Set, Relations and Functions Solved Problems

Page # 7 SET, RELATIONS AND FUNCTIONS SOLVED PROBLEMS Ex.1 Sol. If a set A = {a, b, c} then find the number Case-I of subsets of the set A and also mention When n(A  B) is minimum, i.e., the set of all the subsets of A. n(A  B) = 0 Since n(A) = 3 This is possible only when A  B = . In this  number of subsets of A is 23 case, = 8 and set of all those subsets is P(A) named n(A  B) = n(A) + n(B) – 0 = n(A) + as power set n(B) = 3 + 6 = 9. P(A) : {, {a}, {b}, So, maximum number of elements in A  B is 9. {c}, Case-II {a,b},{b,c},{a,c},{a,b,c}} Ex.2 Show that n {P[P()]} = 4 Sol. We have P() = {}  P(P()) = {, {}}  P[P(P())] = { , {}, {{}}, When n(A  B) is maximum. This is possible only when A  B. In this case, n(A  B) = 3  n(A  B) = n(A) + n(B) – n(A  B) = (3 + 6 – 3) = 6 {, {}}}. So, minimum number of elements in A  B is Hence, n{P[P()]} = 4 Ex.3 If A = {x : x = 2n + 1, n  Z and 6. Ex.6 B = {x : x = 2n, n  Z}, then find A  B. Sol. 11, 13} then find A  B = {x : x is an odd integer}  {x : x is an even integer} = {x : x is an Sol. If A = {x : x = 3n, n  Z} and Ex.7 If the number of elements in A is m and B = {x : x = 4n, n  Z} then find number of element in B is n then find A  B. (i) The number of elements in the power We have, set of A × B. Z (ii) number of relation defined from A to B Sol. (i) Since n(A) = m; n(B) = n then n(A × B )= mn  x is a multiple of 3 and x is a multiple of 4 So number of subsets of A × B = 2mn  x is a multiple of 3 and 4 both   x is a multiple of 12  x = 12n, (ii) number of relation defined from A to B = n  Z 2mn A to set B will be subset of A × B If A and B be two sets containing 3 and 6 elements respectively, what can be the A  B? Find also, the maximum number of We have, maximum according as n(A  B) is maximum or minimum respectively. A × B is largest possible relation  no. of relation from A  B = no. of subsets of set (A × B) Ex.8 Let A and B be two non-empty sets having elements in common, then prove that A × n(A  B) = n(A) + n(B) – n(A  B). This shows that n(A  B) is minimum or  A  B minimum number of elements in elements in A  B. n (P(A × B)) = 2mn Any relation which can be defined from set Hence A  B = {x : x = 12n, n  Z} Sol. A – B = {2, 4, 6} & B – A = {9, 11, 13) x  A  B  x = 3n, n  Z and x = 4n, n  Ex.5 A – B and B – A.. Sol. integer} = Z Ex.4 If A = {2, 3, 4, 5, 6, 7} and B = {3, 5, 7, 9, B and B × A have n2 elements in common. Sol. We have (A × B)  (C × D) = (A  C)× (B  D) Page # 8 SET, RELATIONS AND FUNCTIONS On replacing C by B and D by A, we get  Ex.12 If x = {a, b, c} and y = (a, b, c, d, e} then (A × B)  (B × A) = (A  B) × which of the following are transitive relation. (B  A) (a) R1 = { } It is given that AB has n elements so (A  B) × (B  A) has n2 (b) R2 = {(a, a)} elements (c) R3 = {(a, a}.(c, d)} But (A × B)  (B × A) (d) R4 = = (A  B) × (B  A)  Ex.9 (A × B)  (B × A) has n2 elements {(a, b), (b, c)(a, c),(a, a), (c, a)} Sol. R1 is transitive relation because it is null relation. in common. (b) Let R be the relation on the set N of natural singleton relations are transitive. numbers defined by R : {(x, y)} : x + 3y = 12 x  N, y  N} Find (i) R Sol. (a) Hence A × B and B × A have n2 elements (i) (ii) Domain of R (iii) Range of R Putting y = 1, 2, 3, we get R3 is transitive relation (d) R4 is also transitive relation numbers defined by xRy  x divides y  for all x, y  N. Sol. This relation is an antisymmetric relation on set N. Since for any two numbers a, b  x = 9, 6, 3 respectively N. a | b and b | a For y = 4, we get x = 0  N. Also for andb R a y > 4, x  N  (c) Ex.13 Let R be a relation on the set N of natural We have, x + 3y =12  x = 12 – 3y R2 is transitive relation because all  a = b, i.e., a R b  a = b. It should be noted that this relation is not R = {(9, 1), (6, 2), (3, 3)} antisymmetric on the set Z of integers, (ii) Domain of R = {9, 6, 3} because we find that for any non zero integer (iii) Range of R = {1, 2, 3} a a R (a) and (–a) R a, but a  – a. Ex.10 If X = {x1, x2, x3} and y = (x1, x2, x3,x4,x5} Ex.14 Prove that the relation R on the set Z of all then find which is a reflexive relation of the integers numbers defined by (x, y)  R  x – following : y is divisible by n is an equivalence relation on (a) R1 : {(x1, x1), (x2, x2) (b) R1 : {(x1, x1), (x2, x2), (x3, x3) Sol. We observe the following properties (c) R3 : {(x1, x1), (x2, x2),(x3, x3),(x1, x3),(x2, Reflexivity : x4) For any a  N, we have (d) R3 : {(x1, x1), (x2, x2),(x3, x3),(x4, x4) a – a = 0 × n  a – a is divisible by (a) non-reflexive because (x3, x3)  R1 n  (a, a)  R (b) Reflexive Thus (a, a) R for all Z. so, R is reflexive on (c) Reflexive (d) non-reflexive because x4  X Ex.11 If x = {a, b, c} and y = {a, b, c, d, e, f} then find which of the following relation is symmetric relation : R1 : { } i.e. void relation R2 : {(a, b)} R3 : {(a, b), (b, a)(a, c)(c, a)(a, a)} Sol. Z. Sol. R1 is symmetric relation because it has no Z. Symmetry : Let (a, b)  R. Then (a, b)  R  (a – b)is divisible by n  (a – b) = np for some p  Z  b – a = n(–p)  b – a is divisble by n  (b, a)  R element in it. Thus (a, b)  R  (b, a)  R for all a, b,  R2 is not symmetric because (b, a)  R2 & Z. R3 is symmetric. So R is symmetric on Z. Page # 9 SET, RELATIONS AND FUNCTIONS Transitivity : Ex.16 Let A be the set of first ten natural numbers Let a, b, c  Z such that (a, b)  R and and let R be a relation on A defined by (x, (b, c)  R. Then y)  R  x + 2y = 10 i.e., R = {(x, y) : (a, b)  R  (a – b) is divisible by x  A, y  A and x + 2y = 10}. Express n  a – b = np for some p  Z R and R–1 as sets of ordered pairs. Determine (b, c)  R  (b – c) is divisible by also : n  b – c = np for some q  Z (i) Domains of R and R–1  (a, b)  R and (b, c)  R  a – b = np (ii) Range of R and R–1 and b – c = nq  (a – b) + (b – c) = np + nq  a – c = n(p Sol. We have (x, y)  R  x + 2y = 10  y = + q) x, y  A  where A = {1,2,3,4,5,6,7,8,9,10} a – c is divisible by n 10  x , 2  (a, c)  R Thus (a, b)  R and (b, c)  R  (a, c)  R for all a, b, c  Z. So R is 10  1 9   A 2 2 transitive relation on Z. This shows that 1 is not related to any Thus, R being reflexive, symmetric and element in A. Similarly we can observe that transitive 3, 5, 7, 9 and 10 are not is an equivalence relation on Z. Ex.15 Let a relation R1 on the set R of real numbers Further we find that all a, b  R. Show that R1 is reflexive and for x = 2, y = We observe the following properties : Let a be an arbitrary element of R. Then a  R  1 + a . a = 1 + a2 > 0  (a, a)  R1 Thus (a, a)  R1 for all a  R. So R1 is 10  2 = 4  A 2  (2, 4)  R for x = 4, y = Reflexivity : related to any element of a under the defined relation. be defined as (a, b)  R1  1 + ab > 0 for symmetric but not transitive. Sol. Now, x = 1  y = 10  4 = 3  A 2  (4, 3) R for x = 6, y = 10  6 = 2  A 2  (6, 2)  R 10  8 = 1  A 2 reflexive on R. for x = 8, y = Symmetry :  (8, 1)  R Let (a, b)  R. Then Thus R = {(2, 4), (4, 3), (6, 2), (8, 1)} (a, b)  R1  1 + ab > 0   1 + ba > 0 R–1 = {(4, 2), (3, 4), (2, 6), (1, 8)} Clearly, Dom (R) = {2, 4, 6, 8} = Range  (b, a)  R1 Thus (a, b)  R1  (b, a)  R1 for all a, b  R So R1 is symmetric on R Transitive : We observe that (1, 1/2)  R1 and (1/2, –1)  R1 but (1,–1)  R1 because 1 + 1 × (–1) = 0  / 0. So R1 is not transitive on R. 1 (R–1) and Range (R) = {4, 3, 2, 1} = Dom (R–1)