Maths

4,475 views

Published on

Published in: Education
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
4,475
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
Downloads
125
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

Maths

  1. 1. E-528-529, sector-7, Dwarka, New delhi-110075 (Nr. Ramphal chowk and Sector 9 metro station) Ph. 011-47350606, (M) 7838010301-04 www.eduproz.inEducate Anytime...Anywhere..."Greetings For The Day"About EduprozWe, at EduProz, started our voyage with a dream of making higher education available for everyone. Since itsinception, EduProz has been working as a stepping-stone for the students coming from varied backgrounds. The bestpart is – the classroom for distance learning or correspondence courses for both management (MBA and BBA) andInformation Technology (MCA and BCA) streams are free of cost. Experienced faculty-members, a state-of-the-art infrastructure and a congenial environment for learning - are the fewthings that we offer to our students. Our panel of industrial experts, coming from various industrial domains, leadstudents not only to secure good marks in examination, but also to get an edge over others in their professional lives.Our study materials are sufficient to keep students abreast of the present nuances of the industry. In addition, we giveimportance to regular tests and sessions to evaluate our students’ progress. Students can attend regular classes of distance learning MBA, BBA, MCA and BCA courses at EduProz withoutpaying anything extra. Our centrally air-conditioned classrooms, well-maintained library and well-equipped laboratoryfacilities provide a comfortable environment for learning.Honing specific skills is inevitable to get success in an interview. Keeping this in mind, EduProz has a careercounselling and career development cell where we help student to prepare for interviews. Our dedicated placementcell has been helping students to land in their dream jobs on completion of the course.EduProz is strategically located in Dwarka, West Delhi (walking distance from Dwarka Sector 9 Metro Station and 4-minutes drive from the national highway); students can easily come to our centre from anywhere Delhi andneighbouring Gurgaon, Haryana and avail of a quality-oriented education facility at apparently no extra cost.Why Choose Edu Proz for distance learning?
  2. 2. • Edu Proz provides class room facilities free of cost. • In EduProz Class room teaching is conducted through experienced faculty. • Class rooms are spacious fully air-conditioned ensuring comfortable ambience. • Course free is not wearily expensive. • Placement assistance and student counseling facilities. • Edu Proz unlike several other distance learning courses strives to help and motivate pupils to get high grades thus ensuring that they are well placed in life. • Students are groomed and prepared to face interview boards. • Mock tests, unit tests and examinations are held to evaluate progress. • Special care is taken in the personality development department. "HAVE A GOOD DAY" Karnataka State Open University(KSOU) was established on 1st June 1996 with the assent of H.E. Governor of Karnatakaas a full fledged University in the academic year 1996 vide Government notificationNo/EDI/UOV/dated 12th February 1996 (Karnataka State Open University Act – 1992).The act was promulgated with the object to incorporate an Open University at the State level forthe introduction and promotion of Open University and Distance Education systems in theeducation pattern of the State and the country for the Co-ordination and determination ofstandard of such systems. Keeping in view the educational needs of our country, in general, andstate in particular the policies and programmes have been geared to cater to the needy.Karnataka State Open University is a UGC recognised University of Distance Education Council(DEC), New Delhi, regular member of the Association of Indian Universities (AIU), Delhi,permanent member of Association of Commonwealth Universities (ACU), London, UK, AsianAssociation of Open Universities (AAOU), Beijing, China, and also has association withCommonwealth of Learning (COL).Karnataka State Open University is situated at the North–Western end of the Manasagangotricampus, Mysore. The campus, which is about 5 kms, from the city centre, has a sereneatmosphere ideally suited for academic pursuits. The University houses at present theAdministrative Office, Academic Block, Lecture Halls, a well-equipped Library, Guest House
  3. 3. Cottages, a Moderate Canteen, Girls Hostel and a few cottages providing limitedaccommodation to students coming to Mysore for attending the Contact Programmes or Term-end examinations.BT0063-Unit-01-Set TheoryUnit 1 Set TheoryStructure1.1 Introduction Objectives1.2 Sets and Their Representations1.3 The Empty Set1.4 Finite and Infinite Sets1.5 Equal and Equivalent Sets1.6 Subsets1.7 Power Set1.8 Universal Set1.9 Venn Diagrams1.10 Complement of a Set1.11 Operations on Sets1.12 Applications of Sets1.13 Cartesian Product of Sets1.14 Summary1.15 Terminal Questions1.16 Answers
  4. 4. 1.1 Introduction The concept of set is basic in all branches of mathematics. It has proved to be of particularimportance in the foundations of relations and functions, sequences, geometry, probability theoryetc. The study of sets has many applications in logic philosophy, etc. The theory of sets was developed by German mathematician Georg Cantor (1845 – 1918A.D.). He first encountered sets while working on problems on trigonometric series. In this unit,we discuss some basic definitions and operations involving sets.Objectives:At the end of the unit you would be able to • understand the concepts of sets • perform the different operations on sets • write the Power set of a given set1.2 Sets and their Representations In every day life, we often speak of collection of objects of a particular kind such as pack ofcards, a herd of cattle, a crowd of people, cricket team, etc. In mathematics also, we come acrossvarious collections, for example, collection of natural numbers, points in plane, prime numbers.More specially, we examine the collections: 1. Odd natural numbers less than 10, i.e., 1, 3, 5, 7, 9 2. The rivers of India 3. The vowels in the English alphabet, namely a, e, I, o, u 4. Prime factors of 210, namely 2, 3, 5 and 7 5. The solutions of a equation x2 – 5x + 6 = 0 viz, 2 and 3We note that each of the above collections is a well defined collection of objects in the sense thatwe can definitely decide whether a given object belongs to a given collection or not. Forexample, we can say that the river Nile does not belong to collection of rivers of India. On theother hand, the river Ganga does belong to this collection. However, the following collectionsare not well defined: 1. The collection of bright students in Class XI of a school 2. The collection of renowned mathematicians of the world 3. The collection of beautiful girls of the world 4. The collection of fat peopleFor example, in (ii) above, the criterion for determining a mathematician as most renowned mayvary from person to person. Thus, it is not a well defined collection.We shall say that a set is a well defined collection of objects. The following points may be noted:
  5. 5. 1. Objects, elements and members of a set are synonymous terms. These are undefined 2. Sets are usually denoted by capital letters A, B, C, X, Y, Z etc. 3. The elements of a set are represented by small letters a, b, c, x, y, z etc.If a is an element of a set A, we say that ‘a belongs to A’. The Greek symbol is used to denotethe phrase ‘belongs to’. Thus, we write . If b is not an element of a set A, we writeand read ‘b does not belong to A’. Thus, in the set V of vowels in the English alphabet, but . In the set P of prime factors of but . There are two methods of representing a set: i) Roster or tabular form ii) Set builder form. i) In roster form, all the elements of a set are listed, the elements being separated by commas and are enclosed within braces { }. For example, the set of all even positive integers less than 7 is described in roster form as {2, 4, 6}. Some more examples of representing a set in roster form are given below: a) The set of all natural numbers which divide 42 is {1, 2, 3, 6, 7, 14, 21, 42}. Note that in roster form, the order in which the elements are listed is immaterial. Thus, the above set can also be represented as {l, 3, 7, 21, 2, 6, 14, 42}. b) The set of all vowels in the English alphabets is {a, e, i, o, u}. c) The set of odd natural numbers is represented by {1, 3, 5,. . .}. The three dots tell us that the list is endless. It may be noted that while writing the set in roster form an element is not generally repeated, i.e., all the elements are taken as distinct. For example, the set of letters forming the word “SCHOOL” is {S, C, H, O, L}. ii) In set builder form, all the elements of a set possess a single common property which is not possessed by any element outside the set. For example, in the set “{a, e, i, o, u}” all the elements possess a common property, each of them is a vowel in the English alphabet and no other letter possesses this property. Denoting this set by V, we write V = {x : x is a vowel in the English alphabet}.It may be observed that we describe the set by using a symbol x for elements of the set (any othersymbol like the letters y, z etc. could also be used in place of x). After the sign of ‘colon’ write
  6. 6. the characteristic property possessed by the elements of the set and then enclose the descriptionwithin braces. The above description of the set V is read as ‘The set of allx such that x is a vowel of the English alphabet’. In this description the braces stand for ‘the setof all’, the colon stands for ’such that’.For example, the following description of a setA = {x : x is a natural number and 3 < x < 10)is read as “the set of all x such that x is a natural number and 3 < x < 10″. Hence, the numbers 4,5, 6, 7, 8 and 9 are the elements of set A.If we denote the sets described above in (a), (b) and (c) in roster form by A, B and C,respectively, then A, B and C can also be represented in set builder form as follows A = {x : x is a natural number which divides 42} B = {y : y is a vowel in the English alphabet} C = {z : z is an odd natural number}.Example: Write the set of all vowels in the English alphabet which precede q.Solution: The vowels which precede q are a, e, i, o. Thus A = {a, e, i, o} is the set of all vowelsin the English alphabet which precede q.Example: Write the set of all positive integers whose cube is odd.Solution: The cube of an even integer is also an even integer. So, the members of the requiredset can not be even. Also, cube of an odd integer is odd. So, the members of the required set areall positive odd integers. Hence, in the set builder form we write this set as {x : x is an oddpositive integer} or equivalently as{2k + 1 : k ≥ 0, k is an integer}Example: Write the set of all real numbers which can not be written as the quotient of twointegers in the set builder form.Solution: We observe that the required numbers can not be rational numbers because a rationalnumber is a number in the form , where p, q are integers and q ≠ 0. Thus, these must be realand irrational. Hence, in set builder form we write this set as{x : x is real and irrational}
  7. 7. Example: Write the set in the set builder form.Solution: Each member in the given set has the denominator one more than the numerator. Also,the numerators begin from 1 and do not exceed 6. Hence, in the set builder form the given set isExample: Match each of the sets on the left described in the roster form with the same set on theright described in the set builder form: i) { L, I, T, E) a) {x : x is a positive integer and is a divisor of 18} ii) {0) b) {x : x is an integer and x2 – 9 = 0} iii) {1, 2, 3, 6, 9, 18} c) {x : x is an integer and x + 1 = 1} iv) {3, – 3} d) {x : x is a letter of the word LITTLE}Solution: Since in (d), there are six letters in the word LITTLE and two letters T and L arerepeated, so (i) matches (d). Similarly (ii) matches (c) as x + 1 = 1 implies x = 0. Also, 1, 2, 3, 6,9, 18 are all divisors of 18. So,(iii) matches (a). Finally, x2 – 9 = 0 implies. x = 3, –3. So, (iv) matches (b).Example: Write the set {x : x is a positive integer and x2 < 40} in the roster form.Solution: The required numbers are 1, 2, 3, 4, 5, 6. So, the given set in the roster form is {1, 2, 3,4, 5, 6}.1.3 The Empty SetConsider the setA = {x : x is a student of Class XI presently studying in a school}We can go to the school and count the number of students presently studying in Class XI in theschool. Thus, the set A contains a finite number of elements.Consider the set {x : x is an integer, x2 + 1 = 0}. We know that there is no integer whose squareis –1. So, the above set has no elements.We now define set B as follows:
  8. 8. B = {x : x is a student presently studying in both Classes X and XI}.We observe that a student cannot study simultaneously in both Classes X and XI. Hence, the setB contains no element at all.Definition: A set which does not contain any element is called the empty set or the null set or thevoid set.According to this definition B is an empty set while A is not. The empty set is denoted by thesymbol ‘ . We give below a few examples of empty sets. i) Let P = {x: 1 < x < 2, x is a natural number }. Then P is an empty set, because there is no natural number between 1 and 2. ii) Let Q = {x : x2 – 2 = 0 and x is rational}. Then, Q is the empty set, because the equation x2 - 2 = 0 is not satisfied by any rational number x. iii) Let R = {x : x is an even prime number greater than 2} Then R is the empty set, because 2 is the only even prime number. iv) Let S = {x : x2 = 4, and x is an odd integer}. Then, S is the empty set, because equation x2 = 4 is not satisfied by any value of x which is an odd integer.1.4 Finite and Infinite SetsLet A = {1, 2, 3, 4, 5), B = {a, b, c, d, e, f} and C = {men in the world}.We observe that A contains 5 elements and B contains 6 elements. How many elements does Ccontain ? As it is, we do not know the exact number of elements in C, but it is some naturalnumber which may be quite a big number. By number of elements of a set A, we mean thenumber of distinct elements of the set and we denote it by n(A). If n(A) is a natural number, thenA is a finite set, otherwise the set A is said to be an infinite set. For example, consider the set, N,of natural numbers. We see that n(N), i.e., the number of elements of N is not finite since there isno natural number which equals n(N). We, thus, say that the set of natural number is an infiniteset.Definition: A set which is empty or consists of a definite number of elements is called finite.Otherwise, the set is called infinite.We shall denote several set of numbers by the following symbols:
  9. 9. N : the set of natural numbers Z : the set of integers Q : the set of rational numbers R : the set of real numbers Z+ : the set of positive Integers Q+ : the set of positive rational numbers R+ : the set of positive real numbersWe consider some examples: 1. Let M be the set of days of the week. Then M is finite. 2. Q, the set of all rational numbers is infinite. 3. Let S be the set of solution (s) of the equation x2 – 16 = 0. Then S is finite. 4. Let G be the set of all points on a line. Then G is infinite.When we represent a set in the roster form, we write all the elements of the set within braces { }.It is not always possible to write all the elements of an infinite set within braces { } because thenumber of elements of such a set is not finite. However, we represent some of the infinite sets inthe roster form by writing a few elements which clearly indicate the structure of the set followed(or preceded) by three dots.For instance, {1, 2, 3, 4, … } is the set of natural numbers, {1, 3, 5, 7, 9, .. . } is the set of oddnatural numbers and {…, – 3, –2, –1, 0, 1, 2, 3, … } is the set of integers. But the set of realnumbers cannot be described in this form, because the elements of this set do not follow anyparticular pattern.1.5 Equal and Equivalent SetsGiven two sets A and B. If every element of A is also an element of B and if every element of B isalso an element of A, the sets A and B are said to be equal. Clearly, the two sets have exactly thesame elements.Definition: Two sets A and B are said to be equal if they have exactly the same elements and wewrite A = B. Otherwise, the sets are said to be unequal and we write A ≠ BWe consider the following examples: 1. Let A = {1, 2, 3, 4, } and B = {3, 1, 4, 2). 2. Then A = B.
  10. 10. 3. Let A be the set of prime numbers less than 6 and P the set of prime factors of 30. Obviously, the set A and P are equal, since 2, 3 and 5 are the only prime factors of 30 and are less than 6.Let us consider two sets L = {1, 2, 3, 4} and M = {1, 2, 3, 8}. Each of them has four elements butthey are not equal.Definition: Two finite sets A and B are said to be equivalent if they have the same number ofelements. We write A ≈ B.For example, let A = {a, b, c, d, e} and B = {1, 3, 5, 7, 9}. Then A and B are equivalent sets.Obviously, all equal sets are equivalent, but all equivalent sets are not equal.Example: Find the pairs of equal sets, if any, giving reasons:A = {0}, B = {x : x > 15 and x < 5}, C = {x : x – 5 = 0}, D = {x:x2 = 25}E = {x : x is a positive integral root of the equation x2 – 2x – 15 = 0}Solution: Since 0 ∈ A and 0 does not belong to any of the sets B, C, D and E. Therefore, A B,A C, A D, A E. B = but none of the other sets are empty. Hence B C, B D and B E. C= {5} but , hence C D. Since E = {5}), C = E. D = {–5, 5} and E = {5}. Therefore DE. Thus, the only pair of equal sets is C and E.1.6 SubsetsConsider the sets S and T, where S denotes the set of all students in your school and T denotesthe set of all students in your class. We note that every element of T is also an element of S. Wesay that T is a subset of S.Definition: If every element of a set A is also an element of a set B, then A is called a subset of Bor A is contained in B. We write it as A B.If at least one element of A does not belong to B, then A is not a subset of B. We write it as AB.We may note that for A to be a subset of B, all that is needed is that every element of A is in B. Itis possible that every element of B may or may not be in A. If it so happens that every element ofB is also in A, then we shall also have B A. In this case, A and B are the same sets so that wehave A B and B A which implies A = B.
  11. 11. It follows from the definition that every set A is a subset of itself, i.e., A A. Since the empty set has no elements, we agree to say that is a subset of every set. We now consider someexamples 1. The set Q of rational numbers is a subset of the set R of real numbers and we write Q R. 2. If A is the set of all divisors of 56 and B the set of all prime divisors of 56, then B is a subset of A, and we write B A. 3. Let A = {1, 3, 5} and B = {x : x is an odd natural number less than 6}, then A B and B A and hence A = B. 4. Let A = {a, e, i, o, u}, B = {a, b, c, d}. Then A is not a subset of B. Also B is not a subset of A. We write A B and B A. 5. Let us write down all the subsets of the set {1, 2}. We know is a subset of every set. So is a subset of {1, 2}. We see that {1}, {2} and {l, 2} are also subsets of {1,2}. Thus the set {1,2} has, in all, four subsets, viz. , {1}, {2} and {1,2}.Definition: Let A and B be two sets. If A B and A ≠ B, then A is called a proper subset of B andB is called a superset of A. For example, A= {1, 2, 3} is a proper subset of B = {1, 2, 3, 4}.Definition: If a set A has only one element, we call it a singleton. Thus {a } is a singleton.1.7 Power SetIn example (v) of Section 1.6, we found all the subsets of the set {1, 2}, viz., , {1}, {2} and {1,2}. The set of all these four subsets is called the power set of {1, 2}.Definition: The collection of all subsets of a set A is called the power set of A. It is denoted byP(A). In P(A), every element is a set.Example (v) of section 1.6, if A = {1, 2}, then P(A)={ , {1}, {2}, {1,2}. Also, note that, n[P(A)]= 4 = 22.In general, if A is a set with n(A) = m, then it can be shown thatn[P(A)] = 2m > m = n(A).1.8 Universal Set
  12. 12. If in any particular context of sets, we find a set U which contains all the sets under considerationas subsets of U, then set U is called the universal set. We note that the universal set is not unique.For example, for the set Z of all integers, the universal set can be the set Q of rational numbersor, for that matter, the set R of real numbers.For another example, in the context of human population studies, the universal set consists of allthe people in the world.Example: Consider the following sets : , A = {1, 3), B = {1, 5, 9},C = {1, 3, 5, 7, 9}, Insert the correct symbol or between each pair of sets(i) — B, (ii) A — B (iii) A — C (iv) B — C.Solution: 1. B as is a subset of every set. 2. A B as 3 A and 3 B.. 3. A C as 1, 3 A also belongs to C. 4. B C as each element of B also belongs to C.Example: Let A = {1, 2, 3, 4}, B = {1, 2, 3} and C = {2, 4}. Find all sets X such that(i) X B and X C (ii) X A and X B.Solution: i) X B means that X is a subset of B, and the subsets of B are , {1}, {2}, {3}, {1,2}, {1,3}, {2,3} and {1,2,3} . X C means that X is a subset of C, and the subsets of C are , {2}, {4} and {2, 4}. Thus, we note that X B and X C means that X is a subset of both B and C. Hence, X = , {2}. ii) X A, X B means that X is a sub set of A but X is not a subset of B. So, X is one of these {4}, {1,2,4}, {2,3,4}, {l,3,4}, {1,4}, {2,4}, {3,4}, {1,2,3,4}.Note:A set can easily have some elements which are themselves sets. For example, {1, {2,3}, 4} is a sethaving {2,3} as one element which is a set and also elements 1,4 which are not sets.
  13. 13. Example: Let A, B and C be three sets. If A B and B C, is it true thatA C? If not, give an example.Solution: No. Let A = {1}, B = C = { { 1 }, 2}. Here A B as A = {1} andB = C implies B C. But A C as 1 A and 1 C.Note that an element of a set can never be a subset of it.�1.9 VennDiagramsMost of the relationships between sets can be represented by means of diagrams. Figuresrepresenting sets in the form of enclosed region in the plane are called Venn diagrams namedafter British logician John Venn (1834—1883 A.D.). The universal set U is represented by theinterior of a rectangle.Other sets are represented by the interior of circles.Fig. 1.1Fig. 1.1 is a Venn diagram representing sets A and B such that A ⊂ B.Fig. 1.2
  14. 14. In Fig.1.2, U = {1, 2, 3, …, 10} is the universal set of which A = {2,4,6,8,10} and B = {4,6} aresubsets. It is seen that B A. The reader will see an extensive use of the Venn diagrams when wediscuss the operations on sets.�1.10 Complement of a SetLet the universal set U be the set of all prime numbers. Let A be the subset of U which consistsof all those prime numbers that are not divisors of 42. Thus A = {g x : x U and x is not a divisorof 42}. We see that 2 U but 2 A, because 2 is a divisor of 42. Similarly 3 U but 3 A, and 7 U but7 A. Now 2, 3 and 7 are the only elements of U which do not belong to A. The set of these threeprime numbers, i.e., the set {2, 3, 7} is called the complement of A with respect to U, and isdenoted by A′. So we haveA′ = {2, 3, 7}. Thus, we see that A′={x : x U and x A). This leads to the following definition.Definition: Let U be the universal set and A is a subset of U. Then the complement of A withrespect to (w.r,t.) U is the set of all elements of U which are not the elements of A. Symbolicallywe write A′ to denote the complement of A with respect to U. Thus A′ = {x:x U and x A}. Itcan be represented by Venn diagram asFig. 1.3The shaded portion in Fig. 1.3 represents A′.Example: Let U = {1,2,3,4,5,6,7,8,9,10} and A= {1,3,5,7,9}. Find A′.Solution: We note that 2, 4, 6, 8, 10 are the only elements of U which do not belong to A. HenceA′ = {2, 4, 6, 8, 10}.
  15. 15. Example: Let U be the universal set of all the students of Class XI of aco-educational school. Let A be the set of all girls in the Class Xl. Find A′.Solution: As A is the set of all girls, hence A′ is the set of all boys in the class.1.11 Operations on SetsIn earlier classes, you learnt how to perform the operations of addition, subtraction,multiplication and division on numbers. You also studied certain properties of these operations,namely, commutativity, associativity, distributivity etc. We shall now define operations on setsand examine their properties. Henceforth, we shall refer all our sets as subsets of some universalset.a) Union of Sets:Let A and B be any two sets. The union of A and B is the set which consists of all the elements ofA as well as the elements of B, the common elements being taken only once. The symbol ‘∪‘ isused to denote the union. Thus, we can define the union of two sets as follows.Definition: The union of two sets A and B is the set C which consists of all those elements whichare either in A or in B (including those which are in both).Symbolically, we write = {x:x A or x B} and usually read as‘A union B‘.The union of two sets can be represented by a Venn diagram as shown in Fig. 1.4.Fig. 1.4The shaded portion in Fig. 1.4 represents A B.Example: Let A = {2, 4, 6, 8} and B = {6, 8, 10, 12}. Find A B.
  16. 16. Solution: We have A B = {2, 4, 6, 8, 10, 12}.Note that the common elements 6 and 8 have been taken only once while writing A B.Example: Let A = {a, e, i, o, u} and B = {a, i, u}. Show that A B = A.Solution: We have A B = {a, e, i, o, u} = A.This example illustrates that the union of a set A and its subset B is the set A itself, i.e., if ,then A B = A.Example: Let X = {Ram, Shyam, Akbar} be the set of students of Class XI who are in the schoolHockey team. Let Y = {Shyam, David, Ashok} be the set of students from Class XI who are inthe school Football team. Find and interpret the set.Solution: We have = {Ram, Shyam, Akbar, David, Ashok}. This is the set of studentsfrom Class XI who are either in the Hockey team or in the Football team.Example: Find the union of each of the following pairs of sets:Solution: 1. A B = {1, 2, 3, 4, 5} 2. A = {3, 4, 5,… }, B = {1, 2, 3}. So, A B = {1, 2, 3, 4, 5,… } = Z+ 3. A={1, 2, 3,. . .}, B ={x:x is a negative integer}. So A B={x:x ∈ Z, x ≠ 0}. 4. A = {2, 3, 4}, B = {5, 6, 7, 8}. So, A B = {2, 3, 4, 5, 6, 7, 8}.b) Intersection of Sets: The intersection of sets A and B is the set of all elements which arecommon to both A and B. The symbol ∩ is used to denote the intersection.Thus, we have the following definition.
  17. 17. Definition: The intersection of two sets A and B is the set of all those elements which belong toboth A and B. Symbolically, we write A ∩ B ={x:x ∈ A and x ∈ B} and read as ‘A intersection B’.The intersection of two sets can be represented by a Venn diagram as shown in Fig. 1.5.�Fig. 1.5The shaded portion represents A B.If A B = φ, then A and B are said to be disjoint sets. For example, letA = {2, 4, 6, 8} and B = {1, 3, 5, 7}. Then, A and B are disjoint sets, because there is no elementwhich is common to A and B. The disjoint sets can be represented by Venn diagram as shown inFig. 1.6.Fig. 1.6
  18. 18. Example: Let A = {2, 4, 6, 8} and B = {6, 8, 10, 12}. Find A B.Solution: We see that 6, 8 are the only elements which are common to both the sets A and B.Hence A B = {6, 8}.Example: Consider the sets X and Y of Example 17. Find X Y.Solution: We see that the element “Shyam” is the only element common to both the sets X andY. Hence, X Y = { Shyam }.SAQ 1: Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = {2, 3, 5, 7}. Find and prove that = B.SAQ 2: Let A = A = {x : x ∈ Z+} ; B = {x : x is a multiple of 3, }:C = {x:x is a negative integer}; D = {x:x is an odd integer}. Find (i) A B,(ii) , (iii) , (iv) , (v) , (vi) .c) Difference of Sets: The difference of sets A and B, in this order, is the set of elements whichbelong to A but not to B. Symbolically, we writeA — B and read as ‘A difference B’. Thus A — B = {x : x ∈ A and x ∉ B} and is represented byVenn diagram in Fig.1.7. The shaded portion representsA — B.Fig. 1.7SAQ 3: Let V = {a, e, i, o, u} and B = {a, i, k, u}. Find V – B and B – V.
  19. 19. SAQ 4: Let A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8}. Find A – B andB – A.1.12 Applications of SetsLet A and B be finite sets. If , then = n(A) + n(B) (1) �The elements in A B are either in A or in B but not in both as . So (1) follows immediately.In general, if A and B are finite sets, thenn(A B) = n(A) + n(B) – n(A B) (2)Fig. 1.8Note that the sets A – B, A B and B – A are disjoint and their union isA B (Fig 1.8). Thereforen(A B) = n(A – B) + n(A B) + n(B – A) = n(A – B) + n(A B) + n(B – A) +n(A B) – n(A B) = n(A) + n(B) – n(A B).which verifies (2).
  20. 20. If A, B and C are finite sets, thenn(A B C) = n(A) + n(B) + n(C) –n(A B) –n(B C) – n(A C) + n(A B C) (3)In fact, we haven(A B C) = n(A) + n(B C) – n(A (B C )) [by (2)] = n(A) + n(B) + n(C) – n(B C) – n (A (B C)) [by (2)]Since A = (A B) (A C), we get = n(A B) + n (A C) – n[A B A C)] = n(A B) + n (A C) – n[A B C)]Thereforen(A B C) = n(A) + n(B) + n(C) – n(B C) – n(A B) – n(A C) + n(A B C).This proves (3).Example: If X and Y are two sets such that n(X ∪ Y) = 50, n(X) = 28 and n(Y) = 32, find n(XY).Solution: By using the formula ,we find that
  21. 21. = 28 + 32 –50 = 10..Alternatively, suppose , thenFig. 1.9n(X – Y) = 28 – k, n(Y – X) = 32 – k. (by Venn diagram in Fig 1.9)This gives 50 = = (28 – k) + k + (32 – k).Hence, k = 10Example: In a school there are 20 teachers who teach mathematics or physics. Of these, 12 teachmathematics and 4 teach physics and mathematics. How many teach physics?Solution Let M denote the set of teachers who teach mathematics andP denote the set of teachers who teach physics. We are given that = 4. Therefore = 20 – 12 + 4 = 12.Hence, 12 teachers teach physics.SAQ 5: In a group of 50 people, 35 speak Hindi, 25 speak both English and Hindi and all thepeople speak at least one of the two languages. How many people speak only English and notHindi ? How many people speak English?1.13 Cartesian Product of SetsLet A, B be two sets. If a ∈ A, b ∈ B, then (a, b) denotes an ordered pair whose first componentis a and the second component is b. Two ordered pairs (a, b) and (c, d) are said to be equal if andonly if a = c and b = d.
  22. 22. In the ordered pair (a, b), the order in which the elements a and b appear in the bracket isimportant. Thus (a, b) and (b, a) are two distinct ordered pairs if a ≠ b. Also, an ordered pair (a,b) is not the same as the set {a, b}.Definition: The set of all ordered pairs (a, b) of elements is called the CartesianProduct of sets A and B and is denoted by A x B. ThusLet A = {a1, a2}, B = {b1, b2, b3}. To write the elements of A x B, take a1∈ A and write all elements of B with a1, i.e., (a1, b1), (a, b2), (a1, b3). Now takea2ε A and write all the elements of B with a2, i.e., (a2, b1), (a2, b2), (a2, b3). Therefore, A x B willhave six elements, namely, (a1, b1), (a1, b2), (a1, b3), (a2, b1), (a2, b2), (a2, b3).Remarks: 1. If A = or B = , then A × B = 2. If A ≠ and B ≠ , then . Thus, if and only if A and B ≠ . Also, A B B A. 3. If the set A has m elements and the set B has n elements, then A × B has mn elements. 4. If A and B are non-empty sets and either A or B is an infinite set, so is A x B. 5. If A = B, then A B is expressed as A2. 6. We can also define, in a similar way, ordered triplets. If A, B and C are three sets, then (a ,b, c), where a ∈ A, b ∈ B and c ∈ C, is called an ordered triplet. The Cartesian Product of sets A, B and C is defined asA B C = {(a, b, c): a ∈ A, b ∈ B, c ∈ C}. An ordered pair and ordered triplet are also called 2-tuple and 3-tuple, respectively. In general, ifA1, A2,.. ., An are n sets, then (a1,a2,…, an) is called an n-tuple whereai∈ Ai, i = 1, 2 n and the set of all such n-tuples, is called the Cartesian product of A1, A2, ……, An.It is denoted by A1 x A2 x. . .x An. Thus A1× A2 × ….. × An = {(a1, a2, …. an): a1 ∈ A1, 1≤ i ≤ n}}.Example: Find x and y if (x + 2, 4) = (5, 2x + y).Solution: By definition of equal ordered pairs, we have
  23. 23. x+2=5 (1) 2x + y = 4 (2)Solving (1) and (2), we get x = 3, y = –2.Example: Let A = {1, 2, 3} and B = {4, 5}. Find A x B and B x A and show that A × B ≠ B × A.Solution: We have = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)} and = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)}Note that and (1, 4) ∉ B × A. Therefore, .Example: Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find i) ii) iii) iv)Solution: i) We have . Therefore, = {(1, 4), (2, 4), (3, 4)}. ii) We note that = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} and = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} Therefore = {(1, 4), (2, 4), (3, 4)}. iii) Clearly = {3, 4, 5, 6}. Thus = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}
  24. 24. iv) In view of (ii), we see that = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}.In view of the assertion in Example 3 above, we note thatandSAQ 6: Let A and B be two sets such that n(A) = 5 and n(B) = 2.If (a1, 2), (a2, 3), (a3, 2), (a4, 3), (a5, 2) are in A × B and a1, a2, a3, a4 anda5 are distinct. Find A and B.1.14 SummaryThis unit tells us about sets and their representations. We study the concepts of Empty sets,Finite and Infinite sets, Equal sets. All the concepts discussed is well illustrated by standardexamples. The different operations on sets like complement of Set, Operation on Sets andApplications of sets is discussed here.1.15 Terminal Questions1. Which of the following pairs of sets are equal ? Justify your answer. i) A, the set of letters in “ALLOY” and B, the set of letters in “LOYAL” ii) A= and B = .2. State which of the following sets are finite and which are infinite:
  25. 25. 3. If A and B are two non-empty sets such that , show thatA=B1.16 AnswerSelf Assessment Questions1. We have A ∩ B = {2, 3, 5, 7} = B. We note that if B ⊂ A , then A ∩ B = B.2. A = {x:x is a positive integer}, B = {3n : n ∈ Z}; 1. A B = {3, 6, 9, 12,…} = {3n:n ∈ Z+}. 2. A C = 3. A D = {1, 3, 5, 7,…} 4. B C = {– 3, –6, –9,. . . } = {3n : n is a negative integer} 5. B D = {. . ., –15, –9, –3, 3, 9, 15,…} 6. C D = {–1, –3, –5, –7,…} 3. We have V – B = {e, o}, since the only elements of V which do not belong to B are e and o. Similarly B – V = {k} 4. We have A – B = {1, 3, 5}, as the only elements of A which do not belong to B are 1, 3 and 5. Similarly, B – A = {8}. We note that 5. Let H denote the set of people speaking Hindi and E the set of people speaking English. We are given that = 50, n(H) = 35, = 25. Now = n(H) + n(E – H). So 50 = 35 + n(E – H), i.e. , n(E – H) = 15. Thus, the number of people who speak only English but not Hindi is 15. Also, n(H ∪ E) = n(H) + n(E) – n(H E) implies
  26. 26. 50 = 35 + n(E) – 25, which gives n(E) = 40. Hence, the number of people who speak English is 40. 6. Since and n(A) = 5, A = {a1, a2, a3, a4, a5}. Also and n(B) = 2. Therefore, B = {2, 3}. Terminal Quesitons 1. i) A = {A, L, L, O, Y}, B = {L, O, Y, A, L}. Then A, B are equal sets as repetition of elements in a set do not change a set. Thus A = {A, L, O, Y} = B. ii) A = {–2, –1, 0, 1, 2,}, B = (1, 2). Since 0 ∈ A and 0∉ B, A and B are not equal sets. 2. i) Given set = {1, 2}. Hence, it is finite. ii) Given set = {2}. Hence, it is finite. iii) Given set = . Hence, it is finite. 1. The given set is the set of all prime numbers and since the set of prime numbers is infinite, hence the given set is infinite. 2. Since there are infinite number of odd numbers, hence the given set is infinite 3. Let a ∈ A. Since B ≠ , there exists . Now, implies . Therefore, every element in A is in B giving . Similarly, . Hence A=BBT0063-Unit-02-Mathematical LogicUnit 2 Mathematical LogicStructure2.1 Introduction
  27. 27. Objectives2.2 Statements2.3 Basic Logical Connectives2.4 Conjunction2.5 Disjunction2.6 Negation2.7 Negation of Compound Statements2.8 Truth Tables2.9 Tautologies2.10 Logical Equivalence2.11 Applications2.12 Summary2.13 Terminal Questions2.14 Answers2.1 IntroductionLogic is the study of general patterns of reasoning, without reference to particular meaning orcontexts. If an object is either black or white, and if it is not black, then logic leads us to theconclusion that it must be white. Observe that logical reasoning from the given hypothesescannot reveal what ‘black’ or ‘white’ mean, or why an object can not be both.Logic can find applications in many branches of sciences and social sciences. Logic, infact is thetheoretical basis for many areas of computer science such as digital logic circuit design,automata theory and artificial intelligence.In this chapter, we shall learn about statements, truth values of a statement, compoundstatements, basic logical connectives, truth tables, tautologies, logical equivalence, duality,
  28. 28. algebra of statements, use of Venn diagrams in logic and finally, some simple applications oflogic in switching circuits.Objectives:At the end of the unit you would be able to• understand the ideas in Mathematical Logic• identify a proposition• apply the concept of Mathematical Logic in circuits2.2 StatementsA statement is a sentence which is either true or false, but not both simultaneously.Note: A sentence which is both true and false simultaneously is not a statement, rather, it is aparadox.Example:(a) Each of the following sentences: i) New Delhi is in India. ii) Two plus two is four. iii) Roses are red. iv) The sun is a star. v) Every square is a rectangle. is true and so each of them is a statement.(b) Each of the following sentences: i) The earth is a star. ii) Two plus two is five. iii) Every rectangle is a square. iv) 8 is less than 6. v) Every set is a finite set.
  29. 29. is false and so each of them is a statement.Example:a) Each of the sentences: i) Open the door. ii) Switch on the fan. iii) Do your homework. can not be assigned true or false and so none of them is a statement. Infact, each of them is a command.b) Each of the sentences: i) Did you meet Rahman? ii) Where are you going? iii) Have you ever seen Taj Mahal? can not be assigned true or false and so none of them is a statement. Infact, each of them is a question.c) Each of the sentences: i) May you live long! ii) May God bless you! can not be assigned true or false and so none of them is a statement. Infact, each of them is optative.d) Each of the sentences: i) Hurrah! We have won the match. ii) Alas! I have failed. can not be assigned true or false and so none of them is a statement. In fact, each of them is
  30. 30. exclamatory.e) Each of the sentences: i) Good morning to all. ii) Wish you best of luck. can not be assigned true or false and so none of them is a statement. In fact, each them is a wish.f) Each of the sentences: i) Please do me a favour. T ii) Give me a glass of water. can not be assigned true or false and so none of them is a statement. In fact, each them is a request.g) Each of the following sentences: i) x is a natural number ii) He is a college student. is an open sentence because the truth or falsity of (xv) depends on x and that of xvi) depends on the reference for the pronoun he. We may observe that for some values of x like x = 1, 2,….. etc, (xv) may be true and for some other values like etc, (xv) is false. Similarly, (xvi) may be true or false. However, at a particular point of time or situation they are either true or false. Since, we are interested only in the fact that it is true or false, sentences (xv) and (xvi) can be considered as statements.Note: The statements (xv) and (xvi) in Example 2 are also called open statements.It is useful to have some notation to represent statements. Let us represent the statements bylower case letter like p, q, r, s, ….. Thus, a statement ‘New Delhi is city may be represented ordenoted by p and we write
  31. 31. p : New Delhi is a city.similarly, we may denote a statement ‘2 + 3 = 6’ by q and writeq : 2 + 3 = 6.Truth value of a statement: The truth or falsity of a statement is called its truth value. Everystatement must be either true or false. No statement can be both true and false at the same time.If a statement is true, we say that its truth value is TRUE or T and if it is false we say that itstruth value is FALSE or F.Example: The statements in Example 1(a) have the truth value T while the statements inExample 1(b) have the truth value F.Compound statements: A statement is said to be simple, if it cannot be broken down into two ormore sentences. The statements that we considered in Example 1(a) and (b) are all simplestatements.New statements that can be formed by combining two or more simple statements are calledcompound statements. Thus, a compound statement is the one which is made up of two or moresimple statements.Example:a) The statement “Roses are red and Violets are blue” is a compound statement which is acombination of two simple statements “Roses are red” and “Violets are blue”.b) The statement “Gita is sick or Rehana is well” is a compound statement made up of two simple statements “Gita is sick” and “Rehana is well”.c) The statement “It is raining today and 2 + 2 = 4” is a compound statement composed of two simple statements “It is raining today” and “2 + 2 = 4”.Simple statements, which on combining, form compound statements, are called sub-statementsor component statements of the compound statements. The compound statements S consisting of
  32. 32. sub-statementsp, q, r,… is denoted by S (p, q, r,…).A fundamental property of a compound statements is that its truth value is completelydetermined by the truth value of each of its sub-statements together with the way in whichthey are connected to form the compound statement.2.3 Basic Logical ConnectivesThere are many ways of combining simple statements to form compound statements. The wordswhich combine simple statements to form compound statements are called connectives. In theEnglish language, we combine two or more statements to form a new statement by using theconnectives ‘and’, ‘or’, etc. with a variety of meanings. Because the use of these connectives inEnglish language is not always precise and unambiguous, it is necessary to define a set ofconnectives with definite meanings in the language of logic, called object language. We nowdefine connectives for object language which corresponds to the connectives discussed above.Three basic connectives (logical) are conjunction which corresponds to the English word ‘and’ ;disjunction which corresponds to the word ‘or’ ; and negation which corresponds to the word‘not’.Throughout we use the symbol ‘ ’ to denote conjunction ; ‘∨’ to denote disjunction and thesymbol ‘~‘ to denote negation.Note:. Negation is called a connective although it does not combine two or more statements. Infact, it only modifies a statement.2.4 ConjunctionIf two simple statements p and q are connected by the word ‘and’, then the resulting compoundstatement “p and q” is called a conjunction of p and q and is written in symbolic form as “p ∧q“.Example: Form the conjunction of the following simple statements: p : Dinesh is a boy.
  33. 33. q : Nagma is a girl.Solution: The conjunction of the statement p and q is given by p ∧ q : Dinesh is a boy and Nagma is a girl.Example: Translate the following statement into symbolic form “Jack and Jill went up the hill.”Solution: The given statement can be rewritten as “Jack went up the hill and Jill went up the hill”Let p : Jack went up the hill and q : Jill went up the hill.Then the given statement in symbolic form is p ∧ q.Example: Write the truth value of each of the following four statements: i) Delhi is in India and 2 + 3 = 6. ii) Delhi is in India and 2 + 3 = 5. iii) Delhi is in Nepal and 2 + 3 = 5. iv) Delhi is in Nepal and 2 + 3 = 6.Solution: In view of (D1) and (D2) above, we observe that statement (i) has the truth value F asthe truth value of the statement “2 + 3 = 6” is F. Also, statement (ii) has the truth value T asboth the statement “Delhi is in India” and “2 + 3 = 5” has the truth value T. Similarly, the truthvalue of both the statements (iii) and (iv) is F.2.5 DisjunctionIf two simple statements p and q are connected by the word ‘or’, then the resulting compoundstatement “p or q” is called disjunction ofp and q and is written in symbolic form as “p ∧ q”.
  34. 34. Example: Form the disjunction of the following simple statements: p : The sun shines. q : It rains.Solution: The disjunction of the statements p and q is given by p ∨ q : The sun shines or it rains.Example: Write the truth value of each of the following statements: i) India is in Asia or 2 + 2 = 4. ii) India is in Asia or 2 + 2 =5. iii) India is in Europe or 2 + 2 = 4. iv) India is in Europe or 2 + 2 = 5.Solution: In view of (D3) and (D4) above, we observe that only the last statement has truth valueF as both the sub-statements “India is in Europe” and “2 + 2 = 5” have the truth value F. Theremaining statements (i) to (iii) have the truth value T as at least one of the sub-statements ofthese statements has the truth value T.2.6 NegationAn assertion that a statement fails or denial of a statement is called the negation of the statement.The negation of a statement is generally formed by introducing the word “not” at some properplace in the statement or by prefixing the statement with “It is not the case that” or “It is falsethat”.The negation of a statement p in symbolic form is written as “~ p”.Example: Write the negation of the statement p : New Delhi is a city.Solution: The negation of p is given by
  35. 35. ~ p : New Delhi is not a cityor ~ p : It is not the case that New Delhi is a city.or ~ p : It is false that New Delhi is a cityExample: Write the negation of the following statements: p : I went to my class yesterday. q:2+3=6 r : All natural numbers are integers.Solution: Negation of the statement p is given by ~ p : I did not go to my class yesterday. or It is not the case that I went to my class yesterday. or It is false that I went to my class yesterday. or I was absent from my class yesterday.The negation of the statement q is given by ~q : 2 + 3 ≠ 6 or It is not the case that 2 + 3 = 6
  36. 36. or It is false that 2 + 3 = 6The negation of the statement r is given by ~ r : Not all natural numbers are integers. or There exists a natural number which is not an integer. or it is not the case that all natural numbers are integers. or It is false that all natural numbers are integers.Regarding the truth value of the negation ~ p of a statement p. we have(D5) : ~ p has truth value T whenever p has truth value F.(D6) : ~ p has truth value F whenever p has truth value T.Example: Write the truth value of the negation of each of the following statements:: i) p : Every square is a rectangle. ii) q : The earth is a star. iii) r :2 + 3 < 4Solution: In view of (D5) and (D6), we observe that the truth value of ~p is F as the truth valueof p is T. Similarly, the truth value of both ~q and ~r is T as the truth value of both statements qand r is F2.7 Negation of compound statements
  37. 37. I) Negation of conjunction: Recall that a conjunction p ∧ q consists of two sub-statements p and q both of which exist simultaneously. Therefore, the negation of the conjunction would mean the negation of at least one of the two sub-statements. Thus, we have (D7): The negation of a conjunction p ∧ q is the disjunction of the negation of p and the negation of q. Equivalently, we write ~ ( p ∧ q) = ~ p v ~ q Example: Write the negation of each of the following conjunctions: a) Paris is in France and London is in England. b) 2 + 3 = 5 and 8 < 10.Solution: (a) Write p : Paris is in France and q : London is in England. Then, the conjunction in (a) is given by p ∧ q. Now ~ p : Paris is not in France, and ~ q : London is not in England. Therefore, using (D7), negation of p ∧ q is given by ~ p ∧ q = Paris is not in France or London is not in England. (b) Write p : 2+3 = 5 and q :8 < 10. Then the conjunction in (b) is given by p ∧ q. Now ~ p : 2 + 3 ≠ 5 and Then, using (D7), negation of p ∧ q is given by
  38. 38. ~ p ∧ q = 2 + 3 ≠ 5 or(II) Negation of disjunction: Recall that a disjunction p ∨ q is consisting of two sub-statements p and q which are such that either p or q or both exist. Therefore, the negation of the disjunction would mean the negation of both p and q simultaneously. Thus, in symbolic form, we have (D8): The negation of a disjunction p ∨ q is the conjunction of the negation of p and the negation of q. Equivalently, we write ~ (p∨ q) = ~ p ∧ ~ q Example: Writ the negation of each of the following disjunction: a) Ram is in class X or Rahim is in Class XII b) 7 is greater than 4 or 6 is less than 7Solution: a) Let p : Ram is in class X and q : Rahim is in Class XII. Then, the disjunction in (a) is given by p ∨ q. Now ~ p : Ram is not in Class X. ~ q : Rahim is not in Class XII. Then, using (D8), negation of p ∨ q is given by ~ p ∨ q : Ram is not in Class X and Rahim is not in Class XII. b) Write p : 7 is greater than 4, and q : 6 is less than 7. Then, using (D8), negation of p ∨ q is given by
  39. 39. ~ p ∨ q : 7 is not greater than 4 and 6 is not less than 7.(III) Negation of a negation: As already remarked the negation is not a connective but a modifier. It only modifies a given statement and applies only to a single simple statement. Therefore, in view of (D5) and (D6), for a statement p, we have (D9) : Negation of negation of a statement is the statement itself Equivalently, we write ~ (~p) = p Example: Verify (D9) for the statement p : Roses are red. Solution: The negation of p is given by ~ p : Roses are not red. Therefore, the negation of negation of p is ~ (~ p) : It is not the case that Roses are not red. or It is false that Roses are not red. or Roses are red. Many statements, particularly in mathematics, are of the type “If p then q”. Such statements are called conditional statements and are denoted by p → q read as ‘p implies q’. Another common statement is of the form “p if and only if q”. Such statements are called bi- conditional statements and are denoted by p ↔ q.
  40. 40. Regarding the truth values of p → q and p ↔ q , we have a) the conditional p → q is false only if p is true and q is false. Accordingly, if p is false then p → q is true regardless of the truth value of q. b) the bi-conditional p ↔ q is true whenever p and q have the same truth values otherwise it is false. One may verify that p → q = (~ p) ∨ q2.8 Truth TablesA truth table consists of rows and columns. The initial columns are filled with the possible truthvalues of the sub-statements and the last column is filled with the truth values of the compoundstatement S (the truth values of S depends on the truth values of the sub-statements entered in theinitial columns)Example: Construct the truth table for ~p.Solution: Note that one simple statement ~p is consisting of only one simple statement p.Therefore, there must be 2’ (= 2) rows in the truth table. It is necessary to consider all possibletruth values of p.In view of (D5) above, recall that p has the truth value T if and only if ~p has the truth value F.Therefore, the truth table for ~p is given byTable 21 Truth table for ~ pp ~pT FF TExample: Construct the truth table for p ∧ (~p)Solution: Note that the compound statement p ∧ (~p) is consisting of only one simple statementp. Therefore, there must be 2’ (= 2) rows in the truth table. It is necessary to consider all possibletruth values of p.
  41. 41. Table 2.2 p ~p p ∧ (~p)TFStep 1: Enter all possible truth values of p. namely, T and F in the first column of the truth table(Table 2.2).Table 2.3p ~p p ∧ (~ p)T FF TStep 2: Using (D5) and (D6), enter the truth values of ~ p in the second column of the truth table(Table 2.3).Table 2.4 p ~p p ∧ (~ p) T F F F T FStep 3: Finally, using (D2) enter the truth values of p ∧ (~ p) in the last column of the truth table(Table 2.4)Example: Construct the truth table for p ∧ q.Solution: The compound statement p∧q is consisting of two simple statements p and q.Therefore, there must be 22(= 4) rows in the truth table of p ∧q. Now enter all possible truthvalues of statements p and q namely TT, TF, FT and FF in first two columns of Table 2.5.Table 2.5
  42. 42. P q p∧q T T T F F T F FThen, in view of (D1) and (D2) above, enter the truth values of the compound statement p ∧ q inthe truth table (Table 18.6) to complete the truth table.Table 2.6: Truth table for p ∧ q P q p∧q T T T T F F F T F F F FExample: Construct the truth table for p ∨ q.Table 2.7: Truth table for p ∨ q. P q p∨q T T T T F T F T T F F FSolution: In view of (D3) and (D4) above, recall that the compound statement p ∨ q has the truthvalue F if and only if both p and q have the truth value F; otherwise p ∨ q has truth value T.Thus, the truth table for p ∨ q is as given in Table 2.7.a) ~ [p ∧ (~q)]b) (p ∧q) ∧ (~ p)c) ~[(~p) ∨ (~q)]
  43. 43. Solution:a) Truth table for ~ [p ∧ (~ q)] is given byTable 2.8: Truth table for ~ [p ∧ (~ q)] p q ~p p ∧ (~q) ~[p ∧ (~q)] T T F F T T F T T F F T F F T F F T F Tb) Truth table for (p ∧ q) ∧ (~p) is given byTable 2.9: Truth table for (p ∧ q) ∧ (~ p) p q p∧q ~p (p∧q) ∧ (~ p) T T T F F T F F F F F T F T F F F F T Fc) Truth table for ~ [(~p) v (~q)] is given byTable 2.10 : Truth table for ~ [(~p) ∨ (~q)] ~ [( ~ p)] ∧ p q ~p ~q (~ p) ∨ (~ q) [(~q)] T T F F F T T F F T T F F T T F T F F F T T T F2.9 TautologiesA statement is said to be a tautology if it is true for all logical possibilities. In other words, astatement is called tautology if its truth value is T and only T in the last column of its truth table.Analogously, a statement is said to be a contradiction if it is false for all logical possibilities. In
  44. 44. other words, a statement is called contradiction if its truth value is F and only F in the lastcolumn of its truth table. A straight forward method to determine whether a given statement istautology (or contradiction) is to construct its truth table.Example: The statement p ∨ (~p) is a tautology since it contains T in the last column of its truthtable (Table 2.11)Table 2.11: Truth table for p ∨ (~p) p ~p p ∨ (~p) T F T F T TExample: The statement p ∧ (~p) is a contradiction since it contains F in the last column of itstruth table (Table 2.12)Table 2.12: Truth table for p ∧ (~ p) p ~p p ∧ (~p) T F F F T FRemark: The negation of a tautology is a contradiction since it is always false, and the negationof a contradiction is a tautology since it is always true.SAQ 1: Show thata) ~ [p∨ (~p)] is a contradiction.b) ~ [p ∧ (~p)] is a tautology.Example: Show thata) (p ∨ q) ∨ (~ p) is a tautology.b) (p ∧ q) ∧ (~ p) is a contradiction.
  45. 45. Solution:a) The truth table for (p ∨ q) ∨ (~ p) is given byTable 2.15: Truth table for (p ∨ q) ∨ (~ p) P q p∨q ~p (p ∨ q) ∨ (~ p) T T T F T T F T F T F T T T T F F F T T Since the truth table for (p ∨ q) ∨ (~ p) contains only T in the last column, it follows that (p ∨ q) ∨ (~ p) is a tautology.b) Recall Table 2.9 which is the truth table for (p ∧ q) ∧ (~ p) and observe that it contains only F in the last column. Therefore, (p ∧ q) ∧ (~ p) is a contradiction.2.10 Logical EquivalenceTwo statements S1 (p, q, r, …) and S2 (p, q, r, …) are said to be logically equivalent, or simplyequivalent if they have the same truth values for all logical possibilities is denoted byS1 (p, q, r,…) ≡ S2 (p, q, r,…).In other words, S1 and S2 are logically equivalent if they have identical truth tables (by identicaltruth tables we mean the entries in the last column of the truth tables are same).Example: Show that ~ p ∧ q is logically equivalent to (~p) ∨ (~ q).Solution: The truth tables for both the statements areTable 2.16: Truth table for ~ (p ∧ q) Table 2.17: Truth table for (~ p) ∨ (~q) p q p∧q ~(p ∧ q) p q ~p ~q (~ p) ∨ (~q)
  46. 46. T T T F T T F F F T F F T T F F T T F T F T F T T F T F F F T F F T T TNow, observe that the entries (truth values) in the last column of both the tables are same. Hence,the statement ~(p ∧ q) is equivalent to the statement (~ p) ∨ (~q).Remark: Consider the statements: p : Mohan is a boy. q : Sangita is a girl.Now, we have~(p ∧ q) ≡ (~ p) ∨ (~q).Therefore, the statement “It is not the case that Mohan is a boy and Sangita is a girl”has the same meaning as the statement “Mohan is not a boy or Sangita is not a girl”.Example: Let p : The South-West monsoon is very good this year and q : Rivers are rising.Give verbal translation of ~ [(~p) ∨ (~q)].Solution: we have ~(p ∧ q) ≡ (~ p) ∨ (~q)
  47. 47. Therefore, the statement ~ [(~p) ∨ (~q)] is the same as the negation of the statement ~(p ∧ q)which is the same as the conjunction p ∧ q. Thus, the verbal translation for ~ [(~p) ∨ (~q)] is“The South-West monsoon is very good this year and rivers are rising”Example: Prove the following:a) ~ [p ∨ (~ q)] ≡ (~p) ∧qb) ~ [(~ p) ∧ q] ≡ p ∨ (~q)c) ~ (~p) ≡ pSolution:a) The truth tables for ~ [p ∨ (~q)] and (~p) ∧ q are given byTable 2.18: Truth table for~ [p ∨ (~q)] Table 2.19: Truth table for (~p) ∧ q ~ [p ∨ (~ p q ~q p ∨ (~q) p q ~p (~p) ~ q q)] T T F T F T T F F F T T T F T F F F F T F F T F T T T F F T T F F F T FThe last column of the two tables are the same.b) It follows in view of the truth Table 2.20Table 2.20: Truth table for p ∨ (~q) and ~ [(~ p) ∧ q]
  48. 48. c) The assertion follows in view of Table 2.21Table 2.21: Truth table for ~(~p)2.11 ApplicationsThe logic that we have discussed so far is called two-value logic because we have consideredonly those statements which are having truth values True or False. A similar situation exists invarious electrical and mechanical devices. Claude Shannon, in late 1930’s, was first to notice ananalogy between the operations of switching devices and the operations of logical connectives.He used this analogy with great success to solve problems of circuit design.Observe that an electric switch which is used for turning ‘on’ and ‘off’ an electric light is a two-state device. We shall now explain various electric networks with the help of logicalconnectives. For this, first we discuss how an electric switch works. Observe that, in Fig. 2.1, wehave shown two positions of a simple switch.Fig. 2.1In (a) when switch is closed (i.e. on), current can flow from one terminal to the other. In (b),when the switch is open (i.e. off), current can not flow.
  49. 49. Let us now consider the example of an electric lamp controlled by switch. Such a circuit is givenin Fig. 2.2.Fig. 2.2Observe that when the switch s is open, no current flows in the circuit and therefore, the lampis ‘off’. But when switch s is closed, the lamp is ‘on’. Thus the lamp is on if and only if the switchs is closed.If we denote the statements as p : The switch s is closed l : The lamp l is ‘on’then, by using logic, the above circuit can be expressed as p ≡ l.Next, consider an extension of the above circuit in which we have taken two switches s1 and s2 inseries as shown in Fig. 2.3.Fig. 2.3here, observe that the lamp is ‘on’ if and only if both the switches s1 and s2 are closed.If we denote the statements as:
  50. 50. p : the switch s1 is closed. q : the switch s2 is closed. l : the lamp l is ’on’.then the above circuit can be expressed as p ∧ q ≡1.Now, we consider a circuit in which two switches s1 and s2 are connected in parallel (Fig. 2.4).Fig. 2.4SAQ 2: Express the following circuit in Fig. 2.5 in symbolic form of logic.Fig. 2.52.12 SummaryIn this unit we study the truth values of a statements. The different basic logical connectives arediscussed in detail with some standard examples. Compound statements and the negation are
  51. 51. clearly explained . The concept of Tautology, Contradiction and Logical Equivalence is discussed in detail with example wherever necessary. The applications of mathematical logic to switching circuits is dealt with standard examples. 2.13 Terminal Questions 1. Define Tautology and Contradiction 2. Draw the truth tables of Conjunction, disjunction and Biconditional 2.14 Answers Self Assessment Questions 1. a) The truth table of ~ [p∨ (~p)] is given byTable 2.13: Truth table for ~ [p∨ (~p)] P ~p p ∨ (~p) ~ [p ∨ (~p)] T F T F F T T F Since it contains only F in the last column of its truth table, it follows that ~ [p ∨ (~ p)] is a contradiction. b) The truth table of ~ [p ∧ (~ p)] is given byTable 2.14: Truth table for ~ [p ∧ (~ p)] P ~p p ∧ (~ p) ~ [p ∧ (~ p)] T F F T F T F T Since it contains only T in the last column of its truth table, it follows that ~ [p ∧ (~ p)] is a tautology. 2. Observe that the lamp is ‘on’ if and only if either s1 and s2 both are closed or s1 and s2 both are open or only s1 is closed.
  52. 52. If we denote the statements as p : The switch s1 is closed q : The switch s2 is closed l : The lamp l is ‘on’ then ~p: The switch s1 is open. or The switch s1 is closed. ~ q: The switch s2 is open. or The switch s2 is closed. Therefore, the circuit in Fig. 2.5 in symbolic form of logic may be expressed as p ∨ [(~ p) ∧ (~ q)] ∨ (p ∧ q) ≡1BT0063-Unit-03-Modern AlgebraUnit 3 Modern AlgebraStructure3.1 Introduction Objectives3.2 Binary Operation
  53. 53. 3.3 Addition Modulo n3.4 Multiplication Modulo n3.5 Semigroup3.6 Properties of Groups3.7 Subgroup3.8 Summary3.9 Terminal Questions3.10 Answers3.1 IntroductionThe theory of groups which is a branch of Abstract Algebra is of paramount importance in thedevelopment of mathematics.The idea of group was first given by the French Mathematician Evariste Galois in 1832 who diedat the age of 21 years in a duel. The group theory was later developed by an EnglishMathematician Arthur Cayley. He defined the notion of an abstract group with a generalstructure which could be applied to numerous particular cases. The theory of groups hasapplications in Quantum Mechanics and other branches of mathematics.Objectives:At the end of the unit you would be able to• apply the concepts of Algebraic Structure in practical problems• understand Binary Operations and its applications in group theory3.2 Binary OperationLet G be a non-empty set. Then G × G = {(x, y): x, y ∈ G}. A function ofG × G in to G is said to be a binary operation on the set G. The image of an ordered pair (x, y) under f isdenoted by x f y.
  54. 54. The symbols +, x, 0, *, …. Are very often used as the binary operations on a set.Thus * is a binary operation on the set G if for every a, b∈G implies a * b∈ G.Hence a binary operation * combines any two elements of G to give an element of the same setG.Examples:1. If Z is the set of integers then usual addition (+) is the binary operation on Z. For if Mand n are two integers then m + n is again an integer i.e. for every m, n ∈ Z, m + n ∈ Z. In particular – 5, 3 ∈ Z, implies – 5 + 3 = –2 ∈ Z, etc. Similarly the usual multiplication is the binary operation on the set Q of rationals, forthe product of two rational numbers is again a rational number.2. Let E be the set of even integers. i.e., E = {0, ±2, ±4, ±6, ….} and O be the set of oddintegers i.e. O = {±1, ±3, ±5, ….}. Clearly the usual addition is the binary operation on Ewhereas it is not a binary operation on O. Because the sum of two even integers is even butthe sum of two odd integers is not an odd integer. Also the usual subtraction is not a binary operation on the set N of natural numbers.Algebraic StructureA non-empty set with one or more binary operations is called an algebraic structure. If * is a binaryoperation on G then (G, *) is an algebraic structure.For example the set of integers Z is an algebraic structure with usual addition as the binary operation.Similarly (Q, .), (E, +) are algebraic structures.GroupA non-empty set G is said to be a group with respect to the binary operation * if the following axiomsare satisfied. 1. Closure law. For every a, b ∈ G, a * b ∈ G. 2. Associative law. For every a, b, c ∈ G
  55. 55. a * (b * c) = (a * b) * c 3. Existence of identity element. There exists an element e ∈ G such that a * e = e * a = a for every a ∈ G. Here e is called the identity element 4. Existence of inverse. For every a ∈ G there exists an element b ∈ G such that a * b = b * a = e. Here b is called the inverse of a and is denoted by b = a–1. A group G with respect to the binary operation * is denoted by (G, *). If in a group (G, *), a * b = b * a for every a, b, ∈ G then G is said to be commutative or Abelian group named after Norwegian mathematician Niels Henrik Abel (1802 – 1820).Finite and Infinite GroupsA group G is said to be finite if the number of elements in the set G is finite, otherwise it is said to be aninfinite group. The number of elements in a finite group is said to be the order of the group G and isdenoted by O(G).Example: Prove that the set Z of integers is an abelian group with respect to the usual addition as thebinary operation. 1. Closure law. We know that the sum of two integers is also an integer. Hence for every m, n ∈ Z, m + n ∈ Z. 2. Associative law. It is well known that the addition of integers is associative. Therefore (m + n) + p = m + (n + p) for every m, n, p ∈ Z. 3. Existence of identity element. There exists 0 ∈ Z such that m + 0 = 0 + m = m for every m ∈ Z. Hence 0 is called the additive identity.
  56. 56. 4. Existence of inverse. For every m ∈ Z there exists – m ∈ Z such that m + (–m) = (–m) + m = 0.Here – m is called the additive inverse of m or simply the negative of m. Therefore (Z, +) is agroup. 5. Commutative law. We know that the addition of integers is commutative i.e., m + n = n + m for every m, n ∈ Z. Hence (Z, +) is an abelian group. Since there are an infinite elements in Z, (Z, +) is an infinite group. Similarly we can prove that the set Q of rationals, the set R of reals and the set C of complexnumbers are abelian groups with respect to usual addition.Example: Prove that the set Q0 of all non-zero rationals forms an abelian group with respect to usualmultiplication as the binary operation.Now Q0 = Q – {0}Solution: 1. Closure law. Let a, b ∈ Q0 i.e. a and b are two non-zero rationals. Then their product a b is also a non-zero rational. Hence a b ∈ Q0. Since a, b are two arbitrary elements of Q0, we have for every a, b, ∈ Q0, ab ∈ Q0. 2. Associative law. We know that the multiplication of rationals is associative. i.e.,, a(b c) = (a b) c for every a, b, c ∈ Q0. 3. Existence of identity element. There exists 1 ∈ Q0 such that a.1 = 1 . a = a for every a ∈ Q0. Here 1 is called the multiplicative identity element. 4. Existence of inverse. Let a ∈ Q0. Then a is a non-zero rational. Therefore exists and is also a rational ≠ 0.
  57. 57. Also for every a ∈ Q0. is the multiplicative inverse of a. Therefore (Q0, .) is a group. Further, it is well-known that the multiplication of rationals is commutative i.e., ab = ba forevery a, b ∈ Q0. Hence (Q0, .) is an abelian group. Similarly we can show that the set R0 of non-zero reals and the set C0 of non-zero complexnumbers are abelian groups w.r.t. usual multiplication. 1. The set N of natural numbers is not a group w.r.t. usual addition, for there does not exist the identity element 0 in N and the additive inverse of a natural number is not a natural number i.e., for example 2 ∈ N but – 2 ∉ N. Also N is not a group under multiplication because 5 ∈ N but 2. The set of integers is not a group under multiplication for 2 ∈ Z but 3. The set of rationals, reals and complex numbers (including 0) do not form groups under multiplication for multiplicative inverse of 0 does not exist.SAQ 1: Prove that the fourth roots of unity form an abelian group with respect to multiplication.3.3 Addition Modulo nLet n be a positive integer a and b be any two integers. Then “addition modulo n of two integers a andb”, written a + n b, is defined as the least non-negative remainder when a + b is divided by n. If r is theremainder when a + b is divided by n, then
  58. 58. A + n b = r where 0 ≤ r < n.In other words, if a + b ≡ r (mod n), 0 ≤ r < n. Then a + n b = r.For example,7 + 5 10 = 2 since 7 + 10 = 17 ≡ 2 (mod 5)15 + 7 11 = 5 since 15 + 11 = 26 ≡ 5 (mod 7)17 + 8 21 = 38 since 17 + 21 = 38 ≡ 6 (mod12 + 5 8 = 0 since 12 + 8 = 20 ≡ 0 (mod 5)1 + 7 1 = 2 since 1 + 1 = 2 ≡ 2 (mod 7)Properties:1. Commutative since a + b and b + a leave the same remainder when divided by n, a + n b = b + n a. For example 5 + 7 6 = 4 = 6 + 7 52. Associative since a + (b + c) and (a + b) + c leave the same remainder when divided by n, a + n (b + n c) = (a + n b) + n c. For example 4 +6 (3 + 6 5) = (4 + 6 3) + 6 5Example: Prove that the set Z4 = {0, 1, 2, 3} is an abelian group w.r.t. addition modulo 4.Solution: Form the composition table w.r.t. addition modulo 4 as below: +4 0 1 2 3 0 0 1 2 3 1 1 2 3 0 2 2 3 0 1 3 2 0 1 2
  59. 59. Since 1 + 3 = 4 ≡ 0 (mod 4), 3 + 3 = 6 ≡ 2 (mod 4) 2 + 3 = 5 ≡ 1 (mod 4) etc.1. Closure law. From the above composition table for all a, b ∈ G, a +4 b also belongs to Z4.2. Associative law. Since a + (b + c) and (a + b) + c leave the same remainder when divided by 4, we have (a + 4 (b +4 c) = (a +4 b) +4 c.3. Existence of identity element. From the above table, we observe that 0 ∈ Z4 satisfies a + 4 0 = 0 +4 a = a for every a ∈ Z4. 0 is the identity element.4. Existence of inverse. From the above table, the inverses of 0, 1, 2, 3 are respectively 0, 3, 2, 1 because 0 +4 0 = 0, 1 + 4 3 = 0, 2 +4 2 = 0, and 3 + 41 = 0. Hence (z4, +4) is a group Further, since a + b and b + a leave the same remainder when divided by 4, a + 4 b = b +4 a. (Z4, +4) is an abelian group. Similarly, we can show that the set of remainders of 5 viz. Z5 = {0, 1, 2, 3, 4} from an abelian group under addition (mod 5). In general the set of remainders of a positive integer m. Zm = {0, 1, 2, …. (m –1) form an abelian group under addition (mod m).3.4 Multiplication modulo nLet n be a positive integer an a, b any two integers. Then multiplication modulo n of two integers a andb, written a ×n b, is defined as the least non-negative remainder when ab is divided by n. If r is the
  60. 60. remainder when ab is divided by n. If r is the remainder when ab is divided by n then a ×n b = r, where 0≤ r < n. In other words, if ab ≡ r (mod n), 0 ≤ r < n then a xn b = r.For example,7×5 3 = 1 since 7 . 3 = 21 ≡ 1 (mod 5)9 ×7 5 = 3 since 9 . 5 = 45 ≡ 3 (mod 7)12 ×8 7 = 4 since 12 . 7 = 84 ≡ 4 (mod2 ×7 3 = 6 since 2 . 3 = 6 ≡ 6 (mod 7)14 ×46 = 0 since 14 . 6 = 84 ≡ 0 (mod 4)Properties1. Commutative: Since ab and ba leave the same remainder when divided by n, a ×n b = b ×n a For example 5×74=4×752. Associative: Since a(bc) and (ab)c leave the same remainder when divided by n a ×n (b ×n c) = (a ×n b) ×n c For example 3 ×7 (4 ×7 5) = (3 ×7 4) ×7 5Example: Prove that the set is an abelian group under multiplication modulo 5.Solution: Form the composition table w.r.t. multiplication modulo 5 as below: x5 1 2 3 4 1 1 2 3 4 2 2 4 1 3
  61. 61. 3 3 1 4 2 4 4 3 2 1Since 2 . 3 = 6 ≡ 1 (mod 5)2 . 4 = 8 ≡ 3 (mod 5)4 . 4 = 16 ≡ 1 (mod 5) etc.1. Closure law. Since all the elements entered in the above table are the elements of closure law holds good i.e. for all a, b ∈ G, a ×5 b also belongs to2. Associative law. Since a (bc) and (ab) c leave the same remainder when divided by 5 we have for every a, b, c ∈ a × 5 (b × 5 c) = (a × 5 b) × 5 6.3. Existence of identity element. From the above table, we observe that 1∈ satisfies a ×5 1 = 1 × 5 a = a for every a ∈ . 1 is the identity element.4. Existence of inverse. Also the inverses of 1, 2, 3, 4 are respectively 1, 3, 2, 4 because 1 × 5 1 = 1, 2 × 5 3 = 1, 3 × 5 2 = 1, and 4 × 5 4 = 1. Therefore ( x5) is an abelian group. Similarly, we can show that the non-zero remainders of 7 viz. = {1, 2, 3, 4, 5, 6} form an abelian group under multiplication (mod 7). In general, the non-zero remainders of a positive integer p viz. = {1, 2, 3, …… (p – 1)} form a group under multiplication (mod p) if and only if p is a prime number.
  62. 62. Note: The set = {1, 2, 3, 4, 5} does not form a group under multiplication (mod 6) for 2, 3 ∈G, but 2× 6 3 = 0 ∉ G. This is because 6 is not a prime number.3.5 SemigroupA non-empty set G is said to be a semigroup w.r.t. the binary operation if the following axioms aresatisfied. 1. Closure: For every a, b, ∈ G, a * b ∈ G 2. Associative: For every a, b, c ∈ G, a * (b * c) = (a * b) * c.Examples: 1. The set N of all natural numbers under addition is a semigroup because for every a, b, c ∈ N (i) a + b ∈ N, and (ii) a + (b + c) = (a + b) + c. The set, N is semigroup under multiplication also.2. The set Z of integers is a semigroup under multiplication because for every a, b ∈ Z, a + b ∈Z and for every a, b, c ∈ Z, a(bc) = (ab) c. Note that every group is a semigroup but a semigroupneed not be a group. For example, the set N of all natural numbers is a semigroup undermultiplication (also under addition) but it is not a group. Similarly Z, the set of integers is anexample of a semigroup but not a group under multiplication.3.6 Properties of GroupsFor the sake of convenience we shall replace the binary operation * by dot . in the definition of thegroup. Thus the operation dot . may be the operation of addition or multiplication or some otheroperation. In what follows by ab we mean a . b or a * b. With this convention, we rewrite the definitionof the group.Definition: A non-empty set G is said to be a group w.r.t. the binary operation. if the following axiomsare satisfied.1. Closure property: For every a, b ∈ G, ab ∈ G2. Associative property: For every a, b, c ∈ G, a (bc) = (ab) c.
  63. 63. 3. Existence of identity element: There exists an element e ∈ G such that ae = ea = a for every a ∈ G. Here e is called the identity element.4. Existence of inverse: For every a ∈ G there exists an element b ∈ G such that ab = ba = e. Here b is called the inverse of a i.e., b = a–1 Further,5. If ab = ba for every a, b ∈ G then G is said to be an abelian group or a commutative group.Theorem: The identity element in a group is unique.Proof: Let e and be the two identity elements of a group G. Then by definition, for every a ∈ G. ae = ea = aandSubstitute in (1) and a = e in (2). Then we obtainandHenceThe identity element in a group is unique.Theorem: In a group G the inverse of an element is uniqueProof: Let b and c be the two inverses of an element a in G.Then by definition ab = ba = e ac = ca = eNow consider, b = be = b(ac)
  64. 64. = (ba) c = ec =cTherefore inverse of every element in a group is uniqueTheorem: If a is any element of a group G, then (a–1)–1 = a.Proof: Since a–1 is the inverse of a, we have aa–1 = a–1a = eThis implies that a is an inverse of a–1, but inverse of every element is uniqueThus the inverse of the inverse of every element is the same element.Theorem: If a and b are any two elements of a group G thenProof:Consider, (ab) (b–1 a–1) = = = = aa–1 =eSimilarly we can prove thatHence
  65. 65. Therefore is the inverse of ab,i.e.,Corollary: If a, b, c belong to a group G then (abc)–1 = c–1 b–1 a–1 etc.Note: If (ab)–1 = a–1 b–1 for all a, b ∈ G, the G is abelian.For, (ab)–1 = a–1 b–1 impliesi.e. = ba for all a, b ∈ GHence G is abelian.Theorem: (Cancellation laws).If a, b, c are any three elements of a group G, thenab = ac implies b = c (left cancellation law)ba = ca implies b = c (right cancellation law)Proof: Since a is an element of a group G, there exists a–1 ∈ G there exists a–1 ∈ G such that aa–1 = a–1 a =e, the identity elementNow ab = ac⇒⇒⇒ eb = ec⇒ b=c
  66. 66. Similarly ba = ca⇒⇒⇒ be = ce⇒ b=cTheorem: If a and b are any two elements of a group G, then the equations ax = b and ya = b haveunique solutions in G.Proof: i) Since Now and b ∈ G implies (closure axiom) and Hence x = a–1 b satisfies the equation ax = b and hence is a solution. If x1, x2 are the two solutions of the equation, ax = b then ax1 = b and ax2 = b. ax1 = ax2 x1 = x 2 Hence the solution is unique. ii) Also b ∈ G, a–1 ∈ G implies ba–1 ∈ G and y = ba–1 satisfies the equation ya = b and hence is a solution. If y1, y2 are two solutions of theequation ya = b then y1a = b and y2a = b y1a = y2a y1 = y2
  67. 67. Therefore the solution is uniqueSAQ 2: Prove that in a group G if a2 = a then a = e, the identity element.Note: Any element a which satisfies a2 = a is called the idempotent element in a group. Thus e is theonly idempotent element in G.Example: If in group G, (ab)2 = a2b2 for every a, b ∈ G prove that G is abelian.Solution:Now⇒ (ab) (ab) = (a . a) (b . b)⇒ a[b(ab)] = a[a(bb)] (Associative)⇒ b (ab) = a (bb) (Left cancellation law)⇒ (ba) b = (ab) b (Associative)⇒ ba = ab (Right cancellation law)Hence G is an abelian group.Example: Show that if every element of a group G is its own inverse then G is abelian.Solution: Let a, b ∈ G then a–1 = a and b–1 = bClearly ab ∈ G (ab)–1 = ab by hypothesisi.e. b–1 a–1 = abi.e. ba = ab since b–1 = b, a–1 = a G is abelian.
  68. 68. 3.7 SubgroupA non-empty subset H of a group G is said to be a subgroup of G if under the operation of G, H itselfforms a group.If e be the identity element of a group G, Then H = { e } and H = G are always subgroups of G. These arecalled the trivial or improper subgroups. If H is a subgroup of G and H ≠ {e} and H ≠ G then H is called aproper subgroup.Examples:1. We know that the set Z of integers forms a group under addition. Consider a subset E = {2x : x ∈ Z} = {0, ±2, ±4, …. } of Z. Then E also forms a group under addition. Therefore E is a subgroup of Z. Similarly F = {3x : x ∈ Z} = {0, ±3, ±6, ±9, ….. } is a subgroup of z.2. Clearly the multiplicative group H = {1, –1} is a subgroup of the multiplicative group G = {1 –1, i, –i}.3. Let G = {1, 2, 3, 4, 5, 6} be a subset of G. Now it is clear from the following composition table that H also forms a group under x7. X7 1 2 4 1 1 2 4 2 2 4 1 4 4 1 2 Therefore H is a subgroup of G.Theorem: A non-empty subset H of a group G is a subgroup of G if and only if i) for every a, b ∈ H implies ab ∈ H ii) for every a ∈ H implies a–1 ∈ HNote: Union of two subgroups need not be subgroups for, let H = {0, ±2, ±3, ±4, ….} and K = {0, ±e, ±6,….} be
  69. 69. two subgroups of the group of integers Z, so thatH ∪ K = {0, ±2, ±3, ±4, ±6, …. }.Now 2, 3 ∈ H ∪ K but 2 + 3 = 5 ∉ H ∪ K because 5 is neither a multiple of 2 nor a multiple of 3.3.8 SummaryIn this unit we studied clearly that the rectangular array of numbers is denoted by matrix, also we knowthat determinant is a square matrix which is associated with a real number. Then we studied that a setwhich satisfies certain rules is called as a group. Here we studied sub group, semi group etc. with wellillustrated examples.3.9 Terminal Questions 1. Prove that a non-empty subset H of a group G is a subgroup of G if and only if for every a, b ∈ H implies ab–1 ∈ H. 2. Prove that the intersection of two subgroups of a group is again a subgroup.3.10 AnswersSelf Assessment Questions1. Roots of the equation x4 = 1 are called the fourth roots of unity and they are 1, –1, i, – i. Let G = {1, – 1 i, – i }. From the composition table w.r.t. usual multiplication as follows: . 1 –1 i –i 1 1 –1 i –i –1 –1 1 –i i i i –i –1 1 –i –i I 1 –1 1. Closure Law. Since all the elements written in the above composition table are the elements of G, we have for all a, b, ∈ G, ab ∈ G.
  70. 70. 2. Associative Law. We know that the multiplication of complex numbers is associative and G is a subset of the set of complex numbers Hence a(bc) = (ab) c for all a, b, c ∈ G. 3. existence of identity element. From the composition table it is clear that there exists 1 ∈ G satisfying a . 1 = 1 . a = a for every a ∈ G. Therefore 1 is the identity element. 4. Existence of inverse. From the composition table we observe that the inverses of 1, – 1, i – i are 1, -1, -i, i. Thus for every a ∈ G there exists a–1 ∈ G such that a a–1 = a–1 a = 1, the identity element. Hence (G, .) is a group. Further multiplication of complex numbers is commutative. Therefore ab = ba for every a, b ∈ G. Also we observe that the elements are symmetric about the principal diagonal in the above composition table. Hence commutative law holds good. Therefore (G, .) is an abelian group. Note that G is a finite group of order 4.2. Now since a = ae ⇒ a = eBT0063-Unit-04-TrigonometryUnit 4 TrigonometryStructure4.1 Introduction Objectives4.2 Radian or Circular Measure4.3 Trigonometric Functions4.4 Trigonometrical ratios of angle when is acute4.5 Trigonometrical ratios of certain standard angles4.6 Allied Angles
  71. 71. 4.7 Compound Angles4.8 Multiple and Sub-multiple angle4.9 Summary4.10 Terminal Questions4.11 Answers4.1 IntroductionThis unit of Trigonometry gives us an idea of circular measure. The different Trigonometricfunctions are studied here. Some of the standard angles and their Trigonometric ratios arediscussed in detail. The basic knowledge allied angles and compound angles are explained in asimple manner.Objectives: At the end of the unit you would be able to • understand the concepts of Trigonometrical functions • use allied and compound angles in calculations4.2 Radian or Circular MeasureA radian is the angle subtended at the centre of a circle by an arc equal to the radius of the circle.O is the centre of a circle. A and B are points on the circle such that arc AB = radius OA. Then is called one radian or one circular measure. We write
  72. 72. Radian is a constant angle andConsider a circle whose centre is O and radius r. A and B are points on the circle such that arc AB= OA = r. Join OA, OB and draw OC ⊥ to OA. , right angle and arc AB = r.We know that arc (circumference of the circle) = . In a circle thearcs are proportionated to the angles subtended by them at the centre.1c = 2/π ×1 right angle, which is constantRadian is a constant angleFurther we have, π × 1C = 2 × 1 right angleπC = 2 × 90° = 180°Note:i) πC = 180° mans π radians are equal to 180° Hereafter, this is written as π = 180°. For example and so on. In each of these cases the unit ‘radians’ on the left side is understood.
  73. 73. ii) (nearly) Here π is the real number which is the ratio of circumference of a circle to its diameter.Its approximate value is 22/7.1 radian = (approximately)Clearly 1 radian is < 60°Examples:1) Express 2.53 radians in degrees π radians = 180°2) Express 144° into radians For 180° = π radiansIt is better to remember the following:1) radians2) x radians =Length of an arc of a circle
  74. 74. Consider a circle who centre is O and radius r. A and B are points on the circle such that arc AB= r. P is a point on the circle such that arc PA = s and Hence the length of an arc of a circle is equal to the product of the radius of the circle andthe angle in radians subtended at the centre by the arc. Note:s = the arc length of the circle; r = the radius of the circleθ = angle in radians subtended by s at the centreArea of a sector of a circleThe portion of the circle bounded by two radii, say, OA, OB and the arcAB is called the sector . Consider a circle whose centre is O and radius r. Let AOB be thesector of angle

×