Linear Algebra by Nayak

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PREFACE

Linear Algebra plays an important role in the spheres of Mathematics, Physics, andEngineering due to their inherent viabilities. The aim of this text book is to give rigorousand thorough analysis and applications of various aspects of Linear algebra and analysis withapplications. Also, the present book has been designed in a lucid and coherent manner sothat the Honours and Postgraduate students of various Universities may reap considerablebenefit out of it. I have chosen the topics with great care and have tried to present themsystematically with various examples.

The author expresses his sincere gratitude to his teacher Prof. S. Das, Departmentof Mathematics, R. K. Mission Residential College, Narendrapur, India, who taught himthis course at the UG level. Author is thankful to his friends and colleagues, especially,Dr. S. Bandyopadhyay, Mr. Utpal Samanta and Mr. Arup Mukhopadhyay of BankuraChristian College, Dr. Jayanta Majumdar, Durgapur Govt. College, Pratikhan Mandal,Durgapur Govt. College, for their great help and valuable suggestions in the preparationof the book. Author also extends his thanks to Prof. (Dr.) Madhumangal Pal, Dept. ofApplied Mathematics, Vidyasagar University, for his encouragement and handy suggestions.

This book could not have been completed without the loving support and encouragementof my parents, wife (Mousumi) and son (Bubai). I extend my thanks to other well wishersrelatives and students for embalming me to sustain enthusiasm for this book. Finally, Iexpress my gratitude to Books and Allied (P) Ltd., specially Amit Ganguly, for bringingout this book.

I would like to thank to Dr. Sk. Md. Abu Nayeem of Aliah University, West Bengal andmy student Mr. Buddhadeb Roy for support in writing/typing in LaTex verision.

This book could not have been completed without the loving support and encouragementof my parents, wife (Mousumi) and son (Bubai). I extend my thanks to other well wishersrelatives and students for embalming me to sustain enthusiasm for this book. Finally, Iexpress my gratitude to Asian Books Private Limited, Delhi, for bringing out this book.

Critical evaluation, suggestions and comments for further improvement of the book willbe appreciated and gratefully acknowledged.

Prasun Kumar Nayak ,(nayak [email protected])

Bankura Christian College,Bankura, India, 722 101.

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Dedicated to my parentsSankar Nath Nayak and Mrs. Indrani Nayakfor their continuous encouragement and support..

Contents

1 Theory of Sets 11.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Description of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1.2 Types of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Algebraic Operation on Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2.1 Union of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2.2 Intersection of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2.3 Disjoint Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.4 Complement of a Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.5 Difference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2.6 Symmetric Difference . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.3 Duality and Algebra Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.4 Cartesian Product of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.5 Cardinal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.6 Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.6.1 Equivalence Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.7 Equivalence Class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

1.7.1 Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311.8 Poset . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

1.8.1 Dual Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351.8.2 Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361.8.3 Universal Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361.8.4 Covering Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371.8.5 Maximal and Minimal Elements . . . . . . . . . . . . . . . . . . . . . 381.8.6 Supremum and Infimum . . . . . . . . . . . . . . . . . . . . . . . . . . 40

1.9 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421.9.1 Lattice Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441.9.2 Sublattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451.9.3 Bounded Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451.9.4 Distributive Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451.9.5 Trivially Complement . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

1.10 Mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471.10.1 Types of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481.10.2 Composite mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

1.11 Permutation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 631.11.1 Equal permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 631.11.2 Identity permutation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 631.11.3 Product of permutations . . . . . . . . . . . . . . . . . . . . . . . . . . 631.11.4 Inverse of permutations . . . . . . . . . . . . . . . . . . . . . . . . . . 64

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1.11.5 Cyclic permutation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 651.12 Enumerable Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

2 Theory of Numbers 832.1 Number System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

2.1.1 Non-positional Number System . . . . . . . . . . . . . . . . . . . . . . 832.1.2 Positional Number System . . . . . . . . . . . . . . . . . . . . . . . . 84

2.2 Natural Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 852.2.1 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 852.2.2 Well Ordering Principle . . . . . . . . . . . . . . . . . . . . . . . . . . 852.2.3 Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . . 86

2.3 Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 892.3.1 Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 902.3.2 Division Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

2.4 Common Divisor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 942.4.1 Greatest Common Divisor . . . . . . . . . . . . . . . . . . . . . . . . . 95

2.5 Common Multiple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 972.5.1 Lowest Common Multiple . . . . . . . . . . . . . . . . . . . . . . . . . 97

2.6 Diophantine Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 992.6.1 Linear Diophantine Equations . . . . . . . . . . . . . . . . . . . . . . . 99

2.7 Prime Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1022.7.1 Relatively Prime Numbers . . . . . . . . . . . . . . . . . . . . . . . . . 1032.7.2 Fundamental Theorem of Arithmetic . . . . . . . . . . . . . . . . . . . 108

2.8 Modular/Congruence System . . . . . . . . . . . . . . . . . . . . . . . . . . . 1112.8.1 Elementary Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 1112.8.2 Complete Set of Residues . . . . . . . . . . . . . . . . . . . . . . . . . 1172.8.3 Reduced Residue System . . . . . . . . . . . . . . . . . . . . . . . . . 1212.8.4 Linear Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1222.8.5 Simultaneous Linear Congruences . . . . . . . . . . . . . . . . . . . . 1252.8.6 Inverse of a Modulo m . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

2.9 Fermat’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1312.9.1 Wilson’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

2.10 Arithmetic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1362.10.1 Euler’s Phi Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1362.10.2 The Mobius Function: . . . . . . . . . . . . . . . . . . . . . . . . . . . 1412.10.3 Divisor Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1432.10.4 Floor and Ceiling Functions . . . . . . . . . . . . . . . . . . . . . . . . 1442.10.5 Mod Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

3 Theory of Matrices 1493.1 Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

3.1.1 Special Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1493.1.2 Square Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

3.2 Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1533.2.1 Equality of matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1533.2.2 Matrix Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1533.2.3 Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . 1543.2.4 Transpose of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

3.3 Few Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1613.3.1 Nilpotent Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1613.3.2 Idempotent Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

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3.3.3 Involuntary Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1633.3.4 Periodic Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1633.3.5 Symmetric Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1633.3.6 Skew-symmetric Matrices . . . . . . . . . . . . . . . . . . . . . . . . . 1643.3.7 Normal Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

3.4 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1663.4.1 Product of Determinants . . . . . . . . . . . . . . . . . . . . . . . . . 1713.4.2 Minors and Co-factors . . . . . . . . . . . . . . . . . . . . . . . . . . 1813.4.3 Adjoint and Reciprocal of Determinant . . . . . . . . . . . . . . . . . 1833.4.4 Symmetric and Skew-symmetric Determinants . . . . . . . . . . . . . 1843.4.5 Vander-Monde’s Determinant . . . . . . . . . . . . . . . . . . . . . . . 1863.4.6 Cramer’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

3.5 Complex Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1893.5.1 Transpose Conjugate of a Matrix . . . . . . . . . . . . . . . . . . . . . 1903.5.2 Harmitian Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1903.5.3 Skew-Harmitian Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . 1913.5.4 Unitary Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1923.5.5 Normal Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

3.6 Adjoint of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1923.6.1 Reciprocal of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 1953.6.2 Inverse of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1953.6.3 Singular Value Decomposition . . . . . . . . . . . . . . . . . . . . . . . 201

3.7 Orthogonal Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2023.8 Submatrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2053.9 Partitioned Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

3.9.1 Square Block Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . 2063.9.2 Block Diagonal Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . 2073.9.3 Block Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2083.9.4 Block Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2083.9.5 Inversion of a Matrix by Partitioning . . . . . . . . . . . . . . . . . . . 209

3.10 Rank of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2113.10.1 Elementary Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . 2123.10.2 Row-reduced Echelon Matrix . . . . . . . . . . . . . . . . . . . . . . . 213

3.11 Elementary Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2163.11.1 Equivalent Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2183.11.2 Congruent Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2203.11.3 Similar Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

4 Vector Space 2354.1 Vector Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

4.1.1 Vector Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2404.2 Linear Sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246

4.2.1 Smallest Subspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2474.2.2 Direct Sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247

4.3 Quotient Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2494.4 Linear Combination of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . 251

4.4.1 Linear Span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2524.4.2 Linearly Dependence and Independence . . . . . . . . . . . . . . . . . 257

4.5 Basis and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2624.6 Co-ordinatisation of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277

4.6.1 Ordered Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

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4.6.2 Co-ordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2784.7 Rank of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

4.7.1 Row Space of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 2794.7.2 Column Space of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . 281

4.8 Isomorphic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283

5 Linear Transformations 2935.1 Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293

5.1.1 Kernal of Linear Mapping . . . . . . . . . . . . . . . . . . . . . . . . . 2975.1.2 Image of Linear Mapping . . . . . . . . . . . . . . . . . . . . . . . . . 300

5.2 Isomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3115.3 Vector Space of Linear Transformation . . . . . . . . . . . . . . . . . . . . . . 316

5.3.1 Product of Linear Mappings . . . . . . . . . . . . . . . . . . . . . . . . 3185.3.2 Invertible Mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321

5.4 Singular and Non-singular Transformation . . . . . . . . . . . . . . . . . . . . 3245.5 Linear Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3265.6 Matrix Representation of Linear Transformation . . . . . . . . . . . . . . . . 3275.7 Orthogonal Linear Transformation . . . . . . . . . . . . . . . . . . . . . . . . 3415.8 Linear Functional . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344

5.8.1 Dual Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3455.8.2 Second Dual Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3495.8.3 Annihilators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350

5.9 Transpose of a Linear Mapping . . . . . . . . . . . . . . . . . . . . . . . . . . 354

6 Inner Product Space 3656.1 Inner Product Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365

6.1.1 Euclidean Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3656.1.2 Unitary Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366

6.2 Norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3696.3 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374

6.3.1 Orthonormal Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3756.3.2 Orthogonal Complement . . . . . . . . . . . . . . . . . . . . . . . . . . 3766.3.3 Direct Sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377

6.4 Projection of a Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380

7 Matrix Eigenfunctions 3957.1 Matrix Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395

7.1.1 Polynomials of Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . 3957.1.2 Matrices and Linear Operator . . . . . . . . . . . . . . . . . . . . . . . 396

7.2 Characteristic Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3967.2.1 Eigen Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3987.2.2 Eigen Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3987.2.3 Eigen Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410

7.3 Diagonalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4137.3.1 Orthogonal Diagonalisation . . . . . . . . . . . . . . . . . . . . . . . . 417

7.4 Minimal Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4207.5 Bilinear Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425

7.5.1 Real Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . 4257.6 Canonical Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427

7.6.1 Jordan Canonical Form . . . . . . . . . . . . . . . . . . . . . . . . . . 4357.7 Functions of Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439

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7.7.1 Powers of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4407.7.2 Roots of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441

7.8 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4437.8.1 Exponential of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . 4437.8.2 Logarithm of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 445

7.9 Hyperbolic and Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . 446

8 Boolean Algebra 4558.1 Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455

8.1.1 Unary Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4558.1.2 Binary Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455

8.2 Boolean Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4558.2.1 Boolean Algebra as a Lattice . . . . . . . . . . . . . . . . . . . . . . . 4558.2.2 Boolean Algebra as an Algebraic System . . . . . . . . . . . . . . . . . 4568.2.3 Boolean Algebra Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . 4608.2.4 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4618.2.5 Partial Order Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . 466

8.3 Boolean Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4678.3.1 Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4678.3.2 Literal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4678.3.3 Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4688.3.4 Monomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4688.3.5 Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4688.3.6 Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4688.3.7 Boolean Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468

8.4 Truth Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4698.5 Disjunctive Normal Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470

8.5.1 Complete DNF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4708.6 Conjunctive Normal Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472

8.6.1 Complete CNF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4728.7 Switching Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475

Chapter 1

Theory of Sets

George Cantor gave an intuitive definition of sets in 1895. Sets are building blocks of variousdiscrete structures. The theory of sets is one of the most important tools of mathematics.The main aim of this chapter is to discuss some properties of sets.

1.1 Sets

A well defined collection of distinct object is defined as set. Each object is known as anelement or member of a set. The following are some examples of the set.

(i) All integers.(ii) The positive rational numbers less than or equal to 5.(iii) The planet in solar system.(iv) Indian rivers.(v) 4th semester B.Sc. students of Burdwan University.(vi) The peoples in particular locality.(vii) Cricketers in the world.

By the term ‘well defined’, we mean that we are given a collection of objects with certaindefinite property or properties, given in such a way that we are clearly able to distinguishwhether a given object is our collection or not. The following collections are not examplesof set.

(i) Good students of a class, because ’good’ is not well-defined word, a student may begood for particular people, but he/she may not be good for other people.

(ii) Tall students, ’tall’ is not well-defined measurement.

(iii) Girls and boys of a particular locality, because there is no sharp boundary of age forwhich a female can surely identify.

These type of collections are designated as fuzzy sets. The elements of a set must bedistinct and distinguishable. By ’distinct’, it means that no element is repeated, and by’distinguishable’, means there is no doubt whether an element is either in the set or not inthe set.

(i) The standard mathematical symbols used to represent sets are upper-case letters likeA,B,X, · · · etc. and the elements of the set can be written in lower-case letters like a,b, p, q, x, y, etc.

(ii) If an element x is a member of a set A, we write x ∈ A, read as ‘x belongs to A’ or’a is an element of S’ or ’a is in S’. The symbol ∈ is a Greek alphabet epsilon. On

1

2 Theory of Sets

the other hand, if x is not the element of A, then we write x 6∈ A , which is read as ’xdoes not belong to A’ or ’x is not an element of the set A’.

(iii) If A is a set and a is any object, it should be easy to decide whether a ∈ A or a 6∈ A.Only then is A termed well-defined.

For example, A = 1, 2, 3, 4, 5 be a set, has elements 1, 2, 3, 4, 5. Here 1 ∈ A, but 6 6∈ A.Note that, S = 1, 1, 3 is not a set.

1.1.1 Description of Sets

As a set is determined by its elements, we have specify the elements of set A in order todefine A. Five common methods are used to describe the sets, they are (i) roster or list orenumeration or tabular method, and (ii) selector or rule or set-builder property method (iii)The characteristics method and (iv) Diagrammatic method.

(i) Roster method

In this method, all elements are listed explicitly separated by commas and are enclosedwithin braces . Sometimes parenthesis ( ) or square [ ] may also be used.

A set is defined by naming all its members and can be used only for finite sets. Let X,whose elements are x1, x2, · · · , xn is usually written as X = x1, x2, · · · , xn. For examples,the set of all natural numbers less than 5 can be represented as A = 1, 2, 3, 4.

Sometimes, it is not humanly possible to enumerate all elements, but after knowing someinitial elements one can guess the other elements. In this case dots are used at the end withinthe braces. For an example, set of positive integers can be written as A = 1, 2, 3, 4, · · ·,set of all integers B = . . . ,−2,−1, 0, 1, 2, . . ., etc.

It may be noted that the elements of a set can be written in any order, but the nameof an element is listed only once. For example, 2,3,4, 4,3,2, 2,4,3 all represent thesame set. Thus, while we describe a set in this manner, the order of the elements is notimportant.

(ii) Set-builder method

In this method, a set can be specified by stating one or more properties, which uniquelysatisfy by the elements. A set in this method is written as

A = x : P1(x) or P2(x), etc, (1.1)

i.e., x ∈ A if x satisfy the properties P1(x), P2(x), etc. The symbol ’:’ is read as ’suchthat’, it is also denoted by ’—’ or ’/’. For example,

A = x : x is a positive even integers B = x : x is a vowel in English alphabet C = x : x is integer and 1 ≤ x ≤ 10 etc .

It is required that the property P be such that for any given x ∈ U , the universal set, theproposition P (x) is either true of false.(iii) The characteristics method

A set is defined by a function, usually called a characteristic function, that declares whichelements of U are members of the set and which are not. Let U = u1, u2, . . . , un be theuniversal set and A ⊆ U . Then the characteristic function of A is defined as

χA(ui) =

1, if ui ∈ A0, if ui 6∈ A.

(1.2)

Sets 3

i.e., the characteristic function maps elements of U to elements of the set 0, 1, whichis formally expressed by χA : U → 0, 1. For example, let U = 1, 2, 3, . . . , 10 and A =2, 4, 6, 8, 10, then χA(2) = 1, χA(4) = 1, χA(6) = 1, χA(8) = 1, χA(10) = 1 and χA(a) = 0for all other elements of U . It may be observed that χA is onto function but not one-one.

(iv) Diagrammatic method

A set can be represented diagrammatically by closed figures like circles, triangles, rectangles,etc. The point in the interior of the figure represents the elements of the set. Such a repre-sentation is called a Venn diagram or Venn-Euler diagram, after the British mathematicianVenn. In this diagram the universal set U is represented by the interior of a rectangle andeach subset of U is represented by the circle inside the rectangle.

If two sets are equal then they represent by same circle. If the sets A and B are disjointthen the circles for A and B are drawn in such a way that they have no common area, Ifthe sets A and B have a small common area. If A ⊆ B then the circle for A is drawn fullyinside the circle for B. This visual representation helps us to prove the set identities veryeasily.

(v) Recursion method

A set can be described by giving one or more elements of the set and a rule for generatingthe remaining elements. The underlying process is called recursion. For example, (a) theset A = 1, 4, 7, · · · can be described as A = a0 = 1, an+1 = an + 3; (b) F = Fn : F0 =0, F1 = 1, Fn = Fn−1 + Fn−2 is a set described by recursion. This set is called the set ofFibonacci numbers.Some standard sets and their notations

Some sets are frequently used in mathematical analysis or in algebraic structure, which arestated below.

N → The set of all natural numbers.Z → The set of all integers.Z+ → The set of all positive integer.Q → The set of all rational numbers.Q+ → The set of all positive rational numbers.R → The set of all real numbers.R+ → The set of all positive real numbers.C → The set of all complex numbers.

1.1.2 Types of Sets

Null set

A set which contains no element is called null set or empty set or void set and is denoted bythe Greek alphabet φ (read as phi). In Roaster method, it is denoted by . For example,

A = x : x2 + 4 = 0 and x ∈ Ris a null set. To describe the null set, we can use any property, which is not true for anyelement. It may be noted that the set φ or 0 is not a null set.

A set which is not a null set, is called non-empty set.

Singleton set

A set consisting only a single element is called a singleton or unit set. For example, A = 0,B =x: 1 < x < 3, x is integer, the solution set C = x : x − 2 = 0 = 2 etc., areexamples of singleton set. Note that 0 is not a null set, since it contains 0 as its member,it is singleton set.

4 Theory of Sets

Finite and infinite sets

A set containing finite elements is called finite sets, otherwise it is called infinite set, i.e., aset does not contain a definite number of elements.

(i) A =x: x is the consonant in English alphabet,

(ii) B =x: 1 < x < 15, x is integer

are examples of finite sets, whereas

(i) A =x: x is a rational numbers,

(ii) B =x: x is a straight line in space,

are the examples of infinite sets. Here, the process of counting the different elements comesto one end. Here, the process of counting the different elements comes to an end. We let|A| to denote the numbers of elements of a finite set A.

Indexed set

A set, whose elements are themselves sets is often referred to as a family of sets. Let usconsider a family of n sets A1, A2, . . . , An in the form F = Aα|α ∈ I where Aα correspondsto an element α in the set I. I is said to be an indexing set and α is called the set index.In general, I be an arbitrary set then F = Aα|α ∈ I is an arbitrary collection of sets Aα

indexed by I.

Set of sets

If the elements of a set be also the some other sets, then this set is known as a family ofsets, or a set of sets. For example, A =2, 3, 5, 6, 8, z, R+ is a set of sets.

Subset and superset

Let A and B be two given sets. The set B is said to be a subset of A if

x ∈ B =⇒ x ∈ A, (1.3)

i.e., every element of B is an element of A. This is very often denoted by B ⊆ A (writtenas B is contained or included in A). This is called set inclusion. For example,

(i) The set of all integers (Z) is the subset of all rational numbers (Q).

(ii) A =a, b, c, d, B =a, b, c, d, e, f. Here each element of A is also an element of B,thus A ⊆ B.

(iii) A =1, 5, 7 and B =1, 5, 7. Here A ⊆ B and B ⊆ A.

(iv) φ is a subset of every set.

(v) The subsets of A =2, 3, 4 are φ, 2, 3, 4, 2, 3, 2, 4, 3, 4 and 2, 3, 4.

(vi) A =2, 3, 5, B =2, 3, 6. Then A 6⊆ B and B 6⊆ A, because, 5 ∈ A but 5 6∈ B and6 ∈ B but 6 6∈ B.

If B ⊆ A, then A is called the super set of B, which is read as ‘A is a super set of B’.

Sets 5

Proper subset

The set A is called proper subset of B if every element of A is a member of B and there isat least one element in B such that it is not in the set A. It is written as A ⊂ B. Therefore,B is the proper subset of A if

(i) x ∈ B =⇒ x ∈ A

(ii) ∃ y ∈ A such that y /∈ B.

In this case, B ⊆ A and A 6= B and B is said to the proper subset of A and is denoted byB ⊂ A. If B is the subset of A(i.e. B ⊂ A) A is called the super set of B. For example,

(i) 1, 2 is the proper subset of 1, 2, 3, 4.

(ii) The set of vowels is a proper subset of the set of English alphabet

(iii) N (set of natural numbers) is a proper subset of Z (set of integers).

Note the following:

(i) If ∃ even a single element in A which is not in B, then A is not a subset of B and wewrite A 6⊆ B. For example 1, 2 6⊆ 2, 4, 6, 8, 9.

(ii) If A ⊆ B or B ⊆ A, then the sets A and B are said to be comparable. For example,if A = 1, 2, B = 5, 6, 7 then A 6⊂ B and these are not comparable.

(iii) Every set is a subset of itself and every set is a subset of the universal set.

(iv) φ has no proper subset. Also, A ⊆ φ⇒ A = φ.

(v) For any set A, A ⊆ A. This is known as the reflexive law of inclusion.

(vi) If A ⊆ B and B ⊆ C, then A ⊆ C. This is known as transitive law of inclusion.

In Venn diagram, the universal set is usually represented by a rectangular region and itssubset by closed bounded regions inside the rectangular region.

Equality of sets

If A ⊆ B and B ⊆ A, then A and B contain the same members. Two sets A and B aresaid to be equal if every element of A is an element of B and also every element of B is anelement of A. That is, A ⊆ B and B ⊆ A. The equality, of two sets is denoted by A = B.Conversely, if A = B then A ⊆ B and B ⊆ A must be satisfied. For example, A = 1, 4, 9and B = 4, 9, 1 are equal sets. To indicate that A and B are not equal, we write A 6= B.

Theorem 1.1.1 The null set is a subset of every set.

Proof: Let A be an arbitrary set. Then, in order to show that φ ⊂ A, we must show thatthere is no element of φ which is not contained in A. Since φ contains no element at all, nosuch element can therefore be found. Hence, φ ⊂ A.

Theorem 1.1.2 The number of subsets of a given set containing n elements is 2n.

Proof: Let A be an arbitrary set containing n elements. Then, one of its subsets is theempty set. Apart from this,

(i) The number of subsets of A, each containing 1 element =(n1

).

(ii) The number of subsets of A, each containing 2 elements =(n2

).

6 Theory of Sets

(iii) The number of subsets of A, each containing 3 elements =(n3

).

...

(iv) The number of subsets of A, each containing n elements =(nn

).

Therefore, the total number of subsets of A is

= 1 +(n

1

)+(n

2

)+ · · ·+

(n

n

)= (1 + 1)n = 2n.

The number of proper subsets of a set with n elements is 2n − 1.

Power set

A set formed by the all subsets of a given non empty set set S is called the power set of theset S and is denoted by P (S). If S = a, b, c then

P (S) =φ, a, b, c, a, b, a, c, b, c, a, b, c

.

Note that P (φ) = φ. The power set of any given set is always non-empty. The family ofall subsets of P (A) is called a second order power set of A and is denoted by P 2(A), whichstands for P (P (A)). Similarly, higher order power sets P 3(A), P 4(A), . . . are defined.

Order of a set is defined by the numbers of elements of A and is denoted by O(A). Fromthe above property, it is observed that the number of elements of the power set P (A) is 2n

if A contains n elements. For example, if A = 1, 2,−1 then O(A) = 3 and OP (A) = 8.In general, if O(A) = n then OP (A) = 2n and O(P 2(A)) = 22n.

Universal sets

In the theory of set it is observed that all the sets under consideration are the subsets ofa certain set. This set is called the universal set and it is usually denoted by U or S.Conversely, the universal set is the superset of every set. For example, The set of realnumbers < is the universal set for the set of integers Z and set of rational numbers Q.Again, the set of integers Z is the universal set for the sets of even integers, set of positiveintegers, etc.

This set is of all possible elements that are relevant and considered under particularcontext or application from which sets can be formed. The set U is not unique and it is asuper set of each of the given set. In venn diagram, the universal set is usually representedby a rectangular region.

1.2 Algebraic Operation on Sets

Like addition, multiplication and other operations on numbers in arithmetic, there are cer-tain operations on sets, namely union, intersection, complementation, etc. In this section,we shall discuss several ways of combining different sets and develop some properties amongthem.

1.2.1 Union of Sets

Let A and B be two given subsets of an universal set U . Union (or join) of two subsets Aand B, denoted by A ∪B, is defined by

A ∪B = x : x ∈ A or x ∈ B or both, (1.4)

Algebraic Operation on Sets 7

here the ‘or’ is means ‘and/or’, i.e. the set contains the elements which either belong to Aor B or both. The Venn diagram of Fig. 1.1 illustrate pictorially the meaning of ∪, whereU is the rectangular area, and A and B are disks. Union is also known as join or logical sumof A and B. Note that, the common elements are to be taken only once. For example,

A B U

Figure 1.1: Venn diagram of A ∪B (shaded area)

(i) If A =1, 3, 4, 5, a, b and B =a, b, c, 2, 3, 4, 6 thenA ∪B =1, 2, 3, 4, 5, 6, a, b, c.

(ii) If A = [2, 5] and B = [1, 3], then A ∪B = [1, 5] = x : 1 ≤ x ≤ 5.

From the Venn diagram we get the following properties of set union:

(i) Union s idempotent, i.e., A ∪A = A,

(ii) Set union is associative, i.e., (A ∪B) ∪ C = A ∪ (B ∪ C),

(iii) A ∪ U = U : Absorption by U . A ∪ φ = A: identity law

(iv) Set union is commutative, i.e., A ∪B = B ∪A,

(v) A ⊆ A ∪B and B ⊆ A ∪B for sets A and B.

(vi) If A ⊆ U ⇒ A ∪ U = U , U = the universal set and if A ⊆ B then A ∪B = B.

The union operation can be generalized for any number of sets. The union of the subsetsA1, A2, . . . , An is given by,

n⋃i=1

Ai = A1 ∪A2 ∪ . . . ∪An = x : x ∈ Ai; for some i = 1, 2, . . . , n

and for a family of sets Ai; i ∈ I is defined as⋃i∈I

Ai = x : x ∈ Ai, for some i ∈ I.

1.2.2 Intersection of Sets

Let A and B be two given subsets of an universal set U . The intersection of A and B isdenoted by A ∩B and is defined by

A ∩B = x : x ∈ A and x ∈ B. (1.5)

The intersection of the sets A and B is the set of all elements which are in both the sets Aand B. The A∩B is shown in Fig. 1.2. The intersection is also known as meet. A∩B is readas A intersection B or A meet B. For example, (i) let A =2, 5, 6, 8 and B =2, 6, 8, 9, 10then A∩B =2, 6, 8 (ii) if A = [2, 5] and B = [1, 3], then A∩B = [2, 3] = x : 2 ≤ x ≤ 3.From the Venn diagram we get the following properties of intersection:

8 Theory of Sets

A B

Figure 1.2: Venn diagram of A ∩B (shaded area)

(i) Intersection is idempotent, i.e., A ∩A = A, follows from A ⊂ A,

(ii) A ∩B ⊂ A ; A ∩B ⊂ B and A ⊆ B then A ∩B = A.

(iii) A ∩ φ = φ : absorption by φ. A ∩ U = A: identity law.

(iv) Set intersection is commutative, i.e., A ∩B = B ∩A.

(v) Set intersection is associative, i.e., (A ∩B) ∩ C = A ∩ (B ∩ C).

The intersection of the n subsets is given byn⋂

i=1

Ai = A1 ∩A2 ∩ . . . ∩An = x : x ∈ Ai,∀i.

Ex 1.2.1 For given two sets A = x : 2 cos2 x + sinx ≤ 2;B = x : x ∈ [π2 ,

3π2 ], find

A ∩B. [KH 06]

Solution: The solution of the trigonometric relation 2 cos2 x+ sinx ≤ 2 is given by,2 cos2 x+ sinx ≤ 2 ⇒ sinx[1− 2 sinx] ≤ 0

⇒ sinx ≤ 0 and 1− 2 sinx ≥ 0or sinx ≥ 0 and 1− 2 sinx ≤ 0

⇒ sinx ≤ 0 or sinx ≥ 12.

If x ∈ [π2 ,

3π2 ], then the solutions of sinx ≤ 0 are given by π ≤ x ≤ 3π

2 and the solutions ofsinx ≥ 1

2 are given by π2 ≤ x ≤ 5π

6 . Therefore,

A ∩B =[π,

3π2

]∪[π

2,5π6

].

1.2.3 Disjoint Sets

It is sometimes observed that the intersection between two non-empty sets produced a nullset. In this case, no element is common in A and B and these two sets are called disjointor mutually exclusive sets. Thus two sets A and B are said to be disjoint if and only ifA∩B = φ, i.e. they have no element in common. Then Venn diagram of disjoint sets A andB is shown in Fig. 1.3. For example, (i) A = 1, 2, 3 and B = 6, 7, 9 and (ii) if A =x:x is even integer and B =x: x is odd integer then A ∩B = φ, i.e., A and B are disjoint.

When A ∩B 6= φ, the sets A and B are said to be intersecting.

Note 1.2.1 The three relations B ⊂ A, A∪B = A and A∩B = B are mutually equivalent,i.e., one implies the other two.

Algebraic Operation on Sets 9

A B

Figure 1.3: Disjoint sets A and B

1.2.4 Complement of a Set

Let A be a subset of an universal set U . Then complement of a subset A, with respect toU , denoted by A′, Ac, A or −A, is defined by

A′ = x : x ∈ U but x 6∈ A, (1.6)

i.e., the set contains the elements which belong to the universal set U but not elements ofA. The venn diagram of Ac is shown in Fig. 1.4. Clearly, if A′ is the complement of A, then

"!#

A

U

Ac

Figure 1.4: Complement of A

A is a complement of A′. For example, let A = 1, 3, 5, 7, 9, if U = 1, 2, 3, 4, 5, 6, 7, 8, 9then A′ = 2, 4, 6, 8. From Venn diagram we have

(i) (A′)′ = A : involution property.

(ii) U ′ = φ ; φ′ = U .

(iii) If A ⊆ B then Bc ⊆ Ac and conversely, if Ac ⊆ Bc then B ⊆ A.

(iv) A ∪A′ = U : law of excluded middle. A ∩A′ = φ : law of contradiction.

(v) (a) (A ∪B)c = Ac ∩Bc and (b) (A ∪B)c = Ac ∪Bc : De Morgan’s laws.

In particular, for a finite family of subsets F = A1, A2, . . . , An, the De Morgan’s law canbe written as (

n⋃i=1

Ai

)′=

n⋂i=1

A′i and

(n⋂

i=1

Ai

)′=

n⋃i=1

A′i.

Ex 1.2.2 Prove that (A ∩ C) ∪ (B ∩ C ′) = φ⇒ A ∩B = φ.

Solution: The relation (A ∩ C) ∪ (B ∩ C ′) = φ gives A ∩ C = φ and B ∩ C ′ = φ. Now,

B ∩ C ′ = φ⇒ B ⊆ C.

Therefore, A ∩ C = φ⇒ A ∩B = φ.

10 Theory of Sets

1.2.5 Difference

Let A,B be any two subsets of an universal set U . The difference of two subsets A and Bof an universal set U is a subset of A, denoted by A−B or A/B and is defined by

A−B = x : x ∈ A and x /∈ B, (1.7)

i.e., the set containing of those elements of A which are not elements of B. Also

B −A = x : x ∈ B and x 6∈ A. (1.8)

This is also called the relative component of the set B with respect to the set A. A− B iscalled the complement of B relative to A. The differences A − B and B − A are shown in

A B

A−B B −A6 6

Figure 1.5: Set difference A−B and B −A

Fig. 1.5. A − B is read as A difference B or A minus B. For example, if A =2, 4, 5, 8and B =2, 5, 7, 10 then A−B =4, 8 and B −A =7, 10. From the Venn diagram wehave:

(i) A−A = φ, A− φ = A,

(ii) A−B ⊆ A, B −A ⊆ B, and A−B = A if A ∩B = φ.

(iii) Set difference is non-commutative, i.e., A−B 6= B −A,

(iv) A−B = φ when A ⊆ B.

(v) A ∩A = φ then A−B = A and B −A = B,

(vi) A−B = A ∩B′.

(vii) (A−B)∪A = A, (A−B) ∪B = A ∪B and (A−B) ∩B = φ.

(viii) A−B = A if and only if A ∩B = φ.

(ix) A−B, A ∩B and B −A are mutually exclusive.

If the set A is the universal set, the complement is absolute, known as complementation andis usually denoted by B.

Ex 1.2.3 For two subsets A and B of an universal set U , show thatA− (B − C) = (A−B) ∪ (A ∩ C).

Solution: Let x be any element of A− (B − C), then by definition,

A− (B − C) ⇔ x : x ∈ A and x 6∈ (B − C)⇔ x : x ∈ A and (x 6∈ B or x 6∈ C)⇔ x : (x ∈ A and x 6∈ B) or (x ∈ A and x ∈ C)⇔ x : x ∈ (A−B) or x ∈ (A− C) = (A−B) ∪ (A ∩ C).

Hence, A− (B−C) ⊆ (A−B)∪ (A∩C) and (A−B)∪ (A∩C) ⊆ A− (B−C), consequently,A− (B − C) = (A−B) ∪ (A ∩ C).

Duality and Algebra Sets 11

1.2.6 Symmetric Difference

If A and B be two subsets of an universal set U . The symmetric difference, denoted byA∆B or A⊕B, is defined by

A∆B = = x : x ∈ A or x ∈ B but x 6∈ B= x : (x ∈ A and x /∈ B) or (x ∈ B and x /∈ A).

The set (A−B) ∪ (B −A) is also called the symmetric difference of A and B. Thus,

A∆B = (A ∪B)− (A ∩B) = (A ∩B′) ∪ (A′ ∩B)= (A ∪B)4 (A ∩B).

The Venn diagram of A∆B is shown in Fig. 1.6. If A = 1, 2, 4, 7, 9 and B = 2, 3, 7, 8, 9

A B

A4B

Figure 1.6: Symmetric difference of A and B (shaded ares)

then, A∆B = 1, 3, 4, 8. Note that,

(A−B) ∩ (B −A) = (A ∩B′) ∩ (B ∩A′)= A ∩ (B ∩B′) ∩A′

= (A ∩ φ) ∩A′ = A ∩A′ = φ.

Therefore, A 4 B can be considered as the union of disjoint subsets A − B and B − A,provided A−B and B−A are both non empty. From the definition, we have the following,

(i) Symmetric difference is commutative, i.e., A∆B = B∆A,

(ii) Symmetric difference is associative, i.e., (A∆B)∆C = A∆(B∆C),

(iii) A∆φ = A, for all subsets of A.

(iv) A∆A = φ, for all subsets of A.

(v) A∆B = φ iff A = B,

(vi) A ∩ (B 4 C) = (A ∩B)4 (A ∩ C) : Distributive property.

1.3 Duality and Algebra SetsPrinciple of duality

Let E is an equation(or law) of set algebra (involving ∪, ∩, U , φ). If we replace ∪ by ∩, ∩by ∪, U by φ and φ by U in E then we obtain E∗ another law, which is also a valid law.This is known as principle of duality. For example,

12 Theory of Sets

(i) A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C), its dual law is A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C),

(ii) A ∩A = φ its dual is A ∪A = U .

It is a fact of set algebra, called the principle of duality, that, if any equation E is an identity,then its dual E∗ is also an identity.

Algebra of sets

Some commonly used laws of sets are stated below. Note that the law stated in (b) is thedual law of (a) and conversely.

1. Idempotent laws(a) A ∩A = A (b) A ∪A = A.

2. Identity laws(i) (a) A ∪ φ = A (b) A ∩ U = A.(ii) (a) A ∩ φ = φ (b) A ∪ U = U.3. Commutative laws

(a) A ∪B = B ∪A (b) A ∩B = B ∩A.4. Associative laws

(a) (A ∪B) ∪ C = A ∪ (B ∪ C) (b) (A ∩B) ∩ C = A ∩ (B ∩ C).5. Distributive laws

(a) A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C) (b) A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)6. Inverse law

(a) A ∪Ac = U (b) A ∩Ac = φ7. Domination laws

(a) A ∪ U = U (b) A ∩ φ = φ8. Absorption laws

(a) A ∪ (A ∩B) = A (b) A ∩ (A ∪B) = A9. De Morgan’s laws

(a) (A ∪B)c = Ac ∩Bc (b) (A ∩B)c = Ac ∪Bc.Let S1 and S2 be two set expressions. The notation S1 ⇒ S2 as well as S2 ⇒ S1. These

are the main algebraic operations on sets.

Property 1.3.1 Let A,B and C are any three finite sets, then [WBUT 09](i) A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C); (ii) A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C)

Proof: (i) Let x be any element of A ∪ (B ∩ C). Then,x ∈ A ∪ (B ∩ C) ⇔ x ∈ A or x ∈ (B ∩ C).

⇔ x ∈ A or (x ∈ B and x ∈ C)⇔ (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C)⇔ (x ∈ A ∪B) and (x ∈ A ∪ C)⇔ x ∈ (A ∪B) ∩ (A ∪ C).

The symbol ⇔ stands ‘implies and is implied by’. It also stands for ‘if and only if’. HenceA ∪ (B ∩ C) ⊆ (A ∪B) ∩ (A ∪ C) and (A ∪B) ∩ (A ∪ C) ⊆ A ∪ (B ∩ C). Hence

A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C).(ii) Let x be any element of A ∩ (B ∪ C). Then,

x ∈ A ∩ (B ∪ C) ⇔ x ∈ A ∧ x ∈ (B ∪ C)⇔ x ∈ A ∧ (x ∈ B ∨ x ∈ C)⇔ (x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈ C)⇔ (x ∈ A ∩B) ∨ (x ∈ A ∩ C)⇔ x ∈ (A ∩B) ∪ (A ∩ C)

Duality and Algebra Sets 13

Hence, A ∩ (B ∪ C) ⊆ (A ∩B) ∪ (A ∩ C) and (A ∩B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C). Hence,A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C).

Property 1.3.2 Let A,B and C are any three finite sets. Then,

(i) (A ∪B)′ = A′ ∩B′.

(ii) (A ∩B)′ = A′ ∪B′.

(iii) A− (B ∪ C) = (A−B) ∩ (A− C).

(iv) A− (B ∩ C) = (A−B) ∪ (A− C)

Proof: (i) Let x be an arbitrary element of (A ∪B)′. Then,

x ∈ (A ∪B)′ ⇔ x /∈ (A ∪B)⇔ x /∈ A and x /∈ B⇔ x ∈ A′ and x ∈ B′

⇔ x ∈ (A′ ∩B′)

Hence (A ∪B)′ ⊆ A′ ∩B′ and A′ ∩B′ ⊆ (A ∪B)′. Therefore,(A ∪B)′ = A′ ∩B′.

(ii) Similarly, (A ∩B)′ = A′ ∪B′.(iii) Let x be an arbitrary element of A− (B ∪ C). Now,

x ∈ [A− (B ∪ C)] ⇔ [(x ∈ A) and x /∈ (B ∪ C)]⇔ [(x ∈ A) and (x /∈ B and x /∈ C)]⇔ [(x ∈ A and x /∈ B) and (x ∈ A and x /∈ C)]⇔ [x ∈ (A−B)] and [x ∈ (A− C)]⇔ x ∈ [(A−B) ∩ (A− C)]

Thus, A− (B ∪ C) ⊆ (A−B) ∩ (A− C) and (A−B) ∩ (A− C) ⊆ A− (B ∪ C) and so,A− (B ∪ C) = (A−B) ∩ (A− C).

(iv) Similarly, A− (B ∩ C) = (A−B) ∪ (A− C).

Ex 1.3.1 Prove that (A− C) ∩ (B − C) = (A ∩B)− C.

Solution: We shall use suitable laws of algebra in sets. Using the results A− C = A ∩ C ′;B − C = B ∩ C ′, we get,

L.H.S = (A− C) ∩ (B − C) = (A ∩ C ′) ∩ (B ∩ C ′)= (A ∩ C ′) ∩ (C ′ ∩B) = (A ∩ C ′) ∩ C ′ ∩B= (A ∩ C ′) ∩B = (A ∩B) ∩ C ′

= (A ∩B)− C = R.H.S.(proved)

Ex 1.3.2 Show (A−B) ∪ (B −A) = (A ∪B)− (A ∩B), where A,B,C are three sets.

Solution: We shall use suitable laws of algebra in sets. Using the results A−B = A ∩B′;B −A = B ∩A′, we get,

L.H.S = (A−B) ∪ (B −A) = (A ∩B′) ∪ (B ∩A′)= (A ∩B′) ∪B ∩ (A ∩B′) ∪A′= (A ∪B) ∩ (B ∪B′) ∩ (A ∪A′) ∩ (B′ ∪A′)= (A ∪B) ∩ S ∩ S ∩ (B′ ∪A′) ; S being Universal set.

= (A ∪B) ∩ (B′ ∪A′) = (A ∪B) ∩ (A ∩B)′

= (A ∪B)− (A ∩B) = R.H.S.(proved)

14 Theory of Sets

Ex 1.3.3 Show that (A ∩B)− (A ∩ C) = (A ∩B)− C, where A,B,C are sets.

Solution: We shall use suitable laws of algebra in sets.L.H.S = (A ∩B)− (A ∩ C) = (A ∩B) ∩ (A ∩ C)′

= (A ∩B) ∩ (A′ ∪ C ′) = (A ∩B) ∩A′ ∪ (A ∩B) ∩ C ′= φ ∪ (A ∩B) ∪ (A ∩B) ∩ C ′= φ ∪ (A ∩B) ∩ C ′ = (A ∩B) ∩ C ′

= (A ∩B)− C = R.H.S.(proved)

Ex 1.3.4 Show that (A ∩B ∩C) ∪ (A ∩B ∩C ′) ∪ (A ∩B′ ∩C) ∪ (A ∩B′ ∩C ′) = A, whereA,B,C are sets. [ KH 07]

Solution: We shall use suitable laws of algebra in sets.L.H.S = (A ∩B ∩ C) ∪ (A ∩B ∩ C ′) ∪ (A ∩B′ ∩ C) ∪ (A ∩B′ ∩ C ′)

= (X ∩ C) ∪ (X ∩ C ′) ∪ (Y ∩ C) ∪ (Y ∩ C ′) where X = A ∩B , Y = A ∩B′.

= [X ∩ (C ∪ C ′)] ∪ [Y ∩ (C ∪ C ′)]= (X ∩ U) ∪ (Y ∩ U); where U = Universal set.

= X ∪ Y = (A ∩B) ∪ (A ∩B′)= A ∩ (B ∪B′) = A ∩ U = A = R.H.S.(Proved).

Ex 1.3.5 If A∪B = A∪C and A∩B = A∩C simultaneously for subsets A,B,C of a setS, prove that B = C. [ CH 09]

Solution: We shall use suitable laws of algebra in sets.

L.H.S = B = (A ∪B) ∩B = (A ∩B) ∪ (B ∩B)= (A ∩ C) ∪B [as A ∩B = A ∩ C]= (A ∪B) ∩ (C ∪B)= (A ∪ C) ∩ (B ∪ C) [as A ∪B = A ∪ C]= (A ∩ C) ∪ C = C = R.H.S.(Proved).

From this, we conclude only A ∪ B = A ∪ C or only A ∩ B = A ∩ C does not necessarilyimply B = C.

Ex 1.3.6 Define complement of A by A′ such that A ∪A′ = S,A ∩A′ = φ show that(A ∪B)′ = A′ ∩B′.

Solution: Let C = A ∪B and D = A′ ∩B′. We shall show that D = C ′. Now,

C ∪D = (A ∪B) ∪ (A′ ∩B′)= (A ∪B ∪A′) ∩ (A ∪B ∪B′)= (A ∪A′) ∪B ∩ A ∪ (B ∪B′)= S ∪B ∩ A ∪ S = S ∩ S = S.

Also using the relations C = A ∪B and D = A′ ∩B′, we get,

C ∩D = (A ∪B) ∩ (A′ ∩B′)= (A ∩A′ ∩B′) ∪ (B ∩A′ ∩B′)= (φ ∩B′) ∪ (φ ∩A′) = φ ∪ φ = φ.

Hence C ′ = D i.e. (A ∪B)′ = A′ ∩B′.

Cartesian Product of Sets 15

Ex 1.3.7 If A ⊆ B and C is any set then show that A ∪ C ⊆ B ∪ C.

Solution: Let x be any element of A ∪ C. Hence,x ∈ A ∪ C ⇒ x ∈ A or x ∈ C.

Again, x ∈ A⇒ x ∈ B (since A ⊆ B). Therefore,x ∈ A ∪ C ⇒ x ∈ A or x ∈ C

⇒ x ∈ B or x ∈ C⇒ x ∈ B ∪ C.

Again, x ∈ C ⇒ x ∈ B ∪ C. Hence, A ∪ C ⊆ B ∪ C. (Proved)

Ex 1.3.8 Prove that (A′ ∩B′ ∩ C) ∪ (B ∩ C) ∪ (A ∩ C) = C.

Solution: L.H.S. = (A′ ∩B′ ∩ C) ∪ (B ∩ C) ∪ (A ∩ C). Now consider,(B ∩ C) ∪ (A ∩ C) = (C ∩B) ∪ (C ∩A)

= C ∩ (B ∪A) = (A ∪B) ∩ C.

Now again, A′ ∩B′ ∩ C = (A ∪B)′ ∩ C. Hence,

L.H.S. = A ∪B)′ ∩ C ∪ (A ∪B) ∩ C= (A ∪B)′ ∪ (A ∪B) ∩ C = S ∩ C (S = Universal set)= C = R.H.S. (Proved).

Ex 1.3.9 A,B,C are subsets of U , prove that [A ∩ (B ∪ C)] ∩ [A′ ∪ (B′ ∩ C ′)] = φ.

Solution: Using the properties of sets, we get,

LHS = [A ∩ (B ∪ C)] ∩ [A′ ∪ (B′ ∩ C ′)]= [A ∩ (B ∪ C) ∩A′] ∪ [A(B ∪ C) ∩ (B′ ∩ C ′)]= [A ∩A′ ∩ (B ∪ C)] ∪ [A ∩ (B ∪ C) ∩ (B ∪ C)′]= [φ ∩ (B ∪ C)] ∪ [A ∩ φ] = φ ∪ φ = φ.

1.4 Cartesian Product of Sets

Let A and B are two nonempty sets. An order pair consists of two elements, say a ∈ Aand b ∈ B, and it is denoted by (a, b). The element a is called the first element or firstcoordinate and the element b is called the second element or second coordinate. The orderedpairs (a, b) and (b, a) are distinct unless a = b. Thus (a, a) is a well-defined ordered pair.

If a, c ∈ A and b, d ∈ B, two ordered pairs (a, b) and (c, d) are said to be equal, i.e.,(a, b) = (c, d) if and only if a = c and b = d.

An order triple is ordered triple of objects (a, b, c) where a is first, b is second and c isthird element of triple. An order triple can also be written in terms of ordered pairs as(a, b), c. Similarly, ordered quadruple is an ordered pair ((a, b), c), d with first elementas ordered pair.

Definition 1.4.1 Let A and B be any two finite sets. The cartesian product (or crossproduct or direct product) of A and B, denoted by A × B, (read as A cross B), is the setdefined by,

A×B =

(x, y)|x ∈ A and y ∈ B

16 Theory of Sets

i.e., A × B is the set of all distinct order pairs (x, y), the first element of the pair is anelement of A and the second is an element of B. For example, let A =a, b and B =1, 2,3. Then

A×B =(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3) andB ×A =(1, a), (2, a), (3, a), (1, b), (2, b), (3, b).

The geometric representation of A×B is depicted in the Fig. 1.7.

-

6

1

2

3

(a, 1)

(a, 2)

(a, 3)

•

•

•

•

•

•

(b, 1)

(b, 2)

(b, 3)

a b

Figure 1.7: Representation of A×B

From this example, it is observed that A×B 6= B ×A, so in general, A×B 6= B ×A.

Let A1, A2, · · · , An be finite collection of non-empty sets. The cartesian product of thecollection, denoted by A1 ×A2 × · · · ×An, is the set defined by,

n∏i=1

Ai = A1 ×A2 × · · · ×An = (x1, x2, · · · , xn) : xi ∈ Ai.

In particular, if A1 = A2 = · · · = An = A, the cartesian product of the collection of sets,denoted by An, is the set of all ordered n tuples,

An = (x1, x2, · · · , xn) : xi ∈ A.

If A = B = <, the set of all real numbers, then < × < (= <2) is a set of all points in theplane. The ordered pair (a, b) represents a point in the plane. Similarly, <×<×< (= <3)is a set of all points in space, i.e, (a, b, c) ∈ < × <× < is a point in space. Below are someimportant properties of cartesian product

(i) The cartesian product is non-commutative, i.e., in general A×B 6= B ×A.

(ii) n(A) = p and n(B) = q then n(A×B) = n(B ×A) = pq.

(iii) If A = φ or B = φ then A×B = φ.

(iv) If either A or B is infinite and other is empty then A×B = φ.

(v) If either A or B is infinite and other is non-empty then A×B is infinite.

Following are some results on cartesian product.

(i) If A ⊆ B then A× C ⊆ B × C for any sets A, B, C.

(ii) If A ⊆ B and C ⊆ D then A× C ⊆ B ×D.

(iii) If A ⊆ B then A×A = (A×B) ∩ (B ×A).

Cartesian Product of Sets 17

Result 1.4.1 For any non-empty sets A and B, A×B = B ×A iff A = B.

Proof: Let A×B = B ×A. Then let,

x ∈ A⇒ (x, y) ∈ A×B, where y ∈ B⇒ (x, y) ∈ B ×A, since A×B = B ×A⇒ x ∈ B.

Thus A ⊆ B. Similarly, B ⊆ A. Hence A = B. Conversely, let A = B. Then A×B = A×Aand B ×A = A×A. Hence, A×B = B ×A.

Ex 1.4.1 For three non-empty sets A,B,C, prove thatA× (B ∩ C) = (A×B) ∩ (A× C).

Solution: This result is the direct consequence of the definition. Let (x, y) be an arbitraryelement of the set A× (B ∩ C). Then,

(x, y) ∈ A× (B ∩ C) ⇔ x ∈ A and y ∈ (B ∩ C)⇔ x ∈ A and [y ∈ B and y ∈ C]⇔ [x ∈ A and y ∈ B] and [x ∈ A and y ∈ C]⇔ [(x, y) ∈ A×B] and [(x, y) ∈ A× C]⇔ (x, y) ∈ (A×B) ∩ (A× C).

Therefore, A × (B ∩ C) ⊆ (A × B) ∩ (A × C) and (A × B) ∩ (A × C) ⊆ A × (B ∩ C) andconsequently, A× (B ∩ C) = (A×B) ∩ (A× C). Similarly,

A× (B ∪ C) = (A×B) ∪ (A× C).

Ex 1.4.2 For any three sets A,B,C prove thatA× (B − C) = (A×B)− (A× C).

Solution: Let (x, y) be an arbitrary element of A× (B − C). Then,

(x, y) ∈ A× (B − C) ⇔ x ∈ A and y ∈ (B − C)⇔ x ∈ A and [y ∈ B and y 6∈ C]⇔ [x ∈ A and y ∈ B] and [x ∈ A and y 6∈ C]⇔ [(x, y) ∈ (A×B)] and [(x, y) 6∈ (A× C)]⇔ (x, y) ∈ (A×B)− (A× C).

Therefore, A× (B − C) ⊆ (A× B)− (A× C) and (A× B)− (A× C) ⊆ A× (B − C) andconsequently, A× (B − C) = (A×B)− (A× C).

Ex 1.4.3 For any sets A,B,C and D, we have, [ CH 96, 01](A×B) ∩ (C ×D) = (A ∩ C)× (B ∩D).

Solution: Let (x, y) be an arbitrary element of (A×B) ∩ (C ×D). Then,

(x, y) ∈ (A×B) ∩ (C ×D) ⇔ (x, y) ∈ (A×B) and (x, y) ∈ (C ×D)⇔ (x ∈ A and y ∈ B) and (x ∈ C and y ∈ D)⇔ (x ∈ A and x ∈ C) and (y ∈ B and y ∈ D)⇔ x ∈ (A ∩B) and y ∈ (B ∩D)⇔ (x, y) ∈ (A ∩ C)× (B ∩D).

Thus, (A×B)∩ (C ×D) ⊆ (A∩C)× (B ∩D) and (A∩C)× (B ∩D) ⊆ (A×B)∩ (C ×D),so that (A×B) ∩ (C ×D) = (A ∩ C)× (B ∩D). Similarly, for any sets A,B,C and D, wehave A ⊆ B and C ⊆ D ⇒ (A× C) ⊆ (B ×D).

18 Theory of Sets

1.5 Cardinal Numbers

The number of distinct elements in a set A is called the cardinal number of the set and itis denoted by n(A) or |A| or card(A). For example, n(φ) = 0, n(a) = 1, n(a, b) = 2,n(Z) = ∞, etc. Following are the important properties of cardinal number

(i) If A and B are disjoint, then

(a) n(A ∩B) = n(φ) = 0 and (b) n(A ∪B) = n(A) + n(B),

(ii) Let A and B be two finite sets, with A ∩B 6= φ, then

n(A ∪B) = n(A) + n(B)− n(A ∩B),

(iii) If A, B, C be three arbitrary finite sets then

n(A ∪B ∪ C) = n(A) + n(B) + n(C)− n(A ∩B)

−n(B ∩ C)− n(C ∩A) + n(A ∩B ∩ C),

(iv) Suppose we have any finite number of finite sets, say, A1, A2, · · · , Am. Let sk be thesum of the cardinalities, then

n(A1 ∪A2 ∪ · · · ∪Am) = s1 − s2 + s3 − · · ·+ (−1)m−1sm.

(v) n(A ∪A) = n(A) and n(A ∩A) = n(A).

The inclusion and exclusion principle

The number of elements in finite sets such as A ∪ B, A ∩ B, A∆B, etc. are obtained byadding and as well as deleting certain elements. This method of finding the number ofelements in a finite set is known as inclusion and exclusion principle.

Ex 1.5.1 If n(A) and n(B) denote the number of elements in the finite sets A and Brespectively, then prove that n(A) + n(B) = n(A ∪B) + n(A ∩B).

Solution: If A and B are disjoint then n(A ∪ B) is equal to the sum of the elements ofA and the elements of B. That is n(A ∪ B) = n(A) + n(B). If A and B are disjoint then

A B

B − (A ∩B) A− (A ∩B)7 K

A ∩B

6

Figure 1.8:

A ∪ B is express as union of three disjoint sets A ∩ B, A − (A ∩ B) and B − (A ∩ B). Letus draw the Venn diagram showing A ∩B′, A ∩B, A′ ∩B and A′ ∩B′, where A and B aretwo subsets of the universal set U . From the diagram, we have,

A = (A ∩B′) ∪ (A ∩B) and B = (A ∩B) ∪ (A′ ∩B)

and the sets A ∩B′, A ∩B and A′ ∩B are disjoint. Therefore,

n(A) = n(A ∩B′) + n(A ∩B) and n(B) = n(A ∩B) + n(A′ ∩B)⇒ n(A) + n(B) = n(A ∩B′) + 2n(A ∩B) + n(A′ ∩B).

Cardinal Numbers 19

Again, from the diagram, we have,

A ∪B = (A ∩B′) ∪ (A ∩B) ∪ (A′ ∩B),

in which (A ∩B′), (A ∩B) and (A′ ∩B) are disjoint sets and so

n(A ∪B) = n(A ∩B′) + n(A ∩B) + n(A′ ∩B).

Subtracting, we get, n(A) + n(B) = n(A ∪ B) + n(A ∩ B). Let n(A) = r, n(B) = s andn(A ∩B) = t. Then from Fig. 1.8,

n(A− (A ∩B)) = r − t and n(B − (A ∩B)) = s− t.

Therefore, n(A ∪B) = t+ (r − t) + (s− t) = r + s− t

= n(A ∪B) = n(A) + n(B)− n(A ∩B).

Following are some results on Venn diagrams:

(i) n(A ∪B) = n(A) + n(B)− n(A ∩B) and n(A ∪B) = n(A) + n(B),

provided A and B are disjoints, i.e., n(A ∩B) = 0.

(ii) n(A ∩B′) = n(A)− n(A ∩B) and n(A′ ∩B) = n(B)− n(A ∩B).

(iii) n(A4B) = n(A) + n(B)− 2n(A ∩B).

(iv) n(A′ ∪B′) = n(U)− n(A ∩B) and n(A′ ∩B′) = n(U)− n(A ∪B).

Ex 1.5.2 In a canteen, out of 122 students, 42 students buy ice cream, 36 buys buns and10 buy cakes, 15 student buy ice-cream and buns, 10 ice-cream and cakes, 4 cakes and bunshut not ice-cream and buns butt no cakes. Find (i) how many students buy nothing at all.(ii) how many students buy at least two items. (iii) how many students buy all three items.

Solution: Define the sets A,B and C such that, A = Set of students who buy cakes, B =Set of students who buy ice-cream, C = Set of students who buy buns.

According to question, we have, n(A) = 10, n(B) = 42, n(C) = 10, n(B ∩ C) = 15, n(A ∩B) = 10, n[(A ∩ C)−B] = 4, n[(B ∩ C)−A] = 11 and n[A− (B ∪ C)] = 10. Now, we have,

n(B ∪ C) = n(B) + n(C)− n(B ∩ C) = 42 + 36− 15 = 63.n(B ∪ C)− n(B) = 63− 42 = 21.

and n(B ∪ C)− n(C) = 63− 36 = 27.

The above distribution of the student can be illustrated by Venn diagram. Now the totalnumber of students buying something

= 10 + 6 + 21 + 4 + 4 + 11 + 17 = 73.Therefore, the number of students who did not buy anything = 123 − 73 = 50. Number ofstudents buying all three items = 4.

Ex 1.5.3 In a recent survey 500 students in a college it was found that 150 students readnewspaper A and 200 read newspaper B , 80 students read both the newspaper A and B.Find how many read either newspaper.

Solution: Let X and Y denote the set of students who read newspapers A and B respec-tively. It is given that n(A) = 150, n(B) = 200, n(A ∩ B) = 80, n(U) = 500. Number ofstudents who read either A or B is n(A ∪B). Also,

n(A ∪B) = n(A) + n(B) + n(A ∩B) = 150 + 200− 80 = 270.

20 Theory of Sets

Ex 1.5.4 In a class of 42 students, each play at least one of the three games: Cricket, Hokeyand Football. It is found that 14 play Cricket, 20 play Hokey and 24 play Football, 3 palyboth Cricket and Football, 2 play both Hokey and Football and none paly all the three games.Find the number of students who paly Cricket but not Hokey.

Solution: Let C, H and F be the sets of students who play Cricket, Hockey and Football.Given that n(C) = 14, n(H) = 20, n(F ) = 24, n(C∩F ) = 3, n(H∩F ) = 2, n(H∩F ∩C) = 0and n(H ∪ F ∪ C) = 45 (since each student play at least one game). We know,

n(H ∪ F ∪ C) = n(H) + n(F ) + n(C)− n(H ∩ F )−n(H ∩ C)− n(F ∩ C) + n(H ∩ F ∩ C).

or, 42 = 20 + 24 + 14− 2− n(H ∩ C)− 3 + 0⇒ n(H ∩ C) = 11.

Now, the number of students who paly Cricket but not Hockey is n(C −H). Also, we know

n(C) = n(C ∩H) + n(C −H)i.e., n(C −H) = n(C)− n(C ∩H) = 14− 11 = 3.

Hence 3 students play Cricket but not Hockey.

1.6 Relation

Let A and B are two non empty sets and a ∈ A, b ∈ B. A relation or binary relationbetween two sets A and B is a subset of A × B i.e. if (a, b) ∈ ρ, where ρ ⊆ A × B we saythat a stands in the relationship ρ with b and is denoted by aρb i.e.,

(a, b) ∈ ρ⇒ (a, b) ∈ A×B.

For example, let A = 4, 5, 6, 9 and B = 20, 22, 24, 28, 30. Let us define a relation ρ fromA into B by stipulating aρb if and only if a divides b, where a ∈ A and b ∈ B. Then it isclear that

ρ = (4, 20), (4, 24), (4, 28), (5, 20), (5, 30), (6, 24), (6, 30).If n(A) = n and n(B) = m then the number of elements of A× B is mn. It is known thatthe number of elements of the power set 2mn, where A × B is the original set. Again, anysubset of power set is a relation. Thus the total number of binary relations from A to B is2mn.

A relation ρ between the non empty set A and A is also said to be a binary relation onA. If n(A) = n, then there are 2n2

relations on it.

Definition 1.6.1 [Domain and range :] Let ρ be a relation from a set A into the set B.Then the domain of ρ denoted by D(ρ) is the set

D(ρ) = a : (a, b) ∈ ρ, a ∈ ρ, for some b ∈ B. (1.9)

The domain of a binary relation is the set of all first elements of the ordered pairs in therelation. The range or image of ρ denoted by I(ρ) is the set

I(ρ) = b : (a, b) ∈ ρ, b ∈ ρ, for some a ∈ B. (1.10)

The range of a binary relation is the set of all second elements of the ordered pairs in therelation. For example, let A =a, b, c and B =1, 2, 3. Then let a relation ρ =(a, 1),(b, 1), (c, 2), (a, 2). For this relation D(ρ) =a, b, c and I(ρ) =1, 2.

If (a, b) /∈ ρ, i.e., (a, b) ∈ (A×B)−ρ, then we say that a does not stand in the relation ship ρwith b and is denoted by aρb. Let ρ ⊆ N×N be given by ρ = (x, y)|x is a devisor of y.Then xρy, if x is divisor of y. As example 3ρ6 but 2ρ5.

Relation 21

Universal and null relation

We know, (A×A) is a subset of A×A. ρ = A×A is defined as a universal relation on A. Ifρ = A×B then ρ is called universal relation on A and B. Null set φ ⊂ A×A and if ρ = φthen ρ is called null relation or empty relation from A to B. In general, φ ⊆ ρ ⊆ A×A.

For example, in the set Z of integers ρ = (a, b)|a + b is an integer is an universalrelation and ρ = (a, b)|a+ b is not an integer is null relation.

Identity relation

If the set A, the idempotent relation ρ = (x, y) : x ∈ A, y ∈ A, x = y is called the identityor diagonal relation in A and it will be denoted by IA.

Inverse relation

If ρ be a relation from a set A to B. The inverse of ρ denoted by ρ−1, is a relation from aset B to A and defined by

ρ−1 = (y, x) : y ∈ B, x ∈ A, (x, y) ∈ ρ.

For example,

(i) Let ρ = (1, y), (1, z), (3, y) from A = 1, 2, 3 to B = x, y, z, then its inverse is

ρ−1 =

(y, 1), (z, 1), (y, 3).

(ii) Let A =a, b, c and B =1, 2, 3. The inverse of the relation ρ =(a, 1), (b, 1), (c,2), (a, 2) is ρ−1 =(1, a), (1, b), (2, c), (2, a).

Note that, if ρ ⊆ A × B then, ρ−1 ⊆ B × A. It is easy to verify that domain of ρ−1 =D(ρ−1) = I(ρ) = range of ρ and range of ρ−1 = I(ρ−1) = D(ρ) = domain of ρ. Clearly, if ρis any relation, then (ρ−1)−1 = ρ.

Composition of relations

Let ρ1 be a relation from a set A into the set B and ρ2 be another relation from a set Binto C, i.e. ρ1 ⊆ A×B and ρ2 ⊆ B ×C. The composition of ρ1 and ρ2, denoted by ρ1 ρ2,is the relation from A into C and is defined by a(ρ1 ρ2)c, as

a(ρ1 ρ2)c = (a, c) : (a, b) ∈ ρ1, (b, c) ∈ ρ2. (1.11)

if there exists some b ∈ B such that aρ1b and aρ2b for all a ∈ A and c ∈ C. For example, letA =1,2,3 and B = x, y and C = t, so, ρ1 = (1, x), (2, y), (3, y), ρ2 = (x, t), (y, t).Therefore, ρ1 ρ2 = (1, t), (2, t), (3, t). From this definition it follows that ρ1 = ρ, ρn =ρ ρn−1, n > 1.

Definition 1.6.2 [Set operation on relations] Since every binary relation is a set ofordered pair, so the set operations can also be defined on relations.

Let ρ1 and ρ2 be two relations from a set A to a set B. Then ρ1 ∪ ρ2, ρ1 ∩ ρ2, ρ1 − ρ2,ρ′1 are relations given by

(i) Union : a(ρ1 ∪ ρ2)b = aρ1b or aρ2b.

(ii) Intersection : a(ρ1 ∩ ρ2)b = aρ1b and aρ2b.

22 Theory of Sets

(iii) Difference : a(ρ1 − ρ2)b = aρ1b and a 6 ρ2b.

a(ρ2 − ρ1)b = a 6 ρ1b and aρ2b.

(iv) Complement : a(ρ′1)b = a 6 ρ1b.

The relations correspond to the set operations-union, intersection, difference and comple-mentation on sets. For example, let A =a, b, c, B =2, 4, 6, C =a, b, D =2, 4, 5.Let ρ1 be a relation from A into B defined as ρ1 =(a, 2), (b, 4), (c, 6) and ρ2 be anotherrelation from C into D, which is defined by ρ2 =(a, 2), (b, 4), (b, 5). Thus,

(i) ρ1 ∪ ρ2 =

(a, 2), (b, 4), (c, 6), (b, 5)

(ii) ρ1 ∩ ρ2 =

(a, 2), (b, 4)

(iii) ρ1 − ρ2 =

(c, 6)

, ρ2 − ρ1 =

(b, 5)

and ρ′1 = the set of all ordered pairs of A×B those are not in ρ1

= (a, 4), (a, 6), (b, 2), (b, 6), (c, 2), (c, 4).

1.6.1 Equivalence Relation

Let A be a nonempty set and ρ be a relation on A. Then ρ is called

(i) reflexive, if for all a ∈ A, aρa,

(ii) symmetric, if aρb holds ⇒ bρa must hold, for a, b ∈ A,

(iii) antisymmetric, if aρb and bρa hold then a = b,

(iv) transitive, if aρb and bρc hold ⇒ aρc must be hold, for a, b, c ∈ A.

It may be remembered that the relation ρ is not reflexive, if (a, a) 6∈ ρ for at least a ∈ A,not symmetric if (a, a) ∈ ρ but (b, a) 6∈ ρ for at least one pair (a, b) ∈ ρ and not transitive if(a, b) ∈ ρ, (b, c) ∈ ρ but (a, c) 6∈ ρ if a, b, c ∈ A. If a relation which is not symmetric, it isnot necessarily antisymmetric.

Ex 1.6.1 Let ρ be a relation on Z defined by aρb iff ab ≥ 0 for all a, b ∈ Z. Show that ρ isreflexive and symmetric but not transitive.

Solution: (i) Let a ∈ Z. Then obvious a2 ≥ 0, so aρb holds for all a ∈ Z. Therefore ρ isreflexive.

(ii) Let a, b ∈ Z, aρb and bρc hold. Then it does not imply ac ≥ 0 and c ≥ 0. Forexample, if a = −2, b = 0, c = 8 then ab ≥ 0 and c ≥ 0 hold, but ac = −16 6≥ −16. Hence ρis not transitive.

That is, ρ is reflexive and symmetric but not transitive.

Ex 1.6.2 Show that the relation aρb if ab > 0, a, b ∈ < is symmetric, transitive but notreflexive.

Solution: (i) 0 ∈ <, but 0ρ0. Since 0.0 6> 0. So it is not reflexive.(ii) aρb⇒ ab > 0 ⇒ ba > 0 ⇒ bρa. Hence R is symmetric. (iii) Now,

(a, b) ∈ ρ; (b, c) ∈ ρ⇒ ab > 0; bc > 0⇒ ab2c > 0 ⇒ ac > 0; since b2 > 0⇒ (a, c) ∈ ρ.

Hence ρ is transitive. Therefore, ρ is symmetric, transitive but not reflexive.

Relation 23

Ex 1.6.3 Verify wheatear the relations are reflexive, symmetric or transitive on the set <?(i) xρy if |x − y| > 0. (ii) xρy if 1 + xy > 0. (iii) xρy if |x| ≤ y (iv) xρy if

2x+ 3y = 10.

Solution: (i) ρ is not reflexive, as for any x ∈ <, x− x = 0 and hence|x− x| 6> 0, i.e., xρx.

Again, as |x− y| = |y − x|, we have |x− y| > 0 ⇒ |y − x| > 0, whencexρy ⇒ yρx, for all x, y ∈ <.

So ρ is symmetric. Consider 0, 1 ∈ <, then |1− 0| = |0− 1| = 1 > 0 shows that1ρ0 and 0ρ1 but 1ρ1 as |1− 1| 6> 0.

Hence ρ is not transitive.(ii) Since ∀x ∈ <, x2 ≥ 0,⇒ 1 + x2 > 0 and hence xρx,∀x ∈ <, whence ρ is reflexive.Again, ∀x, y ∈ <, if 1 + xy > 0, then 1 + yx > 0 as xy = yx, whence we see that,

xρy ⇒ yρx, for all x, y ∈ <.So ρ is symmetric. Let 3,− 1

9 ,−6 ∈ <, then 1 + 3.(− 19 ) = 2

3 > 0 shows that3ρ(− 1

9 ) and 1 + (− 19 )(−6) = 5

3 > 0 ⇒ (− 19 )ρ(−6).

But, 1 + 3.(−6) = −17 6> 0, we conclude that 3ρ(−6),, whence ρ is not transitive.(iii) Let us consider −2 ∈ <, then | − 2| = 2 6≤ −2, whence (−2)ρ(−2), showing that ρ is

not reflexive. Indeed | − 2| = 2 ≤ 5 and so (−2)ρ5, but |5| = 5 6≤ −2, whence 5ρ(−2) andconsequently ρ is not symmetric.

But if a, b, c ∈ <, then |a| ≤ b, |b| ≤ c gives |a| ≤ b ≤ c, i.e., aρc and so ρ is transitive.(iv) As 2 · 1 + 3 · 1 = 5 6= 10, so 1ρ1. So ρ is not reflexive. As 2 · 1 + 3 · (8/3) = 10, so

1ρ(8/3). But 2 · (8/3) + 3 · 1 = 25/3 6= 10, so (8/3)ρ1. So ρ is not symmetric.As 2 · (1/2) + 3 · 3 = 10, so (1/2)ρ3. As 2 · 3 + 3 · (4/3) = 10, so 3ρ(4/3) but 2 · (1/2) + 3 ·

(4/3) = 5 6= 10, proving (1/2)ρ(4/3). Hence ρ is not transitive and so there is no questionof equivalence relation.

Ex 1.6.4 Verify wheatear the following relation are reflexive, symmetric or transitive(i) In Z, xρy if x+ y is odd. (ii) In Z, xρy if |x− y| ≤ y.

Solution: (i) Since, x+ x = 2x = even, so xρx and ρ is not reflexive. Also,

xρy ⇒ x+ y is odd.⇒ y + x is odd. ⇒ yρx; ∀x, y ∈ Z.

Hence ρ is symmetric. Again, ρ is not transitive, since 1ρ2(as 1+2 is odd) and 2ρ3(as 2+3 isodd) but 1ρ3(since 1+3 is even). Hence ρ is not reflexive but symmetric and not transitive.

(ii) Here, xρy if |x− y| ≤ y, ∀x, y ∈ Z. Now, |x− x| = 0 ≤ x is not true, for negative x.Hence ρ is not reflexive. Now,

1ρ3 since |1− 3| = 2 < 3 but 3ρ1 since |3− 1| = 2 > 1.

Hence, ρ is not symmetric. By definition, xρy, yρz ⇒ |x− y| ≤ y; |y − z| ≤ z. Now,

|x− z| = |x− y + y − z| ≤ |x− y|+ |y − z| ≤ y + z.

This suggests that |x − z| may not be ≤ z. For example, 4ρ7, 7ρ4 but 9ρ4. Hence ρ is nottransitive.

24 Theory of Sets

Result 1.6.1 The following tabular form will give a comprehensive idea:

xρy iff Reflexive Symmetric Transitivey = 4x × × ×x < y × ×

√

x 6= y ×√

×xy > 0 ×

√ √

y 6= x+ 2√

× ×x ≤ y

√×

√

xy ≥ 0√ √ √

x = y√ √ √

where ρ is the binary relation on <.

Definition 1.6.3 [Equivalence relation] Let ρ be a relation on a set A. A binary relationρ on a set A is said to be an equivalence relation(or an RST relation ) if

(i) ρ is defined to be a reflexive if (a, a) ∈ ρ,∀a ∈ A i.e. if aρa,∀a ∈ A.

(ii) ρ is defined to be a symmetric if for any two elements a and b ∈ A,(a, b) ∈ ρ⇒ (b, a) ∈ ρ i.e. aρb⇒ bρa.

(iii) ρ is defined to be transitive if

(a, b) ∈ ρ; (b, c) ∈ ρ⇒ (a, c) ∈ ρ i.e. aρb, bρc⇒ aρc.

For example, let A = a, b, c and a relation ρ = (a, a), (a, b), (b, b), (b, a), (a, c), (c,a), (c, c). It is easy to verify that ρ is reflexive, as (a, a) ∈ ρ, (b, a) ∈ ρ,(a, c) ∈ ρ, (c, a)∈ ρ. Again ρ is transitive. Hence ρ is an equivalence relation.

Ex 1.6.5 In Z, xρy if 3x+ 4y is divisible by 7, verify wheatear the relation is equivalent.

Solution: We know, ∀x ∈ Z, 3x+ 4x = 7x is divisible by 7. Hence xρx, ∀x ∈ Z and so ρis reflexive. As by definition, xρy if 3x+ 4y is divisible by 7, so, xρy ⇒ 3x+ 4y = 7k, wherek is an integer. Now,

(3x+ 4y) + (3y + 4x) = 7(x+ y) = 7k1k ∈ Zor, 3y + 4x = 7(k1 − k); k1 − k ∈ Z.

Hence 3y + 4x is divisible by 7. Thus xρy ⇒ yρx, ∀x, y ∈ Z. So ρ is symmetric. Now,xρy, yρz ⇒ 3x+ 4y = 7k1; 3y + 4z = 7k2.

⇒ 3x+ 4z + 7y = 7(k1 + k2)⇒ 3x+ 4z = 7k; k ∈ Z⇒ xρz; ∀x, y, z ∈ Z.

Hence ρ is transitive. Thus ρ is reflexive, symmetric and transitive i.e. equivalent relation.

Ex 1.6.6 In Z, define aρb iff a− b is divisible by a given ’+’ ve integer n. Show that ρ isan equivalence relation.

Solution: (i) a ∈ Z, and a− a = 0 is divisible by n. Hence aρa; ∀a ∈ Z and hence ρ is areflexive. Using the definition,

aρb⇒ a− b is divisible by n

⇒ b− a is divisible by n

⇒ bρa; ∀a, b ∈ Z.

Relation 25

Hence ρ is a symmetric. Now,

aρb, bρc⇒ a− b, b− c is divisible by n

⇒ (a− b) + (b− c) is divisible by n

⇒ (a− c)is divisible by n

⇒ aρc; ∀a, b, c ∈ Z.

Hence ρ is transitive and consequently, it is an equivalence relation.

Ex 1.6.7 In the set of all lines in a plane aρb if a ‖ b and aρb if a ⊥ b. Verify wheatear therelations are equivalence relation ?

Solution: We know, any line is parallel to itself, so, a ‖ a; ∀a. Hence aρa. Therefore, ρ isreflexive. Also,

aρb⇒ a ‖ b⇒ b ‖ a⇒ bρa.

Hence ρ is symmetric. ρ is transitive, sinceaρb, bρc⇒ a ‖ b, b ‖ c⇒ a ‖ c

Hence ρ is reflexive, transitive and symmetric. In other words it is an equivalent relation.In the set of all lines in a plane aρb if a ⊥ b, is only symmetric.

Ex 1.6.8 Verify wheatear the following relations are reflexive symmetric or transitive onthe set Z? (i) xρy if |x− y| ≤ 3. (ii) xρy if x− y is a multiple of 6.

Solution: (i) Let a ∈ Z. Then |a− a| = 0 < 3. Therefore, aρb holds for for all a ∈ Z, i.e.,ρ is reflexive.

Let a, b ∈ Z and aρb holds, i.e., |a − b| ≤ 3. Then |b − a| = |a − b| ≤ 3, i.e., bρa holds.Therefore, ρ is symmetric.

Let a, b, c ∈ Z and aρb, bρc hold. Then |a− b| ≤ 3 and |b− c| ≤ 3. Now,|a− c| = |a− b+ b− c| ≤ |a− b|+ |b− c| ≤ 3 + 3 = 6.i.e., |a− c| ≤ 3 does not hold for all a, b, c ∈ Z.

Therefore, ρ is not transitive. For example, 2ρ5, 5ρ7 hold, but2ρ7 as |2− 7| = 5 6≤ 3.

(ii) Let us consider any integer x ∈ Z, then x− x = 0 = 0 · 6 shows thataρa,∀a ∈ Z.

Hence ρ is reflexive. Now, let xρy for some x, y ∈ Z, this meansx− y = 6k, k ∈ Z ⇒ y − x = (−k) · 6, where −k ∈ Z

and this shows that yρa, whence ρ is symmetric. Finally, let x, y, z ∈ Z such that xρy andyρz. Then x− y = 6 · k1 and y − z = 6 · k2 for some k1, k2 ∈ Z. Then

x− y + y − z = 6(k1 + k2)⇒ x− z = 6 · (k1 + k2), where k1 + k2 ∈ Z.

This shows that xρz, whence ρ is transitive. Consequently ρ is an equivalence relation.

Ex 1.6.9 The relation ρ on the set N × N of ordered pairs natural numbers is defined as’(a, b)ρ(c, d) iff ad = bc’. Prove that ρ is an equivalence relation.

Solution: (i) Let (a, a) ∈ N ×N . Then (a, b)ρ(c, d) holds as ab = ba, i.e., ρ is reflexive.(ii) Let (a, b) and (c, d) be any elements of N ×N and (a, b)ρ(c, d) holds. Therefore,

ad = bc or, da = cb, i.e., (c, d)ρ(a, b)

26 Theory of Sets

holds. Hence ρ is symmetric.(iii) Let (a, b), (c, d), (e, f) ∈ N ×N and (a, b)ρ(c, d), (c, d)ρ(e, f) hold. Then ad = bc andde = cf . Now,

(bc)(de) = (ad)(cf) or, be = af ; as dc 6= 0⇒ (a, b)ρ(e, f) holds.

Therefore, ρ is transitive. Hence ρ is equivalence relation.

Ex 1.6.10 Show that the following relation ρ on Z is an equivalence relation: ρ = (a, b):a, b ∈ Z and a2 + b2 is a multiple of 2. [WBUT 08]

Solution: (i) Let a ∈ Z. Then a2 + b2 = 2a2, which is multiple of 2. Therefore, aρb holdsfor a ∈ Z. Thus ρ is reflexive.

(ii) Let a, b ∈ Z and aρb. Then a2 + b2 is a multiple of 2, and also b2 + a2 is a multipleof 2. Therefore bρa and hence ρ is symmetric.

(iii) Let a, b, c ∈ Z and aρb, bρc hold. Then a2 + b2 and b2 + c2 both are multiple of 2.Now,

a2 + c2 = (a2 + b2)− (b2 + c2) + 2c2

is a multiple of 2, i.e., aρc holds. Thus ρ is transitive. Hence ρ is an equivalence relation.

Ex 1.6.11 Let S be the set of all lines in 3-space. Let a relation ρ on the set S be definedby lρm if l,m ∈ S and l lies in the plane m. Examine if ρ is an equivalence relation.

Solution: (i) Let l ∈ S. Then l is a coplanar with itself. Therefore lρm holds for all l ∈ S,i.e, ρ is reflexive.

(ii) Let l,m ∈ S and lρm holds. Then obviously, mρl holds. That is, lρm ⇒ mρl.Therefore ρ is symmetric.

(iii) Let l,m, n ∈ S and lρm, mρn both hold. Then l lies in the plane m and m lies inthe plane of n. This does not always imply that l lies in the plane of n. For example, if llies on the xz-plane, m as taken as x-axis and n is a line on the xy-plane. In this case, lρmand mρn hold, but l and n lie on two different planes.

Thus, ρ is not transitive and hence ρ is not an equivalence relation.

Note 1.6.1 It may be noted that reflexive, symmetric and transitive relations are threeindependent relations, i.e., no two of them do not imply the third.

Elements of equivalence relation

In this section, some properties of the relations reflexive, symmetric and transitive arepresented.

Ex 1.6.12 Let A = a1, a2, . . ., an, be a finite set containing n elements. Find how manyrelations are constructed on A and how many of there are reflexive and symmetric.

Solution: Since A has n elements, so A × A has n2 elements. The 2n2relations can be

constructed on A including the null relation and the relation A×A itself.If the relation ρ on A is reflexive then each all n ordered pairs (ai, ai), i = 1, 2, . . . , n

should be in ρ. The remaining n2 − n = n(n − 1) ordered pairs (ai, aj), i 6= j may not bein ρ. Hence by the rule of cartesian product there are 2n(n−1) reflexive relations on A.

To count the number of symmetric relations we write A×A as A1 ∪A2, whereA1 =

(ai, ai): i = 1, 2, . . . , n

and A2 =

(ai, aj): i 6= j; i, j = 1, 2, . . . , n

.

Thus every element of A × A is exactly one element of A1, A2. The set A2 containsn(n− 1)/2 subsets of the form

(ai, aj), (aj , ai), i, j = 1, 2, . . . , n.

Relation 27

To construct a symmetric relation on A, for each ordered pair taken from A2 we have tochoose some number of ordered pairs from A1. Hence by the rule of cartesian product thereare

2n · 2n(n−1)/2 = 2n(n+1)/2

symmetric relations on A. Therefore, the number of relations which are both reflexive andsymmetric is 2n(n−1)/2.

Ex 1.6.13 If R and S are equivalence relations on a set A, prove that R∩S is an equivalencerelation on A.

Solution: Let R and S are equivalence relations on A. Therefore, R ⊆ A×A and S ⊆ A×A.Hence R ∩ S ⊆ A×A, i.e., R ∩ S is a relation on A.

(i) Since R and S are reflexive, (a, a) ∈ R and (a, a) ∈ S for all a ∈ A. Thus (a, a) ∈R ∩ S for all a ∈ R ∩ S. Hence R ∩ S is reflexive.

(ii) Let (a, a) ∈ R ∩ S. Therefore, (a, a) ∈ R and (a, a) ∈ S. Therefore, (b, a) ∈ R ∩ S.Hence R ∩ S is symmetric.

(iii) Let (a, a) ∈ R ∩ S and (a, c) ∈ R ∩ S. Therefore, (a, b) ∈ R, (a, b) ∈ S and (a, c) ∈R, (a, c) ∈ S. Since R is transitive, (a, b) ∈ R, (a, c) ∈ R ⇒ (a, c) ∈ R. Similarly,(a, c) ∈ S. Thus (a, c) ∈ R ∩ S, i.e., R ∩ S is transitive.

Hence R is an equivalence relation. But, union of two equivalence relations is not neces-sarily an equivalence relation. For example, let A = 1, 2, 3 and R = (1, 1), (2, 2), (3,3), (1, 2), (2, 1), S = (1, 1), (2, 2), (3, 3), (2, 3), (3, 2) be two equivalence relations onA. Then

R ∪ S = (1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2).

Here (1, 2), (2, 3) ∈ R∪ S but (1, 3) 6∈ R∪ S, i.e., R∪ S is not reflexive and hence it is notan equivalence relation on A.

Theorem 1.6.1 If R and S are two relations from A into B, then(a) R−1 ⊆ S−1 then R ⊆ S.(b) (R ∩ S)−1 = R−1 ∩ S−1.(c) (R ∪ S)−1 = R−1 ∪ S−1.(d) If R is reflexive, R−1 is also reflexive.(e) R is symmetric iff R = R−1.

Ex 1.6.14 f R is an equivalence relation on a set A, then prove that R−1 is also an equiv-alence relation on A.

Solution: Since R is an equivalence relation on A, R is reflexive, symmetric and transitive.(i) Let a ∈ A and (a, a) ∈ R. Therefore, (a, a) ∈ R−1, i.e., R−1 is reflexive.(ii) Let (a, b) ∈ R−1. Then (b, a) ∈ R, a, b ∈ A.⇒ (a, b) ∈ R, since R is symmetric⇒ (b, a) ∈ R−1.

Thus R−1 is symmetric.(iii) Let (a, b), (b, c) ∈ R−1. Then (b, a), (c, b) ∈ R for a, b, c ∈ A.⇒ (c, b), (b, a) ∈ R⇒ (c, a) ∈ R, since R is transitive⇒ (a, c) ∈ R−1.

Therefore, R−1 is transitive. Hence R−1 is an equivalence relation on A.

28 Theory of Sets

Antisymmetric relation

A relation ρ is said to be antisymmetric if aρb, bρa⇒ a = b. For example,

(i) In <, the relation ≤ is antisymmetric, since a ≤ b, b ≤ a⇒ a = b.

(ii) In the set of all sets, the relation “ is a subset of ” is antisymmetric for A ⊆ B,B ⊆A⇒ A = B.

(iii) Let F consists of all real valued functions f(x) defined on [−1, 1] and let f(x) ≥ g(x)mean that f(x) ≥ g(x) for every x ∈ [−1, 1].

(iv) A relation ρ, defined on Z by aρb if and only if a is the divisor of b is not antisymmetric.It contain pairs of elements x, y which are incomparable, in the sense that neither x ≤ ynor y ≤ x holds.

Partial ordering relation

A binary relation % defined on a non-empty set A is said to be a partial ordering relation, if %is reflexive, antisymmetric and transitive. There is also an alternative notation for specifyingpartial ordering relation.

(i) P1: Reflexive: x ≤ x, ∀x ∈ A.

(ii) P2: Anti symmetric: x ≤ y and y ≤ x⇒ x = y, for x, y ∈ A.

(iii) P3: Transitive: x ≤ y and y ≤ z ⇒ x ≤ z, for x, y, z ∈ A.

If x ≤ y and x 6= y, one writes x < y and says that x ‘is less than’ or ‘properly containedin’ y. The relation x ≤ y is also written y ≥ x and read y contains x (or includes x).Similarly, x < y is also written y > x. Strict inclusion is characterized by the anti-reflexiveand transitive laws.

Digraph of a relation

A relation ρ on a finite set A can be represented by a diagram called digraph or directedgraph. Draw a dot for each element of A. Now, join the dots corresponding to the elementsai and aj (ai, aj ∈ A) by an arrowed are if and only if ai ρ aj . In case of ai ρ ai for someai ∈ A, the arrowed arc from ai should come back to itself and forms a loop. The resulting

ssss

1 2

34@@

@@@

Figure 1.9: Directed graph of the relation ρ

diagram of ρ is called directed graph or digraph, the dots are called vertex and the arrowedarc is called directed edge or an arc. Thus the ordered pair (A, ρ) is a directed graph ofthe relation ρ. Here two vertices ai, aj ∈ A are to be adjacent if aiρaj . For example, letA = 1, 2, 3, 4 and a relation on A be ρ = (1, 1), (2, 2), (4, 4), (2, 3), (3, 2), (3, 4), (4, 1),(4, 2).

Relation 29

The directed graph (A, ρ) is shown in Fig. 1.9. From the digraph representation of arelation R one can test wether it is an equivalence relation or not. The following test are tobe performed to test an equivalence relation.

(i) The relation is reflexive iff there is a loop on each vertex of the digraph.

(ii) The relation is symmetric iff whenever there is an arc from a vertex a to another vertexb (where a, b are two vertices), there should be an arc from b to a.

(iii) The relation is transitive iff whenever there is an arc from vertex a to a vertex b, anarc from b to a vertex c, then there must be an arc from a to c.

Ex 1.6.15 Find the relation determined by Fig. 1.10.

Solution: Since aiρaj iff there is an edge from ai to aj .

s sss n

ma

b

c

d

Figure 1.10:Thus ρ = (a, a), (a, c), (b, c), (c, b), (c, c), (d, c), (d, d).

Matrix of a relation

A relation between two finite sets can also be represented by a matrix. Let A = a1, a2, . . .,am and B = b1, b2, . . ., bn. The matrix for the relation is denoted by Mρ = [mij ]m×n,where,

mij =

1, if (ai, bj) ∈ ρ,0, if (ai, bj) 6∈ ρ.

The matrix Mρ is called the matrix of ρ. From this matrix one check the property of ρ.

Ex 1.6.16 Let A = a, b, c and B = 1, 2, 3, 4. Consider a relation ρ from A into B as ρ= (a, 1), (a, 3), (b, 2), (b, 3), (b, 4), (c, 1), (c, 2), (c, 4). Then the matrix Mρ is

Mρ =

1 2 3 4abc

1 0 1 00 1 1 11 1 0 1

.From the matrix Mρ one can draw the digraph of the relation and conversely, from thedigraph the matrix Mρ can also be obtained.

Ex 1.6.17 Let A = 2, 4, 6 and let R be given by the digraph shown in Fig. 1.11.s ssm

m2 4

6

Figure 1.11:

Find the matrix Mρ and the relation ρ.

30 Theory of Sets

Solution: The matrix Mρ = [mij ] where,

mij =

1, if (ai, bj) ∈ ρ,0, if (ai, bj) 6∈ ρ.

Therefore, Mρ =

0 1 11 1 00 1 1

and the relation is ρ =

(2, 4), (2, 6), (4, 2), (4, 4), (6, 4), (6,

6)

.

1.7 Equivalence Class

Let ρ be an equivalence relation on a non-empty set A. Then for each a ∈ A, the elementx ∈ A satisfying xρa constitute a subset of A. This subset is called an equivalence class orequivalence set of a with respect to ρ. The equivalence class of a is denoted by cl(a), classa or (a) or [a] or Aa or Ca, i.e.,

[a] = x : x ∈ A and (x, a), i.e., xρa ⊂ A. (1.12)

Again, the set of all equivalence classes of elements of A under the equivalence relation ρ onA is called quotient set denoted by A/ρ, i.e.,

A/ρ = [a] : a ∈ A.

For example, let A = 1, 2, 3, 4 and ρ = (1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (4, 3), (4, 4) be anequivalence relation on A. This equivalence relation has the following equivalence classes:

[1] = 1, 2, [2] = 1, 2, [3] = 3, 4, [4] = 3, 4and the quotient set is A/ρ =

[1], [2], [3], [4]

.

Property 1.7.1 Given that ρ is an equivalence relation, i.e., it is reflexive. Therefore,(a, a) ∈ ρ for all a ∈ A. Also,

[a] = x : x ∈ A and (x, a) ∈ ρ.

Therefore, from the definition aρa ⇒ a ∈ [a] for all a ∈ A. Hence [a] 6= φ for all a ∈ ρ. So[a] is a non-empty subset of A.

Property 1.7.2 Let ρ be an equivalence relation on the set A. Then if b ∈ [a] then [a] = [b],where a, b ∈ A.

Proof: Let b ∈ [a]. Then bρa holds. Let x be an arbitrary element of [b], then

x ∈ [b] and b ∈ [a] ⇒ xρb and bρa⇒ xρa; transitive property⇒ aρx; symmetric property⇒ x ∈ [a].

Thus [b] ⊆ [a]. Similarly, it can be prove that [a] ⊆ [b]. Hence we arrive at a conclusion that[a] = [b].

Property 1.7.3 Two equivalence classes are either equal or disjoint.

Equivalence Class 31

Proof: If for any two classes [a] and [b], [a]∩[b] = φ, then the theorem is proved. [a]∩[b] 6= φ,then let x ∈ [a] ∩ [b]. Then x ∈ [a] and x ∈ [b]. Therefore,

x ∈ [a], x ∈ [b] ⇒ xρa and xρb hold⇒ aρx and xρb; symmetric property,⇒ aρb; transitive property,⇒ a ∈ [b].

Hence by the previous property, [a] = [b]. Hence for all a, b ∈ A, either [a] = [b] or [a]∩[b] = φ,i.e., equivalence classes are either equal or disjoint.

Property 1.7.4 aρb if and only iff a, b belong to the same equivalence classes.

Proof: We know, a, b ∈ [α]. Then by definition αρa and αρb. Hence aρα; αρb (bysymmetric property). Hence aρb (by transitive property).

Conversely, let aρb, then by definition, b ∈ [a]. Also a ∈ [a] (since aρa). Hence a, b belongto the same class.

Property 1.7.5 Let ρ be an equivalence relation on the set A. Then A =⋃

a∈A

[a].

Proof: Let a ∈ A. Then a ∈ [a] ⊆⋃

a∈A

, therefore A ⊆⋃

α∈A

[a]. Again, if X =⋃

a∈A

[a] then

all elements of X belong to A. Therefore,X ⊆ A, i.e.,

⋃a∈A

[a] ⊆ A.

Hence A =⋃

a∈A

[a].

1.7.1 Partitions

Let ρ be an equivalence relation on a non empty set S, then the ρ equivalence classes areeach non-empty and pairwise disjoint. Further, the union of the family of the classes is theset S.

Let S = S1 ∪ S2 ∪ . . ., where S1, S2, S3, . . . are the non-empty subsets of S. Precisely, apartition of S is a collection ρ = S1, S2, S3, . . . of nonempty subsets of S such that

(i) Each x in S belongs to one of the Si.

(ii) The sets of Si are mutually disjoint; that is, if

Si 6= Sj then Si ∩ Sj = φ.

The set of all partitions of a set S is denoted by π(S). The disjoint sets S1, S2, S3, . . .are called cells or blocks. For example,

(i) Consider the subcollection1, 3, 5, 2, 4, 6, 8, 7, 9

of subset of S = 1, 2, · · · , 8, 9,

then it is a partition of S.

(ii) Consider the subcollection1, 3, 5, 2, 6, 4, 8, 9

of subset of S = 1, 2, · · · , 8, 9,

then it is not a partition of S since 7 in S does not belong to any of the subsets.

(iii) Consider the subcollection1, 3, 5, 2, 4, 6, 8, 5, 7, 9

of subset of S = 1, 2, · · · , 8, 9,

then it is not a partition of S since 1, 3, 5 and 5, 7, 9 are not disjoint.

32 Theory of Sets

Another examples are

(i) Let Z, Z−, Z+, Zc, Z0 be the set of integers, negative integers, positive integers, eveninteger and odd integers. Then the partitions of Z are Z−, 0,Z+, Zc,Z0.

(ii) Let Z be the set of integers. Consider a relation ρ = (a, b) : (a− b) is divisible by 5.It is shown that ρ is an equivalence relation on Z. This relation partitions the set Zinto five equivalence classes as [a] = x : xρa i.e., x− a is divisible by 5 . Thus,

[0] = x : x ∈ Z and xρ0= x : x− 0 is divisible by 5= x : x− 0 = 5k, k ∈ Z= . . . ,−10,−5, 0, 5, 10, . . .

Similarly,

[1] = x : x− 1 = 5k, k ∈ Z= . . . ,−9,−4, 1, 6, 11, . . .

[2] = . . . ,−8,−3, 2, 7, 12, . . .[3] = . . . ,−7,−2, 3, 8, 13, . . .[4] = . . . ,−5,−1, 4, 9, 14, . . . .

It is observed that [0]∪ [1]∪ [2]∪ [3]∪ [4] = Z and any two of them are disjoint. Thusa partition of Z is [0], [1], [2], [3], [4].

Ex 1.7.1 Find all partitions of S = a, b, c.

Solution: Since S = a, b, c, the partition π(S) is given by

π(S) =a, b, c

,a, b, c

,a, b, c

,a, c, b

,a, b, c

.

Ex 1.7.2 Determine whether the sets φ, 1, 3, 5, 8, 2, 4, 6, 9, 5, 9, 11, 12 is a partition ofthe set S = 1, 2, 3, · · · , 12.

Solution: Let S1 = φ, S2 = 1, 3, 5, 8, S3 = 2, 4, 6, 9, S4 = 5, 9, 11, 12. We see that,S1 ∪ S2 ∪ S3 ∪ S4 = S, but S2 ∩ S4 = 5 6= φ. Hence the given subsets

φ, 1, 3, 5, 8, 2, 4, 6, 9, 5, 9, 11, 12do not form a partition on S.

Theorem 1.7.1 Fundamental theorem of equivalence relation: An equivalence rela-tion ρ on a set A gives a partition of A into mutually disjoint equivalence classes, such thata, b belongs to the same class if and only iff aρb.

Proof: First, we shall define class [a] (for a given a ∈ A) as [a] = x/aρx, x ∈ A. Let Pbe the set of all distinct equivalence classes in A. If a ∈ A, then since a ∈ [a] and [a] ∈ P .Hence a belongs to the union of all members of P . Hence the union of all members of Pis A. Also the members of P are all pairwise disjoint. Hence P is a partition of A. Nowfor two elements a, b ∈ A, it can be shown that aρb, if and only if they belongs to the sameequivalent classes.Converse theorem: A partition P of a set A gives an equivalence relation for which thenumbers of P are the equivalence classes.

We define a relation ρ in A by aρb if and only if a, b belongs to the same class of P . a, abelongs to the same class of the partition P . Hence aρa, ∀a ∈ A, and so ρ is reflexive. Let

Equivalence Class 33

a, b ∈ S and aρb. Now,

aρb⇒ a, b belongs to the same class.

⇒ b, a belongs to the same class.

⇒ bρa.

Hence ρ is symmetric. Let a, b ∈ S and aρb, bρc hold. Then,

aρb, bρc⇒ a, b and b, c belongs to the same class.

Since b belongs to both classes, it follows that these two classes are the same subset of thepartition P and consequently, a, c belong to one and the same subset of P . Thus

a, c belongs to the same class⇒ aρc.

Hence ρ is transitive. Hence ρ is an equivalence relation on A. And the equivalence classesare just the classes of P . This completes the proof.

Ex 1.7.3 In Z, define aρb iff a−b is divisible by an integer 7. Show that ρ is an equivalencerelation. Hence find the corresponding partition of Z.

Solution: (i) a ∈ Z, and a − a = 0 is divisible by 7. Hence aρa; ∀a ∈ Z and hence ρ is areflexive. Using the definition,

aρb⇒ a− b is divisible by 7⇒ b− a is divisible by 7⇒ bρa; ∀a, b ∈ Z.

Hence ρ is a symmetric. Now,

aρb, bρc⇒ a− b, b− c is divisible by 7⇒ (a− b) + (b− c) is divisible by 7⇒ (a− c)is divisible by 7⇒ aρc; ∀a, b, c ∈ Z.

Hence ρ is transitive and consequently, it is an equivalence relation. For this relation, theequivalent classes are

Ep = 7k + p; k ∈ Z and p = 0, 1, 2, 3, 4, 5, 6.

Therefore, the distinct equivalent classes can be written as

E0 = · · · ,−14,−7, 0, 7, 14, · · ·; E1 = · · · ,−13,−6, 1, 8, 15, · · ·E2 = · · · ,−12,−5, 2, 9, 16, · · ·; E3 = · · · ,−11,−4, 3, 10, 17, · · ·E4 = · · · ,−10,−3, 4, 11, 18, · · ·; E5 = · · · ,−9,−2, 5, 12, 19, · · ·E6 = · · · ,−8,−1, 6, 13, 20, · · ·.

Now, we see that, Z = E0 ∪ E1 ∪ · · · ∪ E6 and Ei ∩ Ej = φ, for i 6= j, i.e., the classes aremutually disjoint. Therefore, E0, E1, E2, E3, E4, E5, E6 is a partition of Z.

Ex 1.7.4 Consider the equivalence relation ρ on Z by xρy if and only if x2−y2 is a multipleof 5. Hence find the corresponding partition of Z.

34 Theory of Sets

Solution: If x ∈ Z, then we have x = 5k+r, where 0 ≤ r. Therefore x2 = 25k2+10kr+r2 ⇒x2 ≡ r2(mod5). For,

r = 0, r2 ≡ 0(mod5); r = 1, r2 ≡ 1(mod5)r = 2, r2 ≡ 4(mod5); r = 3, r2 ≡ 9 ≡ 4(mod5)r = 4, r2 ≡ 16 ≡ 1(mod5).

Hence x2 ≡ 02 or 12 or 22(mod5). Consequently, there are only three congruence classes[0], [1] and [2]. Here,

[0] = a ∈ Z : a2 ≡ 0(mod5) = 5k : k ∈ Z = A0(say),[0] = a ∈ Z : a2 ≡ 1(mod5) = 5k + 1 : k ∈ Z ∪ 5k + 4 : k ∈ Z = A1(say),[2] = a ∈ Z : a2 ≡ 4(mod5) = 5k + 2 : k ∈ Z ∪ 5k + 3 : k ∈ Z = A2(say).

Hence A0, A1, A2 is the corresponding partition of Z.

Theorem 1.7.2 The set Z/(n) of residue classes is a finite set of order n.

We shall first show that a−b is divisible by n iff a, b when divided by n have same remainder.Let a = np+ r1 where p, r1 are integers and 0 ≤ r1 < n and let b = nq + r2 where q, r2 areintegers and 0 ≤ r2 < n. Hence,

a− b = n(p− q) + (r1 − r2) (1.13)

If r1 = r2, this relation shows that a− b is divisible by n. Conversely, if a− b is divisible byn, then (1.13) shows that r1 − r2 is divisible by n. But 0 ≤ |r1 − r2| < n. Hence,

r1 − r2 = 0 i.e. r1 = r2.

Hence the result is proved. Now, all the possible remainders are 0, 1, 2, . . . , (n − 1). Ac-cordingly we get n distinct classes in Z/(n). These are denoted by (0), (1), (2), . . . , (n− 1)(called class 0,class 1,. . . etc.) Hence Z/(n) = (0), (1), (2), . . . , (n− 1) and it is a finite setof order n.

1.8 Poset

Let ρ be a relation on a set A satisfying the following three properties:

(i) (Reflexive): For any x ∈ A, we have aρa.

(ii) (Antisymmetric): If aρb and bρa, then a = b.

(iii) (Transitive): If aρb and bρc, then aρc.

Then ρ is called a partial order or, simply an order relation, and ρ is said to define a partialordering of A. The set A with the partial order is called a partially order set or, simply, anorder set or poset. Thus a non-empty set A together with a partial ordering relation ≤ onA is called a partial ordered set (poset) and is usually denoted by (A,≤). For example,

(i) Consider the setN of positive integers. Herem ≤ nmeans “m divides n”, writtenm|n,if there exists an integer p such that mp = n, m, n ∈ N . For example, 2|4, 3|12, 7|21,and so on. Thus relation of divisibility is a partial ordering and (N ,≤) is a poset.

(ii) Let A be the set of all positive divisors of 72, then (A,≤) is a poset, where m ≤ nmeans ‘m is a divisor of n’, for m,n ∈ A.

Poset 35

(iii) Let P be the set of all real valued continuous functions defined on [0, 1]. Let f, g ∈ Pand f ≤ g mean that f(x) ≤ g(x),∀x ∈ [0, 1]. Then, (P,≤) is a poset.

(iv) Let U be a non empty universal set, i.e., collection of sets and A be the set of allproper subsets of U . The relation P ≤ Q means P ⊆ Q of set inclusion, i.e., P ⊆ Q,for P,Q ∈ A is a partial ordering of U . Specially, P ⊆ P for any set P ; if P ⊆ Q andQ ⊆ P then P = Q; and if P ⊆ Q and Q ⊆ R then P ⊆ R. Therefore (A,≤) is aposet.

(v) (<,≤) is a poset, where m ≤ n means ‘m is less than or equal to n’, for m,n ∈ <.

Ex 1.8.1 Let A = 0, 1 and α = (a1, a2, a3), β = (b1, b2, b3) ∈ A3. Define a relation ρ onA3 by αρβ if and only if ai ≤ bi, for i = 1, 2, 3. Prove that (A3, ρ) is a poset.

Solution: Here, A = 0, 1 and the elements of A3 are of the form (a1, a2, a3), so, the ele-ments ofA3 can be written as (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 1, 1).Here the relation is defined as αρβ if and only if ai ≤ bi, for i = 1, 2, 3. Now,

αρα,∀α = (a1, a2, a3) ∈ A3.Hence ρ is reflexive. Let us now assume that α, β ∈ A3 and αρβ, βρα both hold. Then

ai ≤ bi and bi ≤ ai ⇒ ai = bi; i = 1, 2, 3.⇒ αρβ, βρα⇒ α = β.

Hence ρ is antisymmetric. Let α = (a1, a2, a3), β = (b1, b2, b3), γ = (c1, c2, c3) ∈ A3 and αρβand βργ both hold. Then

ai ≤ bi and bi ≤ ci ⇒ ai ≤ ci; for, i = 1, 2, 3.or, αρβ and βργ ⇒ αργ;∀α, β, γ ∈ A3

and so the relation ρ is transitive. As the ρ is reflexive, antisymmetric and transitive, so(A3,≤) is a poset.

1.8.1 Dual Order

Let ≤ be any partial ordering of a set S and (S,≤) be a poset. Let % be a binary relation onS such that for a, b ∈ S, a%b if and only if b ≤ a. Then the relation % is called the converseof the partial ordering relation ≤ and is denoted by ≥ . It may be easily seen that (S,≥) isalso a poset. It follows that we can replace the relation ≤ in any theorem about posets bythe relation ≥ throughout without affecting the truth. This is known as principle of duality.This duality principle applies to algebra, to projective geometry and to logic.

Ex 1.8.2 Let (A,≤) be a poset. Define a relation ≥ on A by a ≥ b if and only if b ≤ a, fora, b ∈ A, then show that (A,≥) is a poset.

Solution: The relation ≥ on A is defined as a ≥ b if and only if b ≤ a, for a, b ∈ A.

(i) Since, a ≤ a,∀a ∈ A, so, a ≥ a,∀a ∈ A and hence ≥ is reflexive.

(ii) Let a, b ∈ A be such that a ≥ b, b ≥ a, then b ≤ a and a ≤ b. Therefore, b = a, as ≤ isantisymmetric. Therefore, a ≥ b, b ≥ a⇒ a = b,∀a, b ∈ A. Hence, ≥ is antisymmetric.

(iii) Let a, b, c ∈ A be such that a ≥ b, b ≥ c, then b ≤ a and c ≤ b, i.e., c ≤ b and b ≤ a.This implies that c ≤ a since ≤ is transitive, i.e., a ≥ c. Therefore, a ≥ b, b ≥ c⇒ a ≥c,∀a, b, c ∈ A. Hence, ≥ is transitive.

As the ≥ is reflexive, antisymmetric and transitive, so (A,≥) is a poset.

36 Theory of Sets

1.8.2 Chain

Let (S,≤) be a poset. Given x, y ∈ S, let x ≤ y or y ≤ x. The poset which satisfies thisrelation is said to be ‘simply’ or ‘totally’ or ‘linearly’ ordered and is called a chain. In otherwords, of any two distinct elements in a chain, one is less and the other greater. A subsetof S is called an antichain if no two distinct elements in the subset are related. A poset(S,≤) is called a totally ordered set or simply an ordered set if S is a chain and in this casethe binary relation ≤ is called a total ordering relation.

(i) Any subset S of a poset P is itself a poset under the same inclusion relation (restrictedto S).

(ii) Every subset of a linearly ordered set S must be linearly ordered i.e., any subset of achain is a chain.

(iii) Although an ordered set S may not be linearly ordered, it is still possible for a subsetA of S to be linearly ordered.

We frequently refer to the number of elements in the chain as the length of the chain.Consider the following examples:

(i) Consider the set N of positive integers ordered by divisibility. Then 21 and 7 are com-parable since 7|21. On the other hand, 3 and 5 are non-comparable, since neither 3|5nor 5|3. Thus N is not linearly ordered by divisibility. Observe that A = 2, 6, 12, 36is a linearly ordered subset of N since 2|6, 6|12 and 12|36.

(ii) The set N of positive integers with the usual order ≤ is linearly ordered and henceevery ordered subset of N is also linearly ordered.

(iii) The power set P (A) of a set A with two or more elements is not linearly ordered byset inclusion. For instance, suppose a, b ∈ A. Then a and b are non-comparable.Observe that the empty set φ, a and A do not form a linearly ordered subset of P (A)since φ ⊆ a ⊆ A. Similarly, φ, b and A do not form a linearly ordered subset ofP (A).

1.8.3 Universal Bounds

In any poset P = (S,≤), the elements O and I of S, when they exist, will be universalbounds of P , if for any element x ∈ S, we have,

O ≤ x and x ≤ I, i.e., O ≤ x ≤ I,∀x ∈ P. (1.14)

We call these elements O and I as the least element and the greatest element of S.Lemma : A given poset (S,≤) can have at most one least element and at most one greatestelement.Proof: Let O and O∗ be universal lower bounds of (S,≤). Then, since O is the universallower bound, we have O ≤ O∗, and since O∗ is the universal lower bound we have O∗ ≤ O.Hence by the hypothesis P2, we have O = O∗ and similarly, I = I∗.Posets need not have any universal bounds. Thus under the usual relation of inequality, thereal numbers form a poset (<,≤) which has no universal bounds (unless −∞ and +∞ areadjoint to form extended reals).

Poset 37

1.8.4 Covering Relation

Let S be a partially ordered set, and suppose a, b ∈ S. We say that a is an immediateprecedence of b, or that b is an immediate successor of a, or that b is a cover of a, writtena << b, if b < a (i.e., b ≤ a and a 6= b), i.e., if there is no element c ∈ S such that b < c < aholds.

For example, (N ,≤) is a poset, where m ≤ n means m|n, for m,n ∈ N . Therefore, in(N ,≤), 6 covers 2, 10 covers 2, but 8 does not cover 2, as 2 < 4 < 8, although 2 < 8.

Suppose S is a finite partially ordered set. Then the order on S is completely knownonce we know all pairs a, b ∈ S such that a << b, that is, once we know the relation <<on S. This follows from the fact that x < y if and only if x << y or there exist elementsa1, a2, · · · , am in s such that

x << a1 << a2 << · · · << am << y.

Hasse diagram

The Hasse diagram of a finite partially ordered set S is a directed graph whose vertices arethee elements of S and there is a directed edge from a to b whenever a << b in S.

A convenient way of displaying the ordering relation among the elements of an orderedset is done by means of a graph whose vertices represent the elements of a set. Thus wedefine a graph, whose vertices are different elements a, b, c, · · · of S in which a and b arejoined by a segment if and only if a covers b or b covers a. If the graph is so drawn thatwhenever a covers b, the vertex a is at a higher level than the vertex b, then the graph iscalled the Hasse diagram of S. The Hasse diagram of a poset S is a picture of S; hence itis very useful in describing types of elements of S. Sometimes we define a partially orderedset by simply presenting its Hasse diagram, and note that they need not be connected.

Ex 1.8.3 Let A = a, b, c, d, e. The diagram in Fig. 1.12 defines a partial order on A in

@@

@

@

@@

AAAA

d

a

e

b c

Figure 1.12: Hasse diagramthe natural way. That is, d ≤ b, d ≤ a, e ≤ c, and so on.

Ex 1.8.4 Let S = a, b, c, then the power set ℘(S), i.e., the set of all subsets of S has

QQQ

@@@

QQQ

@@@a

a, b

a, b, c b, cc

φ

b

a, c

Figure 1.13: Hasse diagramelements φ, a, b, c, a, b, a, c, b, c, a, b, c. The Hasse diagram of this poset isshown in the Fig. 1.13.

Ex 1.8.5 Let S be the set of all positive divisors of 12, i.e., S = 1, 2, 3, 4, 6, 12. Here 1covers 2, 2covers 4, 4 coves 12, 1covers 3, 3 covers 6, 6 covers 12, 2 covers 6, but 2 does notcover 12, as 2 < 6 < 12. The covering diagram of this poset (S,≤) is given in the Fig. 1.14.

38 Theory of Sets

%%%

3

1 2 4

6 12

Figure 1.14: Hasse diagram

Ex 1.8.6 Let S be the set of all positive divisors of 30, i.e., S = 1, 2, 3, 5, 6, 10, 15, 30.

QQQ

@@@

QQQ

@@@1

5

15 30

6

2

10

3

Figure 1.15: Hasse diagramThe covering diagram of this poset (S,≤) is given in the figure 1.15.

Ex 1.8.7 A partition of a positive integer m is a set of positive integers whose sum is m.For instance, there are seven partitions of m = 5 as follows:

5, 3− 2, 2− 2− 1, 1− 1− 1− 1− 1, 4− 1, 3− 1− 1, 2− 1− 1− 1.

We order the partitions of an integer m as follows: A partition P1 precedes a partition P2 ifthe integers in P1 can be added to obtain the integers in P2 or equivalently, if the integers in

Q

QQ

bb

@@

@@

4− 1 3− 2

3− 1− 1 2− 2− 1

5

2− 1− 1− 1

1− 1− 1− 1− 1Figure 1.16: Hasse diagram

P2 can be further subdivided to obtain the integers in P1. For example, 2− 2− 1 precedes3 − 2 as 2 + 1 = 3. On the other hand, 3 − 1 − 1 and 2 − 2 − 1 are non-comparable. Fig.1.16 gives the Hasse diagram of the partitions of m = 5.

1.8.5 Maximal and Minimal Elements

Let S be a partial ordered set. An element a of a poset (S,≤) is minimal, if no other elementof S strictly precedes (is less than) a. Similarly, an element b is called a maximal element ifno element of S strictly succeeds (is larger than) b. For example,

(i) (N ,≤) is a poset, where m ≤ n means m|n, for m,n ∈ N . This poset (N ,≤) containsno greatest element and no maximal element. The least element is 1 and 1 is the onlyminimal element.

(ii) Let U be a non empty universal set and A be the set of all proper subsets of U , then(A,≤) is a poset, where P ≤ Q means ‘P is a subset of Q’, i.e., P ⊆ Q, for P,Q ∈ A.This poset (A,≤) contains no greatest element and no least element. The the minimalelements are 1, 2, 3 and the three maximal elements are 1, 2, 2, 3, 1, 3.

Poset 39

Geometrically speaking, a is a minimal element if no edge enters a (from below), and b is amaximal element if no edge leaves b (in the upward direction). An element a ∈ S is calleda first element if a ≤ x, for every element x ∈ S, i.e., if a precedes every other element inS. Similarly, an element b in S is called a least element if y ≤ b for every y ∈ S, i.e., if bsucceeds every other element in S. Note the following:

(i) If S is infinite, then S may have no minimal and no maximal element. For instance,the set Z of integers with the usual order ≤ has no minimal and no maximal element.

(ii) If S is finite, then S must have a least one minimal element and at least one maximalelement.

(iii) S can have more than one minimal and more than one maximal element. For example,Let X = 1, 2, 3, S = ρ(X) − φ,X; ∀A,B ∈ S,AρB ⇔ A ⊂ B. Here 1, 2, 3are minimal in poset (S, ρ). 1, 2, 2, 3, 1, 3 are maximal in poset (S, ρ).

(iv) S can have at most one first element, which must be a minimal element, and S cangave at most one last element, which must be a maximal element. Generally speaking,S may have neither a first nor a last element, even when S is finite.

(v) The least element in a poset is the minimal element and the greatest element in a

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HHH

b f

h

j

c

a

d ge

i

k

Figure 1.17: Hasse diagram

poset is a maximal element but the converse is not true. Let us consider the posetwhose Hasse diagram is given in Fig. 1.17, a, b, e are minimal elements and j, k aremaximal elements. Here f covers c but f does not cover a.

(vi) Let A = a, b, c, d, e. The diagram in Fig. 1.12 defines a partial order on A in thenatural way. That is, d ≤ b, d ≤ a, e ≤ c, and so on. A has two minimal elements, dand e, and neither is a first element. A has only one maximal element a, which is alsoa least element.

(vii) Let A = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24 be ordered by the relation ‘x divides y’. The Hasse

@@

@@

@

@

TT

!!

24

8

4

2

13

9

18

6

12

Figure 1.18: Hasse diagram

diagram is given in Fig. 1.18. Unlike rooted trees, the direction of a line in the diagramof a poset is always upward. A has two maximal elements 18 and 24 and neither is alast element. A has only one minimal element, 1, which is also a first element.

40 Theory of Sets

v u z y x

Figure 1.19: Hasse diagram

(viii) The diagram of a finite linearly ordered set, i.e., a finite chain, consists simply of onepath. For example Fig. 1.19 shows the diagram of a chain with five elements. Thechain has one minimal element, x, which is a first element, and one maximal element,v, which is a last element.

(ix) Let A be any non empty set and let P (A) be the power set of A ordered by setinclusion. Then the empty set φ is a first element of P (A) since, for any set X, wehave φ ⊆ X. Moreover, A is a least element of P (A) since every element Y of P (A) is,by definition, a subset of A, that is, Y ⊆ A.

1.8.6 Supremum and Infimum

Let A a subset of a partially ordered set (S,≤). An element M ∈ S is called to be a upperbound of a subset A if M succeeds every element of A, i.e., if, for every x ∈ A, we havex ≤M. If an upper bound of A precedes every other upper bound of A, then it is called thesupremum or least upper bound of A and is denoted by sup(A) and we write sup(A) = M.

In particular, let (S,≤) be any poset and let a, b ∈ S be given. Then an element d ∈ S iscalled the glb or meet of a and b, when

d ≤ a, d ≤ b and x ≤ a, x ≤ b⇒ a ≤ d, (1.15)

and we write d = a ∧ b.An element m ∈ S is called to be a lower bound of a subset A of a poset S if m precedes

every element of A, i.e., if, for every y ∈ A, we have m ≤ y. If a lower bound of A succeedsevery other lower bound of A, then it is called the infimum or greatest lower bound of Aand is denoted by inf(A) and we write inf(A) = m.

Dually, an element s ∈ S is called the lub or join of a and b, when

a ≤ s, b ≤ s and a ≤ x, b ≤ x⇒ s ≤ x. (1.16)

In this case, we write s = a ∨ b.Below are some examples:

(i) Let S = a, b, c, d, e, f be ordered as pictured in Fig. 1.20, and let A = b, c, d.

@@

,,

,,

ll

lle f

c d

b

a

Figure 1.20: Hasse diagram

The upper bounds of A are e and f succeed every element in A. The lower boundsof A are a and b since only a and b precede every element of A. Note that e and fare non-comparable; hence sup(A) does not exist. However, b also succeeds a, henceinf(A) = b.

Poset 41

(ii) Let (N ,≤) be a poset, and ordered by divisibility, where m ≤ n means m|n, form,n ∈ N . The greatest common divisor of m and n in N , denoted by gcd(m,n) isthe largest integer which divides mand n. The least common multiple of m and n,denoted by lcm[m,n] is the smallest integer divisible by both m and n. From numbertheory, every common divisor of a and b divides gcd(m,n) and lcm[m,n] divides everymultiple of m and n. Thus

gcd(m,n) = inf(m,n) and lcm[m,n] = sup(m,n).

In other words, inf(m,n) and sup(m,n) do exist for every pair of elements of Nordered by divisibility.

(iii) For any positive integer m, we will let Dm denote the set of divisors of m ordered

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@@@

@@@

JJJ

aaa

@@

bbb

12

4

12

36

18

9

3

6

Figure 1.21: Hasse diagram

by divisibility. The Hasse diagram of D36 = 1, 2, 3, 4, 6, 9, 12, 18, 36 appears in Fig.1.21. Again, inf(m,n) = gcd(m,n) and sup(m,n) = lcm[m,n] exist for any pair m,nin Dm.

(iv) Let X be a non empty set and P (X) be the power set of X. Then, (P (X),≤) is aposet, where A ≤ B means ‘A is a subset of B’, i.e., A ⊆ B, for A,B ∈ P (X). In thisposet (P (X),≤), the lub of A and B is the union of A,B, i.e. A∪B and the glb of Aand B is the intersection of A,B, i.e., A ∩B.

It is important to note that, if the elements a, b in a poset (S,≤) have an upper bound (a lower bound) then they may not has a least upper bound ( a greatest lower bound). For aposet whose Hasse diagram is shown if figure (1.22), the set A = e, d, g has four elements

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aaaa

aaa b c

de f

g

Figure 1.22: Hasse diagram

a, b, c, d as upper bounds and d is the lub of A. But the subset A′ = a, f, c has no upperbound.

Theorem 1.8.1 Let (A,≤) be a poset and a, b ∈ A. Then one of the relations (i) a ≤ b,(ii) a ∧ b = a and (iii) a ∨ b = b implies the other two.

Proof: Given that, (A,≤) be a poset and a, b ∈ A, then a ≤ a and a ≤ b. This means thata is the lower bound od a and b. Let m be any lower bound of a and b, then by definition,

42 Theory of Sets

m ≤ a,m ≤ b. Since a is a lower bound of a, b and any lower bound m ≤ a, a is the greatestlower bound of a, b, i.e., a = a ∧ b and consequently, (i) ⇒ (ii).

Let the relation (ii), i.e., a ∧ b = a holds. Then, by definition, a is the glb of a and b.Hence, a ≤ a, a ≤ b. Also, we have, b ≤ b. Now, a ≤ b and b ≤ b gives b is an upper boundof a and b. Let n be any upper bound of a and b, then a ≤ n and b ≤ n. Since, b is an upperbound of a and b and b ≤ n, for any upper bound, so, b is the lub of a, b, i.e., a ∨ b = b andconsequently, (ii) ⇒ (iii).

Let the relation (iii), i.e., a ∨ b = b holds. Then, by definition, b is the lub of a and b.As, b is an upper bound of a, b, we have a ≤ b and consequently, (iii) ⇒ (i).

By the transitivity of the implication ⇒, it follows that, (i) ⇒ (ii) and (iii); (ii) ⇒ (iii);(iii) ⇒ (i) and (ii). Hence the theorem.

1.9 Lattices

Let L be a nonempty set closed under two binary operations called meet and join, denotedrespectively by ∧ and ∨. Then L is called a lattice if the following axioms hold where a, b, care elements in L:

(i) (Commutative law): a ∧ b = b ∧ a; a ∨ b = b ∨ a.

(ii) (Associative law): (a ∧ b) ∧ c = a ∧ (b ∧ c); (a ∨ b) ∨ c = a ∨ (b ∨ c).

(iii) (Absorption law): a ∧ (a ∨ b) = a; a ∨ (a ∧ b) = a.

We will sometimes denote the lattice by (L,∧,∨) when we want to show which operationsare involved. A chain (L,≤) is a lattice since lub(a, b) = b and glb(a, b) = a when a ≤ b andlub(a, b) = a and glb(a, b) = b when b ≤ a. For example Fig. 1.19 shows the diagram of achain with five elements, which is a lattice. Below are some examples:

(i) (N ,≤) is a poset, wherem ≤ nmeansm|n, form,n ∈ N . This poset (N ,≤) is a lattice,where for any two elements m,n ∈ N , m ∨ n = lcm(m,n) and m ∧ n = gcd(m,n).

(ii) Let X be a non empty set and P (X) be the power set of X. Then, (P (X),≤) is aposet, where A ≤ B means ‘A is a subset of B’, i.e., A ⊆ B, for A,B ∈ P (X). Thisposet (P (X),≤) is a lattice, where for any two elements A,B ∈ P (X), A∨B = A∪Band A ∧B = A ∩B.

(iii) (Z,≤) is a poset, where m ≤ n means ‘m is less than or equal to n’, for m,n ∈ Z. Thisposet (Z,≤) is a lattice, where for any two elements m,n ∈ Z, m ∧ n = minm,nand m ∨ n = maxm,n. This is a chain.

(iv) Then, (P,≤) is a poset, where, P be the set of all real valued continuous functionsdefined on [0, 1] and f ≤ g, f, g ∈ P mean that f(x) ≤ g(x),∀x ∈ [0, 1]. This poset(P,≤) is a lattice, where for any two elements f, g ∈ P,

f ∨ g = maxf(x), g(x); f ∧ g = minf(x), g(x); x ∈ [0, 1].

(v) Let U be a non empty universal set and A be the set of all proper subsets of U , then(A,≤) is a poset, where P ≤ Q means ‘P is a subset of Q’, i.e., P ⊆ Q, for P,Q ∈ A.This poset (A,≤) is a not a lattice, as the pair of elements 1, 2 and 2, 3 has nolub and the pair of elements 1 and 2 has no glb.

(vi) For any positive integer m, we will let Dm denote the set of divisors of m ordered bydivisibility (|). Let D30 = 1, 2, 3, 5, 6, 10, 15, 30 denotes the set of all divisors of 30.The Hasse diagram of the lattice (D30, |) appears in Fig. 1.23. m∧n = lcm(m,n) and

Lattices 43

SS

S

@

@@

\

\\

AAAA

1

2

6

30

15

5

10

3

Figure 1.23: Hasse diagram

m∨n = gcd(m,n). traverse upwards from the vertices representing a and b and reacha meeting point of the two paths. The corresponding element is a ∧ b. By traversingdownwards onwards, we can get a ∨ b similarly.

Ex 1.9.1 Show that the poset (L,≤) represented by its Hasse diagram (Fig.1.24) is a lattice.

@@

@

@@

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c

cc

a

b c

d

e

g

f

Figure 1.24: Hasse diagram of Ex.1.9.1

Solution: We have to prove that each pair of elements of L = a, b, c, d, e, f, g has a luband glb.

a b c d e f ga (a, a) (a, b) (a, c) (a, d) (a, e) (a, f) (a, g)b (a, b) (b, b) (a, d) (b, d) (b, e) (a, g) (b, g)c (a, c) (a, d) (c, c) (c, d) (a, g) (c, f) (c, g)d (a, d) (b, d) (c, d) (d, d) (b, g) (c, g) (d, g)e (a, e) (b, e) (a, g) (b, g) (e, e) (a, g) (e, g)f (a, f) (a, g) (c, f) (c, g) (a, g) (f, f) (f, g)g (a, g) (b, g) (c, g) (d, g) (e, g) (f, g) (g, g)

If the element in a-row and b-column is (x, y), then x = a ∧ b and y = a ∨ b.

Ex 1.9.2 Show that the following posets are not lattices:

(i) L1 = (1, 2, · · · , 12, |).

(ii) L2 = (1, 2, 3, 4, 6, 9, |).

Solution: For (L1, |) 2 ∧ 7 = lcm2, 7 = 14 6∈ L1. So L1 is not a lattice.For (L2, |) 4 ∧ 6 = lcm4, 6 = 12 6∈ L2. So L2 is not a lattice.

Ex 1.9.3 Which of the following posets given in the Fig.1.25 is a lattice? Not a lattice?

Solution: P1 : d, e, f are upper bounds of b and c. f cannot be the lub of b and c sinced ≤ f and d 6= f . d or e cannot be the lub of b and c since d 6≤ e and e 6≤ d. So lub(b, c)

44 Theory of Sets

@@@

l

ll

###

b

bb

bb

AAAAQ

QQ

ee

SS

S

a

b

d

f

e

ca b a b

P1 P2 P3Figure 1.25: Hasse diagram of Ex.1.9.3

does not exist. Hence (L,≤) is not a lattice.P2 : It is not a lattice since there is no lower bound for (a, b) and hence glb(a, b) does notexist.P3 : It is not a lattice since glb(a, b) does not exist.

1.9.1 Lattice Algebra

The binary operations ∧ and ∨ in lattices have important algebraic properties, some of themanalogous to those of ordinary multiplication and addition.

Theorem 1.9.1 In any lattice the following identities hold:

(i) L1: x ∧ x = x and x ∨ x = x: Idempotency

(ii) L2: x ∧ y = y ∧ x and x ∨ y = y ∨ x, Commutativity

(iii) L3: (x ∧ y) ∧ z = x ∧ (y ∧ z) and (x ∨ y) ∨ z = x ∨ (y ∨ z), Associativity

(iv) L4: x ∧ (x ∨ y) = x and x ∨ (x ∧ y) = x, Absorption.

Moreover x ≤ y is equivalent to each of the conditions: x∧y = x and x∨y = y : Consistency.

Proof: By the principle of duality, which interchange ∧ and ∨ it suffices to prove one ofthe two identities in each of L1− L4.L1 : Since x ∧ y ≤ x, we have x ∧ x ≤ x. Also, d ≤ x, d ≤ y ⇒ d ≤ x ∧ y. It follows thatx ≤ x, x ≤ x⇒ x ≤ x ∧ x. Hence it follows that x ∧ x = x.L2 : Since the meaning of glbx, y is not attend by interchanging x and y, it follows thatx ∧ y = y ∧ x.L3 : Since both of x ∧ (y ∧ z) and (x ∧ y) ∧ z represent the glbx, y, z, the result follows.L4 : Since x ∧ (x ∨ y) is the lower bound of x and x ∨ y, we have x ∧ (x ∨ y) ≤ x. Sincex ≤ x, x ≤ x∨ y, it follows that x is the lower bound of x and x∨ y. And since x∧ (x∨ y) isthe glb of x and x ∨ y, we must have x ≤ x ∧ (x ∨ y). Hence it follows that x ∧ (x ∨ y) = x.

Theorem 1.9.2 In a lattice L, y ≤ z implies x ∧ y ≤ x ∧ z and x ∨ y ≤ x ∨ z, ∀x ∈ L.

Proof: Since y ≤ z, we have y = y ∧ z. Therefore,x ∧ y = x ∧ (y ∧ z) = (x ∧ x) ∧ (y ∧ z), as x ∧ x = x

= x ∧ (x ∧ (y ∧ z)); Associative= x ∧ ((x ∧ y) ∧ z); Associative= x ∧ ((y ∧ x) ∧ z); commutative= (x ∧ (y ∧ x)) ∧ z = ((x ∧ y) ∧ x) ∧ z= (x ∧ y) ∧ (x ∧ z).

Since x ∧ y is the glb of x ∧ y and x ∧ z, so x ∧ y ≤ x ∧ z. By the principle of duality,x ∨ y ≤ x ∨ z.

Lattices 45

Theorem 1.9.3 Any lattice satisfies the distributive inequalities (or semi distributive laws):

(i) x ∧ (y ∨ z) ≥ (x ∧ y) ∨ (x ∧ z).

(ii) x ∨ (y ∧ z) ≤ (x ∨ y) ∧ (x ∨ z).

Proof: We have, x ∧ y ≤ x and x ∧ y ≤ y ≤ y ∨ z. Therefore, x ∧ y is a lower bound of xand y ∨ z. Since x ∧ (y ∨ z) is the glb of x and y ∨ z, we have,

x ∧ y ≤ x ∧ (y ∨ z) and similarly, x ∧ z ≤ x(y ∨ z).

These shows that x ∧ (y ∨ z) is an upper bound of x ∧ y and x ∧ z. But (x ∧ y) ∨ (x ∧ z) isthe lub of x ∧ y and x ∧ z. Therefore

x ∧ (y ∨ z) ≥ (x ∧ y) ∨ (x ∧ z).

1.9.2 Sublattices

Let T is a non empty subset of a lattice L (T ⊆ L).We say T is sublattice of L if T is itselfa lattice (with respect to the operations of L). Therefore, T is a sublattice of L if and onlyif T is closed under the operations of ∧ and ∨, i.e.,

a, b ∈ T ⇒ a ∧ b ∈ T and a ∨ b ∈ T.

For example, the set Dm of divisors of m is a sublattice of the positive integers N underdivisibility.

Two lattices L and L′ are said to be isomorphic if there is a one-to-one correspondencef : L→ L′ such that

f(a ∧ b) = f(a) ∧ f(b) and f(a ∨ b) = f(a) ∨ f(b).

1.9.3 Bounded Lattices

A lattice L is said to have a lower bound 0 if for any element x ∈ L we have 0 ≤ x.Analogously, L is said to have an upper bound I if for any x ∈ L we have x ≤ I. We say Lis bounded if L has both a lower bound 0 and upper bound I. In such a lattice we have theidentities

a ∧ I = I, a ∨ I = a, a ∧ 0 = a, a ∨ 0 = 0,

for any element a ∈ L. For example:

(i) The nonnegative integers with the usual ordering 0 < 1 < 2 < · · · have 0 as the lowerbound but have no upper bound.

(ii) The lattice P (U) of all subsets of any universal set U is a bounded lattice with U asan upper bound and the empty set φ as a lower bound.

Therefore every finite lattice is bounded.

1.9.4 Distributive Lattices

A lattice L in which the distributive laws

(i) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z)

(ii) x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z)

46 Theory of Sets

SS

JJJ

@

@

@

@

I

a b c

0

I

0

a

c

b

(a) (b)Figure 1.26: Hasse diagram

hold for all elements x, y, z in L is called a distributive lattice. We note that by the principleof duality the condition (i) holds if and only if (ii) holds. If the lattice L is not distributive,it is said to be non distributive. Fig. 1.26 is a non distributive lattice, since

a ∧ (b ∨ c) = a ∧ 0 = a but (a ∧ b) ∨ (a ∧ c) = I ∨ c = c.

Fig. 1.26(b) is also a non distributive lattice. In fact, we have the following characteriza-tion of such lattices. A lattice L is non distributive if and only if it contains a sublatticeisomorphic to Fig. 1.26(a) or (b).

Theorem 1.9.4 In a distributive lattice, a ∧ x = a ∧ y and a ∨ x = a ∨ y together implyx = y.

Proof: We have, x = x ∨ (x ∧ a), L4= x ∨ (y ∧ a) = (x ∨ y) ∧ (x ∨ a)= (y ∨ x) ∧ (y ∨ a) = y ∨ (x ∧ a)= y ∨ (y ∧ a) = y.

1.9.5 Trivially Complement

The case in which a ∧ x = a ∧ y = O and a ∨ x = a ∨ y = I is the particular interest. Ingeneral, by a complement of an element a in a lattice L with universal bounds O and I, wemean an element x ∈ L such that

a ∧ x = O and a ∨ x = I.

The elements O and I are trivially complementary in any lattice. It is obvious that in anychain elements other than O and I have no complements.

Theorem 1.9.5 In any distributive lattice L, a given element can have at most one com-plement.

Proof: Let the element a have two complements a′ and a′′, thena ∧ a′ = O = a ∧ a′′ and a ∨ a′ = I = a ∨ a′′.

Since the lattice is distributive, we have a′ = a′′.

Theorem 1.9.6 In any distributive lattice, the set of all complemented elements is a sub-lattice.

Proof: Let a, a′ and b, b′ be complementary pairs, then by definition,a ∧ a′ = O = b ∧b′ and a ∨ a′ = I = b ∨ b′. Now,

Mapping 47

(a ∧ b) ∧ (a′ ∨ b′) = (a ∧ b ∧ a) ∨ (a ∧ b ∧ b′)= (a ∧ a′ ∧ b) ∨ (a ∧O)= (O ∧ b) ∨O = O ∨O = O.

Also, (a ∧ b) ∨ (a′ ∨ b′) = (a ∨ a′ ∨ b′) ∧ (b ∨ a′ ∨ b′)= (I ∨ b′) ∧ (a′ ∨ I) = I ∧ I = I.

Hence a ∧ b and a′ ∨ b′ are complementary. Similarly, a ∨ b and a′ ∧ b′ are complementary.Thus if L is a distributive lattice and if S is a subset of L consisting of the complementarypairs of L, then we see that for any two elements a and b (with complements a′ and b′

respectively ) in S a ∧ b and a ∨ b also belong to S. Hence by definition, S is a sublattice.

1.10 Mapping

Let A and B be two non-empty sets. A function f from A to B, which is denoted byf : A→ B, is a relation from A to B with the property that for every a ∈ A, there exactlyone b ∈ B such that (a, b) ∈ f . Functions are called mappings or transformations. Theelement a ∈ A is called an argument of the function f and f(a) is called the value or imageor f-image of a under f .

rr rrrr r r

f range

A: domain B: Co-domain

Figure 1.27: Pictorial representation of domain, co-domain and range

The set A is called the domain of the function f and B is called the co-domain of f .The range of f consists of those elements in B which appears as the image of at least oneelement of A. It is also known as image set. Thus range of f is denoted by f(A), i.e.,f(A) = f(x) : x ∈ A. Obviously, f(A) ⊆ B. For example, (i) let A = 1, 2, 3, 4 andB = x, y, z and let

f = 1, x, 2, x, 3, y, 4, z.i.e., f(1) = x, f(2) = x, f(3) = y, f(4) = z.

Thus for each element of A has a unique value in B, so it is a function.(ii) Let A = 1, 2, 3 and Let B = x, y, z. Consider the relations

f1 = (1, x), (2, x) and f2 = (1, x), (1, y), (2, z), (3, y).

The relation f1 is a function, as f1(1) = x and f1(2) = x, i.e., each element has uniqueimage. But, the relation f2 is not a function as f2(1) = x and f2(1) = y, i.e., 1 ∈ A has twodistinct images x and y.

If x(∈ A) correspondence to y(∈ B), it is said that y is the image of x under the mappingf and it is expressed by writing either xf = y or f(x) = y. In this case x is said to be the orinverse image of y. Sometimes, it is possible to write down a precise formula showing howf(x) is determine by x. For example, f(x) =

√x, f(x) = 2x+ 5, f(x) = ex + 2x, etc. That

is, for the function f(x) = x3 we mean the function f : R → R which associates with anyx ∈ R, its cube x3. In the notation of binary relation f = (x, x3) : x ∈ R.

Thus, a sub set f of A×B is called a function or mapping from A to B if to each a ∈ A,there exists a unique b ∈ B such that the ordered pair (a, b) ∈ f .

48 Theory of Sets

1.10.1 Types of Functions

Constant function

A function f : A → B is said to be a constant function (or a constant mapping) if f mapseach element of A to one and the same element of B, i.e., f(A) is a singleton set. Forexample, f(x) = 5 for all x ∈ R is a constant function.

Identity function

A function f : A → A is said to be the identity function on A if f(x) = x for all x ∈ A. Itis denoted by IA.

Into function

A function f : A → B is said to be an into function if f(A) is a proper subset of B, i.e.,f(A) ⊂ B. In this case, we say that f maps A into B.

Onto function

Let A and B are two non empty subsets. A mapping f : A → B is defined as a surjectiveor surjection if

∀y ∈ B,∃x ∈ A such that f(x) = y. (1.17)

This is also called an onto mapping and is denoted by f(A) = B. In this case, we say thatf maps A onto B. For example,

(i) Let f : Z → Z be given by f(x) = 3x, x ∈ Z. Then f is into function becausef(z) = 0,±3,±6,±9, . . . is a proper subset of Z (co-domain).

(ii) Let f : Z → Z be given by f(x) = x + 2, x ∈ Z. The f is onto function becausef(z) = z (co-domain).

Pre-image

If f : A → B be a function and x ∈ A then f(x) is a unique element in B. The element xis said to be a pre-image (or inverse image) of f(x).

One-to-one function

A function f : A → B is said to be a one-to-one function, if different elements in A havedifferent images in B, i.e., if x1 6= x2 then f(x1) 6= f(x2) for all x1, x2 ∈ A. The one-to-onefunction is also known as one-one or injective or injection. For example

(i) A mapping f : Z+ → Q is defined by n→ n2n+1 is an injective mapping.

(ii) Let A = 1, 2, 3, 4, . . . and B = 1, 12 ,

13 ,

14 , . . . thus the mapping f : x → 1

x . Hereone element x ∈ A maps exactly one element y ∈ B. So this is an one-to-one mapping.

Ex 1.10.1 Two sets are given by X = 1, 2, 3, 4, Y = α, β, γ and f = (1, α), (2, β),(3, β), (4, β), g = (1, α), (2, β), (3, γ). Test whether f, g are functions and if they arefunctions, test whether they are (i) injective, (ii) surjective.

Mapping 49

r rr rr rr rr rr rr r

f g

X XY Y

(a) (b)

1 12 23 34 4

α α

β βγ γ

Figure 1.28: Functions f and g

Solution: The pictorial representation of f is shown in Fig. 1.28(a). From the Fig. 1.28(a),it is seen that f is a function since every element of X has a unique image. But, all elementsof X are not mapped with the elements of Y so f is not surjective. Also, f is not injectivesince the image of the elements 2, 3, 4 have the same image β. The pictorial representationof g is shown in Fig. 1.28(b). g is not a function because all elements of the domain X arenot mapped to the element of co-domain Y .Difference between relation and function : Let A and B be two sets. By definitionof relation every subset of A × B will be a function if for each x ∈ A there is one andonly one ordered pair. Thus, every function is relation, but, converse is not true. Forexample, let A = a, b, c, B = 0, 1, 2 and f = (a, 0), (a, 1), (b, 1), (c, 2) is a relation butnot a function, since two different ordered pairs, viz., (a, 0) and (a, 1) have the same firstcoordinates. If we consider f as (a, 0), (b, 1), (c, 2) then it becomes a function as well as arelation.

Ex 1.10.2 If Z∗ be the set of nonnegative integers and f : Z → Z∗ is defined by f(x) =12 [x+ |x|], test whether it is injective or not. [CH04]

Solution: The mapping f : Z → Z∗ defined by f(x) = 12 [x+ |x|], is given by,

f(x) =12(x+ x) = x; for x ≥ 0

=12(x− x) = 0; for x < 0.

We know, −1,−2 < 0 but f(−1) = f(−2) = 0. Thus f(x1) = f(x2) although x1 6= x2, so isnot injective.

Ex 1.10.3 If Z∗ be the set of nonnegative integers and f : Z∗ → Z is defined by

f(x) =x

2; when x is an even integer

=−x+ 1

2; when x is an odd integer

Is f injective or surjective ? [ CH’03]

Solution: Let x, y ∈ Z∗ and x, y be even integers, then f(x) = x2 , f(y) = y

2 so thatf(x) = f(y) ⇒ x = y. Next, let x, y ∈ Z∗ and x, y be odd integers, then

f(x) =−x+ 1

2and f(y) =

−y + 12

,

then f(x) = x2 , f(y) = y

2 so that f(x) = f(y) ⇒ x = y. So f is injective. Now, 1 ∈ Z, but 1has no preimage under f . Therefore, f is not surjective.

50 Theory of Sets

Bijective mapping

A mapping f : A → B is defined as a bijective mapping or bijection if it is both injectiveand surjective. For example, let f : Z → Z be given by f(x) = x+ 1,∀ x ∈ Z. This is aninjective and surjective mapping.

Ex 1.10.4 Decide whether the following mapping are surjective or injective?(i)f : C → < defined by f(a+ ib) = a2 + b2.(ii)f : Z → Z+ defined by x→ x2 + 1.(iii)f : Z+ → Q defined by x→ x

2x+1 .

Solution: (i) By definition, f(2 + 3i) = 4 + 9 and f(3 + 2i) = 9 + 4. So,

f(2 + 3i) = f(3 + 2i) 6⇒ 2 + 3i = 3 + 2i.

Hence f(a+ ib) is not an injective or one-one mapping. It is not a surjection. Since −3 ∈ <but it has no pre-image in C, Since a2 + b2 ≥ 0,∀ real value of a and b.

(ii) If x = 2 and x = −2 then f(2) = 22 + 1 = 5 and f(−2) = (−2)2 + 1 = 5. The imageis 5 and ±2 ∈ Z. Therefore,

f(x1) = f(x2) 6⇒ x1 = x2.

Hence it has not an injective mapping. It has not a surjection, since 3 ∈ Z+ but 3 = f(n)given, 3 = n2 + 1 i.e. n =

√2 6∈ Z. Hence 3 has no pre-image in Z.

(iii) Let n1 and n2 be such that f(n1) = f(n2). Therefore,

f(n1) = f(n2) ⇒n1

2n1 + 1=

n2

2n2 + 1⇒ 2n1n2 + n1 = 2n1n2 + n2

⇒ n1 = n2

Hence it is an injection. For positive n, we have n2n+1 > 0. Hence negative numbers in Q

have no pre-image. Hence it is not onto.

Ex 1.10.5 Is the mapping f : < → (−1, 1), defined by, f(x) = x1+|x| a bijective mapping?

Justify your answer. R = set of real numbers and (−1, 1) = x : R : −1 < x < 1. [ KH’06]

Solution: Here, the mapping f : < → (−1, 1) is defined by, f(x) = x1+|x| . Since |x| 6= −1,

so the mapping is well defined. Therefore,

f(x) =x

1− x; when, x < 0, i.e.,−1 < f(x) < 0,

=x

1 + x; when, x > 0, i.e., 0 < f(x) < 1,

= 0; when, x = 0.

In the first case, if f(x1) = f(x2), then,

x1

1− x1=

x2

1− x2⇒ x1 = x2.

For the second case, if f(x1) = f(x2), then,

x1

1 + x1=

x2

1 + x2⇒ x1 = x2.

Mapping 51

This shows that f is one-one. Let y ∈ (−1, 1) and y = f(x). When, x < 0, we have,

y =x

1− x⇒ x =

y

1 + y⇒ f

(y

1 + y

)= y, as x ∈ <.

When, x > 0, we have,

y =x

1 + x⇒ x =

y

1− y⇒ f

(y

1− y

)= y, as x ∈ <.

Thus for x ∈ R their pre-image exist and since f is one-one therefore f(x) is onto. Hencef(x) is bijective mapping.

Ex 1.10.6 Let S be the set of all 2×2 real matrices(a bc d

); ad−bc 6= 0 and <∗ denote the set

of all non zero real numbers. Show that the mapping f : S → <∗ defined by f((

a bc d

))=

ad− bc, is surjective but not injective.

Solution: Let us consider two real matrices A =(

2 00 1

)and B =

(−1 00 −2

). Since

2.1 − 0.0 = (−1).(−2) − 0.0 6= 0 so, A,B ∈ S. Now, we see that, A 6= B, although, usingdefinition,

f

((2 00 1

))= 2 = f

((−1 00 −2

)).

Thus, when, A 6= B but still f(A) = f(B). Therefore, the given mapping f is not injective.Also, ∣∣∣∣a bc d

∣∣∣∣ = ad− bc 6= 0,

consequently, the inverse of the mapping exists and belongs to S. Thus for every real number

a, b, c, d with ad− bc 6= 0, ∃ a real matrix(a bc d

)for which the mapping exists. Thus, f is

surjective.

Ex 1.10.7 Discuss the mapping f : R → (−1, 1) defined by f(x) = x1+x2 , x ∈ R, where R

is the set of real numbers and (−1, 1) = x ∈ R : −1 < x < 1.

Solution: Since x ∈ < so, the given mapping is well defined. Take two elements x1, x2 ∈ <.If f(x1) = f(x2), then

x1

x21 + 1

=x2

x22 + 1

⇒ (x1 − x2)− x1x2(x1 − x2) = 0

⇒ either, x1 = x2 or, x1x2 = 1.

Taking x1 = 5 and x2 = 15 , we see that f(x1) = f(x2) = 5

26 . Thus when, x1 6= x2 but stillf(x1) = f(x2). Therefore, f is not injective. Let y an arbitrary element of (−1, 1), then

y =x

x2 + 1⇒ x =

1±√

1− 4y2

2y; −1 < y < 1.

When, y(∈ <) > 12 , we have, 1−4y2 < 0 and so

√1− 4y2 6∈ < consequently, x = 1±

√1−4y2

2y 6∈

<. Similarly, when, y(∈ <) > − 12 , then also

√1− 4y2 6∈ < and so x = 1±

√1−4y2

2y 6∈ <.Therefore,

x =1±

√1− 4y2

2y6∈ <; ∀y ∈ (−1, 1).

This means that − 12 ≤ f(x) ≤ 1

2 , i.e., f does not map entire co-domain (−1, 1). So, it isnot onto.

52 Theory of Sets

Restriction Mapping

Let f : A→ B and A0 ⊂ A. The mapping g : A0 → B, such that g(x) = f(x); ∀ x ∈ A0,is said to be the restriction mapping of f to A0. It is denoted by f |A0 or fA, read as ’frestricted to A0’. f is said to be an extension of g to A. As for examples: f : R → R begiven by f(x) = |x| −x, x ∈ R and g : R+ → R be given by g(x) = 0x ∈ R+ then g = f |R+.

Inverse mapping

Let f : A → B be a mapping and b ∈ B be arbitrary. Let the mapping f : A → B beone-one onto mapping, then corresponding to each element y ∈ B, ∃ an unique elementx ∈ A such that f(x) = y. Thus a mapping, denoted by f−1, is defined as

f−1 : B → A : f−1(y) = x⇔ f(x) = y. (1.18)

The mapping f−1 defined above is called the inverse of f .The pictorial representation of f−1

is shown in Fig. 1.29. If f−1 is the inverse of f and if f(x) = y then x = f−1(y). For example,let A = 1, 2, 3, B = a, b, c and f = (1, a), (2, b), (3, c). Then inverse relation of f isf−1 = (a, 1), (b, 2), (c, 3), which is a function from B to A. Again g = (1, a), (2, a), (3, b)is a function from A to B and its inverse relation g−1 = (a, 1), (a, 2), (b, 3), which is not a

A B

x y = f(x)

f

f−1

Figure 1.29: Inverse function

function since f−1 = 1, 2.

Theorem 1.10.1 The necessary and sufficient condition that a mapping is invertible is thatit is one-one and onto.

Proof: Suppose f is invertible. Then f−1 : B → A is a function. Let b1, b2 ∈ B. Thereexists a1, a2 ∈ A (a1 6= a2) such that f−1(b1) = a1 and f−1(b2) = a2. That is, b1 = f(a1)and b2 = f(a2). Since f−1 is a function, for different b1 and b2 their images must be different,i.e., f(a1) 6= f(a2).

Again, f−1 is a function, all elements of B must be mapped with some elements of Aunder f−1. Thus, f is onto. Thus the condition is necessary.

Conversely, let f is bijective, i.e., one-one and onto. Then for different a1, a2 ∈ A thereexists b1, b2 ∈ B (where b1 6= b2) such that a1 = f−1(b1) and a2 = f−1(b2). Thus fordifferent b1 and b2 there images a1, a2 under f−1 are different.

Again, f is onto, all elements of B are mapped with the elements of A, i.e., each elementof the domain B of f−1 is mapped with each element of A. Hence f−1 is a function.

Theorem 1.10.2 Let f : A→ B is bijective function then f−1 : B → A is also bijective.

Proof: Let b1, b2 be any two elements of B. Since f is one-one, therefore there existunique elements a1, a2 ∈ A such that b1 = f(a1) and b2 = f(a2). Let a1 = f−1(b1) anda2 = f−1(b2). Suppose and

f−1(b1) = f−1(b2) ⇒ a1 = a2

⇒ f(a1) = f(a2) as f is one-one ⇒ b1 = b2.

Therefore f−1(b1) = f−1(b2) iff b1 = b2. Hence f−1 is one-one.

Mapping 53

To prove f−1 onto : Let a be any element of A. Since f is a function from A to B,there exists a unique element b ∈ B such that b = f(a) or, a = f−1(b). That is, image underf−1 of the element b ∈ B is a ∈ A. Hence f−1 is onto.

Ex 1.10.8 Let A = <−− 12, B = <− 1

2 and f : A→ B is defined by f(x) = x−32x+1 ,∀x ∈

A. Does f−1 exists?

Solution: Since A = < − − 12, so, the given mapping f(x) = x−3

2x+1 is well defined. Let,x1, x2 ∈ A. If f(x1) = f(x2), then

x1 − 32x1 + 1

=x2 − 32x2 + 1

⇒ 2x1x2 + x1 − x2 − 3

= 2x1x2 − 6x1 + x2 − 3 ⇒ x1 = x2.

Therefore, f is injective. Let y an arbitrary element of B, then

y =x− 32x+ 1

⇒ x =y + 31− 2y

,

which is defined as y 6= 12 ∈ B. Also,

f

(y + 31− 2y

)=

y+31−2y − 3

2 y+31−2y + 1

=y + 3− 3(1− 2y)2(y + 3) + 1− 2y

= y,

so for each y ∈ B, ∃ an element x = y+31−2y ∈ A such that f(x) = y. Hence f is surjective and

consequently it is bijective. Therefore, f−1 exists.

Ex 1.10.9 If the function f : < → < be defined by f(x) = x2 + 1, then findf−1(−8) and f−1(17).

Solution: We have, from the definition of the inverse mapping,

f−1(−8) = x ∈ <; f(x) = −8= x ∈ <; x2 + 1 = −8= x ∈ <; x = ±3

√−1 = φ,

as ±3√−1 are not real numbers. Again,

f−1(17) = x ∈ <; f(x) = 17= x ∈ <; x2 + 1 = 17= x ∈ <; x = ±4 = 4,−4.

Theorem 1.10.3 The inverse of a bijective mapping is also bijective.

Proof: Let f : A → B be a bijective mapping, then, we are to show that f−1 : B → A isalso bijective. Let y1, y2 ∈ B. As f is bijective, i.e., one-one and onto so, f−1(y1) = x1 andf−1(y2) = x2. Therefore,

f−1(y1) = f−1(y2) ⇒ x1 = x2

⇒ f(x1) = f(x2); as f is one-one⇒ y1 = y2.

Thus f−1 is one-one. Given any element x ∈ A, we can find an element y ∈ B, wheref−1(y) = x. Thus x is the f−1image of y ∈ B. This shows that f−1 is an onto mapping.Hence the theorem.

54 Theory of Sets

Theorem 1.10.4 For a bijective mapping, the inverse mapping is unique.

Proof: Let, if possible, the mapping f : A → B has two inverses, say, g : B → A andh : B → A. Let b be any element of B and g(b) = a1, h(b) = a2, a1, a2 ∈ A. Since f isone-one function, therefore

f(a1) = f(a2) ⇒ a1 = a2 ⇒ g(b) = h(b),

i.e., h = g. Thus the inverse mapping of f : A→ B is unique.

Theorem 1.10.5 If f is a one-one correspondence between A and B then,f−1of = IA and f−1of = IB.

Proof: By definition, IA(a) = b = f(f−1(b)) = (f−1of)(a). Thus, IA = f−1of . Again,IB(b) = b = f(f−1(b)) = (fof−1)(b). Hence IB = fof−1.

Ex 1.10.10 f : Q → Q is define by f(x) = 5x + a where a, x ∈ Q, the set of rationalnumbers, then show that f is one-one and onto. Find f−1.

Solution: Let x1, x2 ∈ Q. If f(x1) = f(x2) then 5x1 + a = 5x2 + a⇒ x1 = x2. Thus f isone-one. Let y = f(x) = 5x + a, then x = 1

5 (y − a). Since x ∈ Q, 15 (y − a) ∈ Q. Again,

f(y−a5 ) = y−a

5 + a = y. Thus, y ∈ Q is the image of the element y−a5 ∈ Q.

Hence f is onto. Since f is bijective, f is invertible and hence x = f−1(y) = 15 (y − a).

Ex 1.10.11 Show that the functions f : R → (1,∞) and g : (1,∞) → R defined byf(x) = 32x + 1, g(x) = 1

2 log3(x− 1) are inverses to each other.

Solution: Let x ∈ R, then (gof)(x) is given by,

(gof)(x) = gf(x) = g(32x + 1)= 1

2 log3[(32x + 1)− 1]

= log332x = 12 .2x = x

Therefore gof = IR. For x ∈ (1,∞), we have

(fog)(x) = fg(x) = f( 12 log3(x− 1))

= 3( 12 log3(x−1)) + 1

= (x− 1) + 1 = x.

Therefore fog = IR and so the functions f : R → (1,∞) and g : (1,∞) → R are inverses toeach other.

Ex 1.10.12 Show that the functions f : [−π2 ,

π2 ] → [−1, 1] such that f(x) = sinx is one-one

and onto. Also find f−1.

Solution: Let x1, x2 ∈ [−π2 ,

π2 ] be any two numbers. Let

f(x1) = f(x2) ⇒ sinx1 = sinx2 ⇒ x1 = x2.

Hence f is one-one. Let y = f(x) = sinx, or, x = sin−1 y ∈ [−π2 ,

π2 ] where y ∈ [−1, 1]. Also,

f(x) = f(sin−1 y) = sin(sin−1 y) = y. Thus f is onto.Hence f is one-one and onto, so f is invertible. Now,

y = sinx⇒ x = sin−1 y

or, f−1(y) = sin−1 y; ∀y ∈ [−1, 1].

Mapping 55

Ex 1.10.13 Show that the mapping f : Z → Z, defined by f(x) = x + 2 is a bijectivemapping.

Solution: Let x1, x2 be two different elements of Z. If possible let f(x1) = f(x2). Thenx1 + 2 = x2 + 2, i.e., x1 = x2, which contradicts x1 6= x2.

Thus f(x1) = f(x2) iff x1 = x2. Hence f is one-one.Let y = f(x) = x+ 2. Then x = y − 2 ∈ Z andf(x) = f(y − 2) = (y − 2) + 2 = y ∈ Z (co-domain).Thus, f is onto.Hence f is a bijective mapping.

Ex 1.10.14 Let f : R → R be a function defined by

f(x) =−1, if x is rational ,1, if x is irrational ,

then show that f is neither injective nor surjective (This function is known as Dirichlet’sfunction).

Solution: The function is not injective since all rational numbers are mapped to 1 and allirrational numbers are mapped to −1.

It is not surjective as only two elements ofR, viz., −1 and 1 are the images of the elementsof R.

But, if we redefine the function f as f : R → −1, 1 then it become an surjectivefunction.

Ex 1.10.15 Show that the mapping f : R → R given by f(x) = |x|, x ∈ R is neitherone-one nor onto.

Solution: The function f is not one-one, since f(−1) = |−1| = 1 and f(1) = |1| = 1. Thatis, for different elements of R have same image.

Also, it is not onto, as f(R) = R+ ∪ 0 6= R, i.e., only the positive numbers of R(co-domain) are mapped with the elements of R (domain).

Theorem 1.10.6 If |A| = n = |B| then number of bijective mapping from A to B is n!.

Proof: Let Xn be the set of all bijective functions on A, when |A| = n. Let A =a1, a2, . . . , an and B = b1, b2, . . . , bn. When n = 1 there is only one bijection (a1, b1).

When n = 2, then there are two bijections, viz., (a1, b1), (a2, b2) and (a1, b2), (a2, b1),i.e., X2 = (a1, b1), (a2, b2), (a1, b2), (a2, b1). There the number of bijections from A toB, when n = 2 is 2 = 2!.

When n = 3, we construct all bijections starting from a bijective function of X2 as follows.Let (a1, b1), (a2, b2) ∈ X2. One can add new elements a3 ∈ A and b3 ∈ B to this

bijective in three different ways shown below.(a1, b1), (a2, b2), (a3, b3), (a1, b3), (a2, b2), (a3, b1), (a1, b1), (a2, b3), (a3, b2).Similarly, from second element of X2 one can generate other three bijective functions.Hence X3 contains 6 = 3! bijective functions when |A| = |B| = 3.We assume that Xn has n! bijections when |A| = |B| = n. Let A = a1, a2, . . . , an+1

and B = b1, b2, . . . , bn+1 and Xn+1.be the set of all bijections from A onto B.Let (a1, b1), (a2, b2), ldots, (an, bn) be a bijective function of Xn. By introducing an+1 ∈

A and bn+1 ∈ B, one can generates the following bijections starting from the above bijection

(a1, b1), (a2, b2), . . . , (an, bn), (an+1, bn+1)

56 Theory of Sets

rrrrrr rr

f

A B

(a) One-one into function

rrrrr rr

f

A B

(b) Many-one into function

rrrrrr

f

A B

(c) One-one onto function

rrrrrrr

f

A B

(d) Many-one onto function

Figure 1.30: Pictorial representation of different type of function

(a1, bn+1), (a2, b2), . . . , (an, bn), (an+1, b1)(a1, b1), (a2, bn+1), . . . , (an, bn), (an+1, b2)

......

...(a1, b1), (a2, b2), . . . , (an, bn+1), (an+1, bn).

Thus starting from a single bijective function of Xn we generate (n+ 1) bijective functionsby increasing only one element each of A and B. Hence the number of bijective functionsXn+1 is n!× (n+ 1) = (n+ 1)!.

Hence by mathematical induction the number of different bijective functions from A toB is n! when |A| = |B| = n.

If f : A→ B is a bijective function from A onto B then |A| = |B|, i.e., both the sets havesame number of elements. If |A| 6= |B| then f can not be bijective.

Result 1.10.1 Let A and B be two non-empty sets with cardinality m and n respectively.The number of possible relations from A to B is 2mn − 1.

Many-one function

A function f : A → B is said to be many-one function if two or more elements of Acorrespond to the same element of B. The constant function is an example of many-onefunction. Foe example, let A = 1, 2, 3, 4, B = 0, 1, 2. Then

(i) f = (1, 0), (2, 0), (3, 1), (4, 2) is a many-one into function.

(ii) f = (1, 0), (2, 1), (3, 1), (4, 2) is a many-one onto function.

Let A = 1, 2, 3, B = 0, 1, 2, 3 and f = (1, 0), (2, 0), (3, 3) is one-one and into function.

Equality of two function

Two functions f : A → B and g : C → D are said to be equal if A = C, B = D andf(x) = g(x) for all x ∈ A. It is written as f = g.

Ex 1.10.16 Suppose A = 1, 2, 3 and B = 8, 9. Examine whether the following subsetsof A×B are functions from A to B.

Mapping 57

(i) f1 = (1, 8), (1, 9), (2, 8), (3, 9)

(ii) f2 = (1, 9), (2, 9), (3, 9)

(iii) f3 = (1, 8), (2, 9), (3, 9).

How many mapping are there from A into B ? Identify the one-one and onto mapping.

Solution: Here A and B are domain and co-domain respectively. Then,

(i) f1 is not a function since 1 ∈ A has two different images 8 and 9.

(ii) f2 is a function, particularly it is a constant function.

(iii) f3 is also a function.

The possible mapping from A to B areg1 = (1, 8), (2, 8), (3, 8), g2 = (1, 9), (2, 9), (3, 9),g3 = (1, 8), (2, 8), (3, 9), g4 = (1, 8), (2, 9), (3, 8),g5 = (1, 9), (2, 8), (3, 9), g6 = (1, 9), (2, 9), (3, 8),g7 = (1, 9), (2, 8), (3, 9), g8 = (1, 9), (2, 8), (3, 8).

Therefore, there are 8 mappings from A to B. The mappings are many-one and none ofthem are one-one. The onto mappings are g3, g4, g5, g6, g7, g8.

1.10.2 Composite mapping

Let A,B,C be any non empty sets and let f : A → B and g : B → C be two functions. Ifa function h is defined in such a way that h : A → C by h(x) = gf(x), x ∈ A, then h is

r r rf g

h = gof

A B C

x f(x) gf(x)

Figure 1.31: Composite function gof

called the product or composite function of f and g. It is denoted by gof or gf . Thus, theproduct or composite mapping of the mappings f and g, denoted by gof : A→ C is definedby

(g0f)(x) = g[f(x)], for all x ∈ A. (1.19)

Under the mapping f , an element x ∈ A is mapped to an element y = f(x) ∈ B. Again, yis mapped by g to an element z ∈ C such that z = g(y) ∈ C and hence z = g(y) = g[f(x)].Obviously, the domain of (gof) is A and co-domain is C. For example, let f : R → R andg : R → R be two functions where f(x) = 3x+ 2 and g(x) = x2 + 1. Now,

(fog)(x) = f(g(x)) = f(x2 + 1) = 3(x2 + 1) + 2 = 3x2 + 5 and(gof)(x) = g(f(x)) = g(3x+ 2) = (3x+ 2)2 + 1 = 9x2 + 12x+ 5.

For this example, it is seen that (fog)(x) 6= (gof)(x). Thus in general fog 6= gof , i.e.,product of the function is non-commutative.

58 Theory of Sets

Ex 1.10.17 Let f, g : < → < be two mappings, defined by f(x) = |x| + x, x ∈ < andg(x) = |x| − x, x ∈ <. Find fg and gf .

Solution: Here the two mappings f, g : < → < is defined by f(x) = |x| + x, ; g(x) =|x| − x, x ∈ <, i.e.,

f(x) = 2x; if x ≥ 0 g(x) = −2x; if x < 0= 0; if x < 0 = 0; if x ≥ 0. .

Now, let x ≥ 0, then,

(fg)(x) = f(g(x)) = f(0) = 0 and (gf)(x) = g(f(x)) = g(2x) = 0.

Now, let x < 0, then,

(fg)(x) = f(g(x)) = f(−2x) = −4x and (gf)(x) = g(f(x)) = g(0) = 0.

Therefore, (gf)(x) = 0, for all x ∈ < and

(fg)(x) = 0; if x ≥ 0= −4x; if x < 0.

Theorem 1.10.7 The product of any function with the identity function is the functionitself.

Proof: Let f : X → Y and let us denote by Ix and Iy the identity functions on X andY respectively. Then we must show that IY of = f and foIX = f . Since f : X → Y andIY : Y → Y, so IY of : X → Y. Now, let x be an arbitrary element of X and let f(x) = y.Then,

(IY of)(x) = IY [f(x)] = IY (y) = y = f(x)⇒ IY of = f.

Again, since IX : X → X and f : X → Y , so foIX : X → Y. Now, for arbitrary x ∈ X, wehave,

(foIX)(x) = f [IX(x)] = f(x)⇒ foIX = f.

Therefore, the product of any function with the identity function is the function itself.

Theorem 1.10.8 The product of any invertible mapping f with its inverse mapping f−1,is an identity mapping.

Proof: Let f be an one-one mapping of X onto Y and let IX and IY be the identitymappings on X and Y respectively. Clearly, f−1 is a one-one mapping of Y onto X. Now,

f : X → Y and f−1 : Y → X ⇒ f−1of : X → X.

Now, let x be an arbitrary element of X and let f(x) = y so that, x = f−1(y). Therefore,

(f−1of)(x) = f−1[f(x)] = f−1(y) = x = IX(x)⇒ f−1of = IX .

Again, f−1 : Y → X and f : X → Y , so fof−1 : Y → Y. Now, for an arbitrary y ∈ Y , thereis associated an unique x ∈ X such that f(x) = y or x = f−1(y). Therefore,

(fof−1)(y) = f [f−1(y)] = f(x) = y = IY (y)⇒ fof−1 = IY .

Hence, f−1of = IX and fof−1 = IY . Therefore, the product of any invertible mapping fwith its inverse mapping f−1, is an identity mapping.

Mapping 59

Theorem 1.10.9 Composites of functions is associative.

Proof: Let X,Y, Z, T be the four non-empty sets. Let f : X → Y, g : Y → Z andh : Z → T be the three mappings. Then, we are to show that, (hog)of = ho(gof). Now,hog : Y → T, gof : X → Z and so (hog)of : X → T, ho(g)f) : X → T. Let x be an arbitraryelement of X. Then,

[(hog)of ](x) = (hog)[f(x)] = h[gf(x)]= h[(gof)(x)] = [ho(gof)](x); ∀x ∈ X.

Hence (hog)of = ho(gof) and so composites of functions is associative.

Theorem 1.10.10 Let f : X → Y and g : Y → X. If gof is an identity function on Xand fog is an identity function on Y , then g = f−1.

Proof: Being given that, gof = IX and fog = IY . We are to show that g = f−1. We firstprove that f is invertible, i.e., it is one-one and onto. Now,

f(x1) = f(x2) ⇒ g[f(x1)] = g[f(x2)]⇒ (gof)(x1) = (gof)(x2)⇒ IX(x1) = IX(x2) ⇒ x1 = x2.

So, f is one-one. In order to show that f is onto, let y be an arbitrary element of Y and letg(y) = x. Then,

g(y) = x⇒ f [g(y)] = f(x)⇒ (fog)(y) = f(x)⇒ IY (y) = f(x) ⇒ y = f(x).

Thus, for each y ∈ Y , there is an x such that f(x) = y. So, f is onto. Now,

fog = IY ⇒ f−1(fog) = f−1oIY

⇒ (f−1of)og = f−1; associativity⇒ IY og = f−1 ⇒ g = f−1.

Theorem 1.10.11 Let f : A→ B and g : B → C be invertible. Then gof is invertible and(gof)−1 = f−1og−1.

Proof: Since f and g are invertible, they are bijective. Let a1, a2 ∈ A. Now,

(gof)(a1) = (gof)(a2) ⇒ gf(a1) = gf(a2)⇒ f(a1) = f(a2); as g is one-one⇒ a1 = a2; as f is one-one.

Hence (gof) is one-one. Let c ∈ C. Since g is onto, every c has one pre-image, say b ∈ B,such that g(b) = c. Again, since f is onto, every b ∈ B has a pre-image a ∈ A, such thatf(a) = b. Now,

(gof)(a) = gf(a) = g(b) = c.

Thus, for every element c ∈ C, there is an element a ∈ A, such that (gof)(a) = c, and hencegof is onto. Thus (gof) is bijective and so it is invertible. Now,

(gof)(a) = c⇒ (gof)−1(c) = a.

Also, f−1og−1(c) = f−1g−1(c) = f−1(b) = a

as g(b) = c and f(a) = b.

⇒ (gof)−1(c) = f−1og−1(c), i.e., (gof)−1 = f−1og−1.

60 Theory of Sets

Therefore, if f and g be one-one mappings of A onto B and B onto C respectively, so thatf and g are both invertible, then (goh) is also invertible and (goh)−1 = f−1og−1.

Theorem 1.10.12 Let f : A → B and g : B → C be both injective then gof : A → C isalso injective.

Proof: Let x1, x2 ∈ A.Now,

(gof)(x1) = (gof)(x2) ⇒ gf(x1) = gf(x2)⇒ f(x1) = f(x2)( since g is injective )⇒ x1 = x2( since f is injective ).

Hence gof is injective.

Theorem 1.10.13 Let f : A → B and g : B → C be two surjective functions then gof :A→ C is surjective.

Proof: Let z ∈ C. Since g is surjective, there exist y ∈ B such that g(y) = z. Again, sincef is surjective, there exist x ∈ A such that f(x) = y. Now,

(gof)(x) = gf(x) = g(y) = z,

and it is true for arbitrary z. Hence gof is surjective.

Theorem 1.10.14 If f : A→ B and g : B → C be two mappings and gof is bijective, thenf is one-one and g is onto.

Proof: Let any two elements x1, x2 ∈ A be such that f(x1) = f(x2). Now,

(gof)(x1) = g[f(x1)] = g[f(x2)] = (gof)(x2).

Now, since gof is one-one, therefore,

(gof)(x1) = (gof)(x2) ⇒ x1 = x2.

Therefore, f(x1) = f(x2) ⇒ x1 = x2.

Hence f is one-one. Again, since gof : A → C is onto, for all z ∈ C, ∃ an element a ∈ Asuch that (gof)(x) = z. Therefore, z = g[f(x)],∀z ∈ C. Now, for each x ∈ A, we havef(x) = y ∈ B. Thus, for each z ∈ C, there is an element y ∈ B such that z = g(y). Henceg : B → C is onto.

Theorem 1.10.15 The inverse of the inverse of a function is the function itself, i.e.,(f−1)−1 =f.

Proof: Let a mapping f : A→ B be invertible. Then ∃ a function g = f−1 : B → A suchthat

f(x) = y ⇒ x = g(y), x ∈ A and y ∈ B.

Again, f , being invertible, is one-one and onto and so g is one-one and onto, i.e., g isinvertible and g−1 exists. Now,

(fog)(y) = f [g(y)] = f(x),i.e., (fog)(y) = y, as f(x) = y.

This shows that (fog) = IB , identity mapping in B. Thus, f is the inverse of g, i.e., f = g−1.This gives (f−1)−1 = f.

Mapping 61

Definition 1.10.1 Images and inverse images of sets under a mapping : Let X andY be any two non-empty sets and f be a mapping of X into Y . Let A ⊆ X and B ⊆ Y.Then, we define

f(A) = y ∈ Y : y = f(x) for some x ∈ A (1.20)and f−1(B) = x ∈ X : f(x) ∈ B. (1.21)

Thus, y ∈ f(A) ⇒ y = f(x), for some x ∈ A and x ∈ f−1(B) ⇒ f(x) ∈ B.

Note : Here note that, f(x) ∈ f(A) not necessarily implies that x ∈ A, for example, if weconsider the mapping

f : < → < : f(x) = x2,∀x ∈ <

and if A = [0, 1] is a subset of <, then obviously, f(A) = [0, 1]. Also by definition, of f , wehave, f(−1) = 1 ∈ [0, 1] = f(A) but (−1) 6∈ A. However x ∈ A⇒ f(x) ∈ f(A).

Theorem 1.10.16 If X and Y are two non-empty sets and f be a mapping of X into Y ,then for any subsets A and B of X,

(i)f(A ∪B) = f(A) ∪ f(B), (ii)f(A ∩B) ⊆ f(A) ∩ f(B).

Proof: (i) Let y be an arbitrary element of f(A ∪B), then

y ∈ f(A ∪B) ⇔ y = f(x) for some x ∈ (A ∪B)⇔ y = f(x) for some x ∈ A or x ∈ B⇔ y = f(x), for f(x) ∈ f(A) or f(x) ∈ f(B)⇔ y ∈ f(A) or y ∈ f(B) ⇔ y ∈ f(A) ∪ f(B).

Consequently, f(A∪B) ⊆ f(A)∪ f(B) and f(A)∪ f(B) ⊆ f(A∪B) and hence, f(A∪B) =f(A) ∪ f(B).(ii) Let y be an arbitrary element of f(A ∩B), then,

y ∈ f(A ∩B) ⇒ y = f(x) for some x ∈ (A ∩B)⇒ y = f(x) for some x ∈ A and x ∈ B⇒ y = f(x), such that f(x) ∈ f(A) and f(x) ∈ f(B)⇒ y ∈ f(A) and y ∈ f(B) ⇒ y ∈ f(A) ∩ f(B).

Consequently, f(A∩B) ⊆ f(A)∩f(B). Note that, the relation can not in general be replacedby equality. For example, if a = [−1, 0] and B = [0, 1] are any two subsets of the set < ofall real numbers and

f : < → < : f(x) = x2,∀x ∈ <,

then clearly, f(A) = [0, 1] and f(B) = [0, 1] so that f(A) ∩ f(B) = [0, 1] and since A ∩B =0, so f(A ∩ B) = 0. Thus, f(A ∩ B) 6== f(A) ∩ f(B). Also it may be noted here thatf(A ∩B) ⊆ f(A) ∩ f(B). Thus in general, f(A ∩B) ⊆ f(A) ∩ f(B).

Theorem 1.10.17 If X and Y be two non-empty sets and f be a mapping of X into Y ,then for any subsets A and B of Y ,

(i)f−1(A ∪B) = f−1(A) ∪ f−1(B) and (ii)f−1(A ∩B) = f−1(A) ∩ f−1(B).

62 Theory of Sets

Proof: (i) Let x be an arbitrary element of f−1(A ∪B), then

x ∈ f−1(A ∪B) ⇔ f(x) ∈ (A ∪B)⇔ f(x) ∈ A or f(x) ∈ B⇔ x ∈ f−1(A) or x ∈ f−1(B)⇔ x ∈ f−1(A) ∪ f−1(B).

Consequently, f−1(A ∪ B) ⊆ f−1(A) ∪ f−1(B) and f−1(A) ∪ f−1(B) ⊆ f−1(A ∪ B) andhence, f−1(A ∪B) = f−1(A) ∪ f−1(B).(ii) Let x be an arbitrary element of f−1(A ∩B), then

x ∈ f−1(A ∩B) ⇔ f(x) ∈ (A ∩B)⇔ f(x) ∈ A and f(x) ∈ B⇔ x ∈ f−1(A) and x ∈ f−1(B)⇔ x ∈ f−1(A) ∩ f−1(B).

Consequently, f−1(A ∩ B) ⊆ f−1(A) ∩ f−1(B) and f−1(A) ∩ f−1(B) ⊆ f−1(A ∩ B) andhence, f−1(A ∩B) = f−1(A) ∩ f−1(B).

Theorem 1.10.18 If X and Y be two non-empty sets and f be a mapping of X into Y ,then, (i) for any subset A of X, A ⊆ f−1[f(A)] and in general, A 6= f−1[f(A)]. (ii) Forany subset B of Y , f [f−1(B)] ⊆ B and further if B is a subset of the range of f , thenf [f−1(B)] = B.

Proof: (i) Let A be any subset of X. If A = φ, then the result is obvious. So, let A 6= φand let x be an arbitrary element of A, then

x ∈ A⇒ f(x) ∈ f(A) ⇒ x ∈ f−1[f(A)].

Hence, A ⊆ f−1[f(A)]. Now, in order to show that, in general, A 6= f−1[f(A)], considerthe mapping, f : < → < : f(x) = x2,∀x ∈ <. Now, let A = [−1, 0] be any subset of<, then obviously, f(A) = [0, 1]. Therefore, f−1[f(A)] = [−1, 1] 6= A. Thus in general,A 6= f−1[f(A)].(ii) Let B be any subset of X and let y be an arbitrary element of f [f−1(B)], then

y ∈ f [f−1(B)] ⇒ y = f(x) for some x ∈ f−1(B)⇒ y = f(x) such that f(x) ∈ B⇒ y ∈ B ⇒ f [f−1(B)] ⊆ B.

Further, if B is a subset of the range of f , then for each y ∈ B,∃ an x ∈ f−1(B) such thaty = f(x) and so

y ∈ B ⇒ y = f(x) for some x ∈ f−1(B)⇒ y = f(x) for f(x) ∈ f [f−1B]⇒ y ∈ f [f−1B] ⇒ B ⊆ f [f−1(B)].

Hence it follows that, f [f−1B] = B. Note here that f [f−1B] = B holds only when B is asubset of the range of f and so in general, f [f−1B] 6= B. For example, consider the mapping,f : < → < : f(x) = x2,∀x ∈ <. Now, let B = [−1, 0] then f−1(B) = 0 and therefore,f [f−1(B)] = 0 6= [−1, 0] = B. Thus in general, f [f−1B] 6= B.

Permutation 63

1.11 Permutation

Permutation of a non-empty finite set is defined to be a bijective mapping of a finite setonto itself. Let a, b, c, · · · , k be any arrangement of the set of positive integers 1, 2, · · · , n.The one-one mapping p : S → S of the finite set S = 1, 2, · · · , n onto itself

1 2 3 · · · n↓ ↓ ↓ ↓a b c · · · k

where p(1) = a, p(2) = b, · · · , p(n) = k, denoted by the symbol

p =(

1 2 3 · · · np(1) p(2) p(3) · · · p(n)

)=(

1 2 3 · · · na b c · · · k

)(1.22)

is known as the permutation of degree n or n symbols. Obviously, the order of the columnin the symbol is immaterial so long as the corresponding elements above and below in thatcolumn remain unchanged. The order, in which the first row is written, does not matter,

what actually matters is which element is replaced by which. Thus,(

1 2 3a b c

),

(2 1 3b a c

)and(

2 3 1b c a

), are the same. In the standard form, the elements in the top row are in natural

order. If p be a permutation of n symbols, then the set of all permutations, denoted by Pn,will contain n! distinct elements, as n distinct elements can be arranged in n! ways and isknown as symmetric set of permutations.

Ex 1.11.1 Construct the symmetric set of permutations P3.

Solution: The symmetric set of permutations P3 contains 3! = 6 elements, where eachpermutation has 3 symbols. Therefore,

P3 =(

1 2 31 2 3

),

(1 2 31 3 2

),

(1 2 32 1 3

),

(1 2 32 3 1

),

(1 2 33 1 2

),

(1 2 33 2 1

).

1.11.1 Equal permutations

Two permutations p and q of degree n are said to be equal, if p(a) = q(a), for all a ∈ S. For

example, the permutations p =(

1 2 3 42 3 4 1

)and q =

(2 4 3 13 1 4 2

)are equal permutations.

1.11.2 Identity permutation

If S = a1, a2, · · · , an, then the bijective mapping I : S → S, defined by I(ai) = ai, iscalled an identity permutation of degree n. For example,

I =(a b ca b c

)or(

1 2 3 · · · n1 2 3 · · · n

)are identity permutations. Thus, if there be no change of the elements, i.e., if each elementbe replaced by itself, then it is called identity permutation and is denoted by I.

1.11.3 Product of permutations

Since permutation is just a bijective mapping, the product of two permutations is just theproduct of two mappings. Let S = a1, a2, · · · , an and let p : S → S and q : S → S be two

64 Theory of Sets

permutations of S. Since range of p = domain of q, the composite mapping p0q : S → S isdefined. Since the permutations p and q are bijective, p0q is also bijective. Hence p0q is apermutation of S. The product of two permutations p : S → S and q : S → S, denoted byp0q or simply pq is defined by,

pq =(

a1 a2 a3 · · · an

p[q(a1)] p[q(a2)] p[q(a3)] · · · p[q(an)]

).

Since, composition of mappings is non-commutative, so pq 6= qp, in general. Also, sincecomposition of mappings is associative, so p(qr) = (pq)r. Let p be a permutation on a finiteset of degree n, then we define,

pn = p.p · · · p(n factors), ∀n ∈ N

with p0 = I. Also for all integral values of m,n, we have the index laws (i) pmpn = pm+n

and (ii) (pm)n = pmn. As, pq 6= qp, in general, so (pq)n = pnqn does not hold. If ∃ a leastpositive value integer k, such that pk = I, then k is called the order of the permutation p.

Ex 1.11.2 If p = (1 2 3 4 5), q = (2 3)(4 5), find pq.

Solution: Using the definition of product of permutations we have,

pq =(

1 2 3 4 52 3 4 5 1

)(1 2 3 4 51 3 2 5 4

)The product is given by the following mapping procedure, from the right, follows from thedefinition of mapping f(g)(x) = f(g(x)).

1 2 3 4 5↓ ↓ ↓ ↓ ↓1 3 2 5 4↓ ↓ ↓ ↓ ↓2 4 3 1 5

Therefore the product of p and q is given by(

1 2 3 4 52 4 3 1 5

)= (1 2 4). Similarly,

qp =(

1 2 3 4 51 3 2 5 4

)(1 2 3 4 52 3 4 5 1

)=(

1 2 3 4 53 2 5 4 1

)= (1 3 5).

1.11.4 Inverse of permutations

Let S = a1, a2, · · · , an and let f : S → S be a permutation of S. As p : S → S is abijective mapping, it has unique inverse which is also bijective. Let p−1 be the inverse, thenp−1 : S → S is the permutation of S and is defined by

p−1 =(p(a1) p(a2) p(a3) · · · p(an)a1 a2 a3 · · · an

).

The important property is that pp−1 = p−1p = I.

Ex 1.11.3 If p = (1 2 4 3), q = (1 4 3 2), show that, (pq)−1 = q−1p−1.

Permutation 65

Solution: Here the product of p and q is given by,

pq =(

1 2 3 42 4 1 3

)(1 2 3 44 1 2 3

)=(

1 2 3 43 2 4 1

)= (1 3 4).

(pq)−1 =(

3 2 4 11 2 3 4

)=(

1 2 3 44 2 1 3

)= (1 4 3).

As, p = (1 2 4 3), q = (1 4 3 2), so the inverses are given by

p−1 =(

2 4 1 31 2 3 4

)=(

1 2 3 43 1 4 2

); q−1 =

(1 2 3 42 3 4 1

)⇒ q−1p−1 =

(1 2 3 42 3 4 1

)(1 2 3 43 1 4 2

)=(

1 2 3 44 2 1 3

)= (1 4 3).

Hence, (pq)−1 = q−1p−1 is verified.

1.11.5 Cyclic permutation

Let S = a1, a2, · · · , an. A permutation p : S → S is said to be a cycle of length r, oran r cycle, if there are r elements ai1 , ai2 , · · · , air

in S such that, p(ai1) = ai2 , p(ai2) =ai3 , · · · , p(air−1 = air

, p(air) = ai1 and p(aj) = aj ; j 6= i1, i2, · · · , ir. The cycle is denoted

by (ai1 , ai2 , · · · , air) and the elements appear in a fixed cyclic order ai1 , ai2 , · · · , air

are saidto be the elements of the cycle.

(i) Two cycles p and q on the same set S = a1, a2, · · · , an are said to be disjoint if theyhave no common elements.

(ii) A cycle of length 2 is called transposition. The cycle of length 1 may be ignored.

Theorem 1.11.1 Every permutation on a finite set is either a cycle or it can be expressedas a product of disjoint cycles.

Proof: Let S = a1, a2, · · · , an and p : S → S be a permutation on S. Let us con-sider the elements a1, p(a1), p2(a1), · · ·, all these can not de distinct as all of them be-long to a finite set S. Let k be the least positive integer such that pk(a1) = a1. Then,a1, p(a1), p2(a1), · · · , pk−1(a1) are k distinct elements of S, because, if pr(a1) = ps(a1), forsome r, s such that 0 < p < q < r, then pr−s(a1) = p0(a1) = a1 holds and this contradictsthat k is the least positive integer satisfying pk(a1) = a1.

Let us consider a k cycle p1 =(a1, p(a1), p2(a1), · · · , pk−1(a1)

). If k = n, then p = p1 and

the theorem is proved. If k < n, let al be the first element among a2, a3 · · · , an such that al

does not belong to the cycle p1. Let us consider the elements am, p(am), p2(am), · · ·. Neitherof these belong to p1, because, if pi(a1) = pj(am) for some integers i, j then pi−j(a1) = am,a contradiction. Arguing as before we arrive at a cycle p2 of length m, say. If k + m = nthen p is the product of disjoint cycles p1 and p2. If k +m < n, let at be the first elementamong a3, a4 · · · , an, which does not belong to p1 or p2 and proceed as before.Since S is finite set, this process terminates after a finite number of steps and we arrive atthe decomposition of p as the product p1p2 · · · pr of disjoint cycles.

(i) Let p be a permutation on a finite set S = a1, a2, · · · , an. The order of the permu-tation p : S → S is the least positive integer n such that pn = I, I being the identitypermutation.

(ii) The order of an k− cycle is k.

66 Theory of Sets

(iii) The order of a permutation on a finite set is the l.c.m. of the lengths of its disjointcycles.

(iv) Every permutation on a finite set S = a1, a2, · · · , an, n ≥ 2 can be expressed as aproduct of transpositions.

(v) A permutation p is said to be an even permutation, if it can be expressed as a productof even number of transpositions.

(vi) A permutation p is said to be an odd permutation, if it can be expressed as a productof odd number of transpositions.

(vii) The number of even permutations on a finite set S = a1, a2, · · · , an, n ≥ 2 is equalto the number of odd permutations on it.

Ex 1.11.4 Express p =(

1 2 3 4 5 6 7 83 5 4 1 2 6 8 7

)as the product of disjoint cycles.

Solution: Here p is not a cycle. Note that,

p(1) = 3, p2(1) = p(3) = 4, p3(1) = p(4) = 1.

Thus the first cycle is (1 3 4). Since 2 6∈ (1 3 4), we compute

p(2) = 5, p2(2) = p(5) = 2.

Thus the second cycle is (2 5). Also, as 6 6∈ (1 3 4) and (2 5), and p(6) = 6, so the thirdcycle is (6). Again,

p(7) = 8, p2(7) = p(8) = 7.

Thus the fourth cycle is (7 8). All elements have been exhausted. Therefore,

p = (1 3 4)(2 5)(6)(7 8) = (1 3 4)(2 5)(7 8) = (1 4)(1 3)(2 5)(7 8).

Also, we see that, p is the product of 4(even) number of transpositions, so p is an evenpermutation.

Ex 1.11.5 Describe all the permutations on the set 1, 2, 3 and their respective orders.

Solution: There are possible 3! = 6 permutations on the set S = 1, 2, 3, given by,(1 2 31 2 3

)= I,

(1 2 32 3 1

)= (1 2 3),

(1 2 33 1 2

)= (1 3 2),(

1 2 31 3 2

)= (2 3),

(1 2 33 2 1

)= (1 3),

(1 2 32 1 3

)= (1 2).

Thus the even permutations are I, (1 2 3), (1 3 2) and the odd permutations are(2 3), (1 3), (1 2). As I1 = I, so the order of I is 1. As, (1 2 3), (1 3 2) are the cyclesof length 3, the orders of (1 2 3), (1 3 2) are 3. As, (2 3), (1 3), (1 2) are the cycles oflength 2, the orders of (2 3), (1 3), (1 2) are 2.

Enumerable Set 67

1.12 Enumerable Set

Let S and N be the set of real numbers and natural numbers respectively. The set S isdefined as enumerable or de-enumerable or countable if there is a bijection f : S → N .So corresponding to every positive integer n, there exist one and only one element of anenumerable set. This element may be denoted by an or bn or un etc. Thus a countableset can be written as a1, a2, . . . , an, . . .. For example, the set S = 2n|n ∈ N is anenumerable set.

(i) A countable set is an infinite set.

(ii) Obviously an enumerable set is an infinite set. Obviously every infinite set is not enu-merable. If an infinite set be enumerable then it is sometimes said to be an enumerablyinfinite set. It is needless to say that a non enumerable infinite set can not be writtenas : a1, a2, . . . , an, . . .

(iii) A set S is defined to be almost enumerable if it is either finite or enumerably infinite.

(iv) Any sub set of an enumerable set is almost an enumerable.

(v) Any super-set of non-enumerable set is non enumerable.

Theorem 1.12.1 Union of a finite set and an enumerable set is an enumerable.

Proof: Let A be a finite set which can be written as A = a1, a2, . . . , ar in which theelements are increasing order of magnitude. Let B = b1, b2, . . . , bn, . . . be an enumerableset. If A ∩ B = φ, we can define a bijective mapping f : A ∪ B → N such that f(1) =a1, f(2) = a2, . . . , f(r) = ar and then f(r+ k) = bk for k = 1, 2, . . . . That is A ∪B may bewritten as

A ∪B = a1, a2, . . . , ar+1, ar+2, . . . , ar+k, . . .,

where ar+k = bk; ∀ k. Hence A ∪B is an enumerable set. If A ∩B 6= φ, let, B1 = B −A,then B1 ∪A = A ∪B and B1 ∩A = φ. Now, B1 is an infinite subset of B and therefore, B1

is enumerable. Hence B1 ∪A is enumerable and so, A ∪B is enumerable.

Theorem 1.12.2 Union of finite number of enumerable sets is enumerable.

Proof: Let A1, A2, . . . , Ar be (each of) a finite number of enumerable sets. Let

A1 = a11, a12, a13, . . . , a1n, . . .A2 = a21, a22, a23, . . . , a2n, . . .

......

...Ar = ar1, ar2, ar3, . . . , arn, . . .

We can write the elements ofr⊔

i=1

Ai asr⊔

i=1

Ai = a11, a21, a31, . . . , ar1, a12, a22, a32, . . . ,

ar2, . . . , a1n, a2n, a3n, . . . , arn, . . .. Hencer⊔

i=1

Ai is an enumerable set.

Theorem 1.12.3 Union of enumerable set of enumerable sets is enumerable.

68 Theory of Sets

Proof: Let A1, A2, . . . , An, . . . be an enumerable set where each Ai is an enumerable set.

We are to show that∞⊔

i=1

Ai is an enumerable. Let,

A1 = a11, a12, a13, . . . , a1n, . . .A2 = a21, a22, a23, . . . , a2n, . . .

......

...Ar = ar1, ar2, ar3, . . . , arn, . . .

......

...

The elements∞⊔

i=1

Ai are arranged as:

a11; a12, a21; a13, a22, a31; . . .

in which there are several blocks. kth block contains all aij such that i+ j = k + 1 in eachblock the elements are written in the increasing order of the first suffix. In this arrangementthe element aij occupies in (i+ j − 1)th block and occupies ith position in the block. Alsokth block contains exactly k elements. Hence in the above arrangement aij occurs [1 + 2 +

· · ·+ (i+ j − 2)+ i]th position. Hence∞⊔

i=1

Ai is an enumerable set. The set of all positive

rational numbers can be written as Q+ =∞⊔

i=1

Ai, where Ai = ni ; n ∈ N and it is the

union of enumerable set of enumerable sets and hence enumerable. Now, Q+ is similar toQ−, hence Q = Q+ ∪Q− ∪ 0 is enumerable.

Theorem 1.12.4 The set of real numbers is non enumerable.

Proof: We shall first show that the interval 0 < x ≤ 1 is non-enumerable. If possible , letus assume , the set is an enumerable set. If the real numbers lying in the above interval,then they can be written as a1, a2, . . . , an, . . .. Since a real number can be expressed as aninfinite decimal(if we agree not to use recurring in this can be done in only one way). Let,

a1 = 0.a11a12a13 . . .

a2 = 0.a21a22a23 . . .

......

...an = 0.an1an2an3 . . .

Now we construct a number b = 0.b1b2b3 . . ., where br is different from arr, 0 and 9 for allr. Obviously b is a real number lying in 0 < x ≤ 1, and so must itself appear somewherein the succession a1, a2, . . . , an, . . . if this section is to contain all real numbers between0 and 1. But b is different from every ai, since it differs from ai at least in the ith placeof decimal. This contradict the assumption that the given interval is an enumerable set.Hence 0 < x ≤ 1 is non-enumerable. The whole of real numbers is a super set of thisnon-enumerable set and hence is non-enumerable.

(i) The open interval (0, 1) is an non-enumerable. For if this is enumerable then (0, 1)∪1i.e. 0 < x ≤ 1 is also an enumerable which contradict the above result.

(ii) The close interval [0, 1] i.e. 0 ≤ x ≤ 1 being a super set of the non-enumerable set0 < x ≤ 1, is also non-enumerable.

Enumerable Set 69

(iii) Any interval a ≤ x ≤ b is non-enumerable. First we shall prove 0 < x ≤ 1 is a non-enumerable. Let us define a bijection f : x → x−a

b−a ; (b > a) i.e. f(x) = x−ab−a then if

x goes from a to b, f goes from 0 to 1. Hence the interval a < x ≤ b is similar to0 < x ≤ 1.As 0 < x ≤ 1 is non-enumerable so a < x ≤ b is also non-enumerable. Hence its superset [a, b] is also non-enumerable.

(iv) The open interval (a, b) is non-enumerable.

Exercise 1

Section-A[Multiple Choice Questions]

1. A− φ and φ−A respectively(a) A,A′ (b) φ (c) A,φ (d) A.

2. (A ∩B) ∪ (B ∩ C) is equals to(a) B (b) A ∪B ∪ C (c) A ∪ (B ∩ C) (d) (A′ ∩ C ′)′ ∩B.

3. ((((A ∪B) ∩A) ∪B) ∩A) is equals to(a) A (b) B (c) A ∪B (d) A ∩B.

4. (A−B) ∪ (B −A) ∪ (A ∩B) is equals to(a) A ∪B (b) Ac ∪Bc (c) A ∩B (d) Ac ∩Bc.

5. The number of elements in the power set P (S) of the set S = φ, 1, 2, 3 is(a) 2 (b) 4 (c) 8 (d) None of these.

6. Let A be a finite set of size n, the number of elements in the power set of A×A is(a) 22n

(b) 2n2(c) (2n)2 (d) None of these.

7. The number of binary relations on asset with n elements is(a) 2n (b) 2n2

(c) 2n (d) None of these.

8. Suppose A is a finite set with n elements. The number of elements in the largestequivalence relation of A is(a) 1 (b) n (c) n+ 1 (d) n2.

9. The number of equivalence relations of the set 1, 2, 3, 4 is(a) 4 (b) 15 (c) 16 (d) 24

10. The power set 2S of the set S = 3, 1, 4, 5 is

(a) S, 3, 1, 4, 1, 3, 5, 1, 4, 5, 3, 4, φ(b) S, 3, 1, 4, 5(c) S, 3, 3, 1, 4, 3, 5, φ(d) None of these.

11. If there is no onto function from 1, 2, · · · ,m onto 1, 2, · · · , n, then(a) m = n (b) m < n (c) m > n (d) m 6= n.

12. If |A ∪B| = 12, A ⊆ B and |A| = 3, then |B| is(a) 12 (b) 9 (c) ≤ 9 (d) None of these.

70 Theory of Sets

13. Let P (S) denote the power set of the set S. Which of the following is always TRUE?(a) P (P (S)) = P (S) (b) P (S)∩S = P (S). (c) P (S)∩P (P (S)) = φ (d) S 6∈ P (S).

14. The number of relations from A to B with |A| = m and |B|n is(a) mn (b) 2n (c) 2m (d) 2mn.

15. If mρn if m2 = n, then(a) (−3,−9) ∈ ρ (b) (3,−9) ∈ ρ (c) (−3, 9) ∈ ρ (d) (9, 3) ∈ ρ.

16. The relation ρ on Z defined by mρn if m+ n is even is(a) Reflexive (b) Not reflexive (c) Not symmetric (d) Not antisymmetric

17. φ and A×A are(a) Both reflexive (b) Both symmetric (c) Both antisymmetric (d) Both equivalencerelation.

18. The relation aρb if |a− b| = 2 where a and b are real numbers, is(a) Neither reflexive nor symmetric (b) Neither symmetric nor transitive (c) Anequivalence relation (d) Symmetric but not transitive.

19. The relation defined in Z by aρb if |a− b| < 2 is(a) Not reflexive (b) Not symmetric (c) Not transitive (d) An equivalence relation.

20. The relation defined in N by aρb if m2 = n is(a) Reflexive (b) Symmetric (c) Transitive (d) Antisymmetric.

21. The relation defined in N by aρb if m|n or n|m is(a) Not reflexive (b) Not symmetric (c) Not transitive (d) None of these.

22. A relation defined in N by aρb if m and n are relatively prime is(a) A partial ordering (b) Transitive (c) Not transitive (d) An equivalence relation.

23. The ‘subset’ relation on a set of sets is(a) A partial ordering (b) Transitive and symmetric only (c) Transitive and antisym-metric only. (d) An equivalence relation.

24. The binary relation S = φ on set A = 1, 2, 3 is(a) Neither reflexive nor symmetric (b) Symmetric and reflexive (c) Transitive andreflexive (d) Transitive and symmetric.

25. The less than relation, <, on real is

(a) A partial ordering since it is asymmetric and reflexive

(b) A partial ordering since it is anti-symmetric and reflexive

(c) Not a partial ordering because it is not asymmetric and not reflexive

(d) Not a partial ordering because it is not anti-symmetric and not reflexive.

26. A partial order ≤ is defined on the set S = x, a1, a2, · · · , an, y as x ≤ ai, for all i andai ≤ y for all i, where n ≥ 1. The number of total orders on the set S which containthe partial order ≤ is

(a) 1 (b) n (c) n+ 2 (d) n!.

27. The number of possible partial ordering on a, b, c in which a ≤ b is(a) 3 (b) 4 (c) 5 (d) 6

Enumerable Set 71

QQQ

bbb

a

b

e

c

d

f

g

Figure 1.32: Poset

28. In a lattice defined by the Hasse diagram given below, how many (Fig. 1.32) compo-nents does the element e have?

(a) 2 (b) 3 (c) 0 (d) 1.

29. The maximal and minimal elements of poset given by the Hasse diagram (Fig. 1.33)are

%%

%%%@

@

@@

@@

1

2

34

5

6

Figure 1.33: Poset

(a) Max=5,6; Min.=2 (b) Max.=5,6; Min.=1 (c) Max.=3,5: Min.=1,6 (d) None ofthe above.

30. The greatest and least element of the poset given by the Hasse diagram (Fig. 1.34) are

@@

@1 23

4 5

Figure 1.34: Poset(a) Greatest=4,5; least=1,2 (b) Greatest=5; least=1 (c) Greatest=None; least=1 (d)None of the above.

31. If a ≤ b ≤ c in a lattice L(a) a ∨ b = b ∧ c (b) a ∧ b = b ∨ c (c) a ∨ b = b ∨ c (d) a ∧ b = b ∧ c

32. In a lattice L, a ∨ b = b. Then(a) a ≥ b (b) b ≤ a (c) a ∧ b = a (d) None of these.

33. Let X = 2, 3, 6, 12, 24, let ≤ be the partial order defined by x ≤ y if x divides y.Then number of edges in the Hasse diagram of (X,≤) is

(a) 3 (b) 4 (c) 5 (d) None of these.

34. In a lattice L, ((a ∧ b) ∨ a) ∧ b is(a) a ∧ b (b) a ∨ b (c) (a ∧ b) ∨ a (d) ((a ∨ b) ∧ a) ∨ b).

35. In a lattice, if a ≤ b and c ≤ d, then(a) b ≤ c (b) a ≤ d (c) a ∨ c ≤ b ∨ d (d) None of these.

36. (1, 2, 5, 10, 15, a, |) is a lattice if the smallest value for a is(a) 150 (b) 100 (c) 75 (d) 30.

37. S = 1, 2, 3, 12 and T = 1, 2, 3, 24, then

(a) S and T are sublattice of (D24, |)

72 Theory of Sets

(b) Neither S nor T are sublattices of (D24, |)(c) S and T are sublattices of (1, 2, 3, 12, |)(d) S and T are sublattices of (1, 2, 3, 24, |)

38. S = 1, 2, 4, 8 and T = 1, 3, 9, then

(a) Only S is a sublattice of (D72, |)(b) Only T is a sublattices of (D72, |)(c) Both S and T are sublattices of (D72, |)(d) Neither S nor T is a sublattice of (D72, |).

39. If the posets P1 and P2 are given in Fig. 1.35, then

,,

,@@@

%%llP1 P2

Figure 1.35: Poset for Self-Test(a) P1 and P2 are lattices (b) P1 is lattice (c) P2 is lattice (d) None of them is alattice.

40. (1, 2, 4, 6, 12, 24, |) is(a) Not a poset (b) A lattice (c) A complemented lattice (d) A lattice which is notcomplemented.

41. The lattice given in Fig. 1.36 is

ZZ

ZZ

l

lll

A

AAJ

JJ

Figure 1.36: Hasse diagram(a) Complemented but not distributive (b) Distributive but not complemented (c)Both complemented and distributive (d) Neither complemented nor distributive.

42. The lattice given in Fig. 1.37 is

cc

L

LL

Figure 1.37: Hasse diagram(a) Complemented but not distributive (b) Distributive but not complemented (c)Both complemented and distributive (d) Neither complemented nor distributive.

43. If a, b, c ∈ L, L being a distributive lattice, then(a) (a ∨ b) ∧ c ≤ a ∨ (b ∧ c) (b) (a ∨ b) ∧ c ≤ (a ∧ b) ∧ c (c) (a ∨ b) ∧ c = a ∨ b (d)(a ∨ b) ∧ c = c.

44. (D45, |) is not distributive since(a) 1, 3, 5, 45 is a sublattice of D45 (b) 1, 3, 9, 45 is a sublattice of D45 (c)1, 5, 9, 15, 45 is a sublattice of D45 (d) 1, 5, 15, 45 is a sublattice of D45.

Enumerable Set 73

45. A chain with 3 elements is(a) Complemented but not distributive (b) Distributive but not complemented (c)Both complemented and distributive (d) Neither complemented nor distributive.

46. The lattice given in Fig. 1.38 is

HHH

@

@

"

""bb

Figure 1.38: Hasse diagram(a) Complemented but not distributive (b) Distributive but not complemented (c)Both complemented and distributive (d) Neither complemented nor distributive.

47. In a distributive lattice, if a ∧ b′ = 0, then(a) b ≤ a (b) a ≤ b (c) a′ ∨ b = 0 (d) a ∨ b′ = 1.

48. (1, 2, 3, 6, 12, 30, 60, |) is(a) Complemented but not distributive (b) Distributive but not complemented (c)Both complemented and distributive (d) Neither complemented nor distributive.

49. (1, 1, 2, 1, 3, 1, 2, 3,⊆) is(a) Complemented but not distributive (b) Distributive but not complemented (c)Both complemented and distributive (d) Neither complemented nor distributive.

50. The number of functions from an m element set to an n element set is(a) m+ n (b) mn (c) nm (d) m ∗ n.

51. Let A and B be sets with cardinalities m and n respectively. The number of one-onemappings from A and B, when m < n, is

(a) mn (b) nPm (c) nCm (d) None of these.

52. It is given that there is exactly 97 functions from the set A to B. From this one canconclude that

(a) |A| = 1, |B| = 97 (b) |A| = 97, |B| = 1 (c) |A| = 97, |B| = 97 (d) None ofthese.

53. Let f : <×< → <×< be a bijective function defined by f(x, y) = (x+ y, x− y). Theinverse function of f is given by

(a) f−1(x, y) =( 1x+ y

,1

x− y

)(b) f−1(x, y) = (x − y, x + y) (c) f−1(x, y) =(

x+y2 , x−y

2

)(d) f−1(x, y) =

(2(x− y), 2(x+ y)

)54. The range of g f when f : Z → Z and g : Z → Z are defined by f(n) = n + 1 and

g(n) = 2n is(a) Z (b) Z+ (c) The set of all odd numbers (d) The set of all even numbers.

55. If f, g, h are functions from < → < defined by f(x) = x+1, g(x) = x2+2, h(x) = 2x+1,then (h g f)(2) is(a) 20 (b) 23 (c) 21 (d) 22.

56. If f, g, h are functions from Z → Z defined by f(x) = x−3, g(x) = 2x+3, h(x) = x+3,then g f h is(a) f (b) g (c) h (d) h g f.

74 Theory of Sets

57. If f is a function from Z → Z defined by f(x) = x+ 2, then f−3(10) is(a) 7 (b) 6 (c) 5 (d) 4.

58. If f and g are functions from <+ to <+ defined by f(x) = ex and g(x) = x− 3, then(g f)−1(x) is(a) log(3 + x) (b) log(3− x) (c) e3−x (d) log(x− 3).

59. f(A1 ∩A2) = f(A1) ∩ f(A2) holds(a) if f is injective (b) If f is surjective (c) If f is any function (d) For no function.

60. The relation (x, y) ∈ <2 : ax+ by = c is an invertible function from < → < if(a) a 6= 0 (b) b 6= 0, a 6= 0 (c) c 6= 0 (d) c 6= 0, a 6= 0.

61. The number of invertible functions from 1, 2, 3, 4, 5 to a, b, c, d, e is(a) 55 (b) 25 (c) 5! (d) None of these.

62. The number of odd permutations of the set 1, 3, 5, 7, 9 is(a) 15 (b) 30 (c) 60 (d) 120

63. Which one of the following is an even permutation?(a) f = (1, 2, 3)(1, 2) (b) f = (1, 2)(1, 3)(1, 4)(2, 5) (c) f = (1, 2, 3, 4, 5)(1, 2, 3)(4, 5) (d)None of these

64. Which power multiplying itself of the permutation f =(

1 2 3 41 3 4 2

)gives

(1 2 3 41 2 3 4

)(a) f (b) f2 (c) f3 (d) f4

Section-B[Objective Questions]

1. Let S be a non-empty set and P (S) be its power set. Show that there exists nobijection from S to P (S).

2. Let f : X → Y be a map. Show that f is surjective if there exists a map h : Y → Ysuch that f h = IY (identify map).

3. Let f : X → Y be a map. Show that f is injective if and only if there is a mapg : Y → Y such that g f = IX(identify map).

4. Consider the map defined by f(x, y) = (x, 0). Let A = (x, y) ∈ <2 : x − y = 0 andB = (x, y) ∈ <2 : x− y = 1. Show that f(A ∩B) 6= f(A) ∩ f(B).

5. Show that every infinite set contains a countable subset.

6. Let N be the set of positive integers and a ≤ b be the divisibility relation defined bya ≤ b if and only if b is divisible by a. Show that (N ,≤) is a poset.

7. Let ≤ denote the natural ordering in <. Show that the poset (<,≤) has neither minimalnor maximal elements.

8. Show that the following posets are not lattices:

(a) (2, 3, 5, 30, 60, 120, 360, |)(b) (1, 2, 3, 4, 6, 8, 12, |)(c) (2, 3, 6, 12, 24, 36, |)(d) (1, 2, 3, 6, 12, 30, |).

Enumerable Set 75

Section-C[Long Answer Questions]

1. A and B be any two sets, prove that the sets A − B,A ∩ B and B − A are pairwisedisjoint. [ VH’94, ’95, ’99]

2. (a) If A,B,C be three nonempty sets such that A ∪B = A ∪C and A ∩B = A ∩C,prove that B = C. [VH’00, CH’05, ’01, BH’03 ]

(b) If A,B,C be three nonempty sets such that A∩C = B ∩C and A∩C ′ = B ∩C ′,prove that B = C.

3. For each n ∈ N , let An = [n, , 2n] = x ∈ Z : n ≤ x ≤ 2n. Find the value of8⋃

n=4An.

4. Prove the following set theoretic statement if it true or give counter example to disproveit.

(a) A× (B − C) = (A×B)− (A× C). [CH’09]

(b) (AB)′ = (B −A)′. [CH’08,10]

5. For the subsets A,B,C of an universal set U , prove the following:

(a) A− (B − C) = (A−B) ∪ (A ∩ C).

(b) A− (B ∪ C) = (A−B) ∩ (A− C). [KH’07]

(c) A− (B ∩ C) = (A−B) ∪ (A− C). [VH’96]

(d) A× (B ∪ C) = (A×B) ∪ (A× C).

(e) A× (B ∩ C) = (A×B) ∩ (A× C).

(f) (A−B)× C = (A× C)− (B × C).

(g) (A ∪B)c = Ac ∩Bc and (b) (A ∩B)c = Ac ∪Bc. [KH’08]

6. Prove that

(a) (A ∩B) ∩B = φ

(b) A−B, A ∩B and B −A are mutually disjoint

(c) (A−B) ∪A = A.

(d) If A ⊆ B then show that A ∪ (B −A) = B.

7. (a) (a) Show that A ⊆ B ⇔ A−B = φ.

(b) If A ⊆ B and C is any set then show that nA ∪ C ⊆ B ∪ C.

(c) If A ∩X = A ∩ Y and A ∪X = A ∪ Y then prove that X = Y .

(d) If A ∩ C = B ∩ C and A ∩ C ′ = B ∩ C ′, then prove that A = B.

8. Simplify the following expression by using the laws of algebra of sets.

(a) [(A ∪B) ∩ Cc ∪Bc]c

(b) (Ac ∩Bc ∩ C) ∪ (B ∩ C) ∪ (A ∩ C)

(c) A ∩ (B ∩ C) ∩ (Ac ∩ (Bc ∩ Cc))

(d) (A ∩B)c ∪ (Ac ∩Bc).

(e) (A ∩B′) ∪ (B ∩ C).

76 Theory of Sets

9. Let A = 1, 2, 3, 4. List all subsets B of A such that 1, 2 ⊆ B.

10. Let A = 1, 2, 3 and B = a, b. Find A×B and B×A and verify that A×B 6= B×A.

11. Prove that

(a) (A×B) ∩ (C ×D) = (A ∩ C)× (B ∩D).

(b) A× (B ∩ C) = (A×B) ∩ (A× C).

(c) (A−B)× C = (A× C)− (B ×D).

(d) (A× C)− (B ×D) = (A×B)− (C ×D) ∪ (A×B)× (C −D) ∪ (A−B)× (C ∩D).

12. Prove that

(a) A∆B = A∆C ⇒ B = C

(b) A ∩ (B∆C) = (A ∩B)∆(A ∩ C).

13. If A and B are subsets of a set X, then prove that A ⊆ B ⇔ X −B ⊆ X −A.

14. S × T = T × S ⇔ S = T or one is phi.

15. Find the power set of the set A = a, b, c, 1.

16. (a) If the number of elements of the set A is n then show that the number of elementsof the power set P (A) is 2n.

(b) If A and B are two non-empty sets having n elements in common, then prove thatA×B and B ×A have n2 elements in common.

17. If the set X has 5 elements, then find n(P (X)) and P (P (P (φ))).

18. There are 1000 students in a college studying Physics, Chemistry and Mathemat-ics. 658 study Physics, 418 study Chemistry and 328 study Mathematics. Use Venndiagram to find the number of students studying Physics or Mathematics but notChemistry. [JECA’03]

19. Among 100 students, 32 study Mathematics, 20 study Physics, 45 study Biology, 15study Mathematics and Biology, 7 study Mathematics and Physics, 10 study Physicsand Biology and 30 do not study any of 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.

20. In a city, three daily newspaper A, B and C are established. 42 percent of the peoplein that city read A, 6 percent read B, 60 percent read C, 24 percent read A and B,34 percent read B and C, 32 percent read C and A, 8 percent do not read any of thethree newspapers. Find the percentage of the persons who read all the three papers.

21. Let A = 1, 2, 3, 4, 5, 6. Determine whether or not each of the following is a par-tition of A. (a) P1 = 1, 2, 3, 1, 4, 5, 6 (b) P2 = 1, 2, 3, 5, 6 (c) P3 =1, 3, 5, 2, 4, 6 (d) P4 = 1, 3, 5, 2, 4, 6.

22. Let A, B, C be three finite sets of U . Show that

(a) |A−B| = |A| − |A ∩B|(b) |A ∪B| ≤ |A|+ |B|(c) |A ∪B ∪ C| ≤ |A|+ |B|+ |C|

Enumerable Set 77

(d) |A ∪B| ≤ |A|+ |B| − |A ∩B|(e) |A ∪B ∪ C| ≤ |A|+ |B|+ |C| − |A ∩B| − |A ∩ C| − |B ∩ C|+ |A ∩B ∩ C|.

23. Let A = 1, 2, 3, 4. For each of the following relations on A, decide whether it isreflexive, symmetric, antisymmetric or transitive

(a) (1, 3), (3, 1)(b) (2, 2), (1, 1)(c) (1, 2), (1, 4), (2, 3)(d) (1, 1), (2, 2), (3, 3), (4, 4), (1, 3), (3, 1).

24. The following relations are defined on the set of real numbers. Find whether theserelations are reflexive, symmetric or transitive

(a) aRb iff |a− b| > 0

(b) aRb iff 1 + ab > 0

(c) aRb iff |a| ≤ |b|(d) aRb iff |a| ≥ |b|.

25. In each of the following cases, examine whether the relation ρ is an equivalence relationon the set given below

(a) ρ = (a, b) ∈ Z × Z : |a− b| ≤ 3(b) ρ = (a, b) ∈ Z × Z : a− b is a multiple of 6.(c) xρy if and only if |x− y| ≤ y;x, y ∈ <.

26. Let Z∗ be the set of nonzero integers and S = Z×Z∗. Let ρ = x : x = ((r, s), (t, u)) ∈S × S with ru = st. Prove that ρ is an equivalence relation. [CH‘05]

27. A relation ρ on the set of integers Z is define by ρ = (a, b): a, b ∈ Z and |a− b| ≤ 5.Is the relation reflexive, symmetric and transitive? [WBUT 07]

28. Determine whether the relation ρ on the set A of all triangles in the plane defined byρ = (a, b) : triangle a is similar to the triangle b is an equivalence relation.

29. In the set of all points in a plane show that the relation of equidistance from the originis an equivalence relation.

30. A relation ρ on the set of integers Z is defined as aρb iff (a − b) is divisible by m (apositive integer). Show that ρ is an equivalence relation.

31. Determine whether the relation ρ is an equivalence relation on the set of positiveintegers Z+.

(a) aρb iff a = 4b.

(b) aρb iff a = b2.

32. If A and B be equivalence relation in a set X, show that A ∩ B is an equivalencerelation.

33. Let H be a subgroup of a group G. Show that the relation ρ = (a, b) ∈ G × G :a−1b ∈ H is an equivalence relation on the set G.

34. A relation ρ is defined on a set Z by aρb if and only if 2a + 3b is divisible by 5 fora, b ∈ Z. Prove that ρ is an equivalence relation. [ VH‘97, ‘05]

78 Theory of Sets

35. A relation ρ is defined on a set Z by aρb if and only if either a = b or both a, b arepositive, for a, b ∈ Z. Prove that ρ is an equivalence relation on Z. Write down thedistinct equivalence classes of ρ.

36. A relation ρ is defined on a set Z by aρb if and only if a−b is divisible by 5 for a, b ∈ Z.Prove that ρ is an equivalence relation. Write down the distinct equivalence classes ofρ. [VH‘99, ‘03]

37. A relation ρ is defined on a set Z by aρb if and only if a+ b is even, for a, b ∈ Z. Provethat ρ is an equivalence relation.

38. For natural numbers a and b, define aρb iff a2+b is even. Prove that ρ is an equivalencerelation on N .

39. For a, b ∈ R, define aρb iff a− b ∈ Z. Show that ρ is an equivalence relation.

40. For any integers a, b define

(a) aρ1b iff 2a+ 3b = 5n for some integer n.

(b) aρ2b iff 3a+ 4b is divisible by 7.

41. For a, b ∈ Z define

(a) aρ1b iff a2 − b2 is divisible by 3.

(b) aρ2b iff 3a+ b is multiple of 4.

42. A relation ρ is defined on a set Z by aρb if and only if a2− b2 is an even integer. Provethat ρ is an equivalence relation on Z and write down the equivalence classes. [BH‘02]

43. A relation ρ is defined on a set < by aρb if and only if a− b is rational. Prove that ρ isan equivalence relation on < and the set of equivalence classes is uncountable.[BH‘03]

44. A relation ρ is defined on a set Z by aρb if and only if ma+ nb is divisible by (m+ n)for a, b ∈ Z. Prove that ρ is an equivalence relation. Find out an infinite sequence ofpositive integers lying in the equivalence class containing 0. [ BH‘05]

45. Let A be the set of all straight lines in the plane.

(a) aρ1b iff a2 − b2 a is parallel to b

(b) aρ2b iff 3a+ b a is perpendicular to b.Show that ρ1 is is an equivalence relation but ρ2 is not.

46. Determine which of the following define equivalence relations in <2.

(a) (a, b)ρ(c, d) iff a+ 2b = c+ 2d.

(b) (a, b)ρ(c, d) iff a2 + b = c+ d2.

(c) (a, b)ρ(c, d) iff ab = cd.

(d) (a, b)ρ(c, d) iff ab = c2.

47. Let S be a finite set and let f : S → S. If f is one=to-one then show that f is onto.Examine whether this remains true if the set S is infinite. [ CH: 08]

48. ρ1 is a relation on Z such thatρ1 = (a, b) : a, b ∈ Z; a− b = 5n, n ∈ Z.

Show that ρ1 is an equivalence relation. If ρ2 be another relation defined byρ2 = (a, b) : a, b ∈ Z; a− b = 3n, n ∈ Z.

Show that the relation ρ1 ∪ ρ2 is symmetric but not transitive.

Enumerable Set 79

49. Given A = 1, 2, 3, 4 and B = x, y, z. Let ρ be the relation from A to B defined asρ = (1, x), (2, y), (2, z), (3, z).(a) Find the inverse of the relation ρ−1 of ρ.

(b) Determine the domain and range of ρ.

50. Given A = 1, 2, 3, 4. Let ρ be the relation on A and is defined as

ρ = (1, 1), (2, 2), (2, 3), (3, 2), (4, 1), (4, 4).(a) Draw its digraph, (b) Is ρ is equivalence relation?

51. If ρ is an equivalence relation, then prove that ρ−1 is also an equivalence relation inthe set A.

52. If R and S are equivalence relations in the set A then show that R ∩ S is also anequivalence relation in A.

53. Let A = 1, 2, 3, 4. Consider two equivalence relations

R = (1, 2), (1, 1), (2, 1), (2, 2), (3, 3), (4, 4), (4, 5), (5, 4), (5, 5)and S = (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 3), (3, 1), (4, 5), (5, 4).Determine the partitions corresponding to following relations

(a) R−1, (b) R ∪ S, (c) R ∩ S.

54. Let ρ be an equivalence relation on the set A = a, b, c, d define by the partitionP = a, b, c, d. Determine the elements of equivalence relation and also findthe equivalence classes of ρ.

55. For the partition P = a, b, c, d, e, write the corresponding equivalence relationon the set A = a, b, c, d, e.

56. Let S = n ∈ N : 1 ≤ x ≤ 20. Define a relation ρ on A by aρb iff 5 divides a − bfor all a, b ∈ S. Show that ρ is an equivalence relation on S. Find all the equivalenceclasses.

57. Let A be a finite set with n elements. Prove that the number of reflexive relationsthat can be defined on S is 2(n2−n), the number of symmetric relations is 2n(n+1)/2

and the number of relations that are both reflexive and symmetric is 2n(n−1)/2.

58. Let A and B be two non-empty sets with cardinality m and n respectively. Show thatthe number of possible relations from A to B is 2mn − 1.

59. Let A = 1, 2, 3, 4 and B = a, b, c. Determine whether the relation R from A to Bis a function. If it a function, find its domain and range.

(a) R = (1, a), (2, a), (3, b), (2, b), (b) R = (1, c), (2, a), (3, b),(c) R = (1, a), (2, b), (3, c), (4, b), (1, b), (d) R = (1, c), (2, a), (3, a), (4, c).

60. If A = 2, 3, 4, B = 2, 0, 1, 4 and relation f is defined as f(2) = 0, f(3) = 4,f(4) = 2. Find out whether it depends a mapping.

61. Let f : A → B and g : B → C be two mappings. Show that, if g f is injective, f isinjective but g is not so. [ CH: 09,10]

62. Let f : A → B and g : B → C be both surjective, then prove that the compositemapping g f : A→ C is surjective. Give an example to show that f is not surjectiveif g f : A→ C is surjective.

80 Theory of Sets

63. A mapping f : N ×N → N is defined by f(m.n) = 2m.3n. Show that f is injectivebut not surjective. [ CH: 07]

64. If Z+ is the set of positive integers and f(n) = 2n + 1 then show that f : Z+ → Z+

is one-one into mapping.

65. If R is the set of real numbers and f(x) = x2 + 7 then prove that f : R → R ismany-one into mapping.

66. If A and B be two sets having n distinct elements, show that the number of bijectivemappings from A to B is n!. [ CH: 07]

67. Show that the function f defined by f : Q → Q such that f(x) = 3x+4 for all x ∈ Q isone-one onto, where Q is the set of rational numbers. Also find a formula that definesthe inverse function f−1.

68. A function f : Z → Z is defined by:

f(x) =x

2; if x is even

= 7; if x is odd .

Find the left inverse of f ; if it exists. [ CH: 10]

69. Consider the sets A = k, l,m, n and B = 1, 2, 3, 4. Let f : A → B such that (a)f = (k, 4), (l, 1), (m, 2), (n, 3), (b) f = (k, 1), (l, 2), (m, 1), (n, 2).

Determine whether f−1 is a function.

70. Let f(x) =

1, if x is a rational0, if x is a irrational be a function from R to R. Find f(0.5) and

f(√

2).

71. Is the mapping f : X → Z defined by f(x) = 2x−11−|2x−1| is a bijective? Here z = set of

integers and X = x : 0 < x < 1.

72. If A = 1, 2 and B = a, b, find all relations from A into B. Delete which of theserelations are functions from A to B.

73. Show that the following functions are neither injective nor surjective.

(a) f : R→ R given by f(x) = |x|+ 1 x∀R

(b) f : R→ R given by f(x) = sinx x∀R.

74. Show that the following functions are injective but not surjective.

(a) f : Z → Z given by f(x) = 2x+ 3 x∀Z

(b) f : N → N given by f(x) = sinx x∀Z.

75. Show that the following functions are surjective but not injective

(a) f : Z → 1,−1 given by f(n) = (−1)n, n ∈ Z

(b) f : N → Z10 given by f(n) = [r], where r is the remainder when n is divided by10.

Enumerable Set 81

76. Determine which of the following functions are bijective

(a) f : R→ R where f(x) = |x|+ 1 x∀R(b) f : Z → Q where f(x) = 2x x∀R(c) f : R→ R where f(x) = x2 − 3x+ 4 x∀R(d) f : R→ S where f(x) = x

1+|x| where S = x ∈ R : −1 < x < 1.

77. Let A be a finite set and let f : A→ Bbe a surjective function. show that the numberof elements of B cannot be greater than that of A.

78. Let A = 1, 2, 3. Find all possible bijective functions from A into itself.

79. Let |A| = n. Prove that there can be n! different bijective functions on A.

80. Consider the function f : R → R and g : R → R where f(x) = x + 2 and g(x) = x2.Find fog and gof .

81. Suppose f and g are two functions from R into R such that fog = gof . Does itnecessarily imply that f = g? Justify your answer.

82. Let f, g and h : R→ R defined by f(x) = x+ 2, g(x) = 11+x2 , h(x) = 3.

Compute gof , fog, gohof , gof−1of and f−1ogof .

83. Let A = 1, 2, 3, 4 and define functions f, g : A→ A byf = (1, 3), (3, 2), (3, 1), (4, 2) and g = (1, 4), (2, 3), (3, 1), (4, 2).Find fog, gof , g−1ofog, fog−1og and gog−1of .

84. Let A = a, b, c. Define f : A → A such that f = (a, b), (b, a), (c, c). Find (a) f2,(b) f3, (c) f4. [Hints: f3 = fofof ]

85. Define f : Z → N by f(x) =

2|x|, if x < 02x+ 1, if x ≥ 0. Show that f has an inverse and find

f−1(25), f−1(20).

86. If f : x→ x+ 1 and g : x→ 3x be mappings of the set of integers into itself, examinewhether each of f and g is surjective, injective. Also, show that fg 6= gf. [VH‘95, ‘05]

87. Prove that the mapping f : < → < is defined by f(x) = 2x + 3, x ∈ < is a bijectivemapping. [ VH‘96]

88. Prove that the mapping f : Q → Q is defined by f(x) = 5x + 2, x ∈ Q is a bijectivemapping. [ VH‘03]

89. Test whether the mapping f : C → < defined by f(x) = |x|, x ∈ C is a bijectivemapping. [ VH‘98]

90. Show that the mapping f : < → <, defined by f(x) = x3 − x2 is surjective but notinjective. [ BH‘03]

91. A mapping f : < → < is defined by f(x) = xx2+1 , x ∈ <. Examine whether it is a

bijective mapping. [ VH‘97, CH‘05]

92. A mapping f : Z → Z is defined by f(x) = x2 + x − 2, x ∈ <, find f−1(4) andff(−2). [ VH‘01]

82 Theory of Sets

93. A mapping f : < → < is defined by f(x) = x2 + x − 2, x ∈ <, find f−1(−8) andf−117, 37.

94. For the mappings f(x) = x2 and g(x) = 1 + x, x ∈ <, find the set x ∈ < : fg(x) =gf(x).

95. For the mappings f(x) = |x| + x and g(x) = |x| − x, x ∈ <, find fg, gf and the setx ∈ < : fg(x) = gf(x). BH‘04

96. For the mappings f : N → Q; f(x) = 32x+ 1 and g : Q→ Q; g(x) = 6x,, examine with

justification if fg and gf are defined.

97. Let the mappings f, g : Z → Z be defined by f(x) = (−1)x and g(x) = 2x, x ∈ Z, findgf and fg.

98. Prove that the set of rational numbers in [0, 1] is countable. JECA‘06

99. If f =(

1 2 3 42 4 1 3

)and g =

(1 2 3 44 1 2 3

), find fg, f−1, g−1 and prove that (fg)−1 =

g−1f−1.

100. Examine whether the permutations(

1 2 3 4 5 63 1 5 6 4 2

),(

1 2 3 4 5 6 7 8 94 7 9 1 8 2 6 3 5

)are odd or

even.

101. Let X = a, b, c and f, g : X → X be defined by f(a) = b, f(b) = c, f(c) = a andg(a) = a, g(b) = c, g(c) = b. Show that fg 6= gf.

102. A relation ρ is defined on a set Z by aρb if and only if b is the divisor of a, for a, b ∈ Z.Prove that ρ is an partial order relation.

103. Give an example of a partially ordered set which is a lattice and another which is notlattice. Justify your answer.

104. Let X = 0, 1, 2, · · · , 100, define a binary relation ‘≤’ on X by x ≤ y if and only ifx divides y. (i) Prove that, it is a partially ordered set. Find the least and greatestelement of (X,≤) if they exist. (ii) Is (X,≤) a lattice. Justify the answer.

Chapter 2

Theory of NumbersThe integers are the main elements of mathematics. The theory of numbers is concerned,at least in its elementary aspects, with basic properties of the integers and more particu-larly with the positive integers 1, 2, 3, . . ., known as natural numbers. Here we shall discusssome basic properties of integers including well-ordering principle, mathematical induction,Euclidean algorithm representation of integers etc.

2.1 Number System

Number systems are basically of two types (i) Non-positional number system, (ii) Positionalnumber system.

2.1.1 Non-positional Number System

In this number system, people counted on figures in the early days, when ten figures werenot adequate, small stones, balls, sticks, pebbles were used to indicate values. This methodof counting uses an additive approach or the non-positional number system. Each symbolrepresent the same value regardless of its position in the number and the symbols are simplyadded to find out the value of the particular number. Since it is very difficult to performarithmetic with such a number system, positional number system were developed as thecenturies passed.

(i) In this system, we have symbols (Roman number system) I for 1, II for 2, III for 3etc. and so on.

(ii) An example of earlier types of notation can be found in Roman numerals, which areessentially additive: III = I + I + I, XXV = X +X + V. New symbols X,C,M, . . .etc. were used as the numbers increased in value: thus rather than IIIII is equals to5.

(iii) The only importance of position in Roman numbers lies in whether a symbol precedesor follows another symbol, i.e., IV = 4, while V I = 6.

(iv) The clumsiness of this system can be seen easily if we try to multiply XII by XIV .Calculating with roman numbers was to difficult that early mathematicians were forcedto perform arithmetic operations almost entirely on abaci, or counting boards, trans-lating their results back to Roman numeral form.

Some of such roman number system are given below in the tabular form:

1 2 3 4 5 6 7 8 9 10I II III IV V V I V II V III IX X

83

84 Theory of Numbers

11 · · · 39 40 41 · · · 49 50 51 · · · 89XI · · · XXXIX XL XLI · · · XLIX L LI · · · LXXXIX90 91 · · · 99 100 · · · 200 300 400 500 600XC XCI · · · XCIX C · · · CC CCC CD D DC

700 800 900 1000 1100 1200 1300 1400 1500 1600 1700DCC DCCC CM M MC MCC MCCC MCD MD MDC MDCC

1800 1900 2000 5000 10000 50000 100000 500000 1000000MDCCC MCM MM V X L C D M

Pencil and paper computations are unbelievably intricate and difficult in such systems. Infact the ability to perform such operations as addition and multiplication was considered agreat accomplishment in earlier civilizations.

2.1.2 Positional Number System

In a positional number system, there are only a few symbols called digits and these symbolsrepresent different values depending on the position they occupy in the number. In thisnumber system the position of the digit is very important, the digit will view be it values.The value of each digit in such a number is determined by three considerations

(i) the digit itself

(ii) the position of the digit in the number

(iii) the base of the number system.

The positional number system are groups as(i) Decimal number system

(ii) Binary number system

(iii) Octal number system

(iv) Hexadecimal number system.

There are two characteristic of all number systems that are suggested by the value of thebase

(i) The total number of digits (symbols) available to represent numbers in a positionalnumber system. Commonly base as a subscript notation. In all number system thevalue of the base determines the total number of different symbol or digit available inthe number system.

(ii) The second characteristic is that the maximum value of a single digit is always equalto one less than the value of the base. For example, 0011 base 2 first digit is 0 lessthan base 2.

Decimal Number system : In this number system, the base or radix is equal to 10because there are altogether ten symbols or digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are used. In day today life this number system are more useful. The general rule for representing numbers inthe decimal system by using positional notation as

an−1 × 10n−1 + an−2 × 10n−2 + . . .+ a1 × 10 + a0 (2.1)

Natural Number 85

is expressed as an−1, an−2, . . . , a1, a0 where n is the number of digits to the left of the decimalpoint. In this number system, we can start counting from 0 for the converting purpose, wecan (−1) from the total number of digit of the number. For example,

(2586)10 = 2× 104−1 + 5× 104−2 + 8× 104−3 + 6× 104−4.

The other positional number systems, may consult the Author’s Numerical Book.

2.2 Natural Number

A set N of natural numbers is defined by a set in which the following axioms (known asPeano’s axiom) are satisfied:

(i) every element a ∈ N has a unique successor denoted by a∗, a∗ ∈ N .

(ii) If two natural numbers have equal successor, then they are themselves equal, i.e.,

a∗ = b∗ ⇒ a = b; ∀a, b ∈ N .

(iii) ∃ an unique element (denoted by 1) in N , which has no predecessor.

(iv) If M ⊆ N such that 1 ∈ M and k ∈ M ⇒ k∗ ∈ M , then M = N . This is calledprinciple of mathematical induction or first principle of finite induction.

The set of numbers 1, 2, 3, . . . is called natural numbers and is denoted by N = 1, 2, 3, . . ..

2.2.1 Basic Properties

We are acquainted with the following familiar properties of integers.

(i) Closure law : a+ b ∈ N ; ∀a, b ∈ N

(ii) Associative law: (a+ b) + c = a+ (b+ c); ∀a, b, c ∈ N

(iii) Identity law : a+ 0 = 0 + a, a.1 = 1.a = a; ∀a ∈ N

(iv) Additive inverse law : a+ b = b+ a; ∀a, b ∈ N

(v) Commutative law : a+ b = b+ a; ∀a, b ∈ N

(vi) Distributive law : a.(b+ c) = a.b+ a.c; ∀a, b, c ∈ N

(vii) Cancellation law : a+ b = a+ c⇒ b = c; ∀a, b ∈ N

The set of all natural numbers or is closed with respect to addition and multiplication butnot closed with respect to subtraction and division.

2.2.2 Well Ordering Principle

The well ordering principle plays an important role in the proof of the next sections. Theprinciple states that,

every non empty subset of N , of natural numbers have unique least element.Let S be a non empty subset of the set N of natural numbers. Thus ∃ m ∈ S such that

m ≤ a ,∀a ∈ S; m is called the least element of N .From well ordering principle it follows that, every descending chain of natural numbers

must terminate.

86 Theory of Numbers

Theorem 2.2.1 There is no integer m satisfying 0 < m < 1.

If possible, there lie integers in (0,1), then consider the set S =n ∈ Z : 0 < n < 1

.By

assumption S is non empty subset of N . So by well ordering principle it has least elementsay c so that 0 < c < 1 and c ∈ N . Therefore, 1− c > 0 and also c > 0. Thus,

c(1− c) > 0 ⇒ c− c2 > 0 ⇒ 0 < c2 < c < 1.

Thus c2 ∈ Z and 0 < c2 < 1. Hence c2 ∈ S,but c2 < c which contradict the fact that c isthe least element in S. This contradiction shows that our assumption is wrong. Hence thereis no integer satisfying 0 < m < 1.

2.2.3 Mathematical Induction

Form 1: If M, a set of positive integers, be ⊆ N with two specific properties

(i) the integer 1 ∈M and

(ii) k ∈M ⇒ k + 1 ∈M

then M = N .Proof: Let F = N −M , it is sufficient to show that F = φ, null set. Let us suppose F 6= φ,then F is a set of all positive integers not is M . So by well ordering principle, it has an leastelement q(say), then q ∈ F . Since 1 ∈M , so q 6= 1 so q − 1 ∈ N . Since 0 < q − 1 < q and qis the least element in F . So q − 1 ∈M . The hypothesis (ii) gives

q − 1 ∈M ⇒ (q − 1) + 1 ∈M ⇒ q ∈M,

which is a contradiction. This contradiction shows that F = φ and consequence thatM = N .Form 2: Let P (n) be a mathematical statement involving positive value integer n. If

(i) P (1) is valid,

(ii) validity of P (k) ⇒ validity of P (k + 1),

then P (n) is valid for all ’+’ve integer n.Proof: Let M be the subset of all N , of natural numbers n for which P (n) is true . SinceP (1) is true so 1 ∈M and condition (ii) gives k ∈M ⇒ k + 1 ∈M . Consider F = N −M ,we are to show that F = φ, by the previous, F = φ so M = N , hence P (n) is valid for all’+’ve integer n. We see that all the forms are equivalent.

Result 2.2.1 From the above equivalent forms, we see that, the mathematical inductionconsists of three steps

(i) Basis: Show that P (1) is true

(ii) Inductive hypothesis: Write the inductive hypothesis like, let P (k) be true.

(iii) Inductive step: Show that P (k + 1) is true.

Result 2.2.2 Although mathematical induction provides a standard technique for attempt-ing to prove a statement about the positive integers, one disadvantage is that it gives no aidin formulating such statements.

Ex 2.2.1 Prove that 23n − 1 is divisible by 7 for all n ∈ N .

Natural Number 87

Solution: Let us write P (n) for 23n − 1. We have P (1) = 23− 1 = 7 , which is divisible by7. Thus the proposition is true for n = 1. Let us consider P (m+ 1)− P (m), which is givenby,

P (m+ 1)− P (m) = (23m+1 − 1)− (23m − 1)= 23m+3 − 23m = 23m(8− 1)= 23m.7 = 7p,

where p = 23m = an integer. Hence P (m + 1) is divisible by 7, if P (m) is so. This provesthat the proposition is true for n = m + 1, if it is true for n = m. Hence by principle ofmathematical induction, the proposition is true for all n ∈ N .

Ex 2.2.2 Show that n5 − n is divisible by 30 for all n ∈ N .

Solution: Let us write P (n) for n5 − n. We have,

P (1) = 1− 1 = 0, which is divisible by 30.P (2) = 25 − 2 = 30, which is divisible by 30.

Thus the proposition is true for n = 1, 2. Let P (m) = m5 − m is divisible by 30 i.eP (m) = 30k, where k ∈ N

P (m+ 1) = (m+ 1)5 − (m+ 1)= m5 + 5m4 + 10m3 + 10m2 + 5m+ 1−m− 1= (m5 −m) + 5m(m+ 1)(m+ 2)2 − 15m(m+ 1)2

= 30k + 30q − 30r; q, r ∈ N .

P (m+1) is divisible by 30 if P (m) is divisible by 30. Hence, by the principle of mathematicalinduction, the proposition is true for all n ∈ N .

Ex 2.2.3 Show that n3 − n is divisible by 6 for all n ∈ N .

Solution: Let us write P (n) for n3 − n. We have

P (1) = 1− 1 = 0, which is divisible by 6.P (2) = 23 − 2 = 6, which is divisible by 6 .

Thus the proposition is true for n = 1, 2. Let P (m) = m3 −m is divisible by 6 i.e P (m) =6k, k ∈ N

P (m+ 1) = (m+ 1)3 − (m+ 1) = m3 + 3m2 + 2m= (m3 −m) + 3m(m+ 1) = 6k + 6q; q ∈ N

as product of two consecutive number is divisible by 2. P (m + 1) is divisible by 6 if P (m)is divisible by 6. Hence, by the principle of mathematical induction, the proposition is truefor all n ∈ N .

Ex 2.2.4 Show that 2.7n + 3.5n − 5is divisible by 24 for all n ∈ N .

Solution: Let us write P (n) for 2.7n + 3.5n − 5. We have,

P (1) = 2.7 + 3.5− 5 = 24, which is divisible by 24.

88 Theory of Numbers

Thus the proposition is true for n = 1. Let P (m) be divisible by 24 i.e P (m) = 2.7m +3.5m − 5 = 24q, q ∈ N . Now,

P (m+ 1) = 2.7m+1 + 3.5m+1 − 5= 7[2.7m + 3.5m − 5− 3.5m + 5] + 3.5m+1 − 5= 7(2.7m + 3.5m − 5)− 6.5m + 30= 7.24.q − 6.5(5m−1 − 1)= 7.24.q − 6.5.4(5m−2 + 5m−3 + . . .+ 1)= 24[7.q − 5(5m−2 + 5m−3 + . . .+ 1)].

Therefore, P (m+ 1) is divisible by 24 if P (m) is divisible by 24. Hence, by the principle ofmathematical induction, the proposition is true for all n ∈ N .

Ex 2.2.5 Show that, 34n+2 + 52n+1 is divisible by 14 for all n ∈ N .

Solution: Let us write P (n) for 34n+2 + 52n+1. We have,

P (1) = 36 + 53 = 14.61, which is divisible by 14.

Thus the proposition is true for n = 1. Let P (m) be divisible by 14 i.e P (m) = 34m+2 +52m+1 = 14q, q ∈ N . Thus, 52m+1 = 14.q − 34m+2. Now,

P (m+ 1) = 34(m+1)+2 + 52(m+1)+1

= 34m+2.34 + 52m+1.52

= 34m+2.81 + 25(14q − 34m+2)= 34m+2(81− 25) + 25.14q= 14[4.34m+2 + 25q]; where q ∈ N= 14k; where k = 4.34m+2 + 25q ∈ N .

Therefore, P (m+ 1) is divisible by 14, if P (m) is divisible by 14. Hence, by the principle ofmathematical induction, the proposition is true for all n ∈ N .

Ex 2.2.6 Show that nn > 1.3.5 . . . (2n− 1) for n > 1.

Solution: For n = 2, the LHS = 22 = 4 and RHS = 1.3 = 3. As 4 > 3, the inequality holdsfor n = 2. Let the result holds for n = m i.e.,mm > 1.3.5 . . . (2m− 1). Hence,

(2m+ 1)mm > 1.3.5 . . . (2m− 1)(2m+ 1).

Now, (m+ 1)m+1 − (2m+ 1)mm = mm+1[(1 +1m

)m+1 − (2 +1m

)]

= mm+1

[(1 +m+ 1C1

1m

+m+ 2C2

1m2

+ · · ·+ 1mm+1

)−(

2 +1m

)]= mm+1

[m+ 12m

+ · · ·+ 1mm+1

]> 0.

Hence, (m+ 1)m+1 > (2m+ 1)mm > 1.3.5 . . . (2m+ 1).

Thus the inequalities holds for n = m+ 1 when it holds for n = m. Hence it is true for allpositive value integer n.

Ex 2.2.7 For what natural number n is the inequality 2n > n2 valid.

Integers 89

Solution: We shall prove this by using the principle of mathematical induction. Forn = 1 as 2 > 1 so the inequality is valid.n = 2 as 22 = 22 so the inequality is not valid.n = 3 as 23 6> 32 so the inequality is not valid.n = 4 as 24 = 42 so the inequality is not valid.n = 5 as 25 > 52 so the inequality is valid.

Let 2k > k2, when k > 4 and k ∈ N . Therefore,2k > 2k + 1; for k > 4

⇒ 2k + 2k > k2 + 2k + 1 ⇒ 2k+1 > (k + 1)2.

Thus the inequality is valid for n = k + 1, when it is valid for n = k, and k > 4. Hence bythe principle of mathematical induction the inequality is valid for n = 1 and n > 4.

Ex 2.2.8 Prove that the product of r consecutive numbers is divisible by r!.

Solution: Let pn = n(n+ 1)(n+ 2) · · · (n+ r − 1); n ∈ N , then,

pn+1 = (n+ 1)(n+ 2) · · · (n+ r)⇒ n.pn+1 = (n+ r)pn = n.pn + r.pn

⇒ pn+1 − pn =pn

n× r

= r × product of (r − 1) consecutive natural numbers.

If the product of (r − 1) consecutive natural nos. is divisible by (r − 1)! then,pn+1 − pn = k.r! ; k ∈ N .

Now, p1 = r! so p2, p3, p4, . . . are also multiple of r!. We shall show that product of (r − 1)consecutive natural numbers is divisible by (r−1)! then the product of r consecutive naturalnumbers is divisible by r!. The product of two consecutive natural numbers is divisible by2!, so the product of three consecutive natural numbers is divisible by 3! and so on.

Ex 2.2.9 Prove that (2 +√

3)n + (2−√

3)n is an even integer for all n ∈ N .

Solution: Let pn be the statement that (2 +√

3)n + (2 −√

3)n is an even integer. Since,(2 +

√3)1 + (2 −

√3)1 = 4 = even integer, so p1 is true. Let pk be true, i.e., (2 +

√3)k +

(2−√

3)k is an even integer, then,(2 +

√3)k+1 + (2−

√3)k+1

= ak+1 + bk+1; where, a = 2 +√

3, b = 2−√

3= (ak + bk)(a+ b)− (ak−1 + bk−1)ab= 4(ak + bk)− (ak−1 + bk−1).

This is an even integer, as ak + bk and ak−1 + bk−1 are even integers, by assumption.This shows that Pk+1 is true whenever p1, p2, · · · , pk are true. Therefore, by principle ofmathematical induction, the statement is true for all n ∈ N .

2.3 Integers

The set of all integers, denoted by Z,consists of whole numbers asZ =

0,±1,±2,±3, · · · .

(2.2)

The set of all positive integers is identified with the set of natural number N . We shalluse the properties and principles of N in connection with the proof of any theorem aboutpositive integers.

90 Theory of Numbers

2.3.1 Divisibility

In this section, we define the divisibility and division algorithm, for two give integers, whichare most important and fundamental concept in number theory.

Definition 2.3.1 Let a ∈ Z and x is any member of Z. Then ax is called multiple of a.For example,

(i) 3× 7 = 21, then 21 is called a multiple of 3. Also it is called a multiple of 7.

(ii) The number 0 is multiple of every member of Z, as a.0 = 0, ∀ a ∈ Z.

These exist infinitely many elements which are multiple of a ∈ Z.

Definition 2.3.2 An integer a(6= 0) is said to divide as integer b, if

∃ unique c ∈ Z such that ac = b. (2.3)

This is expressed by saying ’a divides b’ or ’a is the divisor of b’ or ’b is the divisible of a’and is denoted by a|b. We also say that b is a multiple of a, that a is a divisor of b or thata is a factor of b.

For example

(i) 9|63 as 63 = 9.7, where 7 ∈ Z, i.e. 63 is a multiple of 9.

(ii) Also −81 is divisible by 3 as −81 = 3.(−27), and −27 ∈ Z.

(iii) Again 3 6 |16 because for 16 there is no integer x such that 16 = 3 · x.

(±1), (±a) are called the improper divisors of a nonzero integer a. We write, a 6 |b to indicatethat b is not divisible by a. Divisibility establishes a relation between any two integers withthe following elementary properties.

Property 2.3.1 If a|b, then every divisor of a divides b.

Proof: Since a|b, ∃ c ∈ Z such that b = ac. Let m be any divisor of a, thena = md, for some d ∈ Z.

Thus, b = mdc⇒ m|b, as cd ∈ Z.

Property 2.3.2 If a|b and a 6= 0 then (b/a)|b

Proof: From definition, we have,a|b⇒ b = ac⇒ b/a = c, an integer

Now, b =(b

a

)a, a ∈ Z ⇒ (b/a)|b.

b/a is called the devisor conjugate of a.

Property 2.3.3 For integers a, b, c ∈ Z,

(i) a|a, ∀a(6= 0) ∈ Z,(reflexive property)

(ii) 1|a, a|0, −1|a;

(iii) a|b, b|c⇒ a|c (transitive property). The converse of this property need not hold. Forexample, a = 5, b = 10, c = 15, then 5|15 but 10 6 |15 although 5|10.

Integers 91

(iv) a|b and b|a if and only if a = ±b.

These properties are immediately follows from definition.

Property 2.3.4 If a, b ∈ Z then a|b implies a|bm.

Proof: If a, b ∈ Z such that, a|b, by definition, b = ac; c ∈ Z. Therefore,

bm = (ac)m = a(cm); where cm ∈ Z,⇒ a|(bm).

Thus, if a|b, then a|(bm),m ∈∈ Z. The converse is not always true. For example, letb = 5,m = 8 and a = 10, then 10|5.8 i.e., 10|40 but 10 6 |8.

Also, if a|b, then ma|mb, m 6= 0. This is known as multiplication property. Also, thecancellation law states that ma|mb and m 6= 0 implies a|b.

Property 2.3.5 If a|b and a|c, then a|(bm+ cn);m,n being arbitrary integers.

Proof: The relations a|b, a|c ensure that ∃ suitable integers x, y such that b = ax, c = ay.Hence mb = max and nc = nay. Thus whatever the choice of integers m and n,

mb+ nc = max+ nay = a(mx+ ny)⇒ mb+ nc = n(mx+ ny), where m,n, x, y ∈ Z

⇒ a|(bm+ nc).

The converse of this result need not hold. For example, let a = 5, b = 6, c = 7,m = 3 andn = 1, so 5|25, but 5 6 |6 and 5 6 |7.

This property of this theorem extends by induction to sums of more than two terms.That is, if a|bk for k = 1, 2, . . . , n, then

a|(b1x1 + b2x2 + . . . bnxn) = a|n∑

i=1

bixi

for any integers x1, x2, . . . , xn. This is known as linearity property of divisibility.

Property 2.3.6 If a|b, ∃c ∈ Z such that b = ac. Also, b 6= 0 implies c 6= 0. Upon takingabsolute values, we get |b| = |ac| = |a||c|.

Because, c 6= 0, it follows that |c| ≥ 1, whence,

|b| = |ac| = |a||c| ≥ |a| ⇒ |b| ≥ |a|.

Thus, if a|b and b 6= 0, then |a| ≤ |b|. Let b 6= 0, then,

a|b⇒ |a| ≤ |b|.

Again, if a 6= 0, then b|a⇒ |b| ≤ |a|. Therefore, if a 6= 0, b 6= 0, then

a|b and b|a⇒ |a| = |b|.

This is known as comparison property.

Property 2.3.7 If 0 ≤ a ≤ b and b|a, then a = 0. For, let a 6= 0 and

b|a⇒ |b| ≤ |a| ⇒ b ≤ a,

which is a contradiction as a, b are both non negative. This contradiction shows the hypoth-esis that a = 0. If b|a and |a| < |b| then a = 0. For if a 6= 0, and b|a ⇒ |b| ≤ |a| which iscontradictory to the hypothesis and hence a = 0.

92 Theory of Numbers

2.3.2 Division Algorithm

Given integers a and b(b > 0), ∃ unique two integers q and r, such that

a = bq + r; where, 0 ≤ r < b. (2.4)

Proof: Existence : We begin by considering the set of non negative integers, given by,

S = a− bx : x ∈ Z, a− bx ≥ 0.

First, we shall show that S is non empty. To do this, it suffices to exhibit a value of xmaking a− bx nonnegative. Now,

−|a| ≤ a and b ≥ 1 ⇒ |a| ≤ b|a|.Therefore, a ≥ −|a| ≥ −b|a| ⇒ a+ b|a| ≥ 0

⇒ a+ b|a| ∈ S.

For this choice of x = −|a|, S is non empty set of non negative integers. Thus,

(i) either S contains 0 as its least element, or,

(ii) S does not contain 0, so, S is a nonempty subset of N , by well ordering principle, ithas a least element which is positive.

Hence in each case, S has a least element r ≥ 0, (say) and r is of the form a− bq. Thus,

r = a− bq; q ∈ Z⇒ a = bq + r; q ∈ Z and r ≥ 0.

We shall now show that r < b. If possible let r ≥ b then r − b ≥ 0 and

r − b = a− b(q + 1); where (1 + q) ∈ Z,

so that (r − b) ∈ S, smaller than its smallest member r, which is a contradiction. Hence,r < b (b > 0, r − b < r). Thus ∃q, r ∈ Z and 0 ≤ r < b such that a = bq + r.

Uniqueness : To prove the uniqueness of integers q, r; we assume that we can findanother pair q1, r1 ∈ Z such that,

a = bq1 + r1; 0 ≤ r1 < b.

⇒ 0 = b(q − q1) + (r − r1)or, b(q − q1) = r − r1 ⇒ b|(r − r1),

where, |r − r1| < b⇒ (r − r1) = 0 ⇒ r = r1

so, b(q − q1) = 0 ⇒ q = q1; as b > 0.

Thus q and r are unique, ending the proof. Also, it is clear that r = 0, if and only if, b|a.This important theorem is is known as division algorithm. The advantage of this algorithmis that it allows us to prove assertions about all the integers by considering only a finitenumber of cases.

Result 2.3.1 The two integers q and r, termed as quotient and remainder in the divisionof a by b respectively.

Result 2.3.2 Though it is an existence theorem, its proof actually gives us a method forcomputing the quotient q and remainder r.

Integers 93

Theorem 2.3.1 If a and b(> 0) be two integers, then ∃ integers Q and R such that

a = bQ±R; 0 ≤ R <b

2. (2.5)

Proof: For any two integers a and b with b > 0, the division algorithm shows that ∃ q, r ∈ Zsuch that

a = bq + r; 0 ≤ r < b (2.6)

Case1: Let r < b2 . Taking q = Q and r = R in (2.6), we have,

a = bQ+R; 0 ≤ R <b

2

Case2: Let r > b2 . Now, a = bq + r can be written in the form

a = b(q + 1) + r − b = b(q + 1)− (b− r)

Taking q + 1 = Q and b− r = R, we have,

a = bQ−R, 0 ≤ R <b

2, R = b− r < b− b

2=b

2.

Thus combining the Case(1) and (2) we have,

a = bQ±R; 0 ≤ R <b

2.

Case3: Let r = b2 , then a = bq + r can be written in the form a = bQ+ R, where we take

q = Q and r = R = b2 . Again,

a = bq + r = b(q + 1)− (b− r) = bQ+R

where q + 1 = Q and R = −(b− r) = − b2 . Thus, it follows that for r = b

2 , Q and R are notunique. In this case, R is called the minimal remainder, i.e. the absolutely least remainderof a with respect to b.

Theorem 2.3.2 (Generalized division algorithm) : Given integers a and b(b 6= 0), ∃ uniquetwo integers q and r, such that a = bq + r; 0 ≤ r < |b|.

Proof: When b is positive, then it is the previous theorem. So it is enough to consider thecase in which b is negative. When b is negative then |b| > 0 as b 6= 0. By the above theorem,∃ unique integers q1 and r such that

a = |b|q1 + r; 0 ≤ r < |b|= −bq1 + r = bq + r, where q = −q1.

Hence the theorem.

Ex 2.3.1 If n be any positive integer, show that the product of the consecutive naturalnumbers n, n+ 1, n+ 2 is divisible by 6.

Solution: In case of division by 3, one of the numbers 0, 1, 2 will be the remainder and thecorresponding integers of the form 3k, 3k + 1, 3k + 2; k ∈ Z. If

n = 3k, then 3|n; n = (3k + 1), then 3|n+ 2; n = 3k + 2, then 3|n+ 1.

94 Theory of Numbers

Hence for any value of n in Z, 3|n(n+1)(n+2). In case of division by 2, one of the numbers0, 1 will be the remainder and the corresponding integers of the form 2k, 2k + 1; k ∈ Z. If,

n = 2k then 2|n and n = 2k + 1 then 2|n+ 1.

Hence for any n ∈ Z, 2|n(n + 1) i.e. the two consecutive integers n, n + 1 one is even i.e.divisible by 2. Therefore,

2|n(n+ 1)(n+ 2) and 3|n(n+ 1)(n+ 2).

Since (2, 3) = 1, so 6|n(n+1)(n+2). In the above procedure, we can show that the productof m consecutive integers is divisible by m.

Ex 2.3.2 Show that the square of an odd integer is of the form 8k + 1; k ∈ Z.

Solution: By division algorithm, we see that when an integer is divided by 4, the remainderwill be one of 0, 1, 2, 3 and the corresponding integers is of the form 4k, 4k+1, 4k+2, 4k+3.Of those form (4m+ 1) and (4m+ 3) will be odd integers. Now,

(4m+ 1)2 = 8(2m2 +m) + 1

where 2m2 +m ∈ Z, which is of the form 8k + 1 and

(4m+ 3)2 = 8(2m2 + 3m+ 1) + 1,

where 2m2 + 3m+ 1 ∈ Z, which is of the form 8k + 1. Therefore, the square of an oddinteger is of the form 8k + 1; k ∈ Z.Ex 2.3.3 Show that square of any integer is of the form 4n or (4n+ 1), for some n ∈ Z.

By division algorithm, we see that when an integer is divided by 2, the remainder of 0, 1and the corresponding integers of the form 2k, 2k + 1; k ∈ Z. Now,

(2k)2 = 4k2 = 4n; k2 = n ∈ Zso, (2k + 1)2 = 4(k2 + k) + 1

= 4n+ 1; k2 + k = n ∈ Z.

Hence square of any integer is of the form 4n, 4n+ 1; n ∈ Z.

2.4 Common Divisor

Let a and b be given arbitrary integers. If d divides two integers a and b, i.e., if both

d|a and d|b, (2.7)

then d is called a common divisor of a and b. The number of divisors of any non-zero integeris finite. Now

(i) 1 is a common divisor of every pair of integers a and b, so the set of positive commondivisors of integers a and b is non empty.

(ii) Every integer divides zero, so that if a = b = 0, then, every integer serves as a commondivisor of a and b. In this instance, the set of positive common divisors of a and b isinfinite.

(iii) However, when at least one of a and b is different from 0, there are only a finite numberof positive common divisors.

Every pair of integers a and b has a common divisor which can be expressed as a linearcombination of a and b. Every finite set has the largest value. It is defined as the gcd as inthe following definition.

Common Divisor 95

2.4.1 Greatest Common Divisor

For two given integers a and b, with at least one of them different from zero, a positiveinteger d is defined to be the greatest common divisor (gcd) of a, b if,

(i) d be a common divisor of a as well as b i.e., d|a, d|b.

(ii) every common divisor of a, b is a divisor of d ,i.e for an integer c ;i.e.,

c|a, c|b⇒ c|d.

The gcd of a, b is denoted by gcd(a, b) or simply (a, b). For more than two integers it isdenoted by (a1, a2, · · · , an). From definition, it follows that,

(a,−b) = (−a, b) = (−a,−b) = (a, b),

where, a, b are integers, not both zero. For example,

(i) (12, 30) = 6, (9, 4) = 1, (0, 5) = 5 etc.

(ii) (12,−30) = 6 and (−16, 40) = 8.

Result 2.4.1 Let d and d1 be two greatest common divisors of integers a and b. Thenby the definition, we find that d|d1 and d1|d. Hence, there exist integer r and t such thatd1 = dr and d = d1t. Now,

d = d1t = drt, d 6= 0 ⇒ rt = 1.

Thus, r = s = ±1, and hence d = ±d1. So it follows that, two different gcd’s of a and bdiffer in their sign only. But we take the positive value as the gcd.

Theorem 2.4.1 Any two non zero integers a, b, not both of which are zero, have an uniquegcd, which can be written as in the form ma+ nb;m,n ∈ Z.

Proof: Let us consider a set S of all positive linear combinations of a and b as,

S = xa+ yb : x, y ∈ Z, xa+ yb > 0.Also, a.a+ 0.b = a2(> 0) ∈ S.

so, S is non empty subset of N . Therefore, by the well ordering principle, it has an leastelement r (say), which is of the form

r = ma+ nb; m,n ∈ Z.

We shall first show that, r|a and r|b. If r is not a devisor of a, by the division algorithm,∃p, q ∈ Z such that a = pr + q; 0 < q < r, i.e.,

q = a− pr = a− p(ma+ nb)= (1−mp)a+ (−np)b ∈ S;

where, 1−mp,−np ∈ Z and q > 0. Since q < r, this representation would imply that, q isa member of S contradicting the fact that r is the least element in S. Hence q = 0 and r|aand similarly r|b. Next let, c|a, c|b, then

c|a, c|b⇒ a = ck1, b = ck2; k1, k2 ∈ Zso, r = ma+ nb = mck1 + nck2

= c(mk1 + nk2); where mk1 + nk2 ∈ Z.

96 Theory of Numbers

Thus, c|r and so r = (a, b) and r = ma+nb; m,n ∈ Z. To the uniqueness of r, let there beanother gcd of a, b say r1 i.e r1 = (a, b) also. r|r1 and r1|r i.e r, r1 are associates ⇒ r1 = ±r.But as r and r1 are both positive so r = r1. Hence gcd is unique which can be expressedas a linear combination of a and b with integral multiplier m and n. This is the Euclideanalgorithm for existence of gcd. Note the following:

(i) This method involves repeated application of the division algorithm.

(ii) If m,n are integers then (a, b) is the least positive integer of the form ma+ nb, wherem and n range over integers.

(iii) The representation of d as ma+ nb is not unique.

(iv) If a and b are integers, not both of which are zero, we have,

(a, b) = (b, a) = (−a, b) = (a,−b) = (−a,−b) = (a, b+ ax),

for any integer x.

(v) The theorem does not give any algorithm how to express (a, b) in the desired formma+ nb.

(vi) If d = (a1, a2, . . . , ar), ai 6= 0; ∀i then ∃ integers m1,m2, . . . ,mr such that

d = a1m1 + a2m2 + . . .+ armr.

Theorem 2.4.2 (Method of finding gcd ) : For two given positive value integers a, b ifa = bq + r; q, r ∈ Z, 0 ≤ r < b then (a, b) = (b, r).

Proof: Let d = (a, b) and d1 = (b, r). Since d is the gcd of a and b so d|a and d|b i.e.,k1, k2 ∈ Z such that a = dk1, b = dk2. Now,

a = bq + r ⇒ r = a− bq

= dk1 − dk2q = d(k1 − k2q); k1 − k2q ∈ Z.

Thus, d|r also d|b so d|d1. Similarly we can get d1|d. As d1 = (b, r) so d1|b, d1|r. b = d1band r = d1r1 so

a = bq + r = d1b1q + d1r1 = d1(b1q + r1).

Therefore, d1|a also d1|b⇒ d1|d. d = ±d1, as d and d1 are both positive so d = d1.

Ex 2.4.1 Find the gcd of 120 and 275 and express the gcd in the form 120m+275n;m,n ∈ Z.

Solution: To find (120, 275), we have the following table:

3 120 275 2105 240

3 15 35 215 30

5

Hence (120, 275) = 5. Now,15 = 5.3 + 035 = 15.2 + 5 ⇒ 5 = 1.35− 15.2

120 = 35.3 + 15 ⇒ 15 = 1.120− 35.3275 = 120.2 + 35 ⇒ 35 = 275− 120.2

Common Multiple 97

Therefore, the gcd 5 = (120, 275) can be written as

5 = 35− 15.2 = 35− (120− 35.3).2= 7.35− 2.120 = 7(275− 120.2)− 2.120= 7.275 + (−16).120

which is of the form 120m+ 275n where m = −16 and n = 7.

Ex 2.4.2 Show that (a, a+ 2) = 1 or 2 for every integer a.

Solution: Let d = (a, a + 2), then d|a and d|a + 2. Therefore, by linearity property ofdivisibility, we have

d|ma+ n(a+ 2); ∀m,n ∈ Z.

Taking m = −1 and n = 1, it follows that d|2, i.e., d is either 1 or 2.

Ex 2.4.3 Let a, b be two integers not both zero. Prove that (ka, kb) = k(a, b) for any positiveinteger k.

Solution: Let (a, b) = d1, then there exist m,n ∈ Z such that

d1 = ma+ nb, i.e., kd1 = kma+ knb.

Let (ka, kb) = d2, so, d2 divides ka and kb, so that d2 divides the least positive value ofkma + knb, i.e., d2 divides kd1. On the other hand, d1 divides a and b, hence kd1 divideska and kb. But d2 = (ka, kb), consequently, kd1|d2. Hence,

(ka, kb) = d2 = kd1 = k(a, b).

This is actually distributive law.

2.5 Common Multiple

Let a1, a2, · · · , an be integers all different from zero. An integer b is said to be a commonmultiple b of a1, a2, · · · , an if ai|b for i = 1, 2, · · · , n. In fact common multiple do exist. Forexample, 2.3.5 is a common multiple of the integers 2, 3, 5, none of which is zero.

2.5.1 Lowest Common Multiple

Let a, b ∈ Z. Let us consider a set S as

S = x : x ∈ N and x is a common multiple of a and b

= x : x ∈ N such that a∣∣∣x and b|x.

Now, a∣∣∣a gives a| |a| or a

∣∣∣ |a||b| or a∣∣∣|ab| and similarly b

∣∣∣|ab| and |ab| ∈ N , so |ab| ∈ S.Therefore, S is nonempty subset of positive integers. Hence by well ordering principle, Shas a least element say m. This m is called the lowest common multiple (lcm) of a andb, written as [a, b]. The least of the positive common multiples is called the least commonmultiple of a1, a2, · · · , an and is denoted by [a1, a2, · · · , an]. If m = [a1, a2, · · · , an], then thecommon multiples of the integers is the set

0,±m,±2m,±3m, · · ·.

The lcm of any set of nonzero integers is unique. For example, the lcm of 2,3,6 is 6; that of-2,-3,-6 is 6; the lcm of -2,-6,10 is 30.

98 Theory of Numbers

Property 2.5.1 The relation between gcd and lcm is [a, b](a, b) = |ab|.

Proof: It is sufficient if we prove the result for positive integers only. First we consider(a, b) = 1. suppose [a, b] = m, then m = ka for some k. Then b|ka and (a, b) = 1. Therefore,b|k and therefore, b ≤ k, ba ≤ ka. But ba, being a positive common multiple of b and a, cannot be less than the least common multiple, and so b = m.

ba = ka = [a, b].

Now, let us consider the general case, (a, b) = d > 1. Then,(a

d,b

d

)= 1 ⇒

[a

d,b

d

]=a

d.b

dby the above

⇒ [a, b](a, b) = ab.

Hence the theorem. Then if

(a, b) = d, [a, b] =ab

d.

From this, we have, if (a, b) = 1, then [a, b] = |ab|.

Property 2.5.2 If m = [a1, a2, · · · , an], then there exists l such that ai|l; i = 1, 2, · · · , n ifand only if m|l.

Proof: Since, m = [a1, a2, · · · , an], so ai|m; i = 1, 2, · · · , n. First let m|l, then,

ai|m; i = 1, 2, · · · , n and m|l⇒ ai|l; i = 1, 2, · · · , n.

Conversely, let ai|l; i = 1, 2, · · · , n. Suppose m 6 |i, then by division algorithm, l = mq + r,where 0 < r < m, i.e., r = l −mq. Now,

ai|l, ai|m⇒ ai|r; i = 1, 2, · · · , n.

But m is least such that ai|m; i = 1, 2, · · · , n and hence m|l. Therefore, if [a1, a2] =m2, [m2, a3] = m3, [m3, a4] = m4, . . . , [mn−1, an] = mn. Then [a1, a2, . . . , an] = mn.

Property 2.5.3 For k > 0, [ka, kb] = k[a, b].

Proof: Let m = [ka, kb], then by definition ka|m and kb|m, so we have m = kx.If [a, b] = x1, we note that a|x1, b|x1, ak|kx1, bk|kx1 and so kx|kx1. Thus, x|x1. Also,ak|kx, bk|kx, a|x, b|x and so x1|x. Hence x = x1 and therefore,

m = kx = kx1 = k[a, b] ⇒ [ka, kb] = k[a, b].

Hence the result.

Ex 2.5.1 Show that, (a+ b, [a, b]) = (a, b) for any two integers a and b.

Solution: Let d = (a, b) and l = [a, b], then, (a+ b, [a, b]) = (a+ b, l). Now,

d = (a, b) ⇒ d|a, d|b⇒ d|(a+ b).Also, a|l, b|l, so, d|a, a|l⇒ d|l.

If d is prime, then (a + b, l) = d = (a, b). Now, let d is not prime, i.e., d is a compositenumber. Then, say, d = d1.d2; where, (d1, d2) = 1; d1, d2 < d.

Thus ∃ a positive number d1 ( or d2) such that,d1|a+ b, d1|l and d1|d.

So, (a+ b, l) = d, i.e.,(a+ b, [a, b]) = (a, b).

Diophantine Equations 99

2.6 Diophantine Equations

In this section, we are to consider the Diophantine equations, named after the Greek math-ematician Diophantos of Alexandria. We apply the term Diophantine equations in one ormore unknowns with integer coefficients which is to be solved in integers only. Such a equa-tion is called an indeterminate equation (i.e., the number of equations is less than that ofthe unknowns). One of the basic interest in the theory of numbers is to obtain all solutionsin Z of a given algebraic polynomial equation

a0xn + a1x

n−1 + · · ·+ an−1x+ an = 0; ai ∈ Z.

Such a problem is called Diophantine problem and we say we are solving DiophantineEquations. As an example, we have to consider one of the oldest Diophantine Equations:x2 + y2 = z2, where x, y, z are pairwise relatively prime integers and obtained its completesolution of the form

x = a2 − b2, y = 2ab, z = a2 + b2 with (a, b) = 1.

In this type of equations we usually look for the solutions in a restricted class of numberssuch as positive integers, negative integers.

2.6.1 Linear Diophantine Equations

Let us consider a linear diophantine equations in two unknown variables x and y as

ax+ by = c; where, a, b, c ∈ Z (2.8)

with a, b are integers (not both zero). A integer solution of (2.8) is a pair of integers (x0, y0),that, when substituted into the equation, satisfies it, i.e., we ask that ax0 + by0 = c. Infinding the solution of a Diophantine equation ax+by = c, (a, b) = 1, we follows the followingmethods

(i) Substitution method,

(ii) Simple continued function,

(iii) Euclidean algorithm method.

If (a, b) = 1 and if x0, y0 is a particular solution of the linear Diophantine equation ax+by =c, then all solutions are x = x0 + bk; y = y0 − ak, for integral values of k.

(i) A given linear Diophantine equation can have a number of integral solutions, as is thecase with 2x+ 4y = 12, where,

2.4 + 4.1 = 12; 2.2 + 4.2 = 12

or may not have even a single solution.

(ii) Conversely, there are some linear Diophantine equations like 2x+ 6y = 13, which hasno solution, due to the fact that, the LHS is an even integer, whatever, the choice ofwhereas the RHS is not.

So, our first task is to find out the condition for solvability of the linear Diophantine equa-tions. The following theorem tells us when a Diophantine equation has a solution.

Theorem 2.6.1 The necessary and sufficient condition that, ax+ by = c has integral solu-tion if (a, b) divides c, where a, b, c are integers such that a, b are not both zero.

100 Theory of Numbers

Proof: Let d = (a, b) = the greatest common divisor of a and b. If d 6 |c, then there exist nointegers x and y with ax+ by = c. Suppose d|c, in this case first determine integers x0 andy0 so that ax0 + by0 = d. Since (a, b) = d, so d can be expressed in the form d = ma + nb,where, m,n ∈ Z. This can be put in the general form as

d = a(m− kb) + b(n+ ka),

where k ∈ Z. We have d|c i.e c = ld; l ∈ Z. Now,

ld = al(m− kb) + bl(n+ ka)or, c = al(m− kb)+ bc(n+ ka).

Let x0 = l(m − kb) ∈ Z and y0 = l(n + ka) ∈ Z, so that (x0, y0) is an integral solution ofax + by = c. Conversely, let (x0, y0) be an integral solution of the equation ax + by = c.Then ax0 + by0 = c, where, x0, y0 are integers. Let (a, b) = d, then

d|a and d|b⇒ d|(ax0 + by0); i.e., d|c.

Now, if (x0, y0) be any particular solution of ax + by = c, we are to all integral solutions.Let (x1, y1) be an integral solution of ax+ by = c, where, set,

x1 = (c/d)x0, and y1 = (c/d)y0

so that ax1 + by1 = c. Suppose r and s are integers satisfying ar + bs = c, we get

ar + bs = ax1 + by1 = c

⇒ a

d(r − x1) = − b

d(s− y1). (2.9)

Now d = (a, b), so, (ad ,

bd ) = 1, then from (2.9) we conclude that a/d|(s−y1) and b/d|(r−x1)

and hence ∃ an integer t, such that

r = x1 +b

dt and s = y1 −

a

dt; t ∈ Z.

So the linear diophantine equations ax + by = c (a, b, c ∈ Z) has a solution iff d = (a, b)divides c. Moreover for integral solution (x∗, y∗), ∃ an integer t such that,

x∗ = x0 +b

dt and y∗ = y0 −

a

dt.

In fact (x0 + bd t, y0 −

ad t) is an integral solution of the given equation, for any integer t, as

a(x0 +b

dt) + b(y0 −

a

dt) = (ax0 + by0) +

ab

dt− ab

dt

= ax0 + by0 = c.

Hence, if (x0, y0) is an integral solution of the given equation, then all the integral solutionsare given by

x = x0 +b

dt; y = y0 −

a

dt,

where t is any integer. Therefore, there are an infinite number of solutions of the givenequation, one for each value of t.

Diophantine Equations 101

Ex 2.6.1 Find all solutions of the Diophantine equation 108x+ 45y = 81.

Solution: By the Euclidean’s algorithm, which is given by (45, 108) = 9. Because 9|81, aintegral solution to this equation exists. To obtain the integer 9 as a linear combination of108, 45, we work as follows:

9 = 45− 2× 18 = 45− 2(108− 2× 45)= 45× 5− 2× 108.

Upon multiplying this relation by 9, we arrive at

81 = 9.9 = 9.[5.45 + (−2).108] = 108.(−18) + 45.45

so that x = −18 and y = 45 provide one integral solution to the given linear Diophantineequation. Also, the equation can also be written in the form

108x+ 45y = 81 = 108.(−18) + 45.45

or,1089

(x+ 18) =459

(45− y).

Since 1089 and 45

9 are prime to each other, we have,

x+ 1845/9

=45− y

108/9= t say.

Thus other integral solutions can be expressed as

x = −18 +459t, y = 45− 108

9t, where, t ∈ Z

or, x = −18 + 5t, y = 45− 12t; where, t = 0,±1,±2, · · · .

Deduction 2.6.1 All integral solutions of ax + by = c, such that a, b, c ∈ N and(a, b) = 1: Since (a, b) = 1, ∃m,n ∈ Z such that am+ bn = 1. Thus,

ax+ by = c(am+ bn) ⇒ a(x− cm) = −b(y − cn)

⇒ x− cm

−b=y − cn

a= t(say) ∈ Z

⇒ x = cm− bt, y = cn+ at; t ∈ Z

where as (a, b) = 1 so b|x − cm and a|y − cn. This is the general solution in integers. Forpositive integral solution, we must have

cm− bt > 0 and cn+ at > 0 ⇒ −cna< t <

cm

b.

If we take cmb = p+f1 and cn

a = q+f2 where p =[

cmb

], q =

[cna

]are integers and 0 < f1 ≤ 1,

0 ≤ f2 < 1, then t ≤ p and t > q. In this case the total number of solutions in positiveintegers is p− q.

Ex 2.6.2 Find all positive integral solution of 5x+ 3y = 52.

Solution: Here, 5 and 3 are prime to each other, i.e., d = (5, 3) = 1. Thus there existsm,n ∈ Z such that 5m+ 3n = 1. Here, m = 2, n = −3. Thus,

5x+ 3y = 52[5.2 + 3.(−3)]or, 5(x− 104) = −3(y + 156).

102 Theory of Numbers

Since 5 and 3 are prime to each other, x− 104 is divisible by 3 and y+ 156 is divisible by 5and therefore,

x− 104−3

=y + 156

5= t; t ∈ Z

or, x = 104− 3t; y = 5t− 156, where, t = 0,±1,±2, · · · .

This is the general solution of integers. For a positive integral solution, we must have

104− 3t > 0 and 5t− 156 > 0 ⇒ 1565

< t <1043.

The solutions in positive integers corresponds to t = 32, 33 and the solution is x = 8, y = 4and x = 5, y = 9.

Ex 2.6.3 Find all positive integral solution of 5x+ 12y = 80.

Solution: Here, 5 and 12 are prime to each other, i.e., d = (5, 12) = 1. Thus there existsm,n ∈ Z such that 5m+ 12n = 1. Here, m = 5, n = −2. Thus,

5x+ 12y = 80(5.5− 12.2)or, 5(x− 400) = −12(y + 160).

Since 5 and 12 are prime to each other, x− 400 is divisible by 12 and y+ 160 is divisible by5 and therefore, x− 400

−12=y + 160

5= t; t ∈ Z

or, x = 400− 12t; y = 5t− 160, where, t = 0,±1,±2, · · · .

This is the general solution of integers. For a positive integral solution, we must have,400− 12t > 0 and 5t− 160 > 0 ⇒ 32 < t < 100

3 .The only solution in positive integers corresponds to t = 33 and the solution is x = 4, y = 5.

Ex 2.6.4 Find all positive integral solution of 12x− 7y = 8.

Solution: Here, 12 and 7 are prime to each other, i.e., d = (12, 7) = 1. Thus there existsm,n ∈ Z such that 12m+ 7n = 1. Here, m = 3, n = −5. Therefore,

12x− 7y = 8[12.3 + 7.(−5)]or, 12(x− 24) = 7(y − 40).

Since 12 and 7 are prime to each other, x− 244 is divisible by 7 and y− 40 is divisible by 5and so, x− 24

7=y − 40

12= t; t ∈ Z

or, x = 24 + 7t; y = 12t+ 40, where, t = 0,±1,±2, · · · .

This is the general solution of integers. For a positive integral solution, we must have,24 + 7t > 0 and 12t+ 40 > 0 ⇒ t > − 24

7 ; t > − 103 .

The solution in positive integer corresponds to t = −3 and so x = 3, y = 4.

2.7 Prime Numbers

An integer p > 1 is called a prime number, or simply a prime, if there is no positive divisord of p satisfying 1 < d < p, i.e., its only positive divisors are 1 and p. If p > 1 is not primeit is called composite number.

(i) The integer 1 is regarded as neither prime nor composite.

(ii) 2 is the only even prime number. All other prime numbers are necessarily odd.

For example, the prime numbers less than 10 are 2, 3, 5, 7, while 4, 6, 8, 9 are composite.

Prime Numbers 103

2.7.1 Relatively Prime Numbers

Two integers a and b, not both of which are zero, are said to be relatively prime or co-primeif (a, b) = 1. In this case, it is guaranteed the existence of integers m and n such that

1 = ma+ nb.

For example, 4 and 9 are not prime numbers, but they are relatively prime as (4, 9) = 1. Aset of integers a1, a2, . . . , an, not all zero, are said to be relatively prime, if

(ai, aj) = 1; ∀i 6= j = 1, 2, . . . , n. (2.10)

Ex 2.7.1 Prove that, for n > 3, the integers n, n+ 2, n+ 4 cannot be all primes.

Solution: Any integer n is one of the forms 3k, 3k + 1, 3k + 2, where k ∈ Z. If

(i) n = 3k, then n is not a prime.

(ii) n = 3k + 1, then n+ 2 = 3(k + 3) and it is not prime.

(iii) n = 3k + 2, then n+ 4 = 3(k + 2) and it is not prime.

Thus in any case, the integers n, n+ 2, n+ 4 cannot be all primes.

Theorem 2.7.1 If m(6= 0) ∈ Z; then, (ma,mb) = m(a, b), where a, b ∈ Z are not bothzero.

Proof: Let (a, b) = k then a = kA, b = kB; k ∈ Z and (A,B) = 1. Therefore,

ma = mkA; mb = mkB and (A,B) = 1⇒ (ma,mb) = mk = m(a, b).

Theorem 2.7.2 If d = (a, b) > 0 then ad and b

d are integers prime to each other.

Proof: We observe that, although ad and b

d have the appearance of fractions, in fact,they are integers as d is a divisor of both a and b. Since d = (a, b) by existence theorem,∃m,n ∈ Z such that d = ma+ nb. Therefore,

1 =(ad

)m+

(b

d

)n.

Since d|a, d|b, by definition of gcd, ∃ integers u, v where ad = u and b

d = v, such that,1 = um+ vn. Since, u, v are integers, the conclusion is a

d and bd are relatively prime.

Theorem 2.7.3 If (a, b) = 1, then for any integer c, (ac, b) = (c, b).

Proof: Let (ac, b) = d and (c, b) = d1. Since (a, b) = 1,∃m,n ∈ Z such that

am+ bn = 1 ⇒ acm+ bcn = c.

Now, (ac, b) = d⇒ d|ac,d|b⇒ d|ac, d|bc; as b|bc

⇒ d|(acx+ bcy) ⇒ d|c.

Then as (c, b) = d, so d|c, d|b⇒ d|d1. Also,

(c, b) = d1 ⇒ d1|c, d1|b⇒ d1|ac, d1|b; as c|ac⇒ d1|d; as (ac, b) = d.

Thus it follows that d = d1. For example (2, 5) = 1, c = 10, (20, 5) = 5. Also, (10, 5) = 5,therefore (2.10, 5) = (10, 5). Therefore, if (ai, b) > 1, ∀i = 1, 2, . . . , n then (a1a2 · · · an, b) =1.

104 Theory of Numbers

Theorem 2.7.4 If p(> 1) is prime and p|ab, then, p|a, or p|b; where a, b are any twointegers.

Proof: Let p be a prime and a, b are integers such that p|ab. If a = 0 or b = 0, the result istrue. If p|a then the result is also true. Let us assume that p 6 |a. Because the only positivedivisors of p are 1 and p itself, we have, (p, a) = 1, so, ∃ m,n ∈ Z such that 1 = ma + np.Multiplying both sides by b, we get,

b = mab+ npb = (mp)c+ npb; let ab = pc; c ∈ Z= p(mc+ nb) = p× an integer

according as p|b. Similarly if p 6 |b then p|a. Conversely, let us suppose that, the integerp(> 1) satisfies the given condition. Let q be a positive divisor od p such that q < p. Weare to show that p = qr. Since p|p, we have p|qr. Hence either p divides q or p divides r.Since 0 < q < p, p 6 |q, so p|r. Thus, ∃ some k ∈ Z such that r = pk, so that

p = qr = qpk ⇒ qk = 1 ⇒ q = 1

so we conclude that 1 and p are the only positive divisors of p. Hence p is a prime. Thus apositive integer p has the property that, if for any a, b ∈ Z,

p|ab⇒ p|a or p|b,

then p is prime. For example, 12|8.3, but neither 8 and 3 is divisible by 12. Hence 12 isnot prime. This theorem distinguish prime numbers from composite numbers, which is thefundamental problem in number theory. Now,

(i) If p = ab, then at least one of a and b must be less p.

(ii) If a(6= 1) ∈ Z, a must have a prime factor.

Ex 2.7.2 Show that the fraction9n+ 86n+ 5

is irreducible for all n ∈ N .

Solution: It is sufficient if we are to show that (9n+8, 6n+5) = 1. Let a = 9n+8, b = 6n+5,then 2a−3b = 1. Therefore a and b are relatively prime. Hence the fraction 9n+8

6n+5 is irreduciblefor all n ∈ N .

Theorem 2.7.5 If p is prime, and p|a1a2a3 · · · then p|ai for some i with 1 ≤ i ≤ n.

Proof: We shall prove this by use of the principle of mathematical principle on n, thenumber of factors. When n = 1,, i.e., if p|a1, then the result is true. Let n = 2 and ifp 6 |a1,then by the previous theorem, we get,

p|(a2a3 · · · an) = p|a2(a3a4 . . . an).

If p 6 |a2, then p|a3a4 · · · an. Let, as the induction hypothesis that n > 2 and that wheneverp divides a product of less than n factors, it divides at least one of the factors. Now,p|a1a2 . . . an, then either p|an or p|a1a2 . . . an−1, the inductive hypothesis ensures that p|ai

for some choice of i, with 1 ≤ i ≤ n − 1. In any event, p divides one of the integersa1, a2, . . . , an.

Therefore, if p, q1, q2, . . . , qn are all primes and p|q1q2 · · · qn, then p = qk, for some k,where 1 ≤ k ≤ n.

Ex 2.7.3 If a, b are both primes with a ≥ b ≥ 5, show that 24|a2 − b2.

Prime Numbers 105

Solution: Since a and b are primes > 3, both of them are of the form 3k + 1 or 3k + 2,where k ∈ Z. If both a and b are either of the forms then 3|a − b. If one of them is of theform 3k + 1 and the other is of the form 3k + 2 then 3|a+ b. Thus, in any case 3|a2 − b2.

Given that a, b are odd primes, so they are of the form 4k + 1 or 4k + 3, where k ∈ Z. Ifboth a and b are either of the forms 4k + 1 then 2|a+ b and 4|a− b. If both of them are ofthe form 4k + 3, then 4|a+ b and 2|a− b. Thus, in any case 8|a2 − b2.

Since (3, 8) = 1, we have 24|a2 − b2.

Theorem 2.7.6 If (a, b) = 1 and b|ac then b|c.

Proof: Since b|ac so ∃r ∈ Z such that ac = br. Also, (a, b) = 1,∃m,n ∈ Z such that1 = ma+ nb. Multiplication of this equation by c produces,

c = c.1 = c(ma+ nb)= mac+ nbc = mbr + nbc

= b(mr + nc) = b× some integer.

Because b|bc and b|ac, it follows that b|(mac+nbc), shows that b|c. This is known as Euclid’slemma. If ap = bq and (a, b) = 1, then a|q and b|p.

If a and b are not relatively prime, then the result may or may not be true. For example,12|9.8, but 12 6 |9 and 12 6 |8.

Theorem 2.7.7 If a|c, b|c and (a, b) = 1 then ab|c.

Proof: In as much a|c, b|c, ∃k1, k2 ∈ Z such that c = ak1 = bk2. Again, the relation(a, b) = 1 allows us to write ∃m,n ∈ Z, such that ma + nb = 1. Multiplying the equationby c, it appears that,

c = c.1 = c(ma+ nb) = mac+ nbc

= mabk2 + nbak1; as c = ak1 = bk2

= ab(mk2 + nk1) = ab× some integer.

Hence as the divisibility statement ab|c.

Theorem 2.7.8 a|b if and only if ac|bc, where c 6= 0.

Proof: If ac|bc then bc = (ac)q; q ∈ Z. Therefore,

c(b− aq) = 0 ⇒ b− aq = 0 as c 6= 0⇒ b = aq; i.e., a|b.

The converse part is obvious. Without the condition (a, b) = 1, a|c and b|c together maynot imply ab|c. For example, 4|12 and 6|12 do not imply 4.6|12.

Theorem 2.7.9 If (a, b) = 1, then for any integer q, (a+ bq, b) = 1.

Proof: Let (a+ bq, b) = k where k ≥ 1. If k = 1, then the result holds. Let k > 1, then

(a+ bq, b) = k(> 1) ⇒ k|a+ bq, k|b⇒ k|(a+ bq).1 + b(−q) ⇒ k|a.

Therefore, k|a, k|b ⇒ (a, b) 6= 1, which is a contradiction. Hence the theorem is true fork = 1 only. Therefore (a+ bq, b) = 1.

Ex 2.7.4 If (a, b) = 1, prove that (a+ b, a2 − ab+ b2) = 1 or 3.

106 Theory of Numbers

Solution: Let d = (a+ b, a2 − ab+ b2), then,

(a+ b, a2 − ab+ b2) = d⇒ (−2ab+ b2, a+ b) = d

⇒ (−3ab, a+ b) = d⇒ (3a2, a+ b) = d⇒ d|3a2.

Also, (3b2, a+ b) = d⇒ d|3b2. Therefore,

d|3a2 ⇒ d = 3 or, d|a2 and d|3b2 ⇒ d = 3 or, d|b2.⇒ d = 3 or d|a2 and d|b2,⇒ d = 1 = (a2, b2), as (a, b) = 1.

Thus, we have, d = 1 or d = 3.

Theorem 2.7.10 If (a, b) = 1, (a, c) = 1 then (a, bc) = 1.

Proof: Since (a, b) = 1 = (a, c),∃m1,m2, n1, n2 ∈ Z such that,

1 = m1a+ n1b = m2a+ n2c

⇒ (n1b)(n2c) = (1−m1a)(1−m2a) = 1− a[k]; [k] = integer⇒ ak + n1n2bc = 1.

So if r = (a, b, c) then r|1 so r = 1 ⇒ (a, bc) = 1. Thus if a is prime to b and a is prime toc, then c is prime to bc. From this theorem, we have the following results

(i) If (a, x) = 1 ⇒ (a, x2) = 1 and in general (a, xn) = 1.

(ii) If a|xn then (a, x) = 1.

(iii) If (a, c) = (b, c) = 1 then (ab, c) = 1.

Theorem 2.7.11 If (a, b) = 1 and c|a then (c, b) = 1.

Proof: Since (a, b) = 1,∃m,n ∈ Z such that 1 = ma + nb. Also, c|a,∃k ∈ Z such thata = ck. Now, 1 = ma+ nb = mkc+ nb = (mk)c+ nb

⇒ (b, c) = 1; mk, n ∈ Z.

Theorem 2.7.12 If (a, b) = 1 then, (a+ b, ab) = 1.

Proof: Since a is prime to b, ∃m,n ∈ Z such that am+bn = 1. This expression am+bn = 1can be written in the form

a(m− n) + (a+ b)n = 1.

Since m − n and n are integers, it follows that a is prime to a + b. Again, the expressionam+ bn = 1 can be written in the form

(a+ b)m+ b(n−m) = 1.

Since m and n−m are integers, it follows that a+ b is prime to b. Hence a+ b is primeto ab.

Ex 2.7.5 If (a, b) = 1, prove that (a2, b) = 1 and (a2, b2) = 1.

Prime Numbers 107

Solution: Since (a, b) = 1, ∃m,n ∈ Z such that am+ bn = 1. Thus,a2m2 = (1− bn)2 = 1− 2bn+ b2n2

⇒ a2m2 + b(2n− bn2) = 1.

Since m2 and (2n − bn2) ∈ Z, it follows that (a2, b) = 1. As (a2, b) = 1, ∃m1, n1 ∈ Z suchthat

b2n21 = (1− a2m1)2 = 1− 2a2m1 + a4m2

1

⇒ a2(2m1 − a2m21) + b2n2

1 = 1.

Since 2m1 − a2m21 and n2

1 ∈ Z, it follows that (a2, b2) = 1. In general, if d = (a, b), then,d2 = (a2, b2).

Theorem 2.7.13 If c|ab and (a, c) = 1 then c|b.

Proof: Since (a, c) = 1,∃m,n ∈ Z such that 1 = ma+ nc. Also, as c|ab,∃k ∈ Z such thatab = kc. Now,

1 = ma+ nc⇒ b = mab+ nbc

= mkc+ nbc = c(mk + nb)= c× an integer ⇒ c|b.

Theorem 2.7.14 If a|c, b|c, (a, b) = d then ab|cd.

Ex 2.7.6 Prove that√m is irrational for any positive prime m.

Solution: If possible, let√m is a rational number. Then,

√m =

p

q; where, p, q ∈ Z, q > 0, (p, q) = 1

or, m =p2

q2⇒ p2 = q2m = q(qm) ⇒ q|p2.

If q > 1, then by fundamental theorem of arithmetic, ∃ a prime m such that m|q. Thus,

m|q and q|p2 ⇒ m|p2 ⇒ m|a.

Then (p, q) ≥ m > 1, a contradiction arises unless q = 1. When, q = 1, we have, p2 = mwhich is not possible, as the square of any integer cannot be prime. Hence,

√m is irrational

for any prime m.

Ex 2.7.7 Prove that Fermat’s numbers are relatively prime to each other.

Solution: The number of the form Fn = 22n

+ 1 is known as a Fermat’s number. It hasbeen also found that the Fermat’s number Fn is prime n ≥ 0.

Let r be the common divisor of the two Fermat’s numbers Fn and Fn+k. Since Fn andFn+k being odd integers cannot have any even integer as a common divisor, so r is odd.Now,

Fn = 22n

+ 1; Fn+k = 22n+k

+ 1 =(22n)2k

+ 1.

Thus, Fn+k − 2 =(22n)2k

− 1 has a factor 22n

+ 1 = Fn.

or, Fn|Fn+k − 2, also, r|Fn ⇒ r|Fn+k − 2.So, r|Fn+k and r|Fn+k − 2 ⇒ r|Fn+k − Fn+k + 2 ⇒ r|2,

which is not possible, since r is odd. Hence, (Fn, Fn+k) = 1, i.e., Fermat’s numbers arerelatively prime to each other.

108 Theory of Numbers

Theorem 2.7.15 Every positive value integer greater than 1 has a least divisor (other than1) which is prime.

Proof: Let n(> 1) be a positive integer. Let S be the set of positive value divisor of nother than 1. So S is non-empty as n ∈ S (since,n|n). Thus S is a non-empty set of naturalnumber. Hence by well ordering principle it has an least element. Let k be the least elementof S. Then k is the least divisor of n other than 1. We assert that k is prime for if k be notprime then

k = k1k2 where, 1 < k1 < k2,

and k1|n, which is contradiction shows that k is a least divisor. Hence k is prime. Therefore,a composite number has at least one prime divisor.

Ex 2.7.8 If 2n − 1 be prime, prove that n is a prime.

Solution: Let n be composite, then n = pq, where, p and q are integers greater than 1.Now,

2n − 1 = 2pq − 1 = (2p − 1)(2p(q−1) + 2p(q−2) + · · ·+ 2p + 1).

Each of the factor on the right hand side is evidently greater than 1 and therefore, 2n− 1 iscomposite. Therefore, 2n − 1 is a prime, i.e., n is prime.

Ex 2.7.9 Let p be prime and a be a positive integer. Prove that an is divisible by p if andonly if a is divisible by p.

Solution: Let a is divisible by p, then a = pk for some k ∈ Z. Thus,

an = pnkn = p(pn−1kn) = p.m

where, m = (pn−1kn) ∈ Z. This shows that an is divisible by p. Let a is not divisible by p,i.e., (a, p) = 1. Therefore, ∃u, v ∈ Z such that au+ bv = 1. Then,

anun = (1− pv)n = 1− ps; s ∈ Zor, anr + ps = 1; r, s ∈ Z.or, (an, p) = 1.

Therefore an is not divisible by p. Hence a is not divisible by p, i.e., an is not divisible byp. Thus p|an ⇒ p|a. Therefore, an is divisible by p if and only if a is divisible by p.

2.7.2 Fundamental Theorem of Arithmetic

Every integer greater than 1 is either a prime number or can be expressed as a productof finite positive primes upto the order of factors and the expression is unique except therearrangement of factors.Proof: Existence: Let n(> 1) be a given positive integer. Since 2 is prime, so if n is 2 ora prime number there is nothing to prove. If n is a composite number then it has a primefactor n1(> 1) and so, ∃ an integer r1 such that

n = n1r1, where, 1 < r1 < n.

Among all such integers n1, choose r1 to be smallest (it is possible by use of well-orderingprinciple). If r1 is prime, then n is product of two primes and the result is obtained. If r1is not prime, then it has a least prime factor n2(> 1) and

n = n1.n2.r2; 1 < r2 < r1 < n.

Prime Numbers 109

This process of factorizing any composite factor is continued n = n1.n2 . . . nk−1.rk−1 where1 < rk−1 < rk−2 < · · · < r2 < r1 < n given a strictly descending chain of positive integersand the chain 1 < rk−1 < rk−2 < · · · < r1 must terminate after a finite steps. If terminates atfinite number of steps rk−1, then, rk−1 is prime say nk. This leads to the prime factorizationn = n1.n2.n3 . . . nk, where ni’s are all prime.Uniqueness : To prove the uniqueness of representation, let us assume that the integer ncan be represented as a product of primes in two ways, say,

n = n1.n2.n3 . . . nk = p1.p2.p3 . . . pl; k ≤ l

where n’s and p’s are all primes. Let l ≥ k then p1|n1.n2.n3 . . . nk and p1 is prime, so p1|ni

for some i and 1 ≤ i ≤ k. Since ni and pi are both prime we conclude pi = ni for some i;1 ≤ i ≤ k. Without loss of any generality we can say

p1 = n1.n2.n3 . . . nk = p2.p3 · · · pl.

Similar argument shows that n2 = p2, · · · , nk = pk, leaving 1 = pk+1 . . . pl which is absurd(as each pi’s are prime and > 1). Hence l = k and pi = ni; ∀i = 1, 2, 3, · · · , k, making thetwo factorizations of n identical. Thus n > 1 can be expressed as a product of finite positiveprimes, the representation being unique apart from the order of the factors. This is knownas fundamental theorem of arithmetic or unique factorization theorem.

Result 2.7.1 Standard form : In the application of this theorem, we may write, anypositive integer n(> 1) can be expressed uniquely in a canonical factorization as

n = p1α1 .p2

α2 . . . prαr , αi ≥ 0, for i = 1, 2, · · · , r,

where pn is the nth prime with p1 < p2 < · · · < pr and the integers If no αi in thecanonical form of n is greater than 1, then integer n is said to be square free. For example,n = 70 = 2.5.7 is square free number, whereas 140 = 22.5.7 is not square free.

Result 2.7.2 Two integers a and b greater than one, then by fundamental theorem ofarithmetic

a = pα11 pα2

2 · · · pαrr and b = pβ1

1 pβ22 · · · pβr

r

then(a, b) = p

minα1,β11 p

minα2,β22 · · · pminαr,βr

r

[a, b] = pmaxα1,β11 p

maxα2,β22 · · · pmaxαr,βr

r .

For example, let a = 491891400 = 23.33.52.72.111.132 and b = 1138845708 = 22.32.72.112.133.171,then

(a, b) = 22.32.50.71.111.132.170 = 468468 and [a, b] = 23.33.52.72.112.133.171 = 1195787993400.

Theorem 2.7.16 (Euclid’s Theorem): The number of prime numbers are infinite; alterna-tively, there is no greater prime.

Proof: If possible, let the number of primes be finite. Then there is a greater prime say pn

and arrange all the primes are ascending order in magnitude p1 < p2 < · · · < pn. Suppose,there is only a finite number, say p1, p2, · · · , pn. Let,

q = (p1.p2 . . . pn) + 1.

110 Theory of Numbers

Here we see that, q > 1, so q is divisible by some prime p. But p1, p2, · · · , pn are the onlyprime numbers, so that p must be equal to one of p1, p2, · · · , pn. Now,

p|p1.p2 . . . pn and p|q⇒ p|q.p1p2 · · · pn ⇒ p|1.

The only positive divisor of the integer 1 is 1 itself and because p > 1, a contradiction arises.If q is prime, we get a contradiction as q > pn. If q is composite it has a prime factor, butnone of the primes p1, p2, . . . , pn divides q (since 1 is the remainder in each case). So theprime of q must be greater than pn, when is again a contradiction. This shows that, thereis no greatest prime i.e. the number of primes are infinite. Now,

(i) Every positive integer greater than one has a prime divisor(factor).

(ii) If n (integer, greater than one) is not a prime, then n has a prime factor not exceeding√n.

(iii) No rational algebraic formula can represent prime numbers only.

(iv) Consider the following consecutive integers(k + 1)! + 2, (k + 1)! + 3, . . . , (k + 1)! + k + 1.

Each of these numbers in a composite number asn|(k + 1)! + n; if 2 ≤ n ≤ k + 1.

Thus there are arbitrarily large gaps in the series of primes.

Ex 2.7.10 If pn is the nth prime number, then pn ≤ 22n−1.

Solution: Clearly, the equality sign holds for n = 1. As an hypothesis of the induction, weassume that the result holds for all integers up to k > 1. Euclid’s theorem shows that theexpression p1p2 . . . pk + 1 is divisible by at least one prime. If there are several such primedivisors, then pk+1 does not exceed the smallest of these so that

pk+1 ≤ p1p2 . . . pk + 1 ≤ 2.22 . . . 22k−1+ 1

≤ 21+2+22+···2k−1+ 1 = 22k−1

+ 1.

However, 1 ≤ 22k−1for all k, whence,

pk+1 ≤ 22k−1 + 22k−1 = 22k

.

Thus the result is true for n = k + 1, if it is true for n = k. Thus for n ≥ 1, there are atleast (n+ 1) primes less than 22k

.

Ex 2.7.11 Determine which of the following integers are primes 287 and 271.

Solution: First we find all primes p such that p2 ≤ 287. These primes are 2,3,5,7,11,13,17.Now, 7|287, hence 287 is not a prime. The primes satisfying p2 ≤ 271 are 2,3,5,7,11,13,17.None of these divide 271, hence 271 is a prime.

Theorem 2.7.17 If be a positive prime and n be a positive integer then prove that an isdivisible by p iff a be divisible by p where a is any positive integer.

Proof: Since a is a positive integer, by fundamental theorem of arithmetic we geta = p1.p2 . . . pk,

where p1.p2 . . . pk are primes and p1 < p2 < . . . < pk. Now, a divisible by p iff one ofp1, p2, . . . , pk is divisible by p. Also, as an = p1

n.p2n . . . pk

n so an is divisible by p iff a isdivisible by p.

Modular/Congruence System 111

2.8 Modular/Congruence System

C.F.Gauss introduces the remarkable concept of congruence and the notion that makes itsuch a powerful techniques for simplification of many problems concerning divisibility ofintegers.

Let m > 0 be a fixed integer. Then an integer a is said to be congruent to another integerb modulo m, if m|(a− b) i.e if m is a divisor of (a− b). Symbolically, this is expressed as

a ≡ b( mod m). (2.11)

The number m is called the modulus of the congruence, b is called the residue of a modulom. In particular, a ≡ 0(modm) if and only if m|a. Hence

a ≡ b(modm) if and only if a− b ≡ 0(modm).

For example,(i) 15 ≡ 7(mod8), 2 ≡ −1(mod3), 52 ≡ −1(mod2).

(ii) n is even if and only if n ≡ 0(mod2).

(iii) n is odd if and only if n ≡ 1(mod2).

(iv) a ≡ b(mod1) for all a, b ∈ Z, this case (m = 1) is not so useful and interesting.Therefore, m is taken to be positive integer greater than 1.

(v) Let a, b be integers and m a positive integer, then a ≡ b(modm), if and only if a =km+ b, for some integer k.

When m 6 |(a−b), we say that a is incongruent to b modulo m and in this case a 6≡ b(modm).For example, 2 6≡ 6(mod5),−3 6≡ 3(mod5).

Ex 2.8.1 Use the theory of congruences to prove that 7|25n+3 + 52n+3; ∀n(≥ 1) ∈ N .

Solution: 25n+3 + 52n+3 can be written as 8.32n + 125.25n. Now,

32n − 25n ≡ 0(mod7), for all n ≥ 1,⇒ 8.32n − 8.25n ≡ 0(mod7), for all n ≥ 1.

Also, we have 133.(25)n ≡ 0(mod7), for all n ≥ 1 and so 8.32n + 125.25n ≡ 0(mod7) for alln ≥ 1. Consequently, 7|25n+3 + 52n+3; ∀n(≥ 1) ∈ N.

2.8.1 Elementary Properties

The congruence is a statement about divisibility slightly different point of view more thanthe convenient notation. Congruence symbol ‘≡’ may be viewed as a generalized form ofequality sign, in the sense that its behavior with respect to addition and multiplication isreminiscent of ordinary equality. Some of the elementary properties of equality that carryover to congruences appear below.

Property 2.8.1 If a ≡ b (mod m), then a ≡ b (mod n), when n|m, m, n > 0.

Proof: From definition, n|m⇒ m = nk for some k ∈ Z. Given a ≡ b(mod m), so

m∣∣∣(a− b) ⇒ a− b = ml for some l ∈ Z

⇒ a− b = nkl = nr; r = kl ∈ Z

⇒ n∣∣∣(a− b) ⇒ a ≡ b(mod n)

112 Theory of Numbers

Property 2.8.2 a ≡ a (mod m), for any m > 0 and a ≡ 0 (mod m), if m|a.

Property 2.8.3 The relation “congruence modulo m,” defined by a ≡ b (mod m) if m|(a−b), is an equivalence relation in the set of integers.

Proof: If m(> 0) be a fixed positive integer, then we define a relation ρ for any twoelements a, b ∈ Z such that

aρb⇔ a ≡ b(mod m).

We are to show that this relation is an equivalence relation.Reflexivity: Let a be any integer, then, we have, a− a = 0 and m|0 for any m(> 0) ∈ Z.Thus it follows that,

m|(a− a); a ∈ Z ⇒ a ≡ a(modm), for all a ∈ Z.⇒ aρa; ∀a ∈ Z.

Thus the relation ρ is reflexive.Symmetry: Let a, b ∈ Z be such that aρb (mod m). Then,

aρb⇒ a ≡ b(mod m) ⇒ m|(a− b) ⇒ m|(−1)(a− b)⇒ m|(b− a) ⇒ b ≡ a (mod m)

Hence, aρb⇒ bρa; ∀a, b ∈ Z.

Therefore, the relation ρ is symmetric.Transitivity: Let a, b, c ∈ Z such that aρb and bρc. Now,

aρb, bρc⇒ a ≡ b(mod m), b ≡ c(mod m)⇒ m|(a− b) and m|(b− c)⇒ m|[(a− b) + (b− c)]⇒ m|(a− c) ⇒ a ≡ c (mod m) ⇒ aρc.

So the relation ρ is transitive. The relation being reflexive, symmetric and transitive is anequivalence relation. Thus, congruence is an equivalence relation in Z.

Result 2.8.1 Hence the equivalence relation will partition I into equivalence classes orresidue classes modulo m. The number of these classes is m. They are denoted as,

[a] = the class in which all integers ≡ a(modm).Hence, [0] = [m] = [2m] = · · · and[a] = [a+m] = [a+ 2m] = · · · .

So the residue classes modulo 5 are

[0] = · · · ,−10,−5, 0, 5, 10, · · ·; [1] = · · · , ,−9,−4, 1, 6, · · ·[2] = · · · ,−8,−3, 2, 7, · · ·; [3] = · · · ,−7,−2, 3, 8, 13, · · ·[4] = · · · ,−6,−1, 4, 9, 14, · · ·

Property 2.8.4 Two congruences with same modulus can be added, subtracted, or mul-tiplied, member by member, as they were equations. Therefore, if a ≡ b (mod m), c ≡ d(mod m) then, a+ c ≡ b+ d (mod m), a− c ≡ b− d (mod m) and ac ≡ bd (mod m).

Proof: Since a ≡ b(modm); c ≡ d(modm), we have assumed that, a = mq + b andc = ms+ d, for some choice of q, s ∈ Z. Hence, adding these equations, we obtain,

a+ c = m(q + s) + (b+ d)⇒ (a+ c)− (b+ d) = m(q + s).

Modular/Congruence System 113

Since q, s ∈ Z so q + s ∈ Z and as a congruent statement a + c ≡ b + d (mod m). Theconverse is not always true. For example, let a = 10, c = 5, b = 1, d = 2 and m = 4. Then(10 + 5) ≡ (1 + 2)(mod4), but 10 6≡ 1(mod4) and 5 6≡ 2(mod4). Again,

ac = (mq + b)(ms+ d) = m(bs+ qd+ qsm) + bd.

Since b, s, q,m, d ∈ Z, bs + qd + qsm ∈ Z says that ac − bd is divisible by m, whenceac ≡ bd (mod m). The converse is not always true. For example, let 50 ≡ 2(mod4), but10 6≡ 1(mod4) and 5 6≡ 2(mod4). In general, if a1 ≡ b1(mod m), a2 ≡ b2(mod m), . . . ,an ≡ bn(mod m), then

a1.a2 · · · an ≡ b1.b2 · · · bn(modm).

Property 2.8.5 If a ≡ b (mod m), then ∀x ∈ Z, a + x ≡ b + x (mod m), a − x ≡ b − x(mod m) and ax ≡ bx (mod m).

Proof: As a ≡ b (mod m), ∃λ ∈ Z such that,

a− b = λm⇒ (a+ x)− (b+ x) = λm

or, (a+ x) ≡ (b+ x)(modm).

Also as a− b = λm, where λ ∈ Z, m|(a− b) and so

m|(ax− bx), where x ∈ Z ⇒ ax ≡ bx(modm).

The converse of the result a ≡ b(modm) ⇒ ax ≡ bx(modm) is not always true. For example,

2.4 ≡ 2.1(mod6), whence 4 6≡ 1(mod6).

Thus we conclude that, one cannot unrestrictedly cancel a common factor in the arithmeticof congruences. The same holds true for any finite number of congruences with the samemodulus. For example,

3.(−2) ≡ 2(mod8) and 3.14 ≡ 2(mod8) ⇒ −2 ≡ 14(mod8).

Cancellation is allowed however, in some restricted sense, which is provided in the followingtheorem.

Property 2.8.6 If a ≡ b(mod m) and d|m,m > 0, then a ≡ b(mod d).

Proof: Given that a ≡ b(modm) and d|m,m > 0. This implies that there are two integersx and y such that

(a− b) = xm and m = yd.

Now (a− b) = xyd. So

(a− b) = zd where xy = z.

Hence, a ≡ b(mod d). The converse of the result is not always true. For example, leta = 5, b = 2 and m = 3, then 5 ≡ 2(mod3). Again 3|6, but 5 6≡ 2(mod6).

Property 2.8.7 A common factor which is relatively prime to the modulus can always becancelled. Thus, if, m be a positive integer and a, x, y be integers, then

ax ≡ ay(mod m) iff x ≡ y(mod

m

d

); where, d = (a,m).

114 Theory of Numbers

Proof: Let d = (a,m) 6= 0, as m > 0, then by definition, d|a, d|m, so that, ∃ k, l ∈ Z suchthat a = kd, m = ld where k and l are prime to each other. Since ax ≡ ay (mod m), i.e.,m|(ax− ay), so, ∃ q ∈ Z such that ax = mq + ay. Now,

kdx = ldq + kdy ⇒ kx = lq + ky; d 6= 0⇒ k(x− y) = lq ⇒ l|k(x− y).

Since k and l are prime to each other, Euclid’s lemma yields l|(x− y), which may be recastas x ≡ y (mod l), i.e.,

x ≡ y (modm

d) ⇒ x ≡ y

(mod

m

(a,m)

).

Thus, a common factor a can be cancelled provided the modulus is divided by d = (a,m).Conversely, let x ≡ y (mod m

d ), then x− y = tmd , for some integer t. Hence,

ax− ay = tm

da = tm

a

d= tmk = (mk)t

⇒ ax ≡ ay(modm).

This theorem gets its maximum force when the requirement that (a,m) = 1 is added, forthen the cancellation may be accomplished without a change in modulus. From this theoremwe have,

(i) If ax ≡ ay (mod m) and (a,m) = 1, then x ≡ y (mod m).

(ii) If x ≡ y(mod mi); i = 1, 2, · · · , r if and only if x ≡ y(mod[m1,m2, · · · ,mr]).

(iii) If ax ≡ ay(modm) and a|m, then x ≡ y(modma ). For example, 5.7 ≡ 5.10(mod15), as

5|15, we get 7 ≡ 10(mod3).

(iv) When ax ≡ 0(modm), with m a prime, then either a ≡ 0(modm) or x ≡ 0(modm).

(v) If ax ≡ ay (mod m) and m 6 |a, where m is a prime number, then x ≡ y(modm).

(vi) It is unnecessary to stipulate that a 6≡ 0(modm). Indeed, if a ≡ 0(modm), then(a,m) = m and in this case x ≡ y(mod1), for all integers x and y.

Property 2.8.8 Let a, b, c, d are integers and m a positive integer. a ≡ b (mod m) andc ≡ d (mod m) then ax+ cy ≡ (bx+ dy) (mod m), for all integers x and y.

Proof: Since, a ≡ b (mod m), c ≡ d (mod m); so m|(a− b) and m|(c− d), i.e., ∃ λ, µ ∈ Zsuch that a− b = mλ and c− d = mµ. For integers x, y we have,

(ax+ cy)− (bx+ dy) = x(a− b) + y(c− d)= mλx+mµy

= m(λx+ µy)

Since λ, µ, x, y ∈ Z so λx+ µy ∈ Z, we get ax+ cy ≡ bx+ dy (mod m).

Property 2.8.9 For arbitrary integers a and b, a ≡ b (mod m) iff a and b leave the samenonnegative principal remainder on division by m.

Proof: Let a = λ1m+r and b = λ2m+r where r is the common principal remainder whena, b are divided by m when λ1, λ2 ∈ Z and 0 ≤ r < m. Therefore,

a− b = (λ1 − λ2)m = λm, where, λ = λ1 − λ2 ∈ Z.⇒ m|(a− b), i.e., a ≡ b(modm).

Modular/Congruence System 115

Conversely, let, a ≡ b (mod m) and a, b leave the remainders r1 and r2 respectively whendivided by m. Hence,

a = λ1m+ r1 ; 0 ≤ r1 < m and λ1 ∈ Zb = λ2m+ r2 ; 0 ≤ r2 < m and λ2 ∈ Z

⇒ a − b = m(λ1 − λ2) + r1 − r2

⇒ r1 − r2 = (a− b) +m(λ2 − λ1).

As a ≡ b (mod m) so that m|(a− b). Also, as m|(a− b) and also m|(λ2 − λ1)m, therefore,

m∣∣∣[(a− b) + m(λ2 − λ1)] ⇒ m|(r1 − r2)

⇒ r1 − r2 = 0; since 0 ≤ |r1 − r2| < m

⇒ r1 = r2.

Thus, the congruent numbers have the same gcd with m. This theorem provides a usefulcharacterization of congruence modulo m in terms of remainders upon division by m. Forexample, let m = 7. Since 23 ≡ 2(mod7) and −12 ≡ 2(mod7), i.e., 23 and -12 leave the sameremainder upon division by 7, so 23 ≡ −12(mod7).

Property 2.8.10 If a ≡ b (mod m) then an ≡ bn (mod m) where n is a positive integer.

Proof: For n = 1, the theorem is certainly true. We assume that the theorem is true forsome positive integer k, so that ak ≡ bk (mod m). Also a ≡ b (mod m). These two relationstogether imply that,

a.ak ≡ b.bk(modm) ⇒ ak+1 ≡ bk+1(modm),

so that the theorem is seen to be true for the positive integer k + 1, if it is true for n = k.Hence the theorem is true for any positive integer n. The converse of the theorem is nottrue, for an example 52 ≡ 42 (mod 3) but 5 does not congruence to 4 (mod 3). The powerapplications are given below.

Ex 2.8.2 Prove that 1920 ≡ 1(mod181).

Solution: We have, 192 ≡ −1(mod181). Therefore,

1920 ≡ (−1)10(mod77) ≡ 1(mod181).

Ex 2.8.3 What is the remainder, when 730 is divided by 4?

Solution: Let r be the remainder, when 730 is divided by 4. Hence by definition, 730− r isdivisible by 4, where 0 ≤ r < 4 and so 730 ≡ r(mod4). Now,

7 ≡ 3(mod4) ⇒ 72 ≡ 32(mod4).

But, 32 ≡ 1(mod4), which implies that (72)15 ≡ 115(mod4), i.e., 730 ≡ 1(mod4). Hence theremainder is 1.

Ex 2.8.4 Let f(x) = a0 + a1x+ · · ·+ an−1xn−1 + anx

n is a polynomial whose coefficientsai are integral. If a ≡ b (mod m), then f(a) ≡ f(b) (mod m).

Solution: We have a ≡ b (mod m) so ak ≡ bk (mod m), where k ∈ Z. Hence,

aiak ≡ aib

k(modm), where, ai ∈ Z.

116 Theory of Numbers

Putting i = 0, 1, 2, . . . , n respectively and adding the congruences, we get,

(a0 + a1a+ a2a2 + · · ·+ ana

n) ≡ (a0 + a1b+ a2b2 + · · ·+ anb

n)(modm)f(a) ≡ f(b)(modm).

If f(x, y, z) be a polynomial in x, y, z with integral coefficient and x = x′(mod m), y =y′(mod m), z = z′(mod m) then,

f(x, y, z) = f(x′, y′, z′)(mod m).

Deduction 2.8.1 Let n = ak10k +ak−110k−1+· · ·+a2102+a110+a0, where ai are integersand 0 ≤ ai ≤ 9; i = 0, 1, · · · , k be the decimal representation of a positive integer n. LetS = a0 + a1 + · · ·+ ak and T = a0 − a1 + · · ·+ (−1)kak. Then

(i) n is divisible by 2 if and only if a0 is divisible by 2;

(ii) n is divisible by 9 if and only if S is divisible by 9;

(iii) n is divisible by 11 if and only if T is divisible by 11.

Ex 2.8.5 Show that an integer N is divisible by 3, if and only if the sum of the digits of Nis divisible by 3.

Solution: Let the number N can be written as

N = am10m + am−110m−1 + · · ·+ a110 + a0.

Let f(x) = a0 + a1x+ · · ·+ am−1xm−1 + amx

m,

so, f(1) = am + am−1 + · · ·+ a1 + a0.

therefore, f(10) = N and f(1) = sum of the digits of N = S (say). Now,

10 ≡ 1(mod3) ⇒ f(10) ≡ f(1)(mod3)⇒ N ≡ S(mod3) ⇒ 3|(N − S).

Thus 3|N, if and only if 3|S. Thus integer N is divisible by 3, if and only if the sum of thedigits of N is divisible by 3.

Ex 2.8.6 N is divisible by 5 if the last digit is either 0 or 5.

Solution: Taking the last digit as a0, the number N can be written as

N = am10m + am−110m−1 + · · ·+ a110 + a0.

Let f(x) = a0 + a1x+ · · ·+ am−1xm−1 + amx

m,

then f(10) = N and f(0) = a0. We have,

10 ≡ 0(mod5) ⇒ f(10) ≡ f(0)(mod5)⇒ N = a0(mod5) ⇒ 5|N − a0.

Now 5|N, then, 5|N − a0 if and only if 5|a0. Therefore,

f |a0 ⇒ either a0 = 0 or, a0 = 5.

Hence, 5|N if the last digit of N is either 0 or 5.

Ex 2.8.7 Show that the integer 23456785 is divisible by 11.

Modular/Congruence System 117

Solution: Let N = 23456785, then N can be written as

N = 23456785 = 23× (1000)2 + 456× 1000 + 785.

Let f(x) = 23x2 + 456x+ 785, then f(1000) = N and f(−1) = 352. Now,

1000 ≡ −1(mod11) ⇒ f(1000) ≡ f(−1)(mod11)⇒ N = 352(mod11) ⇒ 11|N − 352.

Now 11|N, if 11|352 and this is true. Therefore, 11|N.

Ex 2.8.8 Show that the integer 205769 is not divisible by 3.

Solution: Let N = 205769, then N can be written as

N = 205769 = 20× (100)2 + 57× 100 + 69.

Let f(x) = 20x2 + 57x+ 69, then f(100) = N and f(1) = 146. Now,

100 ≡ 1(mod3) ⇒ f(100) ≡ f(1)(mod3)⇒ N = 146(mod3) ⇒ 3|N − 146.

Now 3|N, if and only if 3|146 and this is not true. Therefore, the integer 205769 is notdivisible by 3.

Property 2.8.11 If a ≡ b (mod m) and d|a, d|b and the integers a, b,m are such that(d,m) = 1, then a

d ≡bd (mod m), d > 0.

Proof: Since d|a, d|b, ∃ a1, b1 ∈ Z such that a = da1, b = db1. Now,

a ≡ b (mod m) ⇒ m|(a− b)⇒ m|d(a1 − b1); d(> 0) ∈ Z⇒ m|(a1 − b1); as, (m, d) = 1

⇒ a1 ≡ b1 (modm) ⇒ a

d≡ b

d(modm).

If a ≡ b(modm) and a ≡ b(modn), where, (m,n) = 1, then a ≡ b(modmn). For exam-ple, 8.7 ≡ 2.7(mod6), where (7, 6) = 1, then 8 ≡ 2(mod6). This is known as restrictedcancellation law.

Property 2.8.12 If a ≡ b (mod m) and if 0 ≤ |a− b| < m then a = b.

Proof: Since m|(a− b), we have, m ≤ |a− b|, unless a− b = 0.

2.8.2 Complete Set of Residues

Consider a fixed modulus m > 0. Given an integer a, let q and r be its quotient andremainder upon division by m, so that,

a = mq + r; 0 ≤ r < m.

Then, a− r = mq ⇒ a ≡ r(modm).

r is called the least residue of a modulo m. For example, 5 ≡ 1(mod4), so 1 is the residue of5 modulo 4. Because there are n choices for r, we see that every integer is congruent modulom to exactly one of the values 0, 1, 2, · · · ,m − 1. In particular, a ≡ 0(modm) if and only

118 Theory of Numbers

if m|a. The set of integers 0, 1, 2, · · · ,m − 1 is called the set of least non negative residuesmodulo m.

For example, let a = 8 and m = 3, then

8 = 3.2 + 2 ⇒ 8− 2 = 3.2 ⇒ 8 ≡ 2(mod3).

Then 2 is the least residue of 8 modulo 3. We consider S = 0, 1, 2 and let a = 7 be anyinteger. Then 7 congruent to modulo 3, to exactly one of the number of S, and that is 1. Ifa = 32, then 32 is congruent to only 2 ∈ S modulo 3. Thus if a be any integer, then it mustbe congruent to mod 3, to exactly one of the members of S. S = 0, 1, 2 is called the setof least non-negative residues of an integer, modulo 3.

The following are the important properties of residue class

(i) If a and b be respectively the residue classes a, b modulo m, then a = b, if and only ifa ≡ b(modm).

(ii) Two integers a and b are in the same residue class, if and only if a ≡ b(modm).

(iii) The m residue classes 1, 2, · · · ,mare disjoint and their union is the set of all integers.

(iv) Any two integers in a residue class are congruent modulo m and any two integersbelonging to two different residue classes are incongruent modulo m.

A set S =r1, r2, . . . , rm

of m integers is called a complete residues system modulo m if

ri 6≡ rj(mod m) for 1 ≤ i < j ≤ m

For example,(i) Let m = 11, S = 0, 1, 2, . . . , 10, then ri 6≡ rj(mod m); ∀i, j = 0, 1, . . . , 10, i 6= j.

So S forms a complete system of residues.

(ii) Let m = 5, then S = −12,−15, 82,−1, 31 forms a complete system of residuesmodulo 5.

Complete residue system of a modulo system is not unique. For example,S1 = 0, 1, · · · , 8and S2 = 5, 6, · · · , 13 are two different complete residue system modulo 9.

If r1, r2, . . . , rm be a complete set of residues modulom and (a,m) = 1 then ar1, ar2, . . . , armis a complete set of residues modulo m, as, ri ≡ rj (mod m); 1 ≤ i < j ≤ m, we have

ari ≡ arj(modm) ⇒ m|a(ri − rj).Now, (a,m) = 1 ⇒ m|(ri − rj) ⇒ ri ≡ rj(modm)

⇒ ari ≡ arj(modm).

A reduced residue system modulo m is a set of integers ri such that (ri,m) = 1, ri ≡ rj(mod m) if i 6= j and such that every x prime to n is congruent modulo m to some integerri of the set.

Property 2.8.13 If (a1, a2, · · · , an) be a complete system of residues modulom and b1, b2, · · · , bnany set of integers such that

bi ≡ ai(modm); i = 1, 2, · · · , n

then (b1, b2, · · · , bn) is also a complete system( mod m).

Modular/Congruence System 119

Proof: Let ri be the least residues of ai, modulo m, for i = 1, 2, · · · , n, then ai ≡ ri(modm).Again given bi ≡ ai(modm), therefore, ri are also the same residues of bi(modm), i.e.,bi ≡ ai(modm).

The relation ai ≡ ri(modm) shows that ri 6≡ rj(mod m), then ri − rj is divisible by m,i.e., ri − rj = mk; for some k ∈ Z.

Now, the relation ai ≡ ri(modm) shows that,

ai − ri = mt1 and aj − rj = mt2; for some t1, t2 ∈ Z,⇒ (ai − aj)− (ri − rj) = m(t1 − t2)⇒ (ai − aj) = m(t1 − t2) +mk = mk1; k1 = t1 − t2 + k ∈ Z⇒ ai ≡ aj(modm),

which is a contradiction. Hence, ri ≡ rj(mod m), for every i and j with i 6= j. Therefore,the relation bi ≡ ai(modm) shows that

bi 6≡ bj(mod m),∀i and j with i 6= j

and hence (b1, b2, · · · , bn) forms a complete residue system( modm). Therefore, if a1, a2, · · · , an

be a complete system of residues modulo m, then,

(i) k + a1, k + a2, · · · , k + an

(ii) ka1, ka2, · · · , kan, (k,m) = 1

also form a complete system of residues modulo m.

Property 2.8.14 A set of m integers which are in arithmetic progression with commondifference d, (d,m) = 1, forms a complete residue system modulo m.

Proof: Let us consider the A.P. of m terms with common difference d as

a, a+ d, a+ 2d, · · · , a+ (m− 1)d; (d,m) = 1.Now, a+ id ≡ a+ jd(modm); i, j = 0, 2, · · · ,m− 1; i 6= j

⇒ ikd ≡ jd(modm) ⇒ i ≡ j(modm); (d,m) = 1,

which contradicts that i 6= j(modm). Therefore,

a+ id 6≡ a+ jd(modm).

Ex 2.8.9 Find the remainder when 273 + 143 is divisible by 11.

Solution: Here,

2 ≡ 2(mod11), 24 ≡ 5(mod11), 28 ≡ 3(mod11)⇒ 210 ≡ 3× 22(mod11) ≡ 1(mod11)⇒ 270 ≡ 1(mod11) ⇒ 273 ≡ 8(mod11).

Again, 143 = (11 + 3)3 ≡ 33(mod11) ≡ 5(mod11). Therefore,

273 + 143 ≡ 8 + 5(mod11) ≡ 2(mod11).

Hence the remainder is 2.

Ex 2.8.10 Find the least positive residues of 244(mod89).

120 Theory of Numbers

Solution: We know, 26 ≡ −25(mod89), i.e.,

26 ≡ −52(mod89) ⇒ (26)2 ≡ (−52)2(mod89)or, 212 ≡ 625− 7× 89(mod89) ⇒ 212 ≡ 2(mod89)

⇒ 211 ≡ 1(mod89); as (2, 89) = 1or, (211)4 ≡ 1(mod89) ⇒ 244 ≡ 1(mod89).

Least positive residue is 1, therefore, 89|244 − 1.

Ex 2.8.11 What if the remainder when 15 + 25 + · · ·+ 1005 is divisible by 4.

Solution: We have the following results

1 ≡ 1(mod4) ⇒ 15 ≡ 1(mod4)3 ≡ −1(mod4) ⇒ 35 ≡ −1(mod4)5 ≡ 1(mod4) ⇒ 55 ≡ 1(mod4)

...99 ≡ −1(mod4) ⇒ 995 ≡ −1(mod4)

Adding the above 50 congruence relations, we get,

15 + 35 + · · ·+ 995 ≡ [1− 1 + · · · − 1](mod4) ≡ 0(mod4).

Again, we have the following results

2 ≡ −2(mod4) ⇒ 25 ≡ −25(mod4) ≡ 0(mod4)4 ≡ 0(mod4) ⇒ 45 ≡ −1(mod4)6 ≡ −2(mod4) ⇒ 65 ≡ 0(mod4)

...100 ≡ 0(mod4) ⇒ 1005 ≡ 0(mod4)

Adding the above 50 congruence relations, we get,25 + 45 + · · ·+ 1005 ≡ 0(mod4).

Adding the results, we get,15 + 25 + 35 + 45 · · ·+ 1005 ≡ 0(mod4).

Thus the given expression is completely divisible by 4 and hence the remainder is zero.,when divisible by 4.

Ex 2.8.12 Show that 3.52n+1 + 23n+1 ≡ 0(mod17) for all n ≥ 1.

Solution: The expression 3.52n+1 + 23n+1 can be written as

3.52n+1 + 23n+1 = 15.52n + 2.23n.

We have the following results,25 ≡ 8(mod17) ⇒ 52 ≡ 8(mod17)

or, (52)n ≡ 8n(mod17) ⇒ 15.52n ≡ 8n.15(mod17).Also, 8 ≡ 8(mod17) ⇒ 23n ≡ 8n(mod17)

⇒ 2.32n ≡ 8n.2(mod17).

Modular/Congruence System 121

Adding the two results, we get,

15.52n + 2.23n ≡ 8n(15 + 2)(mod17) ≡ 0(mod17)or, 3.52n+1 + 23n+1 ≡ 0(mod17).

Ex 2.8.13 For any natural number n, show that (2n+ 1)2 ≡ 1 (mod 8)

Solution: Here, (2n+ 1)2 can be written as

(2n+ 1)2 = 4n2 + 4n+ 1 = 4n(n+ 1) + 1.

Now, as n ∈ Z, n and (n + 1) are two consecutive numbers so 2|n(n + 1) so that ∃ k ∈ Zs.t. n(n+ 1) = 2k. Therefore,

(2n+ 1)2 = 4n(n+ 1) + 1 = 8k + 1⇒ (2n+ 1)2 − 1 = 8k, where k ∈ Z⇒ (2n+ 1)2 ≡ 1(mod8).

Ex 2.8.14 Prove that 1! + 2! + · · ·+ 1000! ≡ 3 (mod 15).

Solution: For n = 0 and for any integer n ∈ Z we have (5 + n)! ≡ 0(mod 15). Now,1! + 2! + 3! + 4! = 33 so that 1! + 2! + 3! + 4! ≡ 3 (mod 15). Hence

1! + 2! + · · ·+ 1000! ≡ 3(mod15).

Hence, we the remainder is 3, when 1! + 2! + · · ·+ 1000! is divisible by 15.

Ex 2.8.15 Find all natural number n ≤ 100 such that n ≡ 0(mod 7).

Solution: Here 100 = 14.7 + 2. Hence the required natural numbers n ≤ 100 such thatn ≡ 0 (mod 7) are 7, 14, 21, . . . , 98.

2.8.3 Reduced Residue System

By a reduced residue system modulo m we mean any set of φ(m) integers, incongruentmodulo m, each of which is relatively prime to m, where φ(m) is Euler’s phi function.Reduced set of modulo m can be obtained by deleting from a complete set of residuesmodulo m those members that are not relatively prime to m. A reduced set of residuestherefore, consists of the numbers of a complete system, which are relatively prime to themodulus. For example, in the modulo 8 system, complete set of residue is 0, 1, 2, · · · , 7,its reduced system is 1, 3, 5, 7.

Ex 2.8.16 Find the reduced residue system of m = 40.

Solution: We note that 40 = 5.23, and the suitable reduced residue system of 5 and 23

respectively are 1, 9, 17, 33; 1, 11, 21, 31. Therefore, residue system of 40 is seen from thefollowing table: can be arranged in n lines, each containing m numbers. Thus,

−− 1 9 17 331 1 9 17 3311 11 11× 9 ≡ 19 11× 17 ≡ 27 11× 33 ≡ 321 21 21× 9 ≡ 29 21× 17 ≡ 37 21× 33 ≡ 1331 31 31× 9 ≡ 39 31× 17 ≡ 7 31× 33 ≡ 23

Therefore, the reduced residue system is 1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, 31, 33, 37, 39modulo 42.

122 Theory of Numbers

Ex 2.8.17 Find the least positive residues 336(mod77).

Solution: We know, 34 ≡ 4(mod77), therefore,

312 ≡ 43(mod77) = −13(mod77)⇒ 324 ≡ 169(mod77) = 15(mod77)⇒ 336 ≡ 15.(−13)(mod77) = 36(mod77).

Hence the least positive residue is 36.

2.8.4 Linear Congruences

Let a, b be two integers and m be a positive integer. An equation in unknown x of the formax ≡ b(modm) is called a linear congruence and by a solution of such an equation we meanan integer x0 for which ax0 ≡ b(modm). By definition,

ax0 ≡ b(modm) ⇒ ax0 − b = mk; for some k ∈ Z⇒ m|(ax0 − b).

In finding the solution we observe that the set of non-negative integers 0, 1, 2, . . . ,m − 1forms a complete residue system modulo m. There exists at least one member of S whichsatisfies the linear congruence ax ≡ b(mod m). Thus, the system has either a single solutionor more than one solution which are incongruent to each other with mod m. Thus, theproblem of finding all integers that will satisfy the linear congruence ax ≡ b(modm) isidentical with that of obtaining all solutions of the linear Diophantine equation ax−my = k.For example,

(i) Consider 4x ≡ 3(mod 5) and S = 0, 1, 2, 3, 4. Hence we see that only 2 ∈ S satisfiesthe linear congruence, so 2 is the only solution of linear congruence. Also we observethat (a,m) = (4, 5) = 1.

(ii) Consider 6x ≡ 3(mod 9) and S = 0, 1, 2, . . . , 8. Note that 2, 5, 8 ∈ S satisfies thelinear congruence.

Thus, the linear congruence system has more than one solution. Also,

2 6≡ 5(mod 9), 5 6≡ 8(mod 9), 2 6≡ 8(mod 9).

Hence the solutions are incongruent to each other. Hens we observe that (a,m) = (6, 9) 6= 1,i.e. not prime to each other. Hence, x ≡ 2, 5, 8(mod 9).Note : Now, x = 2 and x = −4 both satisfy the linear congruence 2x ≡ 4(mod6) as2 ≡ −4(mod6), so they are not counted as different solution. Therefore, when we speak tothe number of solutions of a congruence ax ≡ b(modm), we shall mean number of incongruentintegers satisfying this congruence.

Theorem 2.8.1 If x1 be a solution of the linear congruence ax ≡ b(modm) and if x1 ≡x2(modm), then x2 is also a solution of the congruence.

Proof: Given that x1 be a solution of the linear congruence ax ≡ b(modm), therefore,ax1 ≡ b(modm). Now,

x2 ≡ x1(modm) ⇒ ax2 ≡ ax1(modm)⇒ ax2 ≡ b(modm).

Modular/Congruence System 123

Thus, x2 is a solution of the linear congruence ax ≡ b(modm). From this theorem, we have,if x1 be a solution of the linear congruence ax ≡ b(modm), then

x1 + λm; λ = 0,±1,±2, · · ·

is also a solution. All these solutions belong to one residue class modulo m and these arenot counted as different solutions.

Theorem 2.8.2 Let a, b and m be integers with m > 0 and (a,m) = 1. Then the congruenceax ≡ b(modm) has a unique solution.

Proof: Since (a,m) = 1, ∃u, v ∈ Z such that au+mv = 1 and so

a(bu) +m(bv) = b⇒ a(bu) ≡ b(modm).

This shows that x = bu is a solution of the linear congruence ax ≡ b(modm). Let x1, x2 besolutions of the linear congruence ax ≡ b(modm), then ax1 ≡ b(modm) and ax2 ≡ b(modm).This gives

ax1 ≡ ax2(modm) ⇒ x1 ≡ x2(modm); as (a,m) = 1.

This proves that the congruence has an unique solution. The solutions are written in theform x = bu + λm;λ = 0,±1,±2, · · · and they all belong to one and only one residue classmodulo m.

Result 2.8.2 In finding the solution when (a,m) = 1, let am be expressed as a simple

continued fraction with an even number of quotients and y0x0

be the least convergent butone. Then ax0 −my0 = 1 so that

ax0 ≡ 1(mod m) ⇒ abx0 ≡ b(mod m)⇒ bx0 is the required solution of ax ≡ b(mod m)⇒ x ≡ bx0(mod m)

Ex 2.8.18 Solve the linear congruence 5x ≡ 3(mod 11).

Solution: Since d = (a,m) = (5, 11) = 1, the given linear congruence has unique solution.Since (5, 11) = 1 ∃ integers u, v such that 5u+ 11v = 1. Here u = −2, v = 1, so

5.(−2) + 11.1 = 1,i.e., 5.(−2) ≡ 1(mod 11) ⇒ 5.(−6) ≡ 3(mod 11).

Therefore, the value of x is −6. All the solutions are x ≡ −6(mod11), i.e., x ≡ 5(mod11).All the solutions are congruent to 5(mod11) and therefore, the given congruence has uniquesolution.

Ex 2.8.19 Solve the linear congruence 47x ≡ 11(mod249).

Solution: Since (47, 249) = 1, so the given linear congruence 47x ≡ 11(mod249) has anunique solution. We express it as a simple continued fraction with even no. of quotients.

47249

= 0 +1

5+1

3+1

2+1

1+14

Last but one congruent is p5q5

= 1035

(= y0

x0

). Therefore,

x ≡ 11× 53(mod 249)or, x ≡ 583− 2× 249(mod 249)or, x ≡ 85(mod 249).

124 Theory of Numbers

Theorem 2.8.3 Let a, b and m be integers with m > 0. The linear congruence ax ≡b(modm) has incongruent solutions if and only if d|b, where d = (a,m) 6= 1. Also, ifd|b, then there are exactly d mutually incongruent solutions modulo m.

Proof: The given linear congruence ax ≡ b(modm), i.e., ax− b = my, for some m ∈ Z isequivalent to the linear Diophantine equation ax−my = b. Since (a,m) = d, so,

d|a and d|m⇒ a = da1 and m = dm1

for some integers a1 and m1 with (a1,m1) = 1. Then,

ax ≡ b(mod m) ⇒ da1 ≡ b(mod dm1) (2.12)

We require these values of x for which da1x − b is divisible by dm1. No such value of x isobtained unless d|b. Thus if dm1|da1x− b and d|b, i.e. if b = db1 for some integer b1, then

dm1|da1x− db1 ⇒ m1|a1x− b1

⇒ a1x− b1 is divisible by m1

⇒ a1x ≡ b1(mod m1); where (a1,m1) = 1 (2.13)

which is equivalent form of (2.12). Therefore, the congruence (2.13) has one solution x0 < m′

and the d distinct solutions can be written in the form

x0, x0 +m′, x0 + 2m′, . . . , x0 + (d− 1)m′

i.e. x0, x0 +m

d, x0 + 2

m

d, · · · , x0 + (d− 1)

m

d; m′ =

m

d(2.14)

which is also d incongruence solution of (2.12). We assert that, these integers are incongruentmodulo m, and all other such integers x are congruent to some one of them. If it happensthat,

x0 +m

dk1 ≡ x0 +

m

dk2(modm); where,0 ≤ k1 < k2 ≤ d− 1,

⇒ m

dk1 ≡

m

dk2(modm).

Now, (md ,m) = m

d and so, the factor md can not cancelled to arrive at the congruence

k1 ≡ k2(mod d), 0 ≤ |k1 − k2| < d⇒ d|(k2 − k1),

which is impossible due to the fact that 0 ≤ |k2 − k1| < d. Therefore,

k1 − k2 = 0 ⇒ k1 = k2.

So the d distinct solutions are incongruent to each other modulo m.Now, we are to show that, any other solution x0 + m

d k is congruent modulo m to one ofthe d integers 0, 1, 2, · · · , d − 1, i.e., the congruent equation has no solutions except thoselisted in (2.14). Using division algorithm, we get, k = qd+ r; 0 ≤ r ≤ d− 1, and therefore,

x0 +m

dk = x0 +

m

d(qd+ r) = x0 +mq +

m

dr

≡ x0 +m

dr(mod m); 0 ≤ r < d

with x0 + md r being one of our d selected solutions. Thus, if x0 is any solution of ax ≡

b(modm), then the d incongruent distinct solutions are given by,

x = x0 +m

dk(mod m); k = 0, 1, 2, · · · , d− 1

i.e., x0, x0 +m

d, x0 + 2

m

d, · · · , x0 + (d− 1)

m

d.

Modular/Congruence System 125

Ex 2.8.20 Solve the linear congruence 20x ≡ 10(mod 35).

Solution: Since d = (a,m) = (20, 35) = 5 and 5|10, the given linear congruence has uniquesolution. The congruence 20x ≡ 10(mod 35) is equivalent to the linear congruence

4x ≡ 2(mod 7), where, (4, 7) = 1,

so, 4x ≡ 2(mod 7) has an unique solution. Since (4, 7) = 1 ∃ integers u, v such that4u+ 7v = 1. Here u = 2, v = −1, so

4.2 + 7.(−1) = 1,⇒ 4.2 ≡ 1(mod 7) ⇒ 4.4 ≡ 2(mod 7).

Therefore, the value of x is 4. The incongruent solutions are

x = 4 +355t = 4 + 7t; t = 0, 1, 2, 3, 4.

Ex 2.8.21 Solve : 25x ≡ 15(mod 120).

Solution: Here d = (25, 120) = 5 and 5(= d)|10(= b), the given linear congruence hasunique solution. The congruence 25x ≡ 15(mod 120) is equivalent to the linear congruence

5x ≡ 3(mod 24), where, (5, 24) = 1,or, 5x ≡ −45(mod24) ⇒ x ≡ −9(mod24)or, x ≡ 15(mod24); i.e., x ≡ 15 + 24t; t = 0, 1, 2, 3, 4,

having 5 solutions. Therefore, x = 15, 39, 63, 87, 111 are 5 incongruent solutions.

2.8.5 Simultaneous Linear Congruences

Here we consider to the problem of solving a system of simultaneous linear congruences

a1x ≡ b1 (mod m1), a2x ≡ b2 (mod m2), · · · , akx ≡ bk (mod mk). (2.15)

We assume that, the moduli mr are relatively prime in pairs. Evidently, the system of twoor more linear congruences will not have a solution unless dr|br;∀r, where dr = (ar,mr).When these conditions are satisfied, the factor dr can be cancelled in the kth congruence toproduce a new system having the same set of solutions as the original one

a∗1x ≡ b∗1 (mod n1), a∗2x ≡ b∗2 (mod n2), · · · , a∗kx ≡ b∗k (mod nk),

where nr = mr

drand (ni, nj) = 1, for i 6= j. Also, (a∗r , nr) = 1. The solutions of the individual

congruences assume the form

x ≡ c1 (mod n1), x ≡ c2 (mod n2), · · · , x ≡ ck (mod nk). (2.16)

Thus, the problem is reduced to one of finding a simultaneous solution of a system of con-gruences of this type. The kind of problem that can be solved by simultaneous congruencesis given by the following theorem.

Deduction 2.8.2 Let x ≡ a(modp) and x ≡ b(modp) be two simultaneous congruences andlet (p, q) = d. Since x ≡ a(modp), so x = a+ py, where y is given by

a+ py ≡ b(modq) ⇒ py ≡ b− a(modq).

If b − a is not divisible by d, then the solution does not exist. But if d|b − a, there is onlyone solution y1 of y < q

d , which satisfies the last congruence and the general value of y isgiven by

126 Theory of Numbers

y = y1 +q

dt; t ∈ Z

so that x = x1 +pq

dt; where, x1 = a+ py1.

Hence, x = a+ py1 +pq

dt; t ∈ Z.

Thus a solution x1 of the given congruences exists if and only if b−a is divisible by d = (p, q)and the congruence are equivalent to a single congruence x ≡ x1(modl), where l = [p, q].

Ex 2.8.22 Find the general values of x for x ≡ 1(mod 6) and x ≡ 4(mod 9).

Solution: Here, a = 4, b = 1, so that a − b = 3 and (9, 6) = 3 = d so that d|a − b. So thesolution exists. Now,

x ≡ 1(mod 6) ⇒ x = 1 + 6y,

where, y is given by 1 + 6y ≡ 4(mod9) ⇒ 6y ≡ 3(mod9,

which has a solution y = 2(= y1) < qd = 3 and the general values of y are given by

y = y1 +q

dt = 2 + 3t; t ∈ Z

or, x = 1 + 6(2 + 3t) = 13 + 18t; as x = 1 + 6y

which gives the general values of the given congruences equivalent to a single congruencex ≡ 13(mod18, where [p, q] = [9, 6] = 18.

Theorem 2.8.4 Let m1,m2, · · · ,mr denote r positive integers with (mi,mj) = 1, 1 ≤ i <j ≤ r. Let a1, a2, · · · , ar be arbitrary integers. Then the system of linear congruences

x ≡ a1 (mod m1), x ≡ a2 (mod m2), · · · , x ≡ ar (mod mr) (2.17)

has unique simultaneous solution x0 modulo the product m, m = m1.m2 . . .mr, i.e., x ≡x0(mod m).

Proof: Here we take, m = m1.m2 . . .mr. For each k = 1, 2, · · · , r let us define,

Mk =m

mk= m1.m2 . . .mk−1.mk+1 · · ·mr,

i.e., Mk is the product of all integers mr with the factor mk omitted. By hypothesis, mk’sare respectively prime in pairs, i.e., (mi,mj) = 1 for i 6= j, so that (Mk,mk) = 1. Hence bythe theory of linear congruence, it is therefore possible to find the linear congruence

Mkx ≡ 1(modmk).

Let the unique solution be x0. We are to show that the integer

x0 = a1M1x1 + a2M2x2 + · · ·+ arMrxr

is the simultaneous common solution of the given system of congruences. As, mk|Mi; i 6= k,in this case, Mi ≡ 0(mod mk). Hence

x0 = a1M1x1 + a2M2x2 + · · ·+ arMrxr ≡ akMkxk(modmk).

Modular/Congruence System 127

But the integer xk was chosen to satisfy the congruence Mkx ≡ 1(modmk), which gives,

Mkxk ≡ 1(mod mk) ⇒ akMkxk ≡ ak(mod mk)⇒ x0 ≡ ak.1 = ak(mod mk).

This shows that a solution x0 to the system (2.17) of congruences exists. Let, x∗0 be anyother integer that satisfies these congruences, then,

x0 ≡ ak ≡ x∗0(mod mk); k = 1, 2, · · · , r⇒ mk|(x0 − x∗0); for each value of k.

As (mi,mj) = 1, we have,m1m2 . . .mr|(x0 − x∗0) ⇒ x∗0 ≡ x0(modm).

This shows the uniqueness of the solution. This is the Chinese remainder theorem.

Ex 2.8.23 Solve the simultaneous linear congruence x ≡ 36(mod 41), x ≡ 5(mod 17).

Solution: From the given simultaneous linear congruences, we see that m1 = 41,m2 = 17,so that (m1,m2) = (41, 17) = 1. Let m = m1.m2 = 41.17 = 697 and let

M1 =m

m1=

69741

= 17,M2 =m

m2=

69717

= 41,

then (M1, 41) = 1 and (M2, 1) = 1. Since, (M1, 41) = 1, the linear congruence 17x ≡1(mod 41) has an unique solution. Since,

17.(−12) + 41.5 = 1, i.e., 17.(−12) ≡ 1(mod41),

so the solution is x1 ≡ (−12)(mod 41) ≡ 29(mod 41). Since, (M2, 17) = 1, the linearcongruence 41x ≡ 1(mod 17) has an unique solution. Since,

41.5 + 17.(−12) = 1, i.e., 41.5 ≡ 1(mod 17),

so the solution is x2 ≡ 5(mod 41). Therefore, the common integer solution of the givensystem of linear congruences is given by

x0 ≡ a1M1x1 + a2M2x2 ≡ 36.(17.29) + 5.(41.5)(mod 697)≡ 18773(mod 697) ≡ 36(mod 697).

Ex 2.8.24 Solve the following system of linear congruences x ≡ 2(mod 7), x ≡ 5(mod 19)and x ≡ 4(mod 5).

Solution: Let m = 7.19.5. Now, we consider the following simultaneous linear congruences

m

7x ≡ 1(mod 7),

m

19x ≡ 1(mod 19) and

m

5x ≡ 1(mod 5)

i.e., 95x ≡ 1(mod 7), 35x ≡ 1(mod 19) and 133x ≡ 1(mod 5)i.e., (91 + 4)x ≡ 1(mod 7), (38− 3)x ≡ 1(mod 19) and (130 + 3)x ≡ 1(mod 5).

Now, we consider the system of congruences

4x ≡ 1(mod 7),−3x ≡ 1(mod 19) and 3x ≡ 1(mod 5).

128 Theory of Numbers

Now, x = 2 is a solution of the first linear congruence, x = 6 is a solution of the second−3x ≡ 1(mod 19), and x = 2 is a solution of the third 3x ≡ 1(mod 5). Therefore, thelinear congruences

95x ≡ 1(mod 7), 35x ≡ 1(mod 19) and 133x ≡ 1(mod 5)

are satisfied by x = 2, 6, 2 respectively. Hence a solution of the given system is given by,

x0 = 2.2.95 + 5.6.35 + 4.2.133 = 2494

and the unique solution is given by,

x ≡ 2494(mod 7.19.5) ⇒ x ≡ 499(mod 665).

Ex 2.8.25 If x ≡ a(mod16), x ≡ b(mod5) and x ≡ c(mod11), then show thatx ≡ 385a+ 176b− 560c(mod880).

Solution: Here, m1 = 16,m2 = 5,m3 = 11, and they are relatively prime. Therefore,m = m1m2m3 = 880. Chinese remainder theorem is applicable. Now,

m

m1y1 ≡ 1(mod16) ⇒ 55y1 ≡ 33(mod16)

⇒ 5y1 ≡ 3(mod16) ⇒ y1 ≡ 7(mod16)m

m2y2 ≡ 1(mod5) ⇒ 176y2 ≡ 1(mod5)

⇒ y2 ≡ 1(mod5)m

m3y3 ≡ 1(mod11) ⇒ 80y3 ≡ 1(mod11)

⇒ y3 ≡ 4(mod11) ⇒ y3 ≡ −7(mod11)

The integer solution of the solution is given by

x0 ≡m

m1y1a+

m

m2y2b+

m

m3y3c(mod880)

≡ 55a× 7 + 176b× 1 + 80c× (−7)(mod880)≡ 385a+ 176b− 560c(mod880).

Ex 2.8.26 Find four consecutive integers divisible by 3, 4, 5, 7 respectively.

Solution: Let n, n+ 1, n+ 2 and n+ 3 be four consecutive integers divisible by 3, 4, 5 and7 respectively, then

n ≡ 0(mod3), n+ 1 ≡ 0(mod4), n+ 2 ≡ 0(mod5) and n+ 3 ≡ 0(mod7). · · · (i)

We are to solve simultaneous linear congruence (i) by using the Chinese remainder theorem.For this, let m = 3.4.5.7 as they are prime to each other. Now, let

M1 =m

3= 140,M2 =

m

4= 105,M3 =

m

5= 84 and M4 =

m

7= 60,

where (M1, 3) = 1, (M2, 4) = 1, (M3, 5) = 1 and (M4, 7) = 1.

(i) Since, (140, 3) = 1, the linear congruence 140x ≡ 1(mod 3) has an unique solution(mod 3) and the solution is x = x1 = 2.

(ii) Since, (105, 4) = 1, the linear congruence 105x ≡ 1(mod 4) has an unique solution(mod 4) and the solution is x = x2 = 1.

Modular/Congruence System 129

(iii) Since, (84, 5) = 1, the linear congruence 84x ≡ 1(mod 5) has an unique solution (mod5) and the solution is x = x3 = 4.

(iv) Since, (60, 3) = 1, the linear congruence 60x ≡ 1(mod 7) has an unique solution (mod7) and the solution is x = x4 = 2.

Thus, the common integer solution of the given system of congruences is given by

x0 ≡ a1M1x1 + a2M2x2 + a3M3x3 + a4M4x4(mod 420)≡ 1803(mod 420).

Therefore, the consecutive integers are n, n+ 1, n+ 2 and n+ 3, where,

n = 123 + 420t; t = 0,±1,±2, · · · .

Ex 2.8.27 Find the integer between 1 and 1000 which leaves the remainder 1, 2, 6 whendivided by 9, 11, 13 respectively.

Solution: The required integer between 1 and 1000 is a solution of the system of linearcongruences

x ≡ 1(mod 9), x ≡ 2(mod 11) and x ≡ 6(mod 13).

Now we are to solve these system of linear congruences x ≡ 1(mod 9), x ≡ 2(mod 11) andx ≡ 6(mod 13) by using the Chinese remainder theorem. For this, let M = 9.11.13. Now,we consider the congruences

13.11x ≡ 1(mod 9), 13.9x ≡ 1(mod 11) and 9.11x ≡ 1(mod 13)i.e., 143x ≡ 1(mod 9), 117x ≡ 1(mod 11) and 99x ≡ 1(mod 13)i.e., (144− 1)x ≡ 1(mod 9), (110 + 7)x ≡ 1(mod 11) and (91 + 8)x ≡ 1(mod 13).

Now, we consider the system of congruences

−x ≡ 1(mod 9), 7x ≡ 1(mod 11) and 8x ≡ 1(mod 13).

Notice that, x = 8 is a solution of the first linear congruence −x ≡ 1(mod 9), x = 8 isa solution of the second linear congruence 7x ≡ 1(mod 11), and x = 5 is a solution ofthe third linear congruence 8x ≡ 1(mod 13). Therefore, the linear congruences 143x ≡1(mod 9), 117x ≡ 1(mod 11) and 99x ≡ 1(mod 13) are satisfied by x = 8, 8, 5 respectively.Hence a solution of the given system is given by,

x0 = 1.8.11.13 + 2.8.9.13 + 6.5.9.11 = 5986

and the unique solution is given by,

x ≡ 5986(mod 9.11.13) ⇒ x ≡ 838(mod 1287).

Ex 2.8.28 Solve the linear congruence 32x ≡ 79(mod125) by applying Chinese remaindertheorem.

Solution: The canonical form of 1225 is 1225 = 52.72, and (52, 72) = 1. Thus the solutionof the given linear congruence 32x ≡ 79(mod125) is equivalent to finding a simultaneoussolution of the congruences

32x ≡ 79(mod 25) and 32x ≡ 79(mod 49).equivalently, 7x ≡ 4(mod 25) and 16x ≡ 15(mod 49). · · · (i)

130 Theory of Numbers

We are to solve simultaneous linear congruence (i) by using the Chinese remainder theorem.For this, let m = 25.49 as they are prime to each other. Now, let

M1 =m

25= 49,M2 =

m

49= 25,

where (M1, 25) = 1, (M2, 49) = 1.

(i) Since, (49, 25) = 1, the linear congruence 49x ≡ 1(mod 25) has an unique solution(mod 25) and the solution is x = x1 = 24.

(ii) Since, (25, 49) = 1, the linear congruence 25x ≡ 1(mod 49) has an unique solution(mod 49) and the solution is x = x2 = 2.

Thus, the common integer solution of the given system of congruences is given by

x0 ≡ a1M1x1 + a2M2x2(mod m)≡ 26072(mod 1225) ≡ 347(mod1225).

Ex 2.8.29 Give an example of an congruence which has more roots than its degree.

Solution: We know x2 ≡ 1(mod8) has four distinct solutions x = 1, 3, 5, 7, but x2 ≡1(mod8) is of degree 2.

Result 2.8.3 Symbolic fraction method Let the linear congruence be ax ≡ b(mod m),where (a,m) = 1, then we have

ax ≡ b+mh(mod m)

where h is an arbitrary integer. For examples if we consider 5x ≡ 2 (mod 16) then,

x ≡ 25(mod 16) ⇒ x ≡ 2 + 3.16

5(mod 16) ⇒ x ≡ 10(mod 16).

2.8.6 Inverse of a Modulo m

If (a,m) = 1, then the linear congruence ax ≡ b(mod m) has a unique solution modulo m.This unique solution of ax ≡ 1(mod m) is sometimes called the multiplicative inverse orreciprocal of a modulo m. From the definition, it follows that, if a is the reciprocal of a,then ba is the solution of ax ≡ b(mod m). An element a is said to have an unit element, ifit has an inverse modulo m.

Since (1, 12) = 1 = (5, 12) = (7, 12) = (11, 12), so 1, 5, 7, 11 are units of modulo 12.

Ex 2.8.30 Find the inverse of 12 modulo 17, if it exists.

Solution: Consider the linear congruence 12x ≡ 1(mod17). Since (12, 17) = 1, it followsthat the linear congruence 12x ≡ 1(mod 17) has a solution. Hence there exists an inverseof 12 modulo 17. By division algorithm

17 = 12.1 + 5; 12 = 5.2 + 2 and 5 = 2.2 + 1.Since (17, 12) = 1 so inverse of 12 exists. Now, from above we write

1 = 5− 2.2 = 5− 2.(12− 5.2) = 5− 2.12 + 5.4= 5.5− 2.12 = 5.(17− 12.1)− 2.12= 5.17− 5.12− 2.12 = 12(−7) + 17.5.

This shows that 12(−7) ≡ 1(mod 17). Therefore, −7 is a solution of 12x ≡ 1(mod 17).Hence, −7 is an inverse of 12 modulo 17.

Fermat’s Theorem 131

Ex 2.8.31 If possible, find the inverse of 35 modulo 48.

Solution: Since (35, 48) = 1, so inverse of 35 modulo 48 exists. Now48 = 1× 35 + 13, 35 = 2× 13 + 4, 13 = 1× 9 + 4,1 = 9 + (−2)× 4 = 9 + (−2)[13 + (−1)× 9] = 11× 35 + (−8)× 48.

Hence 11 is the inverse of 35 modulo 35.

2.9 Fermat’s Theorem

Let a be an integer. If p be a prime, or p does not divide a, then respectively

ap−1 ≡ 1(mod p) or ap ≡ a(mod p). (2.18)

Proof: Let us consider the set R of nonzero residue classes of integers modulo p as

R = a, 2a, 3a, · · · , (p− 1)a.

It forms a multiplicative group of order p − 1. We shall first show that, no two distinctmembers of the above (p−1) integers are congruent to each other modulo p. Let if possible,

ra ≡ sa(modp); 1 ≤ s < r ≤ p− 1⇒ (r − s)a ≡ 0(modp) ⇒ p|(r − s)a;⇒ p|(r − s) or p|a; as p is prime .

Since 1 ≤ s < r ≤ p−1, we have p 6 |(r− s) and by hypothesis, p 6 |a. Hence, ra 6≡ sa(modp).Also, we find that ra 6≡ 0(modp) for r = 1, 2, . . . , p− 1. Hence,

ra ≡ k(modp), where, k ∈ Z and 0 < k ≤ p− 1.

Since no two distinct members of R are congruent to each other and there are (p−1) distinctintegers a, 2a, · · · , (p−1)a, it follows that the (p−1) integers in R must be congruent modulop to 1, 2, · · · , (p− 1) taken in some order. Let p be not a divisor of a. Therefore,

a.2a.3a . . . (p− 1)a ≡ 1.2.3 . . . (p− 1)(modp)⇒ ap−1[1.2.3 . . . (p− 1)] ≡ 1.2.3 . . . (p− 1)(modp)⇒ ap−1(p− 1)! ≡ (p− 1)!(modp)⇒ ap−1 ≡ 1(modp); as (p, (p− 1)!) = 1.

Hence the theorem. The converse of this theorem is not always true. For example, 2340 ≡1(mod341), as 341 = 11.31, so 341 is not a prime number.

Result 2.9.1 Let p be a divisor of a, i.e., p|a, then a = pk, for some k ∈ Z. Therefore,

ap − a = a(ap−1 − 1

)= pk

(ap−1 − 1

)= pt,

where, t = k(ap−1 − 1

)∈ Z

⇒ ap − a is divisible by p⇒ ap − a ≡ 0(mod p) ⇒ ap ≡ a(mod p).

Ex 2.9.1 Prove that15n5 +

13n3 +

715n is an integer for every n.

132 Theory of Numbers

Solution: n5 ≡ n(mod5) and n3 ≡ n(mod3), by Fermat’s theorem. Then

5∣∣∣(n5 − n) and 3

∣∣∣(n3 − n) ⇒ n5 = 5t+ n, n3 = 3s+ n,

for some integer t and s.Now

15n5 +

13n3 +

715n = (t+ s) +

7n+ 5n+ 3n15

= (t+ s+ n) = an integer.

Ex 2.9.2 Use Fermat’s theorem to prove that a12 − b12, is divisible by 13 × 7, where, a, bare both prime to 91.

Solution: Since a is prime to 91, a is prime to both 13 and 7. Using Fermat’s theorem,

a12 − 1 ≡ 0(mod13) and a6 − 1 ≡ 0(mod7).

Since a6 − 1 ≡ 0(mod7), it follows that a12 − 1 ≡ 0(mod7). Also, a12 − 1 ≡ 0(mod13) anda12 − 1 ≡ 0(mod7), so,

a12 − 1 ≡ 0(mod91); as (13, 7) = 1.Similarly, b12 − 1 ≡ 0(mod91); as (13, 7) = 1.

⇒ a12 − b12 ≡ 0(mod91)

which is required.

Theorem 2.9.1 If p be prime > 2 then 1p + 2p + · · ·+ (p− 1)p ≡ 0 (mod p).

Proof: We have,1p ≡ 1 (mod p); 2p ≡ 2 (mod p), · · · , (p− 1)p ≡ (p− 1) (mod p).

Adding all the results, we get,

1p + 2p + · · ·+ (p− 1)p ≡ 1 + 2 + · · ·+ (p− 1)(mod p)

≡ 12p(p− 1) (mod p)

≡ 0 (mod p); since p− 1 is even.

Theorem 2.9.2 If p be prime and a be prime to p, then ap2−p ≡ 1 (mod p2).

Proof: The Fermat’s theorem is, if p be prime and a is integer then ap−1 ≡ 1 (mod p).Hence ∃ q ∈ Z such that ap−1 = 1 + qp. Therefore,

ap2−p = (1 + qp)p = 1 + p.qp+p(p− 1)

2!(qp)2 + · · ·+ (qp)p

= 1 + kp2; where k ∈ Z, .

Hence, by definition, ap2−p ≡ 1 (mod p2).

Ex 2.9.3 Show that the prime factor of n2 +1 is of the form 4m+1, where m is an integer.

Solution: Let p be a prime factor of n2 + 1, then p is not the divisor of n and n2 + 1 ≡0(mod p). By Fermat’s theorem, we have,

np−1 ≡ 1(modp) ⇒ (n2)p−12 ≡ 1(mod p); assume p is odd

⇒ (n2)p−12 ≡ (−1)

p−12 (mod p); as n2 ≡ −1 (mod p)

⇒ p− 12

= even integer, i.e.,p− 1

2= 2m (say).

Therefore, p = 4m+ 1, where m is an integer. From this it follows that, no prime factor ofn2 + 1 can be put of the form 4m− 1, where m is an integer.

Fermat’s Theorem 133

Ex 2.9.4 If p is prime to a, then apn−1(p−1) ≡ 1(modpn).

Solution: By Fermat’s theorem, we have ap−1 ≡ 1(modp). Using the theorem, if a ≡1(modp) then ap ≡ 1(modpn+1), we get,

ap(p−1) ≡ 1(modp2), ap2(p−1) ≡ 1(modp3), · · · , apn−1(p−1) ≡ 1(modpn).

Result 2.9.2 If p is prime and p 6= 2, then ap−12 ≡ ±1(modp), when p 6 |a.

Proof: By Fermet’s theorem, we have,

ap−1 − 1 ≡ 0(modp)

or,(a

p−12 − 1

)(a

p−12 + 1

)≡ 0(mod p); p 6= 2.

⇒ p|(a

p−12 − 1

)or, p|

(a

p−12 + 1

); as p = prime.

⇒(a

p−12 − 1

)≡ 0(mod p) or

(a

p−12 + 1

)≡ 0(mod p)

⇒ ap−12 ≡ ±1(mod p).

Ex 2.9.5 Show that square of any integer is of the form 5k ± 1.

Solution: We know, ap−12 ≡ ±1(modp), then,

ap−12 ≡ ±1 + pk; k ∈ Z ⇒ a

p−12 = pk +±1.

When, p = 5, then a2 = 5k ± 1. Thus, square of any integer is of the form 5k ± 1.

Ex 2.9.6 Prove that the eighth power of any integer is of the form 17k or 17k ± 1.

Solution: Let a be an integer, divisible by 17, then a = 17k. If a is not divisible by 17,then (a, 17) = 1. By Fermat’s theorem,

a16 − 1 ≡ 0(mod17) ⇒ (a8 − 1)(a8 + 1) ≡ 0(mod17).Either, a8 − 1 ≡ 0(mod17), or, a8 + 1 ≡ 0(mod17).

a8 − 1 ≡ 0(mod17) ⇒ a8 = 17k + 1a8 + 1 ≡ 0(mod17) ⇒ a8 = 17k − 1.

Hence a8 = 17k or 17k ± 1, where a is an integer.

2.9.1 Wilson’s Theorem

Statement : If p be a prime, then (p− 1)! + 1 ≡ 0(modp).Proof: For the prime number p, the set S of integers which are less than and prime to p

is S = 1, 2, 3, . . . ; p − 1. Let a be one of the integers 1, 2, · · · , p − 1. Then no two of theintegers 1.a, 2.a, · · · , (p− 1). a are congruent modulo p, because, if ra ≡ sa(modp), for someintegers r, s such that

r ≡ s(modp); 1 ≤ r < s ≤ p− 1, as (a, p) = 1

which is a contradiction. Then as (a, p) = 1, a ∈ S, the linear congruence ax ≡ 1(mod p)has unique solution. Also, none of these is divisible by p. This means that the integersa, 2a, · · · , (p − 1)a are congruent to 1, 2, · · · , p − 1 modulo p, taken in some order. So fora ∈ S ∃ unique a′ ∈ S such that aa′ = 1(modp). If a = a′, then

134 Theory of Numbers

a2 ≡ 1(mod p) ⇒ a2 − 1 ≡ 0(mod p)⇔ p|

(a2 − 1

)⇔ p|(a+ 1)(a− 1)

⇔ (a+ 1)(a− 1) ≡ 0(mod p)⇔ either (a− 1) ≡ 0(mod p) or (a+ 1) ≡ 0(mod p)

Now, a2 ≡ 1(modp) holds if p|(a2 − 1) and this happens only when p|(a − 1) or p|(a + 1).Since p is a prime and a < p, it follows that when a − 1 ≡ 0(mod p), a = 1 and whena+ 1 ≡ 0(mod p), a = p− 1.

If we omit integers 1 and p − 1 from S, the remaining p − 3 integers 2, 3, · · · , p − 2 aresuch that they are grouped into p−3

2 pairs (a, a′) satisfying ax ≡ 1(mod p), a 6= a′ and1 < a′ < p− 1. Multiplying 1

2 (p− 3) of such pair congruences, we have,

2.3 · · · p− 2 ≡ (modp) ⇒ (p− 2)! ≡ 1(modp).or, (p− 1)! ≡ (p− 1)(modp) ≡ (−1)(modp)⇒ (p− 1)! + 1 ≡ 0(modp).

The converse of this theorem is also true, i.e., if (p − 1)! + 1 ≡ 0(modp), then p(> 1) is aprime. For, if p be not a prime, p is composite and has a divisor d with 1 < d < p such thatd|(p+ 1)! + 1. Since 1 < d < p, d divides one of the factors of (p+ 1)!. Thus

d|(p+ 1)! + 1 and d|(p+ 1) ⇒ d|1

which is absurd as d 6= 1. Therefore p cannot be composite and so p is prime.This theorem provides necessary and sufficient condition for determining primality of a

positive integer p. When p assumes large values, then (p − 1)! becomes very large and inthis case is impracticable.

Ex 2.9.7 Show that 70! + 1 ≡ 0(mod 71).

Solution: Since 71 is a prime number, by Wilson’s theorem

(71− 1)! 6 (−1)(mod71) ⇒ 70! + 1 ≡ 0(mod 71).

Ex 2.9.8 For a prime p,

(p− 1)! ≡ (−1)p−12

[1.2. · · · p− 1

2

]2(mod p).

Show that the integer(

p−12

)! satisfies the congruence x2 ± 1 ≡ 0(mod p) according as

p = 4k + 1 and p = 4k + 3.

Solution: We consider the set of integers, 1, 2, 3, · · · , p−12 , p+1

2 , · · · , (p− 2), (p− 1) where, pis prime. Now, p− 1 ≡ −1(mod p), p− 2 ≡ −2(mod p), p+1

2 ≡ −p−12 (mod p). Now,

(p− 1)! = 1.2.3. · · · . p− 12

.p+ 1

2. · · · .(p− 2).(p− 1)

≡(

1.2.3. · · · . p− 12

).

(−p− 1

2. · · · .(−2).(−1)

)(mod p)

≡ 1.(−1).2.(−2). · · · .(p− 1

2

).

(−p− 1

2

)(mod p)

≡ (−1)p−12

[1.2. · · · p− 1

2

]2(mod p)

≡ (−1)p−12

[(p− 1

2

)!]2

(mod p);p− 1

2= integer.

Fermat’s Theorem 135

Again by Wilson’s theorem (p− 1)! ≡ −1(modp). Therefore,

−1 ≡ (−1)p−12

[(p− 1

2

)!]2

(mod p);p− 1

2= integer.

Now, if p is a prime of the form 4k + 1 for some k ∈ N , then,

−1 ≡ (−1)4k+1−1

2

[(p− 1

2

)!]2

(mod p)

or, −1 ≡ 1[(

p− 12

)!]2

(mod p)

or,[(

p− 12

)!]2≡ −1(mod p)

Thus x =(

p−12

)! satisfies the congruence x2 + 1 ≡ 0(mod p), when p is of the form 4k + 1.

Again, p is of the form 4k + 3 for some k ∈ N , then,

−1 ≡ (−1)4k+3−1

2

[(p− 1

2

)!]2

(mod p)

or, 1 ≡ 1[(

p− 12

)!]2

(mod p)

or,[(

p− 12

)!]2≡ 1(mod p)

Thus x =(

p−12

)! satisfies the congruence x2 − 1 ≡ 0(mod p).

Ex 2.9.9 If p is odd prime, then show that

12.32.52 · · · (p− 2)2 ≡ (−1)p+12 (mod p).

Solution: We use the result, p− k ≡ −k(mod p) and k ≡ −(p− k)(mod p). Now,

(p− 1)! = 1.2.3. · · · . p− 12

.p+ 1

2. · · · .(p− 2).(p− 1)

= 1.(p− 1)3.(p− 3) · · · 2.(p− 2)≡ 1(−1).3(−3) · · · −(p− 2).(p− 2)(mod p)

≡ (−1)p−12 12.32.52 · · · (p− 2)2(mod p).

Again using Wilson’s theorem (p− 1)! ≡ −1(modp), we have,

−1 ≡ (−1)p−12 12.32.52 · · · (p− 2)2(mod p)

or, 12.32.52 · · · (p− 2)2 ≡ (−1)(−1)p−12 (mod p)

or, 12.32.52 · · · (p− 2)2 ≡ (−1)p+12 (mod p).

Ex 2.9.10 If p be a prime number, then show that

(p− 1)! ≡ p− 1(mod(1 + 2 + · · ·+ (p− 1))).

136 Theory of Numbers

Solution: Using Wilson’s theorem (p− 1)! ≡ −1(modp), we have,

(p− 1)! ≡ (p− 1)(modp),⇒ p|(p− 1)!− (p− 1).

Also,p− 1

2|(p− 1)!− (p− 1); as p− 1 is even.

or, pp− 1

2|(p− 1)!− (p− 1) ⇒ (p− 1)!− (p− 1) ≡ 0

[mod p

p− 12

]or, (p− 1)! ≡ p− 1(mod(1 + 2 + · · ·+ (p− 1))).

as 1 + 2 + · · · (p− 1) = pp− 1

2.

Ex 2.9.11 Prove that 4(29)! + 5! is divisible by 31.

Solution: In Wilson’s theorem, let p = 31(prime), then (30)! + 1 ≡ 0(mod31), i.e.,

(31− 1)(29)! + 1 ≡ 0(mod31)⇒ −(29)! + 1 ≡ 0(mod31) ⇒ 4(29)!− 4 ≡ 0(mod31)⇒ 4(29)!− 4 + 124 ≡ 0(mod31) ⇒ 4(29)! + 120 ≡ 0(mod31).

Thus 4(29)! + 5! is divisible by 31.

2.10 Arithmetic Functions

An arithmetic or a number-theoretic function is a real or complex valued function whosedomain is the set of positive integers. If f is an arithmetic function we write f(n) for itsvalues. For example, f : N → N , defined by f(n) = n or n2, are arithmetic functions, butf(n) = log n is not an arithmetic function. The following are arithmetic functions:

(i) f(n) = 2n for all n ∈ N .

(ii) f(n) = 1n for all n ∈ N .

(iii) f(n) = n+ 1n for all n ∈ N .

An arithmetic function not identically zero is said to be normalized if f(1) = 1. Severalarithmetical functions plays an important role in the study of divisibility properties of inte-gers and the distribution of primes. Here we shall discus two important number theoreticfunctions

(i) Euler-Totiant function or phi function,

(ii) Mobius function.

2.10.1 Euler’s Phi Function

Let n ∈ N . Then the number of positive integers less than n and prime to n(i.e., the numberof divisors of n) is denoted by φ(n) with φ(1) = 1. Thus the function

φ : N → N defined by φ(n) =n′∑

k=1

1 (2.19)

is known as Euler’s phi function where ‘′’ indicates the sum is extended over those k(< n)satisfying (k, n) = 1. For example, let n = 12, then k = 1, 5, 7, 11, where k < n and(k, n) = 1. Thus

Arithmetic Functions 137

φ(12) =12′∑k=1

1 = 1 + 1 + 1 + 1 = 4.

A short table for φ(n) is given as follows:

n : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15φ(n) : 1 1 2 2 4 2 6 4 6 4 10 4 12 6 8

The function φ is a number-theoretic function.

Result 2.10.1 Let n be a prime integer. If p is a positive integer such that p < n, then(p, n) = 1. Hence the number of positive integers not exceeding n and relatively prime to nis n− 1, so that, φ(n) = n− 1.

Result 2.10.2 The followings are important:

(i) If n = pα11 · pα2

2 · · · pαk

k , where pi(1 ≤ i ≤ k) are distinct primes and αi ∈ N1 ≤ i ≤ k,then the number of prime divisors of n is (1 + α1)(1 + α2) · · · (1 + αk).

(ii) The highest power of a prime p contained in n! is denoted by k(n!), where

k(n!) =[np

]+[ np2

]+[ np3

]+ · · ·

Ex 2.10.1 Find the highest power of 3 contained in 100!

Solution: The highest power of a prime p contained in n! is denoted by k(n!), where

k(n!) =[np

]+[ np2

]+[ np3

]+ · · ·

=[100

3

]+[100

32

]+[100

33

]+[100

34

]+[100

35

]+ · · ·

= 33 + 11 + 3 + 1 + 0 = 48.

Theorem 2.10.1 If p is prime then φ(pk) = pk(1− 1

p

)where k is a positive integer.

Proof: When k = 1, then φ(p) = p− 1 as p is prime. If k > 1, let us arrange the integersfrom 1 to pk in the following way

1 2 · · · p− 1 pp+ 1 p+ 2 · · · p+ (p− 1) 2p2p+ 1 2p+ 2 · · · 2p+ (p− 1) 3p

...... · · ·

......

(pk−1 − 1)p+ 1 (pk−1 − 1)p+ 2 · · · (pk−1 − 1)p+ (p− 1) pk−1p

Thus if, k > 1, we find that q(≤ pk), a positive integer, then (q, pk) 6= 1, if and only if q is oneof p, 2p, 3p, . . . , pkp and their numbers are pk−1. If q is not equal to one of p, 2p, 3p, . . . , pkp,then (q, pk) = 1. Now there are pk integers 1 to pk. Among these integers, there are pk−1

integers which are not relatively prime to pk. Hence the remaining integers are relativelyprime to k. The total number is pk − pk−1 and so

φ(pk) = pk − pk−1 = pk

(1− 1

p

). (2.20)

138 Theory of Numbers

If the integer n(> 1) is of the form n = pα11 pα2

2 · · · pαrr , where p1, p2, · · · , pr are prime to one

another, i.e., (pαi1 p

αj

2 ) = 1 for all i, j. Thenφ(n) = φ (pα1

1 )φ (pα22 ) · · ·φ (pαr

r )

= pα11

(1− 1

p1

)pα22

(1− 1

p2

)· · · pαr

r

(1− 1

pr

)= pα1

1 pα22 · · · pαr

r

(1− 1

p1

)(1− 1

p2

)· · ·(

1− 1pr

)= n

(1− 1

p1

)(1− 1

p2

)· · ·(

1− 1pr

)= n

r∏α=1

(1− 1

pα

). (2.21)

Ex 2.10.2 Find φ(191) and φ(260).

Solution: First we are to test whether 191 is a prime number or not. For this, we findall primes p satisfying p2 ≤ 191. These primes are 2, 3, 5, 7, 11, 13 and 17. But none of theprimes divide 191, so 191 is a prime.

Therefore, by definition, φ(191) = 191− 1 = 190.By using the unique factorization theorem, we get 260 = 22.5.13. Therefore,

φ(260) = φ(22.5.13) = 260(1− 12)(1− 1

5)(1− 1

13)

= 260.12.45.1213

= 96.

Theorem 2.10.2 Let m and n be two positive integers. If m,n are relatively prime, thenφ(mn) = φ(m)φ(n), i.e., φ(n) is multiplicative.

Proof: Given that (m,n) = 1. We consider the product mn. Then the first mn numberscan be arranged in n lines, each containing m numbers. Thus,

1 2 · · · k · · · mm+ 1 m+ 2 · · · m+ k · · · m+m2m+ 1 2m+ 2 · · · 2m+ k · · · 2m+m

...... · · ·

......

(n− 1)m+ 1 (n− 1)m+ 2 · · · (n− 1)m+ k · · · (n− 1)m+m

Now, we consider the vertical column beginning with k. If (k, a) = 1, all the terms of thiscolumn will be prime to m, but if k and m have a common divisor, no number in the columnwill be prime to m. Now the first row contains φ(m) numbers prime to n, therefore, φ(m)vertical columns in each of which every term is prime to n. Let us suppose that, the verticalcolumn which begins with k is one of these. This column is in arithmetic progression, theterms of which when divided by n leaves remainders

0, 1, 2, 3, · · · , n− 2, n− 1.

Hence, the column contains φ(m) integers prime to n. Thus in the table, there are φ(m)φ(n)integers, which are prime to m and also n and therefore to mn; i.e., φ(mn) = φ(m)φ(n).This theorem can be extended to a finite number of positive integers as

φ(m1m2 · · ·mr) = φ(m1)φ(m2) · · ·φ(mr),

where m1,m2, · · · ,mr are prime to one another. If (m,n) = d, then

φ(m,n) = φ(m) φ(n)d

φ(d).

Arithmetic Functions 139

Theorem 2.10.3 If n is a positive integer, then,

φ(2n) = φ(n); n is odd= 2φ(n); n is even.

Proof: When n is odd and 2, n are prime to each other so

φ(2n) = φ(2)φ(n) = 1.φ(n) = φ(n)

When n is even, let n = 2k.p where p is an odd integer.Therefore,

φ(n) = φ(2k.p) = 2k.(1− 12)φ(p) = 2k−1φ(p)

φ(2n) = φ(2k+1)φ(p) = 2kφ(p)⇒ φ(2n) = 2φ(n).

If p = 1, then also φ(2n) = 2φ(n), since φ(1) = 1.

Ex 2.10.3 Find all integers n such that

(i) φ(n) =n

2, (ii) φ(n) = φ(2n) and (iii) φ(n) = 12.

Solution: (i) Let n be a prime integer. Hence the number of positive integers not exceedingn and relatively prime to n is n− 1, so that, φ(n) = n− 1. Therefore,

n− 1 =n

2⇒ n

2= 1 ⇒ n = 2.

Also, n2 can be written as n(1− 1

2 ), so that the values of n are of the form n = 2α;α ∈ N .(ii) The given equation φ(n) = φ(2n) can be written in the form φ(n) − φ(2n) = 0. So,

using the definition of Euler’s phi function, we have,

φ(n) = 1 or 2 or 4 or 6 · · · i.e., n = 1, 2; 3, 4, 6; 5, 8, 10, 12; 7, 9, 14, 18; · · ·φ(2n) = 1 or 2 or 4 or 6 · · · i.e., 2n = 1, 2; 3, 4, 6; 5, 8, 10, 12; 7, 9, 14, 18; · · ·

Since n ∈ N , so for the second case, 2n 6= 1, 3, 5, 7, · · · and therefore, the common values ofn are n = 1, 3, 5, 7, 9, · · · .

(iii) Case 1: Let n be a prime integer. Hence the number of positive integers notexceeding n and relatively prime to n is n− 1, so that, φ(n) = n− 1, then

φ(n) = n− 1 = 12 ⇒ n = 13.

Case 2: Now, 12 can be written as 12 = 22.3 and so,

n(1− 12)(1− 1

3) = 12 ⇒ n× 1

2× 2

3= 12 ⇒ n = 36.

Thus the values of n satisfying φ(n) = 12 are n = 13, 36.

Ex 2.10.4 Solve for x, y, z ∈ N where φ(x− 5) + φ(3y − 5) + φ(5z − 18) = 3.

Solution: We see that, φ(n) ∈ N , and therefore, the given equation will be satisfied if andonly if φ(x− 5) = 1, φ(3y − 5) = 1, φ(5z − 18) = 1. Now,

φ(x− 5) = 1 ⇒ x− 5 = 1 or 2; i.e., x = 6 or7

φ(3y − 5) = 1 ⇒ 3y − 5 = 1 or 2; i.e., y = 2 but y 6= 73, as y ∈ N

φ(5z − 18) = 1 ⇒ 5z − 18 = 1 or 2; i.e., z = 2 but z 6= 195, as z ∈ N .

Thus the solutions are (6, 2, 4); (7, 2, 4).

140 Theory of Numbers

Theorem 2.10.4 Let a,m(> 0) be integers. If (a,m) = 1 then aφ(m) ≡ 1 (mod m).

Proof: For each positive integer m ≥ 1, φ(m) is the Euler’s phi function, defined as

φ(1) = 1 and φ(m) =∑

1≤k≤m,(k,m)=1

1.

Thus for m = 1, the result holds trivially. Fix a positive integer m and take an inte-ger a, coprime to m. Let r1, r2, . . . , rφ(m) be a reduced residue system mod m. then,ar1, ar2, . . . , arφ(m) is also a reduced residue system mod m in some order. Since each(ari,m) = 1 and they are incongruent to each other. Hence, the product of all the integersin the first set is congruent to the product of those in the second set. Therefore,

ar1.ar2. . . . , arφ(m) ≡ r1.r2. . . . rφ(m)(modm)

⇒ aφ(m)r1.r2 . . . rφ(m) ≡ r1.r2. . . . rφ(m)(modm)

⇒ aφ(m) ≡ 1(modm); as, (r1.r2. . . . rφ(m),m) = 1.

Each ri is relatively prime to m, so we can cancel each ri and obtain the theorem. Thisis known as Euler Fermat theorem. This theorem can be used to calculate the solution oflinear congruence.

Result 2.10.3 If (a,m) = 1, the solution (unique mod m) of the linear congruence ax ≡b(modm) is given by x ≡ baφ(m)−1(modm).

Ex 2.10.5 Solve the linear congruence 5x ≡ 3(mod24).

Solution: Since (5, 24) = 1, there is an unique solution. The solution is given by

x ≡ 3.5φ(24)−1 ≡ 3.57(mod24); as φ(24) = φ(3)φ(8) = 2.4≡ 3.5(mod24); as 52 ≡ 1(mod24) ⇒ 56 ≡ 1(mod24)

⇒ x ≡ 15(mod24).

Ex 2.10.6 Solve the linear congruence 25x ≡ 15(mod120).

Solution: Here d = (25, 120) = 5. As d|15, the congruence has exactly five solutionsmodulo 120. To find them, we are to solve the linear congruence 5x ≡ 3(mod24). Thus thefive solutions are given by

x ≡ 15 + 24k; k = 0, 1, 2, 3, 4or, x ≡ 15, 39, 63, 87, 111(mod120).

Ex 2.10.7 If n > 7 is prime, prove that n6 − 1 is divisible by 504.

Solution: Since 7 is a prime and n is prime to 7, by Fermat’s theorem, n6 − 1 is divisibleby 7. By Euler’s theorem as n is prime to 9, nφ(9) − 1 is divisible by 9. Now,

φ(9) = φ(32) = 9(

1− 13

)= 6.

Therefore, n6 − 1 is divisible by 9. Since n > 7 is an odd prime, n is of the forms 4k + 1 or4k + 3, where k(> 1) ∈ N .

n6 − 1 = (n− 1)(n+ 1)(n4 + n2 + 1).If, n = 4k + 1, then, (n− 1)(n+ 1) = 4k(4k + 2) and,if, n = 4k + 3, then, (n− 1)(n+ 1) = (4k + 2)(4k + 4).

Therefore, in any case n is divisible by 8. Since, the three consecutive integers 7,8,9 arepairwise prime to each other,7.8.9|n6 − 1, i.e., 504|n6 − 1.

Arithmetic Functions 141

Ex 2.10.8 Use Euler-Fermat’s theorem to find the unit digit in 3100. [NET’11]

Solution: Since 3 is prime to 10, by Euler-Fermat’s theorem,

3φ(10) ≡ 1(mod10) where, φ(10) = 4,or, 34 ≡ 1(mod10) ⇒ 3100 = 34.25 ≡ 1(mod10).

Thus the unit digit in 3100 is 1.

Result 2.10.4 The sum of the divisors of a positive integer n is denoted by σ(n), i.e.,σ(n) =

∑d|n

d and it is an arithmetical function. For example, consider the positive integer 4.

The divisors of 4 are 1,2,4. Therefore, σ(4) = 1 + 2 + 4 = 7. Similarly, σ(6) = 12, σ(10) =18, σ(15) = 24. In general, if n = pα1

1 · pα22 · · · pαk

k , then

σ(n) =∑d|n

d =pα1+11 − 1p1 − 1

· pα2+12 − 1p2 − 1

· · ·pαk+1

k − 1pk − 1

. (2.22)

Ex 2.10.9 Find the number of positive divisors of 50000. [NET’12]

Solution: We have, 50000 = 24 × 55. Thus the number of positive divisor of 50000 is

= (4 + 1)(5 + 1) = 5× 6 = 30.

Result 2.10.5 Consider positive integer n and write S = 1, 2, · · · , n. Define ∼ on S bya ∼ b⇔ (an) = (b, n). Then ∼ is an equivalence relation. For a divisor of n,

A(d) =k : (k, n) = d

is an equivalence class. So S =

⋃d|n

A(d). For example, let n = 6, S = 1, 2, · · · , 6. Then,

divisors of 6 are 1,2,3,6. Now

A(1) = 1, 5, A(2) = 2, 4, A(3) = 3, A(6) = 6.

Note that these sets A(1), A(2), A(3), A(6) are disjoint and of is S.

2.10.2 The Mobius Function:

The Mobius function µ(n) is defined as,

µ(n) = 1; n = 1= (−1)r; if n = p1.p2. . . . pr (pi‘s are distinct prime.)= 0; if a2|n for some a > 1, i.e.n has a square factor > 1.

The following is a table, showing some values of µ(n) :

n : 1 2 3 4 5 6 7 8 9 10µ(n) : 1 −1 −1 0 −1 1 −1 0 0 1

The Mobius Function arises in many different places in number theory. One of its funda-mental properties is a remarkably simple formula for the divisor sum

∑d|n

µ(d), extended over

the positive divisor of n.

142 Theory of Numbers

Theorem 2.10.5 If n ≥ 1, then∑d|n

µ(d) = [1n

] = 1;n = 1

= 0;n > 1

Proof: The formula is clearly true if n = 1. Assume, then, n > 1.Case 1: Let n = pα, then∑

d|pα

µ(d) = µ(1) + µ(p) + µ(p2) + · · ·+ µ(pα)

= 1 + (−1) + 0 + · · ·+ 0 = 1− 1 = 0.

Case 2: Assume that the result be true for integers with at most k distinct prime factors,i.e., let n = apα, where a is an integer with k prime factors and p 6 |a. Now,∑

d|n

µ(d) =∑d|a

µ(d) +∑d|a

µ(pd) +∑d|a

µ(p2d) + · · ·+∑d|a

µ(pαd)

=∑d|a

µ(d) +∑d|a

µ(p)µ(d) +∑d|a

µ(p2)µ(d) + · · ·+∑d|a

µ(pα)µ(d)

=∑d|a

µ(d)−∑d|a

µ(d) = 0.

Case 3: Let n = pa11 .p

a22 . . . par

r > 1, be a standard factorization of n. In the sum∑d|n

µ(d)

the only nonzero terms come from d = 1 and from those divisor of n which are products ofdistinct primes. Hence,∑

d|n

µ(d) = µ(1) +∑

1≤i≤r

µ(pi) +∑

1≤i<j≤r

µ(pipj) + · · ·+ µ(p1. . . . pr)

= µ(1) + µ(p1) + · · ·+ µ(pr) + µ(p1p2) + · · ·+ µ(pr−1pr) + · · ·+ µ(p1 . . . pr)

= 1 +(r

1

)(−1) +

(r

2

)(−1)2 + · · ·+ (−1)r

= (1− 1)r = 0

Thus the proof will be induction on different prime factors for n > 1.

Theorem 2.10.6 If F (n) =∑d|n

f(d) for every positive integer n, then

f(n) =∑d|n

µ(d)F (n/d).

Proof: By using the definition,∑d|n

µ(d)F (n/d) =∑

δ|n/d

µ(d)f(δ) =∑δ|n

µ(d)f(δ)

=∑δ|n

f(δ)∑

d|n/δ

µ(d) = f(n)

This is called Mobius inversion formula. If f(n) and g(n) are two arithmetic functionssatisfying the condition f(n) =

∑d|n

g(n), then f(n), g(n) is a Mobius pair. For example,

n, φ(n) is Mobius pair.

Arithmetic Functions 143

2.10.3 Divisor Function

Let n(> 1) be a positive integer. The divisor function, i.e., the number of positive divisorsof a positive integer n, τ : N → N , denoted by τ(n), n ∈ N is given by

τ(n) = 1; if n = 1= 2; if n = p( a prime )> 2; if n is composite.

Note that, τ(n), 1 is Mobius pair. Let a positive integer n(> 1) be expressed in a canonicalform as

n = p1α1 .p2

α2 . . . prαr , αi ≥ 0, for i = 1, 2, · · · , r,

where pn is the nth prime with p1 < p2 < · · · < pr. If m be a positive divisor of n, then mis of the form p1

α1 .p2α2 . . . pr

αr , where,

0 ≤ u1 ≤ α1, 0 ≤ u2 ≤ α2, · · · , 0 ≤ ur ≤ αr.

Thus the positive divisors of n in one-one correspondence with the totality of r tuples(u1, u2, · · · , ur), satisfying the above inequality. The number of such r tuples is (α1 +1)(α2 + 1) · · · (αr + 1). Hence the total number of positive divisors of n is

τ(n) = (α1 + 1)(α2 + 1) · · · (αr + 1).

The total number of positive divisors τ(n) include both the divisors 1 and n. For example,

τ(4) = τ(22) = 2 + 1 = 3; τ(12) = τ(22.3) = (2 + 1)(1 + 1) = 6.

The sum of all positive divisors of a positive integer n is denoted by σ(n). Every positivedivisor of n in the canonical form is a term in the product

(1 + p1 + · · ·+ pα11 )(1 + p2 + · · ·+ pα2

2 )(1 + p1 + · · ·+ pαrr )

and conversely, each term in the product is a divisor of n. Thus the sum of all positivedivisors of n = p1

α1 .p2α2 . . . pr

αr is

σ(n) = (1 + p1 + · · ·+ pα11 )(1 + p2 + · · ·+ pα2

2 )(1 + p1 + · · ·+ pαrr )

=pα1+11 − 1p1 − 1

.pα2+12 − 1p2 − 1

. . . . .pαr+1

r − 1pr − 1

with σ(1) = 1. The functions τ and σ are examples of number-theoretic functions. Both ofτ and σ are multiplicative functions, i.e.,

τ(mn) = τ(m)τ(n) and σ(mn) = σ(m)σ(n).

A positive integer n is said to be a prefect number , if σ(n) = 2n, i.e., if n be the sum of allits positive divisors excluding itself. For example, 6, 28 etc. are perfect number.

Ex 2.10.10 Find τ(360) and σ(360).

Solution: The number 360 can be written in canonical form as 360 = 23.32.5. Therefore,

τ(360) = (1 + 3)(1 + 2)(1 + 1) = 24.

σ(360) =24 − 12− 1

.33 − 13− 1

.52 − 15− 1

= 1170.

144 Theory of Numbers

Ex 2.10.11 The total number of positive divisors of a positive integer n is odd if and onlyif n is a perfect square.

Solution: Let a positive integer n(> 1) be expressed in a canonical form as

n = p1α1 .p2

α2 . . . prαr , αi ≥ 0, for i = 1, 2, · · · , r,

where pn is the nth prime with p1 < p2 < · · · < pr. Then each of α1, α2, · · · , αr is an eveninteger and τ(n) is odd. If however, n = 1, a perfect square, then τ(n) = 1 and it is odd.

Conversely, let τ(n) be odd. Then each of the factors α1 + 1, α2 + 1, · · · , αr + 1 must beodd. Consequently, each of α1, α2, · · · , αr must be even and n is therefore a perfect square.This completes the proof.

2.10.4 Floor and Ceiling Functions

Let x be any real number. The floor function of x denoted by bxc and it is the greatestinteger less than or equal to x. That is, bxc : R → Z where bxc = greatest integer less thanor equal to x. For example, b8.25c = 8, b8.75c = 8, b−10.6c = −11, b8c = 8, b−3c = 3,b√

26c = 5, etc.The ceiling function or x ∈ R is denoted by dxe and it is the smallest integer greater than

or equal to x. Thus, dxe : R → Z, where bxc = least integer greater than or equal to x. Forexample, b8.25c = 9, b8.75c = 9, b−4.6c = −4, b−5c = 5, b5c = 5 etc.Properties

1. bxc = n⇔ n ≤ x < n+ 1, where n is an integer.2. dxe = n⇔ n < x ≤ n+ 1, where n is an integer and x is not an integer.3. x− 1 < bxc ≤ x ≤ dxe < x+ 1.4. bm+ nc = bmc+ n, where n is an integer.5. dm+ ne = dme+ n, where n is an integer.6. bxc+ byc 6= bx+ yc when x, y 6∈ Z.7. dxe+ dye 6= dx+ ye when x, y 6∈ Z.8. dmxe = mdxe, where m is an integer.9. bmxc = mbxc, where m is an integer.

2.10.5 Mod Function

Let m be a positive integer. The (mod m) function is defined as fm(a) = b, where b is theremainder when a is divided by m. The function fm(a) = b is also denoted by a ≡ b (modm), 0 ≤ b < m. Also, fm(a) = b when (b − a) is divisible by m. The integer m is calledthe modulus and a ≡ b (mod m) is read as ’a is congruent to b modulus m’. It can also bedefined as, fm(a) is unique integer r such that a = kq + r, 0 ≤ r < m for some integer q.This function is also written as a (mod m). For example,

f7(35) = 0 as 7 divides 35− 0 or 35 = 5× 7 + 0,f5(36) = 0 as 5 divides 36− 1 or 36 = 5× 7 + 1.

Exercise 2

Section-A[Multiple Choice Questions]

1. Fundamental theorem of arithmetic: every positive integer n > 1 can be uniquely asa product of(a) Prime (b) Positive integers (c) Perfect squares (d) None of the above.

Arithmetic Functions 145

2. Division algorithm states as: let a and b integers with b 6= 0, then there exists integersq and r such that(a) a− bq = r (b) a = bq − r (c) a = q ∗ r + b (d) All of the above.

3. Suppose a, b and c are integers, which of the following are true?(a) If a/b and b/c, then a/c (b) If a/b and b/c, then a/(b + c) and a/(b − c) (c) Ifx > 0, then gcd(ax, bx) = x+ gcd(a, b) (d) For any integer x gcd(a, b) = gcd(a, b+ax).

4. gcd(540, 168) =(a) 168 (b) 34 (c) 12 (d) none of the above

5. Two integers a and b are said to relatively prime or coprime if(a) gcd(a, b) = a (b) gcd(a, b) = 1 (c) gcd(a, b) = a ∗ b (d) All of the above.

6. For linear congruence equation ax ≡ b(mod m) where, d = gcd(a,m), if d does notdivide b then the equation has(a) Unique solution (b) Has no solution (c) Two solutions (d) None of the above.

7. Consider congruence equation, 8x ≡ 12(mod 28), then(a) Equation has no solution (b) has unique solution as 5 (c) 5,12,19 and 26 arethe four solutions (d) All of the above.

8. Solution of 235x ≡ 54(mod7) is(a) x ≡ 12(mod7) (b) x ≡ 3(mod7) (c) x ≡ 5(mod7) (d) x ≡ 4(mod7)

9. Remainder of 8103 from Fermat theorem when divided by 103 is(a) 8 (b) 7 (c) 6 (d) 10

10. The unit digit of 2100 is NET(June)11(a) 2 (b) 4 (c) 6 (d) 8

11. The number of elements in the setm : 1 ≤ m ≤ 1000,mand 1000 are relatively prime

is NET(June)11(a) 100 (b) 250 (c) 300 (d) 400

12. The number of positive divisors of 50000 is NET(June)12(a) 20 (b) 30 (c) 40 (d) 50

13. The last digit of (38)1031 is NET(June)12(a) 6 (b) 2 (c) 4 (d) 8

Section-B[Objective Questions]

1. Show that every integer > 1 has a prime factor.

2. If n > 1 is an integer, show that there exists a prime integer p such that p divides n.

3. Explain the fundamental theorem of arithmetic by an example.

4. Let a, b, n be positive integers such that a ≡ b(mod n), show that (a, n) ≡ (b, n).

5. Show that 32n ≡ 1(mod 8), for all integers n ≥ 1.

6. Find all the integral solutions of the equation 5x+ 4y = 9.

7. Explain the Wilson’s theorem for integers by an example.

146 Theory of Numbers

8. Find φ(14), where φ is the Euler function.

9. Define the Mobius µ-function. Find µ(15).

10. Find the highest power of 7 contained in 1000!. Ans: 164

Section-C[Long Answer Questions]

1. Prove the following by mathematical induction:

(a) 1.2 + 2.22 + 3.23 + 4.24 + · · ·+ n.2n = (n− 1)2n+1 + 2,∀n ∈ N .

(b) 1 + 2 + · · ·+ n = n(n+1)2 , ∀n ∈ N . KU(H) :’09

(c) 1.1! + 2.2! + · · ·+ n.n! = (n+ 1)!− 1, ∀ n ≥ 1.

(d) 112 + 1

22 + 132 + · · ·+ 1

n2 ≤ 2− 1n , ∀ n ≥ 1

(e) 12 + 2

22 + 332 + · · ·+ n

n2 = 2− n+22n , ∀ n ≥ 1.

2. Prove by mathematical induction that, for every n ∈ N

(a) 33n+3 − 8n− 7 is divisible by 49. JECA‘06

(b) 32n+1 + 2n+2 is divisible by 7.

(c) 32n − 1 is not exactly divisible by 2n+3.

(d) 22n+1 − 9n2 + 3n− 2 is divisible by 54 KU(H) :’07

(e) 72n + 16n− 1 is divisible by 64.

(f) 32n − 8n− 1 is divisible by 64.

3. Prove the following inequalities by induction on n ∈ N

(a) 2n < n! for all n ≥ 4.

(b) n! > 3n for all integers n ≥ 7.

(c) n2 < n! for all integers n ≥ 4.

4. Use division algorithm, show that

(a) The product of any k consecutive integers is divisible my k.

5. (a) Show that n2 + 2 is not divisible by 4 for any integer n.

(b) If p is a prime greater than 3, show that 24|p2 − 1.

(c) If n is an integer not divisible by 2 or 3, then 32|(n2 + 3)(n2 + 7).

(d) If n is an odd integer, then 24|(n2 + 3).

(e) Prove that 2n! is divisible by n!(n+ 1)!.

6. (a) Let a1, a2, . . . , an be any non-zero integers and d = (a1, a2, . . . , an). Then ∃m1,m2, . . . ,mn ∈ Z such that d = a1m1 + a2m2 + · · ·+ anmn.

(b) Prove that for any two integers u and v where v > 0 there exists two uniqueintegers m and n such that u = mv + n, where 0 ≤ n < v.

(c) If (a, b) = [a, b] for two positive integers a, b, prove that a = b. BH ′97

(d) If a > 1, prove that (am − 1, an − 1) = a(m,n) − 1.

7. Prove that√

2,√

3,√

5 are irrational numbers.

Arithmetic Functions 147

8. Prove that the product of the first n Fermat’s number is 22n − 1.

9. Prove that every square integer is either of the forms 5m, 5m + 1, 5m − 1, m is aninteger.

10. Prove that the prime factor of n2 + 1 is of the form 4m+ 1.

11. Find integers m,n such that

(a) (95, 102) = 95m+ 102n.

(b) (723, 24) = 723m+ 24n.

(c) (426, 246) = 426m+ 246n.

12. If (a, b) = 1, show that

(a) (a+ b, a− b) = 1 or 2.

(b) (a+ b, ab) = 1 and (a2 + b2, a2b2) = 1.

13. If k be a positive integer, then (ka, kb) = k(a, b).

14. (a) Prove that n12 − 1 is divisible by 7, if (n, 7) = 1.

(b) If n and n2 + 8 are both prime numbers, prove that p = 3.

(c) If 2n − 1 be prime, prove that n is a prime.

(d) Prove that n4 + 4n is a composite number for all natural number n > 1.

15. Show that a natural number is divisible by 9 if and only if the sum of its digits isdivisible by 9.

16. Find the integer between 1 and 1000 which leaves the remainder 1, 2, 6 when dividedby 9, 11, 13 respectively.

17. Solve the Diophantine equations:

(a) 56x+ 72y = 40 : x = 20 + 9t, y = −15− 7t.

(b) 8x− 27y = 125 : x = −1169− 27t, y = −351− 8t.

(c) 7x+ 11y = 1 : x = 8− 11t, y = −5 + 7t.

(d) 68x− 157y = 1 : x = −30− 157t, y = −13− 68t.

(e) 13x− 17y = 5 : x = 20− 17t, y = 15− 13t.

18. The sum of two positive integers is 100. If one is divided by 7 the remainder is 1, andif the other is divided by 9 the remainder is 7. Find the numbers. Ans: 57, 43.

19. For any natural number n show that,

(a) (2n+ 1)2 ≡ 1(mod8).

(b) 4.6n + 5n+1 ≡ 9(mod20).

20. Show that

(a) 241 ≡ 3(mod23).

(b) 315 ≡ 1(mod13).

21. Find all the natural numbers n ≤ 100 that satisfy(i) n ≡ 10(mod7) (ii) n ≡ 3(mod17) (iii) n ≡ 10(mod17).

148 Theory of Numbers

22. If a ≡ b(mod m) and x ≡ y(mod m), then prove that

(a) ap+ xq = (bp+ yq)(mod m)

(b) ax ≡ by(mod m).

23. If a ≡ b(mod m) then prove that (a,m) = (b,m), i.e. congruent numbers have thesame GCD with m.

24. Solve the linear congruence :

(a) 7x ≡ 3(mod15) : Ans: x ≡ 9(mod15)

(b) 37x ≡ 7(mod127) : Ans: x ≡ 86(mod127)

(c) 29x ≡ 1(mod 13).

(d) 15x ≡ 9(mod18) : x = −3 + 6t; t = 0, 1, 2.

25. A certain number of sixes and nines are added to give a sum of 126. If the number ofsixes and nines are interchanged, the new sum is 114. How many sixes and nines werethere originally?

26. Show that the solution of the systemx ≡ a(mod 21), x ≡ b(mod 16) is x ≡ 64a− 63b(mod 336).

27. Find the solution of the system with the help of Chinese Method :

(a) x ≡ 5(mod 4), x ≡ 3(mod 7), x ≡ 2(mod 9).

(b) x ≡ 3(mod 6), x ≡ 5(mod 8), x ≡ 2(mod 11).

(c) x ≡ 1(mod 3), x ≡ 2(mod 5), x ≡ 3(mod7). Ans: x ≡ 52(mod 105)

28. Solve the system of congruence

(a) x ≡ 1(mod 3), x ≡ 2(mod 4), x ≡ 3(mod 5).

(b) x ≡ 11(mod 15), x ≡ 6(mod35), Ans: x ≡ 41(mod 105).

29. Use Fermat’s theorem to prove that for two positive integers a, b; a40− b40 is divisibleby 541 if both a and b are prime to 541.

30. Use Fermat’s theorem to prove that

(a) 1! + 2! + 3! + · · ·+ 79! + 80! ≡ 1(mod 80).

(b) 1p−1 + 2p−1 + 3p−1 + · · ·+ (p− 1)p−1 ≡ (−1)(mod p)

(c) 1p + 2p + 3p + · · ·+ (p− 1)p ≡ 0(mod p)

when p is an odd prime.

31. If p is odd prime, then show that

22.42.62 · · · (p− 1)2 ≡ (−1)p−12 (mod p).

32. Show that 28! + 233 ≡ 0(mod899).

Chapter 3

Theory of Matrices

In this chapter, we are to investigate the concepts and properties matrices and discuss someof the simple operations by which two or more matrices can be combined. Matrices are veryimportant topics in every field of science and engineering.

3.1 Matrix

A matrix is a collection of numbers ordered by rows and columns. It is customary to enclosethe numbers of a matrix in brackets [ ] or parenthesis ( ). For example, the following is amatrix:

A =[

3 5 −30 6 1

].

This matrix has two rows and three columns and it is referred to as a “2 by 3” or 2× 3matrix. Let F be a field of scalars and let the elements aij (i = 1, 2, · · · ,m; j = 1, 2, · · · , n),not necessarily distinct, belong to the field F . If we construct a rectangular array A of mnquantities aij into m rows and n columns, then A is said to be a matrix of order or sizem× n (read as m by n) over the field F and usually written in the form

A =

a11 a12 · · · a1n

a21 a22 · · · a2n

......

...an1 an2 · · · ann

. (3.1)

The mn quantities aij are called elements or constituents or coordinates or entries of thematrix. Frequently, the matrix may be written simply as A = [aij ] or [aij ]m×n or (aij) or(aij)m×n; where aij is the ith element or ij entry appears in the ith row and jth column.The numbers a11, a22, · · · , ann form the main or leading or principle diagonal.

Also, the elements of the the matrix A belong to the field F = <, of real numbers,therefore A is a real matrix.

3.1.1 Special Matrices

Row and column matrices

We draw attention to the fact that each row of an m× n matrix has n components, wheren is the number of columns and each column has m components, where m is the number ofrows. A matrix having a single row (column) is called a row (column) matrix. The ith rowand jth column of the matrix A are

149

150 Theory of Matrices

[ai1 ai2 · · · ain], 1 ≤ i ≤ m;

aij

a2j

...amj

, 1 ≤ j ≤ n. (3.2)

respectively. For example, let us consider the real matrix A =(

1 6 72 4 3

)of order (size) 2×3.

It has two rows [1 6 7] and [2 4 3] and three columns(

12

),(

64

)and

(73

). A 1 × n

or n× 1 is also known as n vector. The row and column matrices are sometimes called rowvectors and column vectors. A matrix having only one row is called row matrix, while amatrix having only one column is called column matrix.

3.1.2 Square Matrix

For an m × n matrix [aij ]m×n if m = n, i.e., the number of rows equal to the number ofcolumns, then the matrix is said to be a square matrix. A n × n square matrix is said tobe of order n and is sometimes known as n square matrix. The elements a11, a22, · · · , ann

are known as diagonal elements of A. For example,

1 6 72 4 34 3 6

is a square matrix of order 3.

with 1, 4, 6 is the leading diagonal.

Null matrix

A matrix whose entries are all zero, i.e., aij = 0, for all pairs of i and j, then the matrixA = [aij ]m×n is said to be a null or zero matrix of order m×n and is denoted by 0m×n. For

example, 0 =(

0 0 00 0 0

)is an example of a 2× 3 null matrix. If any one of aij ’s is non zero,

then A is said to a non-zero matrix.

Diagonal matrix

A square matrix A with all non-diagonal elements as zero, is called a diagonal matrix. For

example,(

1 00 4

),(

8 00 0

),(

0 00 0

)are the examples of diagonal matrices. So, for a diagonal

matrix A = [aij ], aij = 0, for i 6= j and it is denoted by A = diag(d11, d22, · · · , dnn). If ina diagonal matrix, all the elements are equal, then the diagonal matrix is called scalar orconstant matrix. Thus for a scalar matrix A = [aij ], we have,

aij = k; for i = j;= 0; for i 6= j

and is denoted by [k]. For example, the diagonal matrix(

2 00 2

)is scalar matrix.

Ex 3.1.1 If a matrix B commutes with a diagonal matrix, no two diagonal elements ofwhich are equal to each other, show that B must be a diagonal matrix.

Solution: Let A be a diagonal matrix of order n whose elements are

aij = ai δij ; 1 ≤ i, j ≤ n,

Matrix 151

where ai are scalars such that ai 6= aj if i 6= j. Let the ijth element of B be bij . Given thatAB = BA, so taking ijth elements of both sides, we have,

n∑p=1

aip bpj =n∑

p=1

bip apj

or,n∑

p=1

ai δip bpj =n∑

p=1

bip aj δpj

or, ai bij = bij aj ⇒ (ai − aj) bij = 0.

This shows that, if i 6= j, then bij = 0. The only elements of B which are likely to be differentfrom zero are the diagonal elements bii for 1 ≤ i ≤ n, proving that B is a diagonal matrix.

Identity matrix

If in a scalar matrix all the diagonal elements are unity, then it is called identity matrix orunit matrix. The nth order identity matrix is denoted by In and is written as

In =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

. (3.3)

The identity matrix can be written as I = [δij ], where δij is the kronecker delta, defined by,δij = 0, if i 6= j and δij = 1, if i = j. We shall denote the ith column of I by ei. Thus ei

has 1 in the ith position and 0’s elsewhere. A permutation matrix is a square matrix withentries 0’s and 1’s such that each row and each column contains exactly one 1.

Triangular matrix

If in a square matrix A = [aij ], all the elements below the diagonal are zero, i.e., aij = 0,for i > j, then the square matrix is said to be an upper triangular matrix and unit upper

triangular if aii = 1; aij = 0, i > j for all i, j. For example,

−8 4 90 4 70 0 6

is an upper

triangular matrix.If in a square matrix A = [aij ], all the elements above the diagonal are zero, i.e., aij = 0,

for i < j, then the square matrix is said to be an lower triangular matrix. For example,−8 0 02 4 01 3 6

is a lower triangular matrix and unit lower triangular if aii = 1; aij = 0, i < j

for all i, j. A square matrix A = [aij ] is said to be a triangular matrix, if it is either uppertriangular or lower triangular. In a diagonal matrix the non-diagonal elements are all zero,so diagonal matrix is both upper and lower triangular.

A matrix is said to be upper Hessenberg if aij = 0 when i > j + 1 and lower Hessenbergif aij = 0 for i < j − 1.

Ex 3.1.2 Find an upper triangular matrix A such that A3 =(

8 −570 27

).

Solution: Let the required upper triangular matrix be A =(a b0 c

), then,

152 Theory of Matrices

A2 =(a b0 c

)(a b0 c

)=(a2 ab+ bc0 c2

)A3 = AA2 =

(a b0 c

)(a2 ab+ bc0 c2

)=(a3 a2b+ abc+ bc2

0 c3

),

⇒ a3 = 8; c3 = 27; a2b+ abc+ bc2 = −57

⇒ a = 2, c = 3, b = −3,⇒ A =(

2 −30 3

).

Trace of a matrix

The spur or trace of a square matrix A = [aij ]n×n is the sum of the diagonal elements as

trA = a11 + a22 + · · ·+ ann =n∑

i=1

aii. (3.4)

For example, the trace of the above matrices are 5 + 4 = 9 and 1 + 4 + 6 = 11 respectively.If A be an m × n real matrix, then tr(AAT ) ≥ 0, the equality occurs if A is a null matrix.If A and B are square matrices of the same order, then

(i) trA+ trB = tr(A+B).

(ii) trAT = trA.

(iii) tr(BA) = tr(AB).

For an m × n matrix if m 6= n, i.e., the number of rows not equal to the number of

columns, then the matrix is said to be a rectangular matrix. For example, A =(

1 6 72 4 3

)is

a rectangular matrix or order 2× 3.

Ex 3.1.3 If A and B are any two 2 × 2 matrics, show that AB − BA = I2 cannot holdunder any circumstances.

Solution: If possible, let, AB −BA = I2, then

tr(AB −BA) = tr(I2)or, tr(AB)− tr(BA) = tr(I2) = 1 + 1 = 2.

But tr(AB) = tr(BA); hence this cannot hold. Therefore, AB−BA = I2 cannot hold underany circumstances.

Band matrix

A real matrix A = [aij ]m×n is said to be band matrix with bandwidth k if

aij = 0 for |i− j| > k. (3.5)

If k = 1, then the matrix is called tridiagonal and if k = 0, then it is called diagonal. It iscalled diagonally dominant if

|aii| ≥n∑

j=1;i 6=j

|aij |; i = 1, 2, ..., n. (3.6)

Equations containing a diagonal matrix can be easily solved and hence some algorithms forsolution of linear equations actually try to transform the original matrix to an equivalentdiagonal form.

Matrix Operations 153

Filled and sparse matrix

If most elements of a matrix are nonzero, then it is said to be filled, while if most of theelements are zero, then it is said to be sparse.

3.2 Matrix Operations

In this section, we are to define the algebraic operations on matrices that will produce newmatrices out of given matrices. These operations are useful in application of matrices.

3.2.1 Equality of matrices

Two matrices A = [aij ]m×n and B = [bij ]m×n are said to be equal iff they are of the sameorder and each element of A is equal to the corresponding element of B, i.e., aij = bij for all

i and j. For example, A =(

23 52

32 43

)and B =

(8 259 64

)are equal matrices. Two matrices

are said to be comparable, if they are of the same type.

Ex 3.2.1 Find the values of x, y, z and u which satisfy the matrix equation(x+ 3 2y + xz − 1 4u− 6

)=(

0 −73 2u

).

Solution: Since the matrices are equal, x + 3 = 0, 2y + x = −7, z − 1 = 3, 4u − 6 = 2u.Solution of these equations is x = −3, z = 4, y = −2 and u = 3. Hence the required valuesof x, y, z, u are −3,−2, 4, 3 respectively.

3.2.2 Matrix Addition

For addition of two matrices, the matrices must be of same order. Let A = [aij ]m×n andB = [bij ]m×n be two given matrices of the same order. The sum of A and B, denoted byA+B, is obtained by adding the corresponding elements of A and B as

A+B = C = [cij ]m×n,

then the elements of C can be written as

cij = aij + bij ; 1 ≤ i ≤ m, 1 ≤ j ≤ n.

Let, A =(

5 67 8

), B =

(3 12 0

)and C =

(2 1 39 6 −1

)be three matrices of order 2×2, 2×2, 2×3

respectively. As A and B are in same order, A+B is defined and

A+B =(

5 67 8

)+(

3 12 0

)=(

5 + 3 6 + 17 + 2 8 + 0

)=(

8 79 8

).

Since the two matrices A and C are not of same order, then they are not conformable foraddition, i.e., A+C and hence B+C are not defined. Matrix subtraction works in the sameway, except that the elements are subtracted instead of added. For example, if

A =[a1 b1 c1a2 b2 c2

]and B =

[x1 y1 z1x2 y2 z2

]then,

A−B =[a1 − x1 b1 − y1 c1 − z1a2 − x2 b2 − y2 c2 − z2

].

154 Theory of Matrices

3.2.3 Matrix Multiplication

Multiplication of matrices by a scalar

If A is a matrix [aij ]m×n and k is a scalar quantity, then the product kA or Ak is the matrix[bij ]m×n where bij = kaij . Thus, if k ∈ F and A = [aij ]m×n be any matrix, then

P = kA = [pij ]m×n; where pij = kaij ; 1 ≤ i ≤ m, 1 ≤ j ≤ n.

Therefore, we see that for scalar multiplication, each element of P is obtained by multiplyingthe corresponding element of A by k. The negative of A is obtained by multiplying by (−1)scalarly. The difference between two matrices A and B of same order m× n is defined as

A−B = A+ (−1)B.

For example, Let, A =(

5 67 8

)and B =

(3 12 0

)be two matrices of order 2× 2 respectively.

So, the scalar multiplication by 3 of A and A−B is given by

3A = 3(

5 67 8

)=(

3.5 3.63.7 3.8

)=(

15 1821 24

)A−B =

(5 67 8

)−(

3 12 0

)=(

5− 3 6− 17− 2 8− 0

)=(

2 55 8

).

Two m× n matrices A and B are equal, if (A− B) equals to the null matrix. Let A,B betwo matrices such that A+B and AB is defined. Then the following properties are satisfied:

(i) kA = Ak.

(ii) k(A+B) = kA+ kB; k ∈ F

(iii) (k + l)A = kA+ lA; k, l ∈ F

(iv) A(kB) = k(AB) = (kA)B.

Thus, the scalar multiplication of matrices is commutative, associative and distributive. IfA1, A2, · · · , Ak are m × n matrices and c1, c2, · · · , ck are scalars, then an expression of theform

c1A1 + c2A2 + · · ·+ ckAk

is called a linear combination of A1, A2, · · · , Ak and c1, c2, · · · , ck are called coefficients.

Theorem 3.2.1 Matrix addition is commutative as well as associative.

Proof: Let A = [aij ]m×n, B = [bij ]m×n and C = [cij ]m×n be three matrices of same order,so that A+B,B + C,A+ C,B +A,C +A,C +B are all defined. Let

X = A+B = [xij ]m×n; where, xij = pij + qij

Y = B +A = [yij ]m×n; where, yij = qij + pij .

Here, X any Y are of same orders and

xij = pij + qij = qij + pij ; as pij , qij ∈ F= yij ; for all 1 ≤ i ≤ m and 1 ≤ j ≤ n.

or, A+B = B +A.

Matrix Operations 155

Hence the matrix addition is commutative. Now,

(A+B) + C = [xij ]m×n + [cij ]m×n

= [rij ]m×n; where, rij = xij + cij = aij + bij + cij ,

A+ (B + C) = [aij ]m×n + [sij ]m×n

= [tij ]m×n; where, tij = aij + sij = aij + bij + cij .

Since, rij = tij for every pair of i and j, we have (A + B) + C = A + (B + C). Therefore,matrix addition is associative.

Since matrix addition is associative, we can define A+B +C as the matrix A+ (B +C)which is the same as (A+B) + C. We can extended it as

n∑i=1

Ai = A1 +A2 + · · ·+An.

Multiplication of a matrix by another matrix

If the number of columns of a matrix A be equal to the number of rows of another matrix B,then the matrices A and B are said to be conformable for the product AB and the productAB is said to be defined.

The number of rows and the number of columns of C are equal to the number of rows ofA and the number of columns of B, respectively. Let

A =[a1 b1 c1a2 b2 c2

]and B =

x1 y1x2 y2x3 y3

, then

AB =−−−−−−−→[a1 b1 c1a2 b2 c2

]yx1 y1x2 y2x3 y3

=[a1.x1 + b1.x2 + c1.x3 a1.y1 + b1.y2 + c1.y3a2.x1 + b2.x2 + c2.x3 a2.y1 + b2.y2 + c2.y3

].

For the product of two matrices A and B, the number of columns of the matrix A mustbe equal to the number of rows of matrix B, otherwise it is impossible to find the product ofA and B. Let A = [aij ]m×p and B = [bij ]p×n be two matrices. Here A,B are conformablefor the product AB. The ijth element is obtained by multiplying the ith row of A by thejth column of B. Hence,

a11 a12 · · · a1p

a21 a22 · · · a2p

......

...am1 am2 · · · amp

b11 b12 · · · b1n

b21 b22 · · · b2n

......

...bp1 bp2 · · · bpn

=

c11 c12 · · · c1n

c21 c22 · · · c2n

......

...cn1 cn2 · · · cnn

.

In the product, the matrix A is called the pre-factor and B is called the post-factor. Clearly,AB is the m× n matrix, whose ijth element

cij = ai1b1j + ai2b2j + · · ·+ aipbpj =p∑

k=1

aikbkj . (3.7)

In the product, we say that B is pre-multiplied by A and B is post-multiplied by B. Inorder that both AB and BA should exist, if A be of order m× n, B be of order n×m.

In general matrix multiplication is not commutative. The difference between the twomatrices AB and BA is known as the commutator of A and B and is denoted by

[A,B] = AB −BA. (3.8)

156 Theory of Matrices

If should be clear that [B,A] = −[A,B]. If, in particular, AB is equal to BA, the twomatrices A and B are said to be commute with each other. AB and BA are equal onlywhen both the matrix A and B are square matrix of same order. The anticommutator ofthe matrices A and B, denoted by A,B is defined by,

A,B = AB +BA. (3.9)

Ex 3.2.2 Consider the matrices A =(

1 53 −2

), B =

(2 3 −14 0 5

). Here A is of order 2 × 2

and B is of order 2× 3. So the product AB is defined and

AB =(

1.2 + 5.4 1.3 + 5.0 1.(−1) + 5.53.2 + (−2).4 3.3 + (−2).0 3.(−1) + (−2).5

)=(

22 3 24−2 9 −13

)which is of order 2× 3. Notice that BA is not defined here.

Ex 3.2.3 Consider the matrices A =(

1 53 −2

)and B =

(−2 14 6

), then,

AB =(

18 31−14 −9

), BA =

(1 −1222 8

).

Hence AB,BA both are defined but AB 6= BA. The commutator of A and B is

[A,B] = AB −BA =(

18 31−14 −9

)−(

1 −1222 8

)=(

17 43−36 −17

).

The anticommutator of A and B is

A,B = AB +BA =(

18 31−14 −9

)+(

1 −1222 8

)=(

19 198 −1

).

Ex 3.2.4 Consider the matrices P =(

2 33 5

)and Q =

(1 00 1

). Here

PQ =(

2 33 5

)(1 00 1

)=(

2 33 5

)= QP

So we can conclude that, if A is an m × p matrix and B is a p × n matrix, then AB is anm×n matrix. BA is not defined if m 6= n. If m = n, then order of AB and BA are differentsizes. Even if both AB and BA are defined they may not be of the same order and hencemay not be equal. Even if AB and BA are defined and are same order they may not beequal.

Result 3.2.1 In ordinary algebra, we know,ab = 0 ⇒ either a = 0 or b = 0.

But in matrix theory, if AB = 0, then it is not necessarily imply that either A = 0 or B = 0.For example, let,

A =(

1 22 4

), B =

(6 −4−3 2

), then

AB =(

1 22 4

)(6 −4−3 2

)=(

0 00 0

).

In this case A is called the left divisor of zero and B is called right divisor of zero.

Matrix Operations 157

Result 3.2.2 In ordinary algebra, we know,ab = ac⇒ either a = 0 or b = c.

But in matrix theory, if AB = AC, then it is not necessarily imply that either A = 0 orB = C. For example, let,

A =(

1 22 4

), B =

(4 2 63 6 9

)and C =

(0 6 85 4 8

),

then AB =(

10 14 2420 28 48

)= AC.

but neither A = 0 nor B = C.

Result 3.2.3 Let us consider the matrix multiplication with special structures.

(i) The multiplication of(−13

)and [2 4] gives

(−13

)[2 4] =

(−2 −46 12

).

(ii) Let us consider AT = [1 0 − 2] and B =

235

, then ATB = [−8].

(iii) Let CT = [1 0 − 3 4], then

BCT =

235

[1 0 − 3 4] =

2 0 −6 83 0 −9 125 0 −15 20

.

Ex 3.2.5 If A =[

3 −41 −1

], prove that An =

[1 + 2n −4nn 1− 2n

], where n is a positive integer.

Solution: We shall prove this by using the principle of mathematical induction. Now,

A2 = A.A =[

3 −41 −1

] [3 −41 −1

]=[

3.3 + (−4).1 3.(−4) + (−4).(−1)1.3 + (−1).1 1.(−4) + (−1).(−1)

]=[

5 −82 −3

]=[

1 + 2.2 −4.22 1− 2.2

]A3 =

[5 −82 −3

] [3 −41 −1

]=[

7 −123 −5

]=[

1 + 2.3 −4.33 1− 2.3

].

Thus the result is true for n = 1, 2, 3. Let the result be true for n = k, then,

Ak+1 = Ak.A =[

1 + 2k −4kk 1− 2k

] [3 −41 −1

]=[

3 + 2k −4− 4kk + 1 −1− 2k

]=[

1 + 2(k + 1) −4(k + 1)k + 1 1− 2(k + 1)

].

Thus the result is true for n = k+1 if it is true for n = k, but it is true for n = 1, 2, 3. Thus

by the principle of mathematical induction, An =[

1 + 2n −4nn 1− 2n

].

Theorem 3.2.2 Matrix multiplication is associative.

158 Theory of Matrices

Proof: Let A = [aij ]m×n, B = [bjk]n×p and C = [ckj ]p×q be three matrices such that theproducts A(BC) and (AB)C are defined. We are to show that A(BC) = (AB)C. Now,

AB = [dik]m×p; BC = [ejl]n×q,

where, dik =n∑

j=1

aijbjk and ejl =p∑

k=1

bjkckl. Now, (AB)C = [uil]m×q, where

uil =p∑

k=1

dikckl =p∑

k=1

n∑j=1

aijbjkckl.

Now, A(BC) = [vil]m×q, where,

vil =n∑

j=1

aijcjl =n∑

j=1

p∑k=1

aijbjkckl.

Since the sums are equal, i.e., corresponding elements in (AB)C and A(BC) are equal, soA(BC) = (AB)C.

Theorem 3.2.3 Matrix multiplication is distributive over addition, i.e., if A,B,C be threematrices such that A(B + C), AB and AC, BC are defined, then

(i) A(B + C) = AB +AC, left distributive,

(ii) (A+B)C = AC +BC, right distributive.

Proof: Let A = [aij ]m×n, B = [bjk]n×p and C = [ckj ]p×q be three matrices such thatA(B + C), AB and AC, BC are defined. Now,

B + C = [djk]n×p, where, djk = bjk + cjk.

Let, A(B + C) = [eik]m×p, then,

eik =n∑

j=1

aijdjk =n∑

j=1

aij(bjk + cjk)

=n∑

j=1

aijbjk +n∑

j=1

aijcjk.

Let AB = [fik]m×p and AC = [gik]m×p, then fik =n∑

j=1

aijbjk, gik =n∑

j=1

aijcjk. If AB+AC =

[hik]m×p, then,hik = fik + gik =

n∑j=1

aijbjk +n∑

j=1

aijcjk.

As the corresponding elements of A(B + C) and AB + AC are equal, so we conclude thatA(B+C) = AB+AC. Similarly, the right distributive (A+B)C = AC +BC. Also, if k bea scalar, then k(AB) = (kA)B = A(kB). Using distributive laws, we can prove that(

k∑i=1

Ai

) l∑j=1

Bj

=k∑

i=1

l∑j=1

AiBj ,

where the summation on the RHS can be taken in any order.

Matrix Operations 159

Definition 3.2.1 Let A be a square matrix. For any positive integer m, Am is defined, as

Am = A.A. · · ·A(m times ).

when A is an n × n non-null matrix, we define, A0 = In, in analogy with real numbers.Using the laws of matrix multiplication, it is easy to see that for a square matrix A,

AmAn = Am+n and (Am)n = Amn

for non-negative integers m and n. It is important that, (AB)n 6= AnBn, in general, theequality holds only when AB = BA. Also, it follows that AmAn = AnAm, i.e., the powersof A commute.

Ex 3.2.6 If

413

A =

−4 8 4−1 2 1−3 6 3

, then find A.

Solution: Let the given equation be of the form XA = B. Since the size of the matrix Xis 3× 1 and that of the matrix B therefore the size of the matrix A should be 1× 3. Hencewe can take A = [a b c]. Now, from the given relation we have4

13

[a b c] =

−4 8 4−1 2 1−3 6 3

or,

4a 4b 4ca b c

3a 3b 3c

=

−4 8 4−1 2 1−3 6 3

.

Equating both sides we get, 4a = −4, 4b = 8, 4c = 4; a = −1, b = 2, c = 1; 3a = −3, 3b =6, 3c = 3. Therefore, a = −1, b = 2, c = 1. Hence the required matrix A is [−1 2 1].

Ex 3.2.7 If n ∈ N and A =[

cos θ sin θ− sin θ cos θ

]then show that An =

[cosnθ sinnθ− sinnθ cosnθ

].

Solution: Here we use the principle of mathematical induction. Now,

A2 =[

cos θ sin θ− sin θ cos θ

] [cos θ sin θ− sin θ cos θ

]=[

cos2 θ − sin2 θ 2 sin θ cos θ−2 sin θ cos θ cos2 θ − sin2 θ

] [cos 2θ sin 2θ− sin 2θ cos 2θ

].

Thus the result be true for n = 2. Let the result be true for n = k. Now,

Ak+1 = AkA =[

cos kθ sin kθ− sin kθ cos kθ

] [cos θ sin θ− sin θ cos θ

]=[

cos θ cos kθ − sin kθ sin θ sin kθ cos θ + sin θ cos kθ− sin kθ cos θ − sin θ cos kθ cos θ cos kθ − sin kθ sin θ

]=[

cos(k + 1)θ sin(k + 1)θ− sin(k + 1)θ cos(k + 1)θ

].

Therefore, the result is true for n = k + 1 if the result is true for n = k. But, the resultis true for n = 2. Hence the result is true for n = 2 + 1 = 3, 3 + 1 = 4, . . .. Thus, byMathematical induction the result is true for n = 2, 3, 4, . . . etc, i.e., for any positive integer.

160 Theory of Matrices

3.2.4 Transpose of a Matrix

Let A = [aij ]m×n be a given matrix. An n×m matrix, obtained by interchanging rows andcolumns of A as AT = [aji]n×m is said to be the transpose of the matrix A. For example,let,

A =(

2 3 63 5 −7

)and B = [2 − 1 5]

then, AT =

2 33 56 −7

and BT =

2−15

.

Thus, transpose of the transpose of a matrix is the given matrix itself, i.e., (AT )T = A.

Theorem 3.2.4 If A and B be two matrices such that A+B is defined, then (A+B)T =AT +BT .

Proof: Let A = [aij ]m×n and B = [bij ]m×n be two given matrices such that A + B isdefined. Also, let, A+B = [cij ]m×n, where, cij = aij + bij . Now,

ijth element of (A+B)T = jith element of (A+B)= jith element of [cij ]m×n = jith element of [aij + bij ]m×n

= jith element of [aij ]m×n + jith element of [bij ]m×n

= ijth element of [aji + ijth element of bji]m×n

= ijth element of AT + ijth element of BT

= ijth element of (A+B)T .

Also, order of (A+B)T is n×m and order of (AT +BT ) is n×m. Hence, (A+B)T = AT +BT .If k be a scalar, then (kA)T = kAT . If A,B be two matrices of same order, then,

(rA+ sB)T = rAT + sBT ,

provided s, t are scalars.

Theorem 3.2.5 If A,B be two matrices of appropriate sizes, then (AB)T = BTAT .

Proof: Let A = [aij ]m×n and B = [bjk]n×p be two given matrices such that AB is definedand the order is m× p. Also, order of AT is n×m and order of BT is p× n, so that orderof BTAT is p×m. Therefore,

order of (AB)T = order of BTAT .

Now, ijth element of AB is obtained by multiplying ith row of A with kth column of B,which is

ai1b1k + ai2b2k + · · ·+ ainbnk.

Also, ikth element of AB = kith element of (AB)T , which is

ai1b1k + ai2b2k + · · ·+ ainbnk.

Also, column k of B becomes kth row of BT and ith row of A becomes ith column of AT .Now,

kith element of BTAT = [b1k b2k · · · bnk][ai1 ai2 · · · ain]T

= b1kai1 + b2kai2 + · · ·+ bnkain.

= ai1b1k + ai2b2k + · · ·+ ainbnk

= kith element of (AB)T .

Few Matrices 161

Therefore, (AB)T = BTAT , i.e., transpose of the product of two matrices is equal to theproduct of their transposes taken in reverse order. This statement can be extended to severalmatrices as

(A.B · · ·KL)T = LTKT · · ·BTAT .

This can be proved by induction. From this result, it follows, if A is a square matrix, then(An)T = (AT )n, n ∈ N .

Ex 3.2.8 Find the matrices A and B such that 2A+ 3B = I2 and A+B = 2AT .

Solution: The given system is 2A + 3B = I2 and A + B = 2AT . Now from the firstequation, we have B = 2AT −A. Therefore, from the first equation we have,

2A+ 3B = I2 ⇒ 2A+ 6AT − 3A = I2

⇒ −A+ 6AT = I2 ⇒ −AT + 6A = I2; taking transpose.

Solving the equations −A+ 6AT = I2,−AT + 6A = I2, we get, A = 15I2. Using the relation

B = 2AT −A, we get, B = 15I2.

Ex 3.2.9 Find the matrices A and B such that

2A+BT =(

2 510 2

)and AT + 2B =

(1 84 1

).

Solution: Taking the transpose of first equation we get,

2AT +B =(

2 105 2

). Also, AT + 2B =

(1 84 1

)⇒ −3B =

(0 −6−3 0

)⇒ B =

(0 21 0

).

From the first given equation we get,

A =12

[(2 510 2

)−BT

]=

12

[(2 510 2

)−(

0 12 0

)]=

12

(2− 0 5− 110− 2 2− 0

)=(

1 24 1

).

3.3 Few Matrices

3.3.1 Nilpotent Matrix

For a least positive integer r, ifAr = 0, the null matrix, (3.10)

then the non-null matrix A is said to be nilpotent matrix of order r. The least value of r iscalled the index of it. For example, let

A =(

2 4−1 −2

), then, A2 =

(0 00 0

).

Therefore, A is a nilpotent matrix of index 2.

Ex 3.3.1 Show that A =(ab b2

−a2 −ab

)is a nilpotent matrix of index 2.

162 Theory of Matrices

Solution: We are to show that, A2 = 0. Now,

A2 =(ab b2

−a2 −ab

)(ab b2

−a2 −ab

)=(

0 00 0

).

Therefore, the given matrix A is a nilpotent matrix of index 2.

Ex 3.3.2 Find non-null real matrices(a bc d

)such that it is a nilpotent matrix of index 2.

Solution: Let A =(a bc d

)be non-null real matrices such that A2 = 0. Therefore,

(a bc d

)(a bc d

)=(a2 + bc ab+ bdac+ cd bc+ d2

)=(

0 00 0

)⇒ a2 + bc = 0; ab+ bd = 0; ac+ cd = 0; bc+ d2 = 0⇒ a2 + bc = 0; a = −d.

Therefore, the non null matrices are given by(a bc −a

); a2 + bc = 0; a, b, c, d ∈ <

.

3.3.2 Idempotent Matrix

A matrix A is said to be idempotent matrix if A2 = A. For example, let

A =

2 −3 −5−1 4 51 −3 −4

, then, A2 =

2 −3 −5−1 4 51 −3 −4

= A.

Therefore, A is an idempotent matrix. Identity matrix is idempotent as I2 = I.

Ex 3.3.3 If A be an idempotent matrix of order n, show that In −A is also idempotent.

Solution: Since A is an idempotent matrix, so by definition, A2 = A. Now,

(In −A)2 = (In −A)(In −A) = I2n − InA−AIn +A2

= I2n − 2AIn +A = In − 2A+A = In −A.

Hence, if A is an idempotent matrix, the matrix In −A is so.

Ex 3.3.4 If A and B are two matrices such that AB = A and BA = B, then show thatAT , BT and A,B are idempotent.

Solution: From the given first relation, AB = A, we have,(AB)T = AT ⇒ BTAT = AT .

Also, as B = BA, we have,(BA)TAT = AT ⇒ ATBTAT = AT

⇒ ATAT = AT ⇒ (AT )2 = AT .

From the relation, BA = B, we have,(BA)T = BT ⇒ ATBT = BT .

Few Matrices 163

Also, as A = AB, we have,(AB)TBT = BT ⇒ BTATBT = BT

⇒ BTBT = BT ⇒ (BT )2 = BT .

Therefore, AT and BT are idempotent. Also,

A = AB = A(BA) = (AB)A = AA = A2

and B = BA = B(AB) = (BA)B = BB = B2.

This shows that A and B are also indempotent.

3.3.3 Involuntary Matrix

A matrix A is said to be involuntary matrix if A2 = I. For example, let

A =

−5 −8 03 5 01 2 −1

, then, A2 =

1 0 00 1 00 0 1

= I.

Identity matrix is also involuntary as I2 = I, and hence, identity matrix is involuntary aswell as idempotent matrix.

Ex 3.3.5 Find all non-null real matrices A =(a bc d

)such that it is an involuntary matrix.

Solution: Here, we are to find all non-null real matrices A such that A2 = I2. Therefore,(a bc d

)(a bc d

)=(a2 + bc ab+ bdac+ cd bc+ d2

)=(

1 00 1

)⇒ a2 + bc = 1; ab+ bd = 0; ac+ cd = 0; bc+ d2 = 1⇒ a = ±1, d = ±1, a+ d = 0, a2 + bc = 0.

Therefore, the non null real matrices are given by(a bc −a

); a2 + bc = 0; a, b, c, d ∈ <

, I2,−I2.

3.3.4 Periodic Matrix

A matrix A is said to be periodic matrix if Ak+1 = A, where k is a positive integer, where

the least value of k is the period of A. For example, let A =

2 −3 −5−1 4 51 −3 −4

, then, A2 = A.

Therefore, A is a periodic matrix of period 2.

3.3.5 Symmetric Matrices

A square matrix A = [aij ]n×n is said to be symmetric, if

AT = A, i.e., aij = aji,∀ pairs of (i, j). (3.11)

For example, the matrix A =

a h gh b fg f c

is a symmetric. In the symmetric matrix A, the

elements of A are symmetric with respect to main diagonal of A. A diagonal matrix is

164 Theory of Matrices

always symmetric. Advantage of working with symmetric matrices A is that only half of Aneeds to be stored and the amount of calculation required is also halved.

A matrix A = [aij ]n×n is said to be pseudo-symmetric if

aij = an+1−j,n+1−i, ∀i and j. (3.12)

Now, note the following:

(i) If A,B be two symmetric matrices of the same order and c is a scalar, then A+B andcA are symmetric matrices.

(ii) However, if A,B are symmetric matrices of the same order, then AB may not be

symmetric. For example, let A =(

2 11 4

), B =

(3 44 1

), then AB =

(10 919 8

), which

is not symmetric. If A,B be two symmetric matrices of the same order then AB issymmetric if and only if AB = BA.

(iii) If A be a symmetric matrix, then An is symmetric for all n ∈ N .

(iv) The product of any matrix with its transpose is symmetric. If A be an m× n matrix,then AT and ATA are symmetric matrices of order m and n respectively.

(v) A matrix A is diagonal if and only if it is symmetric and upper triangular.

3.3.6 Skew-symmetric Matrices

A square matrix A = [aij ]n×n is said to be skew-symmetric, if

AT = −A, i.e., aij = −aji,∀ pairs of (i, j). (3.13)

For a skew symmetric matrix A = [aij ]n×n, we have by definition,

aii = −aii; for i = j ⇒ 2aii = 0, i.e., aii = 0.

So all the diagonal elements in a skew symmetric matrix is zero. For example, A =(

0 2−2 0

)is a skew symmetric matrix of order 2. If A be a skew-symmetric matrix, then An issymmetric or skew-symmetric according as n is even or odd positive integer.

Theorem 3.3.1 Every square matrix can be uniquely expressed as a sum of a symmetricmatrix and a skew symmetric matrix.

Proof: Let A be any given square matrix. Let us write,

A =12(A+AT ) +

12(A−AT ) = B + C, say,

where, B = 12 (A+AT ) and C = 1

2 (A−AT ). Now,

BT =12(A+AT )T =

12[AT + (AT )T

]=

12[AT +A] = B.

Therefore, B is a symmetric matrix. Again,

CT =12(A−AT )T =

12[AT − (AT )T

]=

12[AT −A] = −1

2[A−AT ] = −C.

Few Matrices 165

Therefore, C is a skew-symmetric matrix. So, every square matrix can be expressed as asum of a symmetric matrix and a skew symmetric matrix. Now, we are to show that therepresentation is unique. For this, let A = M + N, where M is symmetric and N is skewsymmetric. Now,

AT = (M +N)T = MT +NT = M −N

⇒ A+AT = 2M ; A−AT = 2N,

⇒ B =12(A+AT ) and C =

12(A−AT ).

Thus, the representation is unique. Therefore, every square matrix can be uniquely expressedas a sum of a symmetric matrix and a skew symmetric matrix.

Ex 3.3.6 Express A =

2 5 −37 −1 1−1 3 4

as a sum of a symmetric and skew symmetric matrix.

Solution: For the given matrix A, we have, AT =

2 7 −15 −1 3−3 1 4

. Now,

A+AT =

2 5 −37 −1 1−1 3 4

+

2 7 −15 −1 3−3 1 4

=

4 12 −412 −2 4−4 4 8

A−AT =

2 5 −37 −1 1−1 3 4

−

2 7 −15 −1 3−3 1 4

=

0 −2 −22 0 −22 2 0

.

Now the symmetric matrix is 12 (A + AT ) and the skew symmetric matrix is 1

2 (A − AT ).Therefore, 2 5 −3

7 −1 1−1 3 4

=12

4 12 −412 −2 4−4 4 8

+12

0 −2 −22 0 −22 2 0

.

This representation is unique. Therefore, the given square matrix A can be uniquely ex-pressed as a sum of a symmetric matrix and a skew symmetric matrix.

Ex 3.3.7 Show that (I3 −A)(I3 +A) is a symmetric matrix, where A is a 3× 3 symmetricor a skew symmetric matrix.

Solution: If A is symmetric, then AT = A and skew-symmetric if AT = −A. Let B =(I3 −A)(I3 +A), then,

B = (I3 −A)(I3 +A) = I3 +A−A−A2 = I3 −A2

BT = [(I3 −A)(I3 +A)]T = [I3 −A2]T = I3 − (AT )2 = I3 −A2,

whatever, A is a symmetric or a skew symmetric matrix. Hence, BT = B and consequentlyB = (I3 −A)(I3 +A) is a symmetric matrix.

3.3.7 Normal Matrix

A real matrix A is normal if it commutes with its transpose AT , i.e., if AAT = ATA. If A issymmetric, orthogonal or skew symmetric, then A is normal. There are also other normal

166 Theory of Matrices

matrices. For example, let, A =(

6 −33 6

), then,

AAT =(

6 −33 6

)(6 3−3 6

)=(

45 00 45

).

Since AAT = ATA, the matrix A is normal.

3.4 Determinants

A very important issue in the study of matrix algebra is the concept of determinant. In thissection, various properties of determinant are studied. The methods for its computationand one of its application are discussed.

The x eliminant of two linear equations a11x+ a12 = 0 and a21x+ a22 = 0 is

−a12

a11= −a22

a21, ; i.e., a11a22 − a12a21 = 0.

Now, the expression (a11a22 − a12a21) can be written in the form∣∣∣∣a11 a12

a21 a22

∣∣∣∣ . Let A = [aij ]

be a square matrix of order n. We define the determinant of A of order n as∣∣∣∣∣∣∣∣∣a11 a12 · · · a1n

a21 a22 · · · a2n

......

...an1 an2 · · · ann

∣∣∣∣∣∣∣∣∣ (3.14)

and it is denoted by detA or |A| or 4. If we consider the matrix A of order 2, then the

determinant of order 2 is |A| =∣∣∣∣a11 a12

a21 a22

∣∣∣∣ , which is the x−eliminant given by the above.

Similarly, for a matrix A of order 3, we have

|A| =

∣∣∣∣∣∣a11 a12 a13

a21 a22 a23

a31 a32 a33

∣∣∣∣∣∣ = a11

∣∣∣∣a22 a23

a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23

a31 a33

∣∣∣∣+ a13

∣∣∣∣a21 a22

a31 a32

∣∣∣∣= a11(a22a33 − a23a32)− a12(a21a33 − a23a31) + a13(a21a32 − a22a31),

which is the x, y eliminant of the system of equations

a11x+ a12y + a13 = 0; a21x+ a22y + a23 = 0; a31x+ a32y + a33 = 0.

An nth order determinant contains n! terms in its expression of which 12n! terms are positive

and remaining 12n! terms are negative. For example, let

A =

1 −2 02 1 3−1 0 2

, then ,

|A| = 1(1.2− 3.0)− (−2)[2.2− (−1).3] + 0[2.0− (−1).1]= 2 + 14 + 0 = 16.

Definition 3.4.1 Let Mnn be the set of all square matrices of order n, whose elementsbelong to a field of scalars F . Then a mapping f : Mnn → F , which assigns to each matrixA ∈ Mnn, which is called the determinant function on the set Mnn and it is denoted by

Determinants 167

detA, or |A| or det(aij). The determinant associated with a square matrix A = [aij ] of ordern× n is scalar (a real or a complex number) defined by

det A = det(aij) =∑

σ

Sgn σ a1σ(1)a2σ(2) · · · anσ(n),

where σ is the permutation(

1 2 · · · nσ(1) σ(2) · · · σ(n)

)and Sgn σ = ±1 according as σ is even or

an odd permutation, the summation extends over all possible permutations σ(1), σ(2), . . . , σ(n)of n second subscripts in a’s. det A is said to be a determinant of order n and is denoted

by vertical bars

∣∣∣∣∣∣∣∣a11 a12 · · · a1n

a21 a22 · · · a2n

· · · · · · · · · · · ·an1 an2 · · · ann

∣∣∣∣∣∣∣∣ or shortly by |aij |n.

The summation∑σ

is said to be the expansion of det A. It contains n! terms as there

are n! permutations of the set 1, 2, . . . , n. Since there are 12n! even and 1

2n! odd permu-tations on the set 1,2, . . ., n, the expansion of det A contains 1

2n! positive terms and 12n!

negative terms. Each term is a product of n elements. The first subscript of a’s run over1, 2, . . . , n in natural order and the second subscript is a permutation σ(1), σ(2), . . . , σ(n)on the set 1, 2, . . . , n, each of which can occur only once. It is observed that each producta1σ(1)a2σ(2) · · · anσ(n) is constituted by taking one and only one element from each row andeach column of A and has a positive or a negative sign depending on whether Sgn σ is evenor odd.

Let A =[a11 a12

a21 a22

]. Then det A =

∑σSgn σ a1σ(1)a2σ(2), where σ is the permutation on

the set 1, 2. There are two permutations on 1, 2, they are σ1 =(

1 21 2

), σ2 =

(1 22 1

),

σ1 is even and σ2 is odd. Therefore,

det A = Sgn σ1 a11a22 + Sgn σ2 a12a21 = a11a22 − a12a22.

Thus det A =∣∣∣∣a11 a12

a21 a22

∣∣∣∣ = a11a22 − a12a21.

Let A =

a11 a12 a13

a21 a22 a23

a31 a32 a33

; det A =∑σSgn σ a1σ(1)a2σ(2) · · · anσ(n), where σ is a permutation

on 1, 2, 3. There are six permutations on 1, 2, 3, they are

σ1 =(

1 2 31 2 3

), σ2 =

(1 2 31 3 2

), σ3 =

(1 2 32 1 3

), σ4 =

(1 2 32 3 1

),

σ5 =(

1 2 33 1 2

), σ6 =

(1 2 33 2 1

).

Among them σ1, σ4, σ5 are even and σ2, σ3, σ6 are odd. Therefore,det A = Sgn σ1 a11a22a33 + Sgn σ2 a11a23a32 + Sgn σ3 a12a21a33

+Sgn σ4 a12a23a31 + Sgn σ5 a13a21a32 + Sgn σ6 a13a22a31

= a11a22a33 − a11a23a32 − a12a21a33 + a12a23a31 + a13a21a32 − a13a22a31.

If the first two columns of A are adjoined to its right, then the expansion of a 3× 3 deter-minant can be obtained as the product of diagonal elements with the assigned signs shownin Fig. 3.1.

Each product of det A is obtained by taking the row subscripts in the natural orderperforming permutation among the column subscript, and hence it is known as the row

168 Theory of Matrices

a11 a12 a13 a11 a12

a21 a22 a23 a21 a22

a31 a32 a33 a31 a32j j j j jj− − − + + +

Figure 3.1: Row expansion of a third order determinant.

expansion of det A. Similarly, det A obtained by taking the column subscripts in a naturalorder and making all possible permutations among the row subscripts. Thus

det A =∑

σ

Sgn σ aσ(1)1aσ(2)2 · · · aσ(n)n. (3.15)

Ex 3.4.1 Find the number of 2 × 2 matrices over Z3( the field with three elements) withdeterminant 1. [IIT-JAM’10]

Solution: Let us take a 2× 2 matrix as A =[a bc d

]where a, b, c, d can take three 0, 1, 2

values. From the given condition |ad− bc| = 1. Thus either ad = 1, bc = 0 or ad = 0, bc = 1.Now bc = 0, is possible as

b = 0, c = 1; b = 0, c = 0; b = 1, c = 0; b = 2, c = 0; b = 0, c = 1and ad = 1 either through a = 1, b = 1; a = 2, b = 2. Therefore total number of such matrixis = 2× 2× 5 = 20. Also when ad = 2, bc = 1 or ad = 1, bc = 2, then |ad− bc| = 1. Numberof such matrix = 2 + 2 = 4.

Therefore, the number of 2 × 2 matrices over Z3( the field with three elements) withdeterminant 1 is 24.

Ex 3.4.2 Let Dn be a determinant of order n in which the diagonal elements are 1 andthose just above and just below the diagonal elements are a and all other elements are zero.Prove that D4 −D3 + a2D2 = 0 and hence find the value of

44 =

∣∣∣∣∣∣∣∣1 1

2 0 012 1 1

2 00 1

2 1 12

0 0 12 1

∣∣∣∣∣∣∣∣ .Solution: According to the definition of Dn, we get the form of D4 as

D4 =

∣∣∣∣∣∣∣∣1 a 0 0a 1 a 00 a 1 a0 0 a 1

∣∣∣∣∣∣∣∣ =∣∣∣∣∣∣1 a 0a 1 a0 a 1

∣∣∣∣∣∣− a

∣∣∣∣∣∣a a 00 1 a0 a 1

∣∣∣∣∣∣= D3 − a.a

∣∣∣∣ 1 aa 1

∣∣∣∣ = D3 − a2D2.

For the particular given determinant, we have,

44 = 43 −(

12

)2

42 =

∣∣∣∣∣∣1 1

2 012 1 1

20 1

2 1

∣∣∣∣∣∣−(

12

)2 ∣∣∣∣ 1 12

12 1

∣∣∣∣=(

1− 14

)− 1

2

(12− 0)− 1

4

(1− 1

4

)=

516.

Determinants 169

Property 3.4.1 The determinant of a matrix and its transpose are equal, i.e., |A| = |AT |.

Proof: First, consider a matrix A of order 2, then∣∣∣∣a11 a12

a21 a22

∣∣∣∣ = a11a22 − a12a21 = a11a22 − a21a12 = |AT |.

Let us consider a matrix of order 3, then

|A| =

∣∣∣∣∣∣a11 a12 a13

a21 a22 a23

a31 a32 a33

∣∣∣∣∣∣ = a11

∣∣∣∣a22 a23

a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23

a31 a33

∣∣∣∣+ a13

∣∣∣∣a21 a22

a31 a32

∣∣∣∣= a11(a22a33 − a23a32)− a12(a21a33 − a23a31) + a13(a21a32 − a22a31),= a11(a22a33 − a23a32)− a21(a13a32 − a12a33) + a31(a12a23 − a22a13),

=

∣∣∣∣∣∣a11 a21 a31

a12 a22 a32

a13 a23 a33

∣∣∣∣∣∣ = |AT |.

This property is true for any order of determinants. For example, let A =(

1 24 5

),

|A| =∣∣∣∣1 24 5

∣∣∣∣ = 1.5− 2.4 = −3

and |AT | =∣∣∣∣1 42 5

∣∣∣∣ = 1.5− 2.4 = −3.

Hence |A| = |AT |. From this property, we can say that a theorem which holds for some rowoperations on A, also holds equally well when corresponding column operations are madeon A.

Property 3.4.2 The interchange of two rows (or columns) of a square matrix A changesthe sign of |A|, but its value remains unaltered.

Proof: Let A =

a11 a12 a13

a21 a22 a23

a31 a32 a33

and A∗ =

a21 a22 a23

a11 a12 a13

a31 a32 a33

be the square matrices of order

3, where A∗ be obtained by interchanging any two rows of A. Therefore,

|A∗| =

∣∣∣∣∣∣a21 a22 a23

a11 a12 a13

a31 a32 a33

∣∣∣∣∣∣ = a11

∣∣∣∣a22 a23

a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23

a31 a33

∣∣∣∣+ a13

∣∣∣∣a21 a22

a31 a32

∣∣∣∣= a11(a22a33 − a23a32)− a12(a21a33 − a23a31) + a13(a21a32 − a22a31),= −a11(a22a33 − a23a32)− a12(a21a33 − a23a31) + a13(a21a32 − a22a31) = −|A|.

Similar proof for any interchange between two columns independently, by considering theequivalent from of the expression of |A|. This is true for any order of square matrices.

Property 3.4.3 If in a square matrix A of order n, two rows (columns) are equal or iden-tical, then the value of |A| = 0.

Proof: Let |A| = 4, the value of the determinant. We know, the interchange of any tworows or columns of a determinant changes the sign of the determinant without changing itsnumerical value. Hence the matrix A remain unchanged. Therefore,

−4 = |A| = 4⇒ 24 = 0 ⇒4 = |A| = 0.

If |A| = 0, then the matrix A is called singular matrix, otherwise it is non-singular.

170 Theory of Matrices

Property 3.4.4 Let the elements of mth row of A are all zero, then if we expand thedeterminant of A with respect to the mth row, each term in the expression, contains a factorzero. Hence the value of |A| is 0. Thus, if a row or a column of any matrix consists entirelyof zeros, then |A| = 0.

Result 3.4.1 If two rows (or columns) of a matrix A become identical, for x = a, then(x − a) is a factor of |A|. Further, if r rows (or columns) become identical for x = a, then(x− a)r−1 is a factor of |A|.

Ex 3.4.3 Prove without expanding that |A| =

∣∣∣∣∣∣a2 a 1b2 b 1c2 c 1

∣∣∣∣∣∣ = −(a− b)(b− c)(c− a).

Solution: Let us consider the elements of A as polynomials in a. When a = b, two rows ofthe matrix A become identical. Therefore a− b is a factor of |A|. Now, let us consider theelements of A as polynomials in b. When b = c, two rows of the matrix A become identical.Therefore b − c is a factor of |A|. Similarly, c − a is a factor of |A|. Also, we see that, theexpression of |A| is a polynomial in a, b and c of degree 3. The leading term in the expansionof |A| is a2b. No other term in the expansion of |A| is a2b. Therefore,

|A| = k(a− b)(b− c)(c− a),

where the constant k is independent of a, b, c. Equating coefficients of a2b from both sidesof this equality, we get

1 = k.1.1.(−1) ⇒ k = −1.

Therefore, |A| = −(a− b)(b− c)(c− a).

Property 3.4.5 If every element of any row (or column) of a matrix A be multiplied by afactor k, then |A| is multiplied by the same factor k.

Property 3.4.6 If we add k times the elements of any row (column) of a matrix A tothe corresponding elements of any other row (column), the value of the determinant of A

remains unchanged. Therefore,∣∣∣∣a bc d

∣∣∣∣, ∣∣∣∣ a bc+ ak d+ bk

∣∣∣∣, ∣∣∣∣a+ bk bc+ dk d

∣∣∣∣ are of the same value.

Also, if in an m× n matrix A, if one row (column) be expressed as a linear combination ofother rows(columns) then |A| = 0.

Property 3.4.7 If every element of any row (or column) of a matrix A can be expressedas the sum of two quantities, then the determinant can also be expressed as the sum of twodeterminants. Thus,∣∣∣∣∣∣∣∣∣

a11 + k1 a12 + k2 · · · a1n + kn

a21 a22 · · · a2n

......

...an1 an2 · · · ann

∣∣∣∣∣∣∣∣∣ =∣∣∣∣∣∣∣∣∣a11 a12 · · · a1n

a21 a22 · · · a2n

......

...an1 an2 · · · ann

∣∣∣∣∣∣∣∣∣+∣∣∣∣∣∣∣∣∣k1 k2 · · · kn

a21 a22 · · · a2n

......

...an1 an2 · · · ann

∣∣∣∣∣∣∣∣∣ .Property 3.4.8 Let f1(x), f2(x), g1(x) and g2(x) are differentiable functions of the realvariable x, then

d

dx

∣∣∣∣f1(x) f2(x)g1(x) g2(x)

∣∣∣∣ =∣∣∣∣ d

dxf1(x)ddxf2(x)

g1(x) g2(x)

∣∣∣∣+ ∣∣∣∣ f1(x) f2(x)ddxg1(x)

ddxg2(x)

∣∣∣∣=∣∣∣∣ d

dxf1(x) f2(x)ddxg1(x) g2(x)

∣∣∣∣+ ∣∣∣∣f1(x) ddxf2(x)

g1(x) ddxg2(x)

∣∣∣∣ .

Determinants 171

This result can be extended to any finite order of determinants.

Ex 3.4.4 If f(x) =

xn sinx cosxn! sin nπ

2 cos nπ2

a a2 a3

, then show that f (n)(0) = 0.

Solution: According to the property of the derivative of the determinant, we get,

f (n) =

n! sin(x+ nπ

2

)cos(x+ nπ

2

)n! sin nπ

2 cos nπ2

a a2 a3

+

xn sinx cosx0 0 0a a2 a3

+

xn sinx cosxn! sin nπ

2 cos nπ2

0 0 0

.

Therefore, [f (n)(x)]x=0 is given by,

[f (n)(x)]x=0 =

n! sin nπ2 cos nπ

2n! sin nπ

2 cos nπ2

a a2 a3

+

xn sinx cosx0 0 0a a2 a3

+

xn sinx cosxn! sin nπ

2 cos nπ2

0 0 0

= 0 + 0 + 0 = 0.

3.4.1 Product of Determinants

The product of two determinants of order n is also a determinant of the order n. Let |aij |and |bij | be two determinants of order n. Then their product is defined by,

|aij |.|bij | = |cij |; where, cij =n∑

k=1

aikbkj , i.e.,

∣∣∣∣∣∣∣∣∣a11 a12 · · · a1n

a21 a22 · · · a2n

......

...an1 an2 · · · ann

∣∣∣∣∣∣∣∣∣ .∣∣∣∣∣∣∣∣∣b11 b12 · · · b1n

b21 b22 · · · b2n

......

...bn1 bn2 · · · bnn

∣∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

n∑k=1

a1kb1k

n∑k=1

a1kb2k · · ·n∑

k=1

a1kbnk

n∑k=1

a2kb1k

n∑k=1

a2kb2k · · ·n∑

k=1

a2kbnk

......

...n∑

k=1

ankb1k

n∑k=1

ankb2k · · ·n∑

k=1

ankbnk

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣.

This rule of multiplication is called the ‘matrix rule’ or the rule of ‘multiplication of rows bycolumns.’ Since interchange of rows and column does not alter the value of the determinant,hence the product can be obtained in other forms also, cij may be taken also as cij =

n∑k=1

aikbjk. This rule of multiplication is called the rule of ‘multiplication of rows.’ Similarly,

we can define ‘multiplication of rows by columns.’ From the definition, if A and B be squarematrices of the same order, then |AB| = |A|.|B|.

Ex 3.4.5 Prove without expanding

∣∣∣∣∣∣∣∣1 a a2 a3 + bcd1 b b2 b3 + cda1 c c2 c3 + dab1 d d2 d3 + abc

∣∣∣∣∣∣∣∣ = 0

172 Theory of Matrices

Solution: The given determinant can be written in the form

∆ =

∣∣∣∣∣∣∣∣1 a a2 a3 + bcd1 b b2 b3 + cda1 c c2 c3 + dab1 d d2 d3 + abc

∣∣∣∣∣∣∣∣ =∣∣∣∣∣∣∣∣1 a a2 a3

1 b b2 b3

1 c c2 c3

1 d d2 d3

∣∣∣∣∣∣∣∣+∣∣∣∣∣∣∣∣1 a a2 bcd1 b b2 cda1 c c2 dab1 d d2 abc

∣∣∣∣∣∣∣∣= ∆1 + ∆2, say.

Now, we simplify ∆2 as

∆2 =1

abcd

∣∣∣∣∣∣∣∣a a2 a3 abcdb b2 b3 abcdc c2 c3 abcdd d2 d3 abcd

∣∣∣∣∣∣∣∣;R′1 = aR1, R

′2 = bR2

;R′3 = cR3, R′4 = dR4

=

∣∣∣∣∣∣∣∣a a2 a3 1b b2 b3 1c c2 c3 1d d2 d3 1

∣∣∣∣∣∣∣∣ = −

∣∣∣∣∣∣∣∣1 a a2 a3

1 b b2 b3

1 c c2 c3

1 d d2 d3

∣∣∣∣∣∣∣∣ = −∆1,

applying three successive interchanges to bring C4 to C1. Therefore ∆ = 0.

Ex 3.4.6 Let m,n ∈ N with m ≥ n− 1 ≥ 1 and(mr

)= mCr. Prove that∣∣∣∣∣∣∣∣∣∣∣

1 1 1 · · · 1(m1

) (m+1

1

) (m+2

1

)· · ·(m+n−1

1

)(m2

) (m+1

2

) (m+2

2

)· · ·(m+n−1

2

)...

......

......(

mn−1

) (m+1n−1

) (m+2n−1

)· · ·(m+n−1

n−1

)

∣∣∣∣∣∣∣∣∣∣∣= 1.

Solution: Let ∆n be the given determinant. Subtracting the preceding column from eachcolumn beginning with the second, we have,

∆n =

∣∣∣∣∣∣∣∣∣∣∣

1 0 0 · · · 0(m1

)1 1 · · · 1(

m2

) (m1

) (m+1

1

)· · ·(m+n−2

1

)...

......

......(

mn−1

) (m

n−1

) (m+1n−1

)· · ·(m+n−2

n−1

)

∣∣∣∣∣∣∣∣∣∣∣.

Expanding in terms of the first row,

∆n =

∣∣∣∣∣∣∣∣∣1 1 · · · 1(m1

) (m+1

1

)· · ·(m+n−2

1

)...

......

...(m

n−2

) (m+1n−2

)· · ·(m+n−2

n−2

)∣∣∣∣∣∣∣∣∣ = ∆n−1.

Therefore ∆n = ∆n−1 = ∆n−2 = · · · = ∆2. But

∆2 =∣∣∣∣ 1 1(

m1

) (m+1

1

) ∣∣∣∣ = 1.

Consequently, ∆n = 1.

Ex 3.4.7 Prove that

∣∣∣∣∣∣b2 + c2 a2 a2

b2 c2 + a2 b2

c2 c2 a2 + b2

∣∣∣∣∣∣ = 4a2b2c2. [WBUT 2007]

Determinants 173

Solution:

∣∣∣∣∣∣b2 + c2 a2 a2

b2 c2 + a2 b2

c2 c2 a2 + b2

∣∣∣∣∣∣ =∣∣∣∣∣∣

0 −2c2 −2b2

b2 c2 + a2 b2

c2 c2 a2 + b2

∣∣∣∣∣∣ [R′1 = R1 −R2 −R3]

= −2

∣∣∣∣∣∣0 c2 b2

b2 c2 + a2 b2

c2 c2 a2 + b2

∣∣∣∣∣∣ = −2

∣∣∣∣∣∣0 c2 b2

b2 a2 0c2 0 a2

∣∣∣∣∣∣ [R′2 = R2 −R1,R′3 = R3 −R1]

= −20− c2(a2b2) + b2(0− a2c2) = −2 (−2 a2 b2 c2) = 4a2b2c2.

Ex 3.4.8 If tan−1

√a− c

c+ x+ tan−1

√a− c

c+ y+ tan−1

√a− c

c+ z= 0, prove that∣∣∣∣∣∣

1 x (a+ x)√c+ x

1 y (a+ y)√c+ y

1 z (a+ z)√c+ z

∣∣∣∣∣∣ = 0.

Solution: Let α = tan−1√

a−cc+x , β = tan−1

√a−cc+y , γ = tan−1

√a−cc+z . Hence,from the given

condition we get,α+ β + γ = 0 ⇒ α+ β = −γ

or, tan(α+ β) = − tan(γ) or,tanα+ tanβ

1− tanα tanβ= − tan γ.

or, tanα+ tanβ + tan γ = tanα tanβ tan γ.

Again let, x + c = X2, y + c = Y 2, z + c = Z2 then, tanα =√

a−cX , tanβ =

√a−cY , tan γ =

√a−cZ . Therefore, from the above relation we get,

√a− c

[1X

+1Y

+1Z

]=

(√a− c)3

XY Z

or, Y Z + ZX +XY = a− c.

Now, the given determinant 4 becomes,

4 =

∣∣∣∣∣∣1 X2 − c (X2 + a− c)X1 Y 2 − c (Y 2 + a− c)Y1 Z2 − c (Z2 + a− c)Z

∣∣∣∣∣∣ ; as a+ x = a+X2 − c, etc.

=

∣∣∣∣∣∣1 X2 (X2 +XY + Y Z + ZX)X1 Y 2 (Y 2 +XY + Y Z + ZX)Y1 Z2 (Z2 +XY + Y Z + ZX)Z

∣∣∣∣∣∣ ; using C ′2 = C2 + cC1

=

∣∣∣∣∣∣1 X2 X2(X + Y + Z) +XY Z1 Y 2 Y 2(X + Y + Z) +XY Z1 Z2 Z2(X + Y + Z) +XY Z

∣∣∣∣∣∣ =∣∣∣∣∣∣1 X2 X2(X + Y + Z)1 Y 2 Y 2(X + Y + Z)1 Z2 Z2(X + Y + Z)

∣∣∣∣∣∣+∣∣∣∣∣∣1 X2 XY Z1 Y 2 XY Z1 Z2 XY Z

∣∣∣∣∣∣= (X + Y + Z)

∣∣∣∣∣∣1 X2 X2

1 Y 2 Y 2

1 Z2 Z2

∣∣∣∣∣∣+XY Z

∣∣∣∣∣∣1 X2 11 Y 2 11 Z2 1

∣∣∣∣∣∣ = (X + Y + Z).0 +XY Z.0 = 0

Ex 3.4.9 Show that the value of

∣∣∣∣∣∣cos(x+ a) sin(x+ a) 1cos(x+ b) sin(x+ b) 1cos(x+ c) sin(x+ c) 1

∣∣∣∣∣∣ independent of x.

Solution: If 4 be the value of the determinant, we have

4 =

∣∣∣∣∣∣cos(x+ a) sin(x+ a) 1

cos(x+ b)− cos(x+ a) sin(x+ b)− sin(x+ a) 0cos(x+ c)− cos(x+ a) sin(x+ c)− sin(x+ a) 0

∣∣∣∣∣∣

174 Theory of Matrices

=∣∣∣∣ cos(x+ b)− cos(x+ a) sin(x+ b)− sin(x+ a)cos(x+ c)− cos(x+ a) sin(x+ c)− sin(x+ a)

∣∣∣∣=∣∣∣∣2 sin 2x+a+b

2 sin a−b2 2 cos 2x+a+b

2 sin b−a2

2 sin 2x+a+c2 sin a−c

2 2 cos 2x+a+c2 sin c−a

2

∣∣∣∣= sin

a− b

2sin

a− c

2

∣∣∣∣2 sin 2x+a+b2 −2 cos 2x+a+b

22 sin 2x+a+c

2 −2 cos 2x+a+c2

∣∣∣∣= 4 sin

a− b

2sin

a− c

2sin

c− b

2,

which is independent of x.

Ex 3.4.10 If A+B + C = π, then show that

∣∣∣∣∣∣sin2A cotA 1sin2B cotB 1sin2 C cotC 1

∣∣∣∣∣∣ = 0.

Solution:∣∣∣∣∣∣sin2A cotA 1sin2B cotB 1sin2 C cotC 1

∣∣∣∣∣∣ =∣∣∣∣∣∣

sin2A cotA 1sin2B − sin2A cotB − cotA 0sin2 C − sin2A cotC − cotA 0

∣∣∣∣∣∣ [R′2 = R2 −R1,R′3 = R3 −R1]

=∣∣∣∣ sin2B − sin2A cotB − cotAsin2 C − sin2A cotC − cotA

∣∣∣∣=∣∣∣∣ sin(B −A) sin(B +A) sin(A−B)/ sinA sinBsin(C −A) sin(C +A) sin(A− C)/ sinA sinC

∣∣∣∣= sin(B −A) sin(C −A)

∣∣∣∣ sin(B +A) −1/ sinA sinBsin(C +A) −1/ sinA sinC

∣∣∣∣ = 0,

as two rows are identical.

Ex 3.4.11 If 2s = a+b+c, show that

∣∣∣∣∣∣a2 (s− a)2 (s− a)2

(s− b)2 b2 (s− b)2

(s− c)2 (s− c)2 c2

∣∣∣∣∣∣ = 2s3(s−a)(s−b)(s−c).

Solution: Let s− a = α, s− b = β, s− c = γ. Hence

α+ β + γ = 3s− (a+ b+ c) = 3s− 2s = s,

β + γ = 2s− (b+ c) = a.

Similarly γ + α = b, α+ β = c. Now,

∆ =

∣∣∣∣∣∣(β + γ)2 α2 α2

β2 (γ + α)2 β2

γ2 γ2 (α+ β)2

∣∣∣∣∣∣=

∣∣∣∣∣∣(β + γ)2 − α2 0 α2

0 (γ + α)2 − β2 β2

γ2 − (α+ β)2 γ2 − (α+ β)2 (α+ β)2

∣∣∣∣∣∣ (C1′ = C1 − C3, C2

′ = C2 − C3)

= (α+ β + γ)2

∣∣∣∣∣∣β + γ − α 0 α2

0 γ + α− β β2

γ − α− β γ − α− β (α+ β)2

∣∣∣∣∣∣= (α+ β + γ)2

∣∣∣∣∣∣β + γ − α 0 α2

0 γ + α− β β2

−2β −2α 2αβ

∣∣∣∣∣∣ (R3′ = R3 −R1 −R2)

Determinants 175

= 2(α+ β + γ)21αβ

∣∣∣∣∣∣αβ + γα− α2 0 α2

0 βγ + αβ − β2 β2

−αβ −αβ αβ

∣∣∣∣∣∣ (C1′′ = C1

′α, C2′′ = C2

′β)

= 2(α+ β + γ)21αβ

∣∣∣∣∣∣αβ + γα α2 α2

β2 βγ + αβ β2

0 0 αβ

∣∣∣∣∣∣ (C1′′′ = C1

′′ + C3, C2′′′ = C2

′′ + C3)

= 2(α+ β + γ)2∣∣∣∣α(β + γ) α2

β2 β(γ + α)

∣∣∣∣ = 2(α+ β + γ)2αβ(β + γ)(γ + α)− αβ

= 2αβγ(α+ β + γ)3 = 2s3(s− a)(s− b)(s− c).

Ex 3.4.12 Show that

∣∣∣∣∣∣(b+ c)2 c2 b2

c2 (c+ a)2 a2

b2 a2 (a+ b)2

∣∣∣∣∣∣ = 2(ab+ bc+ ca)3.

Solution: Using the properties of the determinants, we have,

4 =

∣∣∣∣∣∣(b+ c)2 c2 b2

c2 (c+ a)2 a2

b2 a2 (a+ b)2

∣∣∣∣∣∣ = 1a2b2c2

∣∣∣∣∣∣(ab+ ca)2 b2c2 b2c2

c2a2 (bc+ ab)2 c2a2

a2b2 a2b2 (ca+ bc)2

∣∣∣∣∣∣=

1a2b2c2

∣∣∣∣∣∣(ab+ ca)2 b2c2 − (ab+ ca)2 b2c2 − (ab+ ca)2

c2a2 (bc+ ab)2 − c2a2 0a2b2 0 (ca+ bc)2 − a2b2

∣∣∣∣∣∣ ; C2 − C1, C3 − C1

=(bc+ ca+ ab)2

b2c2

∣∣∣∣∣∣(b+ c)2 bc− ab− ca bc− ab− cac2 bc+ ab− ca 0b2 0 bc− ab+ ca

∣∣∣∣∣∣=

(bc+ ca+ ab)2

b2c2

∣∣∣∣∣∣2bc −2ab −2cac2 bc+ ab− ca 0b2 0 bc− ab+ ca

∣∣∣∣∣∣ ; R1 − (R2 +R3)

=2(bc+ ca+ ab)2

ab3c3

∣∣∣∣∣∣abc −abc −cabc2a bc2 + abc− c2a 0ab2 0 b2c− ab2 + bca

∣∣∣∣∣∣=

2(bc+ ca+ ab)2

ab3c3

∣∣∣∣∣∣abc 0 0c2a bc2 + abc c2aab2 ab2 b2c+ bca

∣∣∣∣∣∣ ; C2 + C1, C3 + C1

=2(bc+ ca+ ab)2

b2c2(bc2 + abc)(bca+ b2c)− a2b2c2 = 2(bc+ ca+ ab)3.

Ex 3.4.13 For a fixed positive integer n, if 4 =

∣∣∣∣∣∣n! (n+ 1)! (n+ 2)!

(n+ 1)! (n+ 2)! (n+ 3)!(n+ 2)! (n+ 3)! (n+ 4)!

∣∣∣∣∣∣, then show that

[ 4(n!)3 − 4] is divisible by n.

Solution: Taking out n!, (n+ 1)! and (n+ 2)! from the first, second and third row respec-tively, we get,

4 = n!(n+ 1)!(n+ 2)!

∣∣∣∣∣∣1 (n+ 1) (n+ 1)(n+ 2)1 (n+ 2) (n+ 2)(n+ 3)1 (n+ 3) (n+ 3)(n+ 4)

∣∣∣∣∣∣= (n!)3(n+ 1)2(n+ 2)

∣∣∣∣∣∣1 (n+ 1) (n+ 1)(n+ 2)0 1 2(n+ 2)0 1 2(n+ 3)

∣∣∣∣∣∣ ; R2 −R1, R3 −R1

176 Theory of Matrices

= (n!)3(n+ 1)2(n+ 2).2

⇒ 4(n!)3

− 4 = 2(n+ 1)2(n+ 2)− 4 = 2n(n2 + 4n+ 5).

Therefore, [ 4(n!)3 − 4] is divisible by n.

Ex 3.4.14 Show that

∣∣∣∣∣∣−bc bc+ b2 bc+ c2

ca+ a2 −ca ca+ c2

ab+ a2 ab+ b2 −ab

∣∣∣∣∣∣ = (ab+ bc+ ca)3.

Solution: Let the value of the determinant be 4, then,

4 =1abc

∣∣∣∣∣∣−abc abc+ ab2 abc+ ac2

bca+ ba2 −bca bca+ bc2

cab+ ca2 cab+ cb2 −cab

∣∣∣∣∣∣=

1abc

∣∣∣∣∣∣abc+ a2b+ a2c abc+ ab2 + b2c abc+ ac2 + bc2

bca+ ba2 −bca bca+ bc2

cab+ ca2 cab+ cb2 −cab

∣∣∣∣∣∣ ;R1 +R2 +R3

=ab+ bc+ ca

abc

∣∣∣∣∣∣a b c

bca+ ba2 −bca bca+ bc2

cab+ ca2 cab+ cb2 −cab

∣∣∣∣∣∣ = (ab+ bc+ ca)

∣∣∣∣∣∣1 1 1

bc+ ab −ca ab+ bcbc+ ac ac+ bc −ab

∣∣∣∣∣∣= (ab+ bc+ ca)

∣∣∣∣∣∣1 0 0

bc+ ab −ca− bc− ab 0ac+ ac 0 −ab− bc− ac

∣∣∣∣∣∣ ;C2 − C1, C3 − C1

= (ab+ bc+ ca)3

∣∣∣∣∣∣1 0 0

bc+ ab −1 0ac+ ac 0 −1

∣∣∣∣∣∣ = (ab+ bc+ ca)3.

Ex 3.4.15 Show that

∣∣∣∣∣∣1 + a2 − b2 2ab −2b

2ab 1− a2 + b2 2a2b −2a 1− a2 − b2

∣∣∣∣∣∣ = (1 + a2 + b2)3.

Solution: Let the value of the determinant be 4, then,

4 =

∣∣∣∣∣∣1 + a2 + b2 0 −2b

0 1 + a2 + b2 2ab(1 + a2 + b2) −a(1 + a2 + b2) 1− a2 − b2

∣∣∣∣∣∣ ;C1 − bC3, C2 + aC3

= (1 + a2 + b2)2

∣∣∣∣∣∣1 0 −2b0 1 2ab −a 1− a2 − b2

∣∣∣∣∣∣= (1 + a2 + b2)2(1− a2 − b2 + 2a2)− 2b(0− b) = (1 + a2 + b2)3.

Ex 3.4.16 Show that

∣∣∣∣∣∣1 + a 1 1

1 1 + b 11 1 1 + c

∣∣∣∣∣∣ = abc(1 + 1

a + 1b + 1

c

).

Solution: Let the value of the determinant be 4, then,

4 = abc

∣∣∣∣∣∣1 + 1

a1b

1c

1a 1 + 1

b1c

1a

1b 1 + 1

c

∣∣∣∣∣∣ = abc

∣∣∣∣∣∣1 + 1

a + 1b + 1

c1b

1c

1 + 1a + 1

b + 1c 1 + 1

b1c

1 + 1a + 1

b + 1c

1b 1 + 1

c

∣∣∣∣∣∣ ;C1 + C2 + C3

Determinants 177

= abc

(1 +

1a

+1b

+1c

) ∣∣∣∣∣∣1 1

b1c

1 1 + 1b

1c

1 1b 1 + 1

c

∣∣∣∣∣∣= abc

(1 +

1a

+1b

+1c

) ∣∣∣∣∣∣1 1

b1c

0 1 00 0 1

∣∣∣∣∣∣ ;R2 −R1, R3 −R1

= abc

(1 +

1a

+1b

+1c

)1− 0 = abc

(1 +

1a

+1b

+1c

).

Ex 3.4.17 Show that

∣∣∣∣∣∣a b ax+ byb c bx+ cy

ax+ by bx+ cy 0

∣∣∣∣∣∣ = (b2 − ac)(ax2 + 2bxy + cy2).

Solution: Let the value of the determinant be 4, then,

4 =

∣∣∣∣∣∣a b 0b c 0

ax+ by bx+ cy −ax2 − 2bxy − cy2

∣∣∣∣∣∣ ;C3 − xC1 − yC2

= (ax2 + 2bxy + cy2)

∣∣∣∣∣∣a b 0b c 0

ax+ by bx+ cy −1

∣∣∣∣∣∣= (ax2 + 2bxy + cy2)−1(ac− b2) = (b2 − ac)(ax2 + 2bxy + cy2).

Ex 3.4.18 Show that

∣∣∣∣∣∣0 (a− b)2 (a− c)2

(b− a)2 0 (b− c)2

(c− a)2 (c− b)2 0

∣∣∣∣∣∣ = 2(b− c)2(c− a)2(a− b)2.

Solution: Let the value of the determinant be 4, then,

4 =

∣∣∣∣∣∣(a− a)2 (a− b)2 (a− c)2

(b− a)2 (b− b)2 (b− c)2

(c− a)2 (c− b)2 (c− c)2

∣∣∣∣∣∣ =∣∣∣∣∣∣a2 a 1b2 b 1c2 c 1

∣∣∣∣∣∣∣∣∣∣∣∣1 −2a a2

1 −2b b2

1 −2c c2

∣∣∣∣∣∣=

∣∣∣∣∣∣a2 a 1

b2 − a2 b− a 0c2 − a2 c− a 0

∣∣∣∣∣∣∣∣∣∣∣∣1 −2a a2

0 −2b+ 2a b2 − a2

0 −2c+ 2a c2 − a2

∣∣∣∣∣∣= (b− a)(c− a)

∣∣∣∣∣∣a2 a 1b+ a 1 0c+ a 1 0

∣∣∣∣∣∣ (b− a)(c− a)

∣∣∣∣∣∣1 −2a a2

0 −2 b+ a0 −2 c+ a

∣∣∣∣∣∣= (b− a)2(c− a)2(b+ a− c− a)(−2c− 2a+ 2b+ 2a) = 2(b− c)2(c− a)2(a− b)2.

Ex 3.4.19 Show that

∣∣∣∣∣∣∣∣x l m 1α x n 1α β x 1α β γ 1

∣∣∣∣∣∣∣∣ = (x− α)(x− β)(x− γ) for any value of l,m, n.

Solution: Let the value of the determinant be 4, then,

4 =

∣∣∣∣∣∣∣∣x l m 1

α− x x− l n−m 0α− x β − l x−m 0α− x β − l γ −m 0

∣∣∣∣∣∣∣∣ ;R2 −R1, R3 −R1, R4 −R1

178 Theory of Matrices

= −

∣∣∣∣∣∣α− x x− l n−mα− x β − l x−mα− x β − l γ −m

∣∣∣∣∣∣ = −(α− x)

∣∣∣∣∣∣1 x− l n−m1 β − l x−m1 β − l γ −m

∣∣∣∣∣∣= −(α− x)

∣∣∣∣∣∣1 x− l n−m0 β − x x− n0 β − x γ − n

∣∣∣∣∣∣ = −(α− x)∣∣∣∣β − x x− nβ − x γ − n

∣∣∣∣= −(α− x)(β − x)

∣∣∣∣1 x− n1 γ − n

∣∣∣∣ = (x− α)(x− β)(x− γ),

which is independent on the value of l,m, n and consequently, 4 = (x−α)(x− β)(x− γ) istrue for any values of l,m, n.

Ex 3.4.20 Prove that,

∣∣∣∣∣∣(b+ c)2 a2 a2

b2 (c+ a)2 b2

c2 c2 (a+ b)2

∣∣∣∣∣∣ = 2abc(a + b + c)3. [WBUT 2004,

2009]

Solution:∣∣∣∣∣∣(b+ c)2 a2 a2

b2 (c+ a)2 b2

c2 c2 (a+ b)2

∣∣∣∣∣∣=

∣∣∣∣∣∣(b+ c)2 a2 − (b+ c)2 a2 − (b+ c)2

b2 (c+ a)2 − b2 0c2 0 (a+ b)2 − c2

∣∣∣∣∣∣ [C ′2 = C2 − C1, C

′3 = C3 − C1]

= (a+ b+ c)2

∣∣∣∣∣∣(b+ c)2 a− (b+ c) a− (b+ c)b2 (c+ a)− b 0c2 0 (a+ b)− c

∣∣∣∣∣∣= (a+ b+ c)2

∣∣∣∣∣∣2bc −2c −2bb2 (c+ a)− b 0c2 0 (a+ b)− c

∣∣∣∣∣∣ [R′1 = R1 −R2 −R3]

= 2(a+ b+ c)2

∣∣∣∣∣∣bc −c −bb2 (c+ a)− b 0c2 0 (a+ b)− c

∣∣∣∣∣∣= 2(a+ b+ c)2

1bc

∣∣∣∣∣∣bc bc −bcb2 bc+ ab− b2 0c2 0 ac+ bc− c2

∣∣∣∣∣∣= 2(a+ b+ c)2

∣∣∣∣∣∣1 −1 1b2 bc+ ab− b2 0c2 0 ac+ bc− c2

∣∣∣∣∣∣[taking common bc from first row]

= 2(a+ b+ c)2

∣∣∣∣∣∣1 0 0b2 bc+ ab b2

c2 c2 ac+ bc

∣∣∣∣∣∣ [C ′2 = C2 + C1, C

′3 = C3 + C1]

= 2(a+ b+ c)2∣∣∣∣ b(a+ c) b2

c2 c(a+ b)

∣∣∣∣ = 2bc(a+ b+ c)2∣∣∣∣a+ c b

c a+ b

∣∣∣∣= 2bc(a+ b+ c)2(ac+ bc+ c2) = 2abc(a+ b+ c)3.

Ex 3.4.21 Show that

∣∣∣∣∣∣a2 + λ ab acab b2 + λ bcac bc c2 + λ

∣∣∣∣∣∣ is divisible by λ2 and find the other factor.

Determinants 179

Solution:

∣∣∣∣∣∣a2 + λ ab acab b2 + λ bcac bc c2 + λ

∣∣∣∣∣∣ = a b c

∣∣∣∣∣∣a+ λ

a b ca b+ λ

b ca b c+ λ

c

∣∣∣∣∣∣[dividing first, second and third rows by a, b, c respectively]

=

∣∣∣∣∣∣a2 + λ b2 c2

a2 b2 + λ c2

a2 b2 c2 + λ

∣∣∣∣∣∣[multiplying first, second and third columns by a, b, c respectively]

=

∣∣∣∣∣∣λ+ a2 + b2 + c2 b2 c2

λ+ a2 + b2 + c2 b2 + λ c2

λ+ a2 + b2 + c2 b2 c2 + λ

∣∣∣∣∣∣ [C ′1 = C1 + C2 + C3]

= (λ+ a2 + b2 + c2)

∣∣∣∣∣∣1 b2 c2

1 b2 + λ c2

1 b2 c2 + λ

∣∣∣∣∣∣ [λ+ a2 + b2 + c2 is taking out]

= (λ+ a2 + b2 + c2)

∣∣∣∣∣∣1 b2 c2

0 λ 00 0 0λ

∣∣∣∣∣∣ [R′2 = R2 −R1, R′3 = R3 −R1]

= (λ+ a2 + b2 + c2) λ2.

Hence λ2 and λ+ a2 + b2 + c2 are only two factors of the given determinant.

Ex 3.4.22 If4∑

i=1

a2i =

4∑i=1

b2i =4∑

i=1

c2i =4∑

i=1

d2i = 1 and

4∑i=1

ai bi =4∑

i=1

bi ci =4∑

i=1

ci di =

4∑i=1

di ai =4∑

i=1

ai ci =4∑

i=1

bi di = 0 then show that∣∣∣∣∣∣∣∣a1 b1 c1 d1

a2 b2 c2 d2

a3 b3 c3 d3

a4 b4 c4 d4

∣∣∣∣∣∣∣∣ = ±1.

Solution: Let the given determinant be ∆. Then

∆2 =

∣∣∣∣∣∣∣∣a1 b1 c1 d1

a2 b2 c2 d2

a3 b3 c3 d3

a4 b4 c4 d4

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣a1 b1 c1 d1

a2 b2 c2 d2

a3 b3 c3 d3

a4 b4 c4 d4

∣∣∣∣∣∣∣∣Using multiplication by column rule, we get,

∆2 =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

4∑i=1

a2i

4∑i=1

ai bi4∑

i=1

ai ci4∑

i=1

ai di

4∑i=1

bi ai

4∑i=1

b2i4∑

i=1

bi ci4∑

i=1

bi di

4∑i=1

ci ai

4∑i=1

ci bi4∑

i=1

c2i4∑

i=1

ci di

4∑i=1

di ai

4∑i=1

di bi4∑

i=1

di ci4∑

i=1

d2i

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣1 0 0 00 1 0 00 0 1 00 0 0 1

∣∣∣∣∣∣∣∣= 1 ⇒ ∆ = ± 1.

Ex 3.4.23 Prove that∣∣∣∣∣∣∣∣a+ 1 a a aa a+ 2 a aa a a+ 3 aa a a a+ 4

∣∣∣∣∣∣∣∣ = 24(1 + a

1 + a2 + a

3 + a4

)[WBUT 2004]

180 Theory of Matrices

Solution:∣∣∣∣∣∣∣∣a+ 1 a a aa a+ 2 a aa a a+ 3 aa a a a+ 4

∣∣∣∣∣∣∣∣ = (1.2.3.4)

∣∣∣∣∣∣∣∣1 + a

1a2

a3

a4

a1 1 + a

2a3

a4

a1

a2 1 + a

3a4

a1

a2

a3 1 + a

4

∣∣∣∣∣∣∣∣[dividing first, second, third and fourth columns by 1, 2, 3 and 4 respectively]

= 24

∣∣∣∣∣∣∣∣1 + a

1 + a2 + a

3 + a4

a2

a3

a4

1 + a1 + a

2 + a3 + a

4 1 + a2

a3

a4

1 + a1 + a

2 + a3 + a

4a2 1 + a

3a4

1 + a1 + a

2 + a3 + a

4a2

a3 1 + a

4

∣∣∣∣∣∣∣∣ [C′1 = C1 + C2 + C3 + C4]

= 24(1 +

a

1+a

2+a

3+a

4

) ∣∣∣∣∣∣∣∣1 a

2a3

a4

1 1 + a2

a3

a4

1 a2 1 + a

3a4

1 a2

a3 1 + a

4

∣∣∣∣∣∣∣∣[taking common 1 +

a

1+a

2+a

3+a

4from first column

]

= 24(1 +

a

1+a

2+a

3+a

4

) ∣∣∣∣∣∣∣∣1 a

2a3

a4

0 1 0 00 0 1 00 0 0 1

∣∣∣∣∣∣∣∣[R′2 = R2 −R1, R

′3 = R3 −R1, R

′4 = R4 −R1]

= 24(1 +

a

1+a

2+a

3+a

4

).1 = 24

(1 +

a

1+a

2+a

3+a

4

).

Ex 3.4.24 If u = ax4 + 4bx3 + 6cx2 + 4dx + e, u11 = ax2 + 2bx + c, u12 = bx2 + 2cx + d,u22 = cx2 + 2dx+ e then prove that∣∣∣∣∣∣∣∣

a b c u11

b c d u12

c d e u22

u11 u12 u22 0

∣∣∣∣∣∣∣∣ = −u

∣∣∣∣∣∣a b cb c dc d e

∣∣∣∣∣∣ .Solution:

∣∣∣∣∣∣∣∣a b c u11

b c d u12

c d e u22

u11 u12 u22 0

∣∣∣∣∣∣∣∣ =1x2

∣∣∣∣∣∣∣∣ax2 bx2 cx2 u11x

2

b c d u12

c d e u22

u11 u12 u22 0

∣∣∣∣∣∣∣∣=

1x2

∣∣∣∣∣∣∣∣ax2 + 2bx+ c bx2 + 2cx+ d cx2 + 2dx+ e u11x

2 + 2xu12 + u12

b c d u12

c d e u22

u11 u12 u22 0

∣∣∣∣∣∣∣∣[ using R′1 = R1 + 2R2x+R3]

=1x2

∣∣∣∣∣∣∣∣u11 u12 u22 ub c d u12

c d e u22

u11 u12 u22 0

∣∣∣∣∣∣∣∣ =1x2

∣∣∣∣∣∣∣∣0 0 0 ub c d u12

c d e u22

u11 u12 u22 0

∣∣∣∣∣∣∣∣ [R′1 = R1 −R4]

= − u

x2

∣∣∣∣∣∣b c dc d eu11 u12 u22

∣∣∣∣∣∣= − u

x2

∣∣∣∣∣∣b c dc d e

u11 − 2xb− c u12 − 2cx− d u22 − 2xd− e

∣∣∣∣∣∣ [R′3 = R3 − 2xR1 −R2]

Determinants 181

= − u

x2

∣∣∣∣∣∣b c dc d eax2 bx2 cx2

∣∣∣∣∣∣ = −u

∣∣∣∣∣∣b c dc d ea b c

∣∣∣∣∣∣ = −u

∣∣∣∣∣∣a b cb c dc d e

∣∣∣∣∣∣by interchanging first and third rows and then first and second rows.

3.4.2 Minors and Co-factors

Let us consider an n× n matrix A = [aij ]. Consider Mij , the (n− 1)× (n− 1) sub-matrixof A, which obtained by deleting the ith row and jth column of A. Now, |Mij | is called theminor of aij. The co-factor Aij of the element aij is defined as

Aij = (−1)i+j |Mij |. (3.16)

For the matrix A =

−1 2 43 −6 −5−2 0 1

, |M22|, A22 and |M31|, A31 are given by,

|M22| =∣∣∣∣−1 4−2 1

∣∣∣∣ = −1 + 8 = 7; A22 = (−1)2+2|M22| = 7.

|M31| =∣∣∣∣ 2 4−6 −5

∣∣∣∣ = −10 + 24 = 14; A31 = (−1)3+1|M31| = 14,

which are respectively the minors and co-factors. It is obvious that if (i+ j) be even, thenminor and co-factor of aij are same. Since each term in the expansion of a determinantcontains one element from any particular row (or column), we can express the expression asa linear function of that row (or column), for

|A| =

∣∣∣∣∣∣a11 a12 a13

a21 a22 a23

a31 a32 a33

∣∣∣∣∣∣ = a11

∣∣∣∣a22 a23

a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23

a31 a33

∣∣∣∣+ a13

∣∣∣∣a21 a22

a31 a32

∣∣∣∣= a11A11 + a12A12 + a13A13,

where A11, A12, A13 are the co-factors of a11, a12 and a13 respectively.Complementary minor, algebraic compliment, principal minor: Let M be aminor of order m in |A| = |aij |n×n. Now, if we delete all the rows and columns formingM , the minor N formed by the remaining rows and columns of order (n−m) is called thecomplementary minor of M . For the determinant

4 =

∣∣∣∣∣∣∣∣−1 2 4 63 −6 −5 0−2 0 1 92 7 5 4

∣∣∣∣∣∣∣∣ ;∣∣∣∣−1 2

3 −6

∣∣∣∣ and∣∣∣∣1 95 4

∣∣∣∣are complementary. Let M be a minor of order r in which rows i1, i2, · · · , ir and columnsj1, j2, · · · , jr are present. Now, the algebraic complement of M is

(−1)i1+i2+···+ir+j1+j2+···+jr × M ′; (3.17)

where M ′ is the algebraic complement of M . Algebraic complement of an element aij in|aij | is the co-factor of aij in |aij |. In the above example, 4 algebraic complement of∣∣∣∣−1 2

3 −6

∣∣∣∣ is (−1)1+2+1+2

∣∣∣∣1 95 4

∣∣∣∣ = ∣∣∣∣1 95 4

∣∣∣∣ .

182 Theory of Matrices

If the row and column indices in a minor are the same or equivalently, then the minor is said

to be principal minor. In the above example, the principal minor of 4 is∣∣∣∣−6 0

7 4

∣∣∣∣ . If we take

the diagonal elements of minor from the diagonal elements of the matrix, then the indicesare equivalent. Since in a principal minor, sum of row and identical column subscripts arealways even, so the sign of minor is always positive. In this case, algebraic complement ofprincipal minor is equal to its complement.

Ex 3.4.25 If ∆ =

∣∣∣∣∣∣h a 01h

1b

1f

0 c f

∣∣∣∣∣∣ and ∆′ =

∣∣∣∣∣∣1bc − 1

ch1

fh

af −fh ch1

fh −1

af1ab −

1h2

∣∣∣∣∣∣, find ∆′

∆2 in its simplest form.

Solution: Hereadj ∆ =

∣∣∣∣∣∣fb −

cf −

fh

ch

−af fh −chaf −h

fhb −

ah

∣∣∣∣∣∣ = −cafh

∣∣∣∣∣∣1bc −

1f2 − 1

ch1

fh

af −fh ch1

fh − 1af

1ab −

1h2

∣∣∣∣∣∣taking out fc, −1, ah from 1st, 2nd and 3rd row respectively. Hence

∆2 = −cafh∆′, or,∆′

∆2= − 1

cafh.

Theorem 3.4.1 Laplace theorem : In an n order square matrix A, |A| can be expressedas the aggregate of the products of all minors of order r formed from any r selected rows ofA and corresponding algebraic complement of them.

According to this theorem, we expand |A| = |aij |4×4. Let the first two rows be selected. Soif we expand the determinant in terms of minors of ordered 2, we get,

|A| =∣∣∣∣a11 a12

a21 a22

∣∣∣∣ (−1)1+2+1+2

∣∣∣∣a33 a34

a43 a44

∣∣∣∣+ ∣∣∣∣a11 a13

a21 a23

∣∣∣∣ (−1)1+2+1+3

∣∣∣∣a32 a34

a42 a44

∣∣∣∣+∣∣∣∣a11 a14

a21 a24

∣∣∣∣ (−1)1+2+1+4

∣∣∣∣a32 a33

a42 a43

∣∣∣∣+ ∣∣∣∣a12 a13

a22 a23

∣∣∣∣ (−1)1+2+2+3

∣∣∣∣a31 a34

a41 a44

∣∣∣∣+∣∣∣∣a12 a14

a22 a24

∣∣∣∣ (−1)1+2+2+4

∣∣∣∣a31 a33

a41 a43

∣∣∣∣+ ∣∣∣∣a13 a14

a23 a24

∣∣∣∣ (−1)1+2+3+4

∣∣∣∣a31 a32

a41 a42

∣∣∣∣ .Ex 3.4.26 Using Laplace’s theorem, show that,

|A| =

∣∣∣∣∣∣∣∣a b c d−b a d −c−c −d a b−d c −b a

∣∣∣∣∣∣∣∣ = (a2 + b2 + c2 + d2)2.

Solution: Using Laplace’s theorem, we get,

|A| = (−1)1+2+1+2

∣∣∣∣ a b−b a

∣∣∣∣ ∣∣∣∣ a b−b a

∣∣∣∣+ (−1)1+2+1+3

∣∣∣∣ a c−b d

∣∣∣∣ ∣∣∣∣−d bc a

∣∣∣∣+(−1)1+2+1+4

∣∣∣∣ a d−b −c

∣∣∣∣ ∣∣∣∣−d ac −b

∣∣∣∣+ (−1)1+2+2+3

∣∣∣∣ b ca d∣∣∣∣ ∣∣∣∣−c b−d a

∣∣∣∣+(−1)1+2+2+4

∣∣∣∣ b da −c

∣∣∣∣ ∣∣∣∣−c a−d −b

∣∣∣∣+ (−1)1+2+3+4

∣∣∣∣ c dd −c

∣∣∣∣ ∣∣∣∣−c −d−d c

∣∣∣∣= (a2 + b2)(a2 + b2) + (ad+ bc)(ad+ bc) + (ac− bd)(ac− bd)

+(ac− bd)(ac− bd) + (ad+ bc)(ad+ bc) + (c2 + d2)(c2 + d2)= (a2 + b2)2 + 2(a2d2 + b2c2 − 2abcd+ a2c2 + b2d2 − 2abcd) + (c2 + d2)2

= (a2 + b2)2 + 2(a2 + b2)(c2 + d2) + (c2 + d2)2 = (a2 + b2 + c2 + d2)2.

Determinants 183

From this we conclude that, if a, b, c and d are real numbers, then the given determinant isnon-singular if and only if at least one of a, b, c, d is non-zero.

3.4.3 Adjoint and Reciprocal of Determinant

Let A = [aij ] be a square matrix of order n and Aij be the cofactor of aij in |A|. Now, |Aij |is called the adjoint or adjugate of |A|. Similarly, reciprocal or inverse is defined by

|A|′ =1|A|

|Aij |; where |A| 6= 0. (3.18)

Theorem 3.4.2 Jacobi’s theorem : Let A = [aij ] be a non-singular matrix of order nand Aij be the cofactor of aij in |A|, then |Aij | = |A|n−1.

Proof: Let A = [aij ]n×n be a square matrix of order n. Let Aij denotes the cofactor ofijth element of aij in detA. Now, |A|.|Aij |

=

∣∣∣∣∣∣∣∣∣a11 a12 · · · a1n

a21 a22 · · · a2n

......

...an1 an2 · · · ann

∣∣∣∣∣∣∣∣∣

∣∣∣∣∣∣∣∣∣A11 A21 · · · An1

A12 A22 · · · An2

......

...A1n A2n · · · Ann

∣∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

n∑k=1

a1kA1k

n∑k=1

a1kA2k · · ·n∑

k=1

a1kAnk

n∑k=1

a2kA1k

n∑k=1

a2kA2k · · ·n∑

k=1

a2kAnk

......

...n∑

k=1

ankA1k

n∑k=1

ankA2k · · ·n∑

k=1

ankAnk

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣∣|A| 0 · · · 00 |A| · · · 0...

......

0 0 · · · |A|

∣∣∣∣∣∣∣∣∣ ; asn∑

k=1

aikAjk = |A|, if i = j and = 0, if i 6= j

= |A|n ⇒ |Aij | = |A|n−1, as |A| 6= 0.

Ex 3.4.27 If the adjugate of 4 =

∣∣∣∣∣∣a h gh b fg f c

∣∣∣∣∣∣ be 4′ =

∣∣∣∣∣∣A H GH B FG F C

∣∣∣∣∣∣, prove that BC−F 2

a =

CA−G2

b = AB−H2

c = 4 andGH −AF

F=HF −BG

G=FG− CH

H= 4.

Solution: Let us consider the following product∣∣∣∣∣∣1 0 0H B FG F C

∣∣∣∣∣∣×∣∣∣∣∣∣a h gh b fg f c

∣∣∣∣∣∣ =

∣∣∣∣∣∣1.a+ 0.h+ 0.g 1.h+ 0.b+ 0.f 1.g + 0.f + 0.cH.a+B.h+ F.g H.h+B.b+ F.f H.g +B.f + F.cG.a+ F.h+ C.g G.h+ F.b+ C.f G.g + F.f + C.c

∣∣∣∣∣∣=

∣∣∣∣∣∣a h g0 4 00 0 4

∣∣∣∣∣∣ = a42

⇒ (BC − F 2).4 = a42, i.e.,BC − F 2

a= 4.

Similarly, by considering the products∣∣∣∣∣∣A H G0 1 0G F C

∣∣∣∣∣∣×∣∣∣∣∣∣a h gh b fg f c

∣∣∣∣∣∣ and

∣∣∣∣∣∣A H GH B F0 0 1

∣∣∣∣∣∣×∣∣∣∣∣∣a h gh b fg f c

∣∣∣∣∣∣

184 Theory of Matrices

we get,BC − F 2

a=CA−G2

b=AB −H2

c= 4 respectively. Now we consider the product∣∣∣∣∣∣

A H G0 0 1G F C

∣∣∣∣∣∣×∣∣∣∣∣∣a h gh b fg f c

∣∣∣∣∣∣ =∣∣∣∣∣∣4 0 0g f c0 0 4

∣∣∣∣∣∣⇒ (HG−AF )4 = f42 ⇒ HF −AF

f= 4.

Similarly, by considering the products∣∣∣∣∣∣A H GH B F1 0 0

∣∣∣∣∣∣×∣∣∣∣∣∣a h gh b fg f c

∣∣∣∣∣∣ and

∣∣∣∣∣∣0 1 0H B FG F C

∣∣∣∣∣∣×∣∣∣∣∣∣a h gh b fg f c

∣∣∣∣∣∣we get, HF−BG

G = FG−CHH = 4 respectively.

Ex 3.4.28 Prove that

∣∣∣∣∣∣bc− a2 ca− b2 ab− c2

ca− b2 ab− c2 bc− a2

ab− c2 bc− a2 ca− b2

∣∣∣∣∣∣ = (a3 + b3 + c3 − 3abc)2.

Solution: Let us consider the determinant ∆ =

∣∣∣∣∣∣a b cb c ac a b

∣∣∣∣∣∣ and its value is −(a3+b3+c3−3abc)

obtained by direct expansion. Now, adjoined of ∆ is ∆′ =

∣∣∣∣∣∣A B CB C AC A B

∣∣∣∣∣∣ where A,B,C are

cofactor of a, b, c in ∆. Therefore, A = bc − a2, B = ac − b2, C = ab − c2. By Jacobi’stheorem, ∆′ = ∆3−1 = ∆2,

or,

∣∣∣∣∣∣bc− a2 ca− b2 ab− c2

ca− b2 ab− c2 bc− a2

ab− c2 bc− a2 ca− b2

∣∣∣∣∣∣ = (a3 + b3 + c3 − 3abc)2.

3.4.4 Symmetric and Skew-symmetric Determinants

If A be a symmetric matrix of order n, then |A| is said to be symmetric determinant of ordern. If A be a skew-symmetric matrix of order n, Then |A| is said to be skew-symmetric deter-

minant of order n.

∣∣∣∣∣∣a h gh b fg f c

∣∣∣∣∣∣ and

∣∣∣∣∣∣0 a b−a 0 c−b −c 0

∣∣∣∣∣∣ are examples of symmetric and skew-symmetric

determinants respectively. Now,

(i) The adjoint of a symmetric determinant is symmetric.

(ii) The square of any determinant is symmetric.

(iii) In a skew-symmetric determinant |A| = |aij |n×n, Aij = (−1)n−1Aji, where Aij andAji are the cofactors of aij and aji in |A| respectively.

(iv) The adjoint of a skew-symmetric determinant of order n is symmetric or skew-symmetricaccording as n is even or odd.

Theorem 3.4.3 The value of every skew-symmetric determinant of odd order is zero.

Determinants 185

Proof: Let A be a skew-symmetric matrix of order n, where n is odd number. Hence bydefinition, AT = −A. Therefore,

|AT | = | −A| ⇒ |A| = (−1)n|A| = −|A|⇒ 2|A| = 0 ⇒ |A| = 0.

Therefore, the value of every skew-symmetric determinant of odd order is zero.

Theorem 3.4.4 The value of every skew-symmetric determinant of even order is a perfectsquare.

Proof: Let A = [aij ] be a skew-symmetric determinant of n, then by definition aij = −aji

and aii = 0. Let,

An =

0 a12 · · · a1n

a21 0 · · · a2n

......

...an1 an2 · · · 0

.

Now let, Aij be the cofactor of aij in |aij |. Then, Aij be a determinant of order (n − 1).Now, if we transform Aij in Aji, then every rows of it must be multiplied by (−1), therefore,Aij = (−1)n−1Aji. Thus,

adj|An| =

∣∣∣∣∣∣∣∣∣0 A12 · · · A1n

−A12 0 · · · A2n

......

...−A1n −A2n · · · 0

∣∣∣∣∣∣∣∣∣ .By Jacobi’s theorem, we have,∣∣∣∣ 0 A12

−A12 0

∣∣∣∣ = |An|

∣∣∣∣∣∣∣∣∣0 a34 · · · a3n

−a34 0 · · · a4n

......

...−a3n −a4n · · · 0

∣∣∣∣∣∣∣∣∣ = |An||An−2|.

Therefore, |An||An−2| = A212, which is a perfect square. For n > 2, it is true. When n = 0,

the result is obvious. Now,|A2| =

∣∣∣∣ 0 a12

−a12 0

∣∣∣∣ = a212,

which is a perfect square. If |A2| is perfect square, then |A4| is also a perfect square, byusing the relation, |An||An−2| = A2

12. Let the result be true for n = m, then it is true form + 2 as |Am+2||Am| is a perfect square. Also, it is true for n = 2, 4. Hence by method ofinduction the result is true for any even positive integer n.

Ex 3.4.29 Show that the value of 4 =

∣∣∣∣∣∣∣∣0 a b c−a 0 d e−b −d 0 f−c −e −f 0

∣∣∣∣∣∣∣∣ is a perfect square.

Solution: Expanding 4 by the minor of the elements of the first column, we get,

4 = a

∣∣∣∣∣∣a b c−d 0 f−e −f 0

∣∣∣∣∣∣− b

∣∣∣∣∣∣a b c0 d e−e −f 0

∣∣∣∣∣∣+ c

∣∣∣∣∣∣a b c0 d e−d 0 f

∣∣∣∣∣∣= af(af − be+ cd)− be(af − be+ cd) + cd(af − be+ cd)= (af − be+ cd)(af − be+ cd) = (af − be+ cd)2.

Since the given determinant 4 is a skew-symmetric determinant of even order, it is verifiedalso that, its value must be a perfect square.

186 Theory of Matrices

3.4.5 Vander-Monde’s Determinant

The Vander-Monde’s determinant is defined by∣∣∣∣∣∣x2

0 x21 x

22

x0 x1 x2

1 1 1

∣∣∣∣∣∣ = (x0 − x1)(x0 − x2)(x1 − x2) =2∏

i<j=0

(xi − xj).

The difference product of 3 numbers x0, x1, x2 is defined as

D.P.(x0, x1, x2) = (x0 − x1)(x0 − x2)(x1 − x2) =2∏

i<j=0

(xi − xj).

Similarly the other DP are defined by

D.P.(x0, x1, x2, x3) = (x0 − x1)(x0 − x2)(x0 − x3)(x1 − x2)(x1 − x3)(x2 − x3)

=3∏

i<j=0

(xi − xj)

D.P.(x0, x1, ..., xn) =n∏

i<j=0

(xi − xj).

Therefore, the Vander-Monde’s determinant can be written in D.P. form as∣∣∣∣∣∣x2

0 x21 x

22

x0 x1 x2

1 1 1

∣∣∣∣∣∣ =2∏

i<j=0

(xi − xj) = D.P.(x0, x1, x2).

Hence the Vander-Monde’s determinant can be written in DP form. In general,∣∣∣∣∣∣∣∣x3

0 x31 · · · xn

n

· · · · · · · · · · · ·x0 x1 · · · xn

1 1 · · · 1

∣∣∣∣∣∣∣∣ =n∏

i<j=0

(xi − xj) = D.P.(x0, x1, x2, ..., xn).

3.4.6 Cramer’s Rule

Let us consider a system of n linear algebraic equations relating in n unknowns x1, x2, . . . , xn

of the explicit forma11x1 + a12x2 + . . .+ a1nxn = b1a21x1 + a22x2 + . . .+ a2nxn = b2

......

an1x1 + an2x2 + . . .+ annxn = bn

(3.19)

where the n2 coefficients aij and the n constants b1, b2, ..., bn are given real numbers. The(3.19) can be written in the matrix notation as Ax = b where the real n × n coefficientmatrix is A in which aij is the coefficient of xj in the ith equation, bT = [b1, b2, ..., bn] isa column n vector which are prescribed and xT = [x1, x2, ..., xn] is the unknown n columnvector to be computed. Equations (3.19) are said to be homogeneous system if bi = 0;∀iand non homogeneous system if at least one bi 6= 0.

This is simplest method for the solution of a nonhomogeneous system (3.19) of n linearequations in n unknowns. Here it will assume that ∆ = det(A) 6= 0, so that unique solutionfor x1, x2, ..., xn exists. From the properties of determinant, we have,

Determinants 187

x14 = x1

∣∣∣∣∣∣∣∣a11 a12 · · · a1n

a21 a22 · · · a2n

· · · · · · · · · · · ·an1 an2 · · · ann

∣∣∣∣∣∣∣∣ =∣∣∣∣∣∣∣∣x1a11 a12 · · · a1n

x1a21 a22 · · · a2n

· · · · · · · · · · · ·x1an1 an2 · · · ann

∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣a11x1 + a12x2 + · · ·+ a1nxn a12 · · · a1n

a21x1 + a22x2 + · · ·+ a2nxn a22 · · · a2n

· · · · · · · · · · · ·an1x1 + an2x2 + · · ·+ annxn an2 · · · ann

∣∣∣∣∣∣∣∣C ′

1 = C1 + x2C2 + · · ·+ xnCn.

=

∣∣∣∣∣∣∣∣b1 a12 · · · a1n

b2 a22 · · · a2n

· · · · · · · · · · · ·bn an2 · · · ann

∣∣∣∣∣∣∣∣ [Using (3.19)] = 41(say). (3.20)

Therefore, x1 = 41/4. In general, let Aij be the cofactor of aij in ∆, then multiplyingboth sides of the ith equation by Aij for i = 1, 2, . . . , n and then adding we have,

(n∑

i=1

ai1 Aij) x1 + (n∑

i=1

ai2 Aij) x2 + · · ·+ (n∑

i=1

aij Aij) xj + · · ·+ (n∑

i=1

ain Aij) xn

=n∑

i=1

bi Aij .

we multiply each equation by the cofactor in A of the coefficient of xk in that equationand add the results we have,

4.xk =n∑

j=1

Ajkbj ; k = 1, 2, ..., n

where, Aij is the cofactor of aij in det(A) of order n − 1. Hence from equation (3.19) wehave n∑

k=1

aikxk = bi; i = 1, 2, ..., n

⇒ 1∆

n∑k=1

n∑j=1

Ajkbj =1∆

n∑j=1

[n∑

k=1

aikAjk]bj .

Since the RHS would reduce to 4 if b1, b2, ..., bn were replaced by a1k, a2k, ..., ank, so let∆i = determinant obtained from ∆, replacing ith column by the column vector b then theunique solution of (3.19) is given by

xi =∆i

∆= 4−1

n∑j=1

Aijbj ; i = 1, 2, ..., n. (3.21)

This is the well-known Cramer’s rule. Various methods have been devised to evaluate thevalue of a numerical determinants. The following cases may arise:

1. The homogeneous system bi = 0,∀i of equations has a trivial solution x1 = x2 = · · · =xn = 0, if ∆ 6= 0 and an non trivial solution ( at least one of x’s is non-zero) existwhen ∆ = 0. When 4 = 0, the solution is not unique.

2. The non homogeneous system of equations is said to be consistent if 4 6= 0. In thiscase the system (3.19)has an unique solution.

3. If 4 = 0 and all of 4i’s be 0 then the system (3.19) may or may not be consistent. Ifconsistent it admits of infinite number of solutions.

188 Theory of Matrices

4. If4 = 0 and at least one of4i(i = 1, 2, ..., n) be nonzero then the system is inconsistentor incompatible and (3.19) has no solution.

Cramer’s rule have the following serious drawbacks

(i) If the size of (3.19) is large (n ≥ 4), Crammer’s rule requires enormous amount ofcomputation for evaluation of determinants and no efficient method exists for evalua-tion of determinants. Hence this is purely theoretical but certainly not from numericalpoint of view.

(ii) For large systems, this method is not useful as it involves a large number of manipula-tions. The Cramer’s rule involves to compute (n+1) determinants each of order n (fora system (3.19) of n equations and n variables). To evaluate a determinant of order nrequires (n!−1) additions and n!(n−1) multiplications. The total number of multipli-cations and divisions required to solve (3.19) by this method = (n− 1)× (n+ 1)! + n.

The scenario is shown in Figure 3.2.

?4 = 0 4 6= 0

[Consistent with unique solution]

?

?

A system of linear equations

? ?41 = 42 = · · · = 4n = 0

[Consistent withinfinitely many solutions]

At least one 4i’s non-zero[Inconsistent]

?

Figure 3.2: Different cases for existence of solutions.

Ex 3.4.30 Solve the following system of equations by Cramer’s rule

x+ 2y + 3z = 6, 2x+ 4y + z = 7, 3x+ 2y + 9z = 14

Solution: The given system of linear equations can be written in the form Ax = b, where,

A = the coefficient matrix =

1 2 32 4 13 2 9

, bT = (6 7 14) and xT = (x1 x2 x3). Here the

determinant of the coefficient matrix A is given by

4 =

∣∣∣∣∣∣1 2 32 4 13 2 9

∣∣∣∣∣∣ = 1(4.9− 1.2) + 2(1.3− 2.9) + 3(2.2− 4.3) = −20(6= 0).

Here the present system has unique solution. The determinant 4i obtained by replacingthe ith column of 4 by constant vector, given by

41 =

∣∣∣∣∣∣6 2 37 4 114 2 9

∣∣∣∣∣∣ = −20;42 =

∣∣∣∣∣∣1 6 32 7 13 14 9

∣∣∣∣∣∣ = −20;43 =

∣∣∣∣∣∣1 2 62 4 73 2 14

∣∣∣∣∣∣ = −20.

Complex Matrices 189

Hence by Cramer’s rule solution exists. So the unique solution is

x =−20−20

= 1; y =−20−20

= 1, z =−20−20

= 1.

Now the sum of the three equations is 6x+ 8y+ 13z = 27. Substituting the values of x, y, zwe get LHS = 27, which is the check solution.

Ex 3.4.31 Find for what values of a and b, the system of equations

x+ 4y + 2z = 1, 2x+ 7y + 5z = 2b, 4x+ ay + 10z = 2b+ 1

has (i) an unique solution, (ii) no solution and (iii) infinite number of solution over thefield of rational numbers.

Solution: The given system of equations can be written in the form Ax = b, where,

A =

1 4 22 7 54 a 10

, bT = (1 2b 2b+ 1) and xT = (x1 x2 x3).

Hence,

4 =

∣∣∣∣∣∣1 4 22 7 54 a 10

∣∣∣∣∣∣ = 1(7.10− 5.a) + 4(5.4− 10.2) + 2(2.a− 7.4) = 14− a

41 =

∣∣∣∣∣∣1 4 22b 7 5

2b+ 1 a 10

∣∣∣∣∣∣ = 7b− 5a− 68b+ 4ab.

(i) When, 14− a 6= 0, i.e., a 6= 14, then 4 6= 0, in this case, the solution will be unique.(ii) When, 14− a = 0, i.e., a = 14, then 4 = 0, and then 41 = 6 − 12b. If 6 − 12b 6= 0,

i.e., b 6= 12 , then the system has no solution.

(iii) If a = 14 and b = 12 , then the system of linear equations becomes x + 4y + 2z =

1, y − z = 0. Thus the system is consistent and have infinite number of solutions over thefield of rational numbers.

3.5 Complex Matrices

A matrix A is said to be complex, if its elements may be the complex numbers. If A bea complex matrix, then it can be expressed as A = P + iQ, i =

√−1, where P,Q are real

matrices. The matrix A =(−1 + 3i −i3 + i 4

)can be written as

A =(−1 + 3i −i3 + i 4

)=(−1 03 4

)+ i

(3 −11 0

)= P + iQ

so A is a complex matrix. If each element of the matrix A be replaced by its conjugate,then the matrix, so obtained is called the conjugate of A and is denoted by A. Thus, ifA = P + iQ, then A = P − iQ. Thus if A = [aij ], then A = [bij ], where bij = aij . Therefore,

for the above example, A =(−1− 3i i3− i 4

). A matrix A is real if and only if A = A.

Property 3.5.1 If A and B be conjugate of the matrices A and B respectively, then,

(i) A = A.

190 Theory of Matrices

(ii) A+B = A+B; A,B are conformable for addition.

(iii) kA = k A; k being a complex number.

(iv) AB = A B; A,B are conformable for product.

3.5.1 Transpose Conjugate of a Matrix

The transpose of the conjugate of a matrix A is called the transpose conjugate of A and isdenoted by A∗. Thus,

A∗ = (AT ) = (A)T . (3.22)

It is also called as tranjugate of a matrix. For example, let A =(

3 + 2i 3i 3 + 4i

), then,

A =(

3− 2i 3−i 3− 4i

), and so,

A∗ = (A)T =(

3− 2i −i3 3− 4i

)= (AT ).

As A∗ = (AT ) = (A)T , so A is a transpose conjugate matrix. Note that, if A is real, thenA∗ = AT .

Property 3.5.2 If A∗ and B∗ be tranjugates of the matrices A and B respectively, then,

(i) [A∗]∗ = A.

(ii) [A+B]∗ = A∗ +B∗; A,B are conformable for addition.

(iii) [kA]∗ = k A∗; k being a complex number.

(iv) [AB]∗ = B∗ A∗; A,B are conformable for product.

Consider a complex matrix A. The relationship between A and its conjugate transpose A∗

yields following important kinds of complex matrices.

3.5.2 Harmitian Matrix

A complex matrix of order n× n is said to be Harmitian, if, A∗ = A. From the definition,it is clear that, a square complex matrix A = [aij ] is Harmitian if and only if symmetricelements are conjugate, i.e., aij = aji, in which each diagonal element aii must be real. Forexample, consider the following complex matrix

A =

7 1− i 3 + 2i1 + i 2 1 + 2i3− 2i 1− 2i 1

By inspection, the diagonal elements of A are real and symmetric elements 1− i and 1 + i,3 + 2i and 3− 2i, 1 + 2i and 1− 2i are conjugate. Thus A is Harmitian matrix.

Ex 3.5.1 Write down the most general symmetric and hermitian matrix of order 2.

Solution: The most general symmetric complex matrix of order 2 is

As =(a+ ib e+ ife+ if c+ id

); i =

√−1

Complex Matrices 191

which has real independent parameters. The most general hermitian matrix of order 2 canbe written interms of four independent parameters in the form

Ah =(

a c+ idc+ id b

); i =

√−1.

Ex 3.5.2 If A be a square matrix, then show that AA∗ and A∗A are Hermitian.

Solution: Let A∗ and B∗ be the transposed conjugates of A and B respectively, then byproperty (AB)∗ = B∗A∗. Here we have,

[AA∗]∗ = [A∗]∗A∗ = AA∗.

Hence AA∗ is Hermitian. Similarly,[A∗A]∗ = A∗[A∗]∗ = A∗A.

Therefore, A∗A is Hermitian.

3.5.3 Skew-Harmitian Matrix

A complex matrix of order n × n is said to be skew-Harmitian, if, A∗ = (A)T . A squarecomplex matrix A = [aij ] is skew-Harmitian if and only if it is skew-symmetric and eachdiagonal element aii is either zero or purely imaginary. For example, the following complexmatrix

A =

0 2− i 6− 3i−2− i 0 1 + 5i−6− 3i −1 + 5i i

is skew-Harmitian matrix.

Ex 3.5.3 If S = P + iQ be a skew Hermitian matrix, then show that P is a skew symmetricmatrix and Q is real skew symmetric matrix.

Solution: Let S = P + iQ be a skew Hermitian matrix, where P and Q are real matrices.Now, S = P − iQ and (S)T = PT − iQT .

Since, S is skew Hermitian, by definition, (S)T = S, i.e.,

PT − iQT = P + iQ⇒ PT = P and QT = −Q.

Therefore, P is a skew symmetric matrix and Q is real skew symmetric matrix.

Ex 3.5.4 If A be a Hermitian matrix, then show that iA is a skew-Hermitian.

Solution: Since A be a Hermitian matrix, by definition, A∗ = A. Now,

[iA]∗ = i A∗ = −i A∗ = −(iA).

Therefore, iA is skew hermitian.

Ex 3.5.5 Let A be an n×n matrix whose elements are complex numbers. Show that A+A∗

is Harmitian and A−A∗ is skew Harmitian.

Solution: Let P = A+A∗, then using the property (3.5.1), we get,

P = A+A∗ = A+A∗ = A+AT .

Therefore, (P )T = (A)T +A = A+A∗ = P.

Hence Z is Harmitian. Let Q = A−A∗, then using the property (3.5.1), we get,

Q = A−A∗ = A−A∗ = A−AT .

Therefore, (Q)T = (A)T −A = A∗ −A = −Q.

Hence Q is skew Harmitian.

192 Theory of Matrices

3.5.4 Unitary Matrix

A complex square matrix A of order n is said to be unitary, if

A∗A−1 = A−1A∗ = In; i.e., A∗ = A−1. (3.23)

Thus A must be necessarily be square and inverible. We note that a complex matrix A isunitary if and only if its rows (columns) from an orthogonal set relative to the dot productof complex vectors. For example, let,

A =12

(1 + i 1− i1− i 1 + i

); then, A∗ =

12

(1− i 1 + i1 + i 1− i

)so, AA∗ =

12

(1 + i 1− i1− i 1 + i

)12

(1− i 1 + i1 + i 1− i

)=(

1 00 1

)= I2

=12

(1− i 1 + i1 + i 1− i

)12

(1 + i 1− i1− i 1 + i

)= A∗A.

Since A∗A−1 = A−1A∗ = I2, so A is an unitary matrix. Note that, when a matrix A is real,hermitian is same as symmetric and unitary is the same as orthogonal.

Theorem 3.5.1 For an unitary matrix A, |A| = 1.

Proof: Let A be an unitary matrix of order n, then by definition, A∗A = In. Therefore,

|A∗A| = |In| ⇒ |AT ||A| = 1⇒ |A||A| = 1 ⇒ |A||A| = 1⇒ |A|2 = 1 ⇒ |A| = 1.

Therefore, for an unitary matrix A, |A| = 1.

Ex 3.5.6 If A be an unitary matrix and I +A is non-singular

Solution:

3.5.5 Normal Matrix

A complex square matrix A is said to be normal if it commutes with A∗, i.e., if AA∗ = A∗A.For example, let,

A =(

2 + 3i 1i 1 + 2i

), then, A∗ =

(2− 3i −i

1 1− 2i

)AA∗ =

(2 + 3i 1i 1 + 2i

)(2− 3i −i

1 1− 2i

)=(

14 4− 4i4 + 4i 6

)=(

2− 3i −i1 1− 2i

)(2 + 3i 1i 1 + 2i

)= A∗A.

Since AA∗ = A∗A, the complex matrix A is normal.This definition reduces to that for realmatrices when A is real.

3.6 Adjoint of a Matrix

Let A = [aij ]n×n be a square matrix of order n. Let Aij be the cofactor of the ijth elementaij in detA. Then the square matrix (Aij)T is said to be the adjugate or adjoint of A and

Adjoint of a Matrix 193

is denoted by adjA. So, we first find the cofactor of ijth element in detA. Then, the adjoint

of A is obtained by transposing the cofactor. For example, let , A =[

1 −23 4

], then,

A11 = (−1)1+1 |4| = 4, A12 = (−1)1+2 |3| = −3,A21 = (−1)2+1 | − 2| = 2, A22 = (−1)2+2 |1| = 1.

Therefore, adj A =(A11 A12

A21 A22

)T

=(

4 2−3 1

).

For the matrix A =

1 0 2−1 1 0

2 0 1

, we get,

A11 = (−1)1+1

∣∣∣∣1 00 1

∣∣∣∣ = 1, A12 = (−1)1+2

∣∣∣∣−1 02 1

∣∣∣∣ = 1,

A13 = (−1)1+3

∣∣∣∣−1 12 0

∣∣∣∣ = −2, A21 = (−1)2+1

∣∣∣∣0 20 1

∣∣∣∣ = 0,

A22 = (−1)2+2

∣∣∣∣1 22 1

∣∣∣∣ = −3, A23 = (−1)2+3

∣∣∣∣1 02 0

∣∣∣∣ = 0,

A31 = (−1)3+1

∣∣∣∣0 21 0

∣∣∣∣ = −2, A32 = (−1)3+2

∣∣∣∣ 1 2−1 0

∣∣∣∣ = −2,

A33 = (−1)3+3

∣∣∣∣ 1 0−1 1

∣∣∣∣ = 1.

Therefore, the adjugate of A is given by

adj A =

A11 A12 A13

A21 A22 A23

A31 A32 A33

T

=

1 0 −21 −3 −2

−2 0 1

.From definition, we have,

(i) adj(AT ) = (adjA)T , and adj(kA) = kn−1adjA, where k is any scalar.

(ii) If 0 be a zero matrix, which is a square matrix of order n, then adj0 = 0.

(iii) IF I be a unit matrix of order n, then adjI = I.

(iv) If A is symmetric, then adjA is symmetric.

(v) If A is skew-symmetric then adjA is symmetric or skew-symmetric according as theorder of A is odd or even.

(vi) For the matrices A,B, adj (AB) = (adj B) (adj A).

Theorem 3.6.1 If A be a square matrix of order n, then

A.(adjA) = |A| In = (adjA).A. (3.24)

Proof: Let A = [aij ]n×n be a square matrix of order n. Let Aij denotes the cofactor ofijth element of aij in detA. The ijth element of A(adjA) is the inner product of the ith rowof A and the jth column of adjA, as

A(adjA) =

a11 a12 · · · a1n

a21 a22 · · · a2n

......

...an1 an2 · · · ann

A11 A21 · · · An1

A12 A22 · · · An2

......

...A1n A2n · · · Ann

194 Theory of Matrices

=

n∑k=1

a1kA1k

n∑k=1

a1kA2k · · ·n∑

k=1

a1kAnk

n∑k=1

a2kA1k

n∑k=1

a2kA2k · · ·n∑

k=1

a2kAnk

......

...n∑

k=1

ankA1k

n∑k=1

ankA2k · · ·n∑

k=1

ankAnk

=

|A| 0 · · · 00 |A| · · · 0...

......

0 0 · · · |A|

= |A|In, as,n∑

k=1

aikAjk =|A|; if i = j0; if i = j

Similarly, taking the product between adjA andA and proceeding as before we get, (adjA)A =|A|In. Therefore,

A(adjA) = |A|In = (adjA)A.Thus the product of a matrix with its adjoint is commutative and it is a scalar matrix whosediagonal element is |A|.

Theorem 3.6.2 If A be a non-singular matrix of order n, then |adjA| = |A|n−1.

Proof: We know that, A(adjA) = |A|In. Hence,

|A(adjA)| = ||A|In| ⇒ |A||adjA| = |A|n

⇒ |adjA| = |A|n−1; as |A| 6= 0.

Therefore, if A be a non-singular square matrix of order n, then |adjA| = |A|n−1.

Ex 3.6.1 Show that,

∣∣∣∣∣∣bc− a2 ca− b2 ab− c2

ca− b2 ab− c2 bc− a2

ab− c2 bc− a2 ca− b2

∣∣∣∣∣∣ =∣∣∣∣∣∣a b cb c ac a b

∣∣∣∣∣∣2

. KH:09

Solution: If the right hand side is |A|2, the the adjoint of A is given by,

adjA =

bc− a2 ca− b2 ab− c2

ca− b2 ab− c2 bc− a2

ab− c2 bc− a2 ca− b2

.

Let A be a non-singular matrix of order 3, then by theorem (3.6.2), we get,

|adjA| =

∣∣∣∣∣∣bc− a2 ca− b2 ab− c2

ca− b2 ab− c2 bc− a2

ab− c2 bc− a2 ca− b2

∣∣∣∣∣∣ = |A|2 =

∣∣∣∣∣∣a b cb c ac a b

∣∣∣∣∣∣2

.

Theorem 3.6.3 If A be a non-singular matrix of order n, then adj(adjA) = |A|n−2A.

Proof: We know that, A(adjA) = |A|In. Now putting adjA in place of A, we get,

adjA(adj.adjA) = |adjA|Inor, adjA(adj.adjA) = |A|n−1In; as |adjA| = |A|n−1

or, A(adjA)(adj.adjA) = |A|n−1A

or, |A|In(adj.adjA) = |A|n−1A

or, adj.adjA = |A|n−2.A; as |A| 6= 0.

Therefore, if A be a non-singular matrix of order n, then adj(adjA) = |A|n−2A.

Adjoint of a Matrix 195

Ex 3.6.2 Find the matrix A, if adjA =

1 3 −4−2 2 −21 −3 4

.

Solution: Since the adjA is given, so,

adj(adjA) =

∣∣∣∣ 2 −2−3 4

∣∣∣∣ −∣∣∣∣ 3 −4−3 4

∣∣∣∣ ∣∣∣∣3 −42 −2

∣∣∣∣−∣∣∣∣−2 −2

1 4

∣∣∣∣ ∣∣∣∣1 −41 4

∣∣∣∣ −∣∣∣∣ 1 −4−2 −2

∣∣∣∣∣∣∣∣−2 21 −3

∣∣∣∣ −∣∣∣∣1 31 −3

∣∣∣∣ ∣∣∣∣ 1 3−2 2

∣∣∣∣

=

2 0 26 8 104 6 8

.

Now, from the relation, |adjA| = |A|n−1 we have,

|adjA| =

∣∣∣∣∣∣1 3 −4−2 2 −21 −3 4

∣∣∣∣∣∣ = 4 = |A|2 ⇒ |A| = ±2.

Using the relation, adj(adjA) = |A|n−2A, the matrix A is given by,

A =12

2 0 26 8 104 6 8

=

1 0 13 4 52 3 4

.

3.6.1 Reciprocal of a Matrix

For a non-singular square matrix A = [aij ]n×n of order n, the reciprocal matrix of A isdefined by

1|A|

adjA =

1|A|A11

1|A|A21 · · · 1

|A|An11|A|A12

1|A|A22 · · · 1

|A|An2

......

...1|A|A1n

1|A|A2n · · · 1

|A|Ann

. (3.25)

For example, let, A =

6 2 54 2 10 1 −3

, then |A| = 2 and adjA =

−7 11 −812 −18 144 −6 4

. Therefore,

the reciprocal of A is 1|A|

adjA =12

−7 11 −812 −18 144 −6 4

.

3.6.2 Inverse of a Matrix

Let A be a square matrix of order n. For any other square matrix of the same order B, ifA.B = B.A = In, (3.26)

is satisfied then, B is called the reciprocal or inverse of A and is denoted as B = A−1. SinceA and B are conformable from the product AB and BA and AB = BA, A and B are squarematrices of the same order. Thus a matrix A may have an inverse or A may be invertibleonly when it is a square matrix. By the property of adjoint matrix, we have,