Chapter 1.1 Set Theory PDF

60 1.1.1 Definitions. ID A set is often described in the following two ways. E3 A set is a well-defined class or collection of objects. By a well defined collection we mean that there exists a rule with the help of which it is possible to tell whether a given object belongs or does not belong to the...

60 1.1.1 Definitions. ID A set is often described in the following two ways. E3 A set is a well-defined class or collection of objects. By a well defined collection we mean that there exists a rule with the help of which it is possible to tell whether a given object belongs or does not belong to the given collection. The objects in sets may be anything, numbers, people, mountains, rivers etc. The objects constituting the set are called elements or members of the set. U (1) Roster method or Listing method : In this method a set is described by listing elements, separated by commas, within braces {}. The set of vowels of English alphabet may be described as {a, e, i, o, u}. D YG The set of even natural numbers can be described as {2, 4, 6..........}. Here the dots stand for ‘and so on’. Note :  The order in which the elements are written in a set makes no difference. Thus {a, e, i, o, u} and {e, a, i, o, u} denote the same set. Also the repetition of an element has no effect. For example, {1, 2, 3, 2} is the same set as {1, 2, 3} (2) Set-builder method or Rule method : In this method, a set is described by a characterizing property P(x) of its elements x. In such a case the set is described by {x : P(x) holds} or {x | P(x) holds}, which is read as ‘the set of all x such that P(x) holds’. The symbol ‘|’ or ‘:’ is read as ‘such that’. U The set E of all even natural numbers can be written as E = {x | x is natural number and x = 2n for n  N} E = {x | x  N, x = 2n, n  N} or E = {x  N | x = 2n, n  N} ST or The set A  {0, 1, 4 , 9, 16 ,....} can be written as A  {x 2 | x  Z} Note :  Symbols Symbol Meaning  Implies  Belongs to AB  A is a subset of B Implies and is implied by  Does not belong to s.t. Such that  For every  There exists Meaning iff If and only if & And a|b a is a divisor of b Set of natural numbers E3 N I or Z C Set of complex numbers ID Set of real numbers Set of rational numbers The set of intelligent students in a class is (c) A finite set [AMU 1998] (b) A singleton set (d) Not a well defined collection U (a) A null set Solution: (d) Set of integers R Q Example: 1 60 Symbol Since, intelligency is not defined for students in a class i.e., Not a well defined collection. D YG 1.1.2 Types of Sets. (1) Null set or Empty set: The set which contains no element at all is called the null set. This set is sometimes also called the ‘empty set’ or the ‘void set’. It is denoted by the symbol  or {}. A set which has at least one element is called a non-empty set. Let A  {x : x 2  1  0 and x is real) U Since there is no real number which satisfies the equation x 2  1  0 , therefore the set A is empty set. Note : If A and B are any two empty sets, then x  A iff x  B is satisfied because there is ST no element x in either A or B to which the condition may be applied. Thus A = B. Hence, there is only one empty set and we denote it by . Therefore, article 'the' is used before empty set. (2) Singleton set: A set consisting of a single element is called a singleton set. The set {5} is a singleton set. (3) Finite set: A set is called a finite set if it is either void set or its elements can be listed (counted, labelled) by natural number 1, 2, 3, … and the process of listing terminates at a certain natural number n (say). Cardinal number of a finite set: The number n in the above definition is called the cardinal number or order of a finite set A and is denoted by n(A) or O(A). (4) Infinite set: A set whose elements cannot be listed by the natural numbers 1, 2, 3, …., n, for any natural number n is called an infinite set. (5) Equivalent set: Two finite sets A and B are equivalent if their cardinal numbers are same i.e. n(A) = n(B). Example: A  {1, 3, 5, 7} ; B  {10 , 12 , 14 , 16 } are equivalent sets [ O( A)  O(B)  4 ] 60 (6) Equal set: Two sets A and B are said to be equal iff every element of A is an element of B and also every element of B is an element of A. We write “A = B” if the sets A and B are equal and “A  B” if the sets A and B are not equal. Symbolically, A = B if x  A  x  B. The statement given in the definition of the equality of two sets is also known as the axiom of extension. Example: If A  {2, 3, 5, 6} and B  {6, 5, 3, 2}. Then A  B, because each element of A is an E3 element of B and vice-versa. Note :  Equal sets are always equivalent but equivalent sets may need not be equal set. or ID (7) Universal set : A set that contains all sets in a given context is called the universal set. A set containing of all possible elements which occur in the discussion is called a universal set and is denoted by U. U Thus in any particular discussion, no element can exist out of universal set. It should be noted that universal set is not unique. It may differ in problem to problem. D YG (8) Power set : If S is any set, then the family of all the subsets of S is called the power set of S. The power set of S is denoted by P(S). Symbolically, P(S) = {T : T  S}. Obviously  and S are both elements of P(S). Example : Let S = {a, b, c}, then P(S) = {  , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}. Note :  If A   , then P(A) has one element  ,  n[P( A)]  1 U  Power set of a given set is always non-empty.  If A has n elements, then P(A) has 2n elements. ST  P( )  { } P(P( ))  { , { }}  P[P(P( ))]  { , { }, {{ }}, { , { }}} Hence n{P[P(P( ))]}  4. (9) Subsets (Set inclusion) : Let A and B be two sets. If every element of A is an element of B, then A is called a subset of B. If A is subset of B, we write A  B, which is read as “A is a subset of B” or “A is contained in B”. Thus, A  B  a  A  a  B. Note :  Every set is a subset of itself.  The empty set is a subset of every set.  The total number of subset of a finite set containing n elements is 2n. Proper and improper subsets: If A is a subset of B and A  B, then A is a proper subset of B. We write this as A  B. 60 The null set  is subset of every set and every set is subset of itself, i.e.,   A and A A for every set A. They are called improper subsets of A. Thus every non-empty set has two improper subsets. It should be noted that  has only one subset  which is improper. Thus A has two E3 improper subsets iff it is non-empty. All other subsets of A are called its proper subsets. Thus, if A  B, A  B , A   , then A is said to be proper subset of B. ID Example: Let A  {1, 2}. Then A has  ;{1}, {2}, {1, 2} as its subsets out of which  and {1, 2} are improper and {1} and {2} are proper subsets. Which of the following is the empty set [Karnataka CET 1990] (a) { x : x is a real number and x 2  1  0} (b) {x : x is a real number and x 2  1  0} (c) {x : x is a real number and x 2  9  0} (d) {x : x is a real number and x 2  x  2} U Example: 2 Since x 2  1  0, gives x 2  1  x  i D YG Solution: (b)  x is not real but x is real (given)  No value of x is possible. Example: 3 The set A  { x : x  R, x 2  16 and 2 x  6} equals (a)  Solution: (a) [Karnataka CET 1995] (b) [14, 3, 4] (c) (d) x 2  16  x  4 U 2x  6  x  3 There is no value of x which satisfies both the above equations. Thus, A  . If a set A has n elements, then the total number of subsets of A is ST Example: 4 [Roorkee 1991; Karnataka CET 1992, 2000] (a) n (b) n 2 (c) 2 n (d) 2n Solution: (c) Number of subsets of A  n C 0  n C1 .........  n C n  2 n. Example: 5 Two finite sets have m and n elements. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. The values of m and n are [MNR 1998, 91; UPSEAT 1999, 2000] (a) 7, 6 Solution: (b) (b) 6, 3 (c) 5, 1 Since 2 m  2 n  56  8  7  2 3  7  2 (2 n  n  3 and 2 m n  8  2 3  m n  3  m 3  3  m  6 m n  1)  2  7 3 (d) 8, 7  m  6, n  3. The number of proper subsets of the set {1, 2, 3} is (a) 8 (b) 7 (c) 6 3 26. Solution: (c) Number of proper subsets of the set {1, 2, 3) = 2 Example: 7 If X  {8 n  7n  1 : n  N } and Y  {49 (n  1) : n  N }, then (a) X  Y Solution: (a) (c) X  Y (b) Y  X (d) 5 60 Example: 6 (d) None of these Since 8 n  7n  1  (7  1)n  7n  1 = 7 n  n C1 7 n 1  n C 2 7 n  2 .....  n C n 1 7  n C n  7 n  1 ( n C 0  n C n , n C1  n C n 1 etc.) E3 = n C 2 7 2  n C 3 7 3 .....  n C n 7 n = 49 [n C 2  n C 3 (7) ......  n C n 7 n  2 ]  8 n  7n  1 is a multiple of 49 for n  2. ID For n  1 , 8 n  7n  1  8  7  1  0 ; For n  2, 8 n  7n  1  64  14  1  49  8 n  7n  1 is a multiple of 49 for all n  N.  X contains elements which are multiples of 49 and clearly Y contains all multiplies of 49. 1.1.3 Venn-Euler Diagrams. U  X Y. diagrams. D YG The combination of rectangles and circles are called Venn-Euler diagrams or simply Venn- In venn-diagrams the universal set U is represented by points within a rectangle and its subsets are represented by points in closed U curves (usually circles) within the rectangle. If a set A is a subset of a A set B, then the circle representing A is drawn inside the circle U representing B. If A and B are not equal but they have some common elements, then to represent A and B we draw two intersecting circles. Two disjoints sets are ST represented by two non-intersecting circles. 1.1.4 Operations on Sets. (1) Union of sets : Let A and B be two sets. The union of A and B is the set of all elements which are in set A or in B. We denote the union of A and B by A  B U which is usually read as “A union B”. AB symbolically, A  B  {x : x  A or x  B}. A B It should be noted here that we take standard mathematical usage of “or”. When we say that x  A or x B we do not exclude the possibility that x is a member of both A and B. Note : n  If A1 , A 2 ,......, A n is a finite family of sets, then their union is denoted by  Ai or i 1 A1  A 2  A 3......  A n. (2) Intersection of sets : Let A and B be two sets. The intersection of A and B is the set of all 60 those elements that belong to both A and B. U The intersection of A and B is denoted by A  B (read as “A intersection B”) A B Thus, A  B = {x : x  A and x  B}. A B E3 Clearly, x  A  B  x  A and x  B. In fig. the shaded region represents A  B. Evidently A  B  A, A  B  B. :  If A1 , A 2 , A 3......., A n is a finite family of sets, then their intersection is denoted by n  i1 A i ID Note or A1  A 2  A 3 ........  A n. U (3) Disjoint sets : Two sets A and B are said to be disjoint, if A  B = . If A  B  , then A and B are said to be non-intersecting or non-overlapping sets. D YG In other words, if A and B have no element in common, then A and B are called disjoint sets. Example : Sets {1, 2}; {3, 4} are disjoint sets. (4) Difference of sets : Let A and B be two sets. The difference of A and B written as A – B, is the set of all those elements of A which do not U U belong to B. Thus, A – B = {x : x  A and x  B} or A – B = {x  A : x  B} A–B A B–A B A B U Clearly, x  A – B  x  A and x  B. In fig. the shaded part represents A – B. ST Similarly, the difference B  A is the set of all those elements of B that do not belong to A i.e. B  A  {x  B : x  A} Example: Consider the sets A  {1, 2, 3} and B  {3, 4 , 5} , then A  B  {1, 2}; B  A  {4 , 5} As another example, R  Q is the set of all irrational numbers. (5) Symmetric difference of two sets: Let A and B be two sets. The symmetric difference of sets A and B is the set ( A  B)  (B  A) and is denoted by A  B. Thus, A  B = ( A  B)  (B  A)  {x : x  A  B} (6) Complement of a set : Let U be the universal set and let A be a set such that A  U. Then, the complement of A with respect to U is denoted by A or Ac or U C(A) or U – A and is defined the set of all those elements of U which are A A not in A. Thus, A = {x  U : x  A}. Clearly, x  A  x  A Example: Consider U  {1, 2,......,10 } and A  {1, 3, 5, 7, 9}. Then A   {2, 4 , 6, 8, 10 } Given the sets A  {1, 2, 3}, B  {3,4} , C = {4, 5, 6}, then A  (B  C) is CEE 1996] (a) {3} (b) {1, 2, 3, 4} B  C  {4 } ,  A  (B  C) = {1, 2, 3, 4}. Example: 9 If A  B , then A  B is equal to (b) B  A (a) A (c) {1, 2, 4, 5} (c) B Solution: (c) Since A  B  A  B  B. Example: 10 If A and B are any two sets, then A  ( A  B) is equal to Solution: (a) Example: 11 (c) A c (b) B ID (a) A A  B  A. Hence A  ( A  B)  A. (d) None of these (d) B c If A and B are two given sets, then A  ( A  B)c is equal to U 1998; Kurukshetra CEE 1999] (a) A (c)  (b) B A  ( A  B)c  A  ( A c  B c ) = ( A  Ac )  ( A  Bc ) =   ( A  B c )  A  B c. Example: 12 If Na  {an : n  N}, then N 3  N 4  D YG Solution: (d) (a) N 7 Solution: (b) (d) {1, 2, 3, 4, 5, 6} E3 Solution: (b) [MNR 1988; Kurukshetra 60 Example: 8 (b) N12 (c) N 3 [AMU (d) A  B c (d) N 4 N 3  N 4  {3, 6, 9,12,15......}  {4, 8,12,16, 20,.....} = {12, 24, 36......} = N12 Trick: N 3  N 4  N12 Example: 13 [  3, 4 are relatively prime numbers] If aN  {ax : x  N } and bN  cN  dN , where b, c  N are relatively prime, then (b) c  bd (c) b  cd U (a) d  bc Solution: (a) (d) None of these bN  the set of positive integral multiples of b, cN = the set of positive integral multiplies of c. ST  bN  cN = the set of positive integral multiples of bc = b  N [ b, c are prime]  d  bc. Example: 14 If the sets A and B are defined as A  {(x , y ) : y  1 , 0  x  R} x B  {(x , y) : y   x , x  R} , then (a) A  B  A Solution: (c) Since y  1 (b) A  B  B , y   x meet when  x  x (c) A  B   (d) None of these 1  x 2  1 , which does not give any real value of x x Hence A  B  . Example: 15 Let A  [x : x  R,| x |  1]; B  [x : x  R,| x  1 |  1] and A  B  R  D, then the set D is (b) [ x : 1  x  2] (a) [ x : 1  x  2] Solution: (b) (c) [ x : 1  x  2] (d) None of these A  [x : x  R,  1  x  1] B  [ x : x  R : x  1  1 or x  1  1] = [ x : x  R : x  0 or x  2]  AB  R D Example: 16 60 Where D = [x : x  R,1  x  2] If the sets A and B are defined as A  {(x , y ) : y  e x , x  R} B  {(x, y) : y  x, x  R}, then Solution: (c) (c) A  B   (b) A  B Since, y  e x and y  x do not meet for any x  R  AB . If X  {4 n  3n  1 : n  N } and Y  {9(n  1) : n  N }, then X  Y is equal to (a) X (c) N (d) A  B  A [Karnataka CET 1997] (d) None of these Since, 4 n  3n  1  (3  1)n  3n  1 = 3 n  n C1 3 n 1  n C 2 3 n  2 .....  n Cn 1 3  n Cn  3n  1 U Solution: (b) (b) Y ID Example: 17 E3 (a) B  A [UPSEAT 1994, 2002] = n C 2 3 2  n C3.3 3 .....  n Cn 3 n ( n C0 n Cn , n C1 n Cn 1 etc.) D YG = 9[n C 2  n C3 (3) .....  n Cn 3 n 1 ]  4 n  3 n  1 is a multiple of 9 for n  2. For n  1, 4 n  3 n  1 = 4  3  1  0 , For n  2, 4 n  3 n  1 = 16  6  1  9  4 n  3 n  1 is a multiple of 9 for all n  N  X contains elements which are multiples of 9 and clearly Y contains all multiples of 9.  X  Y ,  X Y  Y. U 1.1.5 Some Important Results on Number of Elements in Sets. If A, B and C are finite sets and U be the finite universal set, then ST (1) n(A  B) = n(A) + n(B) – n(A  B) (2) n(A  B) = n(A) + n(B)  A, B are disjoint non-void sets. (3) n(A – B) = n(A) – n(A  B) i.e. n(A – B) + n(A  B) = n(A) (4) n(A  B) = Number of elements which belong to exactly one of A or B = n((A – B)  (B – A)) = n (A – B) + n(B – A) [ (A – B) and (B – A) are disjoint] = n(A) – n(A  B) + n(B) – n(A  B) = n(A) + n(B) – 2n(A  B) (5) n(A  B  C) = n(A) + n(B) + n(C) – n(A  B) – n(B  C) – n(A  C) + n(A  B  C) (6) n (Number of elements in exactly two of the sets A, B, C) = n(A  B) + n(B  C) + n(C  A) – 3n(ABC) (7) n(Number of elements in exactly one of the sets A, B, C) = n(A) + n(B) + n(C) – 2n(A  B) – 2n(B  C) – 2n(A C) + 3n(A  B  C) (8) n(A  B) = n(A  B) = n(U) – n(A  B) (9) n(A  B) = n(A  B) = n(U) – n(A  B) Sets A and B have 3 and 6 elements respectively. What can be the minimum number of elements in A 60 Example: 18 B [MNR 1987; Karnataka CET 1996] Solution: (b) (b) 6 (c) 9 n(A  B) = n(A) + n(B) – n(A  B) = 3  6  n( A  B) Since maximum number of elements in A  B  3 (d) 18 E3 (a) 3  Minimum number of elements in A  B  9  3  6. If A and B are two sets such that n( A)  70 , n(B)  60 and n( A  B)  110 , then n( A  B) is equal to (a) 240 Solution: (d) (b) 50 n( A  B)  n( A)  n(B)  n( A  B)  110 = 70 + 60 – n(A  B) Example: 20 U  n( A  B)  130  110  20. Let n(U)  700 , n( A)  200 , n(B)  300 and n( A  B)  100 , then n( A c  B c )  300. Example: 21 2002] (b) 600 D YG (a) 400 Solution: (c) (c) 300 c If A  [(x , y ) : x 2  y 2  25 ] and B = [(x , y ) : x 2  9 y 2  144 ] , then A  B contains (b) Three points (c) Two points A = Set of all values (x, y) : x 2  y 2  25  5 2 2 2 (d) 200 [AMU 1996; 2 U (12) Clearly, A  B consists of four points. Pb. CET (d) Four points x 2 x y x y B= + 1.   1 i.e., 2 144 16 (12 ) (4 )2 2  y 2 (4 ) 2 1 x2 + y2 = 52 ST In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buy A only is (a) 3100 Solution: (b) [Kurukshetra CEE 1999] c 2 Example: 22 (d) 20 n( A  B ) = n[(A  B)c] = n(U)  n( A  B) = n(U)  [n( A)  n(B)  n( A  B)] = 700 – [200 + 300 – 100] = (a) One point Solution: (d) (c) 40 ID Example: 19 (b) 3300 (c) 2900 (d) 1400 n(A) = 40% of 10,000 = 4,000 n(B) = 20% of 10,000 = 2,000 n(C) = 10% of 10,000 = 1,000 n (A  B) = 5% of 10,000 = 500, n (B  C) = 3% of 10,000 = 300 n(C  A) = 4% of 10,000 = 400, n(A  B  C) = 2% of 10,000 = 200 We want to find n(A  Bc  Cc) = n[A  (B  C)c] = n(A) – n[A  (B  C)] = n(A) – n[(A  B)  (A  C)] = n(A) – [n(A  B) + n(A  C) – n(A  B  C)] = 4000 – [500 + 400 – 200] = 4000 – 700 = 3300. Example: 23 In a city 20 percent of the population travels by car, 50 percent travels by bus and 10 percent travels by both car and bus. Then persons travelling by car or bus is (a) 80 percent Solution: (c) (b) 40 percent (c) 60 percent (d) 70 percent n(C) = 20, n(B) = 50, n(C  B) = 10 Now, n(C  B) = n(C) + n(B) – n(C  B) = 20 + 50 – 10 = 60. Suppose A1 , A2 , A3 ,........, A30 are thirty sets each having 5 elements and B1 , B2 ,......., Bn are n sets each with 3 elements. Let exactly 9 of the B 'j s 30 n i 1 j 1  Ai   B j = S and each elements of S belongs to exactly 10 of the Ai' s and. Then n is equal to (a) 15 (b) 3 (c) 45   1 (5  30 )  15 O(S) = O   Ai    i 1  10 30 Solution: (c) E3 Example: 24 60 Hence, required number of persons = 60%. ID Since, element in the union S belongs to 10 of Ai' s (d) None of these  n  3n n n  ,  Also, O(S) = O   B j    15  n  45. 9 3 3  j 1  In a class of 55 students, the number of students studying different subjects are 23 in Mathematics, 24 in Physics, 19 in Chemistry, 12 in Mathematics and Physics, 9 in Mathematics and Chemistry, 7 in Physics and Chemistry and 4 in all the three subjects. The number of students who have taken exactly one subject is [UPSEAT 1990] Solution: (d) (b) 9 D YG (a) 6 U Example: 25 (c) 7 (d) All of these n(M) = 23, n(P) = 24, n(C)= 19 n(M  P) = 12, n(M  C)= 9, n(P  C)= 7 n(M  P  C) = 4 We have to find n(M  P  C), n(P  M   C ), n ( C  M   P ) Now n (M  P  C) = n[M  (P  C)] = n(M)– n(M  (P  C))  n(M )  n[(M  P)  (M  C)] = n(M) – n(M  P)– n(M  C) + n(M  P  C) = 23 –12 – 9 + 4 = 27 –21 = 6 U n(P  M  C) = n[P  (M  C)] = n(P)– n[P  (M  C)] = n(P)  n[(P  M )  (P  C )] = n(P) – n(P  M) – n(P  C) + n(P  M  C) ST = 24 – 12 – 7 + 4 = 9 n(C  M  P) = n(C) – n(C  P) – n(C  M)+ n(C  P  M) = 19 – 7 – 9 + 4 = 23 – 16 = 7 Hence (d) is the correct answer. 1.1.6 Laws of Algebra of Sets. (1) Idempotent laws : For any set A, we have (i) A  A = A (ii) A  A = A (2) Identity laws : For any set A, we have (i) A   = A (ii) A  U = A i.e.  and U are identity elements for union and intersection respectively. (3) Commutative laws : For any two sets A and B, we have (i) A  B = B  A (ii) A  B = B  A (iii) AB  BA i.e. union, intersection and symmetric difference of two sets are commutative. (iv) A  B  B  A (iv) A  B  B  A (4) Associative laws : If A, B and C are any three sets, then (i) (A  B)  C = A  (B  C) (ii) A  (B  C) = (A  B)  C 60 i.e., difference and cartesian product of two sets are not commutative (iii) ( AB)C  A(BC) (iv) ( A  B)  C  A  (B  C) E3 i.e., union, intersection and symmetric difference of two sets are associative. (v) ( A  B)  C  A  (B  C) i.e., difference and cartesian product of two sets are not associative. ID (5) Distributive law : If A, B and C are any three sets, then (i) A  (B  C) = (A  B)  (A  C) (ii) A  (B  C) = (A  B)  (A  C) i.e. union and intersection are distributive over intersection and union respectively. (iv) A  (B  C)  ( A  B)  ( A  C) U (iii) A  (B  C)  ( A  B)  ( A  C) A  (B  C)  ( A  B)  ( A  C) (v) D YG (6) De-Morgan’s law : If A and B are any two sets, then (i) (A  B) = A  B (ii) (A  B) = A  B (iii) A  (B  C)  ( A  B)  ( A  C) Note :  (iv) A  (B  C)  ( A  B)  ( A  C) Theorem 1: If A and B are any two sets, then (i) A – B = A  B (ii) B – A = B  A A–B=AAB= (iii) (v) (A – B)  B =  (iv) (A – B)  B = A  B U (vi) A  B  B  A ST (viii) (A – B)  (B – A) = (A  B) – (A  B)  Theorem 2 : If A, B and C are any three sets, then (i) A – (B  C) = (A – B)  (A – C) (ii) A – (B  C) = (A – B)  (A – C) (iii) A  (B – C) = (A  B) – (A  C) Example: 26 (iv) A  (B  C) = (A  B)  (A  C) If A, B and C are any three sets, then A × (B  C) is equal to (a) (A × B)  (A × C) (b) (A × B)  (A × C) (c) (A  B) × (A  C) Solution: (b) A × (B  C) = (A × B)  (A × C). It is distributive law. Example: 27 If A, B and C are any three sets, then A × (B  C) is equal to (a) (A × B)  (A × C) (b) (A  B) × (A  C) (c) (A × B)  (A × C) Solution: (a) It is distributive law. Example: 28 If A, B and C are any three sets, then A – (B  C) is equal to (d) (A  B) × (A  C) (d) None of these (a) (A – B)  (A – C) (b) (A – B)  (A – C) (c) (A – B)  C (d) (A – B)  C Solution: (b) It is De' Morgan law. Example: 29 If A  [ x : x is a multiple of 3] and B  [ x : x is a multiple of 5], then A – B is ( A means complement of A) [AMU 1998] (b) A  B (c) A  B Solution: (b) A – B = A  Bc = A  B. Example: 30 If A, B and C are non-empty sets, then (A – B)  (B – A) equals [AMU 1992, 1998; DCE 1998] (A – B)  (B – A) = (A  B) – (A  B). A–B A B B–A 1.1.7 Cartesian Product of Sets. (d) (A  B)  (A  B) ID B A (c) (A  B) – (A  B) U Solution: (c) (b) A – (A  B) E3 (a) (A  B) – B (d) A  B 60 (a) A  B D YG Cartesian product of sets : Let A and B be any two non-empty sets. The set of all ordered pairs (a, b) such that a  A and b  B is called the cartesian product of the sets A and B and is denoted by A  B. Thus, A × B = [(a, b) : a  A and b  B] If A =  or B = , then we define A × B = . Example : Let A = {a, b, c} and B = {p, q}. Then A × B = {(a, p), (a, q), (b, p), (b, q), (c, p), (c, q)} U Also B × A = {(p, a), (p, b), (p, c), (q, a), (q, b), (q, c)} Important theorems on cartesian product of sets : ST Theorem 1 : For any three sets A, B, C (i) A × (B  C) =(A × B)  (A × C) (ii) A × (B  C) =(A × B)  (A × C) Theorem 2 : For any three sets A, B, C A × (B – C) = (A × B) – (A × C) Theorem 3 : If A and B are any two non-empty sets, then A×B=B×AA=B Theorem 4 : If A  B, then A × A  (A × B)  (B × A) Theorem 5 : If A  B, then A × C  B × C for any set C. Theorem 6 : If A  B and C  D, then A × C  B × D Theorem 7 : For any sets A, B, C, D (A × B)  (C  D) = (A  C) × (B  D) Theorem 8 : For any three sets A, B, C (i) A × (B  C) = (A × B)  (A × C) (ii) A × (B  C) = (A × B)  (A × C) B × A have n2 elements in common. If A = {0, 1), and B = {1, 0}, then A × B is equal to (a) {0, 1, 1, 0} {(0,1),(0,0),(1,1),(1,0)} Solution: (d) (b) {(0, 1), (1, 0)} By the definition of cartesian product of sets Clearly, A × B = {(0, 1), (0, 0), (1, 1), (1, 0)}. If A  {2, 4, 5}, B  {7, 8, 9}, then n( A  B) is equal to (a) 6 Solution: (b) (b) 9 (d) (c) 3 (d) 0 A × B = {(2, 7), (2, 8), (2, 9), (4, 7), (4, 8), (4, 9), (5, 7), (5, 8), (5, 9)} U n(A × B) = n(A). n(B) = 3 × 3 = 9. Example: 33 ID Example: 32 (c) {0, 0} E3 Example: 31 60 Theorem 9 : Let A and B two non-empty sets having n elements in common, then A × B and If the set A has p elements, B has q elements, then the number of elements in A × B is (b) p  q  1 D YG (a) p  q (c) pq (d) p 2 Solution: (c) n( A  B)  pq. Example: 34 If A  {a, b}, B  {c, d}, C  {d, e}, then {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e)} is equal to [AMU 2002] (a) A  (B  C) Solution: (c) (b) A  (B  C) (c) A × (B  C) 1999; Him. CET (d) A × (B  C) B  C = {c, d}  (d, e} = {c, d, e}  A × (B  C) = {a, b} × {c, d, e} = {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e)}. If A  { x : x 2  5 x  6  0}, B  {2, 4 }, C  {4 , 5}, then A  (B  C) is U Example: 35 (a) {(2, 4), (3, 4)} (c) {(2, 4), (3, 4), (4, 4)} (d){(2,2), (3,3), (4,4), (5,5)} Clearly, A = {2, 3}, B = {2, 4}, C = {4, 5} ST Solution: (a) (b) {(4, 2), (4, 3)} [Kerala (Engg.) 2002] B  C = {4}  A × (B  C) = {(2, 4), (3, 4)}. Example: 36 If P, Q and R are subsets of a set A, then R × (Pc  Qc)c = (a) (R × P)  (R × Q) (b) (R  Q)  (R  P) (c) (R  P)  (R  Q) Solution: (a, b) R  (P c  Q c )c  R  [(P c )c  (Q c )c ] = R  (P  Q)  (R  P)  (R  Q) = (R  Q)  (R  P) (d) None of these