend of x



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{x

|x is a person


variables x and y .Take a predicate A(x , y ) := y is a close friend of x The sentence becomes:

∀x ∈D (∃y ∈DA(x , y )) or ∀x (∃yA(x , y )) or ∀x ∃yA(x , y )



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{x

|x is a person


variables x and y .Take a predicate A(x , y ) := y is a close friend of x The sentence becomes:

∀x ∈D (∃y ∈DA(x , y )) or ∀x (∃yA(x , y )) or ∀x ∃yA(x , y )

The scope for∃

y is A(x , y ) in which x is still a free variable.The scope for ∀x is ∃yA(x , y ) in which x is no longer a freevariable.



More examples

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Change the following sentences into expressions in predicatelogic.

Jackie has exactly one close friend.Everyone has exactly one close friend.If someone is a woman and has a child, then she is amother.



Summary

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∀xP (x ) :is true iff P (x ) is true for every x ∈D ;is false iff P (x ) is false for some x ∈D .

∃xP (x ):

is true iff P (x ) is true for some x ∈

D ;is false iff P (x ) is false for every x ∈D .

∀x ∀yP (x , y ):is true iff P (x , y ) is true for every pair x ∈D x and y ∈D y is false iff P (x , y ) is false for some pair x ∈D x and y ∈D y

∀y ∀

xP (x , y ):is true iff P (x , y ) is true for every pair y ∈D y and x ∈D x is false iff P (x , y ) is false for some pair y ∈D y and x ∈D x



Summary (2)

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∀x ∃yP (x , y ):is true iff for every x ∈D x , there is one y ∈D y for whichP (x , y ) is true;is false iff there is one x

∈

D x such that P (x , y ) is false forevery y ∈D y .

∃x ∀yP (x , y ):is true iff there is one x ∈D x such that P (x , y ) is true forevery y ∈D y ;

is false iff for every x ∈

D x , there is one y ∈

D y for whichP (x , y ) is false.



Summary (3)

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∃x ∃yP (x , y ):is true iff there is (at least) one pair x ∈D x and y ∈D y suchthat P (x , y ) is true;is false iff for every pair x

∈D x and y

∈D y , P (x , y ) is false.

∃y ∃xP (x , y ):is true iff there is (at least) one pair y ∈D x and x ∈D y suchthat P (x , y ) is true;is false iff for every pair y

∈

D x and x

∈

D y , P (x , y ) is false.



More examples

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Here the domain of all variables are real numbers.Let Q (x , y ) be the relation x + y = y + x . What is the truthvalue of

∀

x

∀

yQ (x , y )?Let Q (x , y ) be the relation x + y = 0. What is the truthvalue of ∀x ∃yQ (x , y ) and ∃y ∀xQ (x , y )?Let Q (x , y , z ) be the relation x + y = z . What is the truthvalue of

∀

x

∀

y

∃

zQ (x , y , z ) and

∃

z

∀

x

∀

yQ (x , y , z )?



Some remarks

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∀x ∀yP (x , y ) ≡ ∀y ∀xP (x , y )

∃x ∃yP (x , y ) ≡ ∃y ∃xP (x , y )

¬∀xP (x ) ≡ ∃x ¬P (x ) (De Morgan’s)

¬∃xP (x ) ≡ ∀x ¬P (x ) (De Morgan’s)

∀x ∃yP (x , y ) is not equivalent with ∃y ∀xP (x , y )If the domain for x is empty , the statement:

∀xP (x ) is true , because we cannot nd anycounterexample for P (x ) (that makes P (x ) false);

∃xP (x ) is false , because we cannot nd any x in thedomain that makes P (x ) true.

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IKI10600I: Discrete Mathematics IRules of Inference

Adila A. Krisnadhi


Valid arguments Fallacies Rules of inference for quantied statements

Rules of InferenceOutline

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8 Valid argumentsModus PonensModus tollensSyllogismsOther rules of inference

ResolutionExamples

9 Fallacies

10 Rules of inference for quantied statementsRules on universal quantiersRules on existential quantiersExamplesUniversal modus ponens & modus tollens


Argument in Logic

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DenitionArgument : a sequence of propositions

All but the nal proposition is called premises , whereas thenal proposition is the conclusion .An argument is valid if the truth of all its premises impliesthat the conclusion is true.


Modus Ponens

Modus Ponens (Law of Detachment)

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Modus ponens

p p

→q

∴ q


Modus Ponens

Modus Ponens (Law of Detachment)

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Modus ponens

p p

→q

∴ q

I f you do a l l t he e xe rc is es , the n you w i l l suceed i n exams .You do a l l t he e xe rc is es .

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−∴ You w i l l succeed in exams .


Modus tollens

Modus Tollens

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Modus tollens

¬q p

→q

∴ ¬p


Modus tollens

Modus Tollens

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Modus tollens

¬q p

→q

∴ ¬p

I f you do a l l t he e xe rc is es , the n you w i l l suceed i n exams .You d id no t succeed in exams .

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−∴ You d id n ot do a l l th e e xe rc is es .


Syllogisms

Syllogisms

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Hypothetical syllogism

p →q q →r

∴ p →r


Syllogisms

Syllogisms

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Hypothetical syllogism

p →q q →r

∴ p →r

Disjunctive syllogism

p

∨

q

¬p ∴ q


Other rules of inference

Other rules

Addition

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p

∴ p ∨q



Other rules

Addition

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p

∴ p ∨q

Simplication

p ∧q

∴ p



Other rules

Addition

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p

∴ p ∨q

Simplication

p ∧q

∴ p

Conjunction

p q

∴ p ∧q


Resolution

Resolution

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p ∨

q

¬p ∨r

∴ q ∨r


Resolution

Resolution

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p ∨

q

¬p ∨r

∴ q ∨r

Resolution is the rule of inference for computer to do anautomatic reasoning.In the above denition, q ∨r is called the resolvent .In using resolution, hypotheses and conclusion must be

expressed as clauses .Clause: disjunction of variables or negations of variables.


Examples

Is the following argument valid?


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If √2 > 32 , then (√2)2 > ( 3

2 )2 . We know that √2 > 32 .

Consequently, (√2)2 = 2 > ( 32 )2 = 9

4 .


Examples



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If √2 > 32 , then (√2)2 > ( 3

2 )2 . We know that √2 > 32 .

Consequently, (√2)2 = 2 > ( 32 )2 = 9

4 .

AnswerLet p be the proposition “ √2 > 3

2 ” and q be the proposition“2 > 3

2 ”. The premises of the argument are p →q and p ,whereas the conclusion is q . Thus, the argument is validbecause it is constructed by using modus ponens, a validargument form.


Examples



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If √2 > 32 , then (√2)2 > ( 3

2 )2 . We know that √2 > 32 .

Consequently, (√2)2 = 2 > ( 32 )2 = 9

4 .

AnswerLet p be the proposition “ √2 > 3

2 ” and q be the proposition“2 > 3

2 ”. The premises of the argument are p →q and p ,whereas the conclusion is q . Thus, the argument is validbecause it is constructed by using modus ponens, a validargument form.However, one of its premises, √2 > 3

2 , is false. Consequently,we cannot conclude that the conclusion is true. Moreover, notethat the conclusion itself is false, because 2 < 9

4


Examples

Show that the hypotheses “If you send me an e-mail, then I will nish writingthe program”, “If you do not send me an e-mail, then I will go to sleep early”,and “If I go to sleep early, then I will wake up feeling refreshed” lead to theconclusion “If I do not nish writing the program, then I will wake up feeling

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refreshed.”


Examples


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refreshed.”SolutionLet p be the proposition “You send me an e-mail” q , the proposition “I willnish writing the program” r , the proposition “I will go to sleep early” s , theproposition “I will wake up feeling refreshed”


Examples


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The hypotheses are p → q , ¬p → r , r → s . The desired conclusion is¬q → s


Examples


f

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1 p → q Hypothesis

3 ¬p → r Hypothesis

5 r → s Hypothesis

6 ¬q → s


Examples


f h d ”

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2 ¬q → ¬p Contrapositive of (1)

3 ¬p → r Hypothesis


6 ¬q → s


Examples


f h d ”

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3 ¬p → r Hypothesis4 ¬q → r Hypothetical syllogism using ( 2) and (3)


6 ¬q → s


Examples


f h d ”

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3 ¬p → r Hypothesis4 ¬q → r Hypothetical syllogism using ( 2) and (3)


6 ¬q → s Hypothetical syllogism using ( 4) an d (5)


Examples

Sh th t th h th ( ) d i l th

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Show that the hypotheses (p ∧

q )∨r and r →s imply theconclusion p ∨s


Examples

Sh th t th h th ( q ) d i l th

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Show that the hypotheses (p ∧

q )∨r and r →s imply theconclusion p ∨s

Solution(p

∧

q )

∨

r can be rewritten as two clauses, p

∨

r and q

∨

r .r →s can also be rewritten as ¬r ∨s . Using resolution on thetwo clauses, p ∨r and ¬r ∨s , we can obtain the conclusionp ∨s .


Fallacy of afrming conclusion

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This is NOT a valid argumentIf you do every problem in this book, then you will learn discrete

mathematics. You learned discrete mathematics.Therefore, you did every problem in this book.


Fallacy of afrming conclusion

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mathematics. You learned discrete mathematics.Therefore, you did every problem in this book.

This argument is not valid because the proposition[(p →q )∧q ] →p is NOT a tautology.


Fallacy of denying the hypothesis

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mathematics. You did not do every problem in this book.Therefore, you did not learn discrete mathematics.


Fallacy of denying the hypothesis

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mathematics. You did not do every problem in this book.Therefore, you did not learn discrete mathematics.

This argument is not valid because the proposition[(p →q )∧¬p ] → ¬q is NOT a tautology.


Rules on universal quantiers

Universal instantiation

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∀x .P (x )∴ P (c )




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∀x .P (x )∴ P (c )

c is a particular member of the domain, given the premise

∀x .P (x )Example: from the statement “All women are wise”, we can

conclude “Lisa is wise”, provided that Lisa is a member ofthe domain of all women.



Universal generalization


P (c ) for an arbitrary c

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∴

∀x .P (x )






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∴

∀x .P (x )

Universal generalization is used when we show that

∀x .P (x ) is true by taking an arbitrary element c from the

domain, and showing that P (c ) is true.Arbitrary means that we have no control over c and cannotmake any assumptions about c other than it comes from

the domain.This rule is often used implicitly in mathematics.Also often used incorrectly, due to making unwarrantedassumptions about the arbitrary element c .


Rules on existential quantiers

Existential instantiation


∃x P (x )

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∃x .P (x )∴ P (c ) for some element c





∃x P (x )

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∃x .P (x )∴ P (c ) for some element c

This rule allows us to conclude that there is an element c in the domain for which P (c ) is true, if we know that

∃x .P (x ) is true.c is NOT arbitrary, but rather it must the c for which P (c ) istrue.Usually we have no knowledge over c , other that it exists.Because it exists, we may give it a name ( c ) and continueour argument.



Existential generalization

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P (c ) for some element c

∴

∃x .P (x )




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P (c ) for some element c

∴

∃x .P (x )

This rule is used to conclude that ∃x .P (x ) is true when weknow a particular element c with P (c ) true.


Examples

Show that the premises “Everyone in this discrete mathematicsclass also takes a course in foundations of programming”, and“Eva is a student is this class” imply the conclusion “Eva takes

a course in foundations of programming”.

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p g g


Examples



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SolutionLet D (x ) denote “ x is in this discrete mathematics class”, andC (x ) denote “ x takes a course in foundations of programming”.


Examples



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SolutionLet D (x ) denote “ x is in this discrete mathematics class”, andC (x ) denote “ x takes a course in foundations of programming”.The hypotheses are

∀x .(D (x ) →C (x )) and D (Eva ). The

desired conclusion is C (Eva )


Examples



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1∀x .(D (x ) →C (x )) Premise


Examples



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2 D (Eva )

→C (Eva ) Universal instantiation from (1)


Examples



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2 D (Eva )


3 D (Eva ) Premise


Examples



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2 D (Eva )


3 D (Eva ) Premise4 C (Eva ) Modus ponens from ( 2) and (3)


Universal modus ponens & modus tollens


Universal modus ponens

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∀x .(P (x ) →Q (x ))P (a ), where a is a particular element in the domain

∴ Q (a )

Universal modus tollens

∀x .(P (x ) →Q (x ))

¬Q (a ), where a is a particular element in the domain

∴ ¬P (a )



Example

Assume that “For all positive integers n, if n is greater than 4,

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Assume that For all positive integers n , if n is greater than 4,then n 2 is less than 2 n ” is true. Show that 100 2 < 2100 .



Example

Assume that “For all positive integers n, if n is greater than 4,

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Assume that For all positive integers n , if n is greater than 4,then n 2 is less than 2 n ” is true. Show that 100 2 < 2100 .

Solution

Let P (n ) denote “ n > 4” and Q (n ) denote “ n 2 < 2n ”. Thepremise can be represented by ∀n .(P (n ) →Q (n )) , where thedomain consists of all positive integers. Note P (100 ) is truebecause 100 > 4. It follows by universal modus ponens thatQ (n ) is true, namely 100 2 < 2100

IKI10600I: Discrete Mathematics IProofs

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Adila A. Krisnadhi


Basic Terminology of Mathematical Proof Proof methods

ProofsOutline

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11 Basic Terminology of Mathematical Proof

12 Proof methodsDirect proofsProof by contraposition


Another example (of previous lecture)


If Superman were able and willing to prevent evil, he would do so. IfSuperman were unable to prevent evil, he would be impotent; if he were

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unwilling to prevent evil, he would be malevolent. Superman does not preventevil. If Superman exists, he is neither impotent nor malevolent. Therefore,Superman does not exist.


Another example (of previous lecture)


If Superman were able and willing to prevent evil, he would do so. IfSuperman were unable to prevent evil, he would be impotent; if he werell l h ld b l l d

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unwilling to prevent evil, he would be malevolent. Superman does not preventevil. If Superman exists, he is neither impotent nor malevolent. Therefore,Superman does not exist.

AnswerSuperman does not prevent evil. By Modus Tollens, we obtain that Supermanis either unable or unwilling to prevent evil. If Superman is unable to preventevil, then he is impotent; similarly, if Superman is unwilling to prevent evil,then he is malevolent. Thus, by Modus Ponens, we have that Superman is

either impotent or malevolent. Since if Superman exists, he is neitherimpotent nor malevolent, by Modus Tollens, we conclude that Supermandoes not exist.Since the conclusion is derived by always using valid rules of inference, weknow that the argument is valid.


Basic Terminology I

Theorem, proposition, lemma : a statement that can be

shown to be true.Theorem is usually reserved for a statement that is

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Theorem is usually reserved for a statement that isconsidered somewhat important; sometimes it is refered asfact or result .

Propositions are typically less important than theorems;lemma are typically less important than propositions.Proof : a valid argument that establishes the truth of atheorem/proposition/lemma.Statements used in a proof can include:

axioms (postulates): statements assumed to be true;premise(s) of the theorem/proposition/lemma that we wantto prove;previously proven theorem/propositio n /lemma.


Basic Terminology II

Lemma (“small” theorem) are usually used to help provingother results (not as itself); complicated proofs are usually

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( ); p p yeasier to understand when they are proved using series oflemmas.

Corollary : a theorem that can be establisheddirectly/straightforwardly from a theorem that has beenproved.

Conjencture : a statement that is being proposed to be atrue statement; its truth is unknown until somebody gives it

a valid proof.


Theorem statement

Many (though not all) theorems assert that a propertyholds for all elements in a domain.

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In such cases, a universal quantier is usually omitted,although it is formally needed for a precise statement.In proving theorems, universal instantiation is often usedimplicitly.


Theorem statement


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In such cases, a universal quantier is usually omitted,although it is formally needed for a precise statement.In proving theorems, universal instantiation is often usedimplicitly.The theorem:“Ifx > y , where x and y are positive real numbers, thenx 2 > y 2”


Theorem statement


h l ll d

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actually means


Theorem statement

Many (though not all) theorems assert that a propertyholds for all elements in a domain.I h i l i i ll i d

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actually means“For all positive real numbers x and y , if x > y , thenx 2 > y 2”


Direct proofs

Direct proof

A direct proof of a statement p →q is constructed asfollows:

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Direct proofs

Direct proof


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rst, assume that p is true;


Direct proofs

Direct proof


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rst, assume that p is true;construct subsequent steps using rules of inference until ...


Direct proofs

Direct proof


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rst, assume that p is true;construct subsequent steps using rules of inference until ...q is obtained as true as the nal step.


Direct proofs

Direct proof


h i

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In intermediate steps, we are also allowed to use axioms,denitions and previously proven theorems.


Direct proofs

Direct proof

A direct proof of a statement p →q is constructed asfollows: h i

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In intermediate steps, we are also allowed to use axioms,denitions and previously proven theorems.A direct proof of a statement ∀x (P (x ) →Q (x )) isconstructed by showing that P (c ) →Q (c ) is true, where c

is an arbitrary element of the domain (and then applyinguniversal generalization).


Direct proofs

Prove the following theorems using direct proofsee the whiteboard for details; take notes if necessary!

Denition

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DenitionAn integer n is even if there exists an integer k such thatn = 2k ; and n is odd if there exists an integer k such thatn = 2k + 1.

TheoremIf n is an odd integer then n 2 is odd.


Direct proofs


TheoremIf m and n are both perfect squares, then then nm is also perfect square where an integer a is a perfect square if there is

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perfect square, where an integer a is a perfect square if there is an integer b such that a = b 2 .

TheoremThe sum of two odd integers is even.The square of an even number is an even number.

TheoremIf m + n and n + p are even integers, where m , n, and p are integers, then m + p is even.


Direct proofs


D i i

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DenitionA real number r is rational if there exists integers p and q withq

=0 such that r

=p

q ; otherwise it is called irrational .

TheoremThe sum of two rational numbers is rational.


Proof by contraposition


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A proof by contraposition for a statement p →q isconstructed as follows:




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assume that ¬q is true;




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assume that ¬q is true;construct subsequent steps using rules of inference until ...




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assume that ¬q is true;construct subsequent steps using rules of inference until ...

¬p is obtained as true as the nal step.



Prove the following theorems by contrapositionsee the whiteboard for details; take notes if necessary!

TheoremIf n is an integer and 3n + 2 is odd, then n is odd.

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TheoremIf n = ab, where a and b are positive integers, then a

≤√n or

b ≤√n.

TheoremIf n is an integer and n 2 is odd, then n is odd.

TheoremIf x is irrational, then 1/ x is irrational.

IKI10600I: Discrete Mathematics IProofs (2)

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Adila A. Krisnadhi


Proof Methods (2) Mistakes in Proof Proof methods (3)

Proofs (2)Outline

13 Proof Methods (2)Vacuous and Trivial ProofsProof by Contradiction

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Proof by ContradictionProofs of equivalenceCounterexamples

14 Mistakes in Proof

15 Proof methods (3)

Exhaustive proofProof by cases


Vacuous and Trivial Proofs

Vacuous proof

We can quickly prove that a conditional statement p →q is truewhen we know that p is false . Why?

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Vacuous proof

We can quickly prove that a conditional statement p →q is truewhen we know that p is false . Why?Because p →q must be true whenever p is false.

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p →q p



Vacuous proof


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p →q p

Show that the proposition P (0) is true, where P (n ) is “If n > 1,

then n 2 > n ” and the domain consists of all integers.



Vacuous proof


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p →q p

Show that the proposition P (0) is true, where P (n ) is “If n > 1,

then n 2 > n ” and the domain consists of all integers.

Vacuous proof

Note that P (0) is “If 0 > 1, then 0 2 > 0”. Then P (0) is true

because the hypothesis 0 > 1 is false.



Trivial proof

We can quickly prove that a conditional statement p →q is truewhen we know that q is true . Why?

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Trivial proof

We can quickly prove that a conditional statement p →q is truewhen we know that q is true . Why?Because p →q must be true whenever q is true (regardless ofthe hypothesis p ).

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Trivial proof


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Let P (n ) be “If a and b are positive integers with a ≥b , thena n

≥b n ” where the domain consists of all integers. Show that

P (0) is true.



Trivial proof


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Let P (n ) be “If a and b are positive integers with a ≥b , thena n

≥b n ” where the domain consists of all integers. Show that

P (0) is true.

Trivial proof

Note that P (0) is “If a

≥b , then a 0

≥b 0”. Because

a 0 = b 0 = 1, the conclusion of the statement P (0) is true.Hence, P (0) itself is trivially true. Note that the hypothesis“a ≥b ” was not needed in this proof.


Proof by Contradiction

Proof by contradiction: idea

The idea of proof by contradiction :Suppose we want to prove that a statement p is true .

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The idea of proof by contradiction :Suppose we want to prove that a statement p is true .Moreover, suppose we can show that there is a statementq such that ¬p →q is true AND q is a contradictorystatement

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statement.




The idea of proof by contradiction :Suppose we want to prove that a statement p is true .Moreover, suppose we can show that there is a statementq such that ¬p →q is true AND q is a contradictorystatement

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statement.Then, we can conclude that ¬p is false , which means that

p is true .




The idea of proof by contradiction :Suppose we want to prove that a statement p is true .Moreover, suppose we can show that there is a statementq such that ¬p →q is true AND q is a contradictorystatement.

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p is true .

How can we nd such a contradiction q that might help usprove that p is true in this way?




The idea of proof by contradiction :Suppose we want to prove that a statement p is true .Moreover, suppose we can show that there is a statementq such that ¬p →q is true AND q is a contradictorystatement.

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p is true .

How can we nd such a contradiction q that might help usprove that p is true in this way?

Because the statement r ∧¬r is always a contradictionwhenever r is a proposition, we can prove that p is true if wecan show that ¬p →(r ∧¬r ) is true for some proposition r .



Proof by contradiction: steps

Suppose that p is a proposition that we want to prove.A proof by contradiction of p is constructed as follows:

Fi t th t i t

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First, assume that ¬p is true.





First assume that p is true

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First, assume that ¬p is true.Derive statements which are true as consequences of thisassumption until ...





First assume that p is true

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First, assume that ¬p is true.Derive statements which are true as consequences of thisassumption until ...we obtain a statement that contradicts previously-derivedstatements, i.e., a statement r whose negation ¬r isalready derived previously.



Prove the following theorems by contradictionsee the whiteboard for details; take notes if necessary!

TheoremAt least four of any 22 days must fall on the same day.

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Theorem

√2 is irrational

TheoremIf 3n + 2 is odd, then n is odd.


Proofs of equivalence

Proof of equivalence

In order to show that a biconditional statement p ↔q is

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p ↔qtrue, we show that p →q and q →p are both true.



Proof of equivalence

In order to show that a biconditional statement p ↔q is

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p ↔qtrue, we show that p →q and q →p are both true.

This approach is valid because it is based on the tautology(p ↔q ) ↔[(p →q )∧(q →p )].



Prove the following theoremssee the whiteboard for details; take notes if necessary!

Theorem

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If n is a positive integer, then n is odd iff n 2 is odd.

TheoremIf n is a positive integer, then n is odd if and only if 5n + 6 is odd.



Proof of equivalence (more than 2 statements)

In order to show that the propositions p 1 , p 2 , . . . ,p n are

equivalent, we show that p 1 ↔p 2 ↔. . . ↔p n is true.One way to show this, is by showing that all conditionalstatements p 1 →p 2 , p 2 →p 3 , . . . , p n →p 1 are true.

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The idea is so that we can establish any chain ofconditional statements from any one of the proposition toany other propositions.

Show the following statements are equivalent1 n is even.2 n −1 is odd.3 n 2 is even.


Counterexamples

Counterexamples

In order to show that a statement of the form ∀xP (x ), we need onlynd a counterexample , i.e., an example x for which P (x ) is false.

Using counterexampleDisprove the statement “Every positive integer is the sum of thesquares of two integers”.

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Counterexamples

Counterexamples

In order to show that a statement of the form ∀xP (x ), we need onlynd a counterexample , i.e., an example x for which P (x ) is false.

Using counterexampleDisprove the statement “Every positive integer is the sum of thesquares of two integers”.

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AnswerTo disprove the universally quantied statement above, we look for acounterexample which is a particular integer that is NOT the sum ofthe squares of two integers. For this statement, a counterexample iseasy to nd. One such counterexample is 3. To show that this is the

case, not that the only two perfect squares not exceeding 3 are02 = 0 and 1 2 = 1. It is easy to see that there is no way to get 3 asthe sum of two terms each of which is 0 or 1. Thus 3 is acounterexample of the statement above.


Fallacy: arming the conclusion

Can you nd what is wrong in the following “proof” of the statement “Ifn 2 is positive, then n is positive”?

Supposed ProofSuppose that n 2 is positive. Because the statement “If n is positive,then n 2 is positive” is true, we can conclude that n is positive.

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Fallacy: arming the conclusion

Can you nd what is wrong in the following “proof” of the statement “Ifn 2 is positive, then n is positive”?

Supposed ProofSuppose that n 2 is positive. Because the statement “If n is positive,then n 2 is positive” is true, we can conclude that n is positive.

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This proof is wrong . Let P (n ) be “n is positive” and Q (n ) be “n 2

is positive”.

Then the statement that we want to prove is ∀n (Q (n ) →P (n )) .Take Q (n ) is the hypothesis.

In the supposed “proof”, we make use of the statement

∀n (P (n ) →Q (n )) .

But from Q (n ) as the hypothesis and the statement

∀n (P (n ) →Q (n )) , we cannot conclude P (n ), since there is novalid rule inference that is usable here.


Fallacy: begging the question (circular reasoning)

Can you nd what is wrong with the following “proof” of thetheorem “ n 2 is an even integer implies n is an even integer”?

Supposed proofSuppose that n 2 is even. Then n 2 = 2k for some integer k . Letn = 2 for some integer . This shows that n is even.

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g


Fallacy: begging the question (circular reasoning)

Can you nd what is wrong with the following “proof” of thetheorem “ n 2 is an even integer implies n is an even integer”?

Supposed proofSuppose that n 2 is even. Then n 2 = 2k for some integer k . Letn = 2 for some integer . This shows that n is even.

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The proof is wrong , because the statement “let n = 2 forsome integer ” occurs in the proof.No argument has been given to show that n can indeed bewritten as 2 for some integer .This is circular reasoning because this statement isequivalent to the statement being proved, i.e., “ n is even”.Note that the theorem itself is actually correct (can yougive a valid proof of it?), but the metho d of proof is wrong.


Exhaustive proof

Some statement can be proved by examining ALL possibleexample. Such proofs is called exhaustive proof .This method is suitable only when the number of examplesare relatively small.

Prove the following resultThe only consecutive positive integers not exceeding 100 thatare perfect powers are 8 and 9 (An integer n is a perfect power

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are perfect powers are 8 and 9. (An integer n is a perfect powerif there exists two integers m and k > 1 such that n = m k ).


Exhaustive proof

Some statement can be proved by examining ALL possibleexample. Such proofs is called exhaustive proof .This method is suitable only when the number of examplesare relatively small.

Prove the following resultThe only consecutive positive integers not exceeding 100 thatare perfect powers are 8 and 9. (An integer n is a perfect power

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are perfect powers are 8 and 9. (An integer n is a perfect powerif there exists two integers m and k > 1 such that n = m k ).

An exhaustive proof of this statement can be done byconstructing a table consisting all possible example ofnumbers not exceeding 100 and of the form m k .

This method is suitable only because the size of the tableis relatively small.If you must prove the result for integers not exceeding1000, this method would be too difcu lt a nd te dious .


Proof by cases

Proof by casessee the whiteboard for details; take notes if necessary!

A proof by cases must cover ALL possible cases that arise

in a theorem.This method works since the statement(p 1∨p 2∨. . .∨p n ) →q is equivalent to(p q )∧(p q )∧ ∧(p q )

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(p 1 →q )∧(p 2 →q )∧. . .∧(p n →q ).

TheoremIf n is an integer, then n 2 ≥n.

TheoremFor every real numbers x and y, |xy | = |x ||y |, where |a | = a if a ≥0 and |a | = −a if a < 0.

IKI10600I: Discrete Mathematics ISet

Adila A. Krisnadhi

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Set: basics Set operations

SetOutline

16 Set: basics

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17 Set operations


Set: denition

DenitionA set is an unordered collection of distinct objects.

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DenitionObjects in a set are called elements or members of the set.Here, the set is said to contain its elements.


Set: notation

Sets are usually denoted by large capitals: A, B , X , Y , . . . ;if necessary with index.Elements of a set are usually denoted by small letters:

b if i h i d

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a , b , x , y , . . . ; if necessary with index.

x ∈

A denotes that x is an element of A, i.e., A contains x as one of its element.x /∈A denotes that x is not an element of A.

∅denotes the empty set.


Set: representation

Representation of sets:using listing:

{x 1 , x 2 , x 3 . . . ,x n }for sets with nite number of elements

{x 1 , x 2 , x 3 . . .}for sets with innite number of elements

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{ }using description (set builder notation) where P is apredicate over elements of the set:

{x : P (x )}or {x | P (x )}{x ∈S : P (x )}or {x ∈S | P (x )}where S is another set inthe context of the discussion, restricting the elements of thedenoted set (sometimes called a universal set ).


Set: examples

The set of natural numbers N = {0, 1, 2, 3, . . . }(Note: sometimes people dene natural numbers starting from 1, not 0)

The set of integersZ = {0, 1, − 1, 2, − 2, . . . } = {. . . , − 2, − 1, 0, 1, 2, . . . }The set of positive integers Z + = {1, 2, 3, . . . }The set of rational numbers Q = {p / q | p ∈Z , q ∈Z , and q = 0}

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Q {p q | p , q , q }

The set of real numbersR

A = {x ∈N | x 2 − 3x + 2 = 0}, set of natural numbers x such thatx 2 − 3x + 2 = 0. A can also be written as {1, 2}.

The set of positive even numbers is written as {2x | x ∈N} or{x | x = 2y , y ∈N} or {x ∈N | x = 2y , y ∈N}.

The set B = {x ∈N | 6 < x < 9} can also be written as B = {7, 8}The set B = {7, 8} can also be written as {x ∈N | 40 < x 2 < 80 } or{x ∈N | x 2 − 15 x + 56 = 0} or {7 + x ∈N | x = 0 or x = 1}


Set equality and subsets

DenitionTwo sets A and B are equal (written A = B ) iffA and B have the same

elements. Otherwise, they are not equal and written as A = B .

A = B iff∀x (x ∈A ↔x ∈B ) is true.

Denition

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The set A is a subset of B (written A

⊆

B ) iff every element of A isalso an element of B .

Here, we also say that B is a superset of A (written B ⊇A)

A⊆B iff∀x (x ∈A →x ∈B ) is true.

DenitionThe set A is a proper subset of B iff A⊆B and A = B


Set equality and subsets (contd.)

{1, 3 , 5}= {3, 5, 1}(the order is not important)

{1, 3 , 5

} ⊆ {3, 5, 1

}{1, 3 , 5} ⊆N{1, 3 , 5} {3, 1}and {1 , 3, 5} {x ∈N | x is even }

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Theorem (proof is for exercise!)For every set S

∅ ⊆S S ⊆S

Theorem (proof is for exercise!)For every two sets A and B, A = B iff A⊆B and B ⊆A


Finite set and cardinality

Denition

Let S be a set.S is a nite set iff it has exactly n elements where n isnonnegative integer.Here, n is called the cardinality of S .

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The cardinality of S is denoted by |S |S is innite iff it is not nite.

Let A be the set of odd positive integer less than 10. ThenA is nite (its cardinality

|A

|is 5).

|∅|= 0.The set of positive integers is innite.


The power set

DenitionLet S be a set. The power set of S is the set of all subsets of S .The power set of S is denoted by P (S ) or sometimes 2 S .

The power set of the set {0, 1 , 2}isP ({0 1 2}) =

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P ({0 , 1, 2}) =

{∅, {0}, {1}, {2}, {0 , 1}, {0 , 2}, {1 , 2}, {0 , 1, 2}}The power set of the empty set ∅is P (∅) = {∅}.The power set of the set {∅}is P ({∅}) = {∅, {∅}}

TheoremIf a set S has n elements then its power set P (S ) has 2n

elements.


Ordered tuple

Sets are unordered ; we need different structure torepresent ordered collection.

DenitionThe (ordered) n-tuple (a 1 , a 2 , . . . ,a n ) is the ordered collectionwith a 1 as its rst element, a 2 as its second element, ..., and a n

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1 2

as its n th element.

2-tuples are usually called ordered pairsTwo n -tuples (a 1 , a 2 , . . . ,a n ) and (b 1 , b 2 , . . . ,b n ) are equaliff a i = b i for i = 1 , 2, . . . ,n (i.e., each corresponding pair oftheir elements are equal).In particular, two ordered pairs (a , b ) and (c , d ) are equaliff a = c and b = d .


Cartesian products

We can use Cartesian product to form a set of ordered pairsfrom two sets.

DenitionLet A and B be sets. The Cartesian product of A and B (denoted by A ×B ), is the set of all ordered pairs (a , b ) wherea ∈A and b ∈B . In other words

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∈ ∈

A ×B = {(a , b ) | a ∈

A∧

b ∈

B }If A = {1, 2, 3}and B = {2, 3}, thenA ×B = {(1, 2), (1, 3), (2, 2), (2, 3), (3, 2), (3, 3)}If A = {x | 2 ≤x ≤3}and B = {x | 1 ≤x ≤2 or 3 ≤x ≤4},then A ×B = {(x , y ) | 2 ≤x ≤3 and 1 ≤y ≤2 or 3 ≤y ≤4}A ×B = B ×A unless A = B or A = ∅or B = ∅


Relation

DenitionA relation R from a set A to a set B is a set of ordered pairs(a , b ) such that a is some element of a and b is some elementB . In other words, R ⊆A ×B .

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⊆ ×Let A = {1 , 2, 3}and B = {2 , 3}. R = {(1, 3), (3, 3)}is arelation from A to B .Let A = {1 , 2, 3}and B = {a , b , c }.R =

{(1 , a ), (1 , b ), (1, c )

}is a relation from A to B .


Cartesian products of more than two sets

DenitionThe Cartesian product of the sets A1 , A2 , . . . ,An , (denoted byA1 ×A2 ×· · ·×An ) is the set of ordered n -tuples(a 1 , a 2 , . . . ,a n ), where a i ∈Ai for i = 1, 2 , . . . ,n . In other words,

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A1 ×A2 ×·· ·×An = {(a 1 , a 2 , . . . ,a n ) | a i ∈

Ai for i = 1 , 2, . . . ,n }Let A = {0, 1}, B = {1, 2}and C = {0, 1, 2}.

A ×B ×C = {(0, 1, 0), (0, 1, 1), (0, 1, 2), (0, 2, 0), (0, 2, 1), (0, 2, 2),

(1, 1, 0), (1, 1, 1), (1, 1, 2), (1, 2, 0), (1, 2, 1), (1, 2, 2)}


Venn diagram

Let V = {a , e , i , o , u }be the set of all vowels. A Venn diagramf V i

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{ }for V is:


Set union, intersection, difference and complement

Denition

The union of the sets A and B isA∪B = {x | x ∈A∨x ∈B }.The intersection of the sets A and B isA ∩B = {x | x ∈A∧x ∈B }.

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∩ { | ∈∧ ∈}Two sets A and B are disjoint iff A ∩B = ∅.The difference of the sets A and B isA −B = A \ B = {x | x ∈A∧x /∈B }.A \ B is also called the complement of B with respect to A.

Assuming that we have a universal set S , the complementof a set A is AC = A = S \ A = {x | x /∈A}.


Venn diagrams of set operations

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Set operations: examples

Let A = {1, 3, 5}and B = {1 , 2, 4}A∪B = {1 , 3, 5, 2 , 4}A ∩B = {1}A \ B = {3, 5}

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\ { }B \ A = {2, 4}A and B are NOT disjoint. But A and {2, 4}are disjoint.If the universal set U = {x ∈N | 1 ≤x ≤10}, thenA = {2, 4, 6 , 7, 8, 9 , 10}


Set identitiesProofs of them are left as exercises

Identity laws: A∪∅= A and A ∩ U = A for a universal set U

Domination laws: A∪U = U for a universal set U and A ∩∅= ∅Idempotent laws: A∪A = A and A ∩ A = A.

Complementation law: A = A.

Complement laws: A∪A = U for a universal set U , and A ∩ A = ∅

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Commutative laws: A∪B = B ∪A and A ∩ B = B ∩ A.Associative laws: A∪ (B ∪C ) = ( A∪B ) ∪C andA ∩ (B ∩ C ) = ( A ∩ B ) ∩ C .

Distributive laws: A∪ (B ∩ C ) = ( A∪B ) ∩ (A∪C ) andA ∩ (B ∪C ) = ( A ∩ B ) ∪ (A ∩ C ).

De Morgan’s laws: A∪B = A ∩ B and A ∩ B = A∪B Absorption laws: A∪ (A ∩ B ) = A and A ∩ (A∪B ) = A.


Generalized unions and intersections

Due to associativity, A∪(B ∪C ) or (A∪B )∪C can bewritten as A

∪

B

∪

C . Similarly, A

∩(B

∩C ) and (A

∩B )

∩C

can be written as A ∩B ∩C .For n ≥3 sets A1 , A2 , . . . ,An , their intersection can bewritten as

n

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n

i = 1Ai = {Ai , i = 1 , 2, . . . ,n }= A1 ∩A2 ∩. . . ∩An

and their union can be written as

n

i = 1Ai = {Ai , i = 1 , 2, . . . ,n }= A1∪

A2∪. . .∪An

IKI10600I: Discrete Mathematics I

Functions, Sequences and Summations

Adila A. Krisnadhi

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Functions Sequences and Summations Cardinality of set: revisited

Functions, Sequences and SummationsOutline

18 Functions

19 Sequences and Summations

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SequencesSummations

20 Cardinality of set: revisited


Function: denitions

DenitionLet A and B be nonempty sets. A function f from A to B (writtenf : A →B ) is an assignment of each element of A to exactly oneelement of B .

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Functions are also called mappings or transformationsWe write f (a ) = b if a ∈A is assigned to a unique b ∈B A function can also be viewed as a relation from A to B thatcontains one and only one ordered pair (a , b ) for evey

element a ∈

A.


Domain, codomain and range

If f is a function from A to B , we say thatA is the domain of f

B is the codomain of f If f (a ) = b , we say thatb is the image of a a is the preimage of b

The range of f is the set of all images of elements of A.

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The range of f is always a subset of the codomain of f If f is a function from A to B , we say that f maps A to B .Two functions are equal if they have the same domain, thesame codomain and map elements of their common

domain to the same elements in their common codomain.Equality of functions can also be seen as equality of sets(recall the denition of function as relation).


More basics on functions

Let f 1 and f 2 be functions from A to R. Then

f 1 + f 2 is a function from A to R dened by(f 1 + f 2)(x ) = f 1(x ) + f 2(x )f 1 f 2 is a function from A to R dened by(f 1 f 2)(x ) = f 1(x )f 2 (x )

Let f be a function from A to B and S ⊆A. The image of S

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⊆under f is the subset of B that consists of the images ofelements of S , i.e.,

{t | ∃s ∈S (t = f (s ))}Note: we can also use a shorthand for this set:{f (s ) | s ∈S }


Increasing and decreasing functions

Let f : A →B be a function.

f is increasing iff∀

x , y ∈

A : x < y →f (x ) ≤ f (y )f is strictly increasing iff∀x , y ∈A : x < y →f (x ) < f (y )f is decreasing iff∀x , y ∈A : x < y →f (x ) ≥ f (y )f is strictly decreasing iff x , y A : x < y f (x ) > f (y )

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∀ ∈ →LemmaLet f : A →B be a function.

If f is strictly increasing, then f is increasing.

If f is strictly decreasing, then f is decreasing.


Surjective, injective and bijective functions

Let f : A

→B be a function.

f is surjective (onto) iff it satises∀

y ∈

B .∃x ∈

A : f (x ) = y f is injective (one-to-one) iff it satises

∀x 1∀x 2(f (x 1) = f (x 2 ) →x 1 = x 2)f is bijective (one-to-one correspondence) iff f is both

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surjective and injective.

TheoremLet f : A →B be a function. If f is either strictly increasing or strictly decreasing, then f is a one-to-one function.


Examples

Let f be a function from {a , b , c , d }to {1 , 2, 3, 4 , 5}withf (a ) = 4, f (b ) = 5, f (c ) = 1 and f (d ) = 3. Then f is

one-to-one, but not onto.Let f : Z →Z be a function such that f (x ) = x 2 . Then

f is not one-to-one, because e.g., f (1) = f (−1) = 1, but1 = −1;f is not onto, because there is no x Z such that f (x ) = 1

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∈ −f is not increasing and not decreasing. Why?Let f : Z →Z be a function such that f (x ) = x + 1. Then

f is one-to-one, because for every x 1 , x 2∈Z,x 1 + 1 = x 2 + 1 implies x 1 = x 2 .f is onto, because for every y

∈

Z, there is an x

∈

Z suchthat f (x ) = y by taking x = y −1.f is bijective. Why?f is strictly increasing. Why?


Inverse function

DenitionLet f : A →B be a bijective function. The inverse function of f isthe function f − 1 : B →A dened such that if f (x ) = y thenf − 1(y ) = x .

− 1

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Do not confuse f with the function 1 / f . They are NOTthe same.Inverse function of f only exists (is well-dened) if f isbijective. Why?If f is bijective, then f has an inverse function, i.e., f isinvertible.


Examples

Let f be the function from

{a , b , c

}to

{1, 2 , 3

}such that

f (a ) = 2, f (b ) = 3 and f (c ) = 1. Then f is invertible (why?)and the inverse of f is the function f − 1 from {1, 2, 3}to

{a , b , c }such that f − 1(1) = c , f − 1(2) = a , and f − 1(3) = b .Let f : Z →Z be such that f (x ) = x + 1. Then f is invertible(why?) and its inverse is the function f− 1 : such that

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(why?) and its inverse is the function f :Z →Z

such thatf − 1 (x ) = x −1.Let f : R →R be such that f (x ) = x 2 . Is f invertible? If it is,what is its inverse?Let f : R+

∪{0

} →R+

∪{0

}be such that f (x ) = x 2 . Is f

invertible? If it is, what is its inverse?


Function composition

DenitionLet f : B →C and g : A →B be functions. The composition of f and g is the function f ◦g : A →C dened as

(f ◦g )(x ) = f (g (x ))

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To nd (f ◦g )(a ), we rst apply g to a obtaining g (a ) andthen we apply f to g (a ) to obtain (f ◦g )(a ).For the function f ◦g to be well-dened, the range of g must be a subset of the domain of f . Why?


Examples

Let g be the function from {a , b , c }to itself such thatg (a ) = b , g (b ) = c and g (c ) = a . Let f be the functionfrom {a , b , c }to {1 , 2, 3}such that f (a ) = 3, f (b ) = 2 andf (c ) = 1. Then

(f ◦g ) is the function from {a , b , c }to {1, 2, 3}such thatf g a f g a 2, f g b f g b 1 and

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(f

◦g

)(a

) =f (g

(a

)) =2,

(f

◦g

)(b

) =f (g

(b

)) =1 and

(f ◦g )(c ) = f (g (c )) = 3Is g ◦ f well-dened? If it is, give the denition of g ◦ f .

Let f , g : Z →Z be functions such that f (x ) = 2x + 3 andg (x ) = 3x + 2. What is f ◦g ? What is g ◦ f ?


Some theorems on composition and inverse

f ◦g does not equal g ◦ f in general.For a function f : A →B , if f − 1 exists, then

f ◦ f − 1 = id A where id : A →A is the identity function on A,i.e., id (x ) = x ;

− 1

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f ◦ f = id B where id : B →B is the identity function on A,i.e., id (x ) = x ;If A = B , then f ◦ f − 1 = f − 1 ◦ f = id A(f − 1 )− 1 = f .


Sequences

Sequences

Denition (Informal denition)

A sequence is an ordered list of terms.

Denition (Formal denition)A sequence is a function from a subset of the set of integers

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(typically the set {0 , 1, 2, . . .}or {1, 2 , 3, . . .}) to a set S .

We use the notation {a n }to denote a sequence (DO NOTconfuse it with sets) where we can either describe the formof a n and/or list the terms of the sequence in order ofincreasing subscripts.


Sequences

Examples

The sequence {a n }where a n = 1/ n for n = 1 , 2, . . . is the

sequence 1 ,12 ,

13 ,

14 , . . . .A geometric progression is the sequence of the form

a , ar , ar 2 , . . . ,ar n , . . . with the initial term a and the ratio r are real numbers. For example:

1, 2, 4, 8, . . . .1 1 1

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1, 2 , 4 , 8 , . . . .1, −1, 1, −1, 1, . . . .An arithmetic progression is the sequence of the forma , a + d , a + 2d , . . . ,a + nd , . . . where the initial term a andthe common difference r are real numbers. For example:

−1, 3, 7, 11 , . . . .7, 4, 1, −2, . . . .


Sequences

Deducing formula for terms of a sequence

When you want to deduce a formula for the terms of asequence, try to answer the following questions:

Are there runs of the same value? I.e., does the samevalue occur many times in a row?Are terms obtained from previous terms by adding thesame amount or an amount that depends on the position in

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the sequence?Are terms obtained from previous terms by multiplying by aparticular amount?Are terms obtained by combining previous terms in a

certain way?Are there cycles among the terms?


Sequences

Examples

Find the formulae for the sequences:1, 1/ 2, 1/ 4, 1/ 8, . . . .1, 3, 5, 7, 9, . . . .1, −1, 1, −1, . . . .

Continue the following sequence with appropriate terms:

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1, 2, 2, 3, 3, 3, 4, 4, 4, 4, . . . .5, 11 , 17 , 23 , 29 , 35 , 41 , 47 , 53 , 59 , . . . .

Guess a simple formula for a n if the rst 10 terms of thesequence {a n }are1, 7, 25 , 79 , 241 , 727 , 2185 , 6559 , 19681 , 59047.


Summations

Summation

We are usually interested in summing a given sequence.Here, we use the summation notation, namely the sigma

notation .Given the terms a m , a m + 1 , . . . ,a n from the sequence {a n },we use the notation

n n n

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j = m a j , j = m a j , or m ≤ j ≤ n a j

to represent a m + a m + 1 + · · ·+ a n

In the notation, j is the index of summation , and we could

also use any other letter such as i or k .Also, m is the lower limit and n is the upper limit of thesummation.


Summations

Example

The sum of the rst 100 terms of the sequences {a n }where a n = 1/ n for n = 1, 2, 3 , . . . is

100 j = 1

1 j .

Calculate 5 j = 1 j 2 .

Calculate 8k = 4 (−1)k .

Index of summation can be shifted, e.g.,5 2 4 (k )2

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5 j = 1 j 2 = 4k = 0(k + 1)2

Prove this result: if a and r ar real numbers, and r = 0,then

n

j = 0

ar j =ar n + 1 − a

r − 1 if r = 1

(n + 1)a if r = 1


Summations

Nested summations and summations over set

Summations can be nested. E.g.:

4

i = 1

3

j = 1

ij =4

i = 1

(i + 2i + 3i )

=4

i=

1

6i

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i 1= 6 + 12 + 18 + 24 = 60


Summations

Nested summations and summations over set

Summations can be nested. E.g.:

4

i = 1

3

j = 1

ij =4

i = 1

(i + 2i + 3i )

=4

i=

1

6i

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i 1= 6 + 12 + 18 + 24 = 60

Given a set S , we can sum the values f (s ) for all members ofS . E.g.:

s ∈{0 , 2 , 4}s

2= 0

2+ 2

2+ 4

2= 20


Summations

Some useful summations

Provided r = 0:n

k = 0

ar k =ar n + 1 − a

r − 1 if r = 1(n + 1)a if r = 1

n

k = 0

(a + kd ) = ( n + 1)a +dn (n + 1)

2n

k ( + 1)

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k = 1

k = n (n + 1)2

n

k = 1

k 2 =n (n + 1)(2n + 1)

6

n

k = 1

k 3 = n 2 (n + 1)2

4


Cardinality of set: revisited

Recall that, cardinality of a set is dened as the number ofelements in the set.Two nite sets A and B are of the same cardinality if thenumber of elements in A is the same as the number of

elements in B ( hich can be co nted)

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elements in B (which can be counted).What if the set is innite? Can we extend the notion ofcardinality to innite sets?


Cardinality of a set: alternative denition

DenitionThe sets A and B have the same cardinality iff there existsa bijection from A to B .A set is countable iff it is either nite or has the same

cardinality with the set of positive integers ( Z+

)

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cardinality with the set of positive integers ( Z ).A set is uncountable iff it is not countable.When an innite set S is countable, we denote thecardinality of S (i.e., |S |) byℵ0 (read as “aleph null”)


Examples

Is the set of odd positive integers is countable? Explain.Is the set {−5 , −4 , . . . ,9, 10}countable? Explain.Is the set of all integers ( Z) is countable? Explain.

Is the set of positive rational numbers is countable?

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Is the set of positive rational numbers is countable?Explain.Is the set of real numbers is countable? Explain.


More examples

Is a subset of a countable set also countable?Suppose A and B are sets, A is uncountable, and A⊆B . IsB countable or uncountable? Prove it.

Suppose A is an uncountable set and B is a countable set

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Suppose A is an uncountable set and B is a countable set.Is A \ B countable or uncountable? Prove it.

IKI10600I: Discrete Mathematics IIntegers

Adila A. Krisnadhi


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The Integers Division Primes and Greatest Common Divisor

Division

Division

From your previous experiences, you should already knowsome properties of integers: on addition, multiplication, etc.We are going to look at an operation on integers in moredetails, namely the division .When an integer is divided by another, nonzero integer, theresult (i e quotient) may or may not be an integer E g



result (i.e., quotient) may or may not be an integer. E.g.,12/4 = 3, whereas 11/4 = 2.75. So, if we require thedivision operation on integers always to give an integer, weneed the following denition


Division

Division

DenitionLet a and b be integers with a = 0.

Then we say, a divides b (written as a | b ) if there is aninteger c such that b = ac .If a | b we say that a is a factor of b and b is a multiple of a

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For (1) only, the rest are for exercise.Suppose a | b and a | c . Then, by denition of division, thereare integers s and t such that b = as and c = at . Hence,

b + c = as + at = a (s + t )

Therefore, a divides b + c , i.e., a | (b + c ).


Division

Division: some results

TheoremIf a,b and c are integers such that a | b and a | c, then a | mb + nc for every integers m and n.

Proof.

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Proof.Left as exercise. (Hint: use the rst part of the previoustheorem).


The Division Algorithm

The division algorithm

TheoremLet a be an integer and d a positive integer. Then there are

unique integers q and r, with 0 ≤ r < d such that a = dq + r .

The proof of this theorem uses the so-called well-orderingproperty of nonnegative integers (optional part of thiscourse).In the above theorem d is called the divisor a is called the

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In the above theorem, d is called the divisor , a is called thedividend , q is called the quotient, and r is called theremainder .The notation used to express the quotient and remainder:

q = a div d , r = a mod d



Example

What are the quotient and remainder when 101 is dividedby 11?What are the quotient and remainder when -11 is dividedby 3.

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Example

What are the quotient and remainder when 101 is divided

by 11?What are the quotient and remainder when -11 is dividedby 3.By the previous denition, remainder cannot be negative.

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Example

What are the quotient and remainder when 101 is divided

by 11?What are the quotient and remainder when -11 is dividedby 3.By the previous denition, remainder cannot be negative.

Theorem

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TheoremAn integer a is divisible (i.e., can be divided) by an integer d if and only if

a mod d = 0


Modular Arithmetic

Modular arithmetic

DenitionLet a and b be integers, and m a positive integer. Then a iscongruent to b modulo m — written as a ≡b ( mod m ) iff

m | a −b

If a and b are not congruent modulo m , we write

a ≡b ( mod m )

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≡ ( )


Modular Arithmetic

Modular arithmetic

DenitionLet a and b be integers, and m a positive integer. Then a is

congruent to b modulo m — written as a ≡b ( mod m ) iff

m | a −b

If a and b are not congruent modulo m , we write

a ≡b ( mod m )

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≡TheoremLet a and b be integers and m a positive integer. Then

a ≡b ( mod m ) iff a mod m = b mod m

Proof of this theorem is left as an exercise.


Modular Arithmetic

Example

Determine whether 17 is congruent to 5 modulo 6.Determine whether 24 and 14 are congruent modulo 6.

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Modular Arithmetic

Modular arithmetic and ordinary arithmetic

TheoremLet m be a positive integer. The integers a and b are congruent

modulo m iff there is an integer k such that a = b + km

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Modular Arithmetic

Modular arithmetic and ordinary arithmetic

TheoremLet m be a positive integer. The integers a and b are congruent

modulo m iff there is an integer k such that a = b + km

Proof.

If a ≡b ( mod m ), then m | (a −b ). Hence, there is an integerk such that a −b = km . This means a = b + km .

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|Conversely, if there is an integer k such that a = b + km , thenkm = a −b . This implies that m divides a −b , which meansthat a ≡b ( mod m ).

Note: the set of all integers congruent to an integer a modulo m is called the congruence class of a modulo m .


Modular Arithmetic

Addition and multiplication with congruencesTheorem (Addition and multiplication with congruences)Let m be a positive integer. If a ≡b ( mod m ) and c

≡d

( modm

), then

a + c ≡b + d ( mod m ) and ac ≡bd ( mod m )

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Modular Arithmetic

Addition and multiplication with congruencesTheorem (Addition and multiplication with congruences)Let m be a positive integer. If a ≡b ( mod m ) and c

≡d

(mod m

), then


Proof.Because a

≡b ( mod m ) and c

≡d ( mod m ), there are

integers s and t such that b = a + sm , and d = c + tm . Hence,b d ( ) ( ) ( ) ( )

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b + d = ( a + sm ) + ( c + tm ) = ( a + c ) + ( s + t )m bd = ( a + sm )(c + tm )

= ac + atm + csm + stm 2 = ac + m (at + cs + stm )Hence,



Modular Arithmetic

Addition and multiplication with congruencesCorollaryLet m be a positive integers and let a and b be integers. Then

(a + b ) mod m = (( a mod m ) + ( b mod m )) mod m

and ab mod m = (( a mod m )(b mod m )) mod m

For the proof, read Rosen’s book page 205.

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For the proof, read Rosen s book page 205.Note: although the we can do addition and multiplicationwith congruences, some properties are NOT valid:

if ac ≡bc ( mod m ), the congruence a ≡b ( mod m ) may

be FALSE.if a ≡b ( mod m ) and c ≡d ( mod m ), the congruencea c ≡b d ( mod m ) may be FALSE.


Modular Arithmetic

Applications of Congruences

Read Rosen’s book page 205 – 208 for more details.Hashing functions: storing records in memory so thatretrieval can be done quickly.Pseudorandom numbers: random numbers generated by

computers.Cryptology: encryption and decryption of messages e g

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Cryptology: encryption and decryption of messages, e.g.with Caesar cipher, shift cipher, etc.


Primes

Primes

DenitionA positive integer p > 1 is called prime if the only positivefactor of p are 1 and p . A pA positive integer that is greater than 1 and is not prime iscalled composite

Remark: an integer n is composite iff there exists an

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Remark: an integer n is composite iff there exists aninteger a such that a | n and 1 < a < n .List all primes less than 100


Primes

Fundamental theorem of arithmetic

Theorem (Fundamental theorem of arithmetic)Every positive integer greater than 1 can be written uniquely as a prime or the product of two or more primes where the prime factors are written in order of nondecreasing size.

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Primes



The prime factorization of 100, 641, 999, 1024 are:

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Primes



The prime factorization of 100, 641, 999, 1024 are:100 = 2 ·2 ·5 ·5 = 2252 .

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Primes




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641 = 641


Primes


Theorem (Fundamental theorem of arithmetic)

Every positive integer greater than 1 can be written uniquely as a prime or the product of two or more primes where the prime factors are written in order of nondecreasing size.


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641 = 641999 = 3 ·3 ·3 ·37 = 33 ·37


Primes


Theorem (Fundamental theorem of arithmetic)

Every positive integer greater than 1 can be written uniquely as a prime or the product of two or more primes where the prime factors are written in order of nondecreasing size.


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641 = 641999 = 3 ·3 ·3 ·37 = 33 ·371024 = 2

·2

·2

·2

·2

·2

·2

·2

·2

·2 = 210


Primes

Prime divisor of a composite integer

TheoremIf n is a composite integer, then n has a prime divisor less than or equal √n

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Primes

Prime divisor of a composite integer

TheoremIf n is a composite integer, then n has a prime divisor less than or equal √n

Proof: look at Rosen’s book page 211.Show that 101 is prime.

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Find the prime factorization of 7007.


Primes

Innitude of primes

TheoremThere are innitely many primes.

Proof: Rosen’s book page 212.Mersenne primes: a prime that has a special form 2 p −1where p is also a prime

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where p is also a prime.


Primes

Distribution of primes

Theorem (The prime number theorem)The ratio of number of primes not exceeding x and x / ln x approaches 1 as x grows without bound, where ln x is the

natural logarithm of x.

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Greatest Common Divisor and Least Common Multiples

Greatest common divisor

DenitionLet a and b be integers, not both zero. The largest integer d such that d | a and d | b is called the greatest common divisorof a and b (written as gcd (a , b )).

What is gcd (24 , 36 )?

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What is gcd (17 , 22 )?



Relatively primes

DenitionThe integers a and b are relatively prime if gcd(a , b ) = 1.

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Relatively primes

DenitionThe integers a and b are relatively prime if gcd(a , b ) = 1.

DenitionThe integers a 1 , a 2 , . . . ,a n are pairwise relatively prime ifgcd (a i , a j ) = 1 for every 1

≤i < j

≤n .

Determine whether 10, 17 and 21 are pairwise relatively

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Determine whether 10, 17 and 21 are pairwise relativelyprime.Determine whether 10, 19 and 24 are pairwise relatively

prime.



Least common multiple

Denition

The least common multiple of positive integers a and b (writtenlcm (a , b )) is the smallest positive integer that is divisible by botha and b .

What is lcm (233572 , 2433)?

Theorem

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Let a and b be positive integers. Then

ab = gcd (a , b )

·lcm (a , b )

IKI10600I: Discrete Mathematics IIntegers and Algorithms

Adila A. Krisnadhi


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Representation of Integers Euclidean Algorithm

Integers and AlgorithmsOutline

23 Representation of Integers

24 Euclidean Algorithm

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Theorem (Base b expansion of n )Let b be a positive integer greater than 1. Then if n is a positive integer, it can be expressed uniquely in the form

n = a k b k + a k − 1b k − 1 + · · ·+ a 1b + a 0

The base b expansion of n is denoted by (a k a k − 1 . . . a 1a 0)b , e.g.,(245 )8 represents 2 ·82 + 4 ·8 + 5 = 165.

The base 10 expansion of integers is called decimal expansion.

The base 2 expansion of integers is called binary expansion.The base 8 expansion of integers is called octal expansion.

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The base 16 expansion of integers is called hexadecimalexpansion (using 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E, and F as digits

where A through F correspond to numbers 10 through 15 indecimal).


Base b expansion to decimal expansion

(101011111 )2 = 1 ·28 + 0 ·27 + 1 ·26 + 0 ·25 + 1 ·24

+ 1 ·23 + 1 ·22 + 1 ·21 + + 1 ·20 = ( 351 )10

(2AE 0B )16 = 2·16 4+ 10 ·16 3+ 14 ·16 2+ 0·16 + 11 = ( 175627 )10

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(3071 )8 = 3 · 83 + 0 · 82 + 7 · 8 + 1 = ( 1593 )10


Decimal expansion to base b expansion

Find octal expansion of 12345

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Decimal expansion to base b expansion

Find octal expansion of 12345Take the remainders of the following divisions:

12345 = 8 ·1543 + 11543 = 8 ·192 + 7

192 = 8 ·24 + 024

=8

·3

+0

3 = 8 ·0 + 3

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Hence, (12345 )10 = ( 30071 )8

Find hexadecimal expansion of (177130 )10 .Find binary expansion of (241 )10 .

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Finding GCD with Euclidean algorithm

Find gcd (91 , 287 ).

287 = 91 ·3 + 1491 = 14 ·6 + 7

14 = 7 ·2 + 0

Hence gcd (91 287 ) = 7



Hence, gcd (91 , 287 ) = 7


Euclidean algorithm

procedure : gcd ( a , b : p o s i t i v e i n t e g e r s )x := a y := b wh i l e y = 0begin

r := x mod y x := y y := r

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end { gcd (a , b ) i s x }


Euclidean algorithm: Why it works?

TheoremLet a = bq + r, where a , b , q, and r are integers. Then gcd (a , b ) = gcd (b , r ).

For example, in the previous slide, it holds that:gcd (91 , 287 ) = gcd (91 , 14 ) = gcd (14 , 7) = gcd (7 , 0) = 7.

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all lectures adila

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