ppt_IT_Discretemaths.pdf

VEL TECH Dr. RR & Dr. SR TECHNICAL UNIVERSITY

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U4MAA06 DISCRETE MATHEMATICS

UNIT I PROPOSITIONAL CALCULUS 9

Propositions – logical connectives – compound propositions – conditional and

biconditional propositions – truth tables – tautologies and contradictions –

contrapositive – logical equivalences and implications– De Morgan’s laws – normal

forms – principal conjunctive and disjunctive normal forms – rules of inference –

arguments – validity of arguments

UNIT II PREDICATE CALCULUS 9

Predicates – statement function – variables – free and bound variables – quantifiers –

universe of discourse – logical equivalences and implications for quantified

statements – theory of inference – the rules of universal specification and

generalization – validity of arguments

UNIT III SET THEORY

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Basic concepts – subsets – algebra of sets – the power set – ordered pairs and

Cartesian product – relations on sets – types of relations and their properties –

matrix representation of a relation – graph of a relation – partitions – equivalence

relations – partial ordering – poset – Hasse diagram – Lattices and their properties –

sub-lattices – Boolean algebra – homomorphism

UNIT IV FUNCTIONS

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Definitions of functions – classification of functions – types of functions – examples –

compositions of functions – inverse functions – binary and n-ary operations –

characteristic function of a set – hashing functions – recursive functions –

permutation functions

UNIT V GROUPS 9

Algebraic systems – definitions – examples – properties – semigroups – monoids –

homomorphism – sub semigroup and submonoids – cosets and Lagrange’s theorem

– normal subgroups – normal algebraic system with two binary operations – codes


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and group codes – basic notions of error correction – error recovery in group codes

TEXT BOOKS

1. Tremblay and Manohar R, Discrete Mathematical Structures with Applications to

Computer Science, Tata McGraw-Hill, New Delhi, 2003.

2. Kenneth H. Rosen, Discrete Mathematics and its Applications, 6th edition, Tata

McGraw Hill, New Delhi, 2008.

REFERENCE BOOKS

1. Bernard Kolman, Robert C. Busby, Sharan Cutler Ross, Discrete Mathematical

Structures, 4th Indian reprint, Pearson Education, New Delhi, 2003.

2. Ralph P. Grimaldi, Discrete and Combinatorial Mathematics, 4th edition, Pearson

Education, New Delhi, 2002.


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UNIT – I

PROPOSITIONAL CALCULUS

PROPOSITIONS

LOGICAL CONNECTIVES

COMPOUND PROPOSITIONS

CONDITIONAL AND BICONDITIONAL PROPOSITION

TRUTH TABLES

TAUTOLOGIES AND CONTRADICTIONS

CONTRA POSITIVE

LOGICAL EQUIVALENCES AND IMPLICATION

DEMORGAN’S LAW

NORMAL FORMS

PRINCIPLE CONJUNCTIVE AND DISJUNCTIVE NORMAL FORMS

RULES OF INFERENCE

ARGUMENTS

VALIDITY OF ARGUMENTS.


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INTRODUCTION

LOGIC:

PROPOSITION: It is a declarative sentence that is either true or false not both.

The area of logic that deals with proposition is called propositional calculus.

Example: Raju is coming everyday in time --- True

Madhu is an animal --- False

APPLICATIONS:

Design of computer circuits.

Construction of computer programs.

Verification of correctness of programs.

System security.

CONNECTIVES:

NEGATION: Let P be a proposition. The statement ‚it is not the case that P‛ is

another proposition, called the negation of P. It is denoted by ‚ P‛.

Example: P = Rose is red.

P = Rose is not red.

Truth Table:

p p

T

F

F

T

CONJUNCTION: Let P & Q be propositions. The propositions ‘P and Q’ (P Q) is

the proposition that is true when both P and Q are true and is false otherwise.

Example: P = Science teacher is absent today.

Q = He is not taking class well.

P Q = Science teacher is absent today and he is not taking class well.


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Truth Table:

p q p q

T

T

F

F

T

F

T

F

T

F

F

F

DISJUNCTION: Let P & Q be propositions. The propositions ‘P or Q’ (P Q) is the

proposition that is false when both P and Q are false and is true otherwise.

Example: P = Students who have studied electronics can take this personality class.

Q = Students who have studied computer science can take this personality

class.

P Q = Students who have studied electronics or computer science can take

this personality class.

Truth Table:

p q p q

T

T

F

F

T

F

T

F

T

T

T

F

EXCLUSIVE OR: Let P and Q be propositions. The exclusive OR of P and Q is

denoted by P Q, is the propositions that is true when exactly one of P and Q true

and is false otherwise.

Example: Taking tea or coffee at a time, not both.

Truth Table:

p q P q

T

T

F

F

T

F

T

F

F

T

T

F


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CONDITIONAL STATEMENTS

IMPLICATIONS: Let P and Q be propositions. The implication P Q is the

proposition that is false when P is true & Q is false and true otherwise.

Example: If I am elected then I will lower taxes.

Truth Table:

p q p q

T

T

F

F

T

F

T

F

T

F

T

T

BICONDITIONAL: Let P and Q be propositions. The bi-conditional P Q is the

proposition that is true when P & Q have the same truth value and it is false

otherwise.

Example: You can take the flight if and only if you buy a ticket.

Truth Table:

p q p q

T

T

F

F

T

F

T

F

T

F

F

T

STATEMENT FORMULA AND TRUTH TABLE

Those statements that do not have connectives are called atomic or simple

statement. On other hand, those statements containing more than one primary

statement and some connectives are called compound statement.

TAUTOLOGY: A compound statement is called a tautology if it is true for all truth

values assigned for its component statements.


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CONTRADICTION: If the compound statement is false for all such assignments,

then it is called contradiction.

EQUIVALENT STATEMENT: Let A & B be two statement formulas. If the truth

value of A is equal to the truth value of B then A and B are said to be equivalent.

Example: P Q P Q

NORMAL FORM:

The solution of decision problem may not be simple & construction of truth

table may not be practical, even with the aid of the computer. We therefore consider

other procedure known as reduction to normal form.

DISJUNCTIVE NORMAL FORM (DNF): A formula which is equivalent to a given

formula and which consists of sum of elementary products is called a DISJUNCTIVE

NORMAL FORM of the given formula.

CONJUNCTIVE NORMAL FORM (CNF): A formula which is equivalent to a given

formula and which consists of product of elementary sums is called a

CONJUNCTIVE NORMAL FORM of the given formula.

PRINCIPLE DISJUNCTIVE NORMAL FORM (PDNF): For the given formula an

equivalent formula consisting of disjunction of min terms only is known as

PRINCIPLE DISJUNCTIVE NORMAL FORM. Such a normal form is also called the

S.O.P canonical form.

PRINCIPLE CONJUNCTIVE NORMAL FORM (PCNF): For the given formula an

equivalent formula consisting of conjunction of max terms only is known as

PRINCIPLE CONJUNCTIVE NORMAL FORM. Such a normal form is also called

the P.O.S canonical form.

INFERENCE THEORY

Valid Conclusion: Let A and B be two statement formula. We say that ‚B logically

follows from A‛ or ‚B is a valid conclusion of the premise A‛ if and only if A B

(A B is a tautology).


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Just as the definition of implication was extended to include a set of formula

rather than a single formula. We say that set of premises {H1, H2……Hm} and a

conclusion ‘C’ follows logically if H1 H2 …... Hm C

Rules of Inference

Rule P : A premise may be introduced at any point in the derivation.

Rule T : A formula may be introduced in the derivation if the formula is

tautologically implied by any one or more of the preceding formula in the

derivation.

Rule CP : If we can derive ‘S’ from ‘R’ and a set of premises then we can derived

‘R S’ from the set of premises alone. i.e.

P R S P R S

FORMULA:

1) IDEMPOTENT LAW:

P P P

P P P

2) ASSOCIATIVE LAW:

(P Q) R P (Q R)

(P Q) R P (Q R)

3) COMMULATIVE LAW:

(P Q) (Q P)

(P Q) (Q P)

4) DISTRIBUTIVE LAW:

P (Q R) (P Q) (P R)

P (Q R) (P Q) (P R)

5) ABSORPTION LAW:

P (P Q) P

P (P Q) P


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6) DE MORGAN’S LAW:

(P Q) P Q

(P Q) P Q

IDENTITIES:

1) P F P; P T P

2) P T T; P F F

3) (P Q) ( P Q)

4) (P Q) (P Q) (Q P)

5) ( P) P

6) P P P ; P P P

7) P P P

8) (P P) Q Q

9) (P P) Q Q

10) P P T; P P F (Inverse Law)

IMPLICATIONS:

1. P Q P

2. P Q Q

3. P P Q

4. Q P Q

5. P (P Q)


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6. Q (P Q)

7. (P Q) P

8. (P Q) Q

9. P, Q P Q

10. P, P Q Q

11. P, P Q Q

12. Q, P Q P


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PART – A

1. Define statement or proposition with example.

A statement or a proposition is a declarative sentence to which it is

meaningful to assign a truth value ‘true’ or ‘false’ but not both simultaneously

Eg: (i) Canada is a country

(ii) Mathematics is an interesting subject

2. Define conjunction of the statement and draw its truth table. Give example.

The process of joining two statements p and q using ‚and‛ produces a new

statement denoted by p q, which has the truth value T whenever both p and q have

truth value T, the truth value F otherwise. The statement p q is called the

conjunction of the statement p and q.

The truth table for p q

P q p q

T

T

F

F

T

F

T

F

T

F

F

F

Example:

The conjunction of the statements

p : Today is Sunday

q : Government offices are working

is p q : Today is Sunday and government offices are working.

3. Define Disjunction of the statement and draw its truth table. Give examples.

The process of joining two statements p and q by ‚or‛ produces a new

statement. The new statement is disjunction of p and q denoted by ‚p q‛.


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The truth table for ‚p q‛.

P q p q

T

T

F

F

T

F

T

F

T

T

T

F

Example:

If p : It is raining

q : 9 is a prime number

then pvq : It is raining or 9 is a prime number

4. Define Negation of a statement and draw its truth table and five examples.

The negation of a statement p is the statement obtained from p by prefixing the

words ‚It is not true that‛. The negation of p is denoted by p and read as ‚Not p‛.

The truth table of p

p p

T

F

F

T

Example: The negation of the statement is the statement ‚John is playing football‛

‚It is not true that john is playing foot ball‛

or simply ‚John is not playing football‛.

5. Define a conditional statement and draw its truth table.

Let p and q be two statements.

The statement ‚if p then q‛ is denoted by p q has the truth value ‘false’ when q

has the truth value ‘false’ and p has the truth value ‘true’. Otherwise p q has the

truth value ‘true’.


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P q p q

T

T

F

F

T

F

T

F

T

F

T

T

Expression: Express in English the statement p q where

p: The sun is shining today

q: 2+7>4

If the sun is shining today, then 2+7>4.

6. Define a bi-conditional statement and draw its truth table.

If p and q are any two statements, then the statement p q has the truth

value T whenever both p and q have identical truth values and in all other cases, the

truth value of p q is F.


P q p q

T

T

F

F

T

F

T

F

T

F

F

T

Example : 7 > 5 if and only if 7-5 is positive.

7. Construct the truth table for (p q) (q p)

p q p q q p (p q) (q p)

T

T

F

F

T

F

T

F

T

F

T

T

T

T

F

T

T

F

F

T


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8. Construct the truth table for (p q) ( p q)

p q p q (p q) p q p q (p q) ( p q)

T

T

F

F

T

F

T

F

T

F

F

F

F

T

T

T

F

F

T

T

F

T

F

T

F

T

T

T

T

T

T

T

9. Define Tautology.

A statement formula which is true regardless of the truth values of the

statements which replace the variables in it is called us ‘universally valid formula’ or

a ‘tautology’ or a ‘logical truth’.

10. Define contradiction

A statement formula which is false regardless of the truth values of the

statements which replace the variables is it is called a contradiction.

11. Prove that the conjunction of any two tautologies is also a tautology.

Proof: let A and B be two statement formulas which are tautologies.

If we assign any truth values to the variables of A and B, then the truth values

of both A and B will be T.

Thus the truth value of A B will be T. So that A B will be a tautologies.

i.e., The conjunction of any two tautologies is also a tautology.

12. Define Equivalence of formulae.

Let A and B be two statement formulae and let p1, p2, ...... pn denote all the

variables occurring in both A and B. Consider an assignment of truth values to p1, p2,

...... pn and the resulting truth values of A and B. If the truth value of A is equal to the

truth value of B for every one of the 2n possible set of truth values assigned to p1, p2,

...... pn then A and B are said to be equivalent.


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13. Prove p q p q by truth table

p q p p q p q

T

T

F

F

T

F

T

F

F

F

T

T

T

F

T

T

T

F

T

T

The column for p q and p q in the above table are identical.

Therefore p q p q

14. Define tautologically implication.

A statement A is said to tautologically imply a statement B if and only is A B

is a tautology.

15. Define an elementary product and an elementary sum.

A product of the variables and their negations in a formula is called an

elementary product. A sum of the variables and their negations is called an

elementary sum.

16. Define Disjunctive Normal Form (DNF)

A formula which is equivalent to a given formula and which consists of a sum

of elementary products is called a disjunctive normal form of the given formula.

17. Define conjunctive Normal form (CNF)

A formula which is equivalent to a given formula and which consists of a

product of elementary sums is called a conjunctive normal form of the given

formula.

18. Obtain disjunctive normal form of p (p q)

Solution:

( ) (replacing by and )

p p (distributive law)

p p q p p q

p q


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19. Obtain conjunctive normal form of p p q

Solution:

(replacing by and )p p q p p q

hence p ( p q) is a required form.

20. Show that the formula q p q p q is a tautology

Solution:

[using distribution law]

q

q

q p q p q q p p q

p p q q

p p q q

since each of the elementary sums is a tautology the given formula is a tautology.

q T T

q T

T

21. Define min terms.

Let p and q be two statement variables. Construct all possible formulae which

consists of conjunction of p or its negation and conjunction of q or its negation, such

conjunctions of p and q are called the min terms of p and q.

For example, min term of 2 variables p and q, there are 22 such formulas given by

, , and p qp q p q p q

22. Define max terms.

For a given number of variables, the max term consists of disjunctions in

which each variable or its negation. But not both, appears only once.


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23. Define PDNF or sum-of-products canonical form.

For a given formula, an equivalent formula consisting of disjunctions of min

term only is known as its principle disjunctive normal form. Also called the sum of

products canonical form.

24. Define PCNF or product-of-sums canonical form.

For a given formula, an equivalent formula consisting of conjunctions of the

max terms only is known as its PCNF. Also called product-of-sums canonical form.

25. Obtain the PDNF of p q.

Solution:

( )

(distributive law)

( commutative law)

p q p q q q p p

p q p q q p q p

p q p q p q

26. Obtain the PDNF of p q p r q r .

Solution:

p q p r q r

p q r r p r q q q r p p

p q r p q r p q r p q r

27 State the rules of inference.

Two rules of inference which are called Rule P and Rule T

Rule P: A premise may be introduced at any point in the derivation

Rule T: A formula S may be introduced in a derivation if S is tautologically implied

by any one or more of the preceding formulas in the derivation.


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28. If P, Q and R are statement variables. Prove that

P P Q P Q R. Nov/Dec - 2006

By using distributive law left hand side can be simplified to

P P (Q Q P P T

F

R F any statement formula .

29. Prove that whenever A B C, we also have A (B C) and vice versa.

Nov/Dec – 2006

Assume that A B C. To prove A (B C) suppose that A is true and B

C is false. Hence B must be true and C must be false. Thus A B is true whereas C

is false, this contradicts our assumption.

Conversely assume that A (B C). Suppose that A B C is false. Hence

A is true and B C is false. This contradicts our assumption.

30. Using truth table verify that the proposition (P Q) (P Q) is a contradiction

Solution :

P Q P Q P Q (P Q) (P Q) (P Q)

T T T T F F

T F F T F F

F T F T F F

F F F F T F

All the entries in the last column are F. Therefore the given proposition is

contradiction.

31. Show that ( P ( Q R)) (Q R) (P R) R (use only the laws)

Solution:

( P ( Q R)) (Q R) (P R) ( P ( Q R)) ((Q P) R) [Distributive Law]

(( P Q) R) ((Q P) R)


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[Associative Law]

( (P Q) R) ((P Q) R) *De Morgan’s Law+

( (P Q) (P Q)) R [A P Q, A A T]

T R

R

32. Define functionally complete set of connectives and give an example.

Any set of connectives in which every formula can be expressed as another

equivalent formula containing connectives from this set is called functionally

complete set of connective.

(or)

A collection of logical operators is called functionally complete if every

compound proposition is logically equivalent to a compound proposition involving

only these logical operators.

Example: The set of connectives { , } and { , } are functionally complete.

{ }, { }, { } or { , } are not functionally complete.

Note: From the five connectives , , , , . we have obtained at least two sets of

functionally complete connectives.

33. Give the converse and the contra positive of the implication ‚If it is raining

then I get wet‛. (AU May 2004)

Solution:

P : It is raining

Q : I get wet‛

Contra Positive

Q P : ‚If I do not get wet, then it is not raining‛

Converse

Q P : ‚If I get wet, then it is raining‛


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34. Show that (P Q) ( P ( P Q)) P Q (use only laws) (AU Mar 2004)

Solution:

(P Q) ( P ( P Q)) (P Q) ( P ( P Q)) [P Q P Q]

(P Q) ( P ( P Q)) [Double negation]

(P Q) (( P P) Q) [Associative law]

(P Q) ( P Q) [( P P) P]

((P Q) P) Q [Associative law]

((P P) (Q P)) Q [Distributive law]

(T (Q P)) Q [P P=T]

(Q P) Q [P T P]

( P Q) Q [commutative law]

( P (Q Q)) [Associative law]

P Q

35. Define contra positive of a statement. (AU Nov 2005)

Solution:

If P Q is an implication, then the converse of P Q is the implication Q P, and

the contra positive of P Q is the implication Q P.

36. What are the contra positive, the converse, and the inverse of the implication?

‚The home team wins whenever it is raining‛

Solution:

‚If it is raining, then the home team wins‛

If P then Q

P : It is raining

Q : Home team wins

Contra positive: Q P

‚If the home team does not win then it is not raining‛.

Inverse: P Q

‚If it is not raining then the home team does not win‛.


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Converse: Q P

‚If the home team wins then it is raining‛.

37. Write an equivalent formula for P (Q R) which is free from bi-conditional as

well as conditional.

Solution:

P (Q R) P [(Q R) (R Q)] P [( Q R) ( R Q)]

38. Show the following equivalences. P (Q P) P (P Q)

Solution:

(i) P (Q P) P ( Q P) [since Q P Q P]

P (P Q) [commutative]

( P P) Q [Associative]

T Q [Negation]

T [since T Q T]

(ii) P (P Q) P ( P Q) [since P Q P Q]

( P) ( P Q) [since P Q P Q]

P ( P Q) [Double negation]

(P P) Q [Associative]

T Q [P P T]

T [T Q T]

From (i) and (ii), we get P (Q P) P (P Q)

39. Write in symbolic form the statement ‚The crop will be destroyed if there is a

flood‛.

Solution:

‚If there is a flood, then the crop will be destroyed‛.

P : There is a flood.

Q : The crop will be destroyed.

If P then Q

i.e., (P Q)


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40. Define tautology and contradiction.

Tautology:

A compound statement is called a tautology, if it is true for all truth values

assigned for its compound statements.

Contradiction:

If a compound statement is false for all such assignments then it is called

contradiction.

41. Define a principal conjective normal form of a statement. (May / Jun 2009).

Statement :

The PCNF of a given formula P is equivalent formula consisting of

conjunction of maxterms only.

42. How many row’s are needed for the truth table of the formula. (May / Jun

2009).

(p q) (( r S) t)?

Ans : The no.of rows will be 32.

43. Prove by truth table that (P ) ( P Q) (P Q) (Apr/May 2008).

Solution :

Truth Table

P Q P Q P Q (P Q) P Q P Q ( P Q) (P Q)

T T F F T F F T F

T F F T F T T T T

F T T F F T T T T

F F T T T F T F F

Hence 6 and 9 columns are same. Hence proved.


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PART – B

1. a. Construct the truth table for the formula

P Q P Q P Q P Q

b. construct the truth table for

(i). P Q P Q

(ii). P Q R P Q P R

Answer: (a)

P Q P Q P Q P Q P Q P Q P Q P Q

P Q P Q

T

T

F

F

T

F

T

F

F

F

T

T

F

T

F

T

T

F

F

F

F

F

T

F

F

T

F

F

F

F

F

T

T

T

T

T

Answer: (b)(i)

P Q P Q (P Q) P Q P Q P Q P Q

T

T

F

F

T

F

T

F

T

F

F

F

F

T

T

T

F

F

T

T

F

T

F

T

F

T

T

T

T

T

T

T

Answer: (b)(ii)

P Q R Q

R

P (Q

R)

(P (Q

R)

(P Q

)

P

R

(P Q)

(P R)

P Q R

P Q P R

T

T

T

T

F

F

F

F

T

T

F

F

T

T

F

F

T

F

T

F

T

F

T

F

T

F

F

F

T

F

F

F

T

T

T

T

T

F

F

F

F

F

F

F

F

T

T

T

T

T

T

T

T

T

F

F

T

T

T

T

T

F

T

F

T

T

T

T

T

F

F

F

F

F

F

F

F

F

F

F


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2. (a) S.T (P Q) (R Q) (P R) Q

(b) S.T ( P ( Q R)) (Q R) (P R) R

Answer: (a)

P Q R Q

P Q R Q replacing P Q P Q

P R Q A C B C A B C

P R Q P R P R

P R Q A B A B

Answer: (b)

P Q R Q R P R

P Q R Q R P R Associative law

P Q R Q R P R Demorgan's law

P Q R Q P R Distributive law

P Q P Q R Distributive law

T R P P T

R T R R

3. Prove that

(i) (P Q) (P Q) (P Q)

(ii) (P Q) (P Q) ( P Q)

Solution:

P Q P Q Q P

P Q Q P

P Q Q P Q P


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P Q Q Q P P Q P

P Q F F Q P

P Q Q P

P Q Q P

P Q P Q i

P Q P Q

P P Q Q P Q

P P P Q Q P Q Q

F P Q Q P F

P Q Q P

P Q P Q ii

This (i) and (ii) are proved

4. (a) P.T. P (Q P) P (P Q)

(b) S.T. Q (P Q) ( P Q) is a tautology.

Answer: (a)

P Q P P Q P

P Q P

P P Q

T Q

T

and

P P Q P P Q

P P Q

P P Q

T Q

T

So P Q P T P P Q


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Answer: (b)

Q P Q P Q Q P Q P Q

Q P Q Q P Q

Q P T P Q

Q P P Q

P Q P Q

T

Hence the given formula is a tautology

5. (a) Establish (P Q) ( P ( P Q)) ( P Q)

(b) Show that ((P Q) ( P ( Q R))) ( P Q) ( P R)

Solution: (a)

P Q P P Q P Q P P Q

P Q P Q

P Q P Q

P P Q P Q

T Q P Q

Q P Q

Q P

Solution: (b)

P Q P Q R P Q P R

P Q P Q R P Q P R

P Q P Q R P Q P R

P Q P Q P R P Q P R

P Q P R P Q R

P Q R P Q R

T

Hence the given formula is a tautology.


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6. Show that the following are equivalent formulas?

(a) P (P Q) P

(b) P ( P Q) P Q

Solution (a):

Write the principal disjunctive normal form of each formula and compare

these normal forms

P P Q P Q Q P Q

P Q P Q P Q

P Q P Q 1

P P Q Q P Q P Q 2

Equation (1) & (2) are equal

P (P Q) P

Solution (b):

P P Q P Q Q P Q

P Q P Q P Q 1

P Q P Q Q Q P P

P Q P Q Q P Q P

P Q P Q P Q 2

equation (1) & (2) are equal

P ( P Q) P Q


28

7. Obtain the principal disjunctive normal forms of

(a) (P Q) ( P R) (Q R)

(b) P ((P Q) ( Q P))

Solution (a):

P Q P R Q R

P Q R R P R Q Q Q R P P

P Q R P Q R P R Q P R Q

Q R P Q R P

P Q R P Q R P R Q P R Q

Solution (b):

P P Q Q P

P P Q Q P using P Q P Q and DeMorgan's law

P P Q Q P

P P Q P Q Q P

P F Q Q P

P F Q P

P Q P

P Q Q Q P

P Q P Q P Q

8. (a) Show that the tautological implication is transitive.

(b) If H1, H2 …… Hm and P imply Q, then H1, H2 …… Hm imply P Q.

Proof: (a)

Assume that A B and B C. Then A B and B C are tautologies.

Hence (A B) (B C) is also a tautology.

But we know that (P Q) (Q R) P R

using this we get

(A B) (B C) (A C)

Hence A C is a tautology

Which means that A C


29

Proof: (b)

From our assumption we have

(H1 H2 …… Hm P) Q

The means

(H1 H2 …… Hm P) Q is a tautology

using the equivalence

P1 (P2 P3) (P1 P2) P3

We get

(H1 H2 …… Hm) (P Q) is a tautology

which completes the proof of the theorem.

9. State and prove De Morgan’s laws

Proof: (i) (P Q) P Q

(ii) (P Q) P Q

(i)

P Q P Q (P Q) P Q P Q

T

T

F

F

T

F

T

F

T

F

F

F

F

T

T

T

F

F

T

T

F

T

F

T

F

T

T

T

This shows that (P Q) ( P Q)

(ii)

P Q P Q (P Q) P Q P Q

T

T

F

F

T

F

T

F

T

T

T

F

F

F

F

T

F

F

T

T

F

T

F

T

F

F

F

T

This shows that (P Q) ( P Q)


30

10. Obtain the principal conjunctive normal for of the formula S given by ( P R)

(Q P).

Solution:

P R Q P

P R Q P P Q

P R Q P P Q

P R Q Q Q P R R P Q R R

P Q R P Q R P Q R P Q R

P Q R P Q R

P Q R P Q R P Q R P Q R P Q R

Now the conjunctive normal form of S can be obtained by writing the

conjunction of the remaining max terms. Thus S has principal conjunctive normal

form.

P Q R P Q R P Q R

by conducting S, we obtain

P Q R P Q R P Q R

P Q R P Q R P Q R

P Q R P Q R P Q R

Which is the principal disjunction normal form of S

11. Obtain PCNF and PDNF for the formula (P Q) ( P R) and represent them

using product and summation notation.

Solution:

To obtain PCNF and its product notation:

P Q P R

P Q P P Q R

P P Q P P R Q R


31

P Q P R Q R

P Q R R P R Q Q Q R P P

P Q R P Q R P R Q P R Q Q R P Q R P

P Q R P Q R P Q R P Q R P Q R P Q R


This is the required PCNF. It is written as 0,2,4,5.

To obtain PDNF and its summation notation

P Q P R

P Q R R P R Q Q



This is the required PDNF. It is written as 1,3,6,7.

12. (a) S.T. S R is tautologically implied by (P Q) (P R) (Q S).

(b) S.T. R (P Q) is a valid conclusion from the premises P Q,

P M and M.

Solution: (a)

{1} (1) P Q P

{1} (2) P Q T, (1), E1 and E16.

{3} (3) Q S P

{1,3} (4) P S T, (2), (3) and I13.

{1,3} (5) S P T, (4), E18 and E1

{6} (6) P R P

{1,3,6} (7) S R T, (5), (6) and I13

{1,3,6} (8) S R T, (7), E16 and E1.

Solution: (b)

{1} (1) P M P

{2} (2) M P

{1,2} (3) P T, (1), (2) and I12

{4} (4) P Q P


32

{1,2,4} (5) Q T, (3), (4) and I10

{6} (6) Q R P

{1,2,4,6} (7) R T, (5), (6) and I11

{1,2,4,6} (8) R (P Q) T, (4), (7) and Ia

13. (a) Show that R S can be derived from the premises

P (Q S), R P and Q.

(b) Show that R S follows logically from the premises

C D, (C D) H, H (A B), and (A B) R S.

Solution (a):

Instead of deriving R S, We shall include R as an additional premise and

show S first.

{1} (1) R P P

{2} (2) R P (assumed Premise)

{1,2} (3) P T, (1),(2) and I10

{4} (4) P (Q S) P

{1,2,4} (5) Q S T, (3),(4) and I11

{6} (6) Q P

{1,2,4,6} (7) S T, (5), (6) and I11

{1,4,6} (8) R S CP

Solution (b):

{1} (1) (C D) H P

{2} (2) H (A B) P

{1,2} (3) (C D) (A B) T, (1), (2) and I13.

{4} (4) (A B) (R S) P

{1,2,4} (5) (C D) (R S) T, (3), (4) and I13.

{6} (6) C D P

{1,2,4,6} (7) R S T, (5), (6) and I11


33

14. Prove that (P Q) (Q R) (P R) (Nov / Dec – 2006)

Solution:

To prove this we have to prove (P Q) (Q R) (P R) is tautology.

P Q R P Q Q R (P Q) (Q R) P R (P Q) (Q R) (P R)

T

T

T

T

F

F

F

F

T

T

F

F

T

T

F

F

T

F

T

F

T

F

T

F

T

T

F

F

T

T

T

T

T

F

T

T

T

F

T

T

T

F

F

F

T

F

T

T

T

F

T

F

T

T

T

T

T

T

T

T

T

T

T

T

(P Q) (Q R) (P R)

15. Find the PCNF and PDNF of the formula.

S [P (Q R)) ( P ( Q R)] (AU/Nov–Dec/2006)

Solution:

(i) Let S P Q R P Q R

P Q R P Q R

P Q P R P Q P R

P Q R R P R Q Q P Q R R P R Q Q



P Q R P Q R P Q R P Q R P Q R P Q R This is the required PCNF (or) POS.

(ii) To Find: PDNF (or) SOP

S P Q R P Q R

S P Q R P Q R

S P Q R P Q R

S P Q R P Q R

This is the required PDNF (or) SOP.


34

16. By using truth tables verify whether the following specifications are

consistent.

Whenever the system software is being upgraded user cannot access the file

system. If user can access the file system, then they can save new files. If users

cannot save new file then the system software is not being upgraded.AU/Nov–

Dec/2006

Solution:

P = The system software is being upgraded.

Q = Users can access the file system

R = Users can save new files

The premises are P Q, Q R, and R P.

Let S = (P Q) (Q R) ( R P)

P Q R P Q R P Q Q R R P S

T

T

T

T

F

F

F

F

T

T

F

F

T

T

F

F

T

F

T

F

T

F

T

F

F

F

F

F

T

T

T

T

F

F

T

T

F

F

T

T

F

T

F

T

F

T

F

T

F

F

T

T

T

T

T

T

T

F

T

T

T

F

T

T

T

F

T

F

T

T

T

T

F

F

T

F

T

F

T

T

The premises are consistent.

17. Show that the following are inconsistent

(i) If jack misses many classes through illness, then he fails high school.

(ii) If jack fails high school, then he is uneducated.

(iii) If jack reads a lot of books, then he is not uneducated.

(iv) Jack misses many classes through illness and reads a lot of books.

Solution:

Let P: Jack misses many classes through illness.

Q: He fails high school.

R: He is uneducated.

S: Jack reads a lot of books.


35

The premises are

H1 : P Q

H2 : Q R

H3 : S R

H4 : P S

(1) P Q Rule P

(2) Q R Rule P

(3) P R Rule T, (1), (2)

(4) P S Rule P

(5) P Rule T, (4)

(6) R Rule T, (3), (5)

(7) S R Rule P

(8) S Rule T, (4)

(9) R Rule T, (7), (8)

(10) R R Rule T, (6), (9)

(11) False

The given premises are inconsistent.

18. Obtain the PCNF and PDNF of P [ P (Q ( Q R))].

Solution:

To find: PCNF (or) POS

Let, S P [ P (Q ( Q R))]

P [ P (Q (Q R))]

P [ P (Q R)]

P [P (Q R)]

P (P Q R)

S P Q R

This is the required PCNF (or) POS

To Find: PDNF (or) SOP

S (P Q R) ( P Q R) ( P Q R) (P Q R)

(P Q R) ( P Q R) ( P Q R)

S (P Q R) (P Q R) (P Q R) ( P Q R)

( P Q R) ( P Q R) ( P Q R)


36

S ( P Q R) (P Q R) (P Q R) ( P Q R)

( P Q R) (P Q R) (P Q R)

This is the required PDNF (or) SOP.

19. (a) Prove that P Q, Q R, P R R by using indirect method.

(b) Derive P (Q S) from the premises P (Q R), Q (R S) by using rule

CP.

Solution (a):

Let us assume the R be one of the premises and arrive a contradiction.

(1) R Rule p

(2) P R Rule p

(3) P Q Rule p

(4) Q R Rule p

(5) P R Rule T, (3), (4)

(6) P Rule T, (1), (5) [ Q, P Q P]

(7) R Rule T, (2), (6) [ P, P Q Q]

(8) R R Rule T, (1), (7)

(9) False

Solution (b):

(1) P (Q R) Rule p

(2) Q (R S) Rule p

(3) P Assumed premise

(4) Q R Rule T, (1), (3)

(5) Q Assumed premise

(6) R S Rule T, (2), (5)

(7) Q S Rule T, (4), (6)

(8) P (Q S) Rule CP


37

20. (a) Show that validity of following arguments:

P Q, Q R, R, P (J S), C:J S.

(b) Derive the following using rule CP if necessary

P Q, Q R, R S P S

Solution (a):

(1) P Q Rule p

(2) Q R Rule p

(3) P R Rule T, (1), (2)

(4) R P contra positive

(5) R Rule P

(6) P Rule T, (4), (5)

(7) P (J S) Rule P

(8) P (J S) Rule T, (7)

(9) J S Rule T, (6), (8)

Solution (b):

(1) P Q Rule P

(2) P Q Rule T, (1)

(3) Q R Rule P

(4) Q R Rule T, (3)

(5) R S Rule P

(6) Q S Rule T, (3), (4)

(7) P S Rule T, (2), (6)

21. Obtain the principal conjunctive Normal form of a statement

P R Q P

(N/D-2008).

Solution:

PCNF:

P R Q Q Q P R R

P R R R


38

P Q R P Q R P Q R P Q R P R R R

S P Q R P Q R P Q R P Q R P Q R P Q R

The RHS is the product of sums form.

22. Show that RVS follows logically from the premises (C D),((C D) H),

H(A B) and (A B) (R S) (Nov/Dec 2008) (May/June-2009)

Solution:

1. (C D) H Rule P

2. H (A B) Rule P

3. (C D) (A B) Rule T,(1),(2)

4. (A B) (RVS) Rule P

5. (C D) (R S) Rule T, (3), (4)

6. (C D) Rule P

7. R S Rule T, (5), (6) and P, P Q P

Hence proved

23. Show that d can be derived from the premises (ab) (aC), (b C), (d a).

(Nov/Dec – 2008)

Sol:

Proof sequence

Steps Premises Rule Reason

(1) (AB) (AC) P Given premises

(2) A (B C) T (1) (AB) (AC) A(B C)

(3) (B C) P Given premise

(4) A T (2),(3) (modus Tollens)

A (B C) (B C) A

(5) D A P Given Premise

(6) D T (4),(5)(P Q) P Q Disjunctive

syllogism

Hence we conclude D from given premises.


39

24. Prove that ( Q P) ( R Q) (pR). [April/May – 2008]

Solution:

Let the formula S: ( Q P) ( R Q) (pR)

To prove: S is Tautology

Truth Table

P Q R P Q R Q

P

R

Q

( Q

P) (

R Q)

PR S

T T T F F F T T T T T

T T F F F T T F F F T

T F T F T F F T F T T

T F F F T T F T F F T

F T T T F F T T T T T

F T F T F T T F F T T

F F T T T F T T T T T

F F F T T T T T T T T

The truth values of (7Q7P) (7R7Q) (P R) are all T.

(7Q7P) (7R7Q) (P R) is a tautology

Hence (7Q7P) (7R7Q) (P R)


40

UNIT – II

PREDICATE CALCULUS

Predicates

Statement Functions

Variables

Free and Bound variables

Quantifiers

Universe of Discourse

Logical Equivalences and Implications for quantified statements

Theory of inference

The rules of Universal specification and generalization

Validity of arguments


41

INTRODUCTION

So far our discussion of symbolic logic has been limited to the consideration

of statements and statement formulas. The inference theory was also restricted in

the sense that the premises and conclusions were all statements. The symbols P, Q,

R … P1, Q1 … were used for statements or statement variables. The statements were

taken as basic units of statements calculus, and no analysis of any atomic statement

was admitted. Only compound formulas were analyzed, and this analysis was done

by studying the forms of the compound formulas, i.e., the connections between the

constituent atomic statements. It was not possible to express the fact that any two

atomic statements have some features in common. In order to investigate questions

of this nature, we introduce the concept of a predicate in an atomic statement. The

logic based upon the analysis of predicates in any statements in any statement is

called predicate logic.

Example:

Represent the following statements in predicate notations:

Sheela is beautiful.

Diana is beautiful.

Solutions:

The part ‚is beautiful‛ is predicate. Let us denote it by B. The above statements

in predicate form is B(s) and B(d) where ‘s’ denotes Sheela and ‘d’ denotes Diana.

Example:

Represent the following statements in predicate notation:

1. Pasupathy is a batsman.

2. Vijay is a batsman.

Solutions:

In the above statements the part ‘is a batsman’ is predicate. Let us denote it by

B. The above statements in predicate notation is B(p) and B(v) where ‘p’ denotes the

subject Pasupathy and ‘v’ denotes the subject Vijay.


42

SIMPLE STATEMENT FUNCTION: It is defined to be an expression a predicate

symbol and variable.

COMPOUND STATEMENT FUNCTION: A compound statement function is

obtained from combining one or more simple statement function and the logic

connectives.

QUANTIFIERS: There are two types of quantifier. They are

(a) Universal quantifier (b) Existential quantifier

UNIVERSAL QUANTIFIER: The universal quantifier of P(x) is a proposition ‚P(x) is

true for all values of x‛ in the universe of discourse.

Notation: ( x)P(x) (or) (x)P(x)

EXISTENTIAL QUANTIFIER: The existential quantifiers of P(x) is a proposition

‚There exists an element x such that P(x) is true‛ in the universe of discourse.

Notation: ( x)P(x)

UNIVERSE OF DISCOURSE: Many mathematical statements assert that a property

is true for all values of a variable in a particular domain is called the universe of

discourse.

BOUND VARIABLE: When a quantifier is a used on the variable ‘x’ or when we

assign a value to the variable, we say that this occurrence of a variable is bound.

FREE VARIABLE: The occurrence of a variable that is not bound by a quantifier or

set equal to the particular value is called a free variable.

SCOPE: The point of a logical expression to which a quantifier is applied is called

scope of a quantifier.

NEGATION OF A QUANTIFIED EXPRESSION: If P(x) is the statement ‚x has

studied computer programming‛, then ( x)P(x) means that ‚every student (in the


43

class) has studied computer programming‛. The negation of this statement is ‚It is

not the case that every student in the class has studied computer programming‛ or

equivalently ‚There is a student in the class who has not studied computer

programming‛ which is denoted by ( x) P(x). Thus we see that ( x)P(x) ( x)

P(x).

INFERENCE THEORY ON PREDICATE CALCULUS:

RULES OF QUANTIFIERS:

Rule US: From ( x)A(x) one can conclude that A(y).

Rule ES: From ( x)A(x) one can conclude that A(y) provided that y is not free in

any given premises, and also not free is any prior stops of the derivation.

Rule EG: From A(x) one can conclude ( y)A(y).

Rule UG: From A(x) one can conclude ( y)A(y) provided that x is not free in any

of the given premises and provided that if x is free in prior step conclude

resulted from use of ‘ES’, but no variables introduced by that use of ES

appear free in A(x).

CONSISTENT:

A set of formulas H1, H2 …… Hm is said to be consistent if their conjunction has the

truth value ‘T’ for assignment of true truth values to the atomic variables appearing

in H1, H2 …… Hm.

INCONSISTENT:

If for every assignment of truth values to the atomic variables, at least one of the

formulas H1, H2 …... Hm is false than the formulas H1, H2 …… Hm are called

Inconsistent.


44

INDIRECT METHOD OF PROOF:

In order to show a conclusion that a conclusion ‘c’ follows logically from the

premises H1, H2 …… Hm, we assume that ‘c’ is false and consider ‘ c’ as an

additional premises, if the new set of premises is inconsistent, so that truly imply a

contradiction. Thus assumption that ‘ c’ is true doesn’t hold simultaneously with

H1 H2 …… Hm being true.

c is true whenever H1 H2 …… Hm is true.

Thus c follows from the premises H1, H2 …… Hm.

PART – A

1. Define predicates.

Predicates is the part of the sentence which predict about the object. The logic

based upon the analysis of predicates in any statement is called predicate logic.

Example: (1) Ram is a boy.

(2) Gopi is a boy.

The part ‚is a boy‛ is called predicate.

2. Define M-Place predicate.

A predicate requiring m (m > 0) names is called as m - place predicates. If S is an

the names of object, then S(a1, a2 ... am) are the names of objects, then S(a1, a2 ... am) is a

statement.

3. Define compound statement function.

Statement functions which are obtained function and the logical connectives are

called compound statement functions.


45

Example. If M(x) : X is a man

H(x) : X is a mortal.

Then the compound statement functions can be formed as

M(x) H(x), M(x) H(x), H(x), M(x) H(x) etc.

4. Define Quantifiers.

The symbol (x) or ( X) are called universal quantifiers. The symbol ( x) is called

existential quantifiers.

(x) or ( x) stands for ‚for all a‛, ‚for every‛ and ( x) stands for ‚there is at least

one x‛ ‚such that, there is exists an x‛ ‚such that for some x‛.

5. Define atomic formula.

Let us denote P(x1, x2 ... xn) for an n-place predicate formula in which P is an n-

Place predicate and x1, x2 ... xn are individual variables. In general P(x1, x2...) will be

called an atomic formula of predicate calculus.

6. State the rules of a well-formed formula of predicate calculus.

i) An atomic formula is a well formed formula.

ii) If A is a well-formed formula, then A is a well formed formula.

iii) If A and B are well-formed formula, then (A B), (A B), (A B),

(A B) are also well-formed formula.

iv) If A is well-formed formula and x is any variable, then (x)A(x) and

( x)A(x) are well-formed formula.

v) Only there formula obtained by using (i) to (iv) are well-formed

formulae.

7. Define free and bound variables.

In a formula of the form (x)P(x) or ( x)P(x), ‚x‛ is called a bound variable while

any occurrence of is not bound, is called free occurrence. P(x) is called the scope of

the quantifiers.


46

Example:

(x)P(x,y)

There P(x,y) is the scope of the quantifier and both occurrences of x are bound

occurrences while the occurrence of y is a free occurrence.

8. Let P(x) : x is a person

F(x, y) : x is the father of y

M(x, y) : x is the mother of y.

Write the predicate ‚x is the father of the mother of y‛. Solution:

In order to symbolize the predicate we name a person called z as the mother

of y. Obviously we want to say that x is the father of z and z the mother of y. It is

assumed that such a person z exists.

We symbolize the predicate

( z)(P(z) F(x,z) M(z,y)).

9. Define universe of discourse or domain.

The variables which are quantified stand for only there objects which are

members of a particular set of class. Such a restricted class is called the universe of

discourse of domain of individual or simply the universe.

10. Symbolize the statement ‚ All men are giants ‚

Solution:

Let G(x) : x is a giant

M(x) : x is a man

Then the given statement is symbolized as (x)(M(x) G(x)).

If we restrict the variable x to the universe which is the class of men, then the given

statement is (x)G(x).


47

11. Symbolize the following statement assuming the real numbers as the universe

of discourse.

(i) There are +ve values of x and y such that xy > 0.

(ii) There is a value of x such that if y is +ve x+y is -ve

(i) ( x)( y)((x>0) (y>0) (xy>0))

(ii) ( x)( y)((y>0) (x+y<0))

12. Define equivalence of two formula.

Let A and B be two predicate formulae defined over a common universe denoted

by E. If for every assignment of object names from the universe of discourse E to

each of the variables appearing in A and B, the resulting statements have the same

truth values, then the predicate formula A and B are said to be equivalent to each

other over E. It is written as A B.

13. When you will say formula A is valid in E.

A formula A is said to be valid in E written ‚T‛ A in E, if for every assignment of

object names from E to the corresponding variables in A and for every assignment of

statements to statement variables, the resulting statements have the truth value T.

14. Prove that ((x)A(x)) ( x)( A(x))

((x)A(x)) (A(a1) A(a2) ... A(an))

(A(a1)) (A(a2)) ... (A(an))

( x)( A(x))

15. Negate the following statements.

(a) Ottawa is a small town.

(b) Every city in Canada is clean.

Solution:

Some possible negations are

(a) ‚It is not the case that Ottawa is a small town‛ (or) ‚Ottawa is not a small

town‛.

(b) ‚It is not the case that every city in Canada is clean‛ (or) ‚every city in

Canada is not clean‛ (or) ‚not every city in Canada is clean‛.


48

16. Show that (x)(H(x) M(x)) H(s) M(s).

Solution:

{1} (1) (x)(H(x) M(x)) P

{1} (2) H(s) M(s) US, (1)

{3} (3) H(s) P

{1, 3} (4) (4) M(s) T, (2), (3), I11

17. Show that P(a,b) follows logically from (x)(y)(P(x,y) W(x,y)) and W(a,b).

Solution:

{1} (1) (x)(y)(P(x,y) W(x,y)) P

{1} (2) (y)(P(a,y) W(a,y)) US, (1)

{1} (3) P(a,b) W(a,b) US, (2)

{1} (4) W(a,b) P

{1, 4} (5) P(a,b) T, (3), (4), I12.

18. Define Rule ES.

From ( x)A(x) one can conclude A(y) provided that y is not free in any given

premise and also free in any prior step of the derivation. These requirements are

easily be made by changing a new variable each time ES is used.

19. Define Rule UG.

From A(x) one can conclude (y)A(y) provided that x is not free in any of the

given premises and provided that if x is free in a prior step which resulted from use

of US, then no variables introduced by that use of US appear free in A(x).

20. Show that Q, P Q P.

Solution:

{1} (1) P Q P

{1} (2) Q P T, (1) and E18

{2} (3) Q P

{1,2} (4) P T, (2), (3) and I11


49

21. Find the truth value of (x)(P Q(x)) ( x)R(x) where P : 2>1, Q(x) : x>3, R(x) : x>4

with the universe of discourse E being E = {2, 3, 4}. (AU/Nov–Dec/2006)

Solution:-

P is true and Q(4) is false. Hence P Q(4) is false. Therefore (x)(P Q(x)) is

false. Since R(2), R(3), R(4) are all false ( x)R(x) is also false. Hence

(x)(P Q(x)) ( x)R(x) is false.

22. Write the following statement in symbolic form ‚Everyone who likes fun will

enjoy each of these plays‛ (May / June 2009)

Let L (x) : x like fun

P (x) : x is a play

E (x,y): x will enjoy y

The given statement can be represented as, ‚For each x, if x likes fun and for each y,

if y is a play then x enjoys y‛.

Symbolic form: ( x) ( y) (L (x) P(y) E (x,y))

23. Determine whether the conclusion C follows logically from the premises H1

and H2 or not H1 : P Q, H2 : P, C:Q (Nov / Dec 2008)

H1: P Q H2 : P C:Q

C H2 H1

Q P P Q

T

F

T

F

T

T

F

F

T

F

T

T

In the third row conclusion H2 is F and H1 is T

The premises H1 and H2 or not.


50

PART – B

1. (a) Show that {(P q) (r s), (q t) (s u), (t u), p r} p

(b) Show that if p q, q r, (p r) and p r, then r.

Solution (a):

{1} (1) (p q) (r s) P

{1} (2) p q T, (1)

{1} (3) r s T, (1)

{4} (4) (q t) (s u) P

{4} (5) q t T, (4)

{4} (6) s u T, (4)

{1,4} (7) p t T, (2), (5), hyp. Syll.

{1,4} (8) t p T, (7) and E16.

{1,4} (9) r u T, (3), (6), hyp. Syll.

{10} (10) p r P

{1,10} (11) p u T, (10), (3), hyp. Syll.

{1,10} (12) u p T, (11) and E18

{1,4,10} (13) ( t u) p T, (8), (12) and I14

{14,10} (14) (t u) p T, (13), De Morgan’s law

{15} (15) (t u) P

{1,4,10,15} (16) p T, (14), (15)

The third and last rule of inference is the rule of conditional proof, which we

call Rule C.P.

Solution (b):

{1}

{2}

{1,2}

{1,2}

{5}

{1,2,5}

{1,2,5}

{8}

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

p q

q r

p r

p r

p r

(p p) r

r

(p r)

P

P

T, (1), (2) and I13

T, (3), and E16

P

T, (4), (5), distributive law

T, (6), (p p) = F

P


51

{8}

{1,2,5,8}

{1,2,5,8}

{1,2,5,8}

{1,2,5,8}

(9)

(10)

(11)

(12)

(13)

p r

r ( p r)

(r p) (r r)

r p

r

T, (8), De morgan’s law

T, (7), (9),

T, (10), distributive law

T, (11), F = (r r)

T, (12), simplifications

2. (a)Using indirect method of proof, derive P S from P Q R, Q P, S R,

P.

(b) Prove by indirect method that ( q), p q, p t t.

(c) Using indirect proof, show that p q, q r, (p r), p r r.

Solution (a):

The desired result is P S. Its negation is P S.

(P S ( P S) (P S) is tautology.

include P S as an additional premise.

{1}

{2}

{1,2}

{4}

{5}

{5}

{4,5}

{1,2,4,5}

{9}

{1,2,4,5,9}

{1,2,4,5,9}

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

(11)

P Q R

P

Q R

S R

P S

S

R

Q

Q P

P

P P

P

P

T, (1), (2), modus ponens

P

P (new premise)

T, (5) and simplification


T, (3), (7 ), I10

P


T, (2), (10) contradiction

Thus additional premise P S and the given premises together lead to a

contradiction.

So (P S) is derivable from P Q R, Q P, S R, P desired result is

(P S), include its negation as a new premise.


52

Solution (b):

{1}

{2}

{1,2}

{4}

{1,2,4}

{6}

{1,2,4,6}

(1)

(2)

(3)

(4)

(5)

(6)

(7)

p t

t

p

p q

q

q

q q

P

P (additional premise)

T, (1), (2).

P


P

T, (5), (6), contradiction

The new premise, together with the given premises leads to a contradiction.

Thus ( q), p q, p t t.

The desired result is t, include t as a new premise.

Solution (c):

{1}

{2}

{1,2}

{4}

{1,2,4}

{6}

{1,2,4,6}

{1,2,4,6}

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

q r

r

q

p q

p

p r

r

r r

P

P (additional premise)

T, (1), (2).

P

T, (3), (4), modus tollens

P

T, (5), (6)

T, (2), (7)

Thus we get a contradiction

we get P q, q r, p r, r

We note that the other premise (p r) will not yield a contradiction with r.


53

3. (a) Determine the validity of the following argument. If two sides of a triangle

are equal, then two opposite angles are equal. Two sides of a triangle

are not equal. Therefore, the opposite sides are not equal.

(b) Determine the validity of the following argument. My father praises me

only if I can be proud of myself. Either I do well in sports or I cannot be

proud of myself. If I study hard, then I cannot do well in sports. Therefore,

if father praises me, then I do not study well.

Solution (a):

Let us indicate the statement as follows

P : Two sides of a triangle are equal.

Q : The two opposite angle are equal.

The given argument in the form

P Q, P Q

Let us construct the truth table for (P Q) ( P) Q.

P Q P Q P (P Q) ( P) Q

T

T

F

F

T

F

T

F

T

F

T

T

F

F

T

T

F

F

T

T

F

T

F

T

This shows that (P Q) ( P) ( Q) is not a tautology. Hence the conclusion

Q is not valid.

Solution (b):

Let us indicate the statement as follows

P : My father praises me

Q : I can be proud of myself

R : I do well in sports

S : I study hard.


54

The given argument is of the form

P Q, R Q, S R P S.

As the desired result in P S, assume that P is true. It is enough to verify

the validity of

P, P Q, R Q, S R S.

From P and P Q, we have Q.

From Q and R Q, We have R.

From R and S R, we have S.

Thus the argument is valid.

4. (a) Show that the following system of premises is inconsistent: ‚The contract

is satisfied if the building is completed by November 30. The building is

completed by November 30 iff the electrical subcontractor completes his

work by Nov 10. The bank loses money iff the contract is not satisfied.

The electrical subcontractor completes his work by Nov 10 if the bank

loses money.‛

(b) Show that the following set of premises is inconsistent. ‚If the contract is

valid, then John is liable for penalty. If John is liable for penalty he will

go bankrupt. If the bank will loan him money, he will not go bankrupt.

As a matter of fact, the contract is valid and the bank will loan him

money.‛

Solution (a):

Let us indicate the statements as follows.

C : The contract is satisfied.

B : The building is completed by November 30.

E : The Electrical subcontractor completes his work by Nov 10.

L : The bank loses money.

The given premises are C B, B E, L C, E L.


55

Now

from C B, B E, we have C E,

from C E and E L, we have C L,

from C L and L C, we have C C.

Which is a contradiction.

Therefore the system is inconsistent.

Solution (b):

We indicate the given statements as follows

V : The contract is valid

L : John in liable for penalty.

M : Bank will loan him money

B : He will go bankrupt

Then the given premises are V L, L B, M B, V M.

{1}

{2}

{1,2}

{4}

{4}

{4}

{1,2,4}

{8}

{4,8}

{1,2,4,8}

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

V L

L B

V B

V M

V

M

B

M B

B

B B

P

P

T, (1), (2), Hypo Syllogism

P

T, (4), Simplification

T, (4), Simplification

T, (3), (5), Modus Ponens

P

T, (6), (8), Modus Ponens

T, (7), (9) Contradiction

Thus the given set of premises leads to a contradiction and hence it is inconsistent.


56

5. (a) Show that x P(x) Q(x) x P(x) x Q(x) is a logically valid

statement.

(b) Show that x P(x) Q(x) x P(x) x Q(x) is valid statement.

Solution (a):

If ( x) P(x) Q(x) is true, then for every c in the universe, P(c) Q(c) is true.

Therefore for each c, P(c) is true and for each c, Q(c)is true. Thus

( x)(P(x)) ( x)(Q(x)) is true.

This shows that ( x) P(x) Q(x) x P(x) x Q(x) is valid.

Conversely, if ( x) P(x) x Q(x) is true, then for each c in the universe, P(c) is

true and for each c in the universe, Q(c) is true.

P(c) Q(c) is true for each object c in the universe.

Thus ( x) P(x) Q(x) is true and

x p(x) x Q(x) x p(x) Q(x) is valid and hence

x p(x) Q(x) x p(x) x Q(x) is logically valid statement.

Solution (b):

We know that x p(x) Q(x) x p(x) x Q(x) is valid

statement.

Hence, x P(x) Q(x) x P(x) x Q(x) is valid.

If two statements always have the same truth value then their negation always have

the same truth value.

Hence x P(x) Q(x) x P(x) x Q(x) is valid.

x P(x) Q(x) x P(x) x Q(x) is a valid statement.


57

6. (a) Show that x P(x) x Qx x P(x) Q(x) is logically valid.

(b) Also show by counter example

x P(x) Q(x) x P(x) x Q(x) is not valid.

Solution (a):

Consider the case when x P(x) x Q(x) is true.

Since this is a disjunction of if–statements, one of the statements x P(x) and

x Q(x) must be true.

If x p(x) is true, then for every object b in the universe P(b) is true, and hence

P(b) Q(b) is true.

Similarly when x Q(x) is true, P(b) Q(b) is true for every object b.

In both cases P(b) Q(b) is true for all b in the universe.

( x) P(b) Q(b) is true and ( x) P(x) ( x) Q(x) ( x) P(x) Q(x) is a

valid statements.

Solution (b):

Consider the following statement ( x) P(x) Q(x)

where P(x) : x is even integer

Q(x) : x is Prime integer

and the universe is {2, 4, 6, 8, 3, 7, 11}

For the universe, the statement ( x) P(x) Q(x) is true. But both ( x) p(x) and

( x) Q(x) are not true.

So ( x) P(x) Q(x) is true, while x P(x) x Q(x) is not true.

Thus ( x) P(x) (Qx) ( x) p(x) ( x) Q(x) is not a valid statement.


58

7. Which of the following preparation are equivalent over the universe of the

integers?

a) 20 n 4 b) 3

0 n 8 c) 0 n 2

Solution:

Let the universe consist of all integers.

Let P(n) : n is an integer such that 20 n 4 .

Q(n) : n is an integer such that 30 n 8.

R(n) : n is an integer such that 0 n 2.

The statement P(n) has the truth value ‚True‛ only when n = 1, 2, -1,-2.

The statement Q(n) has the truth value ‚True‛ only when n = 1, 2.

The statement R(n) has truth value ‚True‛ only when the n = 1, 2.

The truth value of Q(n) and R(n) are same, hence the statement Q(n) and R(n) are

equivalent to each other.

P(n) is not equivalent to either Q(n) or R(n).

8. a) Verify the validity of the following argument?

‚Lions are dangerous animals. There are lions. Therefore there are

dangerous animals.‛

b) Give an argument which will establish the validity of the following inference.

‚All integers are rational numbers some integers are power of 2. Therefore,

some rational numbers are powers of 2.‛

Solution (a):

Let L(x) : x is a Lion.

D(x) : x is a dangerous animal.

Then the inference pattern is : ( x) L(x) D(x) ,( x) L(x) ( x) D(x) .

{1} (1) ( x) L(x) P

{2} (2) L(b) ES, (1)

{3} (3) ( x) L(x) D(x) P


59

{1,3} (4) L(b) D(b) US, (3)

{1,3} (5) D(b) T, (2), (4)

{1,3} (6) ( x) D(x) EG, (5)

Thus the inference is valid

Solution (b):

Let P(x) : x is an integer.

R(x) : x is a rational number.

S(x) : x is a power of 2

Then the given inference pattern is : x P x R x , x P x S x

x R x S x .

{1} (1) x P x S x P

{1} (2) P b S b ES, (1)

{1} (3) P b T, (2)

{1} (4) S b T, (2)

{5} (5) x P x R x P

{5} (6) P b R b US, (5)

{1,5} (7) R b T, (3), (6)

{1,5} (8) R b S b T, (7), (4)

{1,5} (9) x P x R x EG, (8)

9. Show that from ( x)(F(x) S(x)) ( y)(M(y) W(y)), ( y)(M(y) W(y)). The

conclusion ( x)(F(x) S(x)) follows.

Solution:

{1} (1) ( y)(M(y) W(y)) P

{1} (2) M(z) W(z) ES, (1)

{1} (3) (M(z) W(z)) T, (2), E17

{1} (4) ( y) (M(y) W(y)) EG, (3)

{1} (5) ( y)(M(y) W(y)) (4), E26


60

{6} (6) ( x)(F(x) S(x)) ( y)(M(y) W(y)) P

{1,6} (7) ( x)(F(x) S(x)) T, (5),(6), I12

{1,6} (8) ( x) (F(x) S(x)) T,(7), E25

{1,6} (9) ( F(z) S(z)) Us, (8)

{1,6} (10) F(z) S(z) T,(9), E9,E16

{1,6} (11) ( x)(F(x) S(x)) UG, (10)

10. Show that ( x)(P(x) Q(x)) ( x)P(x) ( x)Q(x).

Solution

We shall use the indirect method of proof by assuming (( x)P(x) ( x)Q(x))

{1}

{1}

{1}

{1}

{1}

{1}

{1}

{1}

{1}

{1}

{11}

{11}

{1,11}

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

(11)

(12)

(13)

(( x)P(x) ( x)Q(x))

(( x)P(x)) (( x)Q(x))

(( x)P(x))

( x)( (P(x))

(( x)Q(x))

( x)( Q(x))

P(y)

Q(y)

P(y) Q(y)

(P(y) Q(y))

( x)(P(x) Q(x))

P(y) Q(y)

( P(y) Q(y)) (P(y) Q(y))

P (assumed)

T, (1), Demargan’s law

T, (2), I2

T, (3), E26

T, (2), I2

T, (5), E25

Es, (4)

US, (6)

T, (7) (8), I9

T, (9), Demargan’s law

P

US, (11)

T, (10), (12), I9

Contradiction.

11. Is the following conclusion validly derivable form the premises given?

If ( x)(P(x) Q(x)), ( y)P(y), then ( z)Q(z)?

Solution

We use the indirect method, by assuming that the conclusion ( z)P(z) is false.

{1}

{1}

(1)

(2)

(( z)Q(z))

( z) (Q(z))

P (assumed)

T, (1)


61

{3}

{3}

{1}

{1,3}

{1,3}

{8}

{8}

{1,3,8}

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

( y)P(y)

P(a)

Q(a)

P(a) Q(a)

(P(a) Q(a))

( x)(P(x) Q(x))

P(a) Q(a)

(P(a) Q(a)) (P(a) Q(a))

P

ES, (3)

US, (2)

T, (4), (5)

T, (6)

P

US, (8)

T, (7), (9)

Contradiction.

12. Using CP or otherwise obtain the following implications?

( x)(P(x) Q(x)), ( x)(R(x) Q(x)) ( x)(R(x) P(x))

Solution

{1}

{2}

{2}

{4}

{2,4}

{1}

{1,2,4}

{1,2,4}

{1,2}

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

( x)(P(x) Q(x))

( x)(R(x) Q(x))

R(x) Q(x)

R(x)

Q(x)

P(x) Q(x)

P(x)

R(x) P(x)

( x)(R(x) P(x))

P

P

US, (2)

P (assumed)

T, (3), (4)

US, (1)

T, (5), (6)

CP, (4), ( )

UG, (9)

Hence the argument is valid.

13. Write in the symbolic form and negate the following statements.

(a) Every one who is healthy can do all kinds of work.

(b) Some people are not admired by everyone.

Solution (a):

The given statement is ‚every one who is healthy can do all kinds of work‛.

Let H(x) : x is a healthy person.

W(y) : y is a kind of work.

D(x,y) : x can do y.


62

The statement is ‚for all x, if x is healthy, and for all y, if y is a kind of work, then x

can do y’. So a symbolic form is ( x)( y)(H(x) W(y) D(x,y)).

Its negation is given by (( x)( y)(H(x) W(y) D(x,y)))

ie., ( x)( ( y)(H(x) W(y) D(x,y)))

ie., ( x)( y)( (H(x) W(y) D(x,y)))

ie., ( x)( y)(H(x) H(y) D(x,y))

That is ‚there exists a healthy person and there exists a kind of work such that x

cannot do y‛.

‚there is some healthy person who cannot do some kind of work‛.

Solution (b):

In the universe of people, let

A(x,y) : x admires y

Then the given statement is ‚there is a person who is not admired by some person‛.

So it is ( y)( x)(P(x) P(y) A(x,y)).

Its negation is (( y)( x)(P(x) P(y) A(x,y))).

ie., ( y) (( x)(P(x) P(y) A(x,y)))

ie., ( y)( x)( (P(x) P(y) A(x,y)))

ie., ( y)( x)(P(x) P(y) A(x,y)).

ie., every person is admired by every one.

In the universe, which consists of every thing,

Let P(x) : x is a person

and A(x, y) : x admires y.

Then the given statement is ( y)( x)(P(x) P(y) A(x,y)).

Its negations is ( y)( x)(P(x) P(y) A(x,y)).


63

14. P.T x P x Q x x P x x Q x (Nov / Dec 2008).

Steps Premises Reason

1 x P x Q x Rule P

2 P (y) Q (y) Rule ES, (1) Y fined

3 P (y) Rule T, (2)

4 ( x ) P (x) Rule EG, (3)

5 Q (y) Rule T, (2)

6 ( x) Q (x) Rule EG, (5)

7 ( x) (P(x)) ( x) Q(x) Rule T, (4), (6) P,Q P Q

Hence proved.

15. Show that (x) (P(x) Q (n)) (x) (Q(x) R (x)) (x) (P (x) R (x)) (Nov / Dec

2008)

Solution:

Proof Sequence

Steps Premises Reason

1 ( x) (P(x) Q(x)) P

2 P (y) Q (y) US, (1)

3 ( x) (Q(x) R(x)) P

4 Q (y) R (y) US, (3)

5 P(y) R (y) T (2), (4) Hypothetical syllogism

(P Q) (Q R) P R

6 ( x) (P(x) R(x)) Rule UG, (5)

Hence proved.

16. Show that (x) [ p (x) Q (x) ], (x) [R(x) Q (x)] (x) [R(n) P (x)

](Apr/May)(May/June 2009)

Solution:


1 (x) [ P (x) Q (x) ] P Given premise

2 P(y) Q (y) US (1), (x) P (x) P (y)

3 (x) [R(x) Q (x) P Given Premise


64

4. R(y) Q (y) US (3), (x) P (x) P (y)

5 Q(y) R (y) T (4) P Q Q P

6 P(y) R (y) T (2), (5) (P Q) (Q R) P R

7 R(y) P (y) T (6), P Q Q P

8 (x) [ R (x) P(x)] UG (7), P(y) (x) P (x)

Hence Proved.

17. Using derivation process prove that S Q, SVR, R, ( R Q ) R

(Apr/May 2008)

Solution:

Proof sequence


1 SVR P Given Premise

2 R P Given premise

3 S T (1), (2) (P Q) Q P

4 S Q P Given Premise

5 Q T (3), (4) modus pones P Q, P Q

6 R Q P Given premise

7 ( R Q) (Q R) T (6) ( P Q ) (P Q) (Q P)

8 R Q T ( ) P Q P

9 ( R) T (P Q) Q P

10 R T (9), ( R) P


65

UNIT – III

SET THEORY

BASIC CONCEPTS - NOTATION - SUBSET

ALGEBRA OF SETS - THE POWER SET

ORDERED PAIRS AND CARTESIAN PRODUCT

RELATIONS ON SETS

TYPES OF RELATIONS AND THEIR PROPERTIES

RELATIONAL MATRIX AND THE GRAPH OF A RELATION

PARTITIONS

EQUIVALENCE RELATIONS

PARTIAL ORDERING - POSET

HASSE DIAGRAM

LATTICES AND THEIR PROPERTIES - SUBLATTICES

BOOLEAN ALGEBRA - HOMOMORPHISM


66

BASIC CONCEPTS OF SET THOERY

INTRODUCTION:

The concepts of membership and inclusion are given. Certain special sets such

as the

Universal Set, Empty Set, and the Power Set of a given set are introduced.

SET: A set is a well defined collection of objects. The objects are called the elements

of the set. If an element x belongs to a set A, x is called a member of A and we write

x A. ( is ‘read as belongs to’ and as ‘does not belong to’ ).

Example: Z = ,…, -3, -2, -1, 0, 1, 2, 3 …} set of integers.

FINITE & INFINITE SET: A set is said to be Finite if it contains a finite number of

distinguishable elements. Otherwise a set is Infinite.

Example: The empty set and the set of letters of English alphabets are Finite

sets, whereas the set of even positive integers ,2, 4, 6, 8 …} is Infinite.

NULL SET (EMPTY SET): The set that has no elements in it is called a null set or an

empty set. It is denoted by the symbol or { }.

Example: The square of a real number is always non- negative.

{x | x is a real number and x2 = -1} =

Note: | | = 0 but ,0} ≠ also ≠ , }

CARDINALITY OF A SET: Cardinality is the number of elements in a set and it is

denoted by # (A) or |A| or n(A).

Example: If A = {1, 3, 5, 9} then the cardinality of A is n(A) = 4.

SINGLETON SET: A set consisting of single element is called singleton set.

Example: A = {3}.


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SUBSET: If every element of A is also element of B, that is if whenever x A then x

B, we say that A is a subset of B, or that A is contained in B and we write A B. If

A is not a subset of B, we write A B.

Example:

(1)

Diagram which are used to show relationships between sets, are called Venn

diagrams, after the British logician John Venn.

(2) N = ,1, 2, 3 ……} be the set of all natural numbers, and Z = ,…, -3, -2, -1, 0, 1, 2 …}

be the set of all integers, then N Z.

Note:

(1). If A is any set then, A A. That is every set is subset of itself.

(2). Null set is the subset of all sets.

UNIVERSAL SET: A set is called a universal set if it includes every set under

discussion. A universal set will be denoted by E.

Note: For any set A, we have A E. Thus every element x E, that is (x) (x E) is

identically true. One could specify E in a variety of ways.

Example: E = { x | p(x) p(x)}, where p(x) is any predicate.

EQUALITY OF SETS. Two sets A and B are said to equal iff A B and B A.

i.e., A = B (A B) and (B A)

Note: The following are some of the important properties of an inclusion. For any

sets A, B and C


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A A (Reflexive) (1)

(A B) (B C) (A C) (Transitive) (2)

Example:

1. The following sets are equal

(i) {1, 2, 4} = {1, 2, 2, 4}

(ii) {1, 4, 2} = {1, 2, 4}

(iii),1, 3, 5 …} =,x | x is an odd positive integer}

2. The following sets are not equal

(i) { {1, 2}, 4}={1, 2, 4}

(ii) { {1} } {1} because {1} { {1} }

PROPER SET: A set A is called a proper subset of a set B if A B and A B. It

is written as A B, so that A B (A B A B)

Example: N is a proper subset of Z.

i.e., N Z.

POWER SET. For a set A, a collection or family of all subset of A is called the

power set of A. The power set of A is denoted by (A) or P(A) = 2A, so that (A)= 2A

= {x | x A}

Note:

1. The power set of the null set has only the element , hence ( )={ }

2. For a set S1 = {a}, the power set (S1) = { , {a}} = { , S1}

3. One can easily note that the number of elements of p(A) is 2 A . i.e., p(A) = 2 A .

4. The proper subset of A = p(A) - { , A}

i.e., If a set has n elements, then its power set has 2n elements.


69

Example:

1. What is the power set of a set {0, 1, 2}?

Solution: The power set P({0, 1, 2}) is the set of all subsets of {0, 1, 2}. Hence

P({0, 1, 2}) = { , {0}, {1}, {2}, {3}, {0,1}, {0,2}, {1,2}, {0,1,2}}

2. What is the power set of the empty set? (or) What is the power set of { }?

Solution: The empty set has exactly one subset, namely, itself, consequently,

P( ) = { }

The set has exactly two subsets, namely and the set { } itself.

Therefore, P({ }) = { ,{ }}

3. Give the power set of {a, {b}}

Solution: Let A = {a, {b}}

P(A) = {{a}, {a,{b}}, {{b}}, }

4. Give the power set of {1, }

Solution: Let A = {1, }

P(A) = , 1 , , 1,

INDEX & INDEXED SET: Let J = {s1, s2 ……} and A be a collection of sets A = ,As1,

As2 ……} such that for any si J there corresponds a set Asi A and also Asi = Asj iff

si = sj, then A is called an index set, J the index set, and any subscript such as si of Asi

is called an index.

An indexed family of sets can also be written as A = {Ai}i J

OPERATIONS ON SETS:

INTERSECTION: The intersection of any two sets A and B, written as A B, is the

set consisting of all elements which belong to both A and B.

Symbolically, A B = { x | (x A) (x B) }


70

PROPERTIES OF INTERSECTION OPERATIONS:

Let A, B, C be subsets of U, then

(i) A A = A

(ii) A B = A wherever A B, in particular A U = A

(iii) A B = B wherever B A, in particular A = .

(iv) A B A and A B B

(v) A B = B A

(vi) (A B) C = A (B C) Associative law.

UNION: The union of any two sets A and B, written as A B, is the set consisting of

all elements which belong to either A or B.

Symbolically, A B = { x | (x A) (x B) }

PROPERTIES OF UNION OPERATIONS:

Let A, B, C be subsets of U, then

(i) A A = A

(ii) A = A

(iii) A U = U, A A’ = E

(iv) A A B and B A B

(v) A B = B A (commutative law)

(vi) (A B) C = A (B C) (associative law)

RELATIVE COMPLEMENT OF A SET. Let A and B be any two sets. The relative

complement of B in A (or of B with respect to A), written as A ~ B, is the set

consisting of all elements of A which are not elements of B, that is

A - B = { x | x A x B}

The relative complement of B in A is also called the difference of A and B.


71

PROPERTIES OF THE DIFFERENCE OPERATIONS:

(i) A – B = A B

(ii) A – A =

(iii) A – B B – A

(iv) A – = A

(v) A – B = B – A A = B

(vi) A – B = A A B =

(vii) A B = A B

COMPLEMENT OF A SET: Let E be the universal set. For any set A, the relative

complement of A with respect to E, is E – A and it is called the absolute complement

of A. The absolute complement of a set A is called the complement of A and it is

denoted by ~A symbolically, ~A = E A= {x | x E x A}

= {x | x A}

= {x | (x A)}

PROPERTIES OF THE COMPLEMENT:

(i) ~(~A) = ~ ~A = A

(ii) ~ = E

(iii) ~ E =

(iv) A ~A = E

(v) A ~A =

(vi) ~ (A B) = ~A ~B

(vii) ~ (A B) = ~A ~B

ORDERED PAIR: An ordered pair consists of two objects in a given fixed order.

Note that an ordered pair is not a set consisting of two elements. The ordering of the

two objects is important. The two objects need not be distinct.

CARTESIAN PRODUCT: Let A and B be sets. The Cartesian product of A and B,

denote by A x B, is the set of all ordered pair (x, y) where x A and y B.

Hence A x B = {(x, y) | (x A) (y B)}


72

RELATION ON A SET & ITS PROPERTIES

The word ‚relation‛ suggest some familiar example of relation such as the

relation of father to son mother to son, brother to sister , etc…

Familiar examples in arithmetic relation such as ‚greater than‛, or that of

equality between two real numbers.

RELATION (or) BINARY RELATION: Any set of ordered pairs defines a binary

relation. We shall call a binary relation simplify relation, say (x,y) R where R is a

Relation, by writing xRy which may be read as ‚x is related y by R‛

Example: Let L denote the relation ‚less then or equal to‛ and D denote the relation

‚divides‛ both L and D defined on the set ,1, 2, 3, 6} write L and D. Find L D.

Solution: L = {(1,2), (1,3), (1,6), (1,1), (2,2), (2,3), (2,6), (3,3), (3,6)}

D = {(1,1), (1,2), (1,3), (1,6), (2,2), (2,6), (3,3), (3,6), (6,6)}

L D = {(1,1), (1,2), (1,3), (1,6), (2,2), (2,6), (3,3), (3,6)}

PROPERTIES OF RELATION:

REFLEXIVE: A binary relation R in a set X is reflexive if x X, xRx that is (x,x) R.

That is R is reflexive in X (x)(x X xRx).

SYMMETRIC: A relation R in a set X is a symmetric if for every x and y in X,

whenever xRy then yRx.

That is R is symmetric in X (x)(y)(x X y X xRy yRx)

TRANSITIVE: A relation R in a set X is transitive if for every x, y and z in X,

whenever xRy and yRz, then xRz

That is R is Transitive in X (x)(y)(z)(x X y X z X xRy yRz xRz)

IRREFLEXIVE: A relation R in a Set X is irreflexive if for every x X, (x,x) R.


73

ANTI-SYMMETRIC: A relation R in a set X is anti-symmetric if for every x and y in

X, wherever xRy and yRx, then x = y.

That is R is anti-symmetric in X (x)(y)(x X y X xRy yRx x = y)

PARTITION ON A SET: Let S be a set and A1, A2, A3 … An be a subsets of S. A1, A2,

A3 … An are said to be partition of this set, if (i) n

i

i 1

A = S (ii) Ai Aj = ; i j

EQUIVALENCE RELATION: A relation R on a set X is called an Equivalence

Relation iff it is reflexive, symmetric and transitive.

EQUIVALENCE CLASS: Let R be an Equivalence relation on set X. For any x X,

then, the set [x]R = { y | y X and xRy}. This is called Equivalence Class.

RELATIONAL MATRIX: A relation R from a finite set X to a finite set Y can also be

represented by a matrix called the relational matrix of R.

GRAPH OF A RELATION: A relation can be represented pictorially by a diagram,

called graph. If R is a relation in a set X = {x1, x2 …… xm} each element xi of X is

represented by a point or circle called node or vertex and is labeled by xi itself. If (xi

xj) R then we draw a directed arc from xi to xi. If xi R xi then we draw a directed arc

which starts from node xi, and ended at xi, such an arc is called a loop.

PARTIAL ORDERING & POSET

PARTIAL ORDER RELATION: A binary relation R in a set P is called partial order

relation in P iff R is reflexive, anti-symmetric and transitive.

POSET: A set P together with a partial ordering relation R is called a partially

ordered set or POSET.

HASSE DIAGRAM: A partial ordering ≤ on a set P can be represented by means of

a diagram know as a Hasse diagram or a partially ordered set diagram of (P, ≤ ). In

such a diagram, each element is represented by a small circle or a dot. The circle for

x P is drawn below the circle for y P if x < y, and a line is drawn between x and y


74

if y covers x. If x < y but y does not cover x, then x and y are not connected directly

by a single line. However, they are connected through one or more elements of P. It

is possible to obtain the set of ordered pairs in ≤ from such a diagram.

LATTICE: A lattice is a poset <L, > in which each pair of elements a, b belongs to L

has a g.l.b and l.u.b.

BINARY OPERATION: Let A be a non-empty set. A binary operation * on A is a

function.

* : A × A A

This function * maps each ordered pair (a,b) A × A to a unique element a * b A.

ALGEBRA: A set together with a number of binary operation is called Algebra.

BOOLEAN ALGEBRA: A Boolean Algebra is a complemented, Distributive

lattices. It’s a special lattice, both lattice theory and Boolean algebra have important

applications in the theory and design of computers.

NOTATION:

P NOT P

P Q P AND Q

P Q P OR Q

where P and Q are statements.

STATEMENT ALGEBRA: Set of all statement under the operations , , is an

algebra called statement algebra. Boolean Algebra is a particular case of Statement

Algebra.

PROPERTIES OF LATTICES:

Let < L, > be a lattice. Let * denotes the ‘meet’ and denotes the ‘join’ of the

above lattice. Then for any a, b, c belong to L. We have the following properties

(L-1) a * b = a (L-1)’ a b = a [idempotent]

(L-2) a * b = b * a (L-2)’ a b = b a [commutative]

(L-3) (a * b) * c = a * (b * c) (L-3)’ (a b) c = (a b) c [associative]

(L-4) a * (a b) = a (L-4)’ a (a * b) = a [absorption]


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LATTICE AS ALGEBRAIC SYSTEMS

N – ARY MAPPING: Let X is set. We denote X X X = X3 & so on.

The mapping f : X X … X = Xn X is called n – ary mapping.

ALGEBRAIC SYSTEM: A system consisting of a set and one or more n-array

operations on the set is known as an algebraic system.

(or)

A lattice is a algebraic system <L, *, > with two binary * and on lattice L, which

are commutative, associative, absorption.

Note: The above definition does not ensure relation on L.

SUB – LATTICES: Let <L, *, > be a lattice and let S L be a subset of L. The

algebra < S, *, > is a sub - lattice of <L, *, > iff S is closed under both operation *

and .

BOOLEAN ALGEBRA - HOMOMORPHISM

PRINCIPLE OF DUALITY FOR BOOLEAN ALGEBRA: Let (B, -, , ) be a Boolean

Algebra (under ) and let S be a true statement for (B, -, , ). If S* is obtained from

S by replacing by this is equivalent to turning the graph upside down, by ,

by , 0 by 1 and 1 by 0, then S* is also a true statement.

PROPERTIES OF BOOLEAN ALGEBRA:

A Boolean algebra will generally be denoted (B, *, , ‘, 0, 1) in which (B, *, ) is a

lattice with two binary operations * and called the meet and join respectively, the

corresponding partially ordered set will be denoted by (B, ≤). The bounds of the

lattices are denoted by 0 and 1, where 0 is the least element and 1 is the greatest

element of (B, ≤).

A Boolean Algebra (B, *, , ’, 0, 1) satisfies the following properties in which

a, b and c denote any elements of the set B.


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1.) (B, *, ) is a lattice in which the operations * and satisfy the following

identities:

(L-1) a * a = a (L-1)’ a a = a

(L-2) a * b = b * a (L-2)’ a b = b a

(L-3) (a * b) * c = a * (b * c) (L-3)’ (a b) c = a (b a)

(L-4) a * (a b) = a (L-4)’ a (a * b) = a

2.) (B, *, ) is a distributive lattices and satisfies these identities:

(D-1) a * (b c) = (a * b) (a * c)

(D-2) a (b * c) = (a b) * (a c)

(D-3) (a * b) (b * c) (c * a) = (a b) * (b c) * (c a)

(D-4) a * b = a * c, and a b = a c b = c

3.) (B, *, , ’, 0, 1) is a bounded lattice in which for any a B, the following

hold:

(B-1) 0 a 1

(B-2) a * 0 = 0 (B-2)’ a 1 = 1

(B-3) a * 1 = 1 (B-3)’ a 0 = a

4.) (B, *, , ’, 0, 1) is a uniquely complemented lattice in which the complement

of any element a B is denoted by a’ B and satisfies the following

identities:

(C-1) a * a’ = 0 (C-1)’ a a’ = 1

(C-2) 0’ = 1 (C-2)’ 1’ = 0

(C-3) (a * b)’ = a’ b’ (C-3)’ (a b)’ = a’ * b’

5.) There exists a partial ordering relation on B such that:

(P-1) a * b = GLB {a, b}

(P-2) a b a * b = a a b = b

(P-3) a b a * b’ = 0 b’ a’ a’ b = 1

SUB–BOOLEAN ALGEBRA: Let (B, *, , ’, 0, 1) be a Boolean Algebra and S B. If S

contains the element 0 and 1 and is closed under the operations *, and ‘, then (S, *,

, ’, 0, 1) is called a Sub – Boolean Algebra.


77

JOIN – IRREDUCIBLE: Let (L, *, ) be an lattice. An element a L is called Join -

Irreducible if it cannot be expressed as the join of two distinct elements of L. In other

words a L is Join – Irreducible.

DIRECT PRODUCT:

Let (L, *, ) and (S, , ) be two lattices. Then the direct product of L, and S is

defined by (L S, +, *) where + and * are defined by the following manners.

(a1, b1) + (a2, b2) = (a1 a2, b1 b2)

(a1, b1) . (a2, b2) = (a1 * a2, b1 b2), a1, a2 L b1, b2 S

LATTICES HOMOMORPHISM:

Let (L, , *) and (S, , ) be two lattices. A mapping g : L S is called

Homomorphism if

g(a b) = g(a) g(b)

g(a * b) = g(a) g(b); a, b L.

ISOMORPHISM:

Let <L, *, > and <S, , > be two lattices. If a homomorphism g : L S is one – one

and onto then g is called an Isomorphism.

ENDOMORPHISM:

A homomorphism g : L L where <L, *, > is a lattice is called an Endomorphism.

AUTOMORPHISM:

An Isomorphism g : L L is known as an Automorphism.


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PART – A

1. Define subset.

Let A and B be any two sets. If every element of A is an element of B, then A is

called a subset of B or A is said to be included in B, or B includes A. It is denoted by

A B, or equivalently by B A.

Alternatively, A B (x)(x A x B) B A.

2. Define equal sets.

Two sets A and B are said to be equal iff A B and B A, or symbolically.

A = B (A B B A).

3. Define proper subset.

A set A is called a proper subset of a set B if A B and A B. It is written as A

B, so that A B (A B A B).

4. Define universal set.

A set is called a universal set if it includes every set under discussion. A

universal set will be denoted by E.

E = {x | p(x) p(x)}

5. Define empty set.

A set which does not contain any element is called an empty set or a null set.

It is denoted by .

= {x | p(x) p(x)}

6. Define power set.

For a set A, a collection or family of subsets of A is called the power set of A.

The power set of A is denoted by P(A) or 2A, so that P(A) = 2A = {x | x A}


79

7. Show that for any sets A and B

(i) A – B = A ~B

(ii) A B B A

Solution:

(i) Let x be an arbitrary element.

Then x A–B x {x | (x A) (x B)}

x {x | (x A) (x B)}

A ~B

(ii) x A x B because A B.

A B (x)(x A x B)

(x)( (x B) (x A))

(x)((x B) (x A))

~B ~A.

8. Define an ordered pair and equality of two ordered pairs.

An ordered pair consists of two objects in a given fixed order. The equality of

two ordered pairs <x,y> and <u,v> is defined by

<x,y> = <u,v> ((x = u) (y = v)).

9. Define Cartesian products.

Let A and B be any two sets. The set of all ordered pairs such that the first

member of the ordered pair in an element of A and the second member is an element

of B is called the Cartesian product of A and B and written as A x B.

10. If A={ , } and B={1, 2, 3} what are AxB, BxA, AxA, BxB and (AxB) (BxA)?

Solution:

A x B = {< ,1>, < ,2>, < ,3>, < ,1>, < ,2>, < ,3>}.

B X A = {<1, >, <2, >, <3, >, <1, >, <2, >, <3, >}

A x A = {< , >, < , >, < , >, < , >}

B x B = {<1,1>, <1,2>, <1,3>, <2,1>, <2,2>, <2,3>, <3,1>, <3,2>, <3,3>}.

(A x B) (B x A) =


80

11. If A = and B = {1, 2, 3} what A x B and B x A?

A x B =

B x A = .

12. Give an example of sets A, B, C such that A B=A C, but B C.

Solution:

A = {a1, a2, a3 ... a10}

B = {b1, b2 ... b10}

C = {a1, a2 ... a10, b1, b2 ... b10}

Clearly A B = A C, but B C.

13. Define reflexive.

A binary relation R in a set X is reflexive if x X, xRx.

i.e.) <x,x> R or R is reflexive in X (x)(x X xRx).

14. Define symmetric.

A relation R in a set X is symmetric if for every x and y in X, whenever xRy,

then yRx.

That is R is symmetric in X implies

x y x X y X xRy yRx

15. Define transitive.

A relation R in a set X is transitive for every x, y and z in X, whenever xRy

and yRz, then xRz.

That is R is transitive in X implies

x y z x X y X z X xRy yRz xRz

16. Define anti-symmetric.

A relation R in a set X is anti-symmetric if, for every x and y in X, whenever

xRy and yRx, then x=y,


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That is R is anti-symmetric in X implies

x y x X y X xRy yRx x y .

17. Let X = {1, 2, 3, 4} and R = {<x,y> / x > y}. Draw the graph of R and also give its

matrix.

Solution:

The corresponding relation matrix for the relation R is

R = {<4,1>, <4,2>, <4,3>, <3,1>, <3,2>, <2,1>}

1 2 3 4

1 0 0 0 0

2 1 0 0 0

3 1 1 0 0

4 1 1 1 0

18. Define coverings and partition of S.

Let S be a given set and A = {A1, A2 . . . Am} where each Ai, i = l ... m is a subset of S

and 1

m

ii

A S

Then the set A is called a covering of S and the sets A1, A2 . . . Am are said to

cover S. If, in addition, the elements of A, which are subsets of S are mutually

disjoint, then A is called a partition of S, and the sets A1, A2 . . . Am are called the blocks

of the partition.

19. Define Equivalence relation.

A relation R in a set X is called an equivalence relation if it is reflexive,

symmetric and transitive.

Examples: (i) Equality of numbers on a set of real numbers

(ii) Equality of subsets of a universal set.


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20. Let X = {1, 2, 3, 4} and R = {<1,1>, <1,4>, <4,1>, <4,4>, <2,2>, <2,3>, <3,2>, <3,3>}

Write the matrix of R and sketch its graph.

1 2

1 0 0 1

0 1 1 0

0 1 1 0

1 0 0 1

21. Define partial order relation.

A binary relation R in a set P is called a partial order relation or a partial

ordering in P iff R is reflexive anti-symmetric and transitive.

22. Let X = {2, 3, 6, 12, 24, 36} and the relation x y if x divides y. Draw the Hasse

diagram of <x, >

23. Define Lattice.

A lattice is a partially ordered set <L, > in which every pair of elements a, b

L has a greatest lower bound and a least upper bound.

24. Define Lattices of Algebraic systems.

A lattices is an algebraic system <L, *, > with two binary operations * and

on L which are both commutative and associative and satisfy the absorption laws.

3 4

36 24

12

6

2 3


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25. Define Sub lattices.

Let <L, *, > be a lattices and let S L be a subset of L. The algebra <S, *, > is

a sub lattice of <L, *, > iff S is closed under both operations * and .

26. Define complemented lattice.

A lattice <L, *, , 0, 1> is said to be a complemented lattice if every element of

L has at least one complement.

27. Define distributive lattice.

A lattice <L, *, > is called a distributive lattice if for any a, b, c L.

a * (b c) = (a * b) (a * c)

a (b * c) = (a b) * (a c)

28. Define Boolean algebra and sub-Boolean algebra.

A Boolean algebra is a complemented, distributive lattice.

Let <B, *, , ‘, 0, 1> be a Boolean algebra and S B. If S contains the elements 0

and 1 is closed under the operations *, , and ‘, then <S, *, , ‘, 0, 1> is called a sub-

Boolean algebra.

29. Let A = {a, b, c, d, e} and P = {{a,b}, {c}, {d,e}}. Show that the partition P defines

an equivalence relation on A.

Solution:

1 1 2 2 3 3( )R P P P P P P where 1 2 3

P , , , ,a b P c P d e

, , , , , , , , , , , , , , , , ,R a a a b b b b a c c d e d d e e e d

The above relations is reflective, symmetry and transitive and hence an

equivalent relation.


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30. Give an example of a set such that <P(X), > is a totally ordered set.

Solution:

Consider the singleton set S = {{i}}

Then P(S) = { , S}

Clearly P(S) is a chain

Hence a totally ordered set


85

PART – B

1. (a) Show that for any two sets A and B

( )A A B A B

(b) For any two sets A and B

(i) (A B)'=A' B'

(ii)(A B)'=A' B'

Solution (a):

For any x,

x A A B x x A x A B

x A x A x B

x A x A x B

x A x A x A x B

x x A x B

(b)(i) For any x,

'

and

A' and x B'

' '

x A B x A B

x A x B

x

x A B

Thus (A B)'=A' B'

(b)(ii) For any y,

y A B '

either y A or y B

either y A' or y B'

y A' B'

y A B

Thus (A B)' = A' B'


86

2. If A and B are sets then prove the following identities.

(i) A A

(ii) A A

(iii)A B B A

(iv) ( ) ( )A B A B B A

Proof:

(i) For any x,

x A A x x A A x A A

x x A A x A A

x

x

= A A

(ii) For any x,

x A

x A x x A x A

x A x x x A

x A F

A A

(iii) For any x,

A + B

x x x A B x B A

x x A B x B A

x x B A x A B

x B A

A B B A


87

(iv)

-

-

A B A B A B

A B A B A B A B

A B A B

A B A A B B

A A B A A B B B

A B B A

3. (a) If A B A C and A B A C prove that B C

(b) If A, B, C are sets prove the following

(i) ( - )A B A A B

(ii) -( ) ( - ) ( - )A B C A B A C

Solution (a):

Given that A B A C we have ( ) ( )B A B B A C

(1)

LHS B A B B since B A B

(2)

[ ]

[ ]

[ ]

[ ]

[ ]

B A C B A B C Disributive law

A B B C Commulative law

A C B C as A B A C

A B C Distributive law

A C C as A B A C

C

RHS C

(3)

From (2) & (3) B C


88

Solution (b):

(i)

' [ ']

' [ ]

A B A A B A B A B A

A B A A distributive law

A B U

A B

(ii)

-( ) '= ' ' [1]A B C A B C A B C

( - ) ( - ) ' ' [ - ']

( - ) ( - ) ' ' [2]

A B A C A B A C A B A B

A B A C A B C

From [1] & [2] - -A B C A B A C

4. If A = {c, d}, B = {1, 2}, C = {2, 3}

find (i) A B C

(ii) A B A C

(iii) A B C

(iv) A B A C

Solution:

(i) B C = {1, 2, 3}

A (B U C) = {c, d} {1, 2, 3}

= {<c,1>, <c,2>, <c,3>, <d,1>, <d,2>, <d,3>}

(ii) A B = {<c,1>, <c,2>, <d,1>, <d,2>}

A C = {<c,2>, <c,3>, <d,2>, <d,3>}

(A B) (A C) = {<c,1>, <c,2>, <d,1>, <d,2>, <c,3>, <d,3>}

from (i) & (ii) A (B C)= (A B) (A C)

(iii) B C = {2}

A (B C) = {c, d} {2}

= {<c,2>, <d,2>}

(iv) From (ii), we have

(A B) (A C) = {<c,2>, <d,2>}

from (iii) & (iv) that A (B C) = (A B) (A C)


89

5. If A, B, C are any three sets, prove that A (B C)= (A B) (A C).

Proof:

Now let us show that A (B C) (A B) (A C).

Let (x,y) be any element of A (B C)

Then x A and y (B C)

i.e. x A and y B or y C.

i.e. x A and y B or x A and y C.

A (B C) (A B) (A C)

(1)

Now let us show that (A B) (A C) A (B C)

Let (z,w) be any element of (A B) (A C)

Then (z,w) (A B) or (z,w) (A C)

i.e. z A and w B or z A and w C

i.e. z A and w B or w C

i.e. (z,w) A (B C)

(A B) (A C) A (B C)

(2)

From (1) & (2) we have A (B C) = (A B) (A C).

6. Prove that the relation ‚congruence modulo m‛ over the set of positive integers

is an equivalence relation.

Solution:

Let N be the set of all positive integers and m be a given positive integer.

We define the ‚congruence modulo m‛ relation on N as follows.

For x, y , x y(mod m) iff x-y is divisible by m.

i.e. x-y = km for some integer k .

Let x, y, z .


90

(i) as x-x = 0.m

so x x(mod m) x .

Thus the relation is reflexive.

(ii) x y(mod m) x-y = km for some integer k

y-x = (-k)m

y x(mod m)

so the relation is symmetric.

(iii) x y(mod m) and y z (mod m)

x-y = km and y-z = m for some integer k and

(x-y) + (y-z) = (k+ )m

x-z = (k+ )m

x z(mod m) (as k+ = m is also an integer)

so the relation is transitive.

As it is reflective, symmetric and transitive the relation ‚congruence modulo m‛

is an equivalence relation.

7. Let R denote a relation on the set of all ordered pairs of positive integers, by

(x,y)R(u,v) iff xv = yu. Show that R is an equivalence relation

Solution:

If x, y, u, v are positive integers, it is given that (x,y)R(u,v) iff xv = yu.

(i) As xy = yx is there for all positive integers x and y.

We have (x,y)R(x,y) for all ordered pair (x,y) of positive integers.

So, the relation R is reflective.

(ii) (x,y)R(u,v) xv = yu

yu = xv

uy = vx

(u,v)R(x,y)

So, the relation R is symmetric.

(iii) Let x, y, u, v, m, n be positive integers

(x,y)R(u,v) and (u,v)R(m,n) xv = yu and un = vm

xvun = yuvm


91

xn = ym (by canceling vu)

(x,y)R(m,n)

So, the relation R is transitive.

As R is reflective, symmetric and transitive the relation (x,y)R(u,v) iff xv = yu is

an equivalence relation.

8. Let a relation R be defined on the set of all real numbers by ‚if x, y are real

numbers, xRy x-y is a rational numbers‛ show that the relation R is an

equivalence relation.

Solution:

It is given that xRy iff x-y is a rational numbers.

(i) As x-x = 0 is a rational number for all real numbers x.

So, the relation R is a reflective.

(ii) Let xRy then x-y = z for some rational numbers.

Now y-x = -z is also a rational numbers yRx.

Thus xRy yRx.

So, the relation R is symmetric.

(iii) Let xRy and yRz then x-y = r1 and y-z = r2 for some rational numbers, r1 and r2.

Now x-z = (x-y) + (y-z) = r1 + r2 is also a rational number

xRz

Thus xRy and yRz xRz.

So, the relation R is transitive.

As R is reflective, symmetric and transitive thus it is an equivalence relation.

9. Let A be a finite set and P(A) is its power set. Let be the inclusion relation on

the elements of P(A). Draw Hasse diagram of <P(A), > for (i) A = {a}, (ii) A = {a, b}

(iii) A = {a, b, c}.


92

Solution:

The required Hasse diagrams are given below.

(i) {a} (ii)

{ }

(iii)

10. Every equivalence relation on a set generates a unique partition of the set. The

blocks of this partition correspond to the R - equivalence classes.

Proof:

It is enough to show that R - equivalence classes are disjoint and set of all

equivalence classes cover the whole set.

For any element x X, we have xRx because R is reflexive.

Therefore x [x]R.

Because of the symmetry of R, yRx and x [y]R

Now there is an element z [y]R, then z must be in [x]R because yRz, along with xRy,

implies xRz.

Thus [y]R [x]R.

By symmetry we must also have [x]R [y]R

{a} {b}

{ }

{a,b }

{a,b,c }

{a,b} {a,c}

{b,c}

{a} {b} {c}

{ }


93

Finally from [y]R [x]R and [x]R [y]R

We have [x]R = [y]R.

It is shown that of xRy then [x]R = [y]R. Now show that if xRy, then [x]R and [y]R

must be disjoint.

By assuming that there is at least one element z [x]R and z [y]R

xRz and yRz xRy then from transitivity xRy.

This is a contradiction.

Therefore two R – equivalence classes of X are either the same or disjoint. As

each element of X belongs to some R – equivalence class, union of all R - equivalence

classes cover the whole set X and as all R – equivalence classes disjoint, which can be

called blocks, the partition of X. The uniqueness of the partition is clear because R –

equivalence class of any element is unique.

11. a) For the relation R = {<1,2>, <3,4>, <2,2>} and S = {<4,2>, <2,5>, <3,1>, <1,3>}.

Obtain the relation matrices for R S and S R .

b) Given the relation metrics R S

1 0 1 1 0 0 1 0

M 1 1 0 and M 1 0 1 0 1

1 1 1 0 1 0 1 0

.

Find R S

M , R

M , S

M , RoS

M and show that RoS s RM M

.

Solution (a):

R

0 1 0 0 0

0 1 0 0 0

0 0 0 1 0M

0 0 0 0 0

0 0 0 0 0

S

0 0 1 0 0

0 0 0 0 1

1 0 0 0 0M

0 1 0 0 0

0 0 0 0 0


94

ROS

0 0 0 0 1

0 0 0 0 1

0 1 0 0 0M

0 0 0 0 0

0 0 0 0 0

SOR

0 0 0 1 0

0 0 0 0 0

0 1 0 0 0M

0 1 0 0 0

0 0 0 0 0

Solution (b):

RR

1 1 1

M 0 1 1 transfer of M

1 0 1

SS

1 1 0

0 0 1

0 1 0M transfer of M

1 0 1

0 1 0

ROS

1 1 0 1 0

M 1 0 1 1 1

1 1 1 1 1

RoS

1 1 1

1 0 1

0 1 1M

1 1 1

0 1 1

SOR RoS

1 1 1

1 0 1

0 1 1M M

1 1 1

0 1 1


95

12. Let R and S be relation on A. Then

(i) If R is reflexive, then R S is reflexive.

(ii) If R and S are reflexive, then R S is reflexive.

(iii) If R and S are symmetric, then R S and R S are symmetric.

(iv) If R and S are transitive, then R S is transitive.

(v) If R and S are equivalence relations, then so is R S.

Proof:

(i) If R is reflexive, then a,a a A R so R S.

Thus R S is reflexive.

(ii) If R and S are reflexive, then R and S hence R S.

Thus R S is reflexive.

(iii) Assume that both R and S to be symmetric.

If (a,b) R S, then (a,b) R and (a,b) S

As (a,b) R and as R is symmetric (b,a) R

As (a,b) S and as S is symmetric (b,a) S

Then (a,b) R S (b,a) R S.

So, R S is symmetric.

Similarly, we can show that R S is symmetric.

(iv) Assume that R and S are transitive.

Let (a,b), (b,c) R S

As (a,b), (b,c) R, we have (a,c) R (R is transitive).

As (a,b), (b,c) S, we have (a,c) S (S in transitive).

Thus (a,b) (b,c) R S (a,c) R S.

Thus R S in transitive.

(v) If R and S are equivalence relations on A then by (ii), (iii) and (iv)

It follows that R S is also an equivalence relation.


96

13. Let R be a relation on a set A then

(i) If R is reflexive, then R-1 is also reflexive

(ii) If R is transitive, then R-1 is also transitive

(iii) If R is an equivalence relation, then R-1 is also an equivalence relation.

Proof:

(i) R is reflexive 1 1

1 -1

1

R R

R as

R is reflexive

(ii) R is transitive

1 1

11 1 1 1 1

1

RoR R

RoR R

R oR R as RoS S oR

R is transitive

(iii) If R in an equivalence relation, then it is symmetric.

So, R = R-1

As R-1 = R and R is an equivalence relation thus R-1 is an equivalence relation.

14. (i) Prove that distinct equivalence classes are disjoint.

(ii) In a lattice Show that a b and c d a c b d.

(iii) In a distributive lattice Prove that a b = a c and

a b = a c.

Solution:

Let [x] denote an equivalence class with respect to an equivalence relation R.

(i) If z [x] then [z] = [x].

If [x] and [y] are distinct equivalence classes and z [x] [y]

We obtain a contradiction.


97

z [x] [y] z [x] and z [y]

[z] = [x] and [z] = [y]

[x] = [y]

This is since [x] and [y] are assumed to the district equivalence classes.

Hence [x] [y] = .

(ii) b d is the glb of b and d.

If we show that a c is a lower bound of b and d

It would follow that a c b d.

Now, a c a b and a c c d,

a c is a lower bound of b and d.

(iii) b = b (a b) (by absorption)

= b (a c) (by hypo)

= (b a) (b c) (distributive law)

= (a c) (b c) (by hypo)

= (a b) c (distribution law)

= (a c) c (by hypo)

= c (absorption)

15. (i) Let P = {{1,2}, {3,4}, {5}} be a partition of the set S = {1, 2, 3, 4, 5} contract an

equivalence relation R on S so that the equivalence classes with respect to ‘R’ are

precisely the members of P.

(ii) Show that chain with three or more elements is not complemented

(AU/Nov-Dec-2006)

Solution (i):

Required equivalence relation is given by

R = {(1,1), (2,2), (3,3), (4,4), (5,5), (1,2), (2,1), (3,4), (4,3)}

Proof (ii):

Let L be a chain with 0 and 1 Let 0 < a < 1, we show that ‘a’ has no complement in L.

Let b L and ‘b’ be a complement of a a b = 0 and a b = 1.


98

Since L is a chain, either a b, we have 0 = a b = a.

But a > 0.

Similarly if b a, 1 = a b = a.

But a < 1,

Hence ‘a’ has no complement.

16. Establish De Morgan’s Laws in Boolean Algebra. (AU/Nov-Dec-2006)

Proof:

To show that x = y we need to show that x y = 0 and x y = 1.

To show that (a b)’ = a’ b’.

We need to show that

(a b) (a’ b’) = 0 and (a b) (a’ b’) = 1

(a b) (a’ b’) = ((a b) a’) ((a b) b’)

= ((a a’) b) (a (b b’))

= (0 b) (a 0)

= 0 0

= 0

Similarly

(a b) (a’ b’) = ((a b) a’) ((a b) b’)

= (a a’) (b a’) (a b’) (b b’)

= (1 (b a’)) ((a b’) 1)

= (b a’) (a b’)

= (a a’) (b b’)

= 1 1

= 1

17. Define a relation R on by R = {(a,b) | a – b is a non -negative even integers}.

Verify R is a partial order on .

Solution:

(i) For, a A, a – a = 0 (a, a) R

R is reflexive.


99

(ii) If (a,b), (b,a) R than a – b = 2k

where k is some positive integer.

b – a = - 2k.

But (b,a) R k should be a negative integers

k is both positive and negative then k = 0

b – a = 0 b = a.

R is anti-symmetric.

(iii) If (a,b) (b,c) R then

a – b = 2m; b – c = 2n

a – c = 2(m+n)

(a,c) R

R is transitive.

Since R is reflexive, anti-symmetric and transitive.

R is a partial ordering relation on .

UNIT – IV

FUNCTIONS

INTRODUCTION

REPESENTATION OF A FUNCTION

TYPES OF FUNCTIONS


100

CLASSIFICATIONS OF FUNCTIONS

COMPOSITION OF FUNCTIONS

INVERSE FUNCTIONS

BINARY AND N-ARY OPERATIONS

PROPERTIES OF FUNCTIONS

CHARACTERISTIC FUNCTION OF A SET

HASHING FUNCTIONS

RECURSIVE FUNCTIONS

RECURSION

PRIMITIVE RECURSIVE FUNCTIONS

RECURSIVE RELATIONS AND SETS

PERMUTATION FUNCTIONS


101

INTRODUCTION

We have defined a relation R from X to Y as a subset of the Cartesian product

X Y and noted that domain R = {x X / (x,y) R for some y Y}. If a relation f from X

to Y is also to be a function, the domain of f must be equal to X and if (x,y) f and

(x,z) f, then y must be equal to z. We will discuss the basic concepts involving

function required in discrete structures.

FUNCTION: A relation f from a set X to another set Y is called a function if for every

x X there is unique y Y such that (x,y) f.

If y is the unique element of Y assigned by the function f to the element x of X we

write f(x) = y. If f is a function from X to Y we represent it as f: X Y or X Y.

Some times the term ‘transformation‘, ‘mapping‛ or ‚correspondence’ are also

use in the place of ‘function’.

If y = f(x), x is called an argument or pre-image and y is called image of x

under f or the value of the function f at x.

REPRESENTATION OF FUNCTION: A function can be expressed by means of a

mathematical rule or formula such as y = x3 [= f(x)] or a relation matrix (since a

function is a relation) or a graph

If Df = {a, b, c, d} and f(a) = 2, f(b) = 4, f(c) = 1 and f(d) = 2 then the

pictorial representations of f will be as in

TYPE OF FUNCTION:

ONE-TO-ONE: A function f: X Y is called one-to-one (1-1) or injective or

injection, if distinct element of X are mapped into distinct element of Y in other

words, f is one-to-one if and only if

a

b

c

d

1

2

3

4


102

f(x1) ≠ f (x2), whenever x1 ≠ x2 or equivalently.

f(x1) = f (x2), whenever x1 = x2

EXAMPLE : The function represented by the above diagram is not one-to-one since

f(a) = f(d) = 2, but a ≠ d. The function represented by the following diagram is one-

to-one.

ONTO: A function f: X Y is called onto or surjective or surjection, if the range Rf =

Y; otherwise it is into. In other words f is called onto if and only if for every element

y Y there is an element x X such that f(x) = y.

BIJECTIVE: A function f: X Y is called one-to-one, onto or bijective or bijection or

one-to-one correspondence. If it is both one-to-one and onto

The function represented by is bijective

CLASSIFICATION OF FUNCTIONS:

Functions can be classified mainly into two groups they are

Algebraic function

Transcendental function

a

b

c

d

1

2

3

4

a

b

c

d

1

2

3

4

5


103

ALGEBRAIC FUNCTION:

A function which consists of a finite numbers of terms involving integral and or

fractional powers of the independent variable (argument) a, connected by the four

operation +, -, and is called an algebraic function, three particular cases of

algebraic functions are the following

POLYNOMIAL FUNCTION: A function of form n n 1 n 2

0 1 2 na x a x a x ... a

where n is a position integer and a0, a1, a3 … an are real constants and a0 ≠ 0 is called

polynomial in x of degree n.

EXAMPLE: 2x4 - 3x3 + 2x - 4 is a polynomial in degree 4.

RATIONAL FUNCTION: A function of the form f(x)

g(x), where f(x) and g(x) ≠ 0 are

polynomials is called rational function.

EXAMPLE: f(x) = 3 2

2

x 2x 3x 4

x 3x 1 is rational function.

IRRATIONAL FUNCTION: A function involving radicals, in fractional powers of

polynomials is called an irrational function.

EXAMPLE : f(x) = 2

3

x 1 x

x 1 x is an irrational functions.

TRANSCENDENTAL FUNCTION:

A function which is not algebraic is called a transcendental function.

EXAMPLE: The circular function, inverse circular function exponential function,

logarithmic function, hyperbolic functions are all transcendental function.

FLOOR FUNCTION: If x is a real number, the function that assigns the largest

integer that is less than or equal to x is called the floor function of x or simply the

floor of x and denoted by x .

If x = n, where n is a integer, than n ≤ x < n+1.


104

CEILING FUNCTION: If x is a real number, the function that assigns the smallest

integer that is greater than or equal to x is called the ceiling function of x or simply

the ceiling of x and denoted by x .

If x = n, where n is an integer, then n-1 < x ≤ n.

INTEGER VALUE: The integer value of x, where x is a real number, converts x into

an integer by truncating or deleting the fractional part of number and is denoted by

INT(x).

EXAMPLE: INT(3.25) = 3, INT(-8.54) = -8 and INT(6) = 6.

NOTE: INT(x) = x or x , according as x is positive (or) negative.

ABSOLUTE VALUE FUNCTION: The absolute value of x, where x is a real number

is defined as the greater of x or –x and denoted by ABS(x) = x .

REMAINDER FUNCTION: If a is any integer and m is a positive integer, then the

integer remainder (function) when a is divided by m is denoted by a (mod m) [read

as a modulo m].

EXAMPLE: -30 (mod 5) = 0, -30 (mod 7) = 7 – 2 = 5 and -3 (mod 5) = 5 – 3 = 2.

CONGRUENCE CLASS: The set of all integer that are congruent to a modulo m‘

where a is an integer and m is a positive integer is called the congruence class of ‘a

modulo m’.

COMPOSITION OF FUNCTIONS: If f: A B and g: B C, then the composition

of f and g is a new function from A to C denoted by gof, is given by (gof)(x) = g{f(x)},

x A.

INVERSE OF A FUNCTION: If f: A B and g: B A, then the function g is called

inverse of the function f iff gof = IA, fog = IB.

BINARY AND N-ARY OPERATION: If S is a non empty set and f is a mapping f:

S S S then f is called a binary operation on S. In general, a function f: Sn S is

called an n-ary operation and n is called the order of operation.


105

UNARY OPERATION: When n = 1 the function f: S S is called a unary

operation. When n = 3 the function f: S S S S is called a ternary (3-ary)

operation.

Usually a binary operation (function) is denoted by a symbol such as +, *, -, , ,

, , and the value of the operation by placing the operator between the two

operations.

CHARACTERISTIC FUNCTION OF A SET:

In this section, we shall deal with function from the universal set U to the set {0,

1}, using which statements about sets and their operations can be represented on a

computer in terms of binary numbers and hence can be deal with easily.

CHARACTERISTIC FUNCTION: If A is a subset of a universal set U, the

characteristic function A of A is defined as the function from U to the set {0, 1} such

that

A

1 if x A

0 if x A

HASHING FUNCTIONS:

When records (data) are stored in a direct access file in a computer, the computer

can retrieve a specific record without reading other record first. This is possible only

if the computer can identify the memory locations in which records in the form of

non-negative integers, called keys are stored. A transformation that maps the set of

keys to a set of addresses is called a hashing function. Even though various hashing

functions are used, we will discuss about one of the commonly used hashing

functions obtained by division method or congruence method.

HASHING FUNCTIONS: Hashing function is a function which maps the set of keys

into the set of 'n' addresses. If n is the number of available memory locations and k

is the non-negative integer representing the key, the hashing function h(k)

representing the address of the memory cell in which k is stored is defined as,

h(k) k (mod n)

(i.e.) h(k) is simply the remainder when k is divided by n and it takes the values

from the set {0, 1, 2, ..... n-1} known as the address set.


106

KINDS OF HASHING FUNCTIONS:

DIVISION METHOD: Here we restrict the number of addresses we want to use to a

fixed integer 'n'. For any key 'k' we define,

h(k) r (mod n) : 0 < r < n

FOLDING METHOD: Here the key is split into parts and the parts are added

together to give h(k).

EXAMPLE: h(081 – 37 – 64 – 95) = 081 + 37 + 64 + 95 = 6613

The importance of choosing a hashing function cannot be emphasized enough as

we try to improve efficiency in terms of greater speed and less unused storage.

Using the modular concept, we can develop a hashing function ‘h’ using the

same keys as above.

h(x1x2x3x4x5x6x7x8x9) = y1y2y3

y1 x1 + x2 + x3 (mod 5)

y2 x4 + x5 (mod 3)

y3 x6 + x7 + x8 + x9 (mod 7)

EXAMPLE:

h(081 – 37 – 64 – 95) = 413

COLLISION MANAGEMENT:

A hashing function quite often maps different keys to the same element of the

address set. Thus, the set of records is partitioned into ‘n’ equivalence classes.

Those records, which are mapped to the same address, are in the same equivalence

class.

It is necessary to provide storage space for records and also a method of finding

the collision or overflow records when more than one record has the same address.

There are many collision resolution techniques for this purpose.


107

OPEN ADDRESSING: It probes possible addresses in same reproducible order and

inserts the collision record in the first empty location found. The data structure used

in this case is a vector, each element of which is capable of holding a record.

CHAINING: It use linear linked list to represent each equivalence classes in memory.

In other words, a one-to-one correspondence is established between the equivalence

classes and linked list.

RECURSIVE FUNCTIONS

When it is difficult to define an object explicitly, often it may be easy to define

this object in terms of itself. This process is roughly referred to as recursion using the

concept of recursion, we can define sets, sequences and functions. In this section, we

shall study an important class of functions called recursive functions or recursively

defined functions, whose arguments and values are natural – theoretic functions.

TOTAL FUNCTION: The number – theoretic function f: Nn N is called a total

function, if it is defined for every element of Nn. It is recalled that the total function

f(x1, x2 … xn) was defined as an n-ary operation on the set N natural numbers.

PARTIAL FUNCTION: The function f: D N, where D Nn is called a partial

function.

INITIAL FUNCTIONS: The following functions that are used in defining other

functions by induction are called the initial functions.

(i) The function Z: Z(x) = 0 is called zero function

(ii) The function S: S(x) = x+1 is called successor function.

(iii)The function n

iU : n

iU (x1, x2 … xn) = xi is called the projection function or

generalized function

RECURSION: If a function f(x1, x2 … xn, y) of (n+1) variables is defined, as given

below. In terms of the function g(x1, x2 … xn) of n variables and the function h(x1, x2

… xn, y, z) of (n+2) variables which are assumed to be known, the operation is called

recursion and the function f is said to be defined recursively.

(i) f(x1, x2 … xn, 0) = g(x1, x2 … xn) and

(ii) f(x1, x2 … xn, y+1) = h{x1, x2 … xn, y, f(x1, x2 … xn, y)}


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Since in (ii), the value of f at y+1 is expressed in terms of the value of f at y, y is

referred to as the inductive variable. Since the variables x1, x2 … xn remain fixed

thorough the recursion, they are referred to as the parameter.

PRIMITIVE RECURSIVE FUNCTION: A function f is called primitive recursive, if

and only if it can be constructed from the initial functions and other known

primitive functions by a finite number of operations of composition and recursion

only.

Sl.

No. Name

Short

Notation /

Symbol

Definition / Property

1.

Sign function

Or

Non zero test

function

Sg(x)

0, if x 0Sg(x)

1, if x 0

Or

Sg(0) Z(x) and 2

2Sg(x 1) S Z U x,Sg(x)

2. Zero test function Sg(x)

1, if x 0Sg(x)

0, if x 0

Or

sg(0) S(0) and

sg(x 1) Z(x) 0

3. Predecessor

function P(x)

0, if x 0P(x)

x 1, if x 0

P(0) 0 2

1P(x 1) U x,P(x) x

4.

Odd and Even

parity

Function

Pr(x)

0, if x 0 or evenPr(x)

1, if x is odd

Pr(0) 0 Z(x) 2

2P(x 1) Sg U x,Pr(x)


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5. Proper subtraction

Function −

x y, if x yx y

0, if x y

3

1x 0 U(x,y,x y) x

and x (y 1) P(x y)

6. Absolute value

function | |

x y (x y) (y x) 3

1x 0 U(x,y, x y) x

7. Minimum of x and

y Min(x, y)

min(x, y) = x – (x – y)

min(x, 0) = 0 = Z(x)

8. Maximum of x

and y Max(x, y)

max(x, y) = y + (x – y) 3

1max(x,0) U(x,y,max(x,y)) x

9. Square function Sq(x) 2 1 1

1 1Sq(x) x U(x)* U(x)

Sq(0) Z(0)

10. Compare function

less(x, y)

1, if x yless(x,y)

0, if x y

less(x,y) Sg(y x)

gr(x, y)

1, if x ygr(x,y)

0, if x y

gr(x,y) Sg(x y)

eq(x, y)

1, if x yeq(x,y)

0, if x y

eq(x,y) Sg(x y)

RECURSIVE RELATIONS AND SETS: A set of n-tuples defines an n-ary relation. If

the n-tuples are defined over the set of natural numbers only, then such as n-ary

relation is called number–theoretic. We discuss only number–theoretic function of a

relation R is defined as

1 2 n

R 1 2 n

1 2 n

1, if x ,x ...x Rf x ,x ...x

0, if x ,x ...x R

where R Nn and (x1, x2 … xn) Nn.

Now a relation R is said to be primitive recursive / total recursive / partial recursive.


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If it’s characteristic function is primitive recursive / total recursive / partial recursive.

Similarly a set A of natural numbers or of n-tuples over the natural numbers is said

to be recursive, if its characteristic function is recursive.

PERMUTATION FUNCTION: The set of all one-to-one onto functions from A to A

is called the set of permutation functions from A to A.

If the set A has n elements, there are n! elements in the set of permutation functions

from A to A.

…………………………………………………………………………………………………


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PART – A

1. Define function.

Let X and Y be any two sets. A relation from X to Y is called a function if for

every x X there is a unique y Y such that <x,y> .

Example:

Consider the relation : X Y defined by (a)= a2 for all a X.

Clearly f is function since for every element a X, there is a unique element a2

such that (a)= a2

2. Define Peano’s successor function.

Let P be the set of all positive integers and : P P be such that (n) = n+1

where n P. Obviously (1)=2, (2)=3.... The function is called Peano’s successor

function.

3. Defined restriction and extension of f.

If : X Y and A X then (A Y) is a function from A Y called the

restriction of to A and is sometimes written as |A. If g is a restriction of then is

called the extension of g.

4. Define surjective and give example.

A mapping of : X Y is called onto (Surjective, or surjection) if the range R

= Y. Otherwise it is called into.

5. Define one-to-one function.

A Mapping : X Y is called one-to-one (injective, or 1-1) if distinct elements

of X are mapped into distinct elements of Y. In other words is one-to-one if

x1 x2

(x1) (x2)

Or equivalently

(x1) = (x2)

x1 = x2.


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6. Define bijective mapping.

A mapping : X Y is called one-to-one onto or bijective if it is both one-to-

one and onto. Such a mapping is also called a one-to-one correspondence between X

and Y.

7. Define relative product of function.

Let : X Y and g: Y Z be two functions. The composite relation go such

that

, /( ) ( ) ( )( ), ( ) ( )g f x z x X z Z y y Y y f x z g y

is called the composition of functions or relative product of and g. go is called the

left composition of g with .

8. Let X = {1, 2, 3}, Y = {p, q} and Z = {a, b} Let : X Y be = {<1,p>, <2,p>, <3,p>}

and g: Y Z be given by g ={<p,b>, <q,b>} find go .

Solution:

go = {<1,b>, <2,b>, <3,b>}

9. Let : R R be given by (x) = -x2 and g: R+ R be given g(x) = x where R+ is

the set of nonnegative real numbers and R is the set of all real numbers. Find o .

Is go defined?

Solution:

2

2 2 4

of f x f f x f x x x .

( og)(x) = -x for all x R+.

The function og: R+ R is defined because the range of g is R+ R and R is the

domain of ‘ ’.

The range of is not included in the domain of g.

Therefore go is not defined.

The only element common to R and Dg is 0.


113

10. Define identity map.

A mapping Ix: X X is called an identity map if

Ix = {(x,x)/ x X}

11. Show that the functions (x) = x3 and g(x) = x1/3 for x R are inverse of one

another.

Solution:

( og)(x) = (g(x))

= (x1/3)

= x

= Ix

(go )(x) = g( (x))

= g(x3)

= x

= Ix

then = g-1 or g = 1

12. Define binary, n-ary and unary operation.

Let X be a set and f be a mapping : X X X. Then f is called a binary

operation on X. In general, a mapping f: Xn X is called an n-ary operation and n is

called the order of the operations. For n = 1, f: X X is called a unary operation.

13. Define commutative, associative and distributive of a binary operation.

A binary operation : X X X is said to be commutative if for every x,y,z X,

<x,y> = <y,x>.

A binary operation f: X X X is said to be associative if for every x,y,z X.

< < x, y >,z > = < x, < y, z >>

A binary operation : X X X is said to be distribute over the operation g: X X

X if for every x,y,z X

< x, g < y , z >> = g < < x,y>, < x ,z >>


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14. Define invertible or inverse of an element a with respect to a binary

operations.

Let ‘ ’ be a binary operation on X with the identity ‘e’. An element a X is

said to be left-invertible if there exists an element xl X such that xl a=e, xl is called a

left inverse of a similarly, a X is said to be right-invertible if there exist an element

xr X such that a xr=e. xr is called a right inverse of ‘a’. If an element a X is both left

invertible and right-invertible then ‘a’ is called invertible.

15. Define characteristic function of the set.

Let E be a universal set and A be a subset of E.

The function E {0, 1} defined by.

1 if x A( )

0 if x AAx

is called the characteristic function of the set A.

16. Show that ~~A = A using characteristic function.

~~A ~A

A

A

(x) = 1- (x)

= 1-(1- (x))

= (x)

We have ~~A = A

17. Define primitive recursive.

A function is called primitive recursive iff it can be obtained from the initial

functions by a finite number of operations of composition and recursion.

18. Using recursion function define the multiplication function given by

g<x,y>=x y.

Solution:

Since g<x,0> = 0

We write g<x,0> = Z(x)

g<x,y+1> = g<x,y> + x


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3 3

3 1( , 1) [ ( , , ( , )), ( , , ( , ))]g x y f U x y g x y U x y g x y

where is addition function <x,y> = x + y.

19. Show that {<x,x> / x N} Which defines the relation of equation is primitive

recursive.

Solution:

Obviously <x,y> = sg (|x-y|). Defines a primitive recursive function such

that <x,y> = 1 for x = y otherwise <x,y> = 0.

Thus <x,y> is the required characteristic function which is primitive

recursive.

20. Define recursive function.

A function is said to be recursive iff it can be obtained from the initial

functions by a finite number of applications of the operations of composition,

recursion and minimization over regular functions.

21. Define partial recursive function.

A function is said to be partial recursive iff it can be obtained from the initial

functions by a finite number of applications of the operations of composition,

recursion and minimizations.

22. Show that the sets of even and odd natural numbers are both recursive.

The parity function is the required characteristic function for the set E of even

natural numbers. Hence E is primitive recursive.

Also, the set of odd natural numbers is ~E hence ~E is also primitive recursive.

23. If : A B and g: B C are mappings and go : A C is one – to – one

(injection) prove that is 1-1. (AU/Nov-Dec-2006)

Solution:

If (x) = (y) then

g( (x)) = g( (y))


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(go )(x) = (go )(y)

x = y (since go is 1-1)

Hence is one-to-one.

24. Define Hash function?

Hash functions are used to store or retrieve later from the computers memory

Example: If the cells are indexed from 0 to 20, if we want the hash function

h(x)=n(mod 21).

25. The following is the hash diagram of a partially ordered set. Verify whether it

is a lattice?

Solution:

From the diagram a b does not exist.

For c and e are upper bound of a and b.

‘c’ and ‘e’ cannot be compared and hence

least among these two does not exist.

26. Show that the function defined by (x,y) = xy with (0,0) = 0 is primitive

recursive.

Solution:

Since (x,0)= x0 = 1 for x 0 and we put x0 = 0 for x = 0.

Define the function by (x,0) = Sg(x), where Sg(x) is the non-zero test function.

1

3 3

1 3

( , 1)

*

* ( , )

[ ( , , ( , )), ( , , ( , ))]

y

y

f x y x

x x

x f x y

g U x y f x y U x y f x y

where g is a multiplicative function g(x,y) = x * y

e

c

b a

d


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The given function is expressed in terms of initial and primitive recursive function.

Therefore (x,y) is primitive recursive.

27. Show that the function (x) = x/2 is partial recursive.

Solution:

Define a function g: N N N by g(x,y) = |2y-x|.

This function is not regular because it is not possible to find a y N such that |2y -x|

= 0 x

Define (x) = y{|2y-x|=0}

The least y such that |2y-x|=0 is x/2 for even.

Hence is partial recursive.


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PART – B

1. (a) Prove that the mapping : X X where X = {x R | x 0} defined by (x) =

1/x is one-to-one and onto.

(b) If A has m elements and B has n elements how many functions are there

from A to B?

(c) State which of the following are injections or bijections, from R into R,

where R is the set of all real numbers.

(i) (x) = -2x (ii) g(x) = x2 - 1.

Solution (a):

Let x,y R

If (x) = (y)

1/x = 1/y

then x = y

So is one-to-one.

Let y be a nonzero real number then 1/y is a nonzero real number and hence 1/y X.

So, 1

1y

y

Hence (1/y) is onto so is one-to-one and onto.

Solution (b):

To define a function : A B for each a A, we have to select one element from

B as the image of a. For a given a A, we have n choices viz., n elements of B.

As there are m elements in A and for each elements in a there are n choices, the

number of such choice is nm. Hence the number of distinct functions from A B is

nm.

Solution (c):

(i) Let x,y R

If (x) = (y)


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-2x = -2y

x = y

so in one–to–one.

Also given y R, y y

y ( 2) f2 2

.

Hence is a surjection.

As is one–to–one and onto. It is a bijection.

(ii) g(1) = g(-1) = 0

Hence g is not one–to–one

g is not onto since –2 has no pre-image in R.

For suppose g(x) = -2 then x2 -1 = -2 and so x2 = -1.

But there is no real number x such that X2 = -1.

Thus g in neither an injection nor a surjection

2. (a) Let : A B be a function then -1 is a function from range of to A if is

one–to-one.

(b) Let be a one–to–one function from A onto B then -1: B A is one–to–one.

(c) Let be a one–to–one function from A onto B then ( -1)-1 =

Proof (a):

Let us prove the contrary statement that -1 is not a function from range of into

A iff is not one–to–one.

Only if: Suppose -1 is not a function from range of into A then for some b B, there

exist two pairs (b,a1), (b,a2) in -1 such that (a1) = (a2) = b.

If: Let be not one–to–one. This means that there exist two distinct elements a1 and

a2 such that (a1) = (a2) = b (say).

So (b,a1) and (b,a2) are in -1

Hence -1 is not a function.


120

Proof (b):

Suppose -1(b1) = -1(b2) = a (say).

Then by definition of -1

b1 = (a) = b2

b1 = b2

Hence -1 is one–to–one.

Proof (c):

Given that is one–to–one function.

The inverse relation -1 is a function.

-1 is got by reversing the order of elements in ordered pairs in relation

We know that is one–to–one, -1 is also one–to–one.

Hence ( -1)-1 is a function.

( -1)-1 is got by reversing the order of elements in ordered pairs in -1.

Thus the ordered pairs in ( -1)-1 are precisely the ordered pairs in .

Hence ( -1)-1 = .

3. (a) The inverse of an one–to–one and onto function is one-to–one and onto

function

(b) Let X = {1, 2, 3, 4} and a mapping : X X be given by = {<1,2>, <2,3>, <3,4>,

<4,1>} from the composite functions 2, 3, 4.

Proof (a):

Let : X Y be a one–to–one function from X onto Y.

As is onto, range of is B.

As is one–to–one -1 exists and is a function from B A.

By theorem, -1 is one–to–one.

Now we show that -1 is onto.

Consider a A.

Let b = (a) then a = -1(b).

Hence each element a in A is in range -1 so -1 is onto.


121

Proof (b):

Given that (1) = 2, (2) = 3, (3) = 4 and (4) = 1

As 2: X X is defined by 2(x) = ( (x)) for all x in X.

We have

2(1) = ( (1)) = (2) = 3

2(2) = ( (2)) = (3) = 4

2(3) = ( (3)) = (4) = 1

2(4) = ( (4)) = (1) = 2

So 2 is represented by {<1,3>, <2,4>, <3,1>, <4,2>}.

Now,

3(1) = ( 2(1)) = (3) = 4

3(2) = ( 2(2)) = (4) = 1

3(3) = ( 2(3)) = (1) = 2

3(4) = ( 2(4)) = (2) = 3


Now,

4(1) = ( 3(1)) = (4) = 1

4(2) = ( 3(2)) = (1) = 2

4(3) = ( 3(3)) = (2) = 3

4(4) = ( 3(4)) = (3) = 4


4. (a) Let : A B and g: B C be function then

(i) If both and g are one–to–one (go ) is also one–to–one.

(ii)If both and g are onto (go ) is also onto.

(b) Let : A B and g: B C be both one–to–one and onto functions then (go )-1

= -1og-1.

Proof (a):

(i) To prove that (go ) is one–to–one we must prove, if

(go )(x1) = (go )(x2) then x1 = x2.

Let x1,x2 A then


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1 2

1 2

1 2

1 2

g f x g f x

g f x g f x

f x f x g is one-to-one

x x f is one-to-one

Therefore (go )is one-to-one.

(ii) Let c be an element in C, since g is onto, there exists an element b in B such that

g(b) = c.

As is onto, there is an element a A such that (a) = b.

Now (go )(a) = g( (a)) = g(b) = c.

Thus for an element c C, we have shown that there in an element a A such that

(go )(a) = c.

go is an onto function.

Proof (b):

As and g are both one-to-one and onto, go is also one-to-one and onto from A

onto C.

(go )-1 exists and is also a one-to-one and onto mapping from C to A.

Let c be any element of C, we have to show that

(go )-1(c) = ( -1og-1)(c)

Let (go )-1(c) = a then (go )(a) = c that is g( (a)) = c.

So (a) = g-1(c) (g is invariable)

(a) = g-1(c)

a = -1(g-1(c)) ( is invariable)

Hence a = ( -1og-1)(c)

Thus we have shown that (go )-1= -1og-1.


123

5. (a) If : A B is one-to-one and onto mapping then

(i) o -1 = IB, the identity mapping of B onto itself.

(ii) -1o = IA, the identity mapping of A onto itself.

(b) Let : R R and g: R R when R is the set of real numbers be given by

(x) = x2 - 2 and g(x) = x + 4. Find ( og) and (go ). State whether these function

are injective surjective and bijective.

Proof (a):

Since is a one-to-one mapping from A onto B -1 is also a one-to-one mapping of

B onto A. -1: B A

Hence o -1 is a function from B onto B similarly -1o is a function from A onto A.

(i) Let b be any element of B.

Let -1(b) = a. ( -1 is a function from B onto A)

Then (a) = b

( -1(b)) = (a) = b.

That is o 1(b) = IB(b) b B.

o -1 = IB.

(ii) Let a be any element of A.

Let b = (a).

Then a = -1(b) ( -1 is function from B onto A)

IA(a) = a = -1( (a)) = ( -1o )(a)

Hence -1o = IA.

Proof (b):

Let : R R and g: R R are given by

(x) = x2 - 2 and g(x) = x + 4.

Now ( og): R R is given by

( og)(x) = (g(x)) = (x+4) = (x+4)2 - 2 = x2 + 8x + 14.


124

And (go ): R R is given by

(g0 )(x) = g( (x)) = g(x2-2) = (x2-2) + 4 = x2 + 2.

The function g is injective as

g(x1) = g(x2)

x1 + 4 = x2 + 4

x1 = x2

g is onto as to each x,

g(x-4) = (x-4) + 4 = x

So g is bijective.

The function is not injective as

(2) = (-2) = 4 – 2 = 2

is not onto as there is no real x such that (x) = -4

( og)(x) = (x+4)2 – 2 -2 x R.

( og) is not onto since there is no x R such that ( og)(x) = -4.

( og) is not one-to-one as ( og)(1) = 23 and ( og)(-9) = 23

(go ) is also not one-to-one as (go )(x) = x2 + 2

(go )(1) = (go )(-1) = 3

(go )(x) = x2 + 2 2 x R.

(go ) is not onto since there is no x R such that

(go )(x) = x2 + 2 = 0

6. If :A B and g: B A are mappings such that og = IB and go = IA then and g

are both invertible. Furthermore -1 = g and g-1 = .

Proof:

Let us first show that is one-to–one.

Let x1,x2 A.

Now (x1) = (x2)

g( (x1)) = g( (x2))

(go )(x1) = (go )(x2)

IA(x1) = IA(x2) (as go = IA)

x1 = x2 (as IA(x) = x x A)

(go ) is one-to-one.


125

To show that is onto,

Take any element b in B then b = IB(b) = og(b) ( og = IB)

That is b = (g(b))

Hence is onto.

As is both one-to–one and onto, -1 exists and is a function from B onto A.

Using go = IA.

Now we prove that g-1 exists and it is a function from A to B. Also

g = IA og

= ( -1o )og

= -1o( og)

= -1oIB

= -1

g = -1

Also g = -1 g-1 = ( -1)-1

g-1 =

Hence g-1 = .

7. (a) Show that there exists a one-to-one mapping from A B to B A. Is it also

onto?

(b) Let X = {1, 2, 3, 4}. Define a function : X X such that Ix and is one-to-

one. Find o = 2, 3 = o 2, -1 and o -1. Can you find another function which

in one-to-one g: X X such that g Ix by gog = Ix?

Solution (a):

Let : A B B A be a mapping defined by (<a,b>)= <b,a> for a B.

Clearly is one-to-one.

(<a1,b1>) = (<a2,b2>)

<b1,a1> = <b2,a2>

b1 = b2 and a1 = a2

<a1,b1> = <a2,b2>


126

Clearly is onto because for every element <b,a> B A there is an element <a,b>

A B

Thus ‘ ’ is an bijective map from A B to B A.

Solution (b):

Let : X X defined by (1)=2, (2)=3, (3)=4, (4)=1.

Then 2 = {<1,3>, <2,4>, <3,1>, <4,2>} 3 = {<1,4>, <2,1>, <3,2>, <4,3>} -1 = {<2,1>, <3,2>, <4,3>, <1,4>}

o -1 = {<1,1>, <2,2>, <3,3>, <4,4>}

It is possible to find a one-to-one function g: X X such that g Ix

Take g = {<1,2>, <2,1>, <3,4>, <4,3>}

gog = Ix.

8. (a) Let A be a set with binary operations * and eL and eR be left and right

identities of the set A with respect to the operation *. Then eL = eR = e such

that e * a = a = a * e for a A and in such case e is unique having this property

and is called identity of A with respect to *.

(b) For a set A with binary operation *, which is associative, which has identity

e A, an element a A is invertible, if both left and right inverses are equal.

Such an element is called the inverse of a, because it is unique.

Proof (a):

As eL is the left identify of A with respect to *, we have

eL * eR = eR (1)

and as eR is the right identity of A with respect to *, we also have

eL * eR = eL (2)

From, (1) and (2) it is clear that eL = eR = e.

Suppose we have two identities e1 and e2 implies e1 = e2.

Which is a contraction to the assumption that e1 e2.

The identity of a set with respect to an operation * must be unique.


127

Proof (b):

Let xL, xR be left and right inverse of a A then a * xR = e and xL * a = e.

xL * a * xR = (xL * a) * xR = xL * (a * xR)

e * xR = xL * e

xR = xL

xL = xR.

That is both left and right inverses are equal.

Now to show inverse is unique.

Let us assume that x and y are two distinct inverses of a.

Thus y = y * e = y * (a * x) = (y * a) * x = e * x = x.

This is a contradiction.

Thus inverse of an element a A is unique with respect to the binary operation *.

9. (a) Let g: R’ R’ R’ where R’ is the set of real number and g(x,y) = x * y = x + y -

xy. Show that the binary operation * is commutative and associative. Find

the identity element and indicate the inverse of each element.

(b) Let x * y = least common multiple of x and y where * is an operation on the

set of integers which are greater than zero. Show that * is commutative and

associative. Find the identity and the elements which are idempotent.

Solution (a):

Let g: R’ R’ R’ defined by g(x,y) = x * y = x + y - xy for every x,y R’

g(y,x) = y * x = y + x – yx = x + y – xy = x * y = g(x,y)

Therefore, * is commutative

g(g(x,y),z) = g(x,y) * z

= (x * y) * z

= ((x * y) * z)

= (x + y - xy) + z - (x + y – xy)z

= x + y + z – xy – xz – yz + xyz


128

g(x,g(y,z)) = x * g(y,z)

= x * (y * z)

= x * (y + z - yz)

= (x + y + z - yz) – x(y + z - yz)

= x + y + z – xy – xz – yz + xyz

Thus g(g(x,y),z) = g(x,g(y,x)).

* is associative.

g(x,e) = x * e

= x + e – ex

= x

e = 0

Thus e = 0 is identity of *.

If y is the inverse of x, then g(x,y) = x * y = 0 = x + y - xy

y(1-x) = -x

x 11

xy where

x

Thus inverse of x is 1

x

x with respect to *.

Solution (b):

x * y = l.c.m(x,y)

= l.c.m(y,x)

= y * x

* is commutative.

Also (x * y) * z = l.c.m(l.c.m(x,y), z)

= l.c.m(x, l.c.m(y,z))

= l.c.m(x, (y * z))

= x * (y * z)

Thus * is associative.

Clearly l.c.m(a,b) = b if b > a and a is a divisor of b.

Also 1 is only integer which divides all positive integers.


129

Thus l.c.m(1,m) = m m Z+={1,2 ...}.

Therefore 1 is the identity of Z+ with respect to * and each element Z+ is an

idempotent, because m * m = l.c.m(m,m) = m I

10. (a) Show that the function f<x,y> = x+y is primitive recursive.

(b) Show that f<x,y> = xy is a primitive recursive function.

(c) Show that f<x,y> = y(mod x) is a primitive recursive function.

Solution (a):

We know that (x + y) + 1 = x + (y + 1)

As <x,y> = x + y, it is clear that

<x,y+1> = x + y + 1

= (x + y) + 1

= <x,y> + 1

= S( <x,y>)

As 1

1f x, 0 = x =U (x) [using projective function]

3

3f x, y+1 = S , , ( , )U x y f x y

Hence the given function is expressed in terms of initial function. Hence <x,y> = x +

y is recursive and as it is obtained by finite numbers of operations of composition

and recursion, it is a primitive recursion.

Solution (b):

We know that

0

0

y+1

1 0

0 for x = 0

also x *y

x x

put x

x x

As by definition

<x,0> = 1 if x 0

<x,0> = 0 if x = 0


130

Hence <x,y> = xy is defined as

<x,0> = Sg(x)

<x,y+1> = x * <x,y> = 3 3

1 3, , , * U , , ,U x y f x y x y f x y

Therefore f<x,y> = xy is primitive recursive.

Solution (c):

For y = 0, <x,0> = 0

If y increases by 1, then <x,y> = <x,y-1> + 1, until y becomes equal to x.

When y = x then <x,y> = 0 = <x,0>

This process continues.

As <x,y> is increased by value 1 if y increases by 1, until y = x, one can use successor

function to build the function and as y = x <x,y> = 0, this process continues ever.

Use an addition non-zero test function to represent this function as follows

,0 0

, 1 ( , * ,

f x

f x y S f x y Sg x S f x y

Note That

0 if S(f(x,y))= xSg ,

1 if S(f(x,y)) xf x y

Thus the given function is primitive recursive.


131

11. a. Show that the function 2

x which is equal to the greatest integer which is

2

x.

b. Show that the function f(x) = 2

x is a partial recursive function.

c. Let x be the greatest integer x , show that x primitive recursive.

Solution (a):

Now

10

2

21

2

31

2

42 .,

2etc

x if x is even

2

x-12 if x is odd

2

x

Thus the function should be related to parity function Pr, for odd and even function.

Therefore we define the greatest integer function by

00

2

1

2 2r

x xP x

where Pr(x) is the parity function which is 0, if x is even and 1 if x is odd.

Thus the greatest integer function is primitive recursive function.


132

Solution (b):

Let 2

x = y

2y = x

2y – x = 0 only for all even values of x.

Thus g<x,y> = |2y - x| is not regular as it is 0 only for even values of x, but not all

values of x.

y( ) 2 0f x y x

Thus (x) = 2

x for x is even is a partial recursive.

Solution (c):

Clearly 2

2

2

0 (y+1)(y+1)

0 (y+1)

for xx

for x

2

2

2

1 (y+1) Sg y+1

0 (y+1)

for xx

for x

The smallest value of y for which 2(y+1) is xx

2

y 1 0x Sg y x

As x is total and defined for all values of x, x is primitive recursive function.

12. a. Show that the function when x is even

2( )

x-1when x is odd

2

x

f x is primitive

recursive.

b. Show that the function <x,y> = x - y is partial recursive.

c. Show that the function (x)=x! is primitive recursive where 0!=1 and n!=n(n-1)!

Solution (a): x+1

f(x+1) = if x is odd2


133

x+1 1= if x is even

2 2 2

x

1( 1)

2 2

2

xf x if x is odd

xif x is even

The function 1

( 1) ( ) *2

rf x f x Sg P x defined by non-zero test function

and parity function for odd and even number is primitive recursive.

Solution (b):

Clearly for x,y N, the function is well defined only for x > y.

<x,y> = x - y for only x > y is partial recursive.

Solution (c):

Clearly (0) = 1

and (n) = n!

with (x) = x * (x-1) for x 2

f(x) = x * f x-1 * 1 ( ) 1Sg x Sg x Sg x

will be the required function and hence (x) = x! is primitive.

Note that, when x = 0, the 1st term is 0, and the 2nd term is 1, and 3rd term is 0 thus

(0) = 1.

When x = 1, the 1st term and 2nd term are 0 and the 3rd term is 1 and hence (1) = 1.

For all values of x N with x 2, the 2nd and 3rd terms are zero and the 1st term

becomes

(x) = x * (x-1) * 1

= x * (x-1)

Which is true.


134

13. Find all mappings from A = {1, 2, 3} to B = {4, 5}. Find which of them are one-to-

one and which of them are onto. (AU/Nov-Dec-2006)

Solution:

None of the function is one-to-one. f1 and f7 are not onto.

14. (a) If 1 2 3 4

f3 2 1 4

and 1 2 3 4

g=2 3 4 1

are permutations. Prove that (gof)-

1 = f-1og-1.

(b) If R denotes the set of real number and f: R R is given by f(x) = x3 - 2, find

f-1.

Solution (a):

-11 2 3 4

f3 2 1 4

1

2

3

4

5

f1=

1=

1

2

3

4

5

f2=

=

1

2

3

4

5

f3= 1

2

3

4

5

f4=

1

2

3

4

5

f5= 1

2

3

4

5

f6=

1

2

3

4

5

f7= 1

2

3

4

5

f8=


135

-1

-1 -1

-1

-1 -1 -1

1 2 3 4g

4 1 2 3

1 2 3 4f g

4 3 2 1

1 2 3 4(g f)

4 3 2 1

1 2 3 4(g f)

4 3 2 1

g f f g

Solution (b):

To show that f is one-to-one

(x) = (y) x3 = y3

Both x and y are 0 or both x and y are negative.

x3 = y3 |x| = |y|.

If both x,y 0 then x = y.

If both x,y < 0 –x = -y and so x = y.

Hence is one-to-one.

To show that is onto

Let y be a real number in R

Find (x) = y or x3 - 2 = y or x3 = y + 2.

If y + 2 0, but 1

3x y 2

If y + 2 < 0, 1

3x y 2 hence -1 exist. 1

31

1

3

y 2 if x 2 0f (y)

y 2 if x 2 0


136

15. Let + denote the set of positive integers and denote the set of integers. Let

: + be defined by

nif n is even

2f(n)

1 - nif n is odd

2

Prove that is a bijection and find -1. (AU/Nov-Dec 2006).

Solution:

Let (m) = (n)

If m and n are both even, then m n

2 2 so m = n.

Similarly if m and n are both odd,1-m 1 n

2 2 so m = n.

We show that (m) = (n) m even n odd cannot occur.

Suppose m is even and n is odd then m 1 n

2 2 are m + n = 1.

But m and n are the integers

m + n 2 then is one-to-one is onto.

If n is an integer, we must find a positive integer m such that f(m) = n.

If n > 0 take m = 2n, if n < 0 take m = 1 - 2n.

Hence f is onto.

12n if n 0

f m1 2n if n 0

16. Let A, B, C can be any three non empty sets. Let : A B and g: B C be


137

mappings. If and g are onto, prove that go : A C is onto. Also given an

example to show that go f may be onto but both and g need not be onto.

(AU/Nov-Dec/2006).

Solution:

Let z C, since g: B C is onto, then there exist y B such that g(y) = z.

Since :A B is onto, and y B implies there exist x A such that (x) = y.

z = g(y) = g( (x)) = (go f(x).

Hence go : A C is onto.

Example:

(go ) is onto. But g is onto and is not onto.

17. Let X= {1,2,3,4,5,6,7} and R={(x,y)/x-y is divisible ley 3}. Show that R is an

equivalence relation. Draw the graph of R.

Solution:

Given x={1,2,3,4,5,6,7,8}

Relation R={(x,y)/x-y is divisible by 3}.

R 1,1 , 2,2 , 3,3 , 4,4 , 5,5 , 6,6 , 7,7 , 1,4 , 4,1 , 1,7 , 7,1 , 2,5 , 5,2 3,6 , 6,3 , 4,7 , 7,4

R is an equivalence relation. Since R is reflexive, symmetric and transitive.

Graph of R

1

2

3

4

5

6

7

8

9

10

g

A B C


138

Fig.

UNIT – V


139

ALGEBRAIC SYSTEMS

Algebraic Systems

Definitions & Examples

Properties

Semi Groups

Monoids

Homomorphism

Sub Semi Groups & Sub Monoids

Cosets

Lagrange’s Theorem

Normal Sub Groups

Normal Algebraic System with Two Binary Operations

Codes & Group Codes

Basic Notions Of Error Correction

Error Recovery In Group Codes

ALGEBRIC SYSTEM: A system consisting of a set and one or more n-ary operations

on the set is called an algebric system, or simply an algebric.

The operations and relations on the set S define a structure on the elements of S, an

algebric system is called as an algebric structure.


140

GROUPS: A group (G,*) is an algebraic system with non empty set in which the

binary operation * on G satisfies following conditions.

1. Closure

For all a, b G, a b G

2. Associative

For all x,y,z G x*(y*z) = (x*y)*z

3. Identity:

There exist an element e G such that for any x G,

x e = e x = x where e is identity.

4. Inverse:

For every x G, there exist an element denoted by x-1 G such that

x-1 x = x x-1 = e.

ABELIAN GROUP: For every a,b G, if a b = b a then the group G is called

abelian group.

5. Distributive:

For any a, b, c I

a (b + c) = (a b) + (a c) the operation distributes over +.

6. Cancellation:

For a, b, c I and a ≠ 0

a b = a c b = c

COSET: Let < H, > be a subgroup of < G, > for any a G, the set aH is defined by

aH = {a h / h H} is called left coset of H in G. Similarly right coset can be defined

as Ha = {h a / h H}.

SUBGROUP: Let (G, ) be a group and S G be such that if it satisfies the following

properties


141

1. e S, where e G.

2. a S, a-1 S.

3. a, b S, a b S.

Thus (S,*) is called subgroup of (G, ).

GROUP HOMOMORPHISM: Let (G, ) & (H, Δ) be two groups. A mapping g: G

H is called a group homomorphism from (G, ) to (H, Δ) if for a, b G, g(a b) =

g(a) Δ g(b).

KERNEL OF HOMOMORPHISM: The kernel of homomorphism ‘g’ from a group

(G, ) to (H, Δ) is Ker(g) = ,a G / g(a) = eH}

LANGRANGE’S THEOREM: “The order of a subgroup of a finite group divides the order

of a group”.

NORMAL GROUP: A subgroup (H, ) of (G, ) is called a normal if for any a G,

aH = Ha aHa-1 = H aha-1 H, for every h H.

RING: An algebraic system (S, +, ) is called a ring if the binary operations +

and on S satisfy the following three properties.

1. (S, +) is an abelian group.

2. (S, ) is a semi group.

3. The operation is distributive over + ; i.e. for any a, b, c S,

a (b + c) = a b + a c and (b + c) a = b a + c a.

FIELD: An commutative ring (S, +, ) which has more than one element such

that every non zero element of S has a multiplicative inverse in S called as field.

SUB RING: A subset R S where (S, +, ) is a ring, is called as a sub ring if (R, +,

) is itself a ring with the operations + and restricted to R.

…………………………………………………………………………………………………


142

PART – A

1. Define algebraic system.

A system consisting of a set and one or more n-ary operation on the set is called

an algebraic system or system on algebra. Denote an algebraic system by <S, f1,f2...>

where S is an non empty set and f1,f2... are operation on S.

2. Define homeomorphism.

Let < X, > and < Y,* > be two algebraic systems of the same type in the name that

both and * are binary (n-ary) operations. A mapping g: X Y is called a

homomorphism, or simply morphism from < X, > to < Y,* > if for any

x1,x2 X

g(x1 x2) = g(x1) * g(x2)

In such a functions g exists, then < Y,* > is a homomorphic image of < X, > although

g(X) Y.

3. Define endomorphism monomorphism and isomorphism

Let g be a homomorphism from < X, > to < Y,* >.

If g: X Y is onto, then g is called an epimorphism.

If g: X Y is one-to-one, then g is called a monomorphism.

If g: X Y is one–to–one & onto, then g is called an isomorphism.

4. Define isomorphic.

Let < X, > and < Y,* > be two algebraic system of the same type. If there exists an

isomorphic mapping g: X Y, then < X, > and < Y,* > are said to be isomorphic.

5. Define endomorphism and automorphism.

Let < X, > and < Y,* > be two algebraic system such that Y X. A

homomorphism f from < X, > to < Y,* > in such a case is called on endomorphism. If

Y = X, then an isomorphism from < X, > to < Y,* > is called an automorphism.


143

6. Define semi group.

Let S be a nonempty set and be a binary operation on S. The algebraic system <

S, > is called a semi group if the operation is associative. In other words < S, > is a

semi group if for any x,y,z S,

(x y) z = x (y z).

7. Define monoid.

A semi group < M, > with an identity element with respect to the operation is

called a monoid. In other words, an algebraic system < M, > is called a monoid if for

any x,y,z M.

(x y) z = x (y z)

and there exists an element e M such that for any x M

e x = x e = x

8. Define monoid homomorphism.

Let < M,*,eM > and < T, ,eT > be any two monoids.

A mapping g: M T such that for any two elements a,b M.

g(a * b) = g(a) g(b) and g(eM) = eT

is called a monoid homomorphism.

9. Define sub-monoid.

Let < M,* > be a semi group and T S. If the set T is closed under the operation,

* than < T,* > is said to be a sub semi group of < S,* >.

Similarly, let < M,*,e > be a monoid and T M. If T is closed under the operation

* and e T then < T,*,e > is said to be a sub monoid of < M,*,e >.

10. Define left coset.

Let < H,* > be a sub group of < G,* > for any a G, the set aH defined by

aH = {a*h / h H}

is called the left coset of H in G determined by the element a G. The element ‚a‛ is

called the representative element of the left coset of aH.


144

11. Define normal subgroup.

A sub group <H,*> of < G,*> is called a normal sub group if for any a G, aH=Ha.

12. Define ring.

An algebraic system < S,+,. > is called a ring of the binary operations ‚+‛ and ‚.‛

on S satisfy the following there properties.

(i) < S,+ > is an abelian group.

(ii) < S,. > is a semi-group.

(iii) The operation is distributive over +;

i.e., for any a,b,c S,

a.(a+b) = a.b + a.c and (b+c).a = b.a + c.a.

13. Define field

A commutative ring < S,+,. > which has more than one element such that every

nonzero element of S has a multiplicative inverse in S is called a field.

14. Define sub ring.

A sub set R S where < S,+,. > is a ring is called a sub-ring if < R,+,. > is itself a

ring with the operations + and . restricted to R.

15. Define ring homomorphism.

Let < R,+,. > and < S, , > be rings a mapping g: R S is called a ring

homomorphism from < R,+,. > to < S, , > if for any a,b R,

g(a + b) = g(a) g(b) and g(a . b) = g(a) g(b).

16. Find the direct product of the groups (Z3,+3) and (Z*3, 3).

Solution:

(Z3,+3) consists of the elements [0], [1], [2] and (Z*3, 3) consists of [1], [2].

(Z3,+3) (Z*3, 3) = {<[0],[1]>, <[0],[2]>, <[1],[1]>,

<[1],[2]>, <[2],[1]>, <[2],[2]>}.


145

17. If S denotes the set of positive integers 100 for x,y S, define x y = min{x,y}.

Verify whether (S, ) is a monoid assuming that is associative (AU/Nov-DC/2006)

Solution:

100 is the identity element in (S, ).

Since x 100 = min(x,100) = x.

Since x 100 x S

(S, ) is a monoid.

18. If H is a sub group of the Group G, among the right cosets of H in G. Prove

that there is only one subgroup viz. H. (AU/Nov-dec/2006)

Solution:

Let Ha be a right coset of H in G where a G. If Ha is a subgroup of G, then

e Ha where e is the identity element in G.

Ha is an equivalence class containing ‚a‛ with respect to an equivalence relation. So

e Ha He = Ha. But He = H. Ha = H

19. Find all the non–trivial subgroups of (Z6,+6)

(AU/Nov-Dec -2006)

Solution:

Z6 = {[0], [1], [2], [3], [4], [5]} of H is sub group of Z6, then O(H)|6.

Hence O(H) = 1, 2, 3 or 6 and O(H) = 6 H = Z6.

O(H) = 2 H = {[0], [x]} where [x] + [x] = 0 x = 3.

H = {[0], [3]}.

O(H) = 3 H = {[0], [x], [2x]} and [3x] = 0 x = 2.

H = {[0], [2], [4]}.

20. Define Hamming distance.


146

For x,y Bn the Hamming distance between x and y is the weight of x y.

Ie., H(x,y) = n

i i

i 1

(x y )

Example: Distance between 0110 and 110 is ‘1’

21. Define encoding function?

An (m,n) encoding function is a one-to-one function defined by e: Bm Bn (n > m).

If b Bm then e(b) is the code word representing b.

An (m,m+1) encoding function is defined by e: Bm Bm+1 where e(b1, b2 ... bm) =

b1, b2 ... bm,bm+1

where bm+1 = 0 if |b| is even

= 1 if |b| is odd.


147

PART – B

(1) If

0 1 1 1 0 0 0

1 0 1 0 1 0 0H

1 0 1 0 0 1 0

1 1 0 0 0 0 1

is the parity check matrix, find the Hamming code

generated by H (in which the first three bits represents information portion and

the next four bits are parity check bits). If y = (0, 1, 1, 1, 1, 1, 0) is the received word

find the corresponding transmitted code word. (AU/Nov-Dec -2006)

Solution:

Let n = 7 the Hamming code e is given by T

1 2 ne x (x x...x ): x.H 0(mod 2)

Let x = (x1 x2 x3 x4 x5 x6 x7)

x.HT = 0 x4 = x2 + x3

x5 = x6 = x1 + x3

x7 = x1 + x2.

(x1 = 0 or 1, x2 = 0 or 1, x3 = 0 or 1)

1 2 3 4 5 6 7x x x 0 x x x x 0

1 2 3 4 5 6 7

1 2 3 4 5 6 7

1 2 3 4 5 6 7

x 0,x 0, x 1 x 1,x x 1,x 0

x 0,x 1, x 0 x 1,x x 0,x 1

x 0,x 1, x 1 x 0,x x 1,x 1

1 2 3 4 5 6 7

1 2 3 4 5 6 7

1 2 3 4 5 6 7

1 2 3 4 5 6 7

x 1,x 0, x 0 x 0,x x 1,x 1

x 1,x 0, x 1 x 1,x x 0,x 1

x 1,x 1, x 0 x 1,x x 1,x 0

x 1,x 1, x 1 x 0,x x 0,x 0

Thus the code words are

(0 0 0 0 0 0 0)

(0 0 1 1 1 1 0)

(0 1 0 1 0 0 1)

(0 1 1 0 1 1 1)

(1 0 0 0 1 1 1)


148

(1 0 1 1 0 0 1)

(1 1 0 1 1 1 0)

(1 1 1 0 0 0 0)

Let y = (0 1 1 1 1 1 0) be the received word. To find the corresponding transmitted

word adding this to each code word above we get

e y 0 1 1 1 1 1 0

0 0 1 0 1 1 1

1 1 1 1 0 0 1

1 0 1 0 0 0 0

0 1 0 0 0 0 0

0 0 0 1 0 0 1

1 1 0 0 1 1 1

0 0 0 1 1 1 0

e = word least weight in y =(0 1 0 0 0 0 0)

Transmitted code word = e y = (0 1 0 0 0 0 0) (0 1 1 1 1 1 0)

=(0 0 1 1 1 1 0)

2. Determine the (2, 5) encoding function for the parity check matrix. 1 0 0

1 1 0

1 0 0

0 1 0

0 0 1

H

Solution:

The encoding eH = B2 B5 is defined by

eH(b) = eH(b1, b2) = b1, b2, c1, c2, c3 where

c1 = b11 b1 + b21 b2 = b1 + b2

c2 = b12 b1 + b22 b2 = b2

c3 = b13 b1 + b23 b2 = 0


149

eH(0,0) = 0 0 0 0 0

eH(0,1) = 0 1 1 1 0

eH(1,0) = 1 0 1 0 0

eH(1,1) = 1 1 0 1 0

3. From the encoding function eH for the parity check matrix given by 1 0 0

1 0 0

0 1 1

1 0 0

0 1 0

0 0 1

H

Prove that it is a Group Code.

Solution:

The encoding function eH is defined by

eH(b1 b2 b3) = b1 b2 b3 c1 c2 c3

where c1 = b1 + b2, c2 = b3, c3 = b3,

Hence,

eH(0 0 0) = 0 0 0 0 0 0

eH(1 0 1) = 1 0 1 1 1 1

eH(0 1 1) = 0 1 1 1 1 1

eH(1 0 0) = 1 0 0 1 0 0

eH(1 1 0) = 1 1 0 0 0 0

eH(0 1 0) = 0 1 0 1 0 0

eH(0 0 1) = 0 0 1 0 1 1

eH(1 1 1) = 1 1 1 0 1 1

To show that {eH(b) : b B3} forms a group.

Let us consider from the following table.

Let 1 = 0 0 0 0 0 0,

2 = 0 1 1 1 1 1,

3 = 1 1 0 0 0 0,


150

4 = 0 0 1 0 1 1,

5 = 1 0 1 1 1 1,

6 = 1 0 0 1 0 0,

7 = 0 1 0 1 0 0,

8 = 1 1 1 0 1 1,

1 2 3 4 5 6 7 8

1 1 2 3 4 5 6 7 8

2 2 1 5 7 3 8 4 6

3 3 5 1 8 2 7 6 4

4 4 7 8 1 6 5 2 3

5 5 3 2 6 1 4 8 7

6 6 8 7 5 4 1 3 2

7 7 4 6 2 8 3 1 5

8 8 6 4 3 7 2 5 1

From table we can see that {eH(b) / b B3} is a group.

4. Let f: G G’ be a homomorphism from a group (G, *) to a group (G’, o) then the

Kernel of f is a normal sub group of G.

Proof:

= g G f(g) = e, e is identity in GKerf

Let g1, g2 ker f then 1 -1

1 1 1 1

1

1 1

f(g ) = f(g ) f(g )

= f(g ) f(g )

= e e

= e

g

1

1 1 g g kerf.

Hence ker f is a subgroup of G.


151

For x G, g ker f, we have f(g) = e’

1 1

-1

f(x gx) = f(x )f(g)f(x)

= f(x) e f(x)

= e

i.e., 1kerfx gx .

Hence ker f is a normal sub group.

5. Let (G, ) be a group and H be a sub group of G. Then the following are

equivalent.

(i) aH = Ha, a G

(ii) a-1Ha = H, a G

(iii) a-1Ha = H, a G

Proof:

(i) (ii)

Let aH = Ha, a G

To prove that a-1ha = H

Let a-1ha a-1Ha

Now ha Ha = aH ( H is a normal subgroup)

i.e., ha = ah, for some h1 H ( a-1Ha H)

Now, to prove that H a-1Ha

Let h H, then h = eh = a-1ah = a-1ha a-1Ha

H a-1Ha.

(ii) (iii)

This is also trivial since a-1Ha = H a-1Ha H.

(iii) (i)

Let a-1Ha H, a G.


152

To prove that aH = Ha a G.

Let ah aH then aha-1 aHa-1 H and (aha-1)a Ha

i.e., ah Ha

aH Ha (1)

Conversely if ha Ha, then a-1ha a-1Ha H and a(a-1ha) aH

i.e., ha aH

Ha aH (2)

From (1) and (2)

aH = Ha

INTRODUCTION

Let A be any set. A mapping nf : A A .... A A(or) f : A A is called an

n-binary operation and ‘n’ is called the order of the operation.

For n = 1, f : A A is called an Unary operation

For n = 2, f : A A is called an Binary operation.

A set together with a number of operations (binary) on the set is called an

Algebraic system (or) an Algebra.

Properties of Binary Operations

1. Closure Property: A binary operation * : G G G is said to be closed for all

a, b G, an element a * b = x G.

2. Associative : a * (b * c) = (a * b) * c, for all a, b, c G

3. Existence of Identity : There exist an element e G such that e * a = a * e =a,

4. Existence of Inverse : For a G, there exist an element b G such that a * b =

b * a = e. The element b is called the inverse of a and it is denoted by b = a-1.


153

5. Commutativity: For all a, b G, if a * b = b * a, then * is commutative

(abelian).

6. Distributive Properties:

A *(b . c) = (a * b) . (a*c) (Left distributive law)

(b.c)*a = (b*a) . (c*a) (Right distributive law)

For all a, b, c G

7. Cancellation Properties:

a*b = a* c b = c (Left cancellation law)

b*a = c*a b = c (Right cancellation law)

for all a, b, c G

Semigroups & Monoids

Definition 1: A non-empty set S together with the binary operation *: S S S is

said to be a semigroup if * satisfies the following conditions, namely the closure

property and the associative property. We denote the semigroup by (S, *).

Definition 2: A semigroup (M, *) with an identity element with respect to the

operation * is called a Monoid. In other words, a non – empty set M together with

the binary operation * : M M M is said to be a Monoid if * satisfies the closure

property, associative property and the identity property.

Example 1. Let N = ,1, 2, 3,…..) be the set of natural numbers. Then (N, +) and (N, )

are semigroup under the binary operations addition and multiplication respectively.

Example 2. Let N = ,1, 2, 3, ….) be the set of natural numbers. Then (N, + ) is a

monoid with the identity element 1, but (N, +) is not a monoid since the additivie

identity 0 in not a natural number.

Example 3. Let Z+ = ,0, 1, 2, 3,…} be the set of all non-negative integers. Then

+(Z , ) and (Z , ) for all a, b S.


154

Cyclic Monoids (Cyclic Semigroups)

A Monoid (Semigroup) (M, *) is said to be Cyclic if for some a, M, every

element x M is of the form an, where n is an integer, i.e., x = an.

The cyclic Monoid (M, *) is said to be generated by the element ‘a’, and ‘a’ is

called the generator of the cyclic Monoid.

Theorem 1. Every cyclic monoid (semigroup) is commutative.

Proof. Let (M, *) be a cyclic monoid generated by an element a M. Then for any

two elements x, y M, we have x = an and y = am, m, n are integers

Now

n m n m

m + n m n

x * y a * a a

= a a * a

= y * x

Therefore, (M, *) is abelian or commutative.

Morphism of Semigroups

Let (S, *) and (T, ) be any two semigroups. A mapping f : S T is said to be a

semigroup homomorphism if f(a * b) f(a) f(b), for a, b S.

A one – one semigroup homomorphism is called a semigroup monomorphism.

An onto semigroup homomorphism is called a semigroup epimorphism.

A one – one, onto semigroup homomorphism is called an isomorphism. Two

semigroups are said to be isomorphic if there exist a semigroup isomorphism

between them.

A homomorphism of a semigroup into itself is called a semigroup

endomorphism .

An isomorphism of a semigroup onto itself is called semigroup automorphism.


155

Example Let (N, +) and (Zm, +m) be any two semigroups.

Define a map m mf : N Z by f(a) = [a] ,for all a N.

Then m mf(a b) [a b] [a] [b],for all a N.

Therefore f is a semigroup homomorphism.

Morphism of Monoids

Let (M, *, eM) and (T, , eT) be any two monoids with the identity element EM

and or respectively. A mapping f : M T is said to be a monoid Morphism if for

any two elements a, b M, f ( a*b) = f(a) f(b) and f(eM) = eT.

If f is an injection, it is a monoid monomorphism

If f is an surjection, it is a monoid epimorphism

If f is an bijection, it is a monoid isomorphism.

Theorem 2. Let (S, *), (T, ) and (V, +) be semigroups.

Let f : S T and g : T V be semigroup homomorphism.

Then g f : S V is also a semigroup homomorphism.

Proof: given f: S T be a semigroup homomorphism.

That is f(a * b) f(a) f(b), for all a, b S and g: T V be a semigroup

homomorphism, i.e., g(a b) g(a) g(b), for all a, b, T .

Now define g f : S V by (g f)(a) = g [f (a)], for all a S.

Let a, b S, then

(g f)(a * b) g[f(a * b) g[f(a) f(b)]

= g[f(a)] + g[f(b)]

= (g f) (a) + (g f)(b)

Therefore, g f is a semigroup Morphism.


156

Definition 4. Let (S, *) be a semigroup and R be a congruence relation on (S, *). The

quotient set S/ R is a semigroup (s/R, ) is called the Natural Morphism.

Sub Semigroups and Sub Monoids

Definition 5. Let (S, *) be a semigroup. A subset T of S is said to be a sub semigroup

(T, *) if T is closed under the operation *. i.e., T is itself a semigroup under *.

Definition 6. Let (M, *, e) be a monoid. A subset of T of M is said to be a sub monoid

(T, *, e) if T is closed under the operation * and e T. i.e., T itself a monoid under

the same operation * with the same identity e.

Example 5. Let N = ,1, 2, 3,…} be the set of natural numbers. Then (N, ) and (N, +)

are semigroups. Consider the set E = {2, 4, 6, 8,…} of all even natural numbers.

Clearly (E, ) is a semigroup of (N, ) and (E, +) is a sub semigroup of (N, +).

Groups

Definition 7. A non-empty set G together with the binary operation *, i.e., (G, *) is

called a group if * satisfies the following conditions.

(i) Closure: For every a, b G, a *b G.

(ii) Associative: For every a, b, c G, a *(b*c) = (a* b)*c.

(iii) Identity: There exists an element e G called the identity element such

that a * e e* a a, for all a G .

(iv) Inverse: There exists an element a-1 G called the inverse of ‘a’ such that1 1a * a a * a e , for each a G.

Definition 8: A group (G, *) is said to be an abelian group (Commutative group) if

a * b b* a for all a, b G. Otherwise it is non-abelian group.

Definition 9. The order of a group (G, *), denoted by O(G) or |G|, is the number of

elements of G, when G is finite.

Note: If the number of elements in a group is finite, then the group is said to be a

finite group. If the number of elements in a group. if the number of elements in a

group is infinite group.


157

Example 6: The set of all integer Z with the operation of ordinary addition is an

infinite abelian group.

Example 7: The set of all integers Z is not a group under multiplication. i.e., (Z, .) is

not a group. Because there is no multiplication inverse in Z. But (Z, .) is a monoid

and hence a semigroup.

Example 8. The set of all non-zero real number , under the operation of ordinary

multiplication is a group. An inverse of -11 1a 0 is i.e., a

a a

Example 9. The set of all non-zero real numbers form an infinite abelian group

under the binary operation * defined by a * b = ab

,for all a, b 2

.

Example 10. Prove that the set A = {1, , 2} is an abelian group of order 3 under

multiplication. Where 1, , 2 are cube roots of unity and 3 = 1.

Solution:

Given A = {1, , 2}, with 3 = 1. Let us construct the composition table for the

elements of A, with the binary operation multiplication.

(i) Closure: Since all the entries in the composition table are the elements of

A, it is closed under multiplication.

(ii) Associative: It is true, since the product of complex numbers obey the

associative property.

2 3 3 2 2i.e.,1.( . ) 1. . (1. ).

(iii) Identity: From the table 1 is the identity element.


158

(iv) Inverse: Each element of a posses the multiplicative inverse. i.e., the

inverse of is 2, 1 is 1 etc.

(v) Commutative: From the table we see that the commutative property is

satisfied. 2 3 2i.e., . . etc.

(A, .) is an abelian group under multiplication and O(A) = 3. So that (A, .) is

a finite abelian group of order 3.

Example 11. Let G be a set of all 2 2 matrices with non-zero determinant. Then G is

a group under multiplication. Here the identity element is 1 0

I0 1

.

The inverse of each A = a b

c d is 1

d b1A

c aA. But this is a non-

abelian group. Since matrix multiplication need not be commutative.

Definition 10. For a set S having n elements, let Pn denote the set of n! permutations

on S (the set of all bijections on S.

i.e., nP {f / f is a permutation on S}, which is called the symmetric set of

degree n and nP n! .

Definition 11. The set Pn of all permutations of n elements is a permutation group

(Pn, ) also called the Symmetric group under permutation multiplication . The

group (Pn, ) is or order n! and degree n.

Note 1: The degree of a permutation group is the cardinality of the set on which the

permutation are defined.

Note 2: For n 2, that is 2 1 2 2P (f ,f ),(P , ) an abelian group. But for n > 2, it is always

non-abelian.

Note 3: The set An, of all even permutations of degree n forms a finite group of order n!

2 under permutation multiplication.


159

Note 4: The set Bn of all odd permutations will not be a group. because the product

of two odd permutations is an even permutation, that is, the set of all odd

permutations will not be closed under multiplication.

Example 12. Show that the set P3 of all permutations on three symbols 1, 2, 3, is a

finite non-abelian group of order 6 with respect to permutation multiplication as

composition.

Solution: Let S = {1, 2, 3}. We have 3! Permutations. P3 = 2 3 4 5 6{I,f ,f ,f ,f ,f } where I is

the identity permutation.

Thus P3 having 6 elements, let it be f1 = I, f2 = (1, 2), f3 = (2, 3), f4 = (3, 1), f5 = (1, 2, 3)

and f6 (1, 3, 2)

Now, we prepare a composition table for P3.

(i) Closure Property: Since all the entries in the table are elements of P3, P3 is

closed under multiplication.

(ii) Associative: Multiplication of permutations is an associative operation.

1 2 3 1 2 3

1 6 2 3

6 6

i.e., f .(f .f ) (f .f ).f

f .f f .f

f f


160

(iii) Identity: The identity permutation f1 is the identity element.

(iv) Inverse: Every element in P3 have multiplicative inverse.

1 -1i.e., (1, 2) (2,1) (1,2), (2, 3) (2,3) etc.

(v) Commutative: The composition is not commutative,

2 3 3 2i.e., f .f f .f

P3 is a finite non-abelian group of order 6 degree 3, with respect to

permutation multiplication.

Note: The group formed by the symmetries of a plane (or a solid figure) is known as

symmetric group.

Definition 12: The set of all rigid rotations and reflections of a regular polygon of n

sides under the binary operation is a group (Dn, #), where Dn is of order 2n. This

group (Dn, #), (n = 1, 2, …) is called a Dihedral group.

PROPERTIES OF GROUPS

Property 1. The identity of a group is unique (or) If (G, *) is a group and e is an

identity of G, then no other element of G is an identity of G.

Proof: Suppose that e1 and e2 are two identities of the group (G, *).

Now e1 is the identity, then 1 2 2 1 2e * e e * e e (1)

Again e2 is the identity, then 2 1 1 2 1e * e e * e e (2)

From (1) and (2), we see that e1 = e2

The identity is unique.

Note: The existence of the identity guarantees that no two rows or columns is the

composition table of (G, *) are identical.


161

Property 2: In a group, the left and right cancellation laws are true. That is

a* b a* c b c .

Proof: Let G = (G, *) be a group. For a G, there exists an a – 1 G such that 1 1a * a a * a e

(i) Left Cancellation law:

Let a*b = a * c

1 1 1 1a * (a * b) a * (a * c) (a * a) * b (a * a) * c

e * b = e * c b = c

(ii) Right Cancellation law:

-1 1

1 1

Let b * a = c *a (b* a)*a (c * a) * a

b * (a * a ) c * (a * a )

b*e = c*e b = c

Property 3. The inverse of any element of a group is unique.

Proof: Let G be a group, a G and e be the identity element of G. Suppose that b

and c be two inverses of an element a G.

b * a = e = a * b and c * a = e = a * c.

Now (b * a) *c = e * c = c

(b * a) * c = c b * a(a * c) c

b * e c b c

There is one and only one inverse for each element in a group.

Note (i) Let * be a binary operation on X. If there exists an element x X such that x

* a = a * x = 0, for every a X, then x is called the zero element of X. A group cannot

have a zero element because every element in a group is invertible.

Note (ii) The identity element is its own inverse.

Because e * e = e, e = e-1


162

Property 4: A group cannot have any element which is idempotent except the

identify element.

Proof: Let G be a group. Let us assume that a G is idempotent. Then a * a = a.

-1 1 1Now e = a * a a * (a * a) (a * a) * a

= e * a = a

e = a

Property 5: If a is an element of group G, then 1 1(a ) a (or) If the inverse of a is a-1,

then the inverse of a-1 is a.

Proof: If e is the identity of the group G, then

1 1a * a e a * a , for every a G

Now,

1 1 1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

-1 1

-1 1

(a ) * (a * a) (a ) * e (a )

(a ) * (a * a) (a )

((a ) * a ) * a (a )

e*a = (a )

a = (a )

Property 6: If a has inverse b and b has inverse c, then a = c.

Proof

Given a * b = e = b *a (1)

and b * c = e = c * b (2)

Now a = (a * e = a * (b*c) ((by (2))

= (a * b) * c (by associative)

= e * c

= c (by (1))

i.e., a = c


163

Property 7. The inverse of the product of two elements of a group is the product of

their inverses in reverse order. That is 1 1 1(a * b) b * a

Proof: Let a, b G and 1 1a ,b be their inverses respectively.

-1 1 -1 1 a * a e a * a and b * b e b * b

Now

1 1 1 1

-1 1

-1 1

(a * b) * (b * a ) a * [b * (b * a )]

= a*[(b*(b )* a ]

= a*[e*a ] a * a

= e

Similarly we can prove 1 1(b * a ) * (a * b) e.

-1 1 -1 1 (a * b)* (b * a ) e (b * a ) * (a * b)

We see that -1 1(b * a ) is the inverse of (a*b)

That is 1 -1 1(a * b) b * a

Note: For abelian groups, 1 -1 1(a * b) a * b

Because b-1 * a-1 = a-1*b-1 for abelian group.

Property 8: The existence of a unique inverse of every element of G guarantees the

unique solvability of any equation of the type a *x = b, where a, b G. (OR)

If G is a group and a, b G the equation a *x = b has a unique solution x = a-1 *

b.

Proof: Let G be a group. Then for any a, b G,

We have 1 1 -1 1a * a e a and b*b e b * b.


164

Now

-1

-1

-1

a * x b e * b

= (a*a ) * b

= a*(a * b)

x = a * b (by left cancellation law) To prove this solution is unique, let us suppose x1 and x2 are two solutions of a*x = b.

Then

1 2

1 2

1 2

a * x b and a * x b

a*x = a* x

x = x (by left cancellation law)

1x a * b is the unique solution for a * x =b . Similarly we can prove the

second part.

Property 9: For any group G, if a2 = e with a e, then G is abelian.

(OR)

If every element of a group G is its own inverse, then G is abelian. Is the converse

also true.

Proof: Let G be a group and a, b G, then a * b G.

Given a = a-1 and b = b-1

1 1 1(a * b) (a * b) b * a b * a

a * b = b * a

G is abelian

The converse need not be true, because (Z, +) (or) ( , + ) is an abelian group.

Except 0, there is no element which is its own inverse.


165

Property 10. A group G is abelian if 2 2 2(a * b) a * b

Proof: Let us assume that G is abelian. Hence, for a, b G.

We have

2 2

a * b b * a

Now a * b (a * a) * (b * b)

= a * [a* (b * b)]

= a * [(a*b)*b)]

= a * [[(b*a)*b] ( a * b = b *a)

2

2 2 2

= a * [b* (a*b)]

= (a*b) * (a * b)

= (a*b)

(a*b) a * b

Conversely, assume that (a*b)2 = a2*b2

(a * b) * (a * b) (a * a) * (b * b)

a * [b * (a * a)] a * [a(b * b)]

b*(a*b) = a*(b*b) (Left cancellation law)

(b * a) * b (a * b) * b

b*a = a*b (Right cancellation law)

Note Properties of Exponentiation:

If a G and m, n are integers, then

n 1 1 n

n m n m

n m nm

(i) (a ) (a )

(ii) a * a a

(iii) (a ) a

n, the integers modulo n:


166

The division properties of integers:

If m, n , n > 0, then there exists two unique integers q (quotient) and r

(remainder) such that m = nq + r and 0 r < n

Congruent modulo n:

An integer ‘a’ is said to be congruent to another integer ‘b’ modulo n (n being

fixed positive integer) if a – b is divisible by n.

i.e., (a b)

k or a - b = nk,n

where k is any integer and we express it as a = b

(modulon).

Modular Arithmetic:

1. The addition modulo n is defined by a +nb = the least positive remainder when

(a + b) is divided by n.

2. The multiplication modulo n is defined by a nb = the least positive

remainder when (ab) is divided by n.

Properties of Modular Arithmetic on n

1) Addition modulo n is always commutative and associative. 0 is the identity

for +n and every element of n has an additive inverse.

2) Multiplication modulo n is always commutative and associative and 1 is the

identity for n. Multiplication modulo n is distributive (both right and left)

over addition modulo n. Also every element of n has the multiplicative

inverse.

Example 13. Prove that ( 3, +3) is an abelian group.

Example 14. Prove that ( 5, 5) is an abelian group.

Sub Groups

Definition 13. A non empty subset H of a group G is said to be a subgroup of G if H

itself is a group under the same operation defined on G and with the same identity

element. (OR)


167

Let (G, *) be a group and H G (H ) is said to be a subgroup (H, *) of (G, *)

if

(i) e H, where e is the identity of G

(ii) For any a H, a-1 H.

(iii) For a, b H, a*b H.

Note: For any group (G, *), (i) the subgroups (G, *) and ({e}, *) are called improper

(or) trivial subgroups and (ii) all other subgroups are called the proper or non-trivial

subgroups.

Example 15. (i) The set of all integers is a subgroup of the set of all real numbers

under addition. That is ( , +) is a subgroup of ( , +)

(ii) The set of even integers 2 = 2k/k is a subgroup of ( , +).

(iii) The set of non-negative integers is not a subgroup of ( , +) because, except 0, no

element have the additive inverse.

Theorem 3. The necessary and sufficient condition that a non-empty subset H of a

group G be a subgroup is a H, b H c * b-1 H.

Proof: (Necessary Condition):

Assume that H is a subgroup of G. Since H itself is a group; we have a, b H

(closure). Also b H b-1 H (Inverse).

-1 1 a, b H a,b H a * b H

Sufficient Condition: Let a * b-1 H (1) for all a, b H and H G, we have to

prove that H is a subgroup of G.

For,

(i) Identity: Let a H, a H a * a-1 H

e H (by (1))

Hence the identity e is the element of H.


168

(ii) Existence of inverse:

-1 1Let e H, a H e*a H a H

i.e., Every element of H has an inverse which is in H.

(iii) Closure:

Let b H b-1 H

For a, b H a, b-1 H a*(b-1)-1 H

a * b H. H is closed under the composition.

(iv) Associative: Since H G, the elements of H are also the elements of G. Since the

composition in G is associative, it must also be associative in H.

H itself is a group for the composition in G.

i.e., H is a subgroup of G.

Theorem 4: The intersection of two subgroups of a group is also a subgroup of the

group. (OR) If H1, H2 are two subgroups of a group G, then H1 H2 is also a

subgroup of G.

Proof: Let H1 and H2 be any two subgroups of G, then 1 2H H 0 , because atleast

the identity element is common to both H1 and H2.

To prove 1 2H H be any two subgroup, it is enough if we show that, for a, b

1 2H H a * b-1 1 2H H .

1 2 1 2

1 -11 2 1 2

-11 2

11 2 1 2

1 2

For let a, b H H a,b H and a, b H

a * B H and a*b H ( H and H are subgroups)

a*b H H

i.e., a, b H H a * b H H

H H is a subgroup of G.

Note: The arbitrary intersection of subgroups is also a subgroup.


169

Theorem 5: The union of two subgroups of G need not be a subgroup of G.

Proof: To prove this consider the following example. Let G be a group of integers

under addition.

1

2

1 2

1 2 1 2

i.e.,( , ) (G, )is a group.

Now define H {x x 2n, n } {0, 2, 4, 6.....}

and H {x x 3n, n } {0, 3, 6, 9, 12,....}

we see that H and H are subgroups of G.

Define H H = { x x H or x H } {0, 2, 3, 4,

1 2 1 2

1 2

1 2

1 2

6, 8, 9,.......}

for 2, 9 H H 2 9 11 H H

i.e., H H is not closed under addition.

i.e., H H is not a group

i.e., H H is not a subgroup of G.

Theorem 6: If (G, *) is a finite group, H is a non-empty subset of G and H is closed

under *, then H is a subgroup of G.

Proof: Let (G, *) is a group and O(G) = n (say).

Since H is closed under *, we have a, b H a * b H.

Now a H a2 = a * a H, a3 = (a * a) * a H and so on.

There exists an m, 1 m n such that am = e a-1 H

Again a H am-1 H am*a-1 H e * a-1 H

a-1 H. Thus inverse of an element a H exists and belongs to H.

The subset H satisfies the closure property as given, possesses identity and

inverse also exists for every element belonging to H. The associative property is

implied as the composition in G being associative and H has the same composition

as of G.

H is a subgroup of G.


170

Example 16: Determine whether H1 = (0, 5, 10) and H2 = (0, 4, 8, 12) are subgroups of

15.

Proof: Since all the entries in the addition table for H1 are elements ofH1, we see that

H1 is a subgroup of 15. Again since there is some entries in the addition table for

H2 that are not the elements of H2, H2 is not closed under addition, Hence H2 is not a

subgroup of 15.

Cyclic groups

Definition 14. A group (G, *) is said to be cyclic if there exists a G such that any x

G can be written as either x = an or x = na, where n is some integer. Here the

element ‘a’ is called the ‘generator’ of the cyclic group G. That is the cyclic group

generated by ‘a’ and we denote it by G = (a).

Theorem 7: Every cyclic group is abelian.

Proof: Let (G, *) be a cyclic group generated by an element a G, i.e., G = (a). Then

for any two elements x, y G, we have x = an, y = am, where m, n are integers.

n m n m m n

m n

x * y a * a a a

= a * a =y*x

(G,*) is abelian.

Theorem 8: If ‘a’ is a generator of a cyclic group G, then a-1 is also a generator of G,

(OR) If ‘a’ is an element of group G, then (a) = (a-1).

Proof: Let G = (a) be a cyclic group generated by ‘a’. Then ar G, where r is some

integer we can write ar = (a-1)-r, since –r is also some integer.

Each element of G is generated by a-1.

i.e. a-1 is also a generator of G.

i.e., 1(G) (a) (a ) .


171

Example 17: Prove that the group G = {1, -1, i, -i} is cyclic and find its generators.

Solution: We can write 1 = 4 2 1 31 (i) , 1 (i) , i (i) , i (i)

i.e., All the elements of G can be expressed as integral powers of the element i.

G is a cyclic group generated by i.

Since ‘i’ is the generator of G, (i)-1 is also a generator of G. Hence G is a cyclic

group and its generators are i and (i)-1 = 1

ii

Theorem 9: Every subgroup of a cyclic group is cyclic.

Proof: Let G = (a) is a cyclic group generated by ‘a’ and H is its subgroup. The

elements of H are integral powers of ‘a’. If as H, then the inverse of as i.e., a-s H.

Therefore H contains elements which are positive as well as negative integral powers

of ‘a’. Let m be the least positive integer such that am H. Then we shall H contains

elements which are positive as well as negative integral powers of ‘a’. Let m be the

least positive integral such that am H. Then we shall prove that H = (am). i.e., H is

cyclic and is generated by am.

Let at be any arbitrary element of H. By division algorithm, there exists

integers q and r such that t = mq + r, 0 r < m.

qm m

mq mq

-mq mqt t

t-mq

Now a H (a ) H (by closure property)

a a H

Also a H, a H a * a H

a H

r a H

Now m is the least positive integer such that am H and 0 r < m. Therefore r must

be equal to zero.

Hence t = mq. mq qt ma a (a ) .


172

Thus every element at H is of the form (am)q. Therefore H is cyclic and is

generated by am. i.e., H = (am).

Theorem 10: Let (G, *) be a finite group generated by an element an G. If G is of

order n, that is O(G) = n, then an = e so that G = {a, a2, ……an = e}. Further, n is the

least positive integer for which an = e.

Proof: Let us assume for some positive integer m < n, am = e.

Since G is cyclic, any element of G can be written as ak, for some k . By

division algorithm,

k mq r, where q and 0 r m.

mq r mq q qk r m r r

r r

a a a a (a ) a e a

= e a a

Hence every element of G can be expressed as ar for some 0 r m.

G has at most m distinct elements.

i.e., O(G) = m < n, which is contradiction.

Hence am = e for m < n is not possible.

We now proceed to show that the elements a, a2, a3, ……an are all distinct where an=e.

If possible, let jia a for i < j n.

j ji

i j

a a a a

a e,

Where i – j < n, which is again a contradiction.

Hence jia a for i < j n . Hence the Proof.


173

Note (1) : The above theorem can also be stated as the order of every element of a

finite group is finite and is less that or equal to the order of the group.

Note (2): If g is a cyclic group of order n and ‘a’ is a generator or G, the order of ka is n

d where d is the greatest common divisor of k and n.

Example 18: Let 30 is a cyclic group of order 30 and 1 is a generator of 30 . The

greatest common divisor of 18 and 30 is 6.

Hence the order of 18 = 18 (1) in 30 is 30

6= 5.

i.e., O (18) = 5.

Direct Product: Let (G, *) and (H, ) be two groups. The direct product of these two

groups is the algebraic structure (G H, o) in which the binary operation on G H is

given by

1 1 2 2 1 2 1 2 1 1 2 2(a ,b ) o (a ,b ) (a a ,b b ), for any (a ,b ),(a ,b ) G H

Theorem 11: The director product of two or more groups is again a group.

Proof: Let (G, ) and (H, ) be two group with identities e1 and e2. The direct

product of G and H is defined by G H = 11 1 2{(a ,b ) / a G and b H} .

Now define the binary operator on G H by

1 1 2 2 1 2 1 2 1 1 2 2(a ,b ) o (a ,b ) (a a ,b b ), for any (a ,b ),(a ,b ) G H

Associative: If 1 1 2 2 3 3(a ,b ),(a ,b ),(a ,b ) G H,

1 1 2 2 3 3 1 1 2 3 2 3

1 2 3 1 2 3

1 2 3 1 2 3

(a , b ) (a , b ) (a , b ) (a , b ) (a a , b b )

= [a (a a ), b (b b )]

= [(a a ) a ,(b b ) b ]

1 2 1 2 3 3

1 1 2 2 3 3

= [(a a ) ,(b b )] (a , b )

= [(a , b ) (a , b )] (a , b )


174

Identity: Let e1 be the identity of G and e2 be the identity of H and e = (e1, e2)

1 1 1 1 1 2

1 1 1 2

1 1

Now (a ,b ) e (a ,b ) (e ,e )

=(a e ,b e )

=(a ,b )

Similarly 1 1 1 1e (a ,b ) (a ,b )

Inverse:

1 1 11 1 1 1

1 1 11 1 1 1 1 1 1 1

1 11 1 1 2

1 2

Let (a ,B ) (a b )

(a ,b ) (a ,b ) (a ,b ) (a ,b )

= (a a ,b b )

=(e ,e ) e

Similarly (a – 1, b1)-1 1 1(a ,b ) e

(G H, ) is a group.

Note: If 1 2 nG G G ........ G is a direct product of n groups and (a1, a2, ……an) G,

then

1 1 1 1

1 2 n 1 2 n

m m m m1 2 n 1 2 n

1 2 n j j

(a) (a ,a ,.....a ) (a ,a ,.....a )

(b) (a ,a ,.....a ) (a ,a ,.....a )

(c) The identity of G is (e ,e ,.....e ),where e is the identity of G

(d) G is abelian iff each of the facto 1 2 n

1 2 n 1 2 n

rs G ,G ,.....G is abelian.

(e) If H ,H ,.....H are subgroups of the corresponding factors, then H H ..... H is a subgroup of G,

Morphism of Groups

Definition 15. Let (G, ) and (H, ) be any two groups. A mapping f: G H is said

to be a homomorphism if f (a b) = f(a) f(b), for any a, b G.


175

Example 19: Show that the mapping f of the permutation group Pn on to the

multiplicative group G = {1, -1} defined by

1,if a is even

f(a)-1, if a is odd

is a homomorphism.

Theorem 12: If f is a Morphism of a group G into a group G’, then

(i) Group homomorphism preserves identities

i.e., f(e) = e’, where e is the identity of G and e’ is the identity of G’.

(i) Group homomorphism preserves inverses

i.e., 1 1f(a ) [f(a)] , for all a G.

Proof: (i) Let a G and f is a homomorphism from G into G’, then f(a) G’

f(a) e f(a) f(a e)

= f(a) f(e)

e =f(e) (by Left cancellation law)

(ii) Let a G, then a—1 G and a a-1 = e = a-1 a

-1

-1

1

e = f(e) = f(a a )

= f(a) f(a )

f(a) f(a ) e

We see that f(a-1) is the inverse of f(a) in G’

1 1i.e.,[f(a)] f(a )


176

Theorem 13. Let f be a homomorphism from G into G’. Let f(G) be the

homomorphic image of G into G’, then f(G) is a subgroup of G’.

Proof: Let f(G)= {f(x)/x G}

Clearly f(G) is an non-empty subset of G’.

Now for any a’, b’ f(G), we have f(a) = a’, f(b) = b’, for a, b G.

1 1

-1

-1

a (b ) f(a) [f(b)]

= f(a) f(b )

= f(a b )

But 1 1 1a b G f(a b ) f(G) a (b ) f(G)

i.e., For a’, b’ f(G) a’ (b’)-1 f(G)

f(G) is a subgroup of G’.

Note: Every homomorphic image of an abelian group is abelian.

Theorem 14: Let f : G G’ be a group homomorphism and H is a subgroup of G’,

then f-1 (H) is a subgroup of G.

Proof: Let f-1 (H) = 1{x f (y) G f(x) y H}

Since H is a group of G’, the set f-1 (H) will be a non-empty subset of G.

Now consider x1 = f-1 (y1), x2 = f-1(y2) H with f(x1) = y1 and f(x2) = y2.

11 2 1 2

11 2

11 2

11 2

Let x ,x f (h) f(x ),f(x ) H ( H is a subgroup)

f(x ) [f * (x )] H

f(x f(x ) H

f(x x )H

1 11 2

1 1 11 2 1 2

x x f (H)

i.e., x ,x f (H) x x f (H)

f-1 (H) is a subgroup of G.


177

Kernel of a Homomorphism

Definition 16: Let f: G G’ be a group of homomorphism.

The set of elements of G which are mapped into e’, the identity of G’ is called

the Kernel of f and is denoted by ker (f)

i.e.,ker(f) {x G / f(x) e , the identity of G'}

Theorem 15: The Kernel of a homomorphism f from a group G to G’ is a subgroup of

G.

Proof: ker (f) = {x G/f(x) = e’, where e’ is the identity of G’}

Since f(e) = e’ is true always, atleast e ker(f)

i.e., ker (f) is a non-empty subset of G.

Let a, b ker (f), with f(a) = e’ and f(b) = e’

-1 1

1

1

f(a b ) f(a) f(b ) ( f is a homomorphism)

= f(a) (f(b))

= e e

= e

a b ker(f)

i.e., -1a, b ker(f) a b ker(f)

ker(f) is a subgroup of G.

Definition 18. A mapping f from a group G to a group G’ is said to be an

Isomorphism if

(i) f is a homomorphism. i.e., f(a.b) = f(a). f(b), for any a, b G

(ii) f is one-one. (Injective) i.e., distinct elements of G have different f-images in G'.

(iii) f is onto (Surjective) i.e., every element of G should be the f-image of some element in G. (OR)

A bijective homomorphism is said to be an isomorphism.


178

Note: There could be several different isomorphisms between the same pair of

groups.

Theorem 16. Every cyclic group of order n is isomorphic to the additive group of

residue classes modulo n. (OR) If G is cyclic and O(G) = m. then G is isomorphic to (

n, +n).

Proof: Let G = (a) and O(G) = n. Define

rn n: G by [a ] [r] , for any a G, r I

Since G = (a) is finite, we can use the fact that the elements of (a) are the first n

non-negative multiples of a. From this observation, we see that, is surjection. A

surjection between finite sets and same cardinally must be a bijection.

r s r sn

n

r sn

Now (a a ) (a ) [r s]

= [r]+ [s]

= (a ) (a )

is an isomorphism from G to n.

Theorem 17: If G is an infinite cyclic group, then it is isomorphic to the additive

group of integers.

Proofs: Let rG {a / r Z} and {0, 1, 2, 3,....}

Now define : G by (am) = m, m

The mapping is one-one, because (am) = n m n(a ) m n a a

Obviously is onto. Hence is a bijection.

m n m n

m n

Also (a a ) (a ) m n

= (a ) (a )

is an isomorphism,


179

Theorem 18: (Cayley’s (or) Cayley’s representation theorem)

Every finite group of order ‘n’ is isomorphic to a permutation group of degree

n. (or) Every finite group G is isomorphic to a permutation group.

Proof: Let (G, ) be a group of order n. We know that every row and column of the

composition table of G represents a permutation of the elements of G.

Corresponding to an element a G, we denote by pa the permutation given by the

column under ‘a’ in the composition table.

Thus Pa(c) = c a, for any c G.

For every column, we can define permutation of the elements of G and P be

the set of such permutations. Since G has n elements and we have a permutation

associated with each column, the set P also has n elements.

We now prove that (P, ) is a group, where denotes the right composition of

the permutations of P.

For, since e G, pe P and pe pa = pa pe = pa for any a G.

ep is the identity in (P, ),

Also for any a G a-1 G pa, pa – 1 P

pa – 1 pa = pe

For any pa P Pa – 1 P

Also for a, b G a b G

a b a b a b p ,p P p p p P, because for any

a b

a b

c G,P (c) c a, so that (p - a p )(c) (c a) b

c (a b) P (c). Hence (P, )is a group.

Now consider a mapping f: G P defined by

af(a) p ,for any a G.


180

Since every row (column) in the composition table of G is a permutation of

the elements of G, f is bijective.

Also we have a b a b xp p p . But p f(x) for any x G.

Hence f(a b) = f(a) f(b), for any a, b G

f is a bijective homomorphism. f is an isomorphism.

Every finite group of order n is isomorphic to a permutation group of

degree n.

Definition 18: The permutation group to which G is isomorphic is called a regular

permutation group.

Note: Cayley’s theorem is true even if the group G is not finite. But if G is not finite,

then the word permutation should be omitted from the statement of the theorem. In

that case we should state the theorem as: ‚Every group is isomorphic to a group of

one-one, onto function‛.

COSETS

Definition 19: Let (H, ) be a subgroup of (G, ).

(i) For any a G, the set a H defined by a H = {a h / h H} is called the Left

coset for H in G determined by the element a G.

(ii) For any a G, the set H a defined by H a = {h a / h H} is called the Right

coset of H in G determined by the element a G.

The element a G is called the representative element of the left coset a H and the

right coset H a.

Note:

(i) The right or left coset of H in G is not empty.

(ii) Since e H, e H = H = H e, H itself is a right as well as left coset.

(iii) H a, a H are also subsets of G.

(iv) If G is abelian, then a H = H a.

(v) If all the subsequent discussions in this chapter, aH means a H and Ha

means H a.


181

Example 20: Let G = {1, a, a2, a3} (a4 = 1) be a group and H = {1, a2} is a subgroup of G

under multiplication. Find all the cosets of H.

Solution: Let us find the right cosets of H in G.

2

3

2 2 4 2

3 3 5 3

2 2 3 3

H1 {1,a } H

Ha {1,a },

Ha {a ,a } {a ,1} H

and Ha {a ,a } {a ,a} Ha

H 1 H Ha 1,a and Ha = Ha {a,a }

Are two distinct right cosets of H in G. Similarly we can find the left cosets of H in G.

Example 21:Find the left cosets of {[0],[2]} in the group 4 under +4 (addition mod 4).

The left cosets of H are

[0] H {[0],[2]} H;

[1] H {[1],[3]};

[2] H {[2],[4]} {[2],[0]} {[0],[2]} H

and [3] H {[3],[5]} {[3],[1]} {[1],[3]} [1] H

4[0] H [2] H H and [1] + H=[3] + H are the two distinct left cosets of H in

Note:

(i) The union of all distinct left (right) cosets of H in G is equal to G.

(ii) If H is any subgroup of G and a H, then Ha = H = aH.

i.e., Right and left cosets corresponding to any element a H are H.

Theorem 19: If a Hb then Ha = Hb and if a bH, then aH = bH .

Proof:

-1 1

-1 1

1

1 1

(i) Let a Hb a b Hb b

a b He a b H

H (a b ) H( a H Ha H)

H(a b ) b Hb Ha (b b) Hb

Ha e = Hb Ha = Hb


182

1 1

-1 1

-1 1

-1

(ii) Also Leta bH b a b bH

b a eH b a H

(b a)H H b (b a)H bH

(b b ) aH bH e aH bH

aH = bH

Theorem 20: Any two right (or left) cosets of H in G are either disjoint or identical.

Proof: Let Ha and Hb be two right cosets of a subgroup H of a group G. For the

arbitrary elements a, b G, we have to prove that either.

Ha Hb or Ha = Hb

Suppose that Ha Hb 0 ,

then there exists an x Ha Hb x Ha and x Hb

1 2 1 2h ,h H such that x = h a and x = h b

1 1

1 2 1 1 1 2

-1 11 2 3 3 1 2

h a h b h h a h h b

a=h h b a h b (say h h h )

Since H is a subgroup, 11 2 1 2h ,h H h h H

3 3

3 3

h H Hh H ( a H aH=H)

Ha H(h b) Ha Hh b Ha Hb.

i.e., either Ha Hb 0 or Ha = Hb.

Definition 20. Let (H, ) be a subgroup of the group (G, ). We define the

congruency relation on G called a left coset relation mod H denoted by + such that a,

b G, a is congruent to be mod H written as a + b(mod H) iff b-1 a H.

Theorem 21: The left coset relation mod H on (G, ) defined by, for a, b G, a +

b(mod H) iff b-1 a H is an equivalence relation.


183

Proof: Let (H, ) be a subgroup of the group (G, ).

(i) Let a G

-1 a + a(mod H) a a e H

The relation is reflexive.

(ii) Let a + b (mod H) b-1 a H

1 1

1

(b a) H

a b H

b a(modh)

i.e., a + b (mod H) b a(modH)

The relation is symmetric.

(iii) Let a + b (mod H) and b + c (mod H)

1 -1b a H and c b H

1 1

1 1

1 1

1 1

(c b) (b a) H

c [(b (b a)] H

c [(b b ) a] H

c (e a) H c a H

a + c (mod H)

The relation is transitive.

Since the left coset relation + is reflexive, symmetric and transitive, it is an

equivalence relation.

Remark 1: The left coset equivalence relation partitions G into equivalence classes.

Let [a] be the equivalence class corresponding to any a G.


184

-1

Then [a] = {x G/x + a(mod H)}

= {x G/a x H}

= {a h/h H}

Remark 2: The left and right cosets of H in G are also called residue classes modulo

the subgroup H.

Theorem 22. Let (H, ) be a subgroup of (G, ). The set of left cosets of H in G forms

a partition of G.

Proof: First let us prove that every element of G appears in atleast one left coset.

Let aH = {a h/h H} is a left coset of H in G, for a G.

Now e H a e H a aH

Every element of G appears in atleast one left coset.

Secondly, the left cosets of H in G are either identical or disjoint.

We see that each element of G appears in exactly one and only one left cost

of H in G. Since the union of all distinct left cosets of H in G equals G, the set of left

cosets form a partition of G.

Note: Ever element of G belongs to one and only one right (or left) coset of H in G.

Theorem 23: If (H, ) is a subgroup of group (G, ) and Ha is any right coset of H in

G, then there exists a one-to-one correspondence (bijective mapping) between the

elements of H and Ha (or) O(H) = O(Ha).

Proof: Define the map f: H Ha by

f(h) = h a, for any h H (a G is arbitrary)

(i) For h1, h2 H, f(h1) = f(h2) h1 a = h2 a

1 2h h (by RCL)

The map f is injective


185

(ii) Any h a Ha h H such that f(h) = h a

f(H) = Ha

i.e., by definition itself, f is surjective.

The map f is bijective.

There is a one-to-one correspondence between H and Ha.

Note: There is a one-to-one correspondence between H and aH.

i.e.,0(H) O(aH)

Theorem 24: (Lagrange’s theorem)

The order of each subgroup of a finite group is a divisor of the order of the

group.

Proof: Let (G, ) be a finite group and O (G) = n. Let (H, ) be a subgroup of (G, )

and O(H) = m.

Suppose that h1, h2, h3,… hm are the m members of H. For a G, the right

coset Ha of H in G is defined by

1 2 3 mHa {h a,h a,h a,....h a}

Since there should be a one-to-one correspondence between H and Ha, the members

of Ha are distinct.

Hence each right coset of H in G has m distinct members.

We know that any right cosets of H in G are either disjoint or identical. Since

G is a finite group, the number of distinct right cosets of H in G will be finite, say k.

The union of these k distinct right cosets of H in G is equal to G.

Hence, if Ha1, Ha2, Ha3,…….Hak are the k distinct right cosets of H in G, then

1 2 3 k

1 2 3 k

G Ha Ha Ha ...... Ha

O(G) O(Ha ) O(Ha ) O(Ha ) ...... O(Ha )


186

n = m + m + m +.......+ m ( O(H) = O(Ha) = m)

(k times)

nn = km k m is a divisor of n.

mO(H) is a divisor of O(G).

O(H) divides O(G). Hence the proof.

Note:

1. The converse of the Lagrange’s theorem is not true. That is, if m is a divisor of

n, then it is not necessary that G must have a subgroup of order m. For

example, the alternating group A4 of degree 4 is of order 12. But there is no

subgroup of A4 of order 6, though 6 is a divisor of 12.

2. It is evident from Lagrange’s theorem that if G is a finite group of order n and

if m is not a divisor of n, then there can be no subgroup of G of order m.

Definition 21: Let c and D be left cosets of a subgroup (H, ) of a group (G, ). Then

C D = (c d)H, where c and d are representatives of C and D respectively. The

operation is called the operation induced on left cosets by .

Note: If (H, ) be a subgroup of an abelian group (G, ), then the operation induced

on cosets of H by the operation of G is well defined.

Definition 22: Let (H, ) be a subgroup of a group (G, ). If the operation induced

on left cosets of H by the operation of G is well defined, then the set of left cosets of

H form a group, which is called the factor group of G modulo H and is denoted by

G/H under the operation .

Note: If (G, ) is abelian, then every subgroup of G yields a factor group.

Normal Subgroups

Definition 23: A subgroup (H, ) of a group (G, ) is said to be a normal subgroup of

G, if for every x G and for every h H, xhx-1 H (or) xHx-1 H.


187

Theorem 25: A subgroup H of a group g is normal iff xHx-1 H.

Proof: Let xHx-1 = H xHx-1 H, for all x G

H is a normal subgroup of G.

Conversely, assume that H is a normal subgroup of G, then xHx-1 H, for all

x G (1)

-1

-1 1 1

-1 1 1 1

1 1 1 1

1

Now x G x G, then

x H(x ) H, for all x G

x Hx H x(x Hx)x xHx

x xHxx xHx eHe xHx

H xHx , for all x G (2)

From (1) and (2), we see that

1xHx H, for all x G.

Theorem 26: A subgroup H of G is normal iff each left coset of H in G is equal to the

right coset of H in G.

Proof: Let H be a normal subgroup of G, then

1

1 1

xH x H, for all x G

(xHx )x Hx xHx x Hx

xHe Hx xH Hx, for all x G

each left coset xH is equal to the right coset Hx.

Conversely, assume that each left coset of H in G is a right coset of H in G.

-1 1 1 1

-1

i.e., xH = Hx, for all x G

(xH)x (Hx)x xHx Hxx He

xHx H H is a normal subgroup of G.


188

Theorem 27: The intersection of any two normal subgroups of a group is a normal

subgroup.

Proof: Let H and K be any two normal subgroups of a group G. Obviously H and K

are subgroups of G and H K is a subgroup of G.

Now we shall prove H K is normal.

For, Let x G, h H K x G, h H and x G, h K

1 -1

-1

xhx H and xhx K ( H and K are normal subgroups)

xhx H K

H K is a normal subgroup of G.

Theorem 28. Let G and G’ be any two groups with identities e and e’ respectively. If

f: G G’ be a homomorphism, then ker(f) is a normal subgroup.

Proof. Let e and e’ be the identities of the groups G and G’ respectively.

Let K = ker(f) = {x G/f(x) = e’}.

We know that ker(f) is a subgroup of G. We have to prove ker (f) is normal.

For, x G, h K,

1 1

-1

-1

-1

f(xhx ) f(xh)f(x )

= f(x) f(h) f(x ) ( f is a homomorphism)

= f(x)e f(x )

= f(xx )

=f(e) = e

1 1

1

f(xhx ) e xhx K

x G,h K, xhx K

K = ker (f) is a normal subgroup of G.


189

Theorem 29: Every subgroup of an abelian group is normal.

Proof: Let x be any element of G and h H

1 1

-1 -1 1

-1

xHx x(Hx )

= x(x H) ( G is abelian x Hx xHx )

=(xx )H

= eH = H

i.e., for x G, h H we have xHx-1 = H.

H is a normal subgroup of G.

Note: Since every cyclic group is abelian, every subgroup of a cyclic group is normal.

Theorem 30: If G is a group of Prime order, then G has no proper subgroups.

Proof: Let O(G) = P, where P is a Prime number.

Since P is a prime, the only divisors of P are 1 and P.

By Lagrange’s theorem, we can have subgroups of order 1 and P.

i.e., H = {G} (or) H = {e}. But bothe are not proper subgroups of G.

G has no proper subgroups.

Definition 24: The number of distinct left (or right) cosets of H in G is called the

index of H in G and is denoted by [G.H] = G

O(G)I (H)

O(H).

Example 22: Let G = {i, -1, I, -i} be a group and H = {-1, 1} be a subgroup of G.

O(G) = 4 and O(H) = 2.

The number of distinct right cosets of H in g is


190

G

O(G)I (H)

O(H)

4 =

2 = 2,

Which are H1 = {-1, 1} and Hi = {-i, i}.

Definition 25: Let H be a normal subgroup of a group G.

Then the map f: G G/H such that f(x) = Hx for all x G, is called a Natural

homomorphism of the group G on to the factor (or) quotient group G/H.

Theorem 31: (Fundamental theorem on homomorphism of groups)

Every homomorphic image of a group G is isomorphic to some quotient

group of G.

Proof: Let G’ be the homomorphic image of a group G and f be the corresponding

homomorphism. Then f is a homomorphism of G onto G’. Let K be the Kernal of

this homomorphism. Then K is a normal subgroup of G.

Now we shall prove that G/K G’

If a G, then Ka G/K and f(a) G’.

Consider the mapping : G/K G’ such that (Ka) = f(c), for all a G.

First we shall show that the mapping is well defined i.e., if a, b G and Ka =

Kb, then (Ka) = (Kb).

We have Ka = Kb a b-1 K

1

1

1

1

f(a b ) e ,

f(a) f(b ) e ,

f(a) [f(b)] e ,

f(a) [f(b)] f(b) e f(b),

f(a) f(b)

(Ka) (Kb)

is well defined


191

1 1

-1

-1

1

1

(Ka) (Kb) f(a) f(b)

f(a) f(b ) f(b) f(b )

= f(b b )

= f(e)

f(a) f(b ) e

f(a b ) e

a b K Ka Kb

is one-one.

Also, let y be any element of G’. Then y = f (a) for some a G, because f is

onto G’. Now Ka G/K and we have (Ka) = f(a) = y

is onto.

Finally, (Ka Kb) = (Ka b)

f(a b)

f(a) f(b)

(Ka) (Kb)

Since is a bijective homomorphism, we see that is an isomorphism between G/K

and G’

G/ K G

Rings and Ideals

Rings

Definition 26: Let R be a non-empty set with two binary operation addition and

multiplication denoted by + and respectively. Then R is a ring if the following

axioms are satisfied.

1. (a + b) + c = a + (b + c) for all a, b, c R.

2. There exists an element 0 R called zero element such that a + 0 = 0 + a = a

for every a R.

3. For each a R, there exists an element a R, called the negative of a such

that a + (-a) = (-a) + a = 0.


192

4. a + b = b + a for all a, b R

5. (a . b).c = a. (b . c) for all a, b, c R

6. For a, b, c R

(a) a (b + c) = ab + ac

(b) (b + c)a = ba + ca

i.e., If R is an abelian group under addition with the properties and (6) then R is a

ring.

Definition 27: R is a commutative ring if ab = ba for all a, b R.

Definition 28: A non-zero element 1 R is called an identity element (multiplicative

identity) if a.1 = 1.a = a for all a R.

Definition 29: An element a R is a unit if a has a multiplicative inverse.

i.e., there exist a-1 R such that aa-1 = a-1a = 1

Example 23: consider the set of all integer with the usual addition and

multiplication then.

1. is a commutative ring.

2. The only units in are 1 and -1.

Example 24: Let Mn denotes ring of square matrices of order n with the usual

operations of matrix addition and matrix multiplication.

Then it is easy to see that

1. Mn is not a commutative ring

2. Mn has a multiplicative identity, the unit matrix I of order n.

3. The units in Mn are the non-singular matrices of order n.

Example 25: Find the roots of 26f(x) x 2x 3 0 over

The roots are 1 and 3.


193

Example 26: Find the roots of f(x) = x2 + 5x + 4 = 0 over 10

The roots are 1, 4, 6, and 9.

This shows that a polynomial of degree n can have more than n roots over an

arbitrary ring (This cannot happen if the ring is field).

Definition 30: A non-empty subset S of R is a subring of R if S itself forms a ring

under the binary operations of R.

Clearly S is a subring of R if and only if x, y S x – y S and xy S.

For x, y S x – y S implies that 0 S, -a S a S.

Ideals

Definition 31: A subset I of R is an ideal in R if

0 I,

I is closed under subtraction i.e., x – y I for all x, y I.

I is closed under multiplies from R i.e., mx I for m R, x I.

Note:

(a) I is called a left ideal if only mx I with respect to (iii) above.

(b) I is called a right ideal if only xm I.

(c) The term ideal will mean two sided ideal.

Facts:

1. {0} is an ideal in any ring R.

2. If I and J are ideals in R, then I J is an ideal in R.

3. Let I be an ideal in a ring R with an identity element 1.

4. An ideal I in a ring R is a normal subgroup of the additive group of R. hence

the collection of cosets {x + 1: x R} forms a partition of R.


194

Integral Domains

Definition 32: A non-zero element x R is a zero divisor if there exist a non-zero

element y such that xy = 0.

Definition 33: A commutative ring D with the identity element 1 is an integral

domain if D has no zero divisors.

Example 27: The ring 95 of the integer modulo 95 is not an integral domain.

Note: In fact any k with K composite has divisors of zero. Such K -s are not

integral domain.

Note: If p is a prime then p is an integral domain.

Fact: If D is an integral domain then xy = xz with x 0 implies y = z.

Proof:

xy xz xy xz 0

x(y z) 0

Since x 0 and D has no zero divisors y – z = 0

y = z

i.e., multiplication in D obeys the cancellation law.

Definition 34: Let x be any element in R.

Then the set (x) = {mx: m R} to an ideal. Such an ideal is called the principal

ideal generated by x.

Definition 35: A ring R is a principal ideal domain (PID) if R is an integral domain

and if every ideal in R is principal.


195

Definition 36: An element y R is called associate of x R if y=ux for some unit x R.

Definition 37: A non unit p D in irreducible if p = xy x or y is a unit.

Definition 38: An integral domain D is a unique factorization domain (UFD) if every

unit x D can be written uniquely as a product of irreducible elements.

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ppt_IT_Discretemaths.pdf

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