Discrete Mathematics For Sppu 2 DR H R Bhapkar
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Discrete Mathematics for SPPU 2 - Dr. H. R. Bhapkar
Engineering (National Institute of Technology Raipur)
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SUBJECT CODE : 210241
Strictly as per Revised Syllabus of
Savitribai Phule Pune University
Choice Based Credit System (CBCS)
S.E. (Computer) Semester - I
Discrete Mathematics
(For IN SEM Exam - 30 Marks)
Dr. H. R. Bhapkar
M.Sc. SET, Ph.D. (Mathematics)
MIT Art, Design and Technology University’s
MIT School of Engineering, Lonikalbhor, Pune
Email : drhrbhapkar@gmail.com
Mobile : 9011227141
Dr. Rajesh Nandkumar Phursule
Ph.D. (Computer Science and Engineering)
Associate Professor,
Pimpri Chinchwad College of Engineering
Nigdi, Pune.
® ®
TECHNICAL
PUBLICATIONS
SINCE 1993 An Up-Thrust for Knowledge
(i)
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Discrete Mathematics
(For IN SEM Exam - 30 Marks)
Subject Code : 210241
S.E. (Computer) Semester - I
First Edition : August 2020
ã Copyright with Dr. H. R. Bhapkar
All publishing rights (printed and ebook version) reserved with Technical Publications. No part of this book
should be reproduced in any form, Electronic, Mechanical, Photocopy or any information storage and
retrieval system without prior permission in writing, from Technical Publications, Pune.
Published by :
® ®
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TECHNICAL Pune - 411030, M.S. INDIA, Ph.: +91-020-24495496/97
PUBLICATIONS
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ISBN 978-93-332-0276-3
9 789333 202763 SPPU 19
9789333202763 [1] (ii)
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Preface
The importance of Discrete Mathematics is well known in various engineering fields.
Overwhelming response to our books on various subjects inspired us to write this book.
The book is structured to cover the key aspects of the subject Discrete Mathematics.
The book uses plain, lucid language to explain fundamentals of this subject. The
book provides logical method of explaining various complicated concepts and stepwise
methods to explain the important topics. Each chapter is well supported with necessary
illustrations, practical examples and solved problems. All the chapters in the book are
arranged in a proper sequence that permits each topic to build upon earlier studies. All
care has been taken to make students comfortable in understanding the basic concepts
of the subject.
The book not only covers the entire scope of the subject but explains the philosophy
of the subject. This makes the understanding of this subject more clear and makes it
more interesting. The book will be very useful not only to the students but also to the
subject teachers. The students have to omit nothing and possibly have to cover nothing
more.
We wish to express our profound thanks to all those who helped in making this
book a reality. Much needed moral support and encouragement is provided on
numerous occasions by our whole family. We wish to thank the Publisher and the
entire team of Technical Publications who have taken immense pain to get this book
in time with quality printing.
Any suggestion for the improvement of the book will be acknowledged and well
appreciated.
Authors
Dr. H. R. Bhapkar
Dr. Rajesh N. Phursule
Dedicated to the Readers of the Book
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Syllabus
Discrete Mathematics - (210241)
Credit Scheme Examination Scheme and Marks
03 Mid_Semester (TH) : 30 Marks
Unit I Set Theory and Logic
Introduction and significance of Discrete Mathematics, Sets – Naïve Set Theory
(Cantorian Set Theory), Axiomatic Set Theory, Set Operations, Cardinality of set,
Principle of incl usion and exclusion, Types of Sets - Bounded and Unbounded Sets,
Diagonalization Argument, Countable and Uncountable Sets, Finite and Infinite Sets,
Countably Infinite and Uncountably Infinite Sets, Power set, Propositional Logic -
logic, Propositional Equivalences, Application of Propositional Logic - Translating
English Sentences, Proof by Mathematical Induction and Strong Mathematical
Induction. (Chapters - 1, 2, 3)
Unit II Relations and Functions
Relations and their Properties, n-ary relations and their applications, Representing
relations, Closures of relations, Equivalence relations, Partial orderings, Partitions,
Hasse diagram, Lattices, Chains and Anti-Chains, Transitive closure and Warshall‘s
algorithm. Functions - Surjective, Injective and Bijective functions, Identity function,
Partial function, Invertible function, Constant function, Inverse functions and
Compositions of functions, The Pigeonhole Principle. (Chapters - 4, 5)
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Table of Contents
Unit - I
Chapter - 1 Theory of Sets (1 - 1) to (1 - 48)
1.1 Introduction ......................................................................................................1 - 2
1.2 Sets ...................................................................................................................1 - 2
1.3 Methods of Describing Sets ..............................................................................1 - 2
1.3.1 Roster Method (Listing Method) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 2
1.3.2 Statement Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 2
1.3.3 Set Builder Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 3
1.4 Some Special Sets .............................................................................................1 - 3
1.5 Subsets..............................................................................................................1 - 3
1.5.1 Proper Subset . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 3
1.5.2 Improper Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 4
1.5.3 Equal sets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 4
1.6 Types of Sets .....................................................................................................1 - 4
1.7 Venn Diagrams..................................................................................................1 - 6
1.8 Operations on Sets............................................................................................1 - 6
1.8.1 Union of Two Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 6
1.8.2 Intersection of Two Sets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 7
1.8.3 Disjoint Sets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 8
1.8.4 Complement of a Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 8
1.8.5 Difference of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 8
1.8.6 Symmetric Difference of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 10
1.9 Algebra of Set Operations...............................................................................1 - 11
1.10 Principle of Duality........................................................................................1 - 12
1.11 Cardinality of Sets ........................................................................................ 1 - 23
1.12 The Principle of Inclusion and Exclusion for Sets..........................................1 - 24
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1.13 Multiset.........................................................................................................1 - 45
1.13.1 Multiplicity of an Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 46
1.13.2 Equality of Multisets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 46
1.13.3 Union and Intersection of Multisets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 46
1.13.4 Difference of Multisets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 47
1.13.5 Sum of Multisets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 47
Chapter - 2 Propositional Calculus (2 - 1) to (2 - 38)
2.1 Introduction ......................................................................................................2 - 2
2.2 Statements or Propositions ..............................................................................2 - 2
2.3 Laws of Formal Logic.........................................................................................2 - 3
2.3.1 Law of Contradiction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 3
2.3.2 Law of Excluded Middle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 3
2.4 Connectives and Compound Statements .........................................................2 - 3
2.4.1 Compound Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 3
2.4.2 Truth Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 4
2.4.3 Negation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 4
2.4.4 Conjunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 4
2.4.5 Disjunction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 5
2.4.6 Conditional Statement (If ..... then ....) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 6
2.4.7 Biconditional (If and only if) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 7
2.4.8 Special Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 8
2.5 Propositional or Statement Formula ..............................................................2 - 12
2.6 Tautology ........................................................................................................2 - 12
2.7 Contradiction ..................................................................................................2 - 12
2.8 Contingency ....................................................................................................2 - 12
2.9 Precedence Rule .............................................................................................2 - 16
2.10 Logical Equivalence.......................................................................................2 - 16
2.11 Logical Identities ...........................................................................................2 - 18
2.12 The Duality Principle .....................................................................................2 - 20
2.13 Logical Implication ........................................................................................2 - 20
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2.14 Important Connectives .................................................................................2 - 20
2.15 Normal Forms ...............................................................................................2 - 21
2.15.1 Disjunctive Normal Form (dnf) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 22
2.15.2 Conjunctive Normal Form (cnf) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 22
2.15.3 Principal Normal Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 22
2.16 Methods of Proof..........................................................................................2 - 27
2.16.1 Law of Detachment (or Modus Ponens) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 28
2.16.2 Modus Tollen (Law of Contrapositive) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 29
2.16.3 Disjunctive Syllogism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 29
2.16.4 Hypothetical Syllogism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 29
2.17 Quantifiers ....................................................................................................2 - 32
2.17.1 Universal Quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 33
2.17.2 Existential Quantifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 33
2.17.3 Negation of Quantified Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 33
Chapter - 3 Mathematical Induction (3 - 1) to (3 - 18)
3.1 Introduction ..................................................................................................... 3 - 2
3.2 First Principle of Mathematical Induction Statement.......................................3 - 2
3.3 Second Principle of Mathematical Induction Statement
(Strong Mathematical Induction) ......................................................................3 - 2
Unit - II
Chapter - 4 Relations (4 - 1) to (4 - 64)
4.1 Introduction ......................................................................................................4 - 2
4.2 Cartesian Product ............................................................................................4 - 2
4.3 Relation ............................................................................................................4 - 3
4.4 Matrix Representation of a Relation ................................................................4 - 4
4.4.1 Relation Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 5
4.4.2 Properties of Relation Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 6
4.5 Diagraphs .........................................................................................................4 - 6
4.6 Special Types of Relations ................................................................................4 - 9
4.6.1 Inverse Relation (OR Converse Relation) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 9
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4.6.2 Complement of a Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 10
4.6.3 Composite Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 11
4.7 Types of Relations on Set................................................................................4 - 12
4.7.1 Identity Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 12
4.7.2 Reflexive Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 13
4.7.3 Symmetric Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 14
4.7.4 Compatible Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 14
4.7.4.1 Asymmetric Relation . . . . . . . . . . . . . . . . . . . . . . . . 4 - 15
4.7.5 Antisymmetric Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 15
4.7.6 Transitive Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 15
4.7.7 Equivalence Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 16
4.7.8 Properties of Equivalence Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 16
4.7.9 Equivalence Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 17
4.8 Partitions of a Set............................................................................................4 - 36
4.8.1 Relation Induced by a Partition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 37
4.8.2 Refinement of Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 39
4.8.3 Product and Sum of Partitions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 40
4.8.4 Quotient Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 40
4.9 Closure of a Relation.......................................................................................4 - 40
4.9.1 Reflexive Closure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 40
4.9.2 Symmetric Closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 41
4.9.3 Transitive Closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 42
4.9.4 Warshall's Algorithm to Find Transitive Closure . . . . . . . . . . . . . . . . . . . . . . . . 4 - 42
4.10 Partially Ordered Set.....................................................................................4 - 52
4.10.1 Hasse Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 53
4.10.2 Chains and Antichains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 54
4.10.3 Elements of Poset . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 54
4.10.4 Types of Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 56
4.10.5 Properties of a Lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 58
4.10.6 Types of Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 58
4.11 Principle of Duality........................................................................................4 - 59
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Chapter - 5 Functions (5 - 1) to (5 - 28)
5.1 Introduction ......................................................................................................5 - 2
5.2 Function ............................................................................................................5 - 2
5.2.1 Important Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 2
5.2.2 Partial Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 3
5.2.3 Equality of Two Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 4
5.2.4 Identity Function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 4
5.2.5 Constant Function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 4
5.2.6 Composite Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 4
5.3 Special Types of Functions................................................................................5 - 6
5.4 Infinite Sets and Countability..........................................................................5 - 17
5.4.1 Infinite Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 17
5.4.2 Countable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 18
5.5 Pigeon Hole Principle......................................................................................5 - 20
5.6 Discrete Numeric Functions............................................................................5 - 22
5.6.1 Basic Operations on Numeric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 23
5.6.2 Finite Differences of a Numeric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 - 24
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Unit - I
1 Theory of Sets
Syllabus
Introduction and significance of Discrete Mathematics, Sets – Naïve Set Theory (Cantorian Set
Theory), Axiomatic Set Theory, Set Operatio ns, Cardinality of set, Principle of inclusion and
exclusion, Types of Sets - Bounded and Unbounded Sets, Diagonalization Argument, Countable
and Uncountable Sets, Finite and Infinite Sets, Countably Infinite and Uncountably Infinite Sets,
Power set.
Contents
1.1 Introduction
1.2 Sets
1.3 Methods of Describing Sets
1.4 Some Special Sets
1.5 Subsets
1.6 Types of Sets
1.7 Venn Diagrams
1.8 Operations on Sets
1.9 Algebra of Set Operations
1.10 Principle of Duality . . . . . . . . . . . . . . . . . . Dec.-06, 08, 11, 12, 13, 14, 15, 17, 19
. . . . . . . . . . . . . . . . . . May-05, 06, 08, 14 · · · · · · · · · Marks 6
1.11 Cardinality of Sets
1.12 The Principle of Inclusion and Exclusion for Sets
. . . . . . . . . . . . . . . . . . Dec.-04, 05, 07, 08, 10, 13, 14, 15, 19
. . . . . . . . . . . . . . . . . . May-05, 06, 07, 08,
. . . . . . . . . . . . . . . . . . 14, 15, 17, 19, · · · · · · · · · · · · Marks 13
1.13 Multiset . . . . . . . . . . . . . . . . . . Dec.-13, May-19 · · · · · · · · · · · Marks 4
(1 - 1)
TM
TECHNICAL PUBLICATIONS - An up thrust for knowledge
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Discrete Mathematics 1-2 Theory of Sets
1.1 Introduction
The notion of a set is a fundamental concept to all of Mathematics and every branch
of mathematics can be considered as a study of sets of objects of one kind or another. A
great mathematician Cantor was the founder of the theory of sets. Let us now consider
the idea of a set.
1.2 Sets
A set is a collection of well defined objects. An object in the set is called a member or
element of the set. The objects themselves can be almost anything. e.g. Books, numbers,
cities, countries, animals, etc. In the above definition, the words set and collection for
almost all practical purposes are synonymous. Elements of a set are usually denoted by
lower case letters i.e. a, b, c, ... while sets are denoted by capital letters i.e.
A, B, C, ...
The symbol ' Î' indicates the membership in a set.
If "x is an element of a set A" then we write x Î A.
If "x is not an element of a set A" then we write as x Ï A.
Examples :
1) The set of letters forming the word "MATHEMATICS"
2) The set of students in a class SE Computer Engineering
3) s = {1, 2, 3, 4, 5, 6}
4) The set of professors in SPPU university.
5) The set of all telephone numbers in the directory.
1.3 Methods of Describing Sets
The following are the most useful and common methods of describing sets.
1.3.1 Roster Method (Listing Method)
A set may be described by listing all the members of the set between a pair of braces.
In this method the order in which the elements are listed is immaterial and it is used
for small sets. e.g. A = {1, 2, 3, 4, 5}, B = {a, b, c, d, e, f, g}
1.3.2 Statement Form
In this form, set is formed especially where the elements have a common
characteristic.
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e.g. 1) The set of all even integers
2) The set of all Prime Ministers of India
1.3.3 Set Builder Notation
It is not always possible to describe a set by the listing method or statement form.
A more compact or concise way of describing the set is to specify the common
property of all elements of the set
e.g. A = { x|x is real and x 2 > 100 }
B = { x|x is real and x 20 – x 10 + 1 = 0 }
1.4 Some Special Sets
The following sets are important and occur frequently in our discussion.
1) Set of natural numbers = N = {1, 2, 3, ....}
2) Set of all integers = Z = {...–3, –2, –1, 0, 1, 2, 3 ....}
3) Set of all positive integers = Z + = {0, 1, 2, 3, ...}
p
4) Set of all rational numbers = Q = ìí | p, q Î R, (p, q) =1 q ¹ 0üý
î q þ
5) Set of real numbers = Â
6) Set of complex numbers = C
1.5 Subsets
A set A is said to be a subset of the set B if every element of A is also an element of
B.
We also say that A is contained in B and denoted by A Ì B
If A is a subset of B i.e. A Ì B, then the set
B is called the superset of A.
The set with no elements is called an empty set or null set. It is denoted by f or { }.
If A Ì B then there are two possibilities.
i) A = B
ii) A ¹ B, A Ì B
1.5.1 Proper Subset
A set A is called proper subset of B iff
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i) A Ì B i.e. A is a subset of B.
ii) $ at least one element in B which is not in A.
i.e. B is not subset of A.
e.g. If A = {1, 2, 3} and B = {1, 2, 3, 4, 5}
then A is a proper subset of B.
It is denoted by A Ì B or A Ì B
1.5.2 Improper Subsets
Every set is a subset of itself and null set is a subset of every set.
Subsets A and Q are called improper subsets of A.
1.5.3 Equal sets SPPU : May-18
Two sets A and B are said to be equal sets if A Ì B and BÌ A. We write A = B
e.g. If A = { x | x 2 = 1 } and B = {1, – 1 } then A = B
1.6 Types of Sets
Depending upon some properties there are mainly following types of sets.
1) Universal Set :
A non empty set of which all the sets under consideration are subsets is called
universal set.
It is denoted by U. Universal set is not unique.
2) Singleton Set :
A set having only one element is called a singleton set. e.g. A = {5}, B = {f}
3) Finite Set :
A set is said to be finite if it has finite number of elements.
The number of elements in a set is called the cardinality of that set. It is denoted by
|A| cardinality of set may be finite or infinite.
e.g. {1, 2, 3, 4 , 5} Þ|A| = 5
4) Infinite Set :
A set is said to be infinite if it has infinite number of elements
e.g. N = set of natural numbers
|N| = ¥
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5) Power Set :
The set of all subsets of a set A is called the power set of A.
The power set of A is denoted by P(A)
Hence P(A) = {X | X Ì A}
If A has n elements then P(A) has 2 n elements.
e.g.
i) A = {a, b}, P(A) = { f, {a }, {b}, {a, b}}.
ii) P(A) = f iff A = f
i.e. empty set has only subset f, \ P( f) = {f }
iii) A = {a, b, c}, P(A) = {f, {a }, {b}, {c }, {a, b}, {b, c }, {a, c }, {a, b, c }}.
6) Power set of power set of A :
The power set of the power set of A is denoted by P(P(A)).
e.g. If A = {a, b}, p(A) = { f, {a }, {b }, {a, b }}
P(P(A)) = {f, p(A), {f}, {a}, {b}, {a, b}, {{a}, {b}}{f, {a}}
{{a}, {a, b }}, {f, { b }}, {{b }, {a, b }}
{ f {a, b}}, {f, {a}, {b}}, { f, {a}, {a, b}}
{ f, {b}, {a, b}}, {{a}, {b}, {a, b}}}
If |A| = 2 then |P(A)| = 2 2 , |P(P(A)| = 2 4 = 16
7) A set itself can be an element of some set
Hence one should understand the difference between an element of a set and subset
of a set.
e.g. If A = {x, y}, P(A) = {f, A, {x }, {y }}
Then
i) x Î A is true. But x Ì A is false as x is an element of A not subset of A
ii) {x } Ì A is true as {x} is a subset of A
iii) {x } Î A is false as {x} is not an element of A
iv) {x } Î P(A) is true.
v) f Î P(A), f Ì P(A) are true.
vi) {x } Î P(A) is true.
vii) {{x }} Î P(A) is false but {{x }} Ì P(P(A))
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viii) f Ì f, f Ì {f }, f Î {f }, {f } Ì {f } are true
ix) f Î f is false
x) {a, b} Î {a, b, c, {a, b, c}} is false
xi) {a, b, c } Î {a, b, c, {a, b, c}} is true.
xii) {a, b} Ì {a, b, c, {a, b, c}} is true
xiii) {a, f } Î {f, {a, f }}
1.7 Venn Diagrams
We often use pictures in mathematics. The relation between sets can be conveniently
illustrated by certain diagrams called Venn diagrams.
This representation first time used by the British Logician John Venn.
In Venn diagram
i) Universal set (U) is represented by a large rectangle.
ii) Subsets of U are represented by circles or some closed curve
iii) If AÌ B then the circle representing A lies inside of circle representing B.
iv) If A and B are disjoint sets then circles of A and B do not have common area.
v) If A and B are not disjoint sets then circles of A and B have some common area.
1) A Ì U U 2) A Ì B U
A
A
B
Fig. 1.7.1
1.8 Operations on Sets
To define new sets by combining the given sets, we require set operations. These
operations are analogous to the algebraic operations +, –, × of numbers.
1.8.1 Union of Two Sets SPPU : May-18
Let A and B be two sets. The union of two sets A and B is the set of all those
elements which are either in A or in B or in both sets.
If is denoted by A È B
\ A È B = {x | x Î Aor x Î B }
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By Venn diagram A È B is represented as
A B
AB
Fig. 1.8.1
Examples
1) A = {1, 2, 3, 4}, B = {x, y, z}
A È B = {1, 2, 3, 4, x, y, z}
2) A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, B = {2, 3, 5, 7, 11, 13}
A È B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13}
1.8.2 Intersection of Two Sets
The intersection of two sets A and B is the set of all elements which are in A and
also in B. It is denoted by A Ç B -
\ A Ç B = {x | x Î A and x Î B }
By Venn diagram A Ç B is represented as
A U
AB
AB:
Fig. 1.8.2
Examples :
1) A = {1, 2, 3, 4, 5}, B = {1, 4}
\ A Ç B = {1, 4} = B
2) AÇ f = f
3) A = {a, b, c}, B = {x, y}
AÇ B = f
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1.8.3 Disjoint Sets
I) Two sets A and B are said to be disjoint sets if
AÇ B = f
e.g. A = {1, 2}, B = {x, y} are disjoint sets.
1.8.4 Complement of a Set
Let A be a given set. The complement of A is denoted by A or A' and defined as
A = {x | x Ï A}.
By Venn diagram A' is represented as
A
A or A'
Fig. 1.8.3
Examples :
1) A = {Set of even numbers} and È = IR
then A or A' = set of odd numbers
2) U = {1, 2, 3, 4, 5, .... 15}, A = {1, 2, 3, 4, 5, 6, 7}
A = {8, 9, 10, 11, 12, 13, 14, 15}
Properties of complement of sets
1) U = f and f = U
2) ( A) = A
3) A È A = U, AÇ A=f
4) A È U = U, AÇ U=A
1.8.5 Difference of Sets
Let A and B be two sets. The difference A-B is the set defined as
A – B = {x | x Î A and x Î B }
It is also known as the relative complement of B in A.
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Similarly,
B – A = {x | x Î B and x Ï A}
By Venn diagram it is represented as
A U
B–A
A–B
Fig. 1.8.4
Examples :
1) A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6, 7}
\ A – B = {1, 2}, B – A = {6, 7}
2) A = Set of prime numbers
B = Set of odd integers
A – B = {2}
3) A = {a, b, f, {a, c }}
A – {a, b} = {{a, c }, f }
{a, c} – A = {c}
Properties of difference
Let A and B be two sets. Then
1) A = U–A
2) A – A = f, f– f = f
3) A – A = A, A – A = A, U – f = f,
4) A – f = A, A– f = U – A
5) A – B = A Ç B
6) A – B = B – A iff A=B
7) A – B = A iff AÇ B = f
8) A – B = f iff AÌ B
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1.8.6 Symmetric Difference of Sets
The symmetric difference of two sets A and B is denoted by A Å B and defined as
A Å B = {x | x Î A – B or x ¬ B– A }
i.e. A Å B = (A – B) È (B– A)
If is also denoted by A D B.
By Venn diagram it is represented as
U
A B
AB:
Fig. 1.8.5
Examples :
1) A = {1, 2, 3, 4, 5, 6}, B = {3, 4, 5, 6, 7, 8}
A – B = {1, 2}, B – A = {7, 8}
A Å B = {1, 2, 7, 8}
2) A = f, B = {x, f, {f}}
A Å B = {x, {f}}
3) A = {a}, B = {a, b}
P(A) Å P(B) = {{b}, {a, b}}
Properties of symmetric difference
1) AÅA = f
2) AÅf = A
3) AÅU = U – A = A
4) AÅA = U
5) A Å B = ( A È B) – ( A Ç B)
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1.9 Algebra of Set Operations
The set operations obey the same rules as those of numbers such as commutativity,
associativity and distributivity. In addition there are similar rules to logic such as
Idempotent, Absorption De Morgan's Law's and so on.
Theorem : Let A, B, C be any sets then
1) Commutativity with respect to È and Ç
a) A È B = B È A b) A Ç B = B Ç A
2) Associativity w.r.t. È and Ç
a) A È (BÈ C) = ( A È B) È C
b) A Ç (BÇ C) = ( A Ç B) Ç C
3) Distributivity
a) A È (BÇ C) = ( A È B) Ç (A È C)
b) A Ç (BÈ C) = ( A Ç B) È (A Ç C)
4) Idempotent Laws
a) A È A = A b) A Ç A = A
5) Absorption Laws
a) A È ( A Ç B) = A b) A Ç ( A È B) = A
6) De Morgan's Laws
a) (A È B) = A Ç B b) A Ç B = A È B
7) Double Complement : ( A) = A
Proof : Proofs of 1), 2), 4), 5) are easy, so reader can prove these
3) To Prove A È (BÇ C) = (A È B) Ç ( A È C)
Let x Î A È (BÇ C) Þ x Î A or x Î BÇ C
Þ x Î A or (x Î B and x Î C)
Þ (x Î A or x Î B) and (x Î A or x Î C)
Þ (x Î A È B) and ( x Î A Ç C)
Þ x Î ( A È B) Ç ( A Ç C)
Hence A È (BÇ C) Ì (A È B) Ç (A È C)
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Similarly prove that (A È B) Ç (A È C) Ì A È ( BÇ C)
Hence the proof
Similarly reader can prove
A Ç ( BÈ C) Ì (A Ç B) È ( A Ç C)
6) Prove that AÈ B = AÇ B
Let x Î AÈ B Þ x Ï AÈ B
Þ x Ï A and x Ï B
Þ x Î A and x Î B
Þ x Î AÇ B
Þ AÈ B Ì AÇ B
Similarly AÇ B = AÈ B
Hence AÈ B = AÇ B
Similarly the reader can prove
( A Ç B) = A È B
7) A = {x | x Ï A}
= {x | x Î A} = A
1.10 Principle of Duality SPPU : Dec.-06, 08, 11, 12, 13, 14, 15, 17, 19,
May-05, 06, 08, 14
The principle of duality states that any proved result involving sets and complements
and operations of union and intersection gives a corresponding dual result by replacing
È by f and È by Ç and vice versa.
e.g. The dual of A È A = U is A Ç A = f
Examples
Example 1.10.1 If U = {x Î Z ¢ | – 5 < x < 5} and A = {x Î Z ¢ | – 2 < x < 3} Where Z ¢ = Set of
integers. State the elements of the sets.
A, A Ç A, AÇ U, AÈ U, A Ç U, AÇ A
Solution :
A = {x Î Z ¢ (–5 < x < –2) È (3 < x < 5)}
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AÇ A = A
AÇ U = A , AÈ U = U
AÇ U = A
AÇ A = f
Example 1.10.2 If A = {x , y, {x , z } f }. Determine the following sets
i) A – {x, z} ii) {{x, z} – A
iii) A – {{x, y}} iv) {x, z} – A
v) A – P(A) vi) {x} – A
vii) A – {x} viii) A – f
ix) f – A x) {x , y, f } – A SPPU : Dec.-08, Marks 4
Solution :
i) A – {x, z} = {x, y, f }
ii) {{x, z }} – A = f
iii) A – {{x, y}} = A
iv) {x, z} – A = {z}
v) A – p(A) = {x, y, {x, z}}
vi) {x} – A = f
vii) A – {x} = {y, {x, z }, f }
viii) A – f = {x, y, {x, z}}
ix) f– A = f
x) {x, y, f} – A = {{x, z}}
Example 1.10.3 If A = {f, b}, construct the following sets A – f, {f} – A, A È P ( A)
SPPU : Dec.-19, Marks 03
Solution : We have A = {f, b }
\ A – f = {b}
{f} – A = f and P(A) = {f, {f}, {b}, A}
A È P(A) = {f, {f}, {b}, A}
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Example 1.10.4 Let U = Z ¢ = Set of all integers
A = Set of even integers
B = Set of odd integers
C = Set of prime numbers
Find A – B,, B – A,, C – A, A – C, B – C
Solution :
A – B = B– B = f
B– A = A – A = f
C – A = Set of non prime integers which are not even
A – C = B – C = Set of odd integers which are not prime = { ± 1, ± 9, ± 15, ... }
B– C = A – C = Set of even integers which are not prime = { ± 4 , ± 6, ± 8, ... }
Example 1.10.5 Let A, B, C be three sets and A Ç B = A Ç C
A Ç B = A Ç C is it necessary that B = C ? Justify.
Solution :
Yes, B = C.
Consider B = B Ç U = BÇ ( A È A)
= (B Ç A) È ( B Ç A)
= (A Ç B) È ( A Ç B)
= (A Ç C) È ( A Ç C)
= (A È A) Ç C
= UÇ C
\ B=C
Example 1.10.6 Let A, B, C be three sets
i) Given that A È B = A È C, is it necessary that B = C ?
ii) Given that A Ç B = A Ç C, is it necessary that B = C ?
Solution :
i) No,
Let A = {x, y, z}, B = {x}, C = {z}
AÈ B = A and AÈ C = A
\ AÈ B = AÈ C = A But B¹ C
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ii) No,
Let A = {x, y}, B = {y, z, w}, C = {y, p, q}
A Ç B = {y} = BÇ C But B¹ C
Example 1.10.7 If A Å B = A Å C, is B = C ?
Solution :
Yes, Let x Î B Þ x Î A or x Ï A
Suppose x Î A then x Î A Ç B Þ x Ï A Å B
Hence x Ï A Å C Þ x Î A Ç C Þ x Î C
i.e. if x Î A then BÍ C
Suppose x Ï A then x Ï A Ç B so that x Î A Å B
Þ x Ï A Å C Þ x Ï AÈ B Þ x ÎC
Hence BÍ C
Similarly C Í B
Hence B=C
Example 1.10.8 Let A and B are two sets. If A Í B, then prove that P(A) Í P(B). where
P(A) and P(B) are power sets of A and B sets. SPPU : Dec.-17, Marks 6
Solution : Let A and B be two sets.
The powr set of A is the set of all subsets of a set A.
Let P(A) be the power set of A and P(B) be the power set of B.
If A is a subset of B i.e. A Í B then all elements of A are in B.
If X Í P(A) but A Í B Þ X Î P(B)
Thence P(A) Í P(B) " ´ ÎP(A)
Example 1.10.9 If f is an empty set then find p( f), p( p( f)) p( p( p( f)))
Solution :
p( f) = {f }
p (p( f)) = {f, {f }}
p ( p ( p ( f))) = {f, {f }, {{f }}, {f, {f }}}
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Example 1.10.10 Show that [ A Ç ( B È A )] È B = B
Solution :
[A Ç (BÈ A)] È B = [(A Ç B) È ( A Ç A)] È B
= [(A Ç B) È f] È B
= (A Ç B) È B = B
Example 1.10.11 Prove that A Ç ( B – C )] Ì A – ( B Ç C )
Solution : Let x Î A Ç (B– C) then
Þ x Î A and x Î B– C
Þ x Î A and ( x Î Band x Ï C)
Þ x Î A and x Ï ( BÇ C)
Þ x Î A – ( BÇ C)
\ A Ç (B– C) Ì A – (BÇ C)
Example 1.10.12 Salad is made with combination of one or more eatables, how many different
salads can be prepared from onion, carrot, cabbage and cucumber?
SPPU : Dec.-13, Marks 4
Solution : The number of different salads can be prepared from onion, carrot, cabbage
and cucumber with combination of one or more eatables is 2 4 – 1 = 16 – 1 = 15
Example 1.10.13 Explain the concepts of countably infinite set with example.
SPPU : Dec.-14, Marks 4
Solution : A set is said to be countable if its all elements can be labelled as 1, 2, 3, 4, ...
A set is said to be countably infinite
if, i) It is countable
ii) It has infinitely many elements i.e. It's cardinality is ¥.
For example
1) The set of natural numbers {1, 2, 3, ...} is countably infinite.
2) The set of integers is countably infinite.
3) The set of real numbers is infinite but not countable.
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Example 1.10.14 Draw Venn diagram and prove the expression. Also write the dual of each of
the given statements.
i) ( A È B È C ) C = ( A È C ) C Ç ( A È B ) C
ii) (U Ç A) È (B Ç A) = A SPPU : Dec.-11, Marks 6
Solution : Consider the following Venn diagrams.
ii) Consider the following Venn diagrams.
A B U U
A B
C C
c
1 ABC= 2 (A B C) =
Fig. 1.10.1
A B A B U
C C
c c
3 (A B) = 4 (A C) =
Fig. 1.10.2
A B U
From (2) and (5)
(A B C)c = (A C)c (A B)c
c c
5 (A B) (A C)
Fig. 1.10.3
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A U A U
B B
1 UA= 2 BA=
A U A U
B B
3 (U A) (B A) = 4 A=
Fig. 1.10.4
From Venn diagrams (3) and (4)
(U Ç A) È (BÇ A) = A
Example 1.10.15 Using Venn diagram show that :
A È ( B Ç C) = ( A È B ) Ç ( A È C) SPPU : May-05, Marks 4
Solution : Consider the following venn diagrams
B = {x / x Ï B}
A U A U
B B
C C
1 B= 2 B C=
A U A U
B B
C C
3 A (B C) = 4 AB=
and
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A U A U
B B
C C
5 A C= 6 (A B) (A C) =
From 3 and 6 ,
A È (B Ç C) = (A È B) Ç (A È C)
Example 1.10.16 Using Venn diagram, prove or disprove.
i) A Å (B Å C) = (A Å B) Å C
ii) A Ç B Ç C = A - [( A - B) È ( A - C)] SPPU : May-06, Marks 4
Solution : i) A Å B = (A È B) - (A Ç B)
A Å B : elements which are either in A or in B but not in both A and B.
A X A Y
B B
C C
AÅB= (A Å B) Å C
1 2
A B A B
C C
BÅC= A Å (B Å C)
3 and 4
From 2 and 4 ,
(A Å B) Å C = A Å (B Å C)
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Discrete Mathematics 1 - 20 Theory of Sets
ii) A B C = A [(A B) (A C)]
A B X A B X
C C
ABC= A–B=
1 2
A B X A B X
C C
A–C= (A – B) (A – C) =
3 4
and
A B
A – [(A–B) (A–C)] =
From 1 and 5
A B C = A [(A B) (A C)]
Example 1.10.17 Using Venn diagrams show that
A ( B C) ( A B) ( A C) SPPU : May-08, Dec.-12, Marks 3
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Discrete Mathematics 1 - 21 Theory of Sets
Solution :
A B X A B X
C C
BC= A (B C) =
1 2
A B X A B X
C C
AB= AC=
3 4
A B X
(AB) (AC) =
From 2 and 5 ,
A (B C) = (A B) (A C)
Example 1.10.18 Let A, B, C be sets. Under what conditions the following statements are
true ?
i) ( A B) ( A C) A ii) ( A B) ( A C) SPPU : Dec.-06, 15, Marks 3
Solution : i) (A B) (A C) A
A X A X
B C OR
C
B
1 2
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Discrete Mathematics 1 - 22 Theory of Sets
If B Ì A and C Ì A
then (A – B) È (A – C) = A
Or A, B, C are disjoint sets.
i.e. A Ç B Ç C = Æ
Then (A - B) È (A - C) = A
ii) (A - B) È (A - C) = Æ is true.
If A is empty set i.e. A = Æ
Example 1.10.19 Prove the following using Venn diagram.
A Ç (B Å C) = ( A Ç B) Å ( A Ç C) SPPU : May-08, 14, Dec.-12, Marks 3
Solution :
A X A X
B B
C C
BC= A (B C) =
1 2
A B X A B X
C C
AB= AC=
4
3
A B X
(AB) (AC)=
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Discrete Mathematics 1 - 23 Theory of Sets
From 2 and 5
A (B Å C) = (A B) Å (A C)
1.11 Cardinality of Sets
In the analysis of computer algorithms, we required to count the number of
operations executed by various algorithms. This is necessary to estimate the cost
effectiveness of a particular algorithm. Hence we require to study the cardinality of
finite sets and understand related properties.
Definition :
Let A be any finite set. The number of elements in the set A, is called the cardinality
of the set A. It is denoted by |A| or n(A).
e.g. If A = {x, y, z, p} then |A| = 4
Theorem 1 : If A = f then |A| = 0
Theorem 2 : If A Ì B then |A| £ |B|
Theorem 3 : (Addition Principle)
Let A and B be two finite sets which are disjoint then |A È B| = |A| + |B|
Proof : If A = f, B=f then proof is obvious
Let us assume that A ¹ f, B¹ f
Suppose A = { a 1 , a 2 , .... a n }, B = { b1 , b 2 , .... b m }
\ A Ç B = f, |A| = n, |B| = m
\ A È B = { a 1 , a 2 , ... a n , b1 , b 2 , .... b m }
\ |A È B| = m + n = |A| + |B|. Hence the proof
Theorem 4 : Let A1 , A 2 , A 3 , .... A n be mutually disjoint finite sets then
|A1 È A 2 È A 3 È ... È A n | = |A1| +|A 2|+|A 3|+ ... +|A n |
Theorem 5 : If A is finite set and B is any set then
|A – B| = |A|–|A Ç B| A B
Proof : By Venn diagram
A = (A – B) È (A Ç B)
(A – B) Ç (A Ç B) = f A–B
\ By addition principle |A| = |A – B|+|A Ç B|
Fig. 1.11.1
Þ |A – B| = |A|–|A Ç B|
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Discrete Mathematics 1 - 24 Theory of Sets
1.12 The Principle of Inclusion and Exclusion for Sets
SPPU : Dec.-04, 05, 07, 08, 10, 13, 14, 15, 19
May-05, 06, 07, 08, 14, 15, 17, 19
Theorem 6 : (Principle of Inclusion - Exclusion for 2 sets)
Let A and B be finite sets then
|A B| = |A|+|B|–|A Ç B|
A B
Proof : By venn diagram
A È B = ( A – B) È B
A–B AB
As A–B and B are disjoint sets.
|A È B| = | A – B| +| B|
= | A| – | A Ç B| +| B|
|A È B| = | A| +| B| – | A Ç B| Fig. 1.12.1
Principle of Inclusion-Exclusion for three sets.
Theorem 7 : Let A, B, C be finite sets. Then
|A È BÈ C| = | A| +| B| +|C| – | A Ç B| –|A Ç C| –|BÇ C|
Proof Let D = BÈ C
\ |A È D| = |A|+|D| = |A Ç D|
and |D| = |BÈ C| = |B|+|C|–|BÇ C|
\ |A È BÈ C| = |A È D| = |A|+|B|+|C|–|BÇ C| – | A Ç D|
= |A|+|B|+|C|–|BÇ C| – | A Ç (BÈ C) |
= |A|+|B|+|C|–|BÇ C| – | ( A Ç B) È ( A Ç C) |
= |A|+|B|+|C|–|BÇ C| – | A Ç B| –|A Ç C| +| A Ç BÇ C|
= |A|+|B|+|C|–|A Ç B | –|A Ç C| –|B Ç C| + | A Ç B Ç C|
Theorem 8 : Let A, B, C, D be finite sets then
|A È BÈ C È D| = | A| +| B| +|C| +| D| – | A Ç B| – | A Ç C| – | A Ç D|
– | BÇ C| – | BÇ D| – |C Ç D|
= | A Ç BÇ C| +|A Ç BÇ D| +| A Ç C Ç D| +| BÇ C Ç D|
– | A Ç BÇ C Ç D|
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Discrete Mathematics 1 - 25 Theory of Sets
Examples :
Example 1.12.1 Among the integers 1 to 300 find how many are not divisible by 3, not by 5.
Find also how many are divisible by 3 but not by 7. SPPU : Dec.-08, Marks 6
Solution : Let A denotes the set of integers 1 to 300 divisible
by 3, B denotes the set of integers 1 to 300 divisible
by 5, C denotes the set of integers 1 to 300 divisible.
by 7 |A| = = 100 , |B| = = 60, |C| = = 42, |A Ç B| = é
300 300 300 300 ù
= 20
3 5 7 êë 3 ´ 5úû
Find |A Ç B| and |A - C|
We have A Ç B = A È B = U – (A È B)
\ |A È B| = | A| +| B| – | A Ç B|
= 100 + 60 – 20 = 140
\ |A Ç B| = |U|–|A È B| = 300 – 140 = 160
Hence 160 integers between 1 – 300 are not divisible by 3, not by 5.
| A Ç C| = é
300 ù
|A – C| = | A| – | A Ç C|, = 14
êë 3 ´ 7 úû
|A – C| = 100 – 14 = 86
Hence, 86 integers between 1 – 300 are not divisible by 3 but not by 7.
Example 1.12.2 Out of 30 students in a college, 15 take an art course, 8 take maths course, 6
take physics course. It is know that 3 students take all courses. Show that 7 or more
students take none of the courses?
Solution : Let A be the set of students taking an art course
B be the set of students taking a maths course
C be the set of students taking a physics course
We have |A| = 15, |B| = 8, |C| = 6,
|A Ç BÇ C| = 3
|A È BÈ C| = | A| +| B| +|C| – | A Ç B| – | A Ç C| – | BÇ C| +| A Ç BÇ C|
= 15 +8 +6 – | A Ç B| – | A Ç C| – | BÇ C| +3
= 32 – | A Ç B| – | A Ç C| – | BÇ C|
But |A Ç B| ³ |A Ç BÇ C|, |BÇ C| ³ |A Ç BÇ C|
and |A Ç C| ³ |A Ç BÇ C|
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Discrete Mathematics 1 - 26 Theory of Sets
\ |A Ç B| ³ 3, |BÇ C| ³ 3, |A Ç C| ³ 3
\ |A Ç B| +|A Ç C| +|BÇ C| ³ 3|A Ç BÇ C|
\ |A È BÈ C| ³ 32 – 3| A Ç BÇ C| = 32 – 3 × 3 = 23
Hence |A È BÈ C| ³ 23
Hence the number of students taking at least one course is ³ 23. The students taking
none of the course is £ 30 – 23 = 7
Thus 7 or less students take none of the courses.
Example 1.12.3 How many integers between 1 to 2000 are divisible by 2 or 3 or 5 or 7.
Solution : Suppose set A denotes the number of integers between 1 to 2000 divisible
by 2.
Set B is the number of integers between 1 and 2000 divisible by 3.
Set C is the number of integers between 1 and 2000 divisible by 5.
Set D is the number of integers between 1 and 2000 divisible by 7.
A = é
2000ù
= 1000
êë 2 úû
é 2000ù = 666
B =
êë 3 úû
é 2000ù =
C = 400
êë 5 úû
D = é
2000ù
= 285
êë 7 úû
é 2000ù =
AÇB = 333
êë 2 ´ 3 úû
AÇC = é
2000ù
= 200
êë 2 ´ 5 úû
é 2000ù =
AÇD = 142
êë 2 ´ 7 úû
BÇ C = é
2000ù
= 133
êë 3 ´ 5 úû
é 2000ù =
BÇ D = 95
êë 3 ´ 7 úû
é 2000ù =
CÇ D = 57
êë 5 ´ 7 úû
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Discrete Mathematics 1 - 27 Theory of Sets
é 2000 ù =
A Ç BÇ C = 66
êë 2 ´ 3 ´ 5úû
A Ç BÇ D = é
2000 ù
= 47
êë 2 ´ 3 ´ 7 úû
é 2000 ù =
A Ç CÇ D = 28
êë 2 ´ 5 ´ 7 úû
BÇ CÇ D = é
2000 ù
= 19
êë 3 ´ 5 ´ 7 úû
é 2000 ù =
A Ç BÇ CÇ D = 9
êë 2 ´ 3 ´ 5 ´ 7 úû
Number of elements divisible by 2 or 3 or 5 or 7 are A È B È C È D .
From inclusion exclusion principle
A È BÈ CÈ D = A + B + C + D
– [ A Ç B + BÇ C + A Ç C + A Ç D + BÇ D + CÇ D ]
+ [ A Ç B Ç C + A Ç B Ç D +|A Ç C Ç D|+|B Ç C Ç D|]
– [ A Ç BÇ CÇ D ]
\ A È BÈ CÈ D = 1000 + 666 + 400 + 285 – [333 + 200 + 142 + 133 + 95 + 57]
+ [66 + 47 + 28 + 19] – 9
= 2351 – 960 + 160 – 9
= 1542
Example 1.12.4 In the survey of 260 college students, the following data were obtained :
64 had taken a maths course,
94 had taken a cs course,
58 had taken a business course,
28 had taken both a maths and a business course,
26 had taken both a maths and a cs course,
22 had taken both a cs and a business course,
14 had taken all types of courses.
How many students were - surveyed who had taken none of the three types of courses.
SPPU : May-17, Marks 3
Solution : Let A be the set of students, taken a maths course.
Let B be the set of students, taken a cs course
Let C be the set of students, taken a business course.
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\ We have U = 260, |A| = 64, |B| = 94
| C| = 58, A Ç C| = 28, |A Ç B| = 26
|B Ç C| = 22, |A Ç B Ç C| = 14,
We have,
|A È B È C| = |A| + |B| + |C| – |A Ç B|
– |A Ç C} – |B Ç C| + |A Ç B Ç C|
= 64 + 94 + 58 – 28 – 26 – 22 + 14 = 154
The total number of students taken none of the three types of courses
= |U| – |A È B È C|
= 260 – 154 = 106
Example 1.12.5 Among 100 students, 32 study mathematics, 20 study physics , 45 study
biology, 15 study mathematics and biology, 7 study mathematics and physics, 10 study
physics and biology and 30 do not study any of the three subjects.
a) Find the number of students studying all three subjects.
b) Find the number of students studying exactly one of the three subjects.
Solution : Let A, B, C denotes the set of students studying mathematics, physics and
biology respectively.
And X = 100
A = 32
B = 20
C = 45
AÇC = 15
AÇB = 7
BÇ C = 10
A ¢Ç B¢Ç C¢ = 30
A ¢Ç B¢Ç C¢ = 100 - A È B È C
Or A È B È C = 100 – 30
= 70
a) A È BÈ C = A + B + C -[ A Ç B + A Ç C + BÇ C ] + A Ç BÇ C
70 = 32 + 20 + 45 – [7 + 15 + 10] + A Ç B Ç C
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70 = 97 – [32] + A Ç B Ç C
70 – 65 = A Ç BÇ C
Þ A Ç BÇ C = 5
5 students study all 3 subjects.
b) Number of students studying exactly one subject.
X
A B
Number of students studying only mathematics is
A - A Ç B - A Ç C + A Ç BÇ C
= 32 – 7 – 15 + 5
= 15
Number of students studying only physics is
B - BÇ A - BÇ C + A Ç BÇ C
= 20 – 7 – 10 + 5
= 8
Number of students studying only biology is
C - A Ç C - BÇ C + A Ç BÇ C
= 45 – 15 – 10 + 5
= 25
\ Number of students studying exactly one subject
= 15 + 8 + 25 = 48
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Discrete Mathematics 1 - 30 Theory of Sets
Example 1.12.6 A survey was conducted among 1000 people of these 595 are Democrats, 595
wear glasses, and 550 like icecream. 395 of them are Democrats who wear glasses, 350 of
them are Democrats who like icecream, and 400 of them wear glasses and like icecream 250
of them are Democrats who wear glasses and like icecream. How many of them are not
Democrats do not wear glasses, and do not like icecream ? How many of them are
Democrats who do not wear glasses and do not like icecream ?
Solution : X = 1000
Let A be the number of Democrats.
A = 595
B be the number of people who wear glasses.
B = 595
C be the number of people who like icecream.
C = 550
AÇB = 395
AÇC = 350
BÇ C = 400
A Ç BÇ C = 250
A È BÈ C = A + B + C -[ A Ç B + BÇ C + A Ç C ] + A Ç BÇ C
= 595 + 595 + 550 – [395 + 350 + 400] + 250
A È BÈ C = 845
845 people are either Democrats or wear glasses or icecream.
Þ 1000 – 845 = 155 people are neither Democrats, nor wear glasses nor like
icecream.
1000
A B
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Discrete Mathematics 1 - 31 Theory of Sets
Number of people who are Democrats who do not like icecream and do not wear
glasses.
= A - A Ç B - A Ç C + A Ç BÇ C
= 595 – [395 + 300] + 250
= 845 – 695
= 150
Example 1.12.7 It is known that at the university 60 percent of the professors play tennis, 50
percent of them play bridge. 70 percent jog, 20 percent play tennis and bridge, 30 percent
play tennis and jog, and 40 percent play bridge and jog. If some one claimed that 20
percent of the professors jog and play bridge and tennis, would you believe this claim ?
Why ? SPPU : Dec.-13, Marks 6
Solution :
100
A B
Let A, B, C, denotes the number of professors play tennis, bridge and jog
respectively.
A = 60
B = 50
C = 70
AÇB = 20
AÇC = 30
BÇ C = 40
A Ç BÇ C = 20
A È BÈ C = A + B + C -[ A Ç B + A Ç C + BÇ C ] + A Ç BÇ C
= 60 + 50 + 70 – [20 + 30 + 40] + 20
= 110
which is not possible as A È B È C Ì X and the number of elements in A È B È C
cannot exceed number of elements in the universal set X.
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Discrete Mathematics 1 - 32 Theory of Sets
Example 1.12.8 Among 200 students in a class, 104 students got an 'A', in first examination
and 84 students got 'A' in second examination. If 68 students did not get an 'A' in either
of the examination.
i) How many students got 'A' in both the examination.
ii) If number of students who got an 'A' in the first examination is equal to that who got
an 'A' in second examination. If the total number of students who got 'A' in exactly one
examination is 160 and if 16 students did not get 'A' in either examination. Determine
the number of students who got 'A' in first examination, those who got ‘A’ in second
examination and number of students who got 'A' in both examinations.
SPPU : Dec.-04, Marks 6
Solution : X Universal set
X = 200
F S
Let F denote the set of students who got an 'A' in first examination and S denote the
set of students who got an 'A' in second examination.
68 students did not get A in either examination i.e.
(F È S) ¢ = 68
Þ FÈ S = X - 1 (F È S) ¢
= 200 – 68
Þ FÈ S = 132
i) Number of students who got A in both the examination will be F Ç S .
Now FÈ S = F + S - FÇ S
132 = 104 + 84 – F Ç S
\ FÇ S = 56
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Discrete Mathematics 1 - 33 Theory of Sets
ii) F = S
The total number of students who got A in exactly one examination is F - S È S - F
i.e. F Å S
and |F Å S| = 160
16 students did not get A in either examination
\ (F È S) ¢ = 16
Þ FÈ S = X – (F È S) ¢
Þ FÈ S = 200 – 16 = 184
Now FÈ S = F Å S + FÇ S
As F Å S and F Ç S are pairwise disjoint sets.
Þ 184 = 160 + F Ç S
Þ FÇ S = 184 – 160 = 24
\ FÈ S = F + S - FÇ S
184 = 2 F - 24
2 F = 208
Þ F = 104 and F = S
Þ S = 104
\ Number of students who got A in first examination
= F - FÇ S
= 104 – 24 = 80
Similarly number of students who got A in second examination
= |S| –| F Ç S | = 104 - 24 = 80
Example 1.12.9 Consider a set of integers 1 to 500. Find
i) How many of these numbers are divisible by 3 or 5 or by 11 ? Dec.-14
ii) Also indicate how many are divisible by 3 or by 11 but not by all 3, 5 and 11.
iii) How many are divisible by 3 or 11 but not by 5 ? SPPU : May-05, Marks 6
Solution : Let A denote numbers divisible by 3.
B denote numbers divisible by 5.
C denote numbers divisible by 11.
A denotes cardinality of A similarly B and C denotes cardinality of B and C.
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Discrete Mathematics 1 - 34 Theory of Sets
=
500
A 166
3
=
500
B = 100
5
=
500
C = 45
11
= é
500 ù
AÇB = 33
êë 3 ´ 5úû
= é
500 ù
AÇC = 15
êë 3 ´ 11úû
= é
500 ù
BÇ C =9
êë 5 ´ 11úû
= é
500 ù
A Ç BÇ C =3
êë 3 ´ 5 ´ 11úû
i) A È BÈ C = A + B + C -[ A Ç B + A Ç C + BÇ C ] + A Ç BÇ C
= 166 + 100 + 45 – [13 + 15 + 9] + 3
= 257
ii)
X
A B
Number of integers divisible by 3 or by 11 but not by all 3, 5 and 11.
= A È C - A Ç BÇ C
= [ A + C - A Ç C ] - A Ç BÇ C
= 166 + 45 – 15 – 3 = 193
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Discrete Mathematics 1 - 35 Theory of Sets
iii)
X
A B
Number of integers divisible by 3 or 11 but not by 5.
= A È BÈ C – B
= 257 – 100 = 157
Example 1.12.10 In the survey of 60 people, it was found that 25 read newsweek magazine,
26 read time, 26 read fortune. Also 9 read both newsweek and fortune, 11 read both
newsweek and time, 8 read both time and fortune and 8 read no magazine at all.
i) Find out the number of people who read all the three magazines.
ii) Fill in the correct numbers in all the regions of the Venn diagram.
iii) Determine number of people who reads exactly one magazine.
SPPU : Dec.-05, 10, 19, Marks 6
Solution : X ® Universal set
X = 60
Let N denote the number of people who read newsweek magazine.
\ N = 25
T denote the number of people who read time magazine.
\ T = 26
F denote the number of people who read fortune magazine.
\ F = 26
NÇ T = 11
NÇ F = 9
TÇ F = 8
N¢Ç T¢Ç F¢ = 8
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Using DeMorgan's law
N T F = X N T F
8 = 60 – N T F
N T F = 52
i) N T F = N T F [ N T N F T F ] N T F
52 = 25 + 26 + 26 – [11 9 8] N T F
N T F = 3
ii)
60
N T
8 8 10
3
6 5
12
F
Explanation :
As N T F = 3
and people who read N and T both = 11 which includes people of N T F also.
Hence people who read only N and T but not F will be 11 – 3, hence 8 people read
only N and T similarly T F = 8.
Then people read only T and F = 8 – 3 i.e. 5.
and N F = 9
People read only N and F = 9 – 3 i.e. 6.
Also,
N = 25, which includes N T , N F and N T F .
People who read only N = N – [ N T N F ] N T F
= 25 – [11 + 9] + 3
People who read only N = 8 …(1)
People who read only T = T – [ N T T F ] N T F
= 26 – [11 + 8] + 3
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Discrete Mathematics 1 - 37 Theory of Sets
People who read only T = 10 … (2)
People who read only F = F – [ N F T F ] N T F
= 26 – [9 + 8] + 3
People who read only F = 12 … (3)
iii) Number of people who reads exactly one magazine
= Equation [(1) + (2) + (3) ] = 8 + 10 + 12 = 30
Example 1.12.11 Suppose that 100 out of 120 mathematics students at a college take at least
one of the languages French, German and Russian. Also suppose
65 study French, 20 study French and German
45 study German, 25 study French and Russian
42 study Russian, 15 study German and Russian
i) Find the number of students who study all the three languages.
ii) Fill in correct number of students in each region of Venn diagram.
iii) Determine the number K of students who study
a) Exactly one language.
b) Exactly two languages. SPPU : Dec.-05, Marks 6
Solution : X Universal set
X = 120
Let F be the number of students who study French. G be the number of students
who study German and R be the number of students who study Russian.
| F G R | = 100
F = 65, F G = 20
G = 45, F R = 25
R = 42, G R = 15
i) F G R = F G R [ F G F R G R ] F G R
100 = 65 + 45 + 42 – [20 + 25 + 15] + F G R
100 = 92+ F G R
F G R = 8
Hence 8 students study all three languages.
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ii)
F G
28 12 18
8
17 7
10 R
Explanation
As F G R = 8
Students who study F and G but not all 3 will be
F G – F G R
= 20 – 8 = 12 … (1)
Students who study F and R but not all 3 will be
F R – F G R
= 25 – 8 = 17 … (2)
Students who study G and R but not all 3 will be
G R – F G R
= 15 – 8 = 7 … (3)
Students who study only F but not G and not R.
= F [ F G F R ] F G R
= 65 – [20 + 25] + 8 = 28 … (4)
Students who study only G but not F and not R
= G [ F G G R ] F G R
= 45 – [20 + 15] + 8 = 18 … (5)
Students who study only R but not F and not G.
= R [ F R G R ] F G R
= 42 – [25 + 15] + 8 = 10 … (6)
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iii) a) Number of students who study exactly one language
= Equation (4) + Equation (5) + Equation (6)
= 28 + 18 + 10 = 56
b) Number of students who study exactly two languages.
= Equation (1) + Equation (2) + Equation (3)
= 12 + 17 + 7 = 36
Example 1.12.12 Out of the integers 1 to 1000.
i) How many of them are not divisible by 3, nor by 5, nor by 7 ?
ii) How many are not divisible by 5 and 7 but divisible by 3 ?
SPPU : May-06, 08, 14, Marks 6, May-19, Marks 13
Solution : i) Let A denote numbers divisible by 3.
B denote numbers divisible by 5.
and C denote numbers divisible by 7.
1000
A = = 333
3
1000
B = = 200
5
1000
C = = 142
7
1000
AB = = 66
3 5
1000
AC = = 47
3 7
1000
B C = = 28
5 7
1000
A B C = = 9
3 5 7
A B C = A B C [ A B A C B C ] A B C
= 333 + 200 + 142 – [66 + 47 + 28] + 9
= 543
This show 543 numbers are divisible by 3 or 5 or 7. Hence numbers which are not
divisible by 3, nor by 5, nor by 7.
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= |A B C|
= X – A B C
= 1000 – 543 = 457
ii)
A B
Number of integers divisible by 3 but not by 5 and not by 7 is|A B C|.
|A B C| = |A (B C) |
= A [ A B A C ] A B C
= 333 – [66 + 47] + 9 = 229
Example 1.12.13 In the class of 55 students the number of studying different subjects are as
given below : Maths 23, Physics 24, Chemistry 19, Maths + Physics 12, Maths +
Chemistry 9, Physics + Chemistry 7, all three subjects 4.
Find the number of students who have taken :
i) At least one subject ii) Exactly one subject iii) Exactly two subjects.
SPPU : May-07, Marks 6
Solution : Let X denote universal set.
Let M denote set of students studying mathematics.
P denote set of students studying physics
and C denote set of students studying chemistry.
Then X = 55, M = 23, P = 24, C = 19, M P = 12,
M C = 9, P C = 7
M P C = 4
i) M P C = M P C [ M P M C P C ] M P C
= 23 + 24 + 19 – [12 + 9 + 7] + 4 = 42
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ii) Number of students studying mathematics but not physics and not chemistry.
= M – [ M P M C ] M P C
= 23 – [12 + 9] + 4 = 6 ... (1)
Number of students studying physics but not mathematics and not chemistry.
= P – [ M P P C ] M P C
= 24 – [12 + 7] + 4 = 9 ... (2)
Number of students studying chemistry but not mathematics and not physics.
= C [ M C P C ] M P C
= 19 – [9 + 7] + 4 = 7 … (3)
Hence number of students studying exactly one subject = 6 + 9 + 7 = 22
iii) Number of students studying mathematics and physics but not chemistry
= M P – M P C
= 12 – 4 = 8 … (4)
Number of students studying physics and chemistry but not mathematics
= P C M P C
= 7–4=3 … (5)
Number of students studying mathematics and chemistry but not physics
= M C – M P C
= 9–4=5 … (6)
Hence number of students studying exactly two subjects
= 8 + 3 + 5 = 16
X
M P
6 8 9
4
5 3
7
C
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Example 1.12.14 100 of the 120 engineering students in a college take part in atleast one of
the activity group discussion, debate and quiz.
Also : 65 participate in group discussion, 45 participate in debate, 42 participate in quiz,
20 participate in group discussion and debate 25 participate in group discussion and quiz
15 participate in debate and quiz. Find the number of students who participate in : i) All
the three activities ii) Exactly one of the activities. SPPU : Dec.-07, Marks 4
Solution : Let X denote universal set
X = 120
Let A denote number of students taking part in group discussion.
B denote number of students taking part in debate
and C denote number of students taking part in quiz.
A B C 100, A 65, B 45, C 42,
A B 20, A C 25, B C 15.
i) A B C = A B C [ A B A C B C ] A B C
100 = 65 + 45 + 42 – [20 + 25 + 15] + A B C
A B C = 8
ii) Number of students taking part in group discussion but not in debate and not in
quiz.
= A [ A B A C ] A B C
= 65 – [20 + 25] + 8 = 28 ... (1)
Number of students taking part in debate but not in group discussion and not in
quiz.
= B [ A B B C ] A B C
= 45 – [20 + 15] + 8 = 18 ... (2)
Number of students taking part in quiz but not in group discussion and not in
debate.
= C [ A C B C A B C ]
= 42 – [25 + 15] + 8 = 10 ... (3)
Hence number of students doing exactly one activity
= 28 + 18 + 10 = 56
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A B
28 12
18
8
17 7
10 C
Example 1.12.15 It was found that in the first year computer science class of 80 students, 50
knew COBOL, 55 'C' and 46 PASCAL. It was also know that 37 knew 'C' and COBOL,
28 'C' and PASCAL and 25 PASCAL and COBOL. 7 students however knew none of the
languages. Find
i) How many knew all the three languages ?
ii) How many knew exactly two languages ?
iii) How many knew exactly one language ? SPPU : May-15, Dec.-15, Marks 4
Solution : Let A denote the set of students who know COBOL.
B denote the set of students who know ‘C’.
and C denote the set of students who know PASCAL.
and X denote universal set.
Then X 80, A 50, B 55, C = 46
A B 37, B C 28, A C 25
A B C = 7
Hence (A B C) = 7
Also A B C = X (A B C)
Hence A B C = 80 – 7 = 73
i) A B C = A B C [ A B A C B C ] + A B C
73 = 50 + 55 + 46 – [37 + 28 + 25] + A B C
A B C = 12
ii) Number of students who know only COBOL and 'C' but not PASCAL.
= A B A B C
= 37 – 12 = 25 ... (1)
Number of students who know only COBOL and PASCAL but not 'C'.
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= A C A B C
= 25 – 12 = 13 ... (2)
Number of students who know only 'C' and PASCAL but not COBOL.
= B C A B C
= 28 – 12 = 16 ... (3)
Hence number of students who know exactly two languages.
= 25 + 13 + 16 = 54
iii) Number of students who know only COBOL but not 'C' and not PASCAL.
= A [ A B A C ] A B C
= 50 – [37 + 25] + 12 = 0 ... (4)
Number of students who know only 'C’ but not COBOL and not PASCAL.
= B [ A B B C ] A B C
= 55 – [37 + 28] + 12 = 2 ... (5)
Number of students who know only PASCAL but not COBOL and not 'C'.
= C [ A C B C ] A B C
= 46 – [25 + 28] + 12 = 5 ... (6)
Hence number of students who know only one language = 0 + 2 + 5 = 7
X
B
A 25 2
12
13 16
Example 1.12.16 In a survey of 500 television watchers produced the following information
285 watch football, 195 watch hockey, 115 watch basketball, 45 watch football and
basketball, 70 watch football and hockey, 50 watch hockey and basketball and 50 do not
watch any of the three games.
i) How many people in the survey watch all three games ?
ii) How many people watch exactly one game ? SPPU : May-08, Marks 6
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Solution : Let F denote the set of people who watch football.
H denote the set of people who watch hockey and B denote the set of people who
watch basketball.
X Denote universal set
Then X 500, F = 285, H = 195, B = 115.
F B = 45, F H = 70, H B = 50
F B H = 50
i.e. |(F B H) | = 50
F B H = X ( F B H)
= 500 – 50 = 450
i) Number of people who watch all the three games = F H B .
F H B = F H B [ F H F B H B ] F H B
450 = 285 + 195 + 115 – [70 + 45 + 50] + F H B
F H B = 20
ii) Number of people who watch only football.
= F [ F B F H ] F B H
= 285 – [45 + 70] + 20 = 190 ... (1)
Number of people who watch only hockey.
= H [ H B H F ] F B H
= 195 – [50 + 70] + 20 = 95 ... (2)
Number of people who watch only basketball.
= B [ B H B F ] F B H
= 115 – [50 + 45] + 20 = 40 ... (3)
Number of people who watch exactly one game.
= 190 + 95 + 45 = 325
1.13 Multiset SPPU : Dec.-13, May-19
We know that the set is the collection of well defined distinct objects. But there are so
many practical situations in which objects are not distinct. For example 1) The collection
of alphabets in the word "Missicippi", collection = [m, i, i, i, i, s, s, c, p, p]
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2) Birth month of students studying in S.E. Computer Engineering and so on.
Multiset is the generalization of set.
Definition :
A collection of objects that are not necessarily distinct is called a multiset (or Msets).
To distinguish set and multiset we denote multiset by enclosing elements in a square
brackets e.g. [a, a, b]
1.13.1 Multiplicity of an Element
Let S be a multiset and x S. The multiplicity of x is defined as the number of times
the element x appears in the multiset S. It is denoted by ( x).
For example i) S = [a, b, c, d, d, d, e, e]
( a) = 1, ( b) = 1 , (c) = 1, (d) = 3, (e) = 2
1.13.2 Equality of Multisets
Let A and B be two multisets. A and B are said to be equal multisets iff
( x) = B ( x), x Aor B
e.g. [a, b, a, b] = [a, a, b, b] but [ a, a] [ a]
Subset of Multisets
A multiset 'A' is said to be the multiset of B if multiplicity of each element in A is
less or equal to it's multiplicity in B.
e.g. [a] [a, a], [a, b, a, b, a] [a, a, a, b, b, b].
1.13.3 Union and Intersection of Multisets
If A and B are multisets then A B and A B are also multisets.
The multiplicity of an element x A B is equal to the maximum of the multiplicity
of x in A and in B.
The multiplicity of an element x A B is equal to the minimum of the multiplicity
of x in A and in B.
Example :
A = [a, b, c, a, b, c, a, a, a], B = [a, a, b, b, b, b]
A B = [a, a, a, a, a, b, b, b, b, c, c]
A B = [a, a, b]
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1.13.4 Difference of Multisets
Let A and B be two multisets. The difference A – B is a multiset such that x A – B
if ( A ( x) – B ( x)) 1
e.g. 1) A = [a, a, b, a], B = [a, b]
A – B = [a, a]
2) A = [1, 2, 3, 4, 2, 2, 3, 3], B = [1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4]
A – B = [ ] or
1.13.5 Sum of Multisets
Let A and B be two multisets. The sum of A and B is denoted by A + B and defined
as for each
x A + B, (x) = A ( x) B ( x)
e.g. A = [a, b, c, c], B = [a, a, b, b, c, c]
A + B = [a, a, a, b, b, b, c, c, c, c]
Examples :
Example 1.13.1 Find the multiset to solve equation
A [1, 2, 2, 3, 4] = [1, 2, 3, 4]
A [1, 2, 2, 3] = [1, 1, 2, 2, 3, 3, 4]
Solution : Maximum multiplicity of each element is as follows
(1) = 2, ( 2) = 2, ( 3) = 2, ( 4) = 1
and minimum multiplicity of each element in A is as follows
(1) = 1, ( 2) = 1, ( 3) = 1, ( 4) = 1
Therefore,
A = [1, 2, 3, 4] or A = [1, 1, 2, 2, 3, 3, 4]
Example 1.13.2 Explain examples of multisets with its significance. SPPU : Dec.-13, Marks 4
Solution : 1) Multisets are used to denote roots of the polynomial.
x 3 3x 2 3x +1 = 0
Roots are –1, –1, –1
A = [–1, –1, –1]
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2) Multisets are used to denote prime factors of every non-negative integer.
e.g. prime factors of 72 are 2 × 2 × 2 × 3 × 3
A = [2, 2, 2, 3, 3]
3) Multisets are used to denote zeros or poles of analytic functions.
4) In Computer Science, multisets are applied in a variety of search and sort
procedure.
Example 1.13.3 If P = {a , a , a , c, d , d }, Q = {a , a , b , c, c}. Find union, intersection and
difference of P and Q. SPPU : May-19, Marks 3
Solution : We have
P Q = {a, a, a, b, c, c, d, d}
P Q = {a,a,c}
P – Q = {a, d, d}
qqq
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Unit - I
2 Propositional Calculus
Syllabus
Propositional Logic - logic, Propositional Equivalences, Application of Propositional Logic -
Translating English Sentences.
Contents
2.1 Introduction
2.2 Statements or Propositions
2.3 Laws of Formal Logic
2.4 Connectives and Compound Statements . . . . . Dec.-06, 18, May-05, 07, 08, · · Marks 6
2.5 Propositional or Statement Formula
2.6 Tautology
2.7 Contradiction
2.8 Contingency . . . . . . . . . . . . . . . . . . Dec.-10, 12, · · · · · · · · · · · · · · Marks 6
2.9 Precedence Rule
2.10 Logical Equivalence . . . . . . . . . . . . . . . . . . Dec.-12, · · · · · · · · · · · · · · · · · Marks 3
2.11 Logical Identities
2.12 The Duality Principle
2.13 Logical Implication
2.14 Important Connectives
2.15 Normal Forms . . . . . . . . . . . . . . . . . . Dec.-04, 06, 08, 10, 12, 14
. . . . . . . . . . . . . . . . . . May-17, · · · · · · · · · · · · · · · · · Marks 4
2.16 Methods of Proof . . . . . . . . . . . . . . . . . . Dec.-09, May-06, 10 · · · · · · · · Marks 4
2.17 Quantifiers . . . . . . . . . . . . . . . . . . Dec.-08, 09, 10, 11, 15,
. . . . . . . . . . . . . . . . . . May-15· · · · · · · · · · · · · · · · · · Marks 4
(2 - 1)
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Discrete Mathematics 2-2 Propositional Calculus
2.1 Introduction
A discrete structure is defined by a set of axioms. The properties of structure are
derived from the axioms as theorems. Such theorems are proved using valid rules of
reasoning. The propositional calculus (Mathematical logic) is concerned with all kinds of
reasoning. It has two aspects.
1) It is analytic theory of art of reasoning whose goal is systematic and codify the
principles of valid reasoning.
2) It is inter-related with problems relating the foundation of mathematics.
A great mathematician Frege G. (1884-1925) developed the idea regarding a
mathematical theory as applied branch of logic.
Every student of engineering should learn logic because principles of logic are
valuable to problem analysis, programming, logic designing, code designing and many
more.
2.2 Statements or Propositions
A statement is a declarative sentence which is either true or false but not both. The
truth or falsity of a statement is called its truth value. The truth value of a true
statement is denoted by 'T' and the false statement is denoted by 'F'. They are also
denoted by 1 or 0.
Statements are usually denoted by A, B, C, ....... or a, b, c.
Examples :
1) There are 5 days in a week
2) 2 + 5 = 7
3) y + 3 = 8
4) It will rain tomorrow
5) There are 12 months in a year
Examples (2) and (5) are true statements
Example (1) is false statement
In example (3), it's truth value depends upon the value of y. If y is 5 then sentence is
true and if y ¹ 5 then sentence is false. Therefore (4) is not a statement.
In example (4), it's truth value cannot be predicted at this moment but it can be
definitely determined tomorrow. Hence it is a statement.
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2.3 Laws of Formal Logic
There are two laws of formal logic
2.3.1 Law of Contradiction
It states that the same statement cannot be both true and false
2.3.2 Law of Excluded Middle
It p is a statement then either p is true or p is false and there cannot be middle
ground. If a student appear for the examination of discrete structure then student will be
either pass or fail in that exam. There is no middle stage.
2.4 Connectives and Compound Statements SPPU : Dec.-06,18, May-05,07,08
A statement which cannot be further split into simple sentences, is called primary or
primitive or atomic statements.
In day to day life, we use the words NOT, AND, OR, IF-Then, as well as, BUT,
WHILE to connect two or more sentences. But these connectives are flexible in their
meanings, and lead to inexact and ambiguous interpretations. However, mathematics is a
very precise language and every symbol of mathematics has the unique meaning or
interpretation in mathematical ocean.
Hence we take some special connectives with precise meaning to suit our purpose.
Following are the common connectives with symbols or their rotations.
Sr. No. Name of connectives Symbol
1 Negation ~ or Ø
2 And (Meet) ^
3 Or (Join) v
4 If .... then ®
5 If an only if (iff) «
2.4.1 Compound Statement
A statement which is formed from primary statements by using logical connectives is
called a compound statement.
e.g. If p : I am studying in SE computer class
q : I am learning discrete structure subject
The compound statement is
I am studying in SE computer class and I am learning discrete structure subject.
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2.4.2 Truth Table
A table showing the truth value of a statement formula is called truth table.
2.4.3 Negation
Let p be any simple statement, then the negation of p is formed by writing "It is false
that" before p or introducing the word "not" at the proper place. The negation of p is
also obtained by writing "p is false". The negation of p is denoted by ~ p or Ø p or p. If
the statement p is true, then ~ p is false and if p is false then ~ p is true.
The truth table for negation of p is given below
p ~p
T F
F T
Examples :
Sr. No. Statement p ~p
1. p : Atharva is intelligent boy ~ p : Atharva is not intelligent boy
OR
It is false that Atharva is intelligent boy
2. p : Delhi is the capital of India ~ p : Delhi is not the capital of India
OR
It is false that Delhi is the capital of India
OR
It is not the case that Delhi is the capital
of India
Note : ~ p is the unary connective as only one statement is required to form
negation.
2.4.4 Conjunction
A compound statement which is obtained by combining two primary statements by
using the connective "and" is called conjunction i.e. the conjunction of two statements p
and q is the statement p Ù q. It is read as "p and q" or "p meet q".
The statement p Ù q has the truth value T whenever both p and q have the truth value
T, otherwise p Ù q has the truth value F.
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It's tabular representation is as follows :
p q pÙq
T T T
T F F
F T F
F F F
Example 2.4.1
Sr. No. p q Conjunction p Ùq
1. 3 > 2 (T) 5 is prime (T) 3 > 2 and 5 is prime T
2. Delhi is in India (T) 3 + 30 = 30 (F) Delhi is in India and 3 + 30 = 30 F
3. Construct a truth table for the conjunction of n < 20 and n > 5, n Î N
Solution : When n < 20 and n > 5 are true then the conjunction "n < 20 and n > 5" is
true.
The truth table is as follows :
p : n < 20 q:n>5 p Ù q : n < 20 and n > 5 i.e. 5 < n < 20
T (n = 15) T (n = 15) T (Q 5 < 15 < 20)
T (n = 2) F (n = 2) F (Q n < 5)
F (n = 30) T (n = 30) F (Q n > 20)
F (n = 50) F --- F (There does not exist any number s.t. n > 20 and n < 5)
2.4.5 Disjunction
A compound statement obtained by combining two simple statements by using the
connective "or" is called the disjunction.
i.e. If p and q are simple statements then the compound statement "p or q" is the
disjunction of p and q. It is denoted by p Ú q. p Ú q is read as "p or q" or "p join q".
If p is false and q is false then p Ú q is false. Otherwise p Ú q is true. The truth table of
p Ú q is as follows :
p q pÚq
T T T
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T F F
F T T
F F F
Example 2.4.2 1) If p : The data is wrong, q : There is an error in the program.
Then p Ú q : The data is wrong or there is an error in the program.
The connective 'v' is used in the inclusive sense i.e. at least one possibility exists or even
both possibilities exist.
2)
Sr. p q Disjunction p Ùq
No.
1. 6 < 2 (F) 5 + 3 = 4 (F) 6 > 2 or 5 + 3 = 4 F (Q p and q both are false )
2. Delhi is in India (T) 5.3 = 10 (F) Delhi is in India or 5.3 = 10 T (Q p is true)
3. 6 – 3 = 3 (T) 5 is prime (T) 6 – 3 = 3 or 5 is prime T (Q p and q are true)
3) Construct a truth table for disjunction of p : n is prime, q : n > 10 for nÎ N.
Solution : The truth table for p Ú q is as follows :
p : n is prime q : n > 10 p Ú q : n is prime or n > 10
T (n = 13) T (13 > 10) T (Q n = 13 > 10 or 13 is prime)
T (n = 5) F (5 < 10) T (Q n = 5 is prime)
F (n = 12) T (12 > 10) T (Q n = 12 > 10)
F (n = 6) F (6 < 10) F (Q Both p and q are false)
2.4.6 Conditional Statement (If ..... then ....)
If p and q are two statements, then the statement p ® q which is read as "if p then q"
is called as conditional statement.
The conditional statement p ® q can also be read as i) p only if q ii) p implies q
iii) q if p iv) p is sufficient for q
The statement p is called the hypothesis or antecedent and q is called conclusion or
consequent.
If p is true and q is false then p ® q is false. Otherwise p ® q is true.
The truth table of p ® q is as follows :
p q p®q
T T T
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T F F
F T T
F F T
Examples :
1) Let p : Atharva is a graduate
q:3+5=8
\ p ® q : If Atharva is a graduate then 3 + 5 = 8
Let N be the set of natural numbers and Q be the set of rational numbers. We know
that every natural number is a rational number.
Let p : n is a natural number
q : n is a rational number
p : n ÎN q : n ÎQ p ® q : if n Î N then n Î Q
T (n = 2) T (n Î Q) T (Q n = 2 is rational)
T (n = 5) F (Impossible) F (Q Every natural number is
rational)
æ 3ö æ3 ö T (Q $ number which is not
F çn = ÷ T ç Î Q÷
è 2ø è2 ø natural but it is rational)
F (n = 2 ) F ( 2 ÏQ) T (Q $ number which is
neither natural nor rational)
Note : If p then q
p only if q
q if p
p is sufficient for q
q is necessary for p
Above all are equivalent to p ® q
2.4.7 Biconditional (If and only if)
If p and q are two statements, then the compound statement "p if and only if q" is
called a biconditional statement. It is denoted by P ® or p « q.
¬
The biconditional statement is also read as
"p if and only if q" or "p iff q" or "p implies and implied by q" or "p is necessary and
sufficient for q".
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If p and q have the same truth value then p « q is true. In other cases p « is false.
The truth table for p « q is as follows
p q p«q p®q q ®p (p ® q) Ù (q ® p)
T T T T T T
T F F F T F
F T p T F F
F F T T T T
By above table
p « q and ( p ® q) Ù (q ® p) are equivalent.
Examples
1) Let p : x = 4 q : x + 9 = 13
\ p «q : x = 4 if and only if x + 9 = 13
2) 5 + 6 = 11 iff 11 + 6 = 17
3) We know that an integer n is an even iff n is divisible by 2.
p : n is even q : n is divisible by 2 p « q : n is even iff n is
divisible by 2
T (n = 10) T (2 divides 10) T (Q n = 10 is even and divisible by 2)
T (n = 10) F (Impossible) F (Q There does not exist even integer which is not
divisible by 2)
F (n = 5) T (Impossible) F (Q There does not exist odd integer which is
divisible by 2)
F (n = 5) F (5 is not divisible by 2) T (Q Integer is not even iff it is not divisible by 2)
2.4.8 Special Propositions
If p ® q is a conditional statement, then
i) q ® p is called its converse
ii) ~ p ® ~ q is called its inverse
iii) ~ q ® ~ p is called its contrapositive
The truth table of these propositions are as follows
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p q p®q q®p ~p ~q ~ p®~ q ~q®~p
T T T T F F T T
T F F T F T T F
F T T F T F F T
F F T T T T T T
Example 2.4.3 Let p denote the statement, "The material is interesting'. q denote the
statement, "The exercises are challenging", and r denote the statement, "The course is
enjoyable".
Write the following statements in symbolic form :
i) The material is interesting and exercises are challenging.
ii) The material is interesting means the exercises are challenging and conversely.
iii) Either the material is interesting or the exercises are not challenging but not both.
iv) If the material is not interesting and exercises are not challenging, then the course is
not enjoyable.
v) The material is uninteresting, the exercises are not challenging and the course is not
enjoyable. SPPU : Dec.-06, May-08, Marks 6
Solution :
i) pÙq
ii) (p ® q) Ù (q ® p)
iii) p Å ~ q
iv) (~ p Ù ~ q) ® ~ r
v) ~ p Ù ~ q Ù ~ r
Example 2.4.4 Express following statement in propositional form :
i) There are many clouds in the sky but it did not rain.
ii) I will get first class if and only if I study well and score above 80 in mathematics.
iii) Computers are cheap but softwares are costly.
iv) It is very hot and humid or Ramesh is having heart problem.
v) In small restaurants the food is good and service is poor.
vi) If I finish my submission before 5.00 in the evening and it is not very hot I will go
and play a game of hockey. SPPU : May-05, Marks 6
Solution :
i) p : There are many clouds in the sky
q : It rain
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\ p Ù~q
ii) p : I will get first class
q : I study well
r : Score above 80 in mathematics
\ p « (q Ù r)
iii) p : Computers are cheap
q : Softwares are costly
\ p Ùq
iv) p : It is very hot
q : It is very humid
r : Ramesh is having heart problem
\ (p Ù q) Ú r
v) p : In small restaurant food in good
q : Service is poor
\ p Ùq
vi) p : I finish my submission before 5:00 p.m.
q : It is very hot
r : I will go
s : I will play a game of hockey
\ (p Ù ~ q) ® (r Ù s)
Example 2.4.5 Use p : I will study discrete mathematics, q : I will go to a movie, r : I am in
a good mood. Write the following in English sentence :
a) ~ r ® q b) ~ q Ù p c) q ® ~ p d) ~ p ® ~ r SPPU : Dec.-18, Marks 04
Solution :
a) ~ r ® q means, If I am not in a good mood, then I will go to a movie
b) ~ q Ù p means, I will not go to movie and I will study discrete mathematics.
c) q ® ~ p means, If I will go to a movie then I will not study discrete maths.
d) ~ p ® ~ r means, If I will not study discrete mathematics then I am not in a good
mood.
Example 2.4.6 Express the contrapositive, converse and inverse forms of the following
p 1
statement if 3 < b and 1 + 1 = 2, then sin = .
3 2 SPPU : May-07, Marks 6
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Solution :
p : 3<b
q : 1+1=2
p 1
r : sin =
3 2
Symbolic form : p Ù q ® r
Contrapositive : (~ r ® ~ (p Ù q))
i.e. ~ r ® (~ p Ú ~ q)
p 1
i.e. if sin ¹ then 3 ³ b or 1 + 1 ¹ 2
3 2
Converse : r ® (p Ù q)
p 1
i.e. if sin = then 3 < b and 1 + 1 = 2
3 2
Inverse : ~( p Ù q) ® ~ r
i.e. (~ pÚ ~ q) ® ~ r
p 1
i.e. if 3 ³ b or 1 + 1 ¹ 2 then sin ¹
3 2
Example 2.4.7 Express the contrapositive, converse, inverse and negation forms of the
conditional statement given below.
'If x is rational, then x is real'. SPPU : Dec.-06, Marks 4
Solution :
Let p : x is rational
q : is real
Symbolic form : p ® q
Contrapositive : (~ q ® ~ p)
If x is not real, then x is not rational
Converse : (q ® p)
If x is real then x is rational
Inverse - (~ p ® q)
If x is not rational, then x is not real
Negation : ~ ( p ® q)
º ~ ( p Ú ~ q)
º ~ pÙ~~q
º ~p Ù q
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2.5 Propositional or Statement Formula
A statement formula contains one or more simple statements and some logical
connectives.
A statement formula is a string consisting of variables, parentheses and logical
connective symbols.
It is called a well formed if it can be generated by the following rules :
i) A statement variable p standing alone is a well formed formula.
ii) If p is well formed formula then ~ p is also well formed formula.
iii) If p and q are well formed formulas then p Ù q, p Ú q, p ® q or q ® p and p « q are
well formed formulas.
iv) A string of symbols is well formed formula iff it is obtained by finitely many
applications of rules i), ii) and iii)
\ A statement formula is not a statement and has no truth values. But if put definite
statements in place of variables in a given formula we get a statement. It's truth value
depends upon the truth values of variables.
2.6 Tautology
A statement formula that is true for all possible values of it's propositional variables
is called a Tautology.
e.g. p Ú ~ p is a tautology
2.7 Contradiction
A statement formula that is always false for all possible values of variables is called a
contradiction or absurdity.
Example : p Ù ~ p is a contradiction
2.8 Contingency SPPU : Dec.-10, 12
A statement formula which is neither tautology nor contradiction is called a
contingency.
Example 2.8.1 Prove that p ® p is a tautology.
Solution : We construct truth table for p ® p
p p p®p
T T T
F F T
As p ® p is always true. Hence p ® p is a tautology.
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Example 2.8.2 Prove that p Ú ~ p is a tautology and ~ (p Ú ~ p) is a contradiction.
Solution : Let us construct truth table for the statement ~ (p Ú ~ p)
p ~p p Ú~ p ~ (p Ú ~ p)
T F T F
F T T F
As p Ú ~ p is always true. Hence p Ú ~ p is a tautology and ~ ( p Ú ~ p) is always false.
Hence ~ ( p Ú ~ p) is a contradiction.
Example 2.8.3 Show that p Ù ~ p is a contradiction and ~ ( p Ù ~ p) is tautology.
Solution : We construct truth table for ~ ( p Ù ~ p)
p ~p p Ù~ p ~ (p Ù ~ p)
T F F T
F T F T
As p Ù ~ p is always false. Hence p Ù ~ p is a contradiction. As ~ (p Ù ~ p) is always
true, Hence ~ (p Ù ~ p) is a tautology.
Example 2.8.4 Determine whether each of the following statement formula is a tautology,
contradiction or contingency.
i) ( p Ù q) Ù ~ ( p Ú q)
ii) ( p ® q) « ( q Ú ~ p)
iii) ( p ® ( q ® r)) ® (( p ® q) ® ( p ® r))
iv) [( p ® q) Ù ( q ® r)] ® ( p ® r) SPPU : Dec.-12, Marks 6
Solution : i) Consider the truth table
p q pÙq pÚq ~ (p Úq) (p Ùq) Ù ~ (p Ú 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
Hence ( p Ù q) Ù ~ ( p Ú q) is a contradiction
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ii) Consider the truth table
p q ~p p®q qÚ~ p (p ® q) « (q Ú ~ p)
T T F T T T
T F F F F T
F T T T T T
F F T T T T
As ( p ® q) « (q Ú ~ p) is always true. Hence it is a tautology.
iii) Consider the truth table
p q r p ® q p ® r q ® r p ® (q ® r) (p ® q) ® (p ® r) [ p ® (q ® r) ] ®
[ (p ® q) ® (p ® r) ]
T T T T T T T T T
T T F T F F F F T
T F T F T T T T T
T F F F F T T T T
F T T T T T T T T
F T F T T F T T T
F F T T T T T T T
F F F T T T T T T
Hence [ ( p ® (q ® r) ] ® [ ( p ® q) ® ( p ® r) ] is a tautology.
iv) Consider the truth table
p q r p®q q ®r (p ® q) Ù (q ® r) p®r [ (p ® q) Ù(q ® r) ]
® (p ® r)
T T T T T T T T
T T F T F F F T
T F T F T F T T
T F F F T F F T
F T T T T T T T
F T F T F F T T
F F T T T T T T
F F F T T T T T
Hence given statement formula is a tautology.
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Example 2.8.5 Show that ( p ® q) Ù ~ q ® ~ p is a tautology without using truth table.
Solution : We know that p ® q is true if p is true and q is also true.
\ We need only to show that p ® q and ~ q both are true imply ~ p is true.
As the truth value of ~ q is T, the truth value of q is F. And as p ® q is true, this
means that p is false (Q F ® F is true)
\ The truth value of p is T. Hence the proof.
Example 2.8.6 Prove that [( p ® q) Ù ( r ® s) Ù ( p Ú r)] ® ( q Ú s) is a tautology.
SPPU : Dec.-10, Marks 4
Solution : Consider truth table
p q r s p®q r ®s pÚr (p ® q) Ù (r ® s) Ù (p Úr) q Ús I ® (q Ús)
(I)
T T T T T T T T T T
T T T F T F T F T T
T T F T T T T T T T
T T F F T T T T T T
T F T T F T T F T T
T F T F F F T F F T
T F F T F T T F T T
T F F F F T T F F T
F T T T T T T T T T
F T T F T F T F T T
F T F T T T F F T T
F T F F T T F F T T
F F T T T T T T T T
F F T F T F T F F T
F F F T T T F F T T
F F F F T T F F F T
Hence given statement formula is a tautology.
Example 2.8.7 Prove by truth table p ® (Q Ú R) º ( P ® Q) Ú ( P ® R). SPPU : Dec.-12
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Solution : Consider the truth table
I II
P Q R QÚR P ® (Q Ú R) P®Q P ® R (P ® Q) Ú (P ® R)
T T T T T T T T
T T F T T T F T
T F T T T F T T
T F F F F F F F
F T T T T T T T
F T F T T T T T
F F T T T T T T
F F F F T T T T
From truth table
p ® (Q Ú R) º ( P ® Q) Ú ( P ® R)
2.9 Precedence Rule
The order of preference in which the connectives are applied in a formula of
propositions that has no bracket is
i) ~ ii) Ù iii) Ú and Å iv) ® and «
Remark :
1) 'Ú' includes 'exclusive or' and 'inclusive or'
2) If Ú and Å are present in a statement then first apply the left most one. The same
rule is applicable to ® and «
2.10 Logical Equivalence SPPU : Dec.-12
In real life, we come across several similar things with respect to different aspects.
e.g. Two cars are similar with respect to average, two students are similar with respect
to F.E. marks.
Likewise in logic, we can say that two propositions are similar with respect to their
truth values.
Definition : Two propositions A and B are logically equivalent iff they have the same
truth value for all choices of the truth values of simple propositions involved in it.
Two propositions (formulas) are equivalent even if they have different variables. Two
statement formulas P and Q are equivalent
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Iff P « Q is a tautology.
If p is equivalent to Q then it can be represented as P Û Q or P º Q.
The symbol Û is not a connective
Example 2.10.1 Prove by constructing the truth table p ® ( q Ú r) º ( p ® q) Ú ( p ® r) .
SPPU : Dec.-12, Marks 3
Solution : Consider truth table
p q r q Úr p ® (q Ú r) p®q p®r (p ® q) Ú (p ® r)
T T T T T T T T
T T F T T T F T
T F T T T F T T
T F F F F F F F
F T T T T T T T
F T F T T T T T
F F T T T T T T
F F F F T T T T
In the columns of p ® (q Ú r) and (p ® q) Ú ( p ® r), truth values are same for all
possible choices of truth values of p, q and r. Hence
p ® (q Ú r) º ( p ® q) Ú ( p ® r )
Example 2.10.2 Prove that
p « q º ( p ® q) Ù ( q ® p) º (~ p Ú q) Ù (~ q Ú p).
Solution : Consider the truth table
p q ~p ~q p®q q ® p (p ® q) p « q ~ p Ú q ~ q Ú p (~ p Ú q) Ù (~ q Úp)
Ù (q ® p)
T T F F T T T T T T T
T F F T F T F F F T F
F T T F T F F F T F F
F F T T T T T T T T T
From above table
( p « q) º (p ® q) Ù (q ® p) º (~ p Ú q) Ù (~ q Ú p)
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2.11 Logical Identities
Sr. No. Name of identity Identity
1. Idempotence of Ú p º pÚp
2. Idempotence of Ù p º pÙp
3. Commutativity of Ú pÚq º qÚp
4. Commutativity of Ù pÙq º qÙp
5. Associativity of Ú p Ú ( q Ú r) º ( p Ú q) Ú r
6. Associativity of Ù p Ù ( q Ù r) º ( p Ù q) Ù r
7. Distributivity of Ù over Ú p Ù ( q Ú r) º ( p Ù q) Ú ( p Ù r)
8. Distributivity of Ú over Ù p Ú ( q Ù r) º ( p Ú q) Ù ( p Ú r)
9. Double negation p º ~ (~ p)
10. De Morgan's laws ~ ( p Ú q) º ~ p Ù ~ q
11. De Morgan's laws ~ ( p Ù p) º ~ p Ù ~ q
12. Tautology p Ú ~ q º Tautology
13. Contradiction p Ù ~ p º Contradiction
14. Absorption laws p Ú (p Ù q) º p
15. Absorption laws p Ù (p Ú q) º p
Example 2.11.1 De Morgan's laws
i) ~ ( p Ú q) º ~ p Ù ~ q
ii) ~ ( p Ù q) º ~ p Ú ~ q
Solution : i) ~ (p Ú q) º ~ p Ù ~ q
Consider the truth table
p q ~p ~q pÚq ~ (p Ú q) ~ pÙ~ q
T T F F T F F
T F F T T F F
F T T F T F F
F F T T F T T
From the table, truth values of ~ ( p Ú q) and ~ pÙ ~ q are same for each choice of
p and q
Hence ~ ( p Ú q) º ~ p Ù ~ q
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ii) ~ ( p Ú q) º ~ p Ú ~ q
Consider the table
p q ~p ~q pÙq ~ (p Ù q) ~ pÚ~ q
T T F F T F F
T F F T F T T
F T T F F T T
F F T T F T T
From the table, truth values of ~ ( p Ù q) and (~ pÚ ~ q) are same for each choice of p
and q.
Hence ~ ( p Ù q) º (~ p Ú ~ q)
Example 2.11.2 Absorption laws
i) p Ú ( p Ù q) º p ii) p Ù ( p Ú q) º p
Solution :
i) p Ú (p Ù q) º p
Consider the table
p q pÙq p Ú (p Ù q) p
T T T T T
T F F T T
F T F F F
F F F F F
From the table, in last two columns truth tables of p Ú (p Ù q) and p are same for each
choice of p and q.
Hence p Ú (p Ù q) º p
ii) p Ù (p Ú q) º p
Consider the table
p q pÚq p Ù (p Ú q) p
T T T T T
T F T T T
F T T F F
F F F F F
From the table, last two columns are identical hence p Ù (p Ú q) º p
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Example 2.11.3 p ® q and ~ p Ú q are logically equivalent.
Solution : Consider the table
p q ~p ~ pÚq p®q
T T F T T
T F F F F
F T T T T
F F T T T
Hence ~ p Ú q and p ® q are logically equivalent.
2.12 The Duality Principle
Two statement formulas p and p* are said to be duals of each other if either one can
be obtained from the other by replacing Ù , Ú , T, F by Ú , Ù , F, T respectively.
Example 2.12.1 Write the duals of
i) ~ ( p Ú q) ii) p Ù ( q Ú r) iii) p Ú T
Solution : The duals are i) ~ (p Ù q) ii) p Ú (q Ù r) iii) p Ù F
2.13 Logical Implication
A proposition p is said to logically imply a proposition Q iff one of the following
condition holds :
1) ~ p Ú Q is a Tautology
2) p Ù Q is a Contradiction
3) p ® Q is a Tautology
It is denoted by p Þ Q
e.g. 1) (p Ù q) Ù ~ (q Ú p) is a contradiction
Hence p Ù q Þ q Ú p
The relation p Þ Q is reflexive, antisymmetric and transitive
Note : The symbols ® & Þ are not same
Þ is not connective
2.14 Important Connectives
I) NAND :
The word NAND is a combination of "NOT" and "AND" where "NOT" stands for
negation and "AND" stands for the conjunction. It is denoted by the symbol .
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If P and Q are two formulas then NAND connective is defined as
P Q « ~ (P Ù Q)
The connective has the following equivalence
1) P P « ~ (P Ù P) « ~ P Ú ~ P Û ~ P
\ P P º ~ P
2) ( P Q) ( P Q) « ~ ( P Q) « ~ ~ ( P Ù Q) « P Ù Q
3) ( P P) (Q Q) « ~ P ~ Q « ~(~ P Ù ~ Q) « P Ú Q
Note : NAND connective is commutative but not associative
II) NOR :
The connective "NOR" is a combination of "NOT" and "OR" where "NOT" stands for
negation and "OR" stands for the disjunction. It is denoted by the symbol ¯ and defined
as
P ¯ Q « ~ (P Ú Q) \ P ¯ Q º ~ (P Ú Q)
The connective ¯ has the following equivalences
i) P ¯ P « ~ (P Ú P) « ~ P Ù ~ P « ~ P
ii) (P ¯ Q) ¯ (P ¯ Q) « ~ ( P ¯ Q) « P Ú Q
iii) (P ¯ P) ¯ (Q ¯ Q) « ~ P ¯ ~ Q « ~ (~ P Ú ~ Q) « P Ù Q
iv) The connective ¯ is commutative but not associative
2.15 Normal Forms SPPU : Dec.-04, 06, 08, 10, 12, 14, May-17
If a given statement formula involves n atomic variables, then we have 2 possible n
combinations of truth values of statements replacing the variables. The construction of
the truth table involves finite number of steps but it may not be practical if number of
variables are more. Therefore we reduce the given statement formula to normal form
and find whether a given statement formula is a tautology or contradiction or at least
satisfiable.
A formula which is a conjunction or product of the variables and their p Ù ~ q
negations is called an elementary product. If p and q are statements then p, ~ p, ~ p Ù q,
p Ù ~ q, ~ p Ù ~ q are some examples of elementary products or fundamental conjunctions.
A formula which is a disjunction or sum of the variables and their negations is called an
elementary sum or fundamental disjunctions. p, ~ p, ~ p Ú q, p Ú ~ q, ~ p Ú p Ú ~ q are some
examples of an elementary sum or fundamental disjunctions.
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2.15.1 Disjunctive Normal Form (dnf)
A statement formula which consists of a disjunction ( Ú) of fundamental conjunctions
or elementary product ( Ù). It is abbreviated as dnf.
A disjunctive normal form of a given formula is constructed as follows :
1) Replace ' ®' or ' «' by using logical connectives Ù , Ú & ~ . p ® q º ~ p Ú q,
p « q º (~ p Ú q) Ù (~ q Ú p)
2) Use De Morgan's laws to eliminate '~' before sums or products.
3) Apply distributive laws repeatedly and eliminate product of variables to obtain
the required normal form.
Examples :
1) (p Ù q) Ú (q) Ú (~ q Ù p)
2) (p Ù q Ù r) Ú (p Ù r) Ú (p Ù q) Ú (p Ù r)
3) ( p Ù r ) Ú ( p Ù q)
4) (~ p Ù r ) Ú (~ q Ù r ) Ú (~ r )
All above examples are in disjunctive normal form.
2.15.2 Conjunctive Normal Form (cnf)
A statement formula which consists of a conjunction of the fundamental disjunctions
( Ú). It is denoted as cnf.
Examples :
1) p Ù (p Ú q)
2) ( p Ú q) Ù (~ p Ú q) Ù (~ q)
3) ( p Ú q Ú r ) Ù (~ p Ú r ) Ù ( p Ú ~ q Ú r )
All above examples are in conjunctive normal form.
2.15.3 Principal Normal Form
Let p and q be two statement variables, then p Ù q, p Ù ~ q, ~ p Ù q, ~ p Ù ~ q are called
minterms of p and q. They are also called Boolean conjunctives of p and q. The number
of minterms with n variables is 2 n . None of the minterms should contain both a variable
and it's negation. \ p Ù ~ p is not minterm.
The dual of minterm is called a maxterm.
\ For two statement variables p and q, maxterms are p Ú q, ~ p Ú q, p Ú ~ q and ~ p Ú ~ q.
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I) Principal Disjunctive Normal Form
A statement formula which consists of a disjunction of minterms only is called the
principal disjunctive normal form.
e.g.
1) ( p Ú q) Ù (~ p Ù ~ q), ( p Ù q Ù r ) Ú (~ p Ù q Ù r ) are principal cnf.
2) p Ú (p Ù q), ( p Ù q) Ú (~ p Ù q) Ú (~ q), ( p Ù q Ù r ) Ú ( p Ù q) are not principal dnf.
II) Principal Conjuctive Normal Form
A statement formula which consists of a conjunction of maxterms only is called the
principle conjunctive normal form.
e.g.
1) ( p Ú q) Ù (~ p Ú q), ( p Ú ~ q) Ù (~ pÚ ~ q) are principal cnf.
2) (p Ú q) Ù (~ p) is cnf but not principal cnf.
Example 2.15.1 Obtain the conjunctive normal form and disjunctive normal form of the
following formulae given below :
i) p Ù ( p ® q ) ii) ~ (p Ú q) ® (p Ù q)
¬ SPPU : Dec.-12, Marks 4
Solution :
i) p Ù (p ® q) º p Ù (~ p Ú q) ~ cnf
º (p Ù ~ p) Ú ( p Ù q)
º F Ú (p Ú q)
º (p Ù q) … Definition a single conjunctive
ii) ~ (p Ú q) ®
¬ (p Ù q) º ( ~ ~ (p Ú q) Ú ( p Ù q)) Ù ( ~ (p Ù q) Ú ~ (p Ú q))
º ((p Ú q) Ú ( p Ù q) Ù ((~ p Ú ~ q) Ú (~ p Ù ~ q)
º (p Ú q) Ù ((~ p Ú ~ q Ú ~ p) Ù (~ p Ú ~ q Ú ~ q))
º (p Ú q) Ù (~ p Ú ~ q) Ù (~ p Ú ~ q)
º (p Ú q) Ù (~ p Ú ~ q) … cnf
Further
(p Ú q) Ù (~ p Ú ~ q) º ((p Ú q) Ù ~ p) Ú ((p Ú q) Ù ~ q)
º (p Ù ~ q) Ú (q Ù ~ p) Ú ((p Ù ~ q) Ú (q Ù ~ q)
º F Ú (q Ù ~ p) Ú ( p Ù ~ q) Ú F
º (p Ù ~ p) Ú ( p Ù ~ q) … Definition
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Example 2.15.2 Find the conjunctive normal form and disjunctive normal form for the
following :
i) ( p Ú q) ® q ii) p « ( p Ú q) SPPU : Dec.-06, 14, May-17, Marks 3
Solution :
i) ( p Ú q) ® q º ( p Ú q) Ú q
º ( p Ù q) Ú q
º (p Ù q) Ú q … Definition
(p Ù q) Ú q º ( p Ú q) Ù (q Ú q)
º (~ p Ú q) Ù q … cnf
p ´ (p Ú q) º (p Ú ( p Ú q)) Ù ((p Ú q) Ú p)
º (p Ú p Ú q) Ù ((p Ù q) Ú p)
º (p Ú q) Ù ( p Ú p) Ù (q Ú p)
º (p Ú q) Ù p Ù (q Ú p) … cnf
º ((p Ù p) Ú (q Ù p)) Ù (q Ú p)
º (F Ú (q Ù p)) Ù (q Ú p)
º (q Ù p) Ù (q Ú p)
º (q Ù p Ù q) Ú (q Ù p Ù p)
º (F Ù p) Ú (q Ù p)
º F Ú (q Ù p)
º (q Ù p) … Definition
Example 2.15.3 Find the conjunctive and disjunctive normal forms for the following without
using truth table
i) ( p ® q) Ù ( q ® p) ii) (( p Ù ( p ® q)) ® q) SPPU : Dec.-04, 10, 14, Marks 4
Solution :
i) (p ® q) Ù (q ® p) º (~ p Ú q) Ù (~ q Ú p)
Further, using the distributive law on the above cnf we have
((~ p Ú q) Ù ~ q) Ú ((~ p Ú q) Ù p) = (~ p Ù ~ q) Ú (q Ù ~ q) Ú (~ p Ù p) Ú (q Ù p)
= (~ p Ù ~ q) Ú (q Ù p) … dnf
(Q p Ù ~ p º q Ù ~ q)
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ii) ( p Ù (p ® q)) ® q º ~ ( p Ù ( ~ p Ú q)) Ú q
º ~ p Ú ~ (~ p Ú q) Ú q
º ~ p Ú (p Ù ~ q) Ú q
Example 2.15.4 Obtain cnf of each of the following
i) p Ù ( p ® q ) ii) q Ú ( p Ù ~ q ) Ú ( ~ p Ù ~ q ) SPPU : Dec.-08, Marks 4
Solution :
i) p Ù (p ® q) º ( p Ù (~ p Ú q))
º ( p Ù ~ p) Ú ( p Ù q)
º Ú ( p Ù q)
º ( p Ù q) … cnf
ii) q Ú (p Ù ~ q) Ú (~ p Ù ~ q) º ((q Ú p) Ù (q Ú ~ q) Ú (~ p Ú ~ q)
º (q Ú p) Ù T Ú ( ~ p Ù ~ q)
º (q Ú p) Ú (~ p Ù ~ q)
º (q Ú p Ú ~ p) Ù (q Ù p Ú ~ q)
º (q Ú T) Ù ( p Ú q Ú ~ q)
º T Ù (p Ú T) º T Ù T
º T º (pÚ ~ p) which is the required cnf (single disjunct)
Example 2.15.5 Find DNF of (( p ® q) Ù ( q ® p)) Ú p
SPPU : Dec.-14, Marks 4, May-17, Marks 3
Solution :
((p ® q) Ù (q ® p) Ú p º [(~ p Ú q) Ù (~ q Ú p)] Ú p (Q p ® q º – ~ p Ú q)
º [p Ú ( ~ p Ú q)] Ù [p Ú ( ~ q Ú p)]
º [(p Ú ~ p) Ú q] Ù [p Ú p Ú ~ p] (Q p Ú p = p
p Ú ~ p = T)
º (T Ú q) Ù ( p Ú ~ q)
º T Ù ( p Ú ~ q)
º pÚ ~q which is the required DNF
Example 2.15.6 Obtain the cnf : ( ~ p Ù q Ù r ) Ú ( p Ù q ) SPPU : Dec.-12, Marks 4
Solution :
(~ p Ù q Ù r ) Ú ( p Ù q) º ( p Ù q) Ú (~ p Ù q Ù r )
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º ( p Ú (~ p Ù q Ù r )] Ù [q Ú (~ p Ù q Ù r )]
º [( p Ú ~ p) Ù ( p Ú q)] Ù [ p Ú r)] Ù [ (q Ú ~ p) Ù (q Ú q) Ù (q Ú r )]
º [T Ù ( p Ú q) Ù ( p Ú r )] Ù [(q Ú ~ p) Ù q Ù (q Ú r)]
º ( p Ú q) Ù ( p Ú r ) Ù (q Ú ~ p) Ù q which is the required cnf
Example 2.15.7 Find dnf by using truth table
( p « ( q Ú r)) ® ~ p
Solution : Consider the truth table
p q r ~p q Úr p « (q Ú r) [p « (q Ú r)] ® ~ p
T T T F T T F
T T F F T T F
T F T F T T F
T F F F F F T ¬ 1
F T T T T F T ¬ 2
F T F T T F T ¬ 3
F F T T T F T ¬ 4
F F F T F T T ¬ 5
Consider only 'T' from last column and choose corresponding values of 'T' from p, q
and r. For the first marked row 1 corresponding p is T, q is F and r is F. So take
p Ù ~ q Ù ~ r or p Ù q ¢ Ù r ¢
For 2 nd T ® ~ p Ù q Ù r, 3 rd T ® ~ p Ù q Ù ~ r,
4 th T ® ~ p Ù q Ù r, and 5 th T ® ~ pÙ ~ q Ù ~ r
Hence the logically equivalent dnf form is
[ p « (q Ú r)] ® ~ p º (p Ù ~ q Ù r) Ú (~ p Ù q Ù r) Ú (~ p Ù q Ù ~ r)
Ú (~ p Ù ~ q Ù r) Ú (~ p Ù ~ q Ù ~ r)
Example 2.15.8 Obtain the principal dnf of (~ p Ú ~ q) ® (~ p Ù r)
Solution :
(~ p Ú ~ q) ® (~ p Ù ~ r) º ~ (~ p Ú ~ q) Ú (~ p Ù r)
º (~ (~ p) Ù ~ (~ q)] Ú (~ p Ù r)
º (p Ù q) Ú (~ p Ù r) which is dnf
º (p Ù q Ù (r Ú ~ r)) Ú (~ p Ù r Ù (q Ú ~ q))
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º (p Ù q Ù r) Ú (p Ù q Ù ~ r) Ú (~ p Ù r Ù q) Ú (~ p Ù r Ù ~ q)
which is the required principal dnf.
Example 2.15.9 Find the principal cnf of ( p Ù q) Ú (~ p Ù r)
Solution :
[( p Ù q) Ú ( ~ p Ù r) º [(p Ù q) Ú ~ p] Ù [ (p Ù q) Ú r]
º (p Ú ~ p) Ù (q Ú ~ p) Ù (p Ú r ) Ù ( q Ú r)
º (q Ú ~ p Ú ( r Ù ~ r ) Ù (p Ú r Ú ( q Ù ~ q)) Ù (q Ú r Ú ( p Ù ~ p) )
º (q Ú ~ p Ú r ) Ù (q Ú ~ p Ú ~ r ) Ù (p Ú r Ú q) Ù (p Ú r Ú ~ q) )
Ù (q Ú r Ú p) Ù (q Ú r Ú ~ p)
º (p Ú q Ú r ) Ù (~ p Ú q Ú r ) Ù (~ p Ú q Ú ~ r) Ù (p Ú ~ q Ú r)
which is the required principle cnf.
Example 2.15.10 Find the principal dnf of ~ p Ú q
Solution :
~ p Ú q º [~ p Ù (q Ú ~ q)] Ú [q Ù ( p Ú ~ p)]
º (~ p Ù q) Ú (~ p Ù ~ q) Ú (q Ù p) Ú (q Ù ~ p)
º (~ p Ù q) Ú (~ p Ù ~ q) Ú (p Ù q)
which is the required Principle dnf.
2.16 Methods of Proof SPPU : Dec.-09, May-06, 10
We have learned statements or propositions and their truth values. Now, we will
discuss ways by which statements can be linked to form a logically valid argument.
Whenever an assertion is made, which is claimed to be true, one has to state an
argument which produces the truth of the assertion. To construct a proof we need to
derive new assertions from existing ones by different ways. This is also done by using
valid of inference.
Valid argument : A valid argument is a finite sequence of statements p1 , p 2 , .... p n
called as premises (or assumptions or hypothesis) together with a statement C, called the
conclusion such that p1 Ù p 2 Ù p 3 Ù ... Ù p n ® C is a tautology.
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As mathematical proof is a logical argument that verifies the truth of the theorem.
There are several ways of proving a theorem which are based on one or more rules of
inference.
There are following most commonly used rules of inference.
2.16.1 Law of Detachment (or Modus Ponens)
Whenever the statements p and p q are accepted as true, then we must accept the
statement q as true. This rule is represented in the following form
pq
p
q
The assertions above the horizontal line are called premises OT hypothesis. And the
assertion below the line is called the conclusion.
This rule constitutes a valid argument as (p q) p q is a tautology.
The truth table is as follows
p q pq (p q) p [(p q) p] q
T T T T T
T F F F T
F T T F T
F F T F T
This form of valid argument is called the law of detachment as conclusion q is
detached from premise p q and p.
It is also called as the law of direct inference.
Example
If Sushma gets a first class with distinction in B.E. then she will get a good job easily.
Let p : Sushma gets a first class with distinction in B.E.
q : She will get a good job easily.
pq
The inference rule is p
q
Hence this form of argument is valid.
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2.16.2 Modus Tollen (Law of Contrapositive)
Modus Tollen is a rule of denying. It can be stated as "If p q is true and q is false,
then p is false. This is represented in the following form.
pq
~p
~q
Above argument is valid as ( p q) ~ q ~ p is a tautology. In above example is
b q and ~ q : Sushma will not get a good job easily then ~ p : She has not a first
class with distinction in B.E.
2.16.3 Disjunctive Syllogism
This rule states that "If p q is true and p is false then q is true.
It is represented in the following form as
pq
~p
q
This argument is valid as (p q) ~ p q is a tautology.
2.16.4 Hypothetical Syllogism
It is also known as the transitive rule.
It can be stated as follows
"If ( p q) and (q r ) are true then p r is true.
This rule is presented in the following form.
pq
qr
p r
This argument is valid as ( p q) (q r ) ( p r ) is a tautology.
Example 2.16.1 Determine whether the argument is valid or not. If I try hard and I have
talent then I will become a musician. If I become a musician, then I will be happy.
Therefore if I will not be happy then I did not try hard or I do not have talent.
SPPU : May-10, Marks 4
Solution :
Let p : I try hard
q : I have talent
r : I will become musician
s : I will be happy
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It's symbolic form is as follows
s 1 : ( p q) r
s2 : r s
s 3 : ~s ~ p ~ q
To check validity of this statement, one way is to use truth table or prove logically.
Suppose assignment is invalid. This means that for some assignment of truth values s 1
is T, s 2 is T but s 3 is F. s 3 will have truth value F if ~ s is T and ~ p ~ q is F
i.e. s is F and ~ p is F and ~ q is F
i.e. s is F, and p is T and q is T
As s 2 is T, the truth values of r and s both are F. Since s, is T, r is F implies either p
or q is F. This is contradiction since by assumption both p and q are true.
Hence given statement is valid.
Example 2.16.2 Test the validity of the argument. If a person is poor, he is unhappy. If a person
is unhappy, he dies young. Therefore poor person dies young. SPPU : Dec.-09, Marks 4
Solution :
Let p : Person is poor
q : Person is unhappy
r : Person dies young
In symbolic form argument is
s1 : p q
s2 : q r
s: p r
The above argument is the rule of hypothetical syllogism. Hence it is valid.
Example 2.16.3 Determine the validity of the following argument.
s 1 : If I like discrete structure then I will study
s 2 : Either I will study or I will fail
_________________________________________
s : If I fail then I do not like discrete structure
Solution :
Let p : I like discrete structure
q : I will study
r : I will fail
In symbolic form
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s1 : p q
s2 : q r
s : r ~ p
We know that for the validity of argument s 1 s 2 should logically imply s.
Assign the truth values T, T, T, to p, q, r respectively. Then s 1 is T and s 2 is also T
But s is T F is F
Hence argument is invalid.
Example 2.16.4 Determine the validity of the argument
s 1 : All my friends are musicians
s 2 : John is my friend
s 3 : None of my neighbours are musicians
____________________________________
S : John is not my neighbour
SPPU : May-06
Solution :
Let p : All my friends are musicians
q : John is my friend
r : My neighbours are musicians
s : John is my neighbour
s1 : p
s2 : q
s3 : ~ r
In symbolic form
p
q
~r
~s
As all my friends are musicians and John is my friend John is musician.
p q John is musician
p q ~ r John is musician and my neighbours are not musicians
p q ~ r John is not my neighbour
p q ~ r ~ s is true.
Therefore given argument is valid.
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Example 2.16.5 I am happy if my program runs. A necessary condition for the program to
run is it should be error free. I am not happy. Therefore the program is not error free.
Solution :
Let p : I am happy
q : My program runs
r : Program should be error free
In symbolic form
qp
qr
~p
~r
To check validity, consider the truth table
p q r q p q r ~p (q p) (q r) ~ p (I) ~ r
(I)
F F T T T T T F
Last column is not T
Given argument is not tautology.
Given argument is invalid.
2.17 Quantifiers SPPU : Dec.-08, 09,10, 11, 15, May-15
In grammar a predicate is the word in a sentence which expresses what is said about
object. i.e. properties of an object or relation among objects. For example "is a good
teacher", "is a clever student" are predicates. In logic predicate has a broaden role than
in grammar. Predicate is presented by using a variable x in place of holder. e.g. x is a
prime number.
An assertion that contains one or more variables is called a predicate. It's truth value
is predicated after assigning truth values to its variables.
A predicate p containing n variables x 1 , x 2 , ... x n is called an n-place predicate and
denoted by p (x 1 , x 2 , x 3 , ... x n ). Each variable x i is called as argument.
The values which the variables may assume constitute a collection is called the
universe of discourse or domain of discourse.
There are two types of quantifiers.
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2.17.1 Universal Quantifiers
If p(x) is a predicate with x as an argument then the universal quantifier for p(x) is
the statement.
"For all values of x, p(x) is true". We denote the phrase "For all" by
means for all or for each or for every.
If p(x) is true for all values of x then
x p(x) is true.
For example - p(x) : x 0, where x is any positive integer.
The proposition x p(x) is true.
However, if x is an integer x p(x) is false.
2.17.2 Existential Quantifier
In some situations, we only require that there is at least one value for the predicate is
true. Suppose for the predicate p(x), x p(x) is false, but there exists at least one value of
x for which p(x) is true, then we say that in this proposition, x is bound by existential
quantification.
The phrase "there exists an x" is called an existential quantifier.
The symbol " " is used to denote the logical quantifier "there exists" or "there exists
an x" or "there is x" or "for some x" or "for at least one x".
The existential quantifier for p(x) is denoted by x p(x).
2.17.3 Negation of Quantified Statement
Consider the statement x p(x). It's negation is "It is not the case that for all x, p(x) is
true". This means that for some x = a, p(a) is not true or x s.t. ~ p(x) is true.
Hence the negation of x p(x) is logically equivalent to x [~ p(x)].
Sr. No. Statement Negation
1. x p(x) x [~ p(x)]
2. x [~ p(x)] x p(x)
3. x[~ p(x)] x p(x)
4. x p(x) x[~ p(x)]
I) Equivalence involving quantifiers
1) Distributivity of over
x [p(x) Q(x)] x p(x) x Q(x)
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x [p Q(x)] p ( x Q(x))
2) Distributivity of over
x [p(x) Q(x)] x p(x) x Q(x)
x [p Q(x)] p (x Q(x))
3) x [p Q(x)] p [ x Q(x)]
4) x [p Q(x)] p [x Q(x)]
5) ~ [ xp(x)] x [~ p(x)]
6) ~ [ x p(x)] x [~ p(x)]
7) x p(x) x p(x)
8) x p(x) x Q(x) x ( p(x) Q(x)
9) x(p(x) Q(x)) x p(x) x Q(x)
II) Rules of Inference for addition and deletion of quantifiers
1) Rule 1 : Universal Instantiation
x p (x)
, k is some element of the universe
p ( k)
2) Rule 2 : Existential Instantiation
x p (x)
, k is some element for which p(k) is true.
p ( k)
3) Rule 3 : Universal Generalization
px
x p(x)
4) Rule 4 : Existential Generalization
p(k)
x p(x)
k is some element of the universe
III)
Sr. No. Quantifiers Expression
1. xy p(x, y) There exists a value of x such that for all values of y, p(x, y) is true.
2. y x p(x, y) For each value of y, there exists x such that p(x, y) is true.
3. x y p(x, y) There exist value of x and value of y such that p(x, y) is true.
4. x y p(x, y) For all values of x and y. p(x, y) is true.
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Example 2.17.1 Represent the arguments using quantifiers and find its correctness.
All students in this class understand logic. Ganesh is a student in this class. Therefore
Ganesh understands logic. SPPU : Dec.-11, Marks 4
Solution :
Let C(x) : x is a student in this class
L(x) : x understands logic
In symbolic form
x (C(x) L(x))
C(a)
L(a)
Here a means Ganesh
This is Modus Ponen
Therefore this argument is valid.
Example 2.17.2 Let p(x) : x is even
Q(x) : x is a prime number
R(x, y) : x + y is even
a) Using the information given above write the following sentences in symbolic form.
i) Every integer is an odd integer
ii) Every integer is even or prime
iii) The sum of any two integers is an odd integer. SPPU : Dec.-10, Marks 4
Solution :
i) x [~ p(x)]
ii) x [p(x) Q(x)]
iii) x y [~ R(x, y)]
Example 2.17.3 Using information in example 2.17.2 write an english sentence for each of the
symbolic statement given below
i) x (~ Q ( x ))
ii) y(~ p( y))
iii) ~ [ x ( p( x ) Q ( x ))] SPPU : Dec.-10, Marks 4
Solution :
i) All integers are not prime numbers
ii) At least one integer is not even.
iii) It is not the case that there exists an integer which is even and prime.
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Example 2.17.4 Determine the validity of the following argument
s 1 : All my friends are musician
s 2 : John is my friend
s 3 : Name of my neighbours are musician
s : John is not my neighbour.
Solution : Let the universe of discourse be the set of people.
Let F(x) : x is my friend
M(x) : x is a musician
N(x) : x is my neighbour
It's symbolic form is
s 1 : x [F(x) M(x)]
s 2 : F(a) (a = John]
s 3 : x [N(x) ~ M(x)]
s : ~ N(a)
Suppose ~N(a) has value F.
N(a) is T. Since s 3 is T, we must have ~M(a) is T or M(a) is F. But s 1 is T. Hence
we must have F(a) to be false but this is contradiction. Hence if s is false either of s 1 or
s 3 should be false. Hence argument is valid.
Example 2.17.5 For the universe of all integers. Let
p(x) : x > 0
Q(x) : x is even
R(x) : x is a perfect square
S(x) : x is divisible by 4
T(x) : x is divisible by 7
Write the following statement in symbolic form
i) At least one integer is even
ii) There exists a positive integer that is even
iii) If x is even then x is not divisible by 7
iv) No even integer is divisible by 7
v) There exists an even integer divisible by 7
vi) If x is even and x is perfect square then x is divisible by 4.
Solution :
i) x Q(x)
ii) x [ p(x) Q(x)]
iii) x [Q(x) ~ T(x)]
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iv) x [Q(x) ~ T(x)]
v) x [Q(x) T(x)]
vi) x [Q(x) R(x) s(x)]
Example 2.17.6 Rewrite the following statements using quantifier variables and predicate
symbols.
i) All birds can fly
ii) Not all birds can fly
iii) Some men are genius
iv) Some numbers are not rational
v) There is a student who likes Maths but not Hindi
vi) Each integer is either even or odd SPPU : Dec.-08, Marks 4
Solution :
i) Let B(x) : x is a bird
F(x) : x can fly
Then the statement can be written as
x [B(x) F(x)]
ii) x [B(x) ~ F(x)]
iii) Let M(x) : x is a man
G(x) : x is a genius
The statement in symbolic form as x [M(x) G(x)]
iv) Let N(x) : x is a number
R(x) : x is rational
The statement in symbolic form as x [N(x) ~ R(x)] or ~ [ x (N(x) R(x)]
v) Let S(x) : x is a student
M(x) : x likes Maths
H(x) : x likes Hindi
The statement in symbolic form as x [S(x) M(x) ~ H(x)]
vi) Let I(x) : x is an integer
E(x) : x is even
O(x) : x is odd
The statement in symbolic form as x [I(x) E(x) O(x)]
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Discrete Mathematics 2 - 38 Propositional Calculus
Example 2.17.7 Negate each of the following statements
i) x, | x | = x
ii) x, x 2 = x SPPU : Dec.-09, 15, May-15, Marks 4
Solution :
i) x, x x ii) x , x 2 x
Example 2.17.8 Negate the following
i) If there is a riot, then someone is killed.
ii) It is day light and all the people are arisen. SPPU : May-15, Dec.-15, Marks 4
Solution :
i) It is not the case that if there is a riot then someone is killed.
ii) It is not the case that it is day light and all the people are arisen.
OR
i) Let p : There is a riot
q : Someone is killed
Given statement is p q
Hence ~ ( p q) ~ (~ p q) p ~ q
There is a riot and someone is not killed.
ii) Let p : It is a day light
q : All the people are arisen
Given statement is p q
Hence ~ (p q) = ~ p ~ q
Hence either it is not a day light or all the people are not arisen.
qqq
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Unit - I
3 Mathematical Induction
Syllabus
Proof by Mathematical Induction and Strong Mathematical Induction.
Contents
3.1 Introduction
3.2 First Principle of Mathematical Induction Statement
3.3 Second Principle of Mathematical Induction Statement
(Strong Mathematical Induction) . . . . . . . . . . . Dec.-04, 05, 06, 08, 10, 11,12,
. . . . . . . . . . . . . . . . . . 13, 14, 15, 18, May-05, 06, 07, 08,
. . . . . . . . . . . . . . . . . . 14, 15, 17, 18, 19 · · · · · · · · · · Marks 6
(3 - 1)
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Discrete Mathematics 3-2 Mathematical Induction
3.1 Introduction
Mathematical induction is a powerful technique in applied mathematics especially in
number theory, where many properties of natural numbers are proved by this method.
In day to day life, we are often required to generalise a particular pattern for the
prediction purpose. The generalisation is achieved by using a statement involving a
variable as natural number.
Mathematical induction is very useful technique or tool for the programmers to check
whether a program statement is loop invariant or not.
There are two principles of mathematical induction :
1) First principle of mathematical induction.
2) Second principle of mathematical induction.
3.2 First Principle of Mathematical Induction Statement
Let P(n) be a statement involving a natural number n n 0 such that,
1) If P(n) is true for n = n 0 where n 0 N and
2) Assume that P(k) is true for k = n 0
We prove P(k+1) is also true,
Then P(n) is true for all natural numbers n = n 0 .
Step 1 is called as the basis of induction.
Step 2 is called as the induction step.
3.3 Second Principle of Mathematical Induction Statement
(Strong Mathematical Induction) SPPU : Dec.-04, 05, 06, 10, 11, 12, 13,
14, 15, 18, May-05, 06, 07, 08, 14, 15, 17, 18, 19
Let P(n) be a statement involving a natural number n n 0 such that,
1) If P(n) is true for n = n 0 where n 0 N and
2) Assume that P (n) is true for n 0 n k i.e. P(n 0 1) ( n 0 2)............ P(k) are true.
we prove that P(k + 1) is true,
Then P(n) is true for all natural numbers n n 0 .
Example 3.3.1 n( n 1)
Prove that : 1 + 2 + 3 + 4 … + n =
2
SPPU : May-17, Dec.-18, Marks 4
Solution : Let, P(n) be the given statement
Consider the following steps,
Step 1 : Basis of induction
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Discrete Mathematics 3-3 Mathematical Induction
For n = 1, L.H.S. = 1
1(1 1)
R.H.S. = =1
2
For n = 1, L.H.S. = R.H.S.
P(1) is true.
Step 2 : Assume that P(k) is true
k(k 1)
i.e. 1 + 2 + 3 + ...... + k = … (1)
2
Consider, 1 + 2 + 3 + ...... + k + (k+ 1)
k(k 1) k(k 1) + 2(k 1) (k 1) + (k 2)
= k 1 = =
2 2 2
Hence P(k + 1) is true
By the principle of mathematical induction P(n) is true for all n.
Example 3.3.2 Prove by mathematical induction for n 1.
n (n 1) (n 2)
1 2 2 3 3 4 ... n (n 1)
3 SPPU : May-05, Marks 6
Solution : Let P(n) the given statement
1. Basis of induction
For n 0 = 1 L.H.S. = 1.2 = 2
1( 2) ( 3)
R.H.S. = =2 L.H.S. = R.H.S.
3
Hence P(1) is true.
2. Induction step
Assume that, P(k) is true
k (k 1) (k 2)
i.e. 1 2 2 3 3 4 ... k (k 1) = ... (1)
3
Then we have
[1 2 2 3 3 4 ... k (k 1)]+ (k + 1) ( k + 2)
k (k 1) (k 2)
= (k 1) (k 2) … (Using 1)
3
k ( k 1) ( k +2) (k +3)
= (k + 1) (k + 2) 1 =
3 3
Hence assuming P(k) is true, P(k + 1), is also true. Therefore by mathematical
induction P(n) is true for all n 1.
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Discrete Mathematics 3-4 Mathematical Induction
Example 3.3.3 Show by induction that , n 1
n( 2n 1) ( 2n 1)
1 2 3 2 5 2 ..( 2 n 1) 2
3 SPPU : May-14, Marks 4
Solution : Let P(n) be the given statement,
1. Basis of induction :
For n = 1, L.H.S. = 1 2 = 1,
1 (1) ( 3)
R.H.S. = =1
3
L.H.S. = R.H.S.
Hence P(1) is true
2. Induction step : Assume that P(k) is true.
k (2k 1) (2k 1)
i.e. 1 2 3 2 5 2 ... (2k 1) 2 = ... (1)
3
Hence
k (2k 1) (2 k 1 )
[1 2 3 2 5 2 ... (2k 1) 2 ] (2k 1) 2 = (2k 1) 2 ... (Using 1)
3
(2k 1)
= [2k 2 k + 3 (2k 1)]
3
(2k 1)
= [2k 2 5k 3]
3
(2k 1)
= [2k 2 2k 3k 3]
3
(2k 1)
= [(2k 3) (k 1)]
3
(k 1)(2 k + 1) (2k 3)
=
3
(k 1) [2 (k 1) 1][2 (k 1) 1]
=
3
Hence assuming P(k) is true P(k + 1) is also true. Therefore by mathematical
induction P(n) in true for all n 1.
Example 3.3.4 n 2 (n + 1) 2
Show that 1 3 2 3 3 3 ... n3 = = (1 2 3 ... n) 2
4
SPPU : Dec.-12, May-18, Marks 4
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Discrete Mathematics 3-5 Mathematical Induction
Solution : Let P(n) be the given statement,
2
n(n + 1) n 2 ( n + 1) 2
1 3 2 3 3 3 ... n 3 = =
2 4
1. Basis of induction :
For n = 1, L.H.S. = 1,
1 ( 1 1) 2
R.H.S. = =1
4
L.H.S. = R.H.S.
Hence P(1) is true.
2. Induction step : Assume that P(k) is true.
k 2 (k 1) 2
i.e. 1 3 2 3 3 3 ... + k 3 = … (1)
4
Then we have
(1 3 2 3 3 3 ... + k 3 ) (k 1) 3 = (1 2 3 ... k) 2 (k 1) 3 (Using 1)
2
k (k 1)
= (k 1) 3
2
k2
= (k 1) 2 k 1
4
k 2 + 4k 4
= (k 1) 2
4
(k 1) 2 (k 2) 2
=
4
2
(k 1)(k 2) (k +1) 2 (k 2) 2
= =
2 4
Hence assuming P(k) is true, P(k + 1) is also true. Therefore by mathematical
induction P(n) is true for all n 1.
Example 3.3.5 12 22 n2 n (n 1)
Show that ...
1 3 3 5 (2n 1) (2n 1) 2(2n 1)
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Discrete Mathematics 3-6 Mathematical Induction
Solution : Let P(n) be the given statement,
1. Basis of induction : For n = 1
1 1
We have, L.H.S. = =
1.3 3
1 (2) 1
R.H.S. = =
1(3) 3
L.H.S. = R.H.S.
Hence P(1) is true.
2. Induction step : Assume that P(k) is true
12 22 k2 k (k 1)
i.e. ... = ... (1)
1 3 3 5 (2k 1) (2k + 1) 2 (2k 1)
Then we have,
12 22 k2 (k 1) 2
...
1 3 3 5 (2k 1) (2k 1) [2 (k 1) 1][2 (k 1) 1]
k (k 1) (k 1) 2
= ... (Using 1)
2 (2k 1) (2k 1) (2k 3)
(k 1) k (2k 3) 2 (k 1)
=
(2k 1) 2 (2k 3)
(k 1) 2k 2 5k 2
=
2k 1 2 (2k 3)
(k 1) 2k 2 4k k 2
=
(2k 1) 2 (2k 3)
(k 1) 2k (k 2) 1 (k 2)
=
(2k 1) 2 (2k 3)
(k 1) (k 2)
=
2 (2k 3)
(k 1) [(k 1) 1]
=
2 [2(k 1) 1]
P(k + 1) is true.
Therefore by mathematical induction P(n) is true for all n 1.
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Discrete Mathematics 3-7 Mathematical Induction
1 1 1 1 n
Example 3.3.6 Show that a) ...
1 2 2 3 3 4 n (n 1) n 1
1 1 1 1 n
b) Show that ...
1 3 3 5 5 7 ( 2n 1) ( 2n 1) 2n 1
1 1 1 1 n
c) Show that ...
1 4 4 7 7 10 ( 3n 1) ( 3n 1) 3n 1
SPPU : Dec.-05, Marks 6
Solution : Let P(n) be the given statement,
a) 1. Basis of induction :
1 1
For n = 1 L.H.S. = =
1.2 2
1 1
R.H.S. = =
1+1 2
L.H.S. = R.H.S.
Hence P(1) is true.
2. Induction step : Assume that P(k) is true.
1 1 1 1 k
i.e. ... = ... (1)
1 2 2 3 3 4 k (k 1) k 1
Then we have
1 1 1 1 1 k 1
... = ... (Using 1)
1 2 2 3 3 4 k (k 1) (k 1) (k 2) k 1 (k 1)(k 2)
k (k 2) 1
=
(k 1) (k 2)
k 2 2k 1
=
(k 1) (k 2)
(k 1) 2
=
(k 1) (k 2)
k 1 (k 1)
=
k 2 (k 1) 1
Hence assuming P(k) is true, P(k + 1) is also true. Therefore P(n) in true for all n 1.
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Discrete Mathematics 3-8 Mathematical Induction
1 1 1 1 n
b) Let P(n) : ... =
1 3 3 5 5 7 (2 n 1)(2n 1) 2n 1
1. Basis of induction : For n = 1
1 1 1
L.H.S. = = , R.H.S. =
1.3 3 3
1 1
=
3 2.1+1
L.H.S. = R.H.S.
Hence P(1) is true.
2. Induction step : Assume that P(k) is true
1 1 1 1 k
i.e. .... = ... (1)
1 3 3 5 5 7 (2k 1)(2k 1) 2k 1
Then we have,
1 1 1 1
...
1 3 3 5 (2k 1)(2k 1) (2k 1)(2k 3)
k 1
=
2k 1 (2k 1)(2k 3)
2k 2 3k 1
=
(2k 1) (2k 3)
(2k 1) (k 1)
=
(2k 1)(2k 3)
k 1
=
2k 3
k 1
=
2(k 1) 1
Hence assuming P(k) is true. P(k + 1) is also true. Therefore P(n) is true for all n 1.
1 1 1 1 n
c) Let P(n) : ... =
1 4 4 7 7 10 (3n 2)(3n 1) 3n 1
1. Basis of induction
1 1 1
For n = 1, L.H.S. = = , R.H.S. =
1.4 4 4
1 1
=
1.4 3.1+1
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Discrete Mathematics 3-9 Mathematical Induction
L.H.S. = R.H.S.
Hence P(1) is true.
2. Induction step : Assume that P(k) is true.
1 1 1 1 k
i.e. ... = ... (1)
1 4 4 7 7 10 (3k 2)(3k 1) 3k 1
Then we have,
1 1 1 1
+ ... ... (Using 1)
1 4 4 7 (3k 2)(3k 1) (3k 1) (3k 4)
k 1
=
(3k 1) (3k 1)(3k 4)
3k 2 4k 1
=
(3k 1) (3k 4)
(3k 1) (k 1)
=
(3k 1)(3(k 1) + 1)
k 1
=
3(k 1) + 1
Hence assuming P(k) is true. P(k + 1) is also true. Therefore P(n) is true for all n 1.
Example 3.3.7 1 – a n 1
Prove by induction for n 0. 1 a a 2 ... a n .
1– a
SPPU : Dec.-10, Marks 4
Solution : Let P(n) be the given statement,
1. Basis of induction
1– a
For n = 0, L.H.S. = 1, R.H.S. = =1
1– a
1– a 2
For n = 1, L.H.S. = 1 + a, R.H.S. = =1+a
1– a
For n = 0, 1, L.H.S. = R.H.S.
Hence P(0), P(1) are true.
2. Induction step : Assume that P(k) is true
1– a k+1
1 a + a 2 .... a k = … (1)
1– a
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Discrete Mathematics 3 - 10 Mathematical Induction
Consider,
1 – a k+1
1 a + a 2 .... a k a k+1 = a k+1 … (Using 1)
1– a
1 – a k+1 (1 – a) a k+1
=
1– a
1– a k+1 a k+1 – a k+2
=
1– a
1– a k+2
=
1– a
Hence P(k + 1) is true.
Therefore by the mathematical induction P(n) is true for all n 0.
Example 3.3.8 Use mathematical induction to show that n ( n 2 – 1) is divisible by 24. Where n
is any odd positive number. SPPU : Dec.-14, Marks 4
Solution : If n (n 2 –1) = n3 – n is divisible by 24.
Then n 3 – n = 24 (m) where m is any positive integral.
Let P(n) be the given statement,
1. Induction step : For n = 1,
n(n 2 – 1) = 0 which is divisible by 24.
For n = 3, n n 2 1 = 24 which is divisible by 24.
P(1) and P(3) is true.
2. Induction step :
Assume that P(k) is true.
i.e. k(k 2 – 1) = k 3 – k is divisible by 24.
k (k 2 – 1) = k 3 – k = 24 (m 0 ), m 0 z … (1)
Consider
(k +1) [( k +1) 2 – 1] = ( k +1) 3 – ( k +1)
= k 3 3k 2 3k 1– k –1
= k 3 3k 2 2k
= (k 3 – k) +3k 2 3k … (Using 1)
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Discrete Mathematics 3 - 11 Mathematical Induction
= 24 m 0 3k (k +1) (As k(k + 1) is multiple of 8 for k
odd positive integer and k 3)
= 24m 0 3 ( 8 m 1 )
= 24 ( m 0 m 1 )
= 24 m 2 (Q m 0 m 1 m 2 )
P(k + 1) is true.
By mathematical induction P(n) is true for all n odd positive number.
Example 3.3.9 Show that n 4 4 n 2 is divisible by 3 for all n 2. SPPU : Dec.-15, Marks 4
Solution : Let P(n) be the given statement,
1. Basis of induction
For n = 2
2 4 4 ( 2 2 ) = 16 – 16
= 0 is divisible by 3 as 0 is divisible by every number
P(2) is true.
2. Induction step : Assume that P(k) is true
i.e. k 4 4k 2 is divisible by 3
Then we have,
(k 1) 4 4(k 1) 2 = k 4 4k 3 6k 2 4k 1 4 (k 2 2k 1)
= (k 4 4k 2 ) 4 (k 3 2k) 6k 2 12k 3
k 4 4 k 2 is divisible by 3.
k 3 2k is divisible by 3
Also 6 k 2 12 k 3 = 3 (2k 2 4k 1) is divisible by 3.
Hence (k 1) 4 4 (k + 1) 2 is divisible by 3.
Hence assuming P(k) is true.
P(k + 1) in also true. Therefore P(n) is true for n 2.
Example 3.3.10 Prove that : 7 2n 2 3n– 3 3 n– 1 is divisible by 25 n.
SPPU : May-19, Marks 3
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Discrete Mathematics 3 - 12 Mathematical Induction
Solution : Let P(n) be the given statement.
1) Basis of induction : For n = 1,
7 2 2 0 3 0 = 49 + 1 = 50, which is divisible by 25.
P(1) is true.
2) Induction step :
Assume that P(k) is true.
i.e. 7 2k 2 3k– 3 3 k– 1 = 25 (m)
Consider
7 2(k+1) + 2 3(k+1)– 3 3 k+1– 1 = 7 2k+2 2 3k 3 k
= 497 2k 2 3k 3k
= 49[25 m – 2 3k– 3 3 k– 1 ] 2 3k 3 k
= 25 (49m) – 49 2 3k– 3 3 k– 1 2 3k– 3 3 k– 1 ( 8 3)
= 25 (49m) + 2 3k– 3 3 k– 1 (– 49 + 24)
= 25(49 m) + 2 3k– 3 3 k– 1 ( –25)
= 25 [ 49 m – 2 3k– 3 3 k– 1 ]
= 25 (P) : P N
P(k+1) is true.
By mathematical induction, P(n) is true for all n.
Example 3.3.11 Using mathematical induction, prove that
n( n 1)
1 2 2 2 3 2 4 2 ... ( 1) n 1 n2 ( 1) n 1
2 SPPU : Dec.-04, Marks 6
Solution : Let P(n) be the given statement,
1. Basis of induction
For n = 1 L.H.S = 1, R.H.S. = 1
L.H.S. = R.H.S.
Hence P(1) is true.
2. Induction step : Assume that P(k) in true.
k(k 1)
i.e. 1 2 2 2 3 2 4 2 ... ( 1) k 1 k 2 ( 1) k 1 ... (1)
2
Then we have,
[1 2 2 2 3 2 4 2 ... ( 1) k 1 k 2 ] ( 1) k (k 1) 2 ... (Using 1)
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Discrete Mathematics 3 - 13 Mathematical Induction
k(k 1)
= ( 1) k 1 ( 1) k (k 1) 2
2
k
= ( 1) k (k 1) (k 1)
2
k 2k 2
= ( 1) k (k 1)
2
(k 1)(k 2)
= ( 1) k
2
Hence assuming P(k) is true, P(k + 1) is also true. Therefore P(n) in true for all n 1.
Example 3.3.12 Prove by mathematical induction that for n 1 :
1 1! 2 2! 3 3 ! ... n n ! (n 1) ! 1. SPPU : May-08, 15, Marks 6
Solution : Let P(n) be the given statement,
1. Basis of induction
For n = 1, L.H.S = 1, R.H.S. = 1
L.H.S. = R.H.S.
Hence P(1) is true.
2. Induction step : Assume that, P(k) is true.
i.e. 1 1! 2 2! 3 3! ... k k!
= (k + 1)! – 1 ... (1)
Then we have,
[1 1! 2 2! 3 3! ... k k!] + (k + 1) (k + 1)! ... (Using 1)
= [(k + 1)! – 1] + (k + 1) (k + 1)!
= (k + 1)! + (k + 1) (k + 1)! – 1
= (k + 1)! [k + 1 + 1] – 1
= (k + 2) (k + 1)! – 1
= (k + 2)! – 1
Hence assuming P(k) is true, P(k + 1) is also true. Therefore P(n) is true for n 1.
Example 3.3.13 Prove that for any positive integer n the number n 5 – n is divisible by 5.
SPPU : Dec.-08, Marks 6
Solution : Let P(n) be the given statement,
1. Basis of induction :
For n = 1, 15 1 = 0 is divisible by 5.
As 0 is divisible by every number.
Hence P(1) is true.
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Discrete Mathematics 3 - 14 Mathematical Induction
2. Induction step : Assume that, P(k) is true.
i.e. k5 k is divisible by 5
Then we have
(k 1) 5 (k 1) = (k5 5 C 1 k 4 5 C 2 k 3 5 C 3 k 2 5 C 4 k 5 C5 ) (k 1)
= k5 5 k 4 10 k 3 10 k 2 5k 1 k 1
= (k5 k) 5 [k 4 2k 3 2k 2 k]
k5 k is divisible by 5.
and 5 (k 4 2k 3 2k 2 k) is divisible by 5.
Hence (k 1) 5 (k 1) is divisible by 5.
Hence assuming P(k) is true. P(k + 1) is also true. Therefore P(n) is true for n 1.
Example 3.3.14 Prove that 8 n 3 n is a multiple of 5 by mathematical induction for n 1.
SPPU : May-06, 07, Dec.-13, Marks 6
Solution : Let P(n) be the given statement,
1. Basis of induction :
For n = 1 81 31 = 5
= 5 1
Obviously a multiple of 5.
P(1) is true.
2. Induction step : Assume that, P(k) in true.
i.e. 8 k 3 k is multiple of 5 say 5 r
i.e. 8k 3k = 5 r ... (1)
where r is an integer
Then we have,
8k 1 3k 1 = 8k 8 3k 3
= 8 k (5+ 3) 3 k 3
= 8 k 5+ (8 k 3 3 k 3)
= 8 k 5+ 3(8 k 3 k )
Obviously 8 k 5 is multiple of 5 and also 8 k 3 k is multiple of 5.
Therefore, 8 k 1 3k 1 is multiple of 5.
Hence assuming P(k) is true, P(k + 1) is also true. Therefore P(n) is true for all n 1.
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Discrete Mathematics 3 - 15 Mathematical Induction
Example 3.3.15 Show that the sum of the cubes of three consecutive natural number is
divisible by 9. SPPU : Dec.-06, Marks 6
Solution : Let n, n + 1, n + 2 be three consecutive natural numbers.
We have to show that n 3 (n + 1) 3 (n 2) 3 is divisible by 9.
Let P(n) be the above statement,
1. Basis of induction : For n = 1
1 3 2 3 3 3 = 1 + 8 + 27 = 36 which is divisible by 9.
P(1) is true.
2. Induction step : Assume that P(k) is true.
i.e. k 3 (k 1) 3 (k 2) 3 in divisible by 9.
Then we have,
(k 1) 3 (k 2) 3 (k 3) 3 = [(k 1) 3 (k 2 ) 3 ] [k 3 3 C 1 k 2 (3) 3 C 2 k(3) 2 3 C 3 (3 3 )]
= (k 1) 3 (k 2) 3 k 3 [9 k 2 27 k 27]
= [k 3 (k 1) 3 (k 2) 3 ] 9 [k 2 3k 3]
k 3 (k 1) 3 (k 2) 3 is divisible by 9 and 9 (k 3 3k 3) is divisible by 9.
(k 1) 3 (k 2) 3 (k 3) 3 is divisible by 9.
Hence assuming P(k) in true, P(k + 1) is also true. Therefore P(n) is true for all n 1.
Example 3.3.16 5 n+1 – 1
Using mathematical induction prove that 3 3.5 3.5 2 .....3.5 n .
4
For non-negative number n. SPPU : Dec.-11, Marks 6
Solution : Cancelling 3 from the both sides of given.
Statement, We get
5 n +1 – 1
1 5 5 2 ....5 n = … (1)
5–1
Let P(n) be the above statement.
To prove this refer example 3.3.7 for a = 5.
Example 3.3.17 Let n be a positive integer. Show that any 2 n 2 n chessboard with one
square removed can be covered by L-shaped pieces, where each piece covers three squares at
a time.
Solution : Let P(n) be the proposition that any 2 n 2 n chessboard with one square
removed can be covered using L-shaped pieces.
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Discrete Mathematics 3 - 16 Mathematical Induction
Basis of induction : For n = 1, P(1) implies that any 2 2 chessboard with one square
removed can be covered using L shaped pieces. P(1) is true, as seen below.
Induction step : Assume that, P(k) is true i.e. any 2 k 2 k chessboard with one square
removed can be covered using L-shaped pieces.
Then, we have to show that P(k + 1) is true. For this consider, a 2 k 1 2 k 1
chesshoard with one square removed. Divide the chessboard into four equal halves of
size 2 k 2 k , as shown below.
k k
2 2
C1
k C2 k
2 2
k C3 C4 k
2 2
k k
2 2
The square which has been removed, would have been removed from one of the four
chessboards, say C 1 . Then by induction hypothesis, C 1 can be covered using L-shaped
pieces. Now, from each of the remaining chessboards, remove that particular piece (or
tile), lying at the centre of the large chessboards.
k k
2 2
C2 C1
k k
2 2
k k
2 C3 C4 2
k k
2 2
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Discrete Mathematics 3 - 17 Mathematical Induction
Then by induction hypothesis, each of these 2 k 2 k chessboards with a piece (or tile)
removed can be covered by the L-shaped pieces. Also the three tiles removed from the
centre can be covered by one L-shaped piece. Hence the chessboard of 2 k 1 2 k 1 can
be covered by L-shaped pieces.
Hence proved.
Example 3.3.18 Suppose we have unlimited stamps of two different denominations, 3 rupees
and 5 rupees. We want to show that it is possible to make up exactly any postage of 8
rupees or more using stamps of these two denominations.
Solution : For k = 8, we have one 5 rupees stamp and one 3 rupees stamp.
For k = 9, replace 5 rupees stamp by two 3 rupees stamp, similarly for k = 10,
replace all 3, 3 rupees stamps by, two 5 rupees stamp an so on.
Hence let us assume that, it is possible to make up k rupees stamp using 3 rupees
and 5 rupees stamps (for k 8).
Now we have to show that it is also possible to make up (k + 1) rupees stamps using
3 rupees and 5 rupees stamps.
We examine two cases :
1) Suppose we make up stamps of k rupees using at least one 5 rupees stamp.
Replacing a 5 rupees stamp by two 3 rupees stamp, we can make up k + 1
rupees stamps.
2) Suppose we make up a stamp of k rupees using 3 rupees only. Since k 8 we
must have at least 3, 3 rupees stamps. Replacing these 3, 3 rupees stamps by two
five rupees stamps. We can make up stamps of k + 1 rupees.
Hence proved.
Example 3.3.19 The king summoned the best mathematicians in the kingdom to the palace to
find out how smart they were. The king told them
"I have placed white hats on some of you and black hats on the others. You may look at,
but not talk, to, one another. I will leave now and will come back every hour on the hour.
Every time I return, I want those of you who have determined that you are wearing white
hats to come up and tell me immediately."
As it turned out, at the nth hour every one of the n mathematician who were given white
hats informed the king that she knew that she was wearing a white hat. Why ?
Solution :
1. Basis of induction : For n = 1, there is only one mathematician wearing a white
hat. Since the king said that white hats were placed on some one of the mathematician
(king never lie). The mathematician who saw that all other mathematicians had on black
hats would realize immediately that she was wearing a white hat. Consequently she
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Discrete Mathematics 3 - 18 Mathematical Induction
would inform the king on the first hour. (when the king returned for the first time) that
she was wearing a white hat.
Now let n = 2, i.e. two mathematicians wearing white hats.
Consider, one of these two mathematicians. She saw that one of her colleagues was
wearing a white hat. She reasoned that if she were wearing a black hat, her colleague
would be the only one wearing a white hat. In that case, her colleague would have
figured out the situation and informed the king on the first hour. That did not happen
shows that she was also wearing a white hat. Conesquently she told the king on the
second hour (and so did the other mathematician with a white hat, since all the
mathematicians are smart).
2. Induction step : Assume that, if there were k mathematicians wearing white hats,
then they would have figured out that they were wearing white hats and informed the
king so on the k th hour. Now, suppose that there were k + 1 mathematicians wearing
white hats.
Every mathematician wearing a white hat saw that k of her colleagues were wearing
white hats. However, that her k colleagures did not inform the king of their findings on
the k th hour can only imply that there were more then k. People wearing white hats.
Consequently she knew that, she must be wearing a white hat also on the (k 1) th hour.
She (together with all other mathematicians wearing white hats) would inform the king
their conclusion.
qqq
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Unit - II
4 Relations
Syllabus
Relations and their Properties, n-ary relations and their applications, Representing relations,
Closures of relations, Equivalence relations, Partial orderings, Partitions, Hasse diagram,
Lattices, Chains and Anti-Chains, Transitive closure and Warshall‘s algorithm.
Contents
4.1 Introduction
4.2 Cartesian Product
4.3 Relation
4.4 Matrix Representation of a Relation
4.5 Diagraphs
4.6 Special Types of Relations
4.7 Types of Relations on Set . . . . . . . . . . . . . . . . Dec.-10, 11, 12, 14, 18, 19,
. . . . . . . . . . . . . . . . . . May-14, 17, 18, 19, · · · · · · · Marks 6
4.8 Partitions of a Set . . . . . . . . . . . . . . . . . . Dec.-11, 12, May-17, · · · · · · Marks 3
4.9 Closure of a Relation . . . . . . . . . . . . . . . . . . Dec.-05, 07, 12, 13, 14, 15, 16
. . . . . . . . . . . . . . . . . . May-06, 07, 08, 15, 18, · · · · Marks 6
4.10 Partially Ordered Set . . . . . . . . . . . . . . . . . . Dec.-06
4.11 Principle of Duality . . . . . . . . . . . . . . . . . . Dec.-10, 11, 13, 14, 15, 16,
. . . . . . . . . . . . . . . . . . May-14, 15, 19 · · · · · · · · · · Marks 6
(4 - 1)
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Discrete Mathematics 4-2 Relations
4.1 Introduction
In the chapter 1, we dealt with sets, elements, different types and general properties
of sets. There may exists various relationships among elements of sets. We will study
relations and functions with various properties and examples.
Relation may involves equality or inequality of elements. In mathematics the
expressions like '' is equal to " is similar to ", " is greater than", "is parallel to " are some
relations.
4.2 Cartesian Product
Let A and B be two non empty sets. The cartesian product of A and B is denoted by
A ´ B and defined as
A ´ B = {(a, b) / a Î A and b Î B}
e.g. If A = {1, 2, 3, 4}, B = {a, b} then
A ´ B = {(1, a), (1, b), (2, a), (2, b), (3, a) (3, b) (4, a) (4, b)}
B ´ A = {(a, 1), (a, 2), (a, 3), (a, 4), (b, 1), (b, 2), (b, 3), (b, 4)}
(a, b) Ï A ´ B or B ´ A
and (1, a) ¹ (a, 1)
Hence A´B ¹ B ´ A
So the cartesian product is not commutative.
Similarly,
A ´ B ´ C = {(a, b, c) / a Î A and b Î B and c ÎC}
In general
A1 ´ A 2 ´ ......´ A n = {a 1, a 2 , a 3 ........... , a n ) / a 1 Î A1 and a 2 Î A 2 , .... and a n Î A n )}
Theorem 1 : If |A| = m and |B| = n then |A ´ B| = |B ´ A| = mn
Proof : The proof is trivial by set theory.
Theorem 2 : If A, B, C are non empty sets then A Í B Þ A ´ C Í B ´ C
Proof : Let (x,y) be any element in A ´ C, then
Þ x Î B and y ÎC
Þ (x, y) Î B ´ C
Hence A ´ C Í B ´ C
Theorem 3 : If A, B, C, are sets then
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Discrete Mathematics 4-3 Relations
i) A ´ (B Ç C) = (A ´ B) Ç ( A ´ C)
ii) (A Ç B) ´ C = (A ´ C) Ç (B ´ C)
iii) A ´ (BÈ C) = (A ´ B ) È (A ´ C)
iv) (A È B) ´ C = (A ´ C) È (B ´ C)
Proof :
i) A ´ (B Ç C) = {(x, y) / x Î A and (y Î B Ç C) }
= {(x, y) / x Î A and / (y Ï B and y Î C)}
= {(x, y) / x Î A, y Î B and y Î A, y ÎC}
= {(x, y) / (x, y) Î A ´ B and (x, y) Î A ´ C}
= {(x, y) / (x, y) Î (A ´ B) Ç (A ´ C)}
A ´ (B Ç C) = (A ´ B) Ç (A ´ C)
Similarly students can prove remaining results.
Note : If  is the set of real numbers then
 ´  =  2 ® Euclidean plane
 ´  ´  =  3 ® 3D space
4.3 Relation
Let A and B be two non empty sets. A relation from A to B is any subset of A ´ B. It
is denoted by R : A ® B
e.g. Let A = {x,y,z), B = {1, 2, 3)
then A´B = {(x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3), (z, 1), (z, 2) (z, 3)}
R 1 = {(x,1), (y, 2), (z,1), )}, R 2 = {(z, 3)}, R 3 = { f }
are relations from A to B
But R 4 = {(1, x), (2, x)}is not relation from A to B. R 4 is the relation from B to A.
Important Results :
1) If (a, b) Î R then it is denoted by a R b.
2) If R is a relation from A to B then R Í A ´ B
3) If R Í A ´ B then R is a relation from A to A and R is called a relation on A.
4) If R is a relation from A to B, then the set of all first elements of the ordered
pairs (a,b) Î R is called the domain of R. It is denoted by D(R) = {a/(a, b) Î R}.
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Discrete Mathematics 4-4 Relations
The range of R is the set of all second co-ordinates of the ordered pairs (a,b) Î R.
It is denoted by R (R) = {(b/(a, b) Î R}.
5) The null set is the subset of A ´ B .
\ f is a relation called null relation or empty relation.
Example :
Let A = {1, 2, 3}, B = {x, y}
then R 1 = {1, x) (1, y), (3, x)} is a relation from A to B
\ D (R 1 ) = {1,3}
R (R 1 ) = {x, y}
4.4 Matrix Representation of a Relation
Let A = {a 1 , a 2 , a 3 .... a n } , B = {b1 , b 2 , b 3 , .... b m } and R Í A ´ B . Then the relation
matrix of R is denoted by M R = [ m ij ] n ´ m and defined by
ì0 if (a i , b j ) Ï R i.e. a i Rb j
m ij = í
î1 if (a i , b j ) Î R i.e a i R b j
Example 4.4.1 Let A = {1, 2, 3, 4} and R = {(x, y)/ x < y } then find M R .
Solution : R = {(1, 2) (1, 3) (2, 3) (1, 4) (2, 4) (3, 4) }
1 2 3 4
é0 1 1 1ù
\ M R = [ M ij ] 4 ´ 4 = ê 0 0 1 1ú
ê ú
ê0 0 0 1ú
ê0 0 0 0úû
ë
Example 4.4.2 If A = {1, 2, 3, 4, 5, 6} and a R b iff a divides b for a, b Î A. Find relation
matrix.
Solution : R = {(1, 2} (1, 3), (1, 4), (1, 5), (1, 6), (2, 4) (2, 6), (3, 6), (1, 1) (2, 2), (3, 3),
(4, 4), (5, 5) (6, 6)}
1 2 3 4 5 6
é1 1 1 1 1 1ù
ê0 1 0 1 0 1ú
ê ú
\ Relation Matrix = M R = ê 0 0 1 0 0 1ú
ê0 0 0 1 0 0ú
ê ú
ê0 0 0 0 1 0ú
ê0 0 0 0 0 1úû 6 ´ 6
ë
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Discrete Mathematics 4-5 Relations
Example 4.4.3 A = {x, y, z} Find the relation matrices of the following relations.
i) R 1 = {(x, x) (y, y) (z, z)} ii) R 2 = {(x, y) (y, x), (y, z), (z, y)} iii) R 3 = {(x, x)
(x, y), (x, z) (y, x) (y, y) (y, z), (z, x) (z, y) (z, z)
Solution : (5,5) (6,6)
x y z
1 0 0ù
i) M R1 = é
ê0 1 0ú
ê ú
êë0 0 1úû 3 ´ 3
x y z
0 1 0ù
ii) M R2 = é
ê1 0 1ú
ê ú
êë0 1 0úû
é1 1 1ù
iii) M R3 = ê1 1 1ú
ê ú
êë1 1 1úû
4.4.1 Relation Matrix Operations
We know that relation matrix is a boolean matrix as all entries are either 0 or 1
I) Let A = [ a ij ] m ´ n , B = [b ij ] m ´ n be two relation matrices then
A + B = [ a ij + b ij ] = [c ij ] m ´ n
where c ij = 1 if a ij = 1, or b ij = 1
= 0 if a ij = 0, and b ij = 0
II) Let A = [ a ij ] m ´ n and B = [b jk ] n ´ k then A × B = [ a ij ; b jk ] = [d jk ]
where d jk = 1 if a ij = 1 and b jk = 1
= 0 if a ij = 0 or b jk = 0
é1 1 1ù é1 1 1ù
e.g. If ê ú
A = 0 1 1 , B = ê 0 1 1ú
ê ú ê ú
êë1 1 0úû êë 0 0 0úû
é1 1 1ù
then A + B = ê 0 1 1ú
ê ú
êë1 1 0úû
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Discrete Mathematics 4-6 Relations
é1 1 1ù
and A B = ê1 1 1ú
ê ú
êë1 1 1úû
4.4.2 Properties of Relation Matrix
Let R 1 be a relation matrix from A to B and R 2 be a relation matrix from B to C
then relation matrices satisfy the following properties :
i) M R1 × R = M R1 × M R 2
2
ii) M = Transpose of M R1
R –1
1
iii) M = M M
(R1R 2 ) –1 R –1 R –1
2 1
4.5 Diagraphs
A relation can be represented pictorially by drawing its graph. Let A be any non
empty set and R be a relation on A. R can be represented by graphically by the
following procedure.
i) The elements of a set A are represented by small circles or point i.e. (o) or ...
These elements are called as vertices or nodes.
ii) If (a, b) Î R or a R b then vertices a and b are joined by a continuous arc with an
arrow from a to b. These arcs are known as edges of the graph i.e. if aRb then
a b
iii) If aRa then the vertex a is joined to itself by a loop around a. e.g. if aRa then
This graphical representation of the relation is called as diagraph or directed graph of
R. Let us consider the following examples.
1) aRa 2) aRb
a b
3) bRa 4) aRb Ù bRa
a b a b
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Discrete Mathematics 4-7 Relations
5) aRa Ù aRb Ù bRa Ù bRb
a b
6) aRb Ù bRc Ù CRa Ù aRa
b
7) aRa Ù bRb Ù cRc
a b c
8) aRb Ù bRa Ù aRc Ù cRa a b
c
Examples :
Example 4.5.1 A = {1,2,3,4,5,6} and a R b iff a/b (a divides b). Draw diagraph of R.
Solution : We have
(1,1) ( 2, 2) ( 3, 3) ( 4 , 4) ( 5, 5) ( 6, 6)ü
ï
R = (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) ý
ï ( 2, 4) ( 2, 6) ( 3, 6) ï
î þ
The diagraph of R is as follows :
1
2
6
3
5 4
Fig. 4.5.1
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Discrete Mathematics 4-8 Relations
Example 4.5.2 é1 0 0 1ù
ê0 1 0 1ú
Let A = {a, b, c, d} and M R =ê ú Draw diagraph of R.
ê1 1 0 0ú
ê1 0 0 0úû
ë
Solution : We have
R = {(a,a) (a,d) (b,b) b,d) (c,a) (c,b) (d,a)}
The digraph of R is as follows :
a b
d c
Fig. 4.5.2
Example 4.5.3 Find the relation from the diagraph.
Fig. 4.5.3
Solution : From the given diagraph, the corresponding relation on A = {1, 2, 3, 4} is
R = {(1, 1) (1, 3) (3, 1) (2, 3) (3, 2), (2, 4) (4, 2) (4, 4) (3, 4) (4, 3)}
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Discrete Mathematics 4-9 Relations
4.6 Special Types of Relations
4.6.1 Inverse Relation (OR Converse Relation)
Let R be a relation from A to B. The inverse relation of R is denoted by R –1 is a
relation from B to A defined as
R –1 = {(y, x) | y Î B, x Î A and (x,y) Î R}
i.e. if x R y then yR –1 x
It is also denoted by R c
It is also known as the converse relation.
Example :
1) If R = {(1,2) (2,3) (3,1) (4,5) } then R –1 or R c = {(2,1) (3,2) (1,3) (5,4)}
é1 0 1 1ù é1 1 0 0ù
ê1 0 1 1 ú ê0 0 0 0ú
2) If M R =ê ú then M –1 = ê ú = [M R ] t
ê0 0 1 1ú R ê1 1 1 0ú
ê0 0 0 1úû ê1 1 1 1úû
ë ë
3) If the diagraph R is given by
\ R = {(1,1) (1,2) (2,3) (4,2) (4,1) (4,3) (3,4) }
1 2
4 3
Fig. 4.6.1
then the diagraph of R –1 is obtained by changing the direction of arrow only which
is given below.
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Discrete Mathematics 4 - 10 Relations
1 2
4 3
Fig. 4.6.2
Theorem 1 : If R, R 1 , R 2 are relations from A to B then
i) (R –1 ) –1 = R
ii) (R 1 È R 2 ) -1 = R –1
2
È R 1–1 = R 1–1 È R –1
2
iii) ( R 1 Ç R 2 ) -1 = R 1–1 Ç R –1
2
4.6.2 Complement of a Relation
Let R be a relation from A to B. The complement of R is denoted by R or R ¢ : A ® B
and defined as R = {(x, y) | (x, y) Ï R, x Î A, y Î B}
i.e. x R y iff x R/ y
Note R is the complement ary set of R w.r.t. universal domain A ´ B.
Examples :
Example 4.6.1 Let A = {1, 2, 3, 4}, B = { a, b},
A ´ B = {(1, a) (1, b) (2, a) (2, b) (3, a) (3, b) (4, a) (4, b)},
R = {(1, a) (2, a) (3, a) (4, a)},
S = {4, a) (4, b) (3, a) (3, b)}.
Find R , S, R Ç S, R È S.
Solution :
R = {(1,b) (2,b) (3,b) (4,b)}
S = {(1, a) (1, b) (2, a) (2, b)}
R Ç S = {(1, b) (2, b)}
R È S = {(1, a) (1, b) (2, a) (2, b) (3, b) (4, b)}
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Discrete Mathematics 4 - 11 Relations
Example 4.6.2 If A = {a, b, c, d} and R = {(a, b) (c,d) (c,c) (d,a) (a,a) (b,b) (d,d) } is a relation
on A. Draw diagraph of R .
Solution :
R = {(a, c) (a, d) (b, a) (b, c) (b, d) (c, a) (c, b) (d, b) (d, c)}
The diagraph of R is as follows :
a b
d c
Fig. 4.6.3
Example 4.6.3 é1 0 1ù
Obtain the matrix of R if M R = ê 0 0 1ú
ê ú
êë1 1 0úû
Solution : The matrix of R is obtained from M R be replacing 0 and 1 by 1 and 0
respectively
é0 1 0ù
M R = ê1 1 0ú
ê ú
êë 0 0 1úû
Note : 1) (R 1 È R 2 ) = R 1 Ç R 2
2) (R 1 Ç R 2 ) = R 1 È R 2
4.6.3 Composite Relation
Let R 1 : A ® B and R 2 : B ® C be two relations.
The composition of R 1 and R 2 is denoted by
R 1 o R 2 or R 1 R 2 : A ® C defined as
R 1 × R 2 = {(x,z) | x R 1 y, yR 2 z i.e. (x, y) Î R 1 and (y, z) Î R 2 for x Î A, z Î C }
Examples :
Example 4.6.4 Let A = (x, y, z, w) and R 1 = {(x, y) (x, x) (y, z) (z, w) }, R 2 = {(x, x)
(y, x) (y, y) (w, w) }, Find R 1 , R 2 and diagraph of R 1 × R 2 .
Solution :
In R 1 In R 2 R1 × R2
1) (x, y) (y, x) (x, x)
(y, y) (x, y)
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Discrete Mathematics 4 - 12 Relations
i.e. x ® y y®x x®x
y®y x®y
2) (x, x) (x, x) (x, x)
3) (y, z) - -
4) (z, w) (w, w) (z, w)
\ R 1 × R 2 = {(x, x)}, (x, y), (z, w)} x y
The diagraph of R 1 × R 2 is,
w z
Fig. 4.6.4
Example 4.6.5 Let A = (1, 2, 3, 4). Let R 1 = {(x, y)/ x+y = 5} and R 2 = (x, y) / (y – x) = 1}
verify ( R 1 × R 2 ) C = R C
2
× R 1C
Solution : We have A = {1, 2, 3, 4}
Rl = {(1,4) (2,3), (3,2) (4,1)}
R 2 = {(1,2) (2,3) (3,4)}
R 1 × R 2 = {(2,4) (3,3) (4,2)}
(R 1 × R 2 ) C = {(4,2) (3,3) (2,4)}
R 1C = {(4,1), (3,2), (2,3), (1,4)}
RC
2
= {(2,1), (3,2) (4,3)}
R C × R C = {(2,4), (3,3) (4,2)} = ( R 1 × R 2 )
2 1
C
Hence ( R 1 × R 2 ) = RC × RC
2 1
4.7 Types of Relations on Set
SPPU : Dec.-10, 11, 12, 14, 18, 19, May-14, 17, 18, 19
Let R be a relation from A to A. Then we can say that R is a relation on set A.
\ R Í A ´ A.
A ´ A is the universal set for R. If R = A ´ A then R is called the Universal Relation on
A.
4.7.1 Identity Relation
Let A be any set. The relation on A is called Identity relation if
I A = {(a, a) | for a Î A}
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e.g. If A = {1, 2, 3, 4} then
R 1 = {(1, 1) (2, 2)}, R 2 = {(4, 4)} are identity relations.
The diagraph of the identity relation is the graph with loops only and it's relation
matrix is the diagonal matrix with 0 and 1.
4.7.2 Reflexive Relation
1) Let R be a relation on set A. A relation R is said to be reflexive relation if
(a, a) Î R, " a Î A.
2) A relation R is said to be not reflexive if $ at least one element a Î A such that
(a, a) Ï R i.e. a R a.
3) A relation R is said to be irreflexive if for every a Î A, (a, a) Ï R i.e. a R
/ a
Notes :
1) If R is a reflexive relation on set A then
i) The relation matrix (M R ) is the identity matrix.
ii) Diagraph of reflexive relation will have loop for every element of A.
2) The relation matrix of irreflexive relation has all diagonal elements zero. It's
diagraph is free from loops.
Examples
Example 1 :
Let A = {1,2, 3,4}. Then
i) R1 = {(1, 1) (2, 2) (3, 3) (4, 4)} is the reflexive relation. It is known as the smallest
reflexive relation on set A.
ii) R 2 = {(1, 1) (2, 2) (3, 3) (4, 4) (2, 4) (1, 2) (3, 2)} is reflexive relation.
iii) R 3 = { (1, 1) (2, 2) (3, 4)} is not reflexive relation as 3 Î A but 3 R 33 .
/ 4 a.
iv) R 4 = {(2, 3) (3, 4)} is irreflexive relation A as " a Î A, a R
Example 2 :
Let A = Set of all straight lines in a plane. x R y iff x is parallel to y.
For any line x Î A, x is parallel to x.
\ x Rx " x Î A
\ R is a reflexive relation.
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4.7.3 Symmetric Relation
Let R be a relation on set A. R is said to be symmetric relation if whenever a R b
then b Ra for a, b Î A.
Notes :
1) A relation is symmetric iff R = R –1
2) The relation matrix of symmetric relation is a symmetric matrix.
Example : Let A = {1,2, 3, 4}. then
i) R1 = {(1, 1) (2, 2) (3, 4) (4, 3)} is a the symmetric relation.
ii) R 2 = {(1, 1) (2, 2) (3, 3) (4, 4)} is a symmetric as well as reflexive relation.
iii) R 3 = { (1, 1) (2, 2) (3, 3) (4, 4) (3, 4)} is reflexive but not symmetric relation.
iv) R4 = {(a, b)} a ¹ b is the smallest non- symmetric relation.
Example 4.7.1 Let A = Set of all straight lines in a plane. x R y iff x || y i.e. x is parallel
to y. S. T. R is a symmetric relation.
Solution : R = {x R y | x is parallel to y}
Suppose x R y then x is parallel to y.
iff y is parallel to x.
iff y R x
i.e. whenever x R y, then y R x.
\ R is a symmetric relation.
4.7.4 Compatible Relation
A relation R on a set A is said to be compatible relation if it is reflexive and
symmetric relation. The relation matrix of a compatible relation is a symmetric matrix
with the diagonal elements 1.
e.g. If A = {x, y, z} then
R = {(x, x) (y, y) (z, z) (x, y) (y, x)} is a compatible relation.
Note :
1) If R is a compatible relation on set A S.T. |A| = n then |R| ³ n
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4.7.4.1 Asymmetric Relation
A relation R on a set A is said to be asymmetric relation if whenever aRb then
! .
bRa
Hence R is not asymmetric relation if for some a and b in A, aRb and bRa.
e.g. If A = {1, 2, 3} R = {(1, 2) (2, 3) (3, 1)} is a symmetric relation. But R 1 = {(1, 2)
(2, 3) (3, 2)} is not asymmetric relation as (2, 3) and (3, 2) Î R 1 .
4.7.5 Antisymmetric Relation
A relation R on a set A is said to be antisymmetric relation if whenever aRb and bRa
then a = b. A relation R is not antisymmetric if $ a, b Î A such that a ¹ b but aRb and
bRa. The relation matrix of antisymmetric relation is never symmetric matrix.
Examples :
1) Let A = R and aRb iff a £ b.
Suppose aRb and bRa then a £ b and b £ a.
Þ a = b.
\ R is antisymmetric relation. It is not symmetric relation.
2) Let A = {1, 2, 3} then
i) R 1 = {(1, 1) (2, 2)} is symmetric, antisymmetric relation.
ii) R 2 = {(1, 2) (2, 1)} is symmetric but not antisymmetric relation.
iii) R 3 = {(1, 1) (1, 2) (2, 2)} is antisymmetric but not asymmetric relation.
4.7.6 Transitive Relation
Let R be a relation on set A. R is said to be transitive relation if aRb and bRc, then
aRc for a,b,c, in R.
\ A relation R is not transitive if $ a, b and c in A.
/
Such that aRb, bRc but aRc
Examples :
1) Let A = R, R be the relation " £ ".
Then R is transitive as a £ b, b £ c Þ a £ c
2) If A = {x, y, z w} then,
i) R 1 = {(x, y) (y, z)} is not transitive relation as x Ry, y Rz but x R/ z.
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ii) R 2 = {(x, y) (y, z) (z, x) (z, y) (x, z) (y, x) (x, x) (y, y) b
(z, z) } is transitive relation.
iii) R 3 = {(x, y) (z, y)} is a transitive relation as there are no a
terms of the form x Ry and y R z,
Diagraph of transitive relation is as follows :
c
4.7.7 Equivalence Relation Fig. 4.7.1
Let R be a relation on a set A. Then R is said to be an equivalence relation on A iff R
is reflexive symmetric and transitive relations. It is denoted by 'R' or '~'.
Example :
Example 4.7.2 Let A = R, set of all straight lines in a plane.
R 1 = {x R 1 y | x is parallel to y}, R 2 = {x R 2 y | x is perpendicular to y } check whether
R 1 and R 2 are equivalence relations or not.
Solution : i) For relation R 1 : we know that if x, y, z are straight lines in a plane, then
x is parallel to x. R 1 is reflexive.
ii) If x R 1 y Þ x is || to y Þ y is parallel to x
Þ y R 1 x Þ R 1 is symmetric.
iii) If x R 1 y, y R 1 z Þ x || y, y || z Þ x || z Þ x R 1 z.
Þ R 1 is transitive relation.
\ R1 is an equivalence relation
ii) For relation R 2 :
Any line is not perpendicular to itself.
\ x R/ 2 x, " x Î A
\ R 2 is not reflexive.
\ R 2 is not equivalence relation.
Example 4.7.3 If A = {x, y, z } and R = {(x, x) (y, y) (z, z)}. It is an equivalence relation ?
Solution : The given relation R is reflexive, symmetric and transitive.
\ R is an equivalence relation.
4.7.8 Properties of Equivalence Relations
1) If R 1 and R 2 are equivalence relations on a set A, then
R 1 Ç R 2 is an equivalence relation.
Þ If R 1 and R 2 are reflexive, symmetric and transitive then R 1 Ç R 2 are also
reflexive, symmetric and transitive.
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R 1 Ç R 2 is an equivalence relation.
2) If R 1 and R 2 are equivalence relations on a set A, then it is not necessary that
R 1 È R 2 is an equivalence relation.
Let A = {a, b, c)
And R 1 = {(a, a) (b, b) (c, c) (a, b) (b, a)}
R 2 = { (a, a) (b, b) (c, c) (a, c) (c, a)} are equivalence relation.
R 1 È R 2 = {(a, a) (b, b) (c, c) (a, b) (b, a) (a, c) (c, a)}
(b, a) and (a, c) Î R 1 È R 2 but (b, c) Ï R 1 È R 2 .
\ R 1 È R 2 is not an equivalence relation.
3) If R 1 and R 2 are equivalence relations then R 1 È R 2 is an equivalence relations
iff R 1 Í R 2 or R 2 Í R 1 .
4.7.9 Equivalence Classes
Let R be an equivalence relation on a set A. The equivalence class of a ÎA is denoted
by [a] R and defined as
[a] R = {x Î A | x R a}
= The set of those elements of A which are related to a.
\ a Î [a] R
Example 1 :
Let A = {x, y, z }
R = A ´ A = {(x, x) (y, y) (z, z) (x, y) (x, z) (y, z) (y, x) (z, x) (z, y)}
is an equivalence relation.
\ The equivalence class of x is
[x] R = {x, y, z} = [y] R = [z] R
All these three equivalence classes are identical.
Example 4.7.4 If A = z, R = {(x, y) | x + y = even, x, y Î z }. Is R equivalence relation ? if
yes, find equivalence class of 1 and 2.
Solution : x + x = even number for any x Îz.
\ x Rx Þ R is reflexive relation.
Suppose x Ry Þ x + y = even
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Þ y + x = even
Þ yRx
\ R is symmetric relation.
Suppose xRy and yRz Þ x + y = even, y + z = even
Þ x + y + y + z = even
Þ x + z = even – 2y = even
Þ x + z = even
Þ xRz
\ R is a transitive relation.
\ R is an equivalence relation.
\ The equivalence class of 1 Î z is
[1] R = {xR 1 / x +1 = even}
= {xR 1 / x = even –1= odd}
= {xR 1 / x is odd}
[1] R = Set of all odd integers = {… –3, –1, 1, 3, 5, …}
[ 2] R = {xR 2 |x +2 = even}
= {xR 2| x = even}
= Set of all even integers
[ 2] R = {… – 4, – 2, 0, 2, 4 …]
Theorem 1 : Any two equivalence classes are either identical or disjoint.
Theorem 2 : For any x Î A, x Î A, x Î [x] R
Theorem 3 : A = U [a]
a ÎA
Example 4.7.5 Which of the following are relations from A to B
where A = {1, 2, 3, 4} ; B = {x, y, z}
a) R 1 = {(1, x), (1, y), (1, z), (4, x)}
(b) R 2 = {(x, 1), (y, 1), (z, 1), (x, 4)}
(c) R 3 = {(1, x), (2, y), (3, z), (4, w)}
(d) R 4 = {(1, 1), (2, 2)}
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Solution : Given that R : A ® B
Where A = Domain set
and B = Co-domain set
A ´ B = {(1, x), (1, y), (1, z), (2, x), (2, y), (2, z), (3, x),
(3, y), (3, z), (4, x), (4, y), (4, z)}
(a) R is a relation from A to B because R 1 Ì A × B
1
(b) R is not a relation from A to B as R Ë A × B
2 2
(c) R is not a relation from A to B as (4, w) Î R but w Ë B
3
(d) R is not a relation from A to B as R Ë A × B
4 4
Example 4.7.6 If A = {1,2,3,4}, B = {1, 4, 6, 8, 9} R is a relation from A to B such that aRb
iff b = a 2 . Find domain, range, relation matrix and diagraph. SPPU : Dec.-18, Marks 6
Solution : We have
R = {(1, 1) (2, 4) (3, 9)}
i) Domain of R is {1, 2, 3}
ii) Range of R is {1, 4, 9}
iii) Relation matrix is
1 4 6 8 9
1 é1 0 0 0 0ù
M(R) = 2 ê 0 1 0 0 0ú
ê ú
3 ê0 0 0 0 1ú
4 êë 0 0 0 0 0úû
iv) Diagraph
2
1
9
Example 4.7.7 If R is a relation from A to B where A = {1, 2, 8}, B = {1, 2, 3, 5} and aRb
iff a < b. Find : a) R in Roster form b) Domain and Range c) Diagraph.
SPPU : May-18, Marks 4
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Solution : Given that A= {1, 2, 8}, B = {1, 2, 3, 5}
a) R in Roster form is R = {(1, 2) (1, 3), (1, 5), ( 2, 3), ( 2, 5)}
b) Domain of R is {1, 2}
Range of R is {2, 3, 5}
c) Diagraph :
1 2
5 3
Example 4.7.8 If R is a relation on set A = {1, 2, 3, 4, 5}, R = {(1, 1) (1, 2) (2, 1) (3, 1)
(4, 1) (5, 2)}. Find domain set, codomain set, range. Draw its diagraph and find relation
matrix.
Solution : Given that A = {1, 2, 3, 4, 5}
and R is a relation on set A.
Domain set = A = Co-domain set
R = {(1, 1) (1, 2) (2, 1) (3, 1) (4, 1) (5, 2)}
Range set = {1, 2} & A
Its diagraph is as follows :
1
2
3
5
4
The matrix form of R is
1 2 3 4 5
1 é1 1 0 0 0ù
2 ê1 0 0 0 0ú
M (R) = ê ú
3 ê1 0 0 0 0ú
4 ê1 0 0 0 0ú
ê ú
5 êë 0 1 0 0 0úû 5 ´ 5
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Example 4.7.9 Let A = {1, 2, 3, 4, 5} and aRb if a < b. Find i) R in Roster form ii) Domain
and Range of R iii) Diagraph of R. SPPU : Dec.-19, Marks 3
Solution : We have
A = {1, 2, 3, 4, 5}
i) By considering the given conditions,
R in Roster form is
R = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3) (2, 4), (2, 5) (3, 4) (3, 5) (4, 5)}
ii) Domain of R is = {1, 2, 3, 4}
Range of R is = {2, 3, 4, 5}
iii) The vertex set of R is {1, 2, 3, 4, 5}
It's diagraph is as follows :
5
1
2 3
Example 4.7.10 Let A = {1, 2, 3, 4, 5, 6} aRb or (a, b) Î R iff a is a multiple of b. Find
(i) Range set and R (6), R (3).
(ii) Find relation matrix.
(iii) Draw diagraph
(iv) Find in and out degree of each vertex.
Solution : By considering given condition.
R = {(1, 1), (2, 1), (2, 2), (3, 1), (3, 3), (4, 1), (4, 2), (4, 4), (5, 1), (5, 5),
(6, 1), (6, 2), (6, 3), (6, 6)}
(i) The range set of R is {1, 2, 3, 4, 5, 6}
R (3) = {1, 3} as (3, 1) and (3, 3) Î R
and R (6) = {1, 2, 3, 6} because (6, 1), (6, 2), (6, 3), (6, 6) Î R
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(ii) The relational matrix is
1 2 3 4 5 6
1 é1 0 0 0 0 0ù
2 ê1 1 0 0 0 0ú
ê ú
M (R) = 3 ê1 0 1 0 0 0ú
4 ê1 1 0 1 0 0ú
ê ú
5 ê1 0 0 0 1 0ú
6 êë1 1 1 0 0 1úû
(iii)The diagraph of given relation is
vertex set = {1, 2, 3, 4, 5, 6}
2
1
3
6
4
5
(iv) The indegree of a vertex in diagraph is the number of edges coming towards that
vertex. The outdegree of a vertex is defined as the number of edges going out from a
vertex. Consider the following table.
Vertex Indegee Outdegree Degree
1 6 1 7
2 3 2 5
3 2 2 4
4 3 1 4
5 2 1 3
6 4 1 5
Example 4.7.11 Let R be a relation on set of natural numbers such that
R = {(x, y) | 2x + 3y and x, y Î N}. Find
(i) The domain and codomain of R. ; (ii) R-1
Solution : Given that x, y Î N
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The smallest values of x and y are 1, 1 respectively.
2x + 3y = 2 (1) + 3 (1) = 5
which is the smallest value of co-domain
x = 1 , y = 2 Þ 2x + 3y = 8
x = 2 , y = 1 Þ 2x + 3y = 7
x = 2 , y = 2 Þ 2x + 3y = 10
x = 2 , y = 3 Þ 2x + 3y = 13
x = 3 , y = 2 Þ 2x + 3y = 12
\ The domain of R is = {x | x Î N}
and the codomain of R is = {y | y Î N}
(ii) The inverse relation is
R -1 = {(y, x) / x, y Î N}
Example 4.7.12 Let A = {1, 2, 3, 4, 6} and let R be defined on A as xRy iff x | y (i.e. x
divides y)
(i) Write R as a set of ordered pairs
(ii) Find R -1 and describe R -1 in words
Solution : Given that,
A = {1, 2, 3, 4, 6} and x R y iff x divides y.
(i) \ R = {(1, 2), (1, 3), (1, 4), (1, 6), (2, 4), (2, 6), (3, 6), (1, 1),
(2, 2), (3, 3), (4, 4), (6, 6)} ... [Here 1|2, 1|1, 2|2, .... ]
(ii) The inverse relation R -1 is
R -1 = {(2, 1), (3, 1), (4, 1), (6, 1), (4, 2), (6, 2), (6, 3), (1, 1), (2, 2),
(3, 3), (4, 4), (6,6)}
Now R-1 is defined as : xR -1 y iff x is divisible by y.
–1
Example 4.7.13 Let A = {1, 2, 3) R is the relation on A whose matrix is :
é1 1 1ù
MR = ê 0 0 1ú
ê ú
êë 0 0 1úû
show that R is transitive. SPPU : May-17, Marks 3
Solution : Given that : A = {1, 2, 3}
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and R = {(1, 1), (1, 2), (1, 3), (2, 3), (3, 3)}
In R, (1, 1), (1, 2) Î R Þ (1, 2) Î R
(1, 1), (1, 3) Î R Þ (1, 3) Î R
(1, 2), (2, 3) Î R Þ (1, 3) Î R
(1, 3), (3, 3) Î R Þ (1, 3) Î R
(2, 3), (3, 3) Î R Þ (2, 3) Î R
Hence for any (a, b), (b, c) Î R Þ (a, c) Î R
Thus R is a transitive relation on A.
Example 4.7.14 Let A = Set of all students in SPPU
xRy iff x and y belongs to same class of SPPU V, y Î A.
Is R an equivalence relation ? Find the equivalence class of Atharva in A.
Solution : Given that A = {Set of all students in SPPU}
and xRy iff x and y belong to same class of SPPU.
(i) Any student x and x belong to same class
\ xRx , " x Î A Þ R is reflexive relation.
(ii) Let xRy Þ x and y belong to same class of SPPU.
Þ y and x belong to same class of SPPU.
Þ yRx
i.e. xRy Þ yRx, " x, y Î R.
Thus R is symmetric relation.
(iii) Let xRy and yRz
Þ x and y belong to same class and y and z belong to same class of SPPU
Þ x, y, z belong to same class of SPPU
ÞxRz
Þ R is transitive relation.
Thus R is an equivalence relation.
Suppose there is one student in SPPU whose name is Atharva.
\ The equivalence class of Atharva is the set of all students who are studying in the
class of Atharva.
Example 4.7.15 Let A = {Set of all lines in a plane} and x, y Î A. Define xRy iff x is
perpendicular to y. Is R transitive, reflexive and symmetric ?
Solution : Given that xRy iff x perpendicular y.
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(i) There is no any line which is perpendicular to itself.
#! any x such that xRx.
Þ R is not reflexive.
(ii) Suppose xRy Þ x is perpendicular to y.
Þ y is perpendicular to x.
Þ yRx
i.e. xRy Þ yRx i.e. R is symmetric relation.
(iii) Suppose xRy and yRz
Þ x is perpendicular to y and y is perpendicular to z.
Þ x and z are parallel lines.
Þ xRz
Þ R is not transitive relation.
Example 4.7.16 Let A = {set of lines in a plane} and x, y Î A. xRy iff x is parallel to y. Is R
equivalence relation ? Find the equivalence class of any line 'L'.
Solution :
(i) We know that every line is parallel to itself.
\ xRx , " x Î A
Þ R is reflexive relation.
(ii) Suppose xRy
Þ x is parallel to y.
Þ y is parallel to x.
Þ yRx Þ R is symmetric relation.
(iii) Suppose xRy and yRz.
Þ x is parallel to y and y is parallel to z.
Þ x is parallel to z.
Þ xRz Þ R is transitive relation.
Thus R is an equivalence relation.
Let L be any line in A.
Equivalence class of L = [L] = {xRL | x is parallel to L}
[L] = {x|x is parallel to L}
= Set of all lines which are parallel to L.
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Example 4.7.17 é1 0 0ù
Let A = {a, b, c} and M (R) = ê 0 1 1ú .
ê ú
êë 0 1 1úû
Determine whether R is an equivalence relation ? Find the equivalence class of b in A.
Solution : From relation matrix, the relation R is
R = {(a, a) (b, b) (c, c) (b, c) (c, b)}
(i) Here a, b, c Î A and (a, a) (b, b) and (c, c) Î R
\ R is reflexive relation.
(ii) As (b, c) Î R Þ (c, b) Î R
R is symmetric relation.
(iii) In relation R, we have,
(b, b), (b, c) Î R Þ (b, c) Î R
(b, c), (c, b) Î R Þ (b, b) Î R
(b, c), (c, c) Î R Þ (b, c) Î R
(c, c), (c, b) Î R Þ (c, b) Î R
(c, b), (b, b) Î R Þ (c, b) Î R
(c, b), (b, c) Î R Þ (c, c) Î R
So R is transitive relation. Thus R is an equivalence relation.
Now, the equivalence class of b in A is,
[b] = {x | (x, b) or (b, x) Î R}
[b] = {b, c}
Example 4.7.18 If R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5) (3, 5), (5, 3), (1, 3), (3, 1)}
is an equivalence relation on X = {1, 2, 3, 4, 5}. Find equivalence
classes. SPPU : May-19, Marks 3
Solution : If R is an equivalence relation on X and a Î X then [a] = a = {x|x~ a }.
Therefore, equivalence classes are as follows
[1] = 1 = {1, 3}
[2] = 2 = {2}, [3] = 3 = {1, 3, 5}
[4] = 4 = {4}
[5] = 5 = {3, 5}
Example 4.7.19 Let A be the set of integers and let R be defined as xRy iff x £ y. Is R
equivalence relation ?
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Soltuion : Given that xRy iff x % y, " x, y Î A
(i) We have, x £ x , " x Î A
Þ xRx, " x Î A
Þ R is reflexive relation.
(ii) If xRy Þ x £ y
then y £/ x Þ y R/ x
Þ R is not symmetric relation.
\ R is not an equivalence relation.
Example 4.7.20 Let A = Z + the set of positive integers, and let
R = {(a, b) Î A ´ A | a divides b}
Is R symmetric, asymmetric or antisymmetric. SPPU : May-17, Marks 3
Ans. : Given that : A = Z + = The set of positive integers
and R = {(a, b)/a divides b}
i) If (a, b) Î R then a divides b
\ b = aK, K Î Z +
1 1
Þ a = b but if K ¹ 1 then is not an integer
K K
\ a does not divides b if
b is not multiples of K
\ (a, b) Î R Þ (b, a) may not be present in R.
\ R is not symmetric relation.
ii) If (a, b) Î R then a/b
\ b = aK, K Î Z +
1
Þ a = b, As K Î Z +
K
b is not divide a
i.e. if (a, b) Î R then (b, a) Ï R.
\ R is asymmetric relation.
iii) As (a, b) Î R, (b, a) Î R
then b = K 1 a, a = K 2 b, K 1 , K 2 Ï Z +
b = K 1 , a = (K 1 K 2 )b
Þ K 1K 2 = 1 Þ K 1 = K 2 = 1
\ a =b
Thus R is antisymmetric relation.
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Example 4.7.21 Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9} and let N be the relation on A ´ A defined
by (a, b) ~ (c, d) iff a + d = b + c.
(i) Prove that ~ is an equivalence relation.
(ii) Find equivalence class of (2, 5) SPPU : Dec.-11
Solution : Given that (a, b) ~ (c, d) iff a + d = b + c, a, b, c, d A
(i) (a) We have a + b = b + a (a, b) ~ (a, b)
'~' is reflexive relation.
(b) If (a, b) ~ (c, d) then a + d = b + c.
b+c=a+dc+b=d+a
(c, d) ~ (a, b) by definition
'~' is symmetric relation.
(c) Suppose (a, b) ~ (c, d) and (c, d) ~ (e, f)
then a+d = b+c
and c+f = d+e
a+d +e
c + f = b + c d
a+f = b+e
(a, b) ~ (e, f)
~ is a transitive relation.
Thus '~' is reflexive, symmetric and transitive.
'~' is an equivalence relation.
(ii) We have (2, 5) A × A
The equivalence class of (2, 5) is the set of elements of A × A which are equivalent to
(2, 5)
[(2, 5)] = {(x, y) | (x, y) ~ (2, 5) , x, y A}
= {(x, y) | x + 5 = y + 2}
= {(x, y) | x - y = -3 or y = x + 3 ; x , y A}
Hence, [(2, 5)] = {(1, 3) (2, 5) (3, 6) (4, 7) (5, 8) (6, 9)} ... [Q (7, 10) A × A]
Example 4.7.22 Prove that in the set N × N, the relation R defined by (a, b) R (c, d) iff
ad = bc is an equivalence relation.
Solution : We know that relation R is an equivalence relation if R is reflexive, symmetric
and transitive.
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(i) We have, ab = ba
(a, b) R (a, b)
R is reflexive relation.
(ii) Suppose (a, b) R (c, d)
ad = bc
bc = ad
cb = da
(c, d) R (a, b)
R is symmetric relation.
(iii) Let (a, b) R (c, d) and (c, d) R (e, f)
ad = bc and cf = ed
adcf = bced
af = be (a, b) R (e, f)
R is transitive relation.
Thus, R is an equivalence relation.
Example 4.7.23 Let A = R × R (R is set of real numbers) and define the following relation on
A. (a, b) R (c, d) iff a 2 b 2 = c 2 d 2
(i) Show that (A, R) is an equivalence relation.
(ii) Find equivalence class of (3, 2)
Solution : Given that (a, b) R (c, d) iff a 2 b 2 = c 2 d 2
(i) (a) We have, a 2 b 2 = a 2 b 2
(a, b) R (a, b)
R is a reflexive relation.
(b) If (a, b) R (c, d) then a 2 b 2 = c 2 d 2
c 2 d2 = a 2 b 2
(c, d) R (a, b)
R is symmetric relation.
(c) Suppose (a, b) R (c, d) and (c, d) R (e, f)
a 2 b 2 = c 2 d 2 and c 2 d 2 = e2 + f2
a 2 b 2 = e2 f 2
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(a, b) R (e, f)
R is transitive relation.
Thus, R is an equivalence relation.
(ii) An equivalence class of (3, 2) is the set of elements of A which are equivalent to
(3, 2)
[(3, 2)] = {(x, y) | (x, y) R (3, 2) ; x, y R}
= {(x, y) | x 2 y 2 = 9 + 4 = 13 ; x, y }
[(3, 2)] = The set of points on circle x 2 y 2 = 13.
Example 4.7.24 Let R be the binary relation defined as R = {(a, b) 2/ (a – b) 3}
(i) Determine whether R is reflexive, symmetric, transitive and antisymmetric relation.
(ii) How many binary relations are there on the finite set
Solution : Given that,
R = {(a, b) R 2 | (a – b) 3}
(i) (a) We have, a – a = 0 3
(a, a) R R is reflexive.
(b) Suppose (a, b) R a – b 3
– (b – a) 3
b–a –3
(b – a) – 3 does not imply b – a 3
e.g. a = – 100 , b = 2
a – b = – 100 – 2 = – 102 < 3
(– 100, 2) R
But b – a = 2 – (– 100) = 2 + 100 = 102 3
Thus R is not symmetric relation.
(c) If (a, b), (b, c) R
then a–b 3, b–c 3
a–b+b–c 6
a–c 6
a – c may or may not be 3.
R is not tansitive relation.
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(d) Suppose aRb and bRa
a–b 3 and b – a 3
e.g. a = 1, b = 3,
a – b = – 2 < 3 and b – a = 2 < 3
but a b.
R is not antisymmetric relation.
2
(ii) Let n be the number of elements in a finite set A, then they are n n binary
relations on set A.
Example 4.7.25 Let S be the set of points in a plane. Let R be a relation such that xRy iff y
is within two centimeter from x. Is R equivalence relation ?
Solution : Given that
xRy iff |x - y| < 2 , x , y A
(i) We have, |x – x| = 0 < 2
xRx, x A
R is reflexive relation
(ii) If xRy |x – y| < 2
|y - x| < 2
yRx R is symmetric relation.
(iii) Suppose xRy and yRz
|x – y| < 2 and |y – z| < 2
|x – y| + |y – z| < 2 + 2 = 4
|x – z| < 4
|x – z| 2
R is not transitive relation.
So R is not an equivalence relation.
Example 4.7.26 Let R be a relation on the set of positive integers such that
R = {(x, y)|x – y is an odd positive integer}
Is R reflexive, symmetric, transitive, antisymmetric, an equivalence relation ?
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Solution : Given that, R = {(x, y) | x – y is an odd positive integer}
(i) For any positive integer x, x – x = 0 which is even integer.
(x, x) R
R is not reflexive relation.
(ii) Suppose (x, y) R
then x – y = Odd positive integer
y – x = Odd negative integer
(y, x) R
R is not symmetric relation.
(iii) Suppose (x, y), (y, z) R
x – y = Odd positive integer = p
y – z = Odd positive integer = q
x–y+y–z = p+q
x – z = p + q which is even positive integer.
{e.g. (5, 4) R , (4, 1) R. But (5, 1) R as 5 – 1 = 4}
(x, z) R
R is not transitive relation.
(iv) Suppose (x, y) and (y, x) R
then x – y = Odd positive integer = p
y – x = Odd positive integer = q
x y and y x
x = y
R is antisymmetric relation.
Example 4.7.27 Show that R = {(a, b) / a b (mod m)} is an equivalence relation on z. Show
that x 1 y 1 and x 2 y 2 then x 1 + x 2 y 1 + y 2
Solution : Given that
R = {(a, b)|a b (mod m)}
We know that a b (mod m) iff m | (a – b)
a – b = mk , k z
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(i) Let a z , a – a = 0 = m (0)
m|(a – a) a a (mod m)
(a, a) R R is reflexive relation.
(ii) Let (a, b) R
a b (mod m)
a – b = mk
b – a = m (– k)
m | b–a
b a (mod m)
(b, a) R
R is symmetric relation.
(iii) Suppose (a, b) (b, c) R
then a b (mod m) and b c (mod m)
a – b = mk1 and b – c = mk2
a – b + b – c = mk1 + mk2
a – c = m(k1 + k2) = mk
m|(a - c) a c (mod m)
(a, c) R
R is transitive relation.
R is reflexive, symmetric and transitive.
Thus R is an equivalence relation,
Now, x1 y1 x1 y1 (mod m)
x2 y2 x2 y2 (mod m)
x1 – y1 = mk1 and x2 – y2 = mk2
x1 – y1 + x2 – y2 = mk1 + mk2
(x1 + x2) – (y1 + y2) = m (k1 + k2)
m|(x1+ x2) – (y1 + y2)
(x1 + x2) (y1 + y2) (mod m)
x1 + x2 y1 + y2 (mod m)
x1 + x2 y1 + y2
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Example 4.7.28 For each of these relations on set A = {1, 2, 3, 4} decide whether it is
reflexive,
symmetric, transitive or antisymmetric.
R1 = {(1, 1) (2, 2) (3, 3) (4, 4)}
R 2 = {(1,1)(1, 2) (2, 2) (2,1) (3, 3) (4, 4)}, R 3 = {(1, 3) (1, 4) (2, 3) (2, 4) (3, 1) (3,4,)}.
SPPU : Dec.-10
Solution : i) For R 1 :
As a A, (a, a) R 1
R 1 is Reflexive, symmetric relation.
(a, b) and (b, c) R 1 R 1 is transitive relation.
(a, b) and (b, a) R 1 R 1 is antisymmetric relation.
R 1 reflexive, symmetric, transitive and anti-symmetric relation.
ii) For R 2 :
As a A (a, a) R 2 and aR 2 b bR 1 a, for a,b A
R 2 reflexive and symmetric relation.
For any aR2b and bR2C aR2C
R2 is transitive realtion
R2 is an equivalence relation
But (1, 2) and (2, 1) R2 and 1 2
R2 is not antisymmetric relation.
iii) For R 3 :
As 2 R 3 but (2, 2) R3
R 3 is not reflexive relation.
As (1, 4) R 3 but (4, 1) R3
R 3 is not symmetric relation.
As (2, 3) and (3, 1) R 3 but (2, 1) R 3.
R 3 is not transitive relation.
As (1, 3) and (3, 1) R3 but 1 3.
R3 is not transitive relation
As (1, 3) and (3, 1) R 3 but 1 3
R3 is not antisymmetric relation
Example 4.7.29 Consider the relation on A = { 1, 2, 3, 4, 5, 6} . R = { (i, j ) | i – j| = 2}. Is
R reflexive ? Is R symmetric ? Is R transitive ? SPPU : Dec.-14
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Solution : Given that A = {1, 2, 3, 4, 5, 6}
R = { (1, 3) (3, 1) (2, 4) (4, 2) (3, 5) (5, 3) (4, 6) (6, 4)}
As 2 Î A but (2, 2) R R is not reflexive.
For any (a, b) R (b,a) R R is symmetric.
As (1, 3) (3, 1) R but (1, 1) R R is not transitive.
Example 4.7.30 Let A = { 1, 2, 3, 4, 5, 6, 7 } and R = {(x, y) | x – y is divisible by 3}. Show
that R is an equivalence relation ? Draw graph of R.
SPPU : May-14, Dec.-12, 19, Marks 6
Solution : We have R = {x, y) | x – y is divisible by 3}
We know that x – x = 0 is divisible by 3
x R x, x A R is reflexive relation.
As x R y x – y is divisible by 3
y – x is also divisible by 3
y – z is divisible by 3
y R x for x, y A
R is a symmetric relation.
As,
x R y and y R x – y and y – z are divisible by 3
(x – y) + (y – z) is also divisible by 3
x – z is divisible by 3
xRz
R is a transitive relation.
R is an equivalence relation.
and 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)}
Its graph is as follows
1 7 3
4
2 5
Fig. 4.7.2
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4.8 Partitions of a Set SPPU : Dec.-11, 12, May-17
We shall now discuss the concept of partitions of a set which is similar to
equivalence class of set.
Definition :
Let A be any non empty set. A set
P = {A1 , A 2 , A 3 , .... A n } of non empty subsets of A is called a partition of set A if
n
i) A1 A2 A3 .... An = A = U Ai
i =1
i.e. Set A is the union of the sets A1 , A 2 , ... A n .
ii) Ai Aj = for i j i.e. All sets A i are mutually disjoints.
The partition of a set A is denoted by .
An element of a partition set is called a block. The rank of a partition is called as the
number of blocks of that partitions. It is denoted by r( ). For any non empty set, its
partitions are not unique. There are different partitions of the same set.
e.g.
1) Let A = Z = Set of integers
A
A1 = Set of even integers A1
A 2 = Set of odd integers
A2
A 3 = {1, 2, 3, 4, ....},
A 4 = {0}
Fig. 4.8.1
A 5 = {–1, –2, –3, –4, .....}
P1 = {A1 , A 2 } and P2 = {A 3 , A 4 , A 5 } are different partitions of set A.
2) If A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and it's subsets
A1 = {1, 4, 9} A 2 = {2, 6, 8, 10}, A 3 = {3, 5, 7},
The set P = {A1 , A 2 , A 3 } is such that
i) A1 , A 2 , A 3 are non empty sets
ii) A = A1 A2 A3
iii) A1 A2 = , A1 A3 = , A2 A3 =
Hence {A1 , A 2 , A 3 } form a partition for set A.
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4.8.1 Relation Induced by a Partition
Let P be the partition of a non empty set A. We can define a relation R on a set A as
x Ry iff x and y belong to the same block of the partition P. This relation R is called
as the relation induced by the partition.
e.g. i) Let A = Z and P = {A1 , A2 }
where A1 = Set of even integers and
A 2 = Set of odd integers
P is a partition for A = Z
Define x Ry Iff x and y belong to the same partition of Z
i.e. x and y belong to either A1 or A 2
i.e. x and y are either even or odd
This relation is an equivalence relation.
Theorem 1 : Let R be an equivalence relation on a set A then the set of equivalence
classes {[ a] R a A}.
Theorem 2 : Let A be any non empty set and let be a partition of A. Then induces
an equivalence relation on set A.
Example 4.8.1 Let A = {x, y, z, u, v}, = {{x, y}, {z}, {u, v}}. Find the equivalence relation
induced by .
Solution :
We have A = {x, y, z, u, v} and Define
x Ry iff x and y belongs to the same block of the partition of A.
= {{x, y}, {z}, {u, v}} has 3 blocks.
The first block {x, y} x, y same block
x Rx, xRy, yRx, yRy
The second block, {z} zRz
The third block, {u, v} uRu, uRv, vRu, vRv
The required relation is
R = {(x,x) (x,y), (y,x), (y,y), (z,z), (u,u), (u,v), (v,u), (v,v)}
The relation R is reflexive, symmetric and transitive
R is an equivalence relation.
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Example 4.8.2 Let A = (1, 2, 3, 4) consider partition.
P = {{1, 2, 3}, {4}}.
of A find the equivalence relation R on A determined by P. SPPU : May-17, Marks 3
Solution : Given that : A = {1, 2, 3, 4} and the partition P = {{1, 2, 3}, {4}}
We know that equivalence classes form a partition for the corresponding set.
P1 = {1, 2, 3} and P2 = {4} are two equivalence classes.
In each equivalence class, every element is related to all elements of that class.
Due to P1 = (1, 1) (1, 2), (1, 3), (2, 2), (2, 3), (2, 1), (3, 3), (3, 1), (3, 2) and for P2 , (4, 4)
Hence R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 4)}
is an equivalence relation with respect to the given partition P.
Example 4.8.3 Let A = {1, 2, 3, 4}, = {{1}, {2, 3}, {4}}. Find the equivalence relation
induced by .
Solution : Given that
= {{1}, {2, 3}, {4}} has 3 blocks
The first block, {1} iR1
The second block,{2, 3} 2R 2 , 2R 3 , 3R 2 , 3R 3
The third block, {4} 4R 4
The required relation is
R = {(1, 1), (2, 2), (2, 3), (3, 2), (3, 3), (4, 4)}
R is reflexive symmetric and transitive
R is an equivalence relation
Example 4.8.4 Define partition of a set. Let x = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Determine whether
or not each of the following is a partition of x :
A = {{2, 4, 5, 8}, {1, 9}, {3, 6, 7}}
B = {{1, 3, 6}, {2, 8}, {5, 7, 9}} SPPU : Dec.-11
Solution : Please refer section 4.8 for definition.
i) For set A :
The set A has 3 blocks,
A1 = {2, 4, 5, 8}, A 2 = {1, 9} A 3 = {3, 6, 7}
· The union of all these blocks is a set X
A1 A2 A3 = X
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· These blocks are mutually disjoints.
The set A forms a partition for the set X
ii) For set B :
The set B has 3 blocks.
B1 = {1, 3, 6}, B2 = {2, 8} B3 = {5, 7, 9}
As B1 B2 B3 X,
B is not a partition of set X.
Example 4.8.5 If S = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Determine whether or not each of the
following is a partition of S.
i) A = {{1, 3, 5}, {2, 6}, {4, 8, 9}}
ii) B = {{1, 3, 5}, {2, 4, 6, 8}, {7, 9}}
iii) C = {{1, 3, 5}}, {2, 4, 6, 8}, {5, 7, 9}}
iv) D = {{5}} SPPU : Dec.-12
Solution : Given that
S = {1, 2, 3, 4, 5, 6, 7, 8, 9}
i) A is not a partition of S because S is not the union of all blocks of A.
i.e. S A1 A2 A3
ii) B is the partition of S as B1 B2 B3 = S and B1 , B2 , B3 are mutually disjoints.
iii) As blocks of set C are not disjoints. The set C is not a partition of S.
iv) D = {{s}} is a partition of S, called as trivial partition.
4.8.2 Refinement of Partitions
Let and be partitions of a non empty set A.
Then is called refinement of if every block (or an element) of is contained in
a block of .
e.g. Let A = {1, 2, 3}, = {{1}, {2, 3}}, = {{1}, {2}, {3}}
{1}, {2}, {3} but {1}, {2, 3}
is a refinement of
Theorem : Let and be partitions of a non empty set A and R 1 , R 2 be the
equivalence relations induced by and respectively. Then refines iff R 2 R 1 .
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4.8.3 Product and Sum of Partitions
I) Let 1 and 2 be two partitions of a non empty set A. The product of 1 and 2
is denoted by 1 – 2 = is a partition of A such that
1) refines both 1 and 2
2) If refines both 1 and 2 then refines .
II) Let 1 and 2 be the partitions of a non empty set A. The sum of 1 and 2 is
denoted by = 1 2 is a partition of A such that
1) Both 1 and 2 refine
2) If 1 and 2 refine then refines
4.8.4 Quotient Set
Let A be any non empty set and R be an equivalence relation on A. The set of
mutually disjoint equivalence classes in which A is partitioned relatively to the
equivalence relation R is called the quotient set of A for relation R. If is denoted by A/R
or A.
e.g. The quotient set of Zl, for an equivalence relation
R = {( a, b) 3|{( a – b) , a, b Î Z } is
Z/R = {[0], [1], [2]}.
Theorem 1 : Let R 1 and R 2 be an equivalence relations induced by partitions 1 and
2 of a non empty set A respectively, then
i) The relation R = R 1 R 2 induces the product partition 1 2
ii) The relation R = (R 1 R 2 {i.e. The transitive closure of R 1
) R 2 } is an
equivalence relation on A and induces the partition 1 2 .
4.9 Closure of a Relation
SPPU : May-06, 07, 08, 15, 18, Dec.-05, 07, 12, 13, 14, 15, 16
Depending upon the nature of relations, there are mainly three types of closures of
relations.
4.9.1 Reflexive Closure
Let R be a relation on a set which is not reflexive relation. A relation R 1 = R is
called the reflexive closure of R if R is the smallest reflexive relation containing R.
If A = {a, b, c, d} then = {(a, a), (b, b), (c, c), (d, d)}
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Example 4.9.1 Let A = {1, 2, 3}. R 1 , R 2 and R 3 are relations on set A. Find the reflexive
closures of R 1 , R 2 and R 3 .
Where R 1 = {(1, 1) (2, 1)}, R 2 = {(1, 1) (2, 2), (3, 3)},
R 3 = {(3, 1) (1, 3), (2, 3)}.
Solution :
We have A = {1, 2, 3} = {(1, 1), (2, 2), (3, 3)}
Then
i) The reflexive closure of R 1 is R = R 1
R = {(1, 1) (2, 2) (3, 3) (2, 1)}
ii) The reflexive closure of R 2 is R = R 2 = R2
iii) The reflexive closure of R 3 is R = R 3
R = {(1, 1) (2, 2) (3, 3) (3, 1) (1, 3), (2, 3)}
4.9.2 Symmetric Closure
Let R be a relation on a set A and R is not symmetric relation.
A relation R 1 = R R –1 is called the symmetric closure of R if R R –1 is the smallest
symmetric relation containing R.
Example 4.9.2 Find the symmetric closure of the following relations. On A = {1, 2, 3}.
R 1 = {(1, 1) (2, 1)}
R 2 = {(1, 2) (2, 1) (3, 2) (2, 2)}
R 3 = {(1, 1) (2, 2) (3, 3)} SPPU : Dec.-12
Solution : Given that
We have A = {1, 2, 3}
i) R 1–1 = {(1, 1) (1, 2)}
R = R 1 R 1–1 = {(1, 1) (1, 2) (2, 1)} is the symmetric closure
of R 1
ii) R –1
2
= {(2, 1) (1, 2) (2, 3) (2, 2)}
R = R 2 R –1
2
= {(1, 2) (2, 1) (3, 2) (2, 3) (2, 2)} is the
symmetric closure of R 2
iii) R 3 is the symmetric relation.
R 3 itself is the symmetric closure.
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4.9.3 Transitive Closure
Let R be a relation on a set A which is not transitive relation. The transitive closure
of a relation R is the smallest transitive relation containing R. It is denoted by R .
The following theorem gives a procedure to find the transitive closure of a given
relation.
Theorem : Let A be any non empty set and |A| = n. Let R be a relation on A. Then
the transitive closure of R is R = R R 2 R 3 ... R n .
Example 4.9.3 If A = {1, 2, 3, 4, 5} and R = {(1, 2), (3, 4), (4, 5), (4, 1), (1, 1)}. Find it's
transitive closure.
Solution : Let R be the transitive closure of given relation R.
R = R R2 R3 R4 R5
Now R 2 = R R = {(3, 5) (3, 1) (4, 2) (4, 1) (1, 2) (1, 1)}
R 3 = R R 2 = {(3, 2) (3, 1) (4, 2) (4, 1) (1, 1) (1, 2)}
R 4 = R R 3 = {(3, 1) (3, 2) (4, 1) (4, 2) (1, 1) (1, 2)} = R 3
R 5 = R R 4 = {(3, 1) (3, 2) (4, 1) (4, 2) (1, 1) (1, 2)} = R 4 = R 3
R = {(1, 2) (3, 4) (4, 5) (4, 1) (1, 1) (3, 5) (3, 1) (4, 2) (3, 2)}
Which is the transitive closure of R.
4.9.4 Warshall's Algorithm to Find Transitive Closure
To find the transitive closure of a relation by computing various powers of R or
product of the relation matrix is quite impractical for large relations. Warshall's
algorithm gives an alternate method for finding transitive closure of R. Warshall's
algorithm is practical and efficient method.
Consider the following steps to find transitive closure of the relation R on a set A.
Step 1 : We have |A| = n
We require Wo , W1 , W2 , ... Wn . Warshall sets
Wo = Relation Matrix of R = M R .
Step 2 : To find the transitive closure of relation R on set A, with |A| = n
Procedure to compute Wk from Wk–1 is as follows
i) Copy 1 to all entries in Wk from Wk–1 , where there is a 1 in Wk–1 .
ii) Find the row numbers p1 , p 2 , p 3 ... for which there is 1 in column k in Wk–1 and
the column numbers q 1 , q 2 , q 3 ... for which there is 1 in row k of Wk–1 .
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iii) Mark entries in Wk as 1 for ( p i , q i ). If there are not already 1.
Step 3 : Stop the procedure when Wn is obtained and it is the required transitive
closure of R.
Examples
Example 4.9.4 Find the transitive closure of R by Warshall's algorithm.
Where A = {1, 2, 3, 4, 5, 6} and R = {(x, y)/(x – y) = 2} SPPU : Dec.-05, 12, 13, 14
Solution :
Step 1 : We have |A| = 6,
R = {(1, 3), (3, 1), (2, 4), (4, 2), (4, 6), (6, 4), (3, 5), (5, 3)}
Thus we have to find Warshall's sets, Wo , W1 , W2 , W3 , W4 , W5 and W6 .
The first set Wo is same as M R . Which is shown below
1 2 3 4 5 6
1 0 0 1 0 0 0
2 0 0 0 0 1 0
3 1 0 0 0 1 0
W0 = MR =
4 0 1 0 0 0 1
5 0 0 1 0 0 0
6 0 0 0 1 0 0
Step 2 : To find W1 :
To find W1 from Wo , we consider the first column and first row.
In a column C1 , 1 is present at R 3
In a row R 1 , 1 is present at C 3
Thus add new entry in W1 , at (R 3 , C 3 ) which is given below
0 0 1 0 0 0
0 0 0 1 0 0
1 0 1 0 1 0
W1 =
0 1 0 0 0 1
0 0 1 0 0 0
0 0 0 1 0 0
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Step 3 : To find W2 :
To find W2 from W1 , we consider the second column C 2 and second row R 2 .
In C 2 , 1 is present at R 4
In R 2 , 1 is present at C 4
Thus add new entry in W2 at (R 4 , C 4 ), which is given below
0 0 1 0 0 0
0 0 0 1 0 0
1 0 1 0 1 0
W2 =
0 1 0 1 0 1
0 0 1 0 0 0
0 0 0 1 0 0
Step 4 : To find W3 :
To find W3 from W2 , we consider the third column and third row.
In C 3 , 1 is present at R 1 , R 3 , R 5
In R 3 , 1 is present at C1 , C 3 , C 5
Thus add new entries in W3 at ( R 1 , C1 ), ( R 1 , C 3 ), ( R 1 , C 5 )
( R 2 , C1 ), ( R 2 , C 3 ), ( R 2 , C 5 ) ( R 3 , C1 ), ( R 3 , C 3 ), ( R 3 , C 5 ) which is given below
1 0 1 0 1 0
0 0 0 1 0 0
1 0 1 0 1 0
W3 =
0 1 0 1 0 1
1 0 1 0 1 0
0 0 0 1 0 0
Step 5 : To find W4 :
To find W4 from W3 , we consider the fourth column and fourth row.
In C 4 , 1 is present at R 2 , R 4 , R 6
In R 4 , 1 is present at C 2 , C 4 , C 6
Thus add new entries in W4 at (R 2 , C 2 ), (R 2 , C 4 ), (R 2 , C 6 ), (R 4 , C 2 ), (R 4 , C 4 ),
(R 4 , C 6 ), (R 6 , C 2 ), (R 6 , C 4 ), (R 6 , C 6 ) which is given below
1 0 1 0 1 0
0 1 0 1 0 1
W4 = 1 0 1 0 1 0
0 1 0 1 0 1
1 0 1 0 1 0
0 1 0 1 0 1
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Step 6 : To find W5 :
To find W5 from W4 , we consider the 5 th column and 5 th row.
In C 5 , 1 is present at R 1 , R 3 , R 5
In R 5 , 1 is present at C1 , C 3 , C 5
Thus add new entries in W5 at (R 1 , C1 ), (R 1 , C 3 ), (R 1 , C 5 ), (R 3 , C1 ), (R 3 , C 3 ),
(R 3 , C 5 ), (R 5 , C1 ), (R 5 , C 3 ), (R 5 , C 5 ) which is given below
1 0 1 0 1 0
0 1 0 1 0 1
W5 = 1 0 1 0 1 0
= W4
0 1 0 1 0 1
1 0 1 0 1 0
0 1 0 1 0 1
Step 7 : To find W6 :
To find W6 from W5 we consider the 6 th column and 6 th row.
In C 6 , 1 is present at R 2 , R 4 , R 6
In R 6 , 1 is present at C 2 , C 4 , C 6
Thus add new entries in W6 at (R 2 , C 2 ), (R 2 , C 4 ), (R 2 , C 6 ), (R 4 , C 2 ), (R 4 , C 4 ),
(R 4 , C 6 ), (R 6 , C 2 ), (R 6 , C 4 ), (R 6 , C 6 ) which is given below
W6 = W5 = W4
Hence W6 is the relation matrix of R
(1, 1), (1, 3) (1, 5) ( 2, 2) ( 2, 4) ( 2, 6) ( 3, 1) ( 3, 3) ( 3, 5)
R =
( 4 , 2) ( 4 , 4) ( 4 , 6) ( 5, 1) ( 5, 3) ( 5, 5) ( 6, 2) ( 6, 4) ( 6, 6)
Example 4.9.5 Let R = {(a, d), (b, a), (b, d), (c, b), (c, d), (d, c)}
Use Warshall's algorithm to find the matrix of transitive closure where A = {a, b, c, d}
SPPU : Dec.-15
Solution : Step 1 : We have
A = {a, b, c, d} |A| = 4
R = {(a, d) (b, a), (b, d) (c, b), (c, d), (d, c)}
Thus we have to find Warshall's sets Wo , W1 , W2 , W3 , W4
The first set Wo = M R
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a b c d
a 0 0 0 1
b 1 0 0 1
W0 = MR =
c 0 1 0 1
d 0 0 1 0
Step 2 : To find W1 :
To find W1 from Wo , we consider the first column and first row.
In C1 , 1 is present at R 2
In R 1 , 1 is present at C 4
Thus add new entry in W1 at (R 2 , C 4 )
0 0 0 1
1 0 0 1
W1 =
0 1 0 1
0 0 1 0
Step 3 : To find W2
To find W2 , from W1 , we consider the second column and second row.
In C 2 , 1 is present at R 3
In R 2 , 1 is present at C1 and C 4
Thus add new entries in W2 at (R 3 , C1 ), (R 3 , C 4 ) which is given below.
0 0 0 1
1 0 0 1
W2 =
1 1 0 1
0 0 1 0
Step 4 : To find W3 :
To find W3 from W2 , we consider the 3 rd column and 3 rd row.
In C 3 , 1 is present at R 4
In R 3 , 1 is present at C1 , C 2 , C 4
Thus add news entries in W3 at (R 4 , C 1 ), (R 4 , C 2 ), (R 4 , C 4 )
Which is given below
0 0 0 1
1 0 0 1
W3 =
1 1 0 1
1 1 1 1
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Step 5 : To find W4 :
To find W4 from W3 , we consider the 4 th column and 4 th row.
In C 4 , 1 is present at R 1 , R 2 , R 3 , R 4
In R 4 , 1 is present at C1 , C 2 , C 3 , C 4
Thus we add new entries in W4 at (R1 , C1 ), (R1 , C2 ), (R1 , C3 ), (R1 , C4 ), (R 2 , C1 ),
(R 2 , C2 ), (R 2 , C3 ), (R 2 , C4 ), (R 4 , C3 ), (R 3 , C1 ), (R 3 , C2 ), (R 3 , C4 ), (R 4 , C1 ), (R 4 , C2 ),
(R 4 , C3 ), (R 4 , C4 )
Which is given below
1 1 1 1
1 1 1 1
W4 =
1 1 1 1
1 1 1 1
Hence W4 is the relation matrix of R .
(a, a), (a, b) (a, c) (a, d) (b, a) (b, b) (b, c) (b, d)
and R =
(c, a), (c, b) (c, c) (c, d) (d, a) (d, b) (d, c) (d, d)
Example 4.9.6 Find the transitive closure of the relation R on
A = {1, 2, 3, 4} defined by
R = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (3, 4), (3, 2), (4, 2), (4, 3)}
SPPU : Dec.-07, May-15, 18, Marks 4
Solution : Step 1 :
We have |A| = 4, Thus we have to find
Warshall's sets, W0 , W1 , W2 , W3 , W4
The first set W0 = M R
1 2 3 4
1 0 1 1 1
2 1 0 1 0
W0 = MR =
3 0 1 0 1
4 0 1 1 0
Step 2 : To find W1 :
To find W1 from W0 , we consider the first column and first row.
In C1 , 1 is present at R 2
In R 1 , 1 is present at C 2 , C 3 , C 4
Thus add new entries in W1 at (R 2 , C2 ), (R 2 , C3 ), (R 2 , C4 ) which is given below
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0 1 1 1
1 1 1 1
W1 =
0 1 0 1
0 1 1 0
Step 3 : To find W2 :
To find W2 from W1 , we consider the 2 nd column and 2 nd row.
In C 2 , 1 is present at R 1 , R 2 , R 3 , R 4
In R 2 , 1 is present at C1 , C 2 , C 3 , C 4
Thus we add new entries in W2 at (R1 , C1 ), (R1 , C2 ), (R1 , C3 ), (R1 , C4 ), (R 2 , C1 ),
(R 2 , C2 ), (R 2 , C3 ), (R 2 , C4 ), (R 3 , C1 ), (R 3 , C2 ), (R 3 , C2 ), (R 3 , C4 ), (R 4 , C1 ), (R 4 , C2 ),
(R 4 , C3 ), (R 4 , C4 )
Which is given below
1 1 1 1
1 1 1 1
W2 =
1 1 1 1
1 1 1 1
All entries in W2 are 1
Hence W2 is the relation matrix of transitive closure of R.
(1, 1), (1, 2) (1, 3) (1, 4) (2, 1) (2, 2) (2, 3) (2, 4)
and R =
(3, 1), (3, 2) (3, 3) (3, 4) (4, 1) (4, 2) (4, 3) (4, 4)
Example 4.9.7 If R = {(a, b), (b, a), (b, c), (c, d), (d, a)} be a relation on the set A = {a, b, c,
d}. Find the transitive closer of R using Warshall's algorithm. SPPU : Dec.-16, Marks 6
Solution :
Step 1 : We have A = {a, b, c, d}
R = {(a, b), (b, a), (b, c), (c, d), (d, a)}
Thus we have to find Warshall's sets W 0 , W 1 , W 2 , W 3 , W 4 .
The first set W0 = MR a b c d
a 0 1 0 0
b 1 0 1 0
W0 = MR = c 0 0 0 1
d 1 0 0 0
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Step 2 : To find W 1
To find W 1 from W 0 , we consider the first column and first row. In C 1 , 1 is present
R 2 and R 4
In R 1 , 1 is present at C 2
Thus add new entries in W 1 at (R 2 , C 2 ) (R 4 , C 2 )
0 1 0 0
W1 = 1 1 1 0
0 0 0 1
1 1 0 0
Step 3 : To find W 2
To find W 2 from W 1 , we consider second column and second row of W 1 .
In C 2 , 1 is present at R 1 , R 2 , R 3
In R 2 , 1 is present at C 1 , C 2 , C 3
Add new entries in W 2 at (R 1 , C 1 ), (R 1 , C 2 ), (R 1 , C 3 ) (R 2 , C 1 ), (R 2 , C 2 ), (R 2 , C 3 )
(R 3 , C 1 ), (R 3 , C 2 ), (R 3 , C 3 )
1 1 1 0
1 1 1 0
W2 = 1 1 1 1
1 1 0 0
Step 4 : To find W 3
To find W 3 from W 2 , we consider the III rd row and III rd column of W 2 .
In C 3 , 1 is present at R 1 , R 2 , R 3
In R 3 , 1 is present at C 1 , C 2 , C 3 , C 4
Add new entries in W 2 at (R 1 , C 1 ), (R 1 , C 2 ), (R 1 , C 3 ), (R 1 , C 4 ), (R 2 , C 1 ), (R 2 , C 2 ),
(R 3 , C 3 ), (R 2 , C 4 ), (R 3 , C 1 ), (R 3 , C 2 ), (R 3 , C 3 ), (R 3 , C 4 )
1 1 1 1
W3 = 1 1 1 1
1 1 1 1
1 1 0 0
Step 5 : To find W 4
To find W 4 from W 3 , we consider 4 th row and 4 th column of W 3 .
In C 4 , 1 is present at R 1 , R 2 , R 3
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In R 4 , 1 is present at C 1 , C 2 , C 3
Add new entries in W 3 at (R 1 , C 1 ), (R 1 , C 2 ), (R 1 , C 3 ), (R 2 , C 1 ), (R 2 , C 2 ) ,(R 2 , C 3 ),
(R 3 , C 1 ), (R 3 , C 2 ), (R 3 , C 3 )
1 1 1 1
1 1 1 1
W4 =
1 1 1 1
1 1 0 0
Hence the W 4 is the relation metrix of R
(a, a) (a, b) (a, c) (a, d) (b, a) (b, b) (b, c)
And R =
(b, d) (c, a) (c, b) (c, c) (c, d) (d, a) (d, b)
Example 4.9.8 Find the transitive closure of R by Warshall's algorithm. A = {Set of positive
integers ! 10}.
R = {(a, b) / a divides b} SPPU : May-07
Solution :
Step 1 : We have |A| = 10. Thus we have to find
Warshall's sets W0 , W1 , W2 , .... W10
R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10),
(2, 2), (2, 4), (2, 6), (2, 8), (2, 10), (3, 3), (3, 6), (3, 9), (4, 4), (4, 8),
(5, 5), (5, 10), (6, 6), (7, 7), (8, 8), (9, 9), (10, 10)}
The first set W0 = M R
1 2 3 4 5 6 7 8 9 10
1 1 1 1 1 1 1 1 1 1 1
2 0 1 0 1 0 1 0 1 0 1
3 0 0 1 0 0 1 0 0 1 0
4 0 0 0 1 0 0 0 1 0 0
W0 = 5 0 0 0 0 1 0 0 0 0 1
6 0 0 0 0 0 1 0 0 0 0
7 0 0 0 0 0 0 1 0 0 0
8 0 0 0 0 0 0 0 1 0 0
9 0 0 0 0 0 0 0 0 1 0
10 0 0 0 0 0 0 0 0 0 1
The relation R itself is a transitive relation on the set of positive integers. Hence
R = R and
W0 = M R is the relation matrix of R
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Example 4.9.9 Use Warshall's Algorithm to find transitive closure of R where
1 0 1
MR = 0 1 0 and A = {1, 2, 3}
1 1 0 SPPU : May-06, May-08
Solution :
Step 1 : We have |A| = 3. Thus we have to find Warshall's sets W0 , W1 , W2 , W3
The first set is
1 0 1
W0 = MR = 0 1 0
1 1 0
Step 2 : To find W1 :
To find W1 from W0 , we consider the first column and the first row.
In C1 , 1 is present at R 1 , R 3
In R 1 , 1 is present at C1 and C 3
Thus add new entries in W1 at (R1 , C1 ), (R1 , C3 ), (R 3 , C1 ), (R 3 , C3 )
Which is given below
1 0 1
W1 = 0 1 0
1 1 1
Step 3 : To find W2 :
To find W2 from W1 , we consider the 2 nd column and 2 nd row.
In C 2 , 1 is present at R 2 , R 3
In R 2 , 1 is present at C 2
Thus add new entries in W2 at (R 2 , C2 ), (R 3 , C2 ) which is given below
1 0 1
W2 = 0 1 0 = W1
1 1 1
Step 4 : To find W3
To find W3 from W2 , we consider the 3 rd column and 3 rd row.
In C 3 , 1 is present at R 1 , R 3
In R 3 , 1 is present at C1 , C 2 , C 3
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Thus add new entries in W3 at (R1 , C1 ), (R1 , C2 ), (R1 , C3 ), (R 3 , C1 ), (R 3 , C2 ),
(R 3 , C3 ) which is given below
1 1 1ù
W3 = ê 0 1 0ú
ê ú
ë1 1 1û
Hence W3 is the relation matrix of R *
and R * = {(1, 1) (1, 2) (1, 3) (2, 2) (3, 1) (3, 2) (3, 3)}
4.10 Partially Ordered Set SPPU : Dec.-06
A relation R on a set A is called a partially ordered relation iff R is reflexive,
anti-symmetric and transitive relation.
The set A together with partially ordered relation is called a partially ordered set or
POSET.
It is denoted by (A, R) or (A, £) where £ is a partially ordered relation.
Examples :
1) (Â, £) (N, £) are Posets.
where '£' is reflexive, antisymmetric and transitive relation.
2) If A = P(S) where S = (a, b, c) and for X, Y Î A, Define X £ Y or XRY iff X Í Y.
As X £ X Þ X £ X. \ ' £ ' is reflexive.
If X £ Y, Y £ Z Þ X Í Y and Y Í X Þ X = Y
\ ' £ ' is antisymmetric relation.
If X £ Y, Y £ Z Þ X Í Y and Y Í Z Þ X Í Z Þ X £ Z
Þ ' £' is transitive relation.
\ (P(S), Í) or (P(S), £) is a poset.
I) Comparable elements : Let (A, £) be a poset. Two elements a, b in A are said to be
comparable elements if a £ b or b £ a. Two elements a and b of a set A are said to be
non-comparable if neither a £ b nor b £ a.
In above example (2),
The comparable elements are
{a} = {a, b}, {b} = {a, b, c} {b, c} Í {a, b, c}
Non comparable elements are
{a} Ë {b } {a} Ë {a, c}
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II) Totally ordered set : Let A be any nonempty set. The set A is called linearly
ordered set or totally ordered set if every pair of elements in A are comparable.
i.e. for any a, b Î A either a £ b or b £ a .
4.10.1 Hasse Diagram
It is useful tool, which completely describes the associated partially ordered relation.
It is also known as ordering diagram.
A diagram of graph which is drawn by considering comparable and non-comparable
elements is called Hasse diagram of that relation. Therefore while drawing Hasse
diagram following points must be followed.
1) The elements of a relation R are called vertices and denoted by points.
2) All loops are omitted as relation is reflexive on poset.
3) If aRb or a £ b then join a to b by a straight line called an edge the vertex b
appears above the level of vertex a. Therefore the arrows may be omitted from
the edges in Hasse diagram.
4) If a £/ b and b £/ a i.e. a and b are non - comparable elements, then they lie on same
level and there is no edge between a and b.
5) If a £ b and b £ c then a £ c. So there is a path a ® b ® c. Therefore do not join a
to c directly i.e. delete all edges that are implied by transitive relation.
Example 4.10.1 Draw Hasse diagram of a poset (P(s), Í) where S = {a, b, c}.
Solution : P (s) = { f, {a}, {b}, {c}, {a, b} {a, c} {b, c} {a, b, c}}
Now find the comparable and non comparable elements.
f Í {a}, fÍ {b}, fÍ {c}, \ {a}, {b}, {c} lie above the level of f.
{a} Í {a, b}, {b}Í {a, b}, {c} Í {a, c} \ {a, b}, {b, c} {a,c} lies above the level of {a}, {b}, {c}.
{a, b} Í S, {b, c} Í S, {a, c} Í S \ S lies above the level of {a, b}, {a, c}, {b, c}
But {a}, {b} {c{ are non comparable \ {a}, {b}, {c} lie on same level.
{a, b}, {a, c}, {b, c} are non comparable \ lie on same level.
By considering the above observations, the Hasse diagram is as follows :
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{a,b,c}
Level 4
Level 3 {a,b} {a,c}
{b,c}
Level 2 {a}
{b} {c}
Level 1
Fig. 4.10.1
4.10.2 Chains and Antichains
Let (A, £) be a poset. A subset of A is called a chain if every pair of elements in the
subset are related.
A subset of A is called antichain if no two distinct elements in a subset are related.
e.g. In above example (4.10.1)
1) The chains are {f, {a }, {a, b}, {a, b, c}}, {{a}, {a, c}, {a, b, c}}, {{b, c}, {a, b, c}}
2) Antichains are {{a}, {b}, {c}}
Note : 1) The number of elements in the chain is called the length of chain.
2) If the length of chain is n in a poset (A, £ ) then the elements in A can be
partitioned into n disjoint antichains.
4.10.3 Elements of Poset
1) Let (A, £ ) be a poset. An element a Î A is called a maximal element of A if there
is no element c Î A such that a £ c.
2) An element b Î A is called a minimal element of A if there is no element c Î A
such that c £ b
3) Greatest element : An element x Î A is called a greatest element of A if for all
a Î A, a Î x. It is denoted by 1 and is called the unit element.
4) Least element : An element y Î A is called a least element of A if for all a Î A,
y £ a.
It is denoted by 0 and is called as zero element.
5) Least upper bound (lub) : Let (A, £ ) be a poset. For a, b, c Î A, an element C is
called upper bound of a and b if a £ c and b £ c C is called as least upper bound
of a and b in A if C is an upper bound a and b there is no upper bound d of a
and b such that d £ c. It is also known as supremum.
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6) Greatest lower bound (glb) : Let (A, £ ) be a poset. for a, b, l Î A, an element l is
called the lower bound of a and b if l £ a and l £ b.
An element l called the greatest lower bound of a and b if l is the lower bound of a
and b and there is no lower bound f of a and b such that · l £ f .
glb is also called as infimum.
Example 4.10.2 Determine the greatest and least elements of the poset whose Hasse diagrams
are shown below.
d
d e d c d
c
c b b c
b
a b a a
a
I II III IV
Fig. 4.10.2
Solution : The Poset shown in Fig. 4.10.2 (I) has neither greatest not least element.
The Poset shown in Fig. 4.10.2 (II), has greatest and a as least element.
The Poset shown in Fig. 4.10.2 (III), has no greatest element but a is the least element.
The Poset shown in Fig. 4.10.2 (IV), has greatest and a as least element.
Example 4.10.3 Find glb, lwb, ub, lb, maximal, minimal, of the poset (A, R), Here aRb if
a / b where. A = { 2, 3, 5, 6, 10, 15, 30, 45}
Solution : We have A = {2, 3, 5, 6, 10, 15, 30, 45} and aRb iff a|b.
Hasse diagram is as follows :
30
Level 3 45
Level 2 6 10
15
Level 1 2
3 5
Fig. 4.10.3
1) Here 10 and 30 are upper bounds of 2 and 5, But 10 is the least upper bound of
2 and 5.
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2) 5, 15, 3 are lower bounds of 30 and 45. But 15 is the greatest lower bound of 30
to 45.
3) This Poset has neither greatest element nor least element.
4) This poset has two maximal elements 45 and 30 as there is no element c such
that 45 £ C and 30 £ C.
5) This poset has three minimal elements 2, 3 and 5. because there is no element
x Î A such that x £ 2, x £ 3 and x £ 5
4.10.4 Types of Lattices
A lattice is a poset in which every pair of elements has a least upper bound (lub) and
a greatest lower (glb).
Let (A, £) be a poset and a, b Î A then lub of a and b is denoted by a Ú b. It is called
the join of a and b.
i.e. a Ú b = lub (a, b)
The greatest lower bound of a and b is called the meet of a and b and it is denoted
by a Ù b
\ a Ù b = glb (a, b)
From the above discussion, it follows that a lattic is a mathematical structure with
two binary operations Ú (join) and Ù (meet). It is denoted by {L, Ú , Ù }.
Examples :
Example 4.10.4 Let A = {1, 2, 3} P (A) = { f, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3 } show that
P(A), Í ) is a lattice
Solution : The Hasse diagram of the poset (P(A), Í) is given below :
{a,b,c}
{a,b} {a,c}
{b,c}
{a}
{b} {c}
Fig. 4.10.4
Here every pair of elements of a poset has lub and glb. Hence (P(A), Í) is a lattice.
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Example 4.10.5 Determine which of the following posets are lattice.
f
6 4
5 d c
3 d e
2 b
2
4 c
3 1
a
1 a b
I II III IV
Fig. 4.10.5
Solution : I) In Fig. 4.10.5 (I), every pair of elements has lub and glb.
\ It is a lattice.
II) In Fig. 4.10.5 (II), every pair of elements has lub and glb.
\ It is a lattice.
II) In Fig. 4.10.5 (III), a Ù b does not exist.
\ It is not a lattice.
IV) In Fig. 4.10.5 (IV), c Ú d does not exist.
\ It is not a lattice.
Example 4.10.6 Let A be the set of positive factors of 15 and R be a relation on A s.t.
R = {xRy| x divides y, x y Î A}. Draw Hasse diagram and give and Ù and Ú for lattice.
SPPU : Dec.-06
Solution : We have A = {1, 3, 5, 15}
R = {(1,1) (1,3) (1,5) (1, 15) (3,15) (5,15) (15,15)}
Hasse diagram of R is : Table for Ù and Ú
Ú 1 3 5 15 Ù 1 3 5 15
15 1 1 3 5 15 1 1 1 1 1
3 3 3 15 15 3 1 3 1 3
5 3
5 5 15 5 15 5 1 1 5 5
1
15 15 15 15 15 15 1 3 5 15
Fig. 4.10.6
Every pair of elements has lub and glb. \ It is a lattice.
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4.10.5 Properties of a Lattice
Let (L, Ù , Ú) be a lattice and a, b, c Ï L. Then L satisfies the following properties.
1) Commutative property
a Ù b = b Ù a and a Ú b = b Ú a
2) Associative law
a Ú (b Ù c) = (a Ú b) Ù (a Ú c)
a Ù (b Ú c) = (a Ù b) Ú (a Ù c)
3) Absorption law
a Ù (a Ú b) = a and av (a Ù b) = a
3) a Ù a = a, a Ú a = a
4) a Ù b = a iff a Ú b = b
4.10.6 Types of Lattices
I) Bounded lattice : A lattice L is called a bounded lattice if it has a greatest element 1
and least element 0.
II) Sublattice : Let, (L, Ú , Ù ) be a lattice. A non empty subset L 1 of L is called a
sublattice of L if L 1 itself is a lattice w.r.t. the operations of L.
III) Distributive lattice : A lattice (L, Ú , Ù ) is called a distributive lattice if for any
elements a,b,c Î L, it satisfies the following properties,
i) a Ú (b Ù c) = ( a Ú b) Ù (a Ú c)
ii) a Ù (b Ú c) = ( a Ù b) Ú (a Ù c)
If the lattice does not satisfy the above properties then it is called a non distributive
lattice.
Theorem : Let, (L, Ù , Ú) be a lattice with universal bounds 0 and 1 then for any
a Î L, a Ú 1 =1, a Ù 1 = a, 0 Ú a = a, 0 Ù a = 0.
30 1
III) Complement lattice : Let (L, Ù , Ú) be a lattice with
universal bounds 0 and 1 for any a Î L, b Î L is said to be
2 3 5
complement of a if a Ú b = 1 and a Ù b = 0.
A Lattice in which every element has a complement in that 10
lattice, is called the complemented lattice.
e.g. 1) The Hasse diagram is here 0 » 1 and 1 » 30. Fig. 4.10.7
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i) 2 Ù 3 = 0 and 2 Ú 3 = 1, 2 Ù 5 = 0 and 2 Ú 5 = 1
\ 2 has two compliments 3 and 5
Hence the complement is not unique.
4.11 Principle of Duality SPPU : Dec.-10, 11, 13, 14, 15, 16, May-14, 15, 19
Any statement about lattice involving Ù , Ú £ , ³ remains true if 'Ù' is replaced by 'Ú',
'Ú by Ù ', ' £ by ³', ' ³ by £', '0 by 1' and '1 by 0'.
e.g. 1) a Ú (b Ù c) = a Ù (b Ú c)
2) a Ù (b Ú 1) = a Ú (b Ù 0)
Examples
Example 4.11.1 Let A = {1, 2, 3, 4, 6, 9, 12} Let a relation R on a set A is R = {(a,b)/ a
divides b " a, b Î A}. Give list of R. Prove that it is a partial ordering relation. Draw
Hasse diagram of the same. Prove or disprove it is a lattice. Give two examples of chain
and antichains. SPPU : Dec.-11, May-19, Marks 6
Solution : We have A = {1, 2, 3, 4, 6, 9, 12}
ì(1,1), (1,2), (1,3), (1,4) (1,6), (1,9), (1,12), (2,2), (2,4), (2,6),(2,12) (3,3), (3,6), (3,9) ü
and R = í ý
î(3,12), (4,4), (4,12), (6,6), (6,12), (9,9), (12,12) þ
We know that for any a Î A, a | a \ aRa
\ R is a reflexive relation.
As a | b and b | a Þ a = b \ R is antisymmetric relation.
As a | b and b | c Þ a | c Þ R is a transitive relation.
\ R is reflexive antisymmetric and transitive
\ (A, R) is a poset and R is a partial ordering relation.
Hasse diagram is as follows :
12
4 6 9
3
2
1
Fig. 4.11.1
In above diagram 6 Ú 9 does not exist. \ It is not a lattice.
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i) Chains are
{1, 2, 4, 12}
{1, 2, 6, 12}
{1, 3, 6, 12}
ii) Antichains are
{6, 9} {9, 12}
{4, 9}
Example 4.11.2 Determine whether the poset represented by each of the Hasse diagram are
lattices. Justify your answer.
6 h
f
e 4 5 c g
f
c d
c
2 3 b d
b
a
1 a
Fig. 4.11.2 SPPU : Dec.-10
Solution :
I) In Fig. 4.11.2 (I), every pair of element has glb and lub. \ It is a lattice.
II) In Fig. 4.11.2 (II), every pair of elements has lub and glb. \ It is a lattice.
III) In Fig. 4.11.2 (III), every pair of elements gas lub and glb. \ It is a lattice.
Example 4.11.3 Show that the set of all divisors of 36 forms a lattice. SPPU : Dec.-14
Solution : Let A = {1, 2, 3, 4, 6, 9,12, 18, 36 } and Let '£' is a divisor of.
It's Hasse diagram is as follows.
36 1
12 18
4 6 9
2 3
10
Fig. 4.11.3
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The universal upper bound 1 is 36 and lower bound 0 is 1. Every pairs of elements
of this poset has lub and glb.
\ It is a lattice.
Example 4.11.4 SLet n be a positive integer, S n be the set of all divisors of n, Let D denote
the relation of divisor. Draw the diagram of lattices for n = 24, 30, 6. SPPU : May-15
Solution : Given that
i) We have S 6 = {1,2,3,6}, D is the relation of divisor.
ii) S 24 = {1, 2, 3, 4, 6, 8, 12, 24}
iii) S 30 = {1, 2, 3, 5, 6, 10, 15, 30}
Diagrams of Lattices are as follows.
24
12 30
6 6 6 10
4 15
2
2 3 2 3 3 5
1 1 1
S6 S24 S30
Fig. 4.11.4
Example 4.11.5 Show that the set of all divisors of 70 forms a lattice.
SPPU : Dec.-13, May-19, Marks 3
Solution : Let A = {1, 2, 5, 7, 10, 14, 35, 70}
and Let ' £' is "a divisor of".
The universal upper bound 1 is 70 and the lower bound 0 is 1.
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It's Hasse diagram is as follows :
70
10 14
35
2
5 7
1
Fig. 4.11.5
Every pair of elements of A has Ù and Ú.
\ It is a lattice [write table of Ù and Ú].
Example 4.11.6 Let A = {1, 2, 3, 4 , 5, 6, 7, 8 ,9, 12, 18, 24} be ordered by the relation x
divides y. Show that the relation is a partial ordering and draw Hasse diagram.
SPPPU : Dec-15
Solution : We have A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 18, 24}
R = {(x,y) | x divides y, for x, y Î A}
R = {(1, 2) (1, 3), (1, 4), (1, 5) (1, 6) (1, 7) (1, 8) (1, 9) (1, 12) (1, 18) (1, 24)
(2, 2) (2, 4) (2, 6) (2, 8) (2, 12) (2, 18) (2, 24) (5, 5) (6, 6) (6, 12) (6, 18)
(6, 24), (7, 7) (8, 8) (8, 24) (9, 9) (9, 18) (12, 12) (12, 24) (18, 18) (4, 24)}
We have for any x Î A, x|x Þ R is a reflexive for x|y and y|x Þ x = 0 Þ R is
antisymmetric. If x|y and y|z Þ x|z \ R is a transitive relation.
\ R is a partial ordering relation, It's Hasse diagram is as follows.
24
8 18
12
4 6 9
5 7
3
2
1
Fig. 4.11.6
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Example 4.11.7 Let x = {2, 3, 6, 12, 24, 36} and x £ y iff x divides y find
i) Maximal element ii) Minimal element iii) Chain iv) Antichain v) Is Poset lattice
SPPU : May-14
Solution : We have x = {2, 3, 6, 12, 24, 36}
The relation 'R' = '£'
\ R = {(2, 2) (2, 6) (2, 12) (2, 24) (2, 36) (3, 3) (3, 6) (3, 12) (3, 24) (3, 36),
(6, 6) (6, 12) (6, 24) (6, 36) (12, 12)(12, 24) (12, 36) (24, 24) (36, 36)}.
It's Hasse diagram is as follows .
36 24
12
2 3
Fig. 4.11.7
i) Maximal elements are 24, 36
ii) Minimal elements are 2, 3
iii) Chain { 2, 6, 12, 24}, { 2, 6, 12, 36}, {3, 6, 12, 24}, {3, 6, 12, 36}
iv) Antichain : {2, 3} {24, 36}
v) The given poset is not a lattice as 2 Ù 3 does not exist.
Example 4.11.8 The following is the Hasse diagram of the poset :
{(a, b, c, d, e), <} Is it a lattice ? Justify SPPU : Dec.-16, Marks 2
b c d
e
Fig. 4.11.8
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Solution : Let
A = {a, b, c, d, e}
From the given Hasse diagram, every pair of element has glb and lub. Hence it is a
lattice.
qqq
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Discrete Mathematics 5-2 Functions
5.1 Introduction
The concept of relation was defined very generally in the preceeding chapter. We
shall now discuss a particular class of relations called functions. Many concepts of
computer science can be conveniently stated in the language of functions.
5.2 Function SPPU : May-07, 08, 15, Dec.-07, 12,
Let A and B be non-empty sets. A function f from A to B is denoted by f : A ® B
and defined as a relation from A to B such that for every a Î A, there exists a unique
b Î B such that (a, b) Î f .
If (a, b) Î f then it can be written as f(a) = b. Functions are also called as Mappings or
Transformations.
e.g.
1) f = {(a, b) (b, c) (c, a)} is a function on set A = {a, b, c}
2) f = {(a, b) (b, c) (b, a)} is not a function as (b, c) and (b, a) Î f .
3) f = {(a, b) (b, c)} is not a function on set A = {a, b, c} as for C Î A $/ any element x
in A such that (c, x) Î f .
5.2.1 Important Definitions
I) Let f : A ® B be a function such that f(a) = b then b Î B is called the image of
a Î A and a is called the pre-image of b. The element a is called argument of f.
II) Let f : A ® B be a function. The set A is known as domain set of f and the set B
is known as co-domain set of f.
III)The range set of a function f : A ® B is denoted by R(f) and defined as
R(f) = {b b Î B and f(a) = b for some a Î A }
In other words, range of f is the set of all images of the elements of A under f.
Examples
Example 5.2.1 If f is a function such that f(x) = x 2 – 1 where x Î R find the values of f(–1),
f(0), f(1), f(3) and range set of f.
Solution : We have f(x) = x 2 – 1
\ f(–1) = (–1) 2 – 1 = 1 – 1 = 0
f(0) = 0 – 1 = –1
f(1) = 1 2 – 1 = 0
f(3) = 3 2 – 1 = 8
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Discrete Mathematics 5-3 Functions
For any x Î Â, x 2 – 1 Î Â
\ The range of f is R(f) = {x –1 £ x < ¥}
Example 5.2.2 Let f : Â ® Â where f is defined by
a) f(x) = x
b) f(x) = x 2
c) f(x) = sin x " x ÎÂ
Solution :
a) We have f(x) = x \ f(4) = 4 = ±2
\ f is not a function. Therefore we can not find range of f.
b) We have f(x) = x 2 , x 2 is positive real number
Hence range of f is R(f) = {x |0 £ x < ¥}
c) " x Î Â, sin x lies between –1 to +1
\ R(f) = {x | – 1 £ x £ 1}
5.2.2 Partial Functions
Let A and B be two non empty sets. A partial function f with domain set A and
codomain set B is any function from A' to B where A' CA. For any x Î A – A' , f(x) is not
defined.
To make the distinction more clear, the function which is not partial is called as a
total function.
e.g.
1)
A f B
1 a
A' 2 b f : A' B is a partial function
3 c
4 d
5 e
Fig. 5.2.1
1
2) The function f : Â ® Â defined as f(x) = is a partial function. As it is not
x
defined for x = 0.
3) The function f : Â ® Â defined as f(x) = x is a partial function, as x is not
defined for x < 0, in Â.
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Discrete Mathematics 5-4 Functions
5.2.3 Equality of Two Functions
Two functions f : A ® B and g : A ® B are said to be equal functions or identical
functions iff.
f(x) = g(x) " x Î A
Note : Two functions f : Z ® Z and g : Â ® Â defined as f(x) = x, " x Î Z and
g(x) = x, " x Î Â are not identical or equal functions because their domains are not
same.
5.2.4 Identity Function
Let A be any non empty set and function f : A ® A is said to be the identity function
if f(x) = x, " x Î A.
e.g.
A f A
1 1
2 2 A is the identity function
f:A
3 3
4 4
Fig. 5.2.2
5.2.5 Constant Function
A function f : A ® B is said to be a constant function if f(x) = constant = k ; " x Î A.
The range set of a constant function consists of only one element.
5.2.6 Composite Function
Let f : A ® B and g : B ® C be two functions.
The composite function of f and g is denoted by gof and defined as gof : A ® C is a
function such that (gof) (a) = g[f(a)] " a Î A.
Note : gof is defined only when the range of f is a subset of the domain of g.
e.g. 1)
2
A f(x) = x B g(x) = x C gof
Therefore gof : A C A C
1 1 1 gof (1) = g[f(1)] = g(1) = 1 1 1
2 4 4 2 4
gof (2) = g[f(2)] = g(4) = 4
3 9 9 3 9
gof (3) = g[f(3)] = g(9) = 9 16
4 16 16 4
25 gof (4) = g[f(4)] = g(16) = 16 25
25
36 36
Fig. 5.2.3 Fig. 5.2.4
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Discrete Mathematics 5-5 Functions
Example 5.2.3 Let f(x) = x+2, g(x) = x – 2, h(x) = 3x, for x Î Â Where  is the set of real
numbers
Find i) gof ii) fog iii) fof iv) hog v) gog vi) foh vii) hof viii) fohog ix) gofoh.
SPPU : May-08, 15, Dec.-12, Marks 4
Solution : Let x Î Â be any real number.
i) gof(x) = g[f(x)] = g[x + 2] = x + 2 – 2 = x
ii) fog(x) = f[g(x)] = f[x – 2] = x – 2 + 2 = x
iii) fof(x) = f[f(x)] = f[x + 2] = x + 2 + 2 = x + 4
iv) hog(x) = h[g(x)] = h[x – 2] = 3(x – 2) = 3x – 6
v) gog(x) = g[g(x)] = g[x – 2] = x – 2 – 2 = x – 4
vi) foh(x) = f[h(x)] = f[3 x] = 3x + 2
vii) hof(x) = h[f(x)] = h[x + 2] = 3(x + 2) = 3x + 6
viii) fohog(x) = f[h(g(x))] = f[h(x – 2)]
= f[3(x – 2)] = f(3 x – 6)
= 3x – 6 + 2 = 3 x – 4
ix) gofoh(x) = g[f(h(x))] = g[f(3 x)]
= g[3 x + 2] = 3 x + 2 – 2 = 3 x
Example 5.2.4 Let functions f and g be defined by f(x) = 2x + 1, g(x) = x 2 – 2
Find a) gof(4) and fog (4)
b) gof(a + 2) and fog(a + 2)
c) fog(5)
d) gof(a + 3) SPPU : May-07, Dec.-07, Marks 4
Solution :
a) gof(4) = g[f(4)] = g[2(4) + 1] = g[9] = 9 2 – 2 = 79
fog(4) = f[g(4)] = f(4 2 – 2) = f(14) = 2(14) + 1 = 29
b) gof(a + 2) = g[f(a + 2)] = g[2(a + 2) + 1] = g[2a + 5]
= ( 2 a +5) 2 – 2 = 4 a 2 +20a + 23
fog(a + 2) = f[g(a + 2)] = f[(a +2) 2 – 2] = f[a 2 + 4a +2]
= 2[(a 2 + 4 a +2] +1 = 2 a 2 +8 a +5
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c) fog(5) = f[g(5)] = f[25 – 2] = f(23) = 2(23) + 1 = 47
d) gof(a + 3) = g[f(a + 3)] = g[2(a + 3) + 1] = g[2 a + 7]
= [2 a + 7] 2 – 2 = 4 a 2 +28 a + 47
5.3 Special Types of Functions SPPU : Dec.-10, 11, 16, 17, May-18
I) Let f : A ® B be a function.
i) Function f is said to be one to one (or Injective) function if distinct elements of A
are mapped into distinct elements of B.
i.e. f is one to one if
a 1 ¹ a 2 Þ f(a 1 ) ¹ f(a 2 )
OR f(a 1 ) = f(a 2 ) Þ a 1 = a 2
ii) Function f is said to be onto function if each element of B has at least one
preimage in A.
OR
A function f : A ® B is called onto (or surjective) function if the range set of f is equal
to B.
iii) A function f is called a objective function if it is both one to one and onto.
iv) A function f is called into function if $ at least one element in B which has no
preimage in A.
i.e. A function f is called into function if R(f) ¹ B
2) Let a function f : A ® B be a bijective function then f –1 : B ® A is called the
inverse mapping of f and defined as f(b) –1 = a iff f(a) = b
It is also known as invertible mapping.
3) Characteristic function of a set :
Let U be a universal set and A be a subset of U.
Then the function y A : U ® {0, 1} defined by
ì1 if x Î A
y A (x) = í
î0 if x Ï A
is called a characteristic function of the set A.
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Discrete Mathematics 5-7 Functions
Examples
Example 5.3.1 Give examples of functions of the following types by diagrams.
a) Injective function but not surjective
b) Surjective but not injective
c) Neither injective nor surjective
d) Injective as well as surjective
e) Into function
f) Inverse function
Solution :
a) Injective but not surjective
2
f(x) = x
A B
1 1
2 4
3 9
4 16
25
Fig. 5.3.1
b) Surjective but not injective
A B
–2
–1 0
0 1
1 4
2
Fig. 5.3.2
c) Neither injective nor surjective
A f B
–2
0
–1
1
0
4
1
9
2
Fig. 5.3.3
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Discrete Mathematics 5-8 Functions
d) Injective as well as surjective i.e. bijective
A f B
1 1
2 4
3 9
4 16
5 25
Fig. 5.3.4
e) Into function
A f B
1 1
2 2
3 3
Fig. 5.3.5
f) Inverse function
–1
A f B f A
1 1 1
2 4 2
3 9 3
4 16 4
Fig. 5.3.6
Example 5.3.2 Determine if each is a function. If yes is it injective, surjective, bijective ?
a) Each person in the earth is assigned a number which corresponds to his age.
b) Each student is assigned a teacher.
c) Each country has assigned it's capital. SPPU : Dec.-11, Marks 4
Solution :
a) Every person has unique age
It is a function. Two person's may have same age. It is not injective. There is no
person whose age is 300 years. It is not surjective. Function is not bijective.
b) It is a function. It is not injective. It is not surjective. It is not bijective.
c) It is a function. It is injective as well as surjective. It is bijective.
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Discrete Mathematics 5-9 Functions
Example 5.3.3 Let A = {a, b, c, d} and B = {1, 2, 3}. Determine whether the relation R from
A to B is a function. Justify. If it is function give the range :
i) R = [(a, 1), (b, 2), (c, 1), (d, 2)]
ii) R = [(a, 1), (b, 2), (a, 2), (c, 1), (d, 2)] SPPU : Dec.-16, Marks 4
Solution : Given that
A = {a, b, c, d}, B = {1, 2, 3}
i) R = {(a, 1), (b, 2), (c, 1), (d, 2)}
Every element of the domain set A is related to the unique element of set B.
R is a funtion from A ® B.
Range of R is {1, 2}.
ii) R = {(a, 1), (b, 2), (a, 2), (c, 1) (d, 2)}
Here a ® 1 and a ® 2 hence a has two images 1 and 2 in set B. i.e. a has no unique
image in set B. Thus given relation R is not a function.
Example 5.3.4 What are relations and functions. Given a relation R = {(1, 4), (2, 2), (3, 10),
(4, 8), (5, 6)} and chek whether the following relations R 1 , R 2 , R 3 and R 4 is a function
or not.
R 1 = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4)}
R 2 = {(1, 2), (2, 4), (2, 10), (3, 8), (4, 6), (5, 4)}
R 3 = {(1, 6), (2, 2), (4, 4), (5, 10)}
R 4 = {(1, 6), (2, 2), (3, 2), (4, 4), (5, 10} SPPU : Dec.-17, Marks 6
Solution : Refer sections 4.3 and 5.2
Domain set = {1, 2, 3, 4, 5}
Codomain = {2, 4, 6, 8, 10}.
Given that R 1 = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4)}
All elements are mapped to 4.
\ R 1 is a constant function.
R 2 = {(1, 2), (2, 4), (2, 10), (3, 8), (4, 6), (5, 4)}
In R 2 2 ® 4 and 2 ® 10 which is not possible in function.
\ R 2 is not a function
R 3 = {(1, 6), (2, 2), (4, 4), (5, 10)}
As 3 Î D but 3 has no any image in R 3
\ R 3 is not a function.
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Discrete Mathematics 5 - 10 Functions
R 4 = {(1, 6), (2, 2), (3, 2), (4, 4), (5, 10)}
Every element of a domain set has unique image in codomain set so R 4 is a function.
Example 5.3.5 Let A = B be the set of real numbers.
f : A ® B given by f(x) = 2x 3 – 1
1 1
g : B ® A given by g(y) = 3 y +
2 2
Show that f is a bijective function and g is also bijective function :
SPPU : Dec.-10, Marks 4
Solution :
1) Suppose f(x 1 ) = f(x 2 )
Þ 2x 13 – 1 = 2x 32 – 1
Þ x 13 = x 32
Þ x1 = x2
\ f is injective mapping
Let y Î B and f(x) = y Þ 2x 3 – 1 = y
1+ y
\ 2x 3 = 1+ y Þ x 3 =
2
3
1+ y 3
y 1
Þ x = = + Î A = Â for any y Î B
2 2 2
\ f is a surjective mapping.
Thus f is injective as well as surjective function.
Hence f is a bijective function.
1 1 1 1
We have f(x) = y Þ x = 3 y + Þ f –1 ( y) = x = 3 y + = g(y)
2 2 2 2
Thus f –1 = g. We know that if f is bijective function then f –1 is also bijective. Hence g
is bijective function.
Example 5.3.6 Explain classification of functions with example.
Solution : Depending upon the nature of function, there are mainly two functions.
1. Algebraic function : A function which consists of a finite number of terms
involving powers and roots of the independent variable and four fundamental
operations addition, subtraction, multiplication and division, is called an algebraic
function.
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Discrete Mathematics 5 - 11 Functions
There are three types of algebraic functions.
(a) Polynomial function : A function of the form
a 0 x n + a 1 x n - 1 + a 2 x n - 2 + ..... + a n , where n is positive integer, a0, a1, a2, .... ,an are real
numbers and a0 ¹ 0, is called polynomial function of x in degree n.
e.g. f(x) = x 3 - 3x 2 + 2x - 5
f(x)
(b) Rational function : A function of the form where f(x) and g(x) are
g(x)
x 2 - 3x + 5
polynomial functions and g(x) ¹ 0 is called rational function e.g.
x2 + 1
(c) Irrational function : A function involving radicals is called irrational function.
e.g. f(x) = x 2/ 3 + 5x 2 + 1, 2 x + 1
2. Transcendental function : A function which is not algebraic is called
transcendental function.
e.g. f(x) = sin x + x3 + 5 x
(a) Trigonometric function : The six functions sin x, cos x, tan x, sec x, cosec x, cot x,
where x is in radians are called trigonometric functions.
e.g. f(x) = sin x + tan x.
(b) Inverse trigonometric function : The six functions sin–1 x, cos–1 x, tan–1 x,
cosec–1 x, sec–1 x and cot–1 x are called inverse trigonometric functions.
e.g. f(x) = cos–1 x + 5 tan–1 x
(c) Exponential function : A function of the form f (x) = ax (a > 0) satisfying
a x × a y = ax + y and a' = a is called exponential fucntion. e.g. f(x) = 5x.
(d) Logarithmic function : The inverse function of the exponential function is called
logarithmic function. e.g. f(x) = log x.
Example 5.3.7 If f : A ® B is bijective function then f–1 is unique.
Solution : Let f A ® B is a bijective function.
Claim : Show that f–1 is unique.
Suppose f–1 is not unique, so there are two inverse functions say, g and h.
Let x1, x2 Î A, $ y in B such that
f–1 (y) = x1 Þ g(y) = x1 Þ f(x1) = y
and f–1 (y) = x1 Þ h(y) = x2 Þ f(x2) = y
This implies f(x1) = f (x2) , but f is 1 – 1 function.
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Discrete Mathematics 5 - 12 Functions
x1 = x 2
Þ g(y) = h (y) " y
Þ g º h i.e. g and h are equal function.
Hence universe of f is unique.
Example 5.3.8 If f : A ® B and g : B ® C are bijective functions the gof is also bijective.
Solution : Let f : A ® B and g : B ® C be two bijective functions then gof A ® C is a
function.
(i) Let x1, x2 Î A and suppose gof(x1) = gof(x2)
g[f (x1)] = g[f(x2)]
Þ f(x1) = f(x2) ... (Q g is 1 – 1 function)
Þ x 1 = x2 ... (Q f is 1 – 1 function)
Þ gof is 1 – 1 function.
(ii) Let z Î C.
\ $ y in B such that g (y) = z and $ x in A such that f(x) = y.
\ g(y) = z Þ z = g (f (x)) = gof (x) where x Î A
Þ gof is onto function.
Hence gof is bijective function.
Example 5.3.9 Determine whether the function is bijective.
i
f : I ® I such that f(i) = if i is even
2
( i - 1)
= if i is odd
2
Solution : Let a and b be any integers such that a ¹ b.
There are three possibilities
Case 1 : If a and b are odd
a -1 b -1
\ f(a) = , f(b) =
2 2
a -1 b -1
f(a) = f(b) Þ =
2 2
Þ a=b
Case 2 : If a and b are even
a a
then f(a) = , f(b) =
2 2
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Discrete Mathematics 5 - 13 Functions
a b
f(a) = f(b) Þ = Þ a=b
2 2
Case 3 : If a is odd and b is even
a -1 b
then f(a) = and f(b) =
2 2
a -1 b a b 1
Now, f(a) = f(b) Þ = Þ - =
2 2 2 2 2
Þ a–b = 1 \ a¹b
In particular a = 7, b = 6
7 -1 6
f(7) = =3 f(6) = =3
2 2
f(7) = f(6) but 6 ¹ 7
Thus f is not one one function. Hence f is not bijective function.
Example 5.3.10 Let f (x) = ax2 + b and g(x) = cx2 + d, where a, b, c, d are constants.
Determine for which values of constants gof (x) = fog (x).
Solution : Given that, f(x) = ax2 + b and g(x) = cx2 + d
Suppose fog (x) = gof (x)
f [g (x)] = g [f (x)]
f [ cx2 + d] = g [ax2 + b]
Þ a [ cx2 + d]2 + b = c [ax2 + b]2 + d
Þ a [c2 x4 + 2cdx2 + d2] + b = c (a2 x4 + 2abx2 + b2) + d
Þ ac2 x4 + 2acdx2 + ad2 + b = ca2 x4 + 2cabx2 + cb2 + d
Þ Coefficient of x4 = coefficient of x4
and Coefficient of x2 = coefficient of x2
Þ ac2 = ca2 Þ a=c
and 2acd = 2acb Þ b=d
Thus a = c and b = d
Example 5.3.11 Show that the functions f(x) = x3 + 1 and g(x) = (x – 1)1/3 are converse to
each other.
Solution : We have, g(x) = (x - 1) 1 3
Let y = g (x) = (x - 1) 1 3
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Discrete Mathematics 5 - 14 Functions
Þ y3 = x – 1
Þ x = y3 + 1 ... (Which is f (x))
Hence f (x) and g (x) are converse to each other.
Example 5.3.12 Let f : X ® Y and X and Y are set of real numbers. Find f–1 if
2x - 1
(i) f(x) = x2 ; (ii) f(x) =
5
Solution : Let f : X ® Y and X, Y Í Â
(i) f(x) = x2
Let f(x) = y
x2 = y
Þ y = x2
Þ y = ± x
Therefore, given function is not one to one. Hence f–1 does not exist.
(ii) Let f(x) = y
2x - 1
= y
5
Þ 2x = 5y + 1
5y + 1
x = " yÎÂ
2
Function f is 1 – 1 and onto.
\ The inverse function of f is given as,
1 + 5y
f–1 (x) =
2
Example 5.3.13 Let X = {a, b, c}. Define f : X ® X such that f = {(a, b) (b, a) (c, c)}.
Find (i) f –1 ; (ii) f 2 ; (iii) f 3 ; (iv) f 4
Solution : Given fucntion f is bijective function.
(i) f–1 = {(b, a) (a, b) (c, c)}
(ii) f2 = fof = {(a, a) (b, b) (c, c)}
fof (a) = f[f (a)] = f(b) = a
fof (b) = f[f (b)] = f (a) = b
fof (c) = f [f (c)] = f (c) = c
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Discrete Mathematics 5 - 15 Functions
(iii) f3 (a) = f [f2 (a)] = f (a) = b
f3 (b) = f [f2 (b)] = f (b) = a
f3 (c) = f [f2 (c)] = f (c) = c
f2 = {(a, b), (b, a), (c, c)}
(iv) f4 = f2 o f2
f4 (a) = f2 [f2 (a)] = f2 (a) = a
f4 (b) = f2 [f2 (b)] = f2 (b) = b
f4 (c) = f2 [f2 (c)] = f2 (c) = c
f4 = {(a, a) (b, b) (c, c)}
Example 5.3.14 Let A = {1, 2, 3}, B = {1, 2, 3} and f 1 and f 2 are functions from A to B.
f 1 = {(1, 2), (2, 3), (3, 1)}
f 2 = {(1, 2), (2, 1), (3, 3)}
Compute f 1 o f 2 and f 2 o f 1 SPPU : May-18, Marks 4
Solution : We have f1 o f2 ( x) = f1 [ f2 ( x)]
f1 [ f2 (1)] = f1 [ 2] = 3
f1 [ f2 ( 2)] = f1 [1] = 2
f1 [ f2 ( 3)] = f1 [ 3] = 1
Now, f2 o f1 ( x) = f2 [f1 (x)]
f2 [ f1 (1)] = f2 [ 2] = 1
f2 [ f1 ( 2)] = f2 [ 3] = 3
f2 [ f1 ( 3)] = f2 [1] = 2
Example 5.3.15 Let X = {1, 2, 3} , Y = {p, q} and z = {a, b}
Let f : X ® Y be {(1, p), (2, p), (3, q)} and g : Y ® Z be
g = {(p, b) (q, b)}. Find gof and show it pictorically.
Solution : Given that, f = {(1, p), (2, p), (3, q)}
g = {(p, b) (q, b)}
and gof (x) = g [ f (x) ] and gof : X ® Z
gof (1) = g [f (1)] = g (p) = b
gof (2) = g [f (2)] = g (p) = b
gof (3) = g [f (3)] = g (q) = b
gof = {(1, b) (2, b) (3, b)}
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Discrete Mathematics 5 - 16 Functions
It's pictorial representation is
f g
X Y Z
1 p a
2
q b
3
gof
Fig. 5.3.7
Example 5.3.16 Let R be the set of real numbers and f, g, h : Â ® Â such that
1
f(x) = x + 2 , g (x) = h (x) = 3. Compute
x2 +1
(i) f–1 g (x) ; (ii) hf (gf–1) [h f (x)]
Solution : (i) Given that f (x) = x + 2
Let f(x) = y = x + 2 Þ x = y–2
f–1 (y) = y – 2
é 1 ù 1
f–1[g (x)] = f -1 ê ú = 2 -2
2
ë x + 1û x +1
=
1 - 2x 2 - 2
=
(
- 2x 2 + 1 )
(x 2 + 1) x2 +1
(ii) We have f(x) = x + 2
h f (x) = h [f (x)] = h [x + 2] = 3
g f–1 (x) = g [f–1 (x)] = g [x – 2]
1 1
= =
( x - 2) 2
+1 x2 - 2x + 5
1
(g f–1) [h f (x)] = g f–1 (3) =
32 - 2(3) + 5
1 1
= =
9-6+6 8
é 1ù é æ 1 öù
\ h f (g f–1) (h f (x)) = h f ê ú = h ê f ç ÷ ú
ë 8û ë è 8 øû
é1 ù é 17 ù
= h ê + 2ú = h ê ú = 3
ë 8 û ë 8û
i.e. h f (g f–1) (h f (x)) = 3
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Discrete Mathematics 5 - 17 Functions
Example 5.3.17 Show that f, g : N ´ N ® N as f (x, y) = x + y. g(x, y) = xy are onto but not
one-one
Solution : (i) Suppose f (x 1 , y 1 ) = f (x 2 , y 2 )
Þ x1 + y 1 = x 2 + y 2
Þ x1 ¹ x2 and y1 ¹ y2
e.g. x1 = 5 , x2 = 2 , y1 = 4 , y2 = 7.
\ f is not one to one.
Now, g (x1, y1) = g (x2, y2)
Þ x1 y1 = x2 y2
Þ x1 may or may not be equal to x2
and y1 may or may not be equal to y2
\ g is not one to one.
(ii) f(x, y) = x + y
\ Every element of N can be written as the sum of two elements of N. Hence f is
onto.
Now, g (x, y) = xy
Every element of can be written as the product of two elements of N.
\ g is onto.
5.4 Infinite Sets and Countability SPPU : May-18, 19
5.4.1 Infinite Set
A set A is said to be an infinite set if there exists an injective mapping (function)
f : A ® A such that f(A) is a proper subset of A.
If no such injective function exists, then set is finite.
Examples
1) Let f : N®N such that f(n) = 2 n, " n Î N = Natural number set.
There range set = f(N) = {Set of positive even natural numbers} C N
\ N is an infinite set.
ìx + 2 if x ³ 0
2) Define f : Â ® Â such that f(x) = í
î x if x < 0
f is an injective mapping and for x ³ 0, f (x) = x + 2
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Discrete Mathematics 5 - 18 Functions
i.e. f(x) ( 2 and for x < 0, f(x) = x < 0
Range set = f(Â) = {y Î Â/f(x) ³ 2 or f(x) < 0}
\ 1 Ï f (Â)
Ì
Thus f (Â) ¹ Â
Hence  is an infinite set.
I) Properties of an infinite sets
1) If A is an infinite set the A ´ A, P (A) are infinite sets.
2) If A and B are non empt sets and either A or B is an infinite set then A ´ B is an
infinite set.
3) If either A or B is an infinite set then A È B is an infinite set.
4) If A Í B and A is an infinite set then B is also an infinite set.
i.e. the superset of an infinite set is an infinite.
5.4.2 Countable Sets
We know that the codinality of a set is the number of elements of that set. If set is
finite then, we can list elements as 1, 2, 3, …. Therefore every finite set is countable. As
|f| = 0, the null set is also countable. A question remains same for an infinite sets. Let
us define countable infinite sets.
Definition :
An infinite set A is said to be countable if there exists a bijection f : N®A
\ A = { f (1), f (2), f (3), …}
A countably infinite set is also known as a denumarable set.
i.e. If A is a denumerable set then we can least elements of A as a 1 , a 2 , a 3 , … a n …
or f (1), f (x) … f (n) … .
e.g. 1) f is countable
2) A = {1, 2, 3, 4, … 1000} is countable as |A| = 1000.
3) As f : N ® N defined as f (x) = n, \ N is countable
4) The set of integers is countable
n+ 1
As f : N ® Z such that f (n) = if n = 1, 3, 5, …
2
-n
= if n = 0, 2, 4, 6, …
2
is bijective mapping.
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Discrete Mathematics 5 - 19 Functions
5) The set of rational numbers is countable.
6) The set of real numbers is not countable
7) The set of complex numbers is not countable.
8) The set of real numbers in [a, b], a < b is not countable.
9) The countable union of countable sets is countable.
Properties of countable sets :
1) A subset of a countable set is countable.
2) Let A and B be countable sets then A * B is countable.
Þ Define f : A È B ® N as f (a i ) = 2 i – 1
and f (b i ) = 2 i
f is bijective \ A È B is countable
3) Prove that the set of rational numbers is countably infinite.
SPPU : May-18, 19, Marks 4
Proof : We know that the countable union of countable sets is countable. Therefore it is
sufficient to prove that the set of rational numbers in [0, 1] is countable.
We have to prove that $ atleast one function f.
f : [0, 1] ® N such that f is injective.
We arrange the rational numbers of the interval according to increasing denominators
as
1 1 2 1 3 1 2 3 4
0, 1, , , , , , , , , ,L
2 3 3 4 4 5 5 5 5
then the one to one correspondence is as follows
0«1
1«2
1
«3
2
1
«4
3
2
«5
3
1
« 6 and so on
4
Hence set of rationals in [0, 1] is countable. Thus the set of rational numbers is
countable and as the set is infinite, it is countably infinite.
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Discrete Mathematics 5 - 20 Functions
4) The set of irrational numbers is uncountable.
Proof : We know that  = Q È Q where Q = set of rational numbers, Q = set of
irrational numbers.
Suppose Q is countable Þ Q È Q = Â is countable which is contradiction.
\ Q is not countable i.e. uncountable.
5) The set of real numbers in (0, 1) is not countable. Assume that the set is
countable. \ A = {x 1 , x 2 , x 3 , L x n , L}.
Proof : Any real number in (0, 1) can be written in a unique decimal without an
infinite string of 9's at the end. i.e. 0.3459999 will be represented as 0.345000. Let the
infinite sequence be given by,
1 ® x 1 = 0 × a 11 a 12 a 13 …
2 ® x 2 = 0 × a 21 a 22 a 23 …
3 ® x 3 = 0 × a 31 a 32 a 33 …
M M
n ® x n = 0.a n1 a n2 a n3 …
M M
Construct a new number y = 0 × b1 b 2 b 3 …
Where b i = 0 if a ii ¹ 0
b i = 1 if a ii = 0
Hence b1 ¹ a 11 , b 2 ¹ a 22 , b 3 ¹ a 33 … b n ¹ a nn
\ b i ¹ a ii , " i
\ y ¹ x1 , y ¹ x i " i
Hence y is not in the list of numbers {x 1 , x 2 , L x n L} Thus y Î(0, 1) and it is
different from elements in {x 1 , x 2 , L x n L} which is contradiction that A is countable.
Hence A is not countable.
Thus the set of real numbers is not countable.
5.5 Pigeon Hole Principle SPPU : Dec.-09, 11, 12
I) This principle states that if there are n + 1 pigeons and only n pigeon holes then
two pigeons will share the same whole.
This principle is stated by using the analogy of the bijective mapping i.e. If A and B
are any two sets such that |A| > |B| then there does not exist bijective mapping from A
to B.
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Examples
Example 5.5.1 If 11 shoes are selected from 10 pairs of shoes then there must be a pair and
matched shoes among the selection.
Solution : In the pigeonhole principle, 11 shoes are pigeons and the 10 pairs are the
pigeon holes.
Example 5.5.2 Show that if seven numbers from 1 to 12 are chosen then two of them will add
upto 13.
Solution : We have A = {1, 2, 3, 4, 5, … 12}
We form the six different sets each containing 2 numbers that add uptot 13.
A1 = {1, 12}, A 2 = {2, 11}, A 3 = {3, 10}, A 4 = {4, 9}, A 5 = {5, 8}, A 6 = {6, 7}
Each of the seven numbers chosen must belong to one of these sets. As there are only
six sets, by pigeonhole principle two of the chosen numbers must belong to the same set
and their sum is 13.
II) The extended pigeon hole principle
If n pigeons are assigned to m pigeon holes, then one of the pigeon holes must be
é n - 1ù
occupied by at least ê + 1 pigeons. It is also known as generalized pigeon hole
ë m úû
é n - 1ù é 9ù é 16ù
principle. Here ê ú is the integer division of n – 1 by m. e.g. ê ú = 4, ê ú = 3,
ë m û 2
ë û ë 5û
é 8ù
êë 3úû = 2.
Examples
Example 5.5.3 Show that 7 colours are used to paint 50 bicycles, then at least 8 bicycles will
be of same colour. SPPU : Dec.-09, 12, Marks 4
é n - 1ù
Solution : By the extended pigeonhole principle, at least ê + 1 pigeons will occupy
ë m úû
one piegeonhole.
Here n = 50, m = 7 and m < n then
é 50 - 1ù
ê 7 ú +1 = 7+1=8
ë û
Thus 8 bicycles will be of the same colour.
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Example 5.5.4 Write generalized pigeonhole principle. Use any form of pigeonhole principle to
solve the given problem.
i) Assume that there are 3 mens and 5 womens in a party show that if these people are
lined up in a row at least two women will be next to each other.
ii) Find the minimum number of students in the class to be sure that three of them are
born in the same month. SPPU : Dec.-11, Marks 4
Solution : Please refer section 5.5 (II) for definition.
i) By using analogy of pigeon hole principle, we get
3 men = pigeonholes and 5 women = pigeon
Pigeons are more than pigeon holes.
At least two pigeons share the same pigeon hole i.e. at least two women in a row
will be next to each other.
ii) Let h = Number of pigeons = Number of students
n = Number of pigeon holes = Number of months = 12
Given that three students in the class are born in the same month.
é n - 1ù
êë m úû + 1 = 3
n- 1
Þ = 3–1=2
12
n = 2 ´ 12 + 1 = 25
Therefore there are 25 minimum number of students in the class.
5.6 Discrete Numeric Functions
A function whose domain is a set of natural numbers including zero and whose
range set is the set of real numbers, is called a discrete numeric function or numeric
function. It is also known as a sequence. If f : N È {0} ® Â is a discrete numeric
function then f(0, f(1), f(2), f(3), … denote the value of function at 0, 1, 2, 3, …
The numeric function f is written as
f = {f 0 , f 1 , f 2 , L }
Hereafter, to denote numeric function, we use
< a r > = {a 0 , a 1 , a 2 , a 3 L a r , L }
e.g. 1) a r = 2 k if 0 £ k £ 3
= k if k > 3
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This discrete function can be written as
a = {1, 2, 4, 8, 4, 5, 6, 7, 8, …}
ì100 + r, 0 £ r £ 5
2) ak = í 2
î r r³ 6
\ This discrete function can be written as
a = {100, 101, 102, 103, 104 , 105, 6 2 , 7 2 , 8 2 , 9 2 , L }
5.6.1 Basic Operations on Numeric Functions
Let a and b be two numeric functions. Then
i) Addition : c = a + b is also a numeric function defined as
c r = a r + br
i.e. c 0 = a 0 + b 0 , c 1 = a 1 + b1 , c 2 = a 2 + b 2 and so on.
ii) Multiplication : d = ab is also a numeric function defined as d r = a r b r ,
" Î N È {0}.
iii) Scaling : Let k be any real number then b = ak is a numeric function and k is
called as scalling factor.
iv) Linearity : Let p and q be any real numbers, then x = pq + qb is a numeric
function where
x r = p a r + q br
v) The convolution of two numeric functions : Let a and b be two numeric
functions. The convolution of two functions a and b is denoted by c = a * b and
defined as
r
c r = a 0 br + a1 br -1 + a 2 br - 2 + … + a r b0 = åan br - n
n= 0
Example 1 :
1 ì 1 1 1 ü
If ar = , r ³ 0 then a = í1, , , , Ký
3r î 3 2
3 3 3
þ
ì 5 5 5 ü
and 5a = í5, , , , Lý
î 3 32 33 þ
1
\ 5 a r = 5×
3r
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Example 2 :
ì 1 if 0 r 2 ì2 r + 1 , 0 £ r £ 1
If ar = í and br = í
î3 r if r³ 3 î r- 5 , r³ 2
then a+b = c
ì2r + 1 + 2 , 0 £ r £ 1
ï
\ cr = a r + b r = í 1 + ( - 3) , r= 2
ï3 r + r - 5 , r³ 3
î
ì2 r + 1 + 2 , 0 £ r £ 1
ï
= í -2 , r= 2
ï 4 r- 5 , r³ 3
î
ì 1 (2 r + 1) , 0 £ r £ 1
ï
and d = ab and d r = a r b r = í 1 ( - 3) , r= 2
ï3 r ( r - 5) , r³ 3
î
ì 2r + 1 , 0£ r£ 1
ï
dr = í -3 , r= 2
ï3 r - 15 r ,
2 r³ 3
î
Example 3 :
If a r = 3r , r ³ 0
br = 5r , r ³ 0
r
then c = a * b Þ cr = å a n br - n = a 0 br + a 1 br - 1 + a 2 br - 2 +L + a r b0
n= 0
5.6.2 Finite Differences of a Numeric Functions
I) Shift of a numeric function
The shift of a numeric function of a r is denoted by E [a r ] and defined as
E [a r ] = a r +1
E - 1 [a r ] = a r - 1
E n [a r ] = a r +n , E - n [a r ] = a r - n
Note : It is also denoted by s n [a r ] = E n [a r ]
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Discrete Mathematics 5 - 25 Functions
Example 1 :
If a r = 5 r , r ³ 0 then E [a r ] = a r + 1 = 5r + 1
E 2 [a r ] = a r + 2 = 5 r + 2 , E - n [a r ] = a r - n = 5 r - n
Example 2 :
ìr 3 - 2r + 5 ; 0 £ r £ 6
If ar = í
î 3 ; r³ 7
ì 0 , 0£ r< 2
3 [a ï 3
then E r ] = í(r + 3) - 2 (r + 3) + 5 , 3 £ r £ 9
ï 3 , r ³ 10
î
ì(r - 2) 3 - 2(r - 2) + 5 ; 0 £ r £ 4
and E - 2 [a r = í
î 3 ; r³ 5
II) Forward difference of a numeric function
The forward difference of a numeric function a r is denoted by D a r and defined as
D a r = a r +1 – a r ; r ³ 0
Example 1 :
If a r = 3 r ; r ³ 0 then
D ar = ar+ 1 – a r = 3r + 1 - 3r
= (3 – 1) 3 r = 2.3 r
Example 2 :
ì 3 ; 0£ r£ 4
If ar = í
î r + 2 ; r³ 5
then D ar = ar+ 1 - ar
ì 3 ; 0£ r+ 1£ 4
\ ar+ 1 = í
î(r + 1) + 2 ; r+ 1³ 5
ì 3 ; 0£ r£ 3
ar+ 1 = í
îr + 3 ; r³ 4
ì 0 ; 0£ r£ 3
ï
\ D ar = í - 3 + (4 + 3) ; r= 4
ï(r + 3) - (r + 2) ; r³ 5
î
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Discrete Mathematics 5 - 26 Functions
ì0 ; 0 £ r £ 3
ï
D ar = í4 ; r= 4
ï1 ; r³ 5
î
III) Backward difference of a numeric function
The backward difference of a numeric function a r is denoted by Ñ a r and defined as
Ñ a r = a r – a r -1 ; r ³ 1
Examples
Example 1 :
Let a r = 6r ; r ³ 0
a r - 1 = 6 r - 1 ; r – 1 ³ 0 i.e. r ³ 1
\ Ñ a r = a r – a r - 1 = 6r - 6r - 1 ; r ³ 1
= (6 – 1) 6 r - 1 ; r ³ 1
Ñ a r = 5 × 6r - 1 ; r ³ 1
Example 2 :
ì1 ; 0£ r£ 2
If ar = í r
î3 ; r³ 3
ì 1 ; 0£ r- 1£ 2
ar- 1 = í r- 1
î3 ; r- 1³ 3
ì 1 ; 1£ r£ 3
= í r- 1
î3 ; r ³4
ì 0 ; r= 0
ï 1 ; r= 1
ïï
Ñ ar = ar – ar- 1 =í 0 ; r= 2
ï 26 ; r= 3
ï
ïî2 ´ 3 r - 1 ; r³ 4
Example 5.6.1 Determine a * b for the following numeric functions
ì1 ; 0 £ r £ 2 ìr + 1 ; 0 £ r £ 2
ar = í and b r = í
î 0 ; r ³ 3 î 0 ; r³3
Solution : By the definition of a r , the numeric function of a is given by
a = {1, 1, 1, 0, 0, 0, …} = {a 0 , a 1 , a 2 , a 3 , …}
b = {1, 2, 3, 0, 0, 0 …} = {b 0 , b1 , b 2 , b 3 , …}
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Discrete Mathematics 5 - 27 Functions
The convolution of a and b is a numeric function c such that
r
c = a*b where cr = åan br - n
n= 0
\ c r = a 0 br + a1 br -1 + a 2 br - 2 + … + a r b0
\ c 0 = a 0 b0 = 1 ´ 1 = 1
c 1 = a 0 b1 + a 1 b 0 = 1.2 + 1.1 = 3
c 2 = a 0 b 2 + a 1 b1 + a 2 b 0 = 1.3 + 1.2 + 1.1 = 6
c 3 = a 0 b 3 + a 1 b 2 + a 2 b1 + a 3 b 0
= 1.0 + 1.3 + 1.2 + 1.0 = 5
c 4 = a 0 b 4 + a 1 b 3 + a 2 b 2 + a 3 b1 + a 4 b 0
c 4 = 0 + 0 + 1.3 + 0 + 0 = 3
c 5 = 0, c 6 = 0, c r = 0 ; r ³ 5
Thus the numeric sequence of c is given by
c = {1, 3, 6, 5, 3, 0, 0, 0, …}
ì1 ; r= 0
ï3 ; r= 1
ï
ï6 ; r= 2
\ cr = í
ï5 ; r= 3
ï3 ; r = 4
ï0 ; r ³ 5
î
qqq
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Notes
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