SPECTRUM DISCRETE MATHE 31 30 AEMAN RELATION AND FUNCTION that AN(BC) = (A N B)(A NC) SETS, Eumple 6. Show R.H.S. (A OB)(A nC) =(A NB)n(ANC=(AN B)n (A° uc) SECTIONII Sol.  [(A B)n A"]u[(AN B)nc]=[A NA')OB] u[A n (B nc RELATIONS (p n B)U [A N(BC)]=¢U[An(B)] 1.12. Ordered Pair An(BC) denotes an ordered pair A and bEB, then (a, b)  L.H.S. and B be two given nonempty sets. If a E Let A is b. whose first component is a and the second component EXERCISE1.3 The ordered pair (a, b) and (b, a) are different unless a =b. Also two ordered pairs (a, b) and (c, d) if a = c and b=d. identities: are only equal if and 1 Verify the following whereas the sets {5, 2} and {2, 5} are equal. AUBnC)=(A UB) N (A UC) () AnBUC)=(A N B) U (A NC) Note 1. Ordered pairs (2, 5) and (5, 2) are not equal also and b may not be distinct i.e., (a, a) and (b, b) are where, A. B, C are three sets defined by Note 2. In the definition of an ordered pair, a ordered pairs. 7 {1.2,4,5 }, B {2,3,5, 6 }, C={4,5, 6, = A 1.13. Cartesian Product of Sets 2. are two sets, then find X n (XU Y). If X and Y Cartesian Product of Two Sets. If A and B are two nonempty sets, then the set of all ordered pairs (a, b) and is denoted by A x B.. Showthat )ACAUB (i) ANBCA. 3. where a E A and b E B is called the cartesian product of sets A and B 4. Prove the following: In symbols, A xB= {(a b): a E A, bE B} (i) ANBCB (i) BCAANB=B ( BCAUB Note 1. A x B and B x A are different sets if A * B. () C and B C D AC AUBC CUD (v)BCC>ANBCAnC 2. AxB= when one orboth of A, B are empty. (v) A =B ACB and BCA. Cartesian Product of Three Sets. The set of all ordered triplets (a, b, c), of elements 5. For any two sets A and B, prove that A nB=¢ * ACB. a EA, b E B, c E C is called the Cartesian product of the three sets A, B, C and is denoted by AXBxC. 6. Prove that in) A (A  B) = A OB We know that ) An(A'UB) =ANB 7. Prove that A\ B' = B\A. A=A XA ={«,y):x,y E A} A=AxA XA={x, y, z):x, y, zE A} 8. Prove the following : (G) An(B A) = ¢ ii) AU (B A) = AUB 1.14. Graphical Representation of A xB ((AB)nB=o Draw two perpendicular lines X'OX and Y'OY intersecting at (i) AB =A  (A NB) ) (AC)U(BC)=(AUB)C 0, where X'OX is horizontal and Y'OY is vertical. Now on  (a, b) (v) (A B)C=A  (B U C) = (A  B) n A C) (vii) IfA C B, then B (BA/" horizontal line X'OX represent the elements of A and on vertical 9. Prove that A n (B\C) (A N line Y'OY, represent the elements of B. = B)MAN C). Now if a E A, bB, draw a vertical line through a and a E ANSWERSs horizontal line through b. The point where they meet represents the ordered pair (a, b), The set of all such obtained 2. points graphically represents A x B. SPECTRUM DISCRETE MATHEM, 33 32 Note 1. If A the of all numbers, then A consists of all points in a line. A x A will consist al SETS, RELATION AND FUNC is set of find G x H and H xG. {5, 4, 2}, {7, 8} and H = tn the plane. IfG = Example 2. Note 2. The ordered pair (a, b} represents a point whose coordinates are (a, 6). {7, 8), H {5,4, 2} (8, 2)} (7, 4), (7, 2). (8, 5),(8, 4), = Sol. G Note 3. Let m(A) denote the number of elements of A. ={(7, 5), GxH (7,8} x {5,4,2 7), (2, 8) 8), (4, 7), (4, 8), (2, Then n (A x B) = n (A) X n (B). HxG (5,4,2} x {7, 8} {(5, 7), (5, and B. (6, )}, find A 1.15. Prove tha Example 3. IfA X B = {(a, x), (a. y). (6, x), A x (BU C) (A x B) U(A x C) () A x (B nC) =(A x B) n(A x C) Sol. A x B= {(a, x). (a, y), (b, x), (6, y)} Proof. ) LH.S. = A x (B UC) A set of first elements= {a, b}  {a.):rE yE(B Uc)} {a,):x*E Aand A and = (y E B ory E C)} and B s e t of second elements = {r, y} found (1, 0) and (0, 1). Find among which are elements A x A has a = {r.1): («E A and y E B) or (r E A andyE C)} Example 4. The cartesian product elements of A x A. the set A and the remaining = {a. ): ay) E (A x B) or (z,) E (A x C)} XA Sol. Since (1, 0) E A X A and (0, 1) EA = {a y):(a)E (A X B) U(A x C)} =(A x B) U (A x C)  1 , 0 EA and 0, 1, E A 1, 0, 1 E A. = R.H.S elements Now AX A has 9 elementsA has 3 Ax (BUC)= (A x B) U(A x C) A={1, 0, 1} () L.H.S. A x (BnC) (0, 0), (1, 1). (1,0), (1, 1) Remaining elements of A X A and (1, 1), (1, 1), (0, 1), a. y):a E A and y E(B nc)} = {«,y):r¬A and (oEBand y EC} Find 5. Let A {1,2, 3}, B {3,4} and C {4, 5,6}. = = Example (BUC) (iv) (A x B) U (A x C). a ,y): (x EA andy E B) and (r E A andy E C)} ()A x(Bnc) (in) (A x B)n(A x C) Gi) A x Sol.Here A= {1,2,3, B= {3,4}, C={4, 5,6} =r. y): (a ) E (A X B) and (x. y) E (A x C)} BUC (3,4} U {4,5, 6} = {3,4,5,6} (r. y):(x,y) E (A X B)n(A x C)} =(A x B) n(A x C) BnC 3,4} {4,5,6} = {4} R.H.S AXB {1,2, 3} x {3,4) ((1, 3), (1,4), (2, 3), (2,4), (3,3), (3,4)} Ax (BnC)= (A B) n(A x x C) AxC {1,2,3} x {4,5,6} {(1,4,(1, 5),(1,6),(2, 4), (2, 5). 2.6,,4,3,5),3.6} A x (Bnc)={1,2,3} x {4)  {(1, 4),.(2, 4), (3, 4)} ILLUSTRATIVEEXAMPLES (in(A x B) n(A x C) = {(1, 3), (1, 4), (2, 3), (2,4), (3, 3), (3, 4)} Example 1. If (r+ 1,y2)=(3, 1). find the values of x Sol. Here and y. n1,4), (0,5),(1,6),(2,4), 2,5),(2, 6),(3,4), (3.5),(3, 6)} «+ 1,y2) =(3, 1) {(1,4), (2, 4), (3, 4)} by definition of equal ordered pairs, (in A x (BUC) = {1,2,3} x {3,4, 5,6} 1=3 y21 {(1,3), (1,4), (1, 5), (1.6), (2, 3), (2,4),(2, 5), (2,6),(3, 3), 3, 4), (3,5), (3,6) () (AxB) U(A x C)= {(1, 3), (1,4),(2, 3), (2,4), (3, 3), (3, 4)} From (1). r=312 From (2), U1, 4), (1, 5), (1, 6). (2, 4), (2, 5), (2, 6), (3,4), (3, 5), (3,6)} y=1+2=3 x2, y=3. {(1.3), (1,4), (1,.5), (1,6),(2,3),(2,4),(2, 5), 2,6), (3, 3), (3,4),(3,5),(3, 6)} SPECTRUMDISCRETEMATHEN EMAN ANDFUNCTION C. Show that B C. SETS, RELATION that A xB=A x ..(1) set such (2.3,5).C (1,21,thenverifythat nonempty Let A be a Eumple6. Let A =2.B Example 8. =A x C A xB Sol. Here =(A *B) U(A C). element of B. * A (BUC) Let b be any EA : of(1] EA xB V a ereA B(2.3.5),Ct2 ( a , b) (a, b) E A xC BUC 2 3,5} U{ 1, 2) = {2,3.5,  1, 2} bEC ...(2) bEB bEC (2.2),(2, 3),(2,5) AB .) BCC Let cbe any element ofC AC 1 2lapa3 (a, c) EA XC V a EA [: of ()) (a, c) EA XB LHS. A B UC)=2 U{2,3,5,1,2 cEB cEC cEB ...3) 2,(2.2),(2,3).(2, 5)(2,),(2,2) CCB From (2) and (3), we get B C. RH.S. (A B) U (A C) that (A B) x C= (A x C) (B x C) For anythree A, B, C prove  sets Example 9. anasa.su}2Jan.aa Sol. L.H.S. (A B) xC (A B) andyEC} {«): («EA and x B) and yEC} £ = ):x E {a,:rEAandy E C) and (x £ B and y C) E 2.2),(2.3),2, 5,20,2,2 {):«, y) E (A C) and(x, y) £ (B C)} x x L.H.S. RH.S. C)(B x C) , ):(, )E(A C)\(B C)} =(A C)\ (B C) =(A x x x x x {1,2,3, B {2,3,4} and C {4,5). Verify that A x (Bnc)=(A B) n(Ax¢ x Example 7. Let A = R.H.S. Sol. A= {1,2,3}, B {2, 3,4), C= {4, 5) C) 2.B(2,3,5},C 1,2), then verify that A B)(A x Example 10. Let A x (BC)(A x BnC 12,3,4} 14,5)= 4 LH.S. A Bnc)= {1,2, 3} x {4) = {(1,4), (2, 4), (3, 4)} Now AB {1,2,3} x {2,3,4} {(1, 2). (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3,4) Sol. Here A2B12,3.51, Ct1,2) AC{1,2,3} * 14,5) 0 ,4),(1, 5), (2,4), 2, 5), 3,4), (3, 5)} B  C 12,3, 5}   1 ,  2 }  {2,3,5) R.H.S. (A B) N(A* C) ={(1,2), (1, 3),(1, 4), (2,2). (2,3), (2, 4), 3,2), (3, 3), (3, 4)} AxB 3.3 Gsaaa».as (1,4), (1, 5), (2, 4), (2, 5), (3, 4),(3. {(1,4), (2, 4), (3, 4)} L.H.S.  R.H.S. Ac (2Jaaa 32 SPECTRUM DISCRETE MATHEMAT. 6 RELATION AND FUNCTION SETS, P. R x R R and R x P x x LHS. A (B C) 4. IfP={1, 2}, form theset P do the cartesian products R x numbers, what set of all real 5. If R is the RH.S. ( 4 B) (A C) represent ? q), (m, r)}, find Aand B. B have ? List them. {(p. 9), (p. r), (m, will A x .a.aucas,es 2an.c.s IFA XB = 6. xB. How many subsets {3, 4). Write A statement is false, rewrite 7. Let A = {1, 2}) and B true or false. If the statements are of the following 8. State whether each (2.2).(2.3),(2. 5) the given statement correctly. PxQ= {(m, n), (n, m)}. IfP= {m, n} and Q= {n, m}, then of ordered pairs (x, y) such () a nonempty set LH.S. RH.S then A x B is If A and B are nonempty sets, Example 11. If A and B be nonempty subsets, then show that A x B = B XA iff A=B (i) that x E A and y E B. xB=B A (Bn$) =Ù. Sol () Assume that A x (ii) IfA = {1, 2), B = {3, 4), then A x (B) =2. Letr be any element of A. 5 and n n (A) = such that distinct. Find A Let A and B be two sets 9. and as are Now r E A y) EA <BVy ¬B in A x B and a, a2, a3, as 2) are If (a1, 2), (a2, 3), (as, 2), (a, 3), (a, , ) EB xA :of (l) and B. 3 and = n (B) = 2. If (r, 1), (, 2), (z, ) are in A x B, r EB such that n (A) 10. Let A and B be two sets distinct elements. r E A x EB find A and B, where , y, z are that {1,2, 3,4), C ={5, 6} and D {5, 6, 7, 8}. Verify ACB Let A {1,2), B = = .. 11. Let: be any element of B ( A x (BnC)=(A x B) n(A xKC) Now: E B (.:) EA XBVrEA C is subset of B x D. i) A x a .)EB XA :of (l) 12. Let A= {1,2,3}, B {4} and C = {5}. Verify that :EA A (BUC)=(A x x B) U(A x C) EB : ¬ A (i) A x (BC)= (A x B)(A x C). BCA 13. Let A {1,2,3,4} = and S= {(a, b) : a ¬ A, b E A, a divides b}. Write S explicitly. From () and (I), we get, x C= (A x C)n (B x C). 14. A, B, C are any three sets, then prove that (A n B) A= B x C =(A x C)U (B x C) 15. IfA, B, C are any three sets, then prove that (A U B) i) Assume that A = B (2) B) (A x C). Forany three sets A, B, C, prove that A (BC)= (A x x  6. AxB=A xA of(2)) Prove that (A B) n(C D) (A NC) (BN D). x = x x 17. and BX A =A x A [: of(2)] 18. Let A and B be two nonempty sets having n elements in common. Prove that A x B and BXA we have, A x B =B x A have n elements in common. Hence A x B B xA iff A = B ANSWERS EXERCISE 14 1. x2, y=1 2. 9 3. No 4. (1, 1, 1), (0, 1,2), (1, 2, 1), (1,2, 2), (2, 1, 1), (2, 1,2), (2, 2, ). (2,2, 2)} find the values of r and y. 5. Coordinates of all the points in two dimensional space 2. If the set A has 3 elements and the set B Coordinates of all the points in three dimensional space = (3, 4, 5), then find the number of elements in (A X B) 3. IfP= {a, b, c} and Q {r}, form the = sets P x Q and Q x P. Are these two products equal 6. A p , m}, B = {4, r} SPECTRUM DISCRETE MATHE WEMAN FUNCTION a relation on a set A SETS, RELATION AND 7. Suhsets of A x B are relation on A. Thus calleda then R is set Ato the set A, 1.3)). (1, 4)1. (2. 3)}. H(2,4)}. 0, 3), (1,4)}. IfR is a relation from a t1, 3), 2.3)1. {(1. 3), (2, 4)}. 1, 4).(2, 3). (1, 4), (2. 4)), as any subset is defined ofA XA. 2. 3). (2.4)):(0.3). (1. 4). (2.3)} (1. 3), (1, 4), (2. 4, Example: Let A = {1,2, 3 (2, 3), (3, ), 3,2), (3, 3)}. 3), (2, 1), (2, 2), Then A x A ={(1, ), (1,2), (1, (1.4). (2. 3). (2, 4)}. {(1, 3). (2, 3), (2, 4)}. A X B (3, 3)}. Let R {(1, 2), (2, 2), (3, 2), the set A. AX B has 16 subsets. R is a relation on Therefore Then RS A XA. 2. 8. ( False therefore I R2 ie, l is R related to Since (1, 2) ER, Here P {m, n}. Q = {n, m} R1 ie, 1 is not R related to 1. Again, since (1,1) £ R so 1 PxQ= {(m, n). (m, m). (n, n), (n, m)} Domain of R ={1,2,3. Rb ifa divides b. is the correct statement. Range of R={2, 3}. define a relation R by a the set of natural numbers, Example: Forany a, b E N, (i) True (iin) True 3), (3, 6),. Then R(1, 1, (1,2), (1, 3),.. 2,2),(2, 4)..3, 9. Afa, a, as, d, as }, B= {2,3} relation on N. x N and hence a Then R is clearly a subset of N 10. A x , y,z}, B = {1,2} (1,2) E R since 1divides 2 13. 1. ),(1.2),(1, 3),(1,4),(2,2),(2,4).G, 3),(4,4)} (2, 1) ER since 2 does not divide 1. Example: Let A {1, 2} and B {3, 4} = = 1.16. Relation A relation R from a nonempty set A to a nonempty set B is a subset of the cartesian product A x Then Ax B={(1,3), (1, 4), (2, 3), (2, 4)} The subset is derived by describing a relationship between the first element and the second element ofs Let R {(1,3),2,4)} ordered pairs in A x B. The second element is called the image of the first element. Then RC A XB and hence R is arelation from A to B. IR3 since (1,3) E R. to b. Let R be a relation from A into B. If (a, b) ER, then we write it as aR band read it, a is in relate 1R 4 since (1,4) E R. Domain ofR = {1,2}. If(a, b) E R, then we write it an a R band read it, a is not related tob by the relation R. Range of R ={3,4}. Example: Let A= {1,2, 3}, B {4,5} 1.18. Representation of Relations AXB={1,2, 3} x 14, 5) {(1,4), (1, 5), (2, 4), (2, 5), (3,4), 3,5)} by the Setbuilder method. A relation may be represented algebraically either by the Roster method or Now any subset of A x B is a relation from A into B. Graphical and Tabular Methods. Consider Another method is that of an arow diagram which is a visual representation of a relation. R {(1, 4), (2, 5), (3, 4)} Let A {1,2, 3} and B {4,5}. Then relation = R (1, 2), (3,5)} R= (2,4), 2, 5),3,5)} from A into B is represented by arrow diagram as shown in the figure given below : Clearly R, is a relation from A into B as Ri is a subset of A x B. A Now 1R 4, 2R 5,3 R4 Again R, is not a relation from A into B as R is not a subset of A x B. Here 1R2. 1.17. Domain and Range of a Relation IfR is a relation from a set A to a set B. Then the set called the domain of R and the set of the second the of first components of the elements of K components of the elements of R is called the range ofk 3 5 Thus, domain of R = { (a, b) E R}, and range of R {b: (a, b) ER}. a: = The whole set B is called the codomain of the relation R. 40 SPECTRUM DISCRETE MATHEN 41 Above relation R be represented in tabular form as follows AEMAT SETS, RELATION AND FUNCTION can 3. Inverse of a Relation B. Then, the inverse of R, denoted by set A to a set relation from a 0 two sets and let R be a Let A, B be relation from B to A and is defined by R , is a {(b, a): (a, b) E R} R In tabular form, if (a, b) E R, then we write 1 and if (a, b) E R, we write 0. Since (1, 4) # t (a, b) E R (b, a) E R write 0 in the row containing 1 and the column containing 4. Also (2, 4) E R, so we write 1 in th Domain of R of R = the n Rand Range Range of of R = and Domain containing 2 and the column containing 4. 4. Identity Relation the identity relation on A. Note: Total Number of Relations E A} on A is called Let A be a set. Then, the relation la = {(a, a): a Let A and B be two nonempt finite sets having m and n elements respectively. Then A x Bhhas elements. Therefore number of subsets of A x B is 2mn. ILLUSTRATIVEEXAMPLES 1.19. Particular Types of Relations 4, 6, 8, 10}. Example 1. Let A {1, 2, 3, 4, 5} and B {2, = domain of R from A into B. Find R. Show that Is a divides b} be a relation AS we know that every subset of A X B is a relation from A into B, so there are 2m n relatione . Let R {(a, b) : a EA, b E B, A and range of R is B. A into B. These relations also include o and A xB. Sol. Here A = {1,2, 3, 4, 5}, B = {2, 4, 6, 8, 10} 1. Empty Relation (2, 8) A XB={1, 2, 3,4, 5} 10) {(1, 2), (1, 4). (1, 6), (1, 8), (1, 10). (2, 2). (2, 4), (2, 6). x {2, 4, 6, 8, IfR = 9, then R is called the empty relation. (2, 10), (3, 2), (3, 4), (3, 6), (3, 8), (3, 10), (4, 2), (4, 4). (4, 6), 2. Universal Relation (4. 8). (4, 10), (5, 2), (5, 4), (5, 6), (5, 8), (5, 10)} Now R {(a, b): a E A, b E B, a divides b} IfR A x B, then R is called the universai relation. R(1.2),0, 4, (1,6. (, 3), (1, 10), 2, 2), (2,4), (2, 6), (2, 8). (2. 10), 3,), 4. 4), (4, 8), (6, 10)) Note: Both the empty relation and the universal relation are called trivial relations. Domain of R {1,2, 3, 4, 5} = A Range o f R = {2, 4 , 6, 8, 10} = B Consider the relation R on the set A = {1,2, 3, 4, 5} defined by Example 2. Let R be a relation from Q into Q defined by R {(a, b):ab= 16} R {(a, b): a, b E Q and a  b E Z}. Show that Now ab* 16 for any two elements of A. ( (a, a) ER for all E Q. () (a, b) E R implies (b, a) ER (a, b) E R for any a, bEA. (i) (a, b) ER, (b, c)ER implies (a, c) E R. Sol. Here R is a relation from Q into Q defined by R does not contain any element of A x A * R is empty set R i s the empty relation on A. () R (a, b) Since a  a : a, b E Q and = 0 bE Z} E Z a Consider the relation R on the set A = {1, 2, 3,4, 5, 6} defined by ( a , a) ER V a EQ (i) (a, b) ER >abEZ ba EZ R {(a, b) ER:ab2 0} Now ab20 for all a, b E A (b, a) ER (ii) (a,b) ER, (b, c) ER a b E Z, bcEZ ( a , b) E R for all (a, b) E A xA * each element of set A is related to every element of set ac=(a b) + (b c) EZ RA X A > R is universal relation on set A. ' ( a , c) E R 42 SPECTRUM DISCRETE MATHEMAT. 4 FUNCTION Exampe3. Let A = {7, 11}. 3. 5} andB = SETS, RELATION AND R defined by Let R t(a. b): a E A. bEB, a b is odd}. Showthat R is an empty relation from A into B. Example 7. Determine the domain and range of the relation 10} Sol. Here A = {3. S}. B {7, 11} ( R {(x, v):rEN, yEN and x+y= R (a. b): a E A. bEB, a  b is odd } (in R{(a,y):x¬N, x <5,y=3. x+y= 10 y= 10x Now 3  7 5  4. 3  11 = 8 , 5  7=2. 5  1 1 = 6 are not odd numbers. Sol. () XI = 1 0  I =9 R is an empty relation. A defined by x =2 y= 102 =8 Eample4. Let A {1.2.3, 4. 6}. Let R be the relation = on 103 7 a.bE A, b is exactly divisible by a} x =3 y= ( a b): x=4 y= 1046 Write R in roster form =5 x =5 y= 105 (i) Find the domain of R x =6 y= 10 6 = 41 (in Find range of R. 7 y=107 3 Sol Here A= {1, 2, 3, 4, 6} r=8 Y=108 =2 R {(a, b): a.bEA, b isexactly divisible by a} x9 y=109=1 () In roaster form = 10 y= 10 10 = 0 EN R (1. ). (1. 2). (1.3). (a, 4). (1. 6), 2. 2),.(2. 4), (2, 6). (3, 3). (3, 6). (4, 4). (6. 6} domain o f R = {1, 2, 3 , , 8, 9} Domain ofR = {1. 2, 3. 4, 6} and range o f R = { 1 , 2 , 3, . . . . . . , 8, 9} (ui) Range o f R = { 1 , 2 . 3, 4 , 6} (in x+y = 10 Example 5. Let A = {1. 2. 3, 4} and B x.y, 7). Let R be a relation from A into B defined by R ((1, 10 When y3, then x +3 = >x=7 (1. :). 3, r). (4. y)}. Represent R in the tabular formn. there is no x E N, x <5, y = 3 which satisfies x +y = 10 Sol Here A = {1,2. 3,4}. B= {.y. domain ofR = ¢, range of R = R is a relation from A into B defined by Example 8. Let A = {1, 2, 3, 4, 5, 6}. R (1. x). (1. z), (3, r). (4. y)} Define a relation R on set A by R={(«, y) :y=x + 1} domain of R = {1, 3, 4} and range of R = {x. y, z3. () Depict this relation using an arrow diagram. Tabular form of relation R is (i) Write down the domain, codomain and range of R. Sol. y=x+1 () Putting x = 1,2, 3, 4, 5, 6, we get, y2,3, 4, 5, 6, 7. Forx =6, we get y =7 which does not belong to set A. 0 R {(1, 2), (2, 3), (3,4), (4, 5), (5, 6)} The arrow diagram representing R is as follows : Example 6. If R is the relation "less than" from A = {1, 2, 3, 4, 5} to B = {1,4, 5}, write down the se ordered pairs corresponding to R. Also find the inverse relation to R. Sol. A {1.2, 3,4, 5), B {1,4, 5} R is the relation less than" from A to B R {(1, 4). (1, 5). (2, 4). (2, 5), (3, 4), (3, 5), (4, 5)} R {(4, 1). (5, 1), (4, 2), (5, 2), (4, 3), (5, 3), (5, 4)} 45 SPECTRUM DISCRETE MAT 44 { I.2, 3, 4, 5} KAEMA SETS, RELATION AND FUNCTION Domain of R = (in and Range of R= {2, 3, 4, 5, 6. EXERCISE 1.5 into B defined by A and B. Write this relation from A below shows a relation R between the sets this relation R rel {1, 2, 3, 4} and B {x, y, z). Let R be a Example 9. The figure given Let A = . Find the domain and range of R. Set builder form (i) Roster form. R {(1,), (1, z), (3, x), (4,y)}. integer. Find the domain ( b): a, b EZ,ab is an 2. Let R be the relation on Z defined by R {(a, and range of R. a, b E N and a = b*}. Are the relation from N into N defined by R {(a, b) : Let R be a following true ? ) (a, a) E R, for all a E N. 251 E R 5 () (a, b) ER implies (6, a) (in (a, b) ER, (6, c) E implies(a, e) R ER What is its domain and range? Justify your answer in each case. Sol. Given figure is {(a, b): E Z, b E Z, a* =b°}. Find the relation Z defined by R = a 4. Let R be on () R (i) domain of R (ii) Range of R. a + 3 b= 12. Find 5. Let R be the relation on the set N of naturals defined by ()R (ii) domain of R (ii) Range of R 6. Let A ={1,2,3, . . , 14}. Define a relation R from Ato Aby R= {(r, y):3xy=0, where x, yE A}. 25 Write down its domain, codomain and range. 7. Define a relation R on the set N of natural numbers by ) Relation R consists ofelements (x, y). where x is the square ofy ie. x = y*. Therefore, relati R r .):y=x +5,x is a natural number lessthan 4 ;x, yEN} in Roster from is Depict this relationship using roster form. Write down the domain and range. R {r,):x= y', xEA, yE B} 8. A {1,2, 3, 5} and B= {4,6,9. Define a relation R from A to B by R{(x. y): the difference between x andy is odd ;rE A,y E B. (in R {9, 3), (9, 3), (4, 2), (4, 2), (25, 5), (25, 5)} Write R in roster form. The domain of R is {9,4, 25} 9. Let A = {1, 2} and B = {3,4}. Find the number ofrelations from A into B. The range of Ris (5,3,2,2,3, 5}. 10. Let A = {1, 2). List all the relation on A. Example 10. Let A be the set of all students of a boys school. Show that the relation R in A gvent 11. R {(a, b) : a is sister ofb} is the empty relation and R' {(a, b): the difference between heights = ofas Let A fx, y. z} and B = {1,2}. Find the number ofrelations from A into B. 12. b is less than 3 meters) is the universal relation. The figure given below shows setbuilder form (i) roaster form. a relationship between the sets P and Q. Write this relation in () Sol. Since the school is boys school i.e. there is not girl student . no student of the school can be sister of any student of the school. R R is the empty relation. Clearly the difference between heights of a and b is less than 3 metres. R' A X A is the universal relation. What is its domain and range? SPECTRUM DISCRETEMATHEMAT 47 46 SETS, RELATION AND FUNCTION b where a, b E R. It Sb ifa is less than ANSWERS Example. We define a relation S on the set for any aER, ofreal numbers R is not less then by a a and hence (a, a) £ S. reflexive relation since a I. Domain = {1,3, 4). Range = { ax, y, z} not a IS not a is reflexive, since every of lines by R1, if l, is parallel to 1 2. Domain = Z, Range =Z Example: The relation R defined on set to itself. line is parallel 3. ( No; (a, a) ER istrue for a=1 Rb ifa> b is not reflexive Example: The relation R defined on set of natural numbers a (i) No: (4,2) E R but (2, 4) £ R not true. R but (16, 2) ¬ R a>a is (iin) No: (16, 4) E R, (4,2) E Irreflexive Relation EZ}U {(a, a):a EZ} 4. R ={(a, a): a a set A is ireflexive if a R a, i.e. (a, a) £ R for every a E A. (ii) Range = Z A relation R on () Domain Z i) Domain = {9,6, 3} R i s irreflexive itf no element is related to itself. 5. ( R {(9. 1), (6, 2). (3, 3)} and let R= {(1, 2), (2, 1)} Example. Let A {1,2} = (üi) Range = {1,2,3} Ris ireflexive as (1, 1) (2, 2) £R. 6. Domain {1, 2, 3, 4}, Codomain= {1, 2,3,. 4}, Range= {3, 6,9, 12} Example. LetA = {1, 2/ and let R = {(1, 2), (2, 2)} 7. R {(1, 6).(2, 7). (3, 8)}, Domain ={1,2,3}, Range = {6, 7, 8} Ris not ireflexive as (2, 2) E R. 9. 16 8. R (1. 4). (1, 6). (2, 9). (3, 4), (3, 6), (6, 4). (5, 6)} Note: Here R is not reflexive as (1, 1) £R. 10. .1,1)),{(22)}).{(1,2)}. {(2,1)}, {(1,1}, {(2,2)}), (1,1), (1,2) Symmetric Relation 1,1), (2. 1)). {(2,2), (1, 2)}). {(2, 2), (2, 1)}, {(0, 2), (2, 1)). A relation R on a set A is called a symmetric relation if aRb ~bRa where a, b E A. ie, if (a, b) E (1, 1).(2,2), (1, 2)}. {(1,1), (2, 2). (2, 1)}. {(1,1), (1,2), (2, 1)}, R(b, )E R where a, b EA. (2.2). (1, 2), (2, 1)}, A X A Example. Let A={1,2,3} 11. 64 ThenAXA= {. D. (1,2),(0,3), 2, ),2, 2), 2, 3),0, ), G,2) 6.3} 12. () R={(x, y):y =x2, xE A, y E B} (in R= {(5, 3). (6, 4), (7, 5)}, Domain= {5, 6,7}, Range = {3,4,5 Let R {(1, 1), (1, 3), (3, 1)} and R{(1,2), (1, 3), (3, 1)} Then R,R, SAXA 1.20. Properties of Relations Therefore R and R, are both relations on A Reflexive Relation Since(x, y) ER° (V,x) ER, therefore R is a symmetric relation on A. A relation R on a set A is called a reflexive relation if (x, x) ER for allx E A ie., if x Rx for even xER. Since (1,2) R, but (2,1)£R, therefore (x, y) E R, (0, 1) E R, does not hold in Ri always. Example. Let A = { 1,2}. R is not a symmetric relation. Then A X A ={(1, 1), (1,2),. (2, 1), (2,2)). Example. For a, b E N, the set of natural numbers define a relation R by a R b if a < b. Then LetR{(1,1),c,2),0,2)}. R{0,2),(1,3).2,3), (2, 4). Then RS A X A and so R is a relation on the set A. Since R C N x N, so R is relation a on N. (1, 2) ER since l is less than 2. Since (x, x) ERVx¬A, so Ris a reflexive relation on A. But(2, 1) E R since 2 is not less than 1. Example. For a, bEN, the set ofnatural numbers, define a Rbifa divides b Therefore R is not a symmetric relation on N. Then R {(1, 1), (1, 2),2,2), (2, 4).. EXample: Relation Rdefined on set of lines by R I, if , is perpendicular to , is symmetric Then RENX Nand so R is a relation on N. Since for any natural numberx, x divides x, so x Rx VXE if then 2 Therefore R is a reflexive relation. SPECTRUM DISCRETE MATHE 49 48 MATN ANDFUNCTION SETS, RELATION Asymmetric Relations {1, 2, 3 A (3, 2, (3, 3)} = A relation R on a set A is asymmetric if whenever a R b, thenb R a. Example. Let (2,2), (2, 3), (3, 1), Then A x A {(1, 1), (1, 2), (1, 3), (2, 1), = and let R be the relation <'. Example. Let A = R, the set of real numbers (2, 1)} Let R {(1, 1), (2, 2),(1, 3), (2, 3), so *<* is asymmetric. Ifa <b. then b a(b is not less than a), 3), 2, 1)} Is R symmetnc, asymmetric or antisymmetrien Let R (1,2), (2,subsets {1.23} and let R {(1.2, (2, 1), (2, 3)}. etric? = Example. Let A of A x A. = both Then R and R, are Sol. Here A={1. 2,3},R={(1,2). (2, 1), (2,3)}. are both relations on A. Therefore, R and R, transitive relation. ER but (3, 2) £R Thus R is a Symmetry: R is not symmetric either since (2, 3) Also (a, b) ER and (b, c) ER> (a, c) ER. since both (1, 2) and (2, 1) ER but (1, 3)ER. Further,(1, 2) and (2, 3) R, E Asymmetry: R is not asymmetric and (2, 1) ER. transitive relation Antisymmetry: R is not antisymeetric since (1, 2) R is not a Thus 10. define a Rb if2a+b= b EN, theset of natural numbers, AntiSymmetric Relation Example. For a, given by: relation 2 +b= 10 a are numbers a andb satisfying the set A is called an antisymmetric relation if a Rb and b R a implies that a = b The natural A relation R on a b =2 8, a=2, b=6, a=3, b= 4, =4, a E R *a=b. a 1,b ie, if(a b) E R and (b, a) R {(1,8), 2,6),(3,4), (4, 2)} OR is not transitive relation. ER but (3, 2) £ R. Therefore R a Since (3, 4) ER and (4, 2) A relation R on aset Ais called antisymmetric if a, bEA (a b) Circular Relation and (a. b)ER ° (6, a) ER. circular if (a, b) E R, (b, c) ER (c, a) E R. A relation R is called A. We define relation R A by L RL and circular is equivalence relation. iffit Show that a relation is reflexive on Example. Let A be the set ofall lines in a plane. Let Li, L, E a Example. different lines Lj and Lz such that L,U L I L ie, if L, is paralel to L2. Since in any plane there exist Sol. Let R be retlexIve and circular but L, * L; i.e, Ly R L, and L, R L, but L, # L2, therefore R is not an antisymmetric relatin and La l L (a, a) E Rand(a, b) E R, (6, c) ER » (c, a) ¬R. have (a, c) ER Example. LetA= { 1,2,3} R and (a, a)ER and R is circular, we Since (c, a) E Then R {(1,1), (1.2), (2,1)} is a relation on the set A. (4, c) E R, which showsthat R is symmetric. ( c , a) ER » antisymmetric relation. (¢, a) E R, since R is circular Since (1.2) ER and (2,1) ER but 1 2, therefore R is not Again (a, b) ER, (6, c) ER E R R is proved to be symmetric But R 3,3) is an antisymmetric relation on A. (a, c) as R is transitive. Example. Fora, bEN the setofnatural numbers define a Rbifasb is an equivalence relation. . Ris reflexive, symmetric and transitive and so Let a, bE N such that a Rband b R a. Conversely, let R be reflexive, symmetric and transitive a s band b s a. a=b. ( of transitivity) (a,b) E R,(b, c) ER * (a, c) ER R is an antisymmetric relation. (c, a) ER ( ofsymmetry) compatible relation if it is Note: Compatible Relation : A relation R in A is said to be retlexive R is circular and so R is reflexive and circular. symmetric. Equivalence Relation Transitive Relation and transitive. A relation R on a set A is called an equivalence relation if R is reflexive, symmetric A relation R on a set A is called a transitive relation if Example. Let X be the set of all triangles in a plane. aRb,bRc aRcVa, b, c E A R A2, ifA, and Az congruent triangles. Then Forany two triangles A and A, in X define A are each A in X. ie, if (a, b)ER () Ris Reflexive. Since each triangle is congruent to itself, so ARA for and (b, c) ¬R (a, c) ER where a, b, c EA. 50 SPECTRUM DISCRETE MATHEM 51 (in) Ris Symmetric. Let A and A: E X such that A, R A. Then A and A are congruent EMATN SETS,RELATION AND FUNCTION Hence AR A triang Proof: LetcEC and a EA SOR (in R is Transitive. Let A. A: A: E X such that A A; . R and A; R Az. Then A. then (c, a) E (SoR)" iff(a, c) E Az are also congruent congruent triangles and so are A: and A;. This implies that the Aj and triang, congruent Now (a, c) E SoR which means there exist b E B Hence , R A R and (b, c) ES So, R is reflenive, symmetric and transitive. such that (a, b) E (c, a) E R"oS" and (c, 6) ES or (c, b) ES' and (6, a) ER' Therefore, R is an equivalence relation on X. (b, a)ER (SoR)= R o s' So Partial Order Relation from B toC and T is a D be sets. Suppose R is a relation from A to B, S is relation A relation R on a set X is said to be a partial order relation if it satisfies the following three conditione. 1.23. Let A, B, C, ()rRx, for every a E X (reflexivity) ditions relation from C to D. Then show that (RoS)oT Ro(SoT). i) xRy and y Rr r=y(antisymmetry) d)E (RoS)oT (iu) x Ry and y R: x R: (ransitivity). x, y, : EX Proof: Let(a, somecEC such that Then there exists Remark: The only equivalence relation on a set X which is also a partial order relation on X is the i identi (a, e)ERoS and (c, d) ET relation Iy. that is, the relation defined by xRY iff x=y. there exist b in B such that 1.21. Composition of Relations Since (a, c) E RoS, so (a, b) ER and (b, c) E S Let R be an relation from a set A to a set B and S be a relation from set B to a set C. Then h composition relation denoted by SoR is a relation from a set A to a set C defined as Now (b, c) ES and (c, d) ET (b, d) E SoT SoR {(a. e):3bEB for which(a, b) E R. (h. c)E S} Also if A be any nonempty set and R. S be any two relations Again (a, b) E R and (b, d) E SOT on A. Then composition of R and denoted by SoR is defined as (a, d)E Ro(SoT) SoR {(a. c):3bEA for which (a, b) ER. (b. c) E S} (Ros)oT C Ro(SoT) Example. Let A = {1,2, 3, 4, 5, 6, 7 Similarly, RofSOT) C (RoS)T .(2) and R{(1,2). (2, 5). (3, 6), (7, 4} From equations (1) and (2) Ro(SoT) = (RoS)oT. S {(1, 4). (7. 5), (3. 7). (4. 3)} be two relations on a set A 1.24. Let R be a relation from X to Y and X, X, be two subsets of X then Then SoR {(7,3)} ( XSX R(X,)S R(X;) (i) R(X, U x) =R(X,) U R(X) RoS {(3,4), (4.6)} (Gi) R(X, nXi) E R(X,) n R(X;) R {(2, ).6, 2). (6, 3), (4. 7) Proof: () Let bE R(X1) S={(4, 1). (5, 7). (7. 3), (3, 4)} Since b E R(X,) there exist a E X, From above example it is clear that such that (a, b) E R(X;) RoS SoR. But X X2, Now we discuss some theorems on composition of relations. aE X 1.22. Let A. B and C be sets, R is relation from A to B, and S is a relation from B to C. Then prove u ds aEX bER(X;) (SoR) = R oS R(X,)C RX,) 52 (in) Let b E R(X, U X,) SPECTRUM DisCRETE MATH 53 there exist some a E X, UX; TA SETS, RELATION AND FUNCTION relation on a set X. Then that R is an equivalence S.t. (a, b) E R(X, UX) SuDpose Now a E X, U 1.26. aEla]Va EX. X; a E Xi or a E X; ifand only if[ a] = [6] Va, bEX. identical. aE[b} or If ( equivalence classes are disjoint aEX bER(X}) n al [b] or[a] n [6]=¢Va,bEXie, any two Similarly if a EX; bE RX:) Proof: () Since R is an equivalence relation on X. so bE R(X) U Ris reflexive. R(X;) aE[a]VaEEX. > RX, UX) S R(X)) U RX) aRaVa EX. (1) b EX such that a E [b] Also we know X, C X, UX2 and X, C X, U X (i) Let a, aRb By part () R(X,)SRX, U X) b R a , since R is equivalence relation ..(2) R(X) CROX, U X) show that [ a ] [b]. = Now we R(X,) U ROX)E RX, U X) Let pE[al1 From (1) and (2) ..3) pRa R b, since R is an equivalence relation R(X, UX)= R(X,) U From, (1) and (3), p R(X;). (ii) Let bE R(X, nX; SopE[a) p¬[b]. Therefore [a]S[b]. there exist some a E X, N X2 Now let g E [b] 5.t. (a. b)E R(OX, n X) qRb. Also b R a(From (2). Now a EX, NX2 aE X1 and a EX2 qRa and hence q E [ al bER(XI) and bE R(X,) b ER(X1) n R(X,) b]Sa] R(X,X) R(X)) O R(X) Hence[a=[b] Conversely let [ a] = [b ] for some a, b E X. 1.25. Equivalence Class Let R be From(),aE[a] equivalence relation an on a nonempty set X. let a E X. Then the equivalence class of a aE[b]. since [ a) = [ b] denoted by [a), is defined as follows Cii) Let a, bEX. al {bEX:bRa}. If[a]n[b]=p, then we have nothingto prove. Example. Let A ={ 1,2, 3. Ifa]tb]*9,thenthere existsp E Xsuchthat p E [a] n[b}. R (1, 1), (2, 1), (1, 2), 2,2). (3, 3)} pE[a] and p E [ b] []={1,2} since only I and 2 are related tol I p l  l a ] and [p] = [b]}, From (i)) Similarly, 2]2,1} and [ 3] ={3} al= [b] we observe that any two equivalence classes are either disjoint or identical. The distinct equivalente 1.27. The distinct equivalence classes of an equivalence relation on a set form a partition of that set. classes are [l ] and [3]. Proof: Let R be an equivalence relation on a set X. Therefore R is also a reflexive relation. Also A =[1]U[3] and [1] O[3]=¢ aRava EX. Then R {(1,1), 2.2), (3,3), (1,2), (2,1)} aE[al Va EX. ..) is an equivalence relation on A. where [a) denotes the equivalence class of a. 55 SPECTRUM DISCRETE MATHEM RELATION AND FUNCTION classes. So, our supposition We prove that X = U[a). SETS, distinct equivalence our supposition that [a) and [b] a r e a EX t this is against 3ut is wrong. Let a EX. [o]#p that(oln aln b]=. distinct equivalence classes are Then a E[a] Thus distinct equivalence classes and any two of X is union From ( Therefore, a EX partition of X. equivalence classes of R forms disjoint. a distinct the sets in the set of classes are XCa EX l] Lence there is an equivalence relation on X whose equivalence 1 For any partition of A, Since ll={bEX:bRa),therefore t h ep a r t i t i o n . Therefore we have be a partition of X. {A}iEn lalCXVa EX Proof: Let 0 X A aEX Ý ifA *u where Aand u E A. From (2) and (3), get X (i) A NA we = U [a]. lf we delete the repetitions from this union, we get X Rb ifand only ifa, b are in the same A. Fora, bEX, define a EX a union of distinct equivalence of X under R. b, c, E A, we have: Then for any a, Now we prove that any two distinct equivalence classes are disjoint. ( 0 R i s reflexive Let la J and [ b] be any two distinct equivalence classes where a, bEX Leta EX. We want to prove that[a ] n[b]=¢. ,U A by () aE 4EA so that aE A, for some 2EA. f possible, let [a] n [b]* o. Therefore aR a. 3rEX such thatx E [ a] n[b] (i) R is symmetric r¬[a ] andx E [b]. Let a Rb. xRaandx Rb EA. a and b belong to A, for some a a Rrand b Rx bRa. b a n d abelong to some Ag. Now we prove that [ a ] = [b]. (i) Let a R b and b R c. Let pEla E A and b, belong to same Ag for some ßE A. a and b belong to same A for some a c pRa. AlsoaRr From (5 bE A, and Ag both i.e., b E A, N A. pRx. Alsox Rb (From(4 aB, Forifa * ß, then by (i) A, N Ap . PRb (Since R is an equivalence relation a and c belong to same A pE [b] aRc. [alSb] Therefore R is reflexive, symmetric as well as transitive relation on X. Hence R is an equivalence Now let q E [6] relation on X. (From(5 9Rb. Also bRx Now we prove that each equivalence class of X is equal to A. gRx.(Since Ris equivalencerelation) Let [ a] denote the equivalence of a for any a EX. Also x R a From(4 Then a]{bEX:bRa}. q R a g E la [ b E X : b and a are in the same A,for some 2 E A} = A Equivalence class of X is equal to Ai. Hence (a]= [6]. 57 S6 SPECTRUM DISCRETE MATHEMA FUNCTION AND the partition. RELATION relation induced by Conversely we prove that each A is equal to some equivalence of X. SETS, let R be the a set with partition tA,,A2,..,A,and Consider any A A be 1.32. Let relation. equivalence Take any a E A Then R is a n that a E A. Suppose Such an a exist, since A, is nonempty. ( for some i. Proof: is a partition of A, So a E A, We prove that A = [ a] Since {A, A2.A * aRa such that a E A, and a E A, Let bE A. Then a and b belong to same A . there is a set A, of the partition bRaand hence bE [ a Ris reflexive bEA r¬A^ * AS[a] (i) Let a, bEA such that a R 6. both a and b are in A t h e r e is a subset A, of the partition such that Now, let x ¬[a] bRa, xRa and hence x, a belong to same A ie both band a belongto some A, But aE A r¬ A Ris symmetric. Let a, b, c E A such that a R b and b R c. (i) and A, of the partition such that a, b, E A, and b, cE A, Hence A= [a] there are subsets A, AlE is the set ofall equivalence classes under the relation R. We claim that A, =A, 1.29. Quotient of A by R Then all sets of the partition are suppose that A, *A, A, nA, as = Ifpossible, Let R be an equivalence relation on A. Then the collection of equivalence classes of the elementsof disjoint. is called Quotient of A by R and is denoted by A  R. AR [a:aEA}. Butbisin both A, and A, Hence A, n A, *p. w e arrive at a contradiction. Therefore our supposition is wrong. 1.30. Fundamental Theorem on Relations A, A, shoing that a, b and c are all in A, In particular a and c are both in A, showing IfR is an equivalence relation on A, then prove that A R is the partition of A. Proof: We know that A  R = { [a]: a E A}. Therefore it is sufficient to show that A is the unionof that a R c and so R is transitive. disjoint equivalence classes. Let PU [a), a E A PSA ILLUSTRATIVE EXAMPLES Example 1.() Give an example of a relation which is reflexive but neither symmetric nor transitive. Also for each a E A, there exists an equivalence class [a] containing a. (i) Give an example of a relation which is symmetric but neither reflexive nor transitive. aEAa E[a] a EP (Gi) Given an example of a relation which is reflexive and ASP symmetric but not transitive. (v) Give an example of arelation which is reflexive and transitive but not symmetric. From (1) and (2), we get () Give an example of a relation which is reflexive and antisymmetric but not transitive. A=P. (vi) Give an example of a relation which is 1.31. Relation Induced by the Partition symmetric and transitive but not reflexive. vi) Give an example of a relation which is reflexive and Let A = A, U A, U . . U A, be a partition of a set A. A relation R on A is said to be relan transitive. antisymmetric but neither symmetric nor Vu) Give an inducedby the partition if for all a, bEA, example of a relation which is neither reflexive, nor symmetric. symmetric, nor transitive nor anti aRb there is subset a A, of the partition such that both a and b are in A x) Give an example ofa relation which is reflexive, symmetric, transitive and antisymmetric. 58 SPECTRUM DISCRETE MATHEN. MAN SETS, RELATION AND FUNCTION Sol. () Let A {2,3,4}. {1, 2, 3). 1),. (3, 2), (3, 3)}. Let A= Then A x A {2.2).(2. 3). (2, 4), (3, 2). (3, 3). (3, 4). (4, 2). (4, 3), (4, 4)} () (2, 1), (2, 2), (2, 3), (3, AXA {(1, 1), (1,2),(1, 3). Then (2, 3)}. Let R {2.2). (3. 3), 4, 4). (2, 3), (4, 3), (3, 4)} R 1), (2, 2), (3, 3), (1, 2), {(1, Let A. Since RC A x A, therefore R is a relation on A as R SAXA. Risa relation on RVa¬A. since (a, a) E R is reflexive since (a, a) ER Va E A. on A Ris areflexive relation A since (a, b) ER and (b, a) E R. R is not symmetric since (2, 3) E R but (3, 2) £ R. antisymmetric relation on Ris also a R is not transitive since (2, 3) and (3, 4) ER but (2, 4) R. a=b. 3) E R but 3 4. but (1, 3) £ R. Further R is not antisymmetric since (3,4) and (4, since (1, 2) and (2, 3) ER ( n L e t A = {1,2 Ris nottransitive (viLet A={1,2,3} (2, 3), (3,1). (3, 2), (3, 3)} Then Ax A= {(1, 1),(1, 2), (2, 1), (2. 2)}. Then AxA={(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), Let R {(1, 2). 2. 1)} R{(1, 1), (1,2), (2, 1), (2, 2)}. Let Then RC A x A and hence R is a relation on the set A. E A and (3,3) £ R. Risnotreflexive 3 as (b, a) E R. R is symmetric since (a, b) ER ER (b, a) ER. Ris symmetric as(a, b) Ris notreflexive since 1 EA but (1, 1) £A. that (a, c) ER. Ris transitive since(a, b) ER and (b, c)ER implies R is not transitive since (1, 2) E R, (2, 1) ER but (1, 1) E R is not antisymmetric since (1, 2) E and I # 2. Ris not antisymmetric since (1,2) and (2,1) both belong to r 1 2. and (2, 1) ER but (vin Let A={1,2,3} (ii) Let A= {1,2,3} Then AXA={(, 1), (1, 2), (1, 3),(2, 1), (2, 2).(2, 3), (3, 1), (3, 2), (3, 3)} Then AxA= {(1, 1), (1,2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)). Let R{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}. Let R={(1, 1), (2,2),(3, 3), (1.,2), (2, 1), (2, 3), (3, 2)}. V aE A. R is a relation on A as RCAXA. Ris reflexive since as (a, a) ER Ris not symmetric since (1, 2) E R and (2, 1) ¢R. RVaEA. Ris reflexive as(a, a) E Also, R is symmetric since (a, b) E R implies that (b, a) ER. Ris antisymmetric as (a, b) ER(6, a) E R»a=b holds in R. Ris nottransitive since (1, 2) and (2, 3) ER but (1, 3) ER. But R is not transitive since (1, 2) E R and (2, 3) ER but (1, 3) ER. 2. (viin Let A= {1,2,3). 1 Moreover, R is not antisymmetric as (1, 2) ER and (2, 1) ER but Then A x A= {(1, 1), (1, 2), (1, 3), (2, 1, (2, 2), (2, 3), (3, D.(3, 2), (3, 3)} (iv) Let A={1,2,3). Then A x A = {(1, 1). (1, 2). (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), 3, 2), (3, 3)}. Let R {(1,2), (2, 1), (2,3)}. Since RC A x A, therefore R is a relation on the set A. Let R{(1, 1, 2,2), (3, 3), (1, 3). Then R is a relation on A as RCA X A Ris not reflexive since 1 E A and (1, 1) ¢R. R is not symmetric since (2,3) E R and (3, 2) £ R. Ris reflexivesince(a,a) ERVaEA. R Ris notsymmetric as (1, 3) ER and (3, 1) £R. is not transitive since (1, 2) and (2, 1) ER and (1, 1) ¬R R iS Ris transitive since (a, b) E R and (b, ) E R implies that (a, c) E R. not antisymmetric since (1, 2) and (2,1) E R but I *2 60 SPECTRUM DISCRETE MATHEMATI ETS, RELATION ANDFUNCTION (i) Let A {1,2 N o w IENbut (1, 1) ¢R Then AXA {(1, D. (1, 2), (2, 1). (2, 2)}. is not reflexive R Let R {(1, 1). (2. 2)}. () (1, 39) ER but(39, 1) ¢R RSAX A. So R is a relation on the set A. not symmetric R is R is reflexive since (a, a) E R V aE A. (i) (20, 1), (1, 39) E R but (20, 39) ER R is symmetric since (a, b) E R ( b , a) ER R i s nottransitive. R is transitive and antisymmetric clearly. reflexive relations on a set then s o are R UR' and RNR'. 4. IfR and R' are Example 2. Check whetherthe relation R defined in the set {1, 2. 3, 4, 5. 6} as R {(a, b): b=at ua are relations on a set A. reflexive, symmetric or transitive. Sol. Since RandR' Sol. Let A  {1.2, 3,4, 5, 6} RCA x Aand R'SAXA. A. R { ( a , b): b = a + 1}= {(a, a + 1)} = {(1, 2). (2, 3), (3,4). (4, 5), (5, 6)} RUR'SAXA and R NR'SA X V aEA RUR' and R OR' are also relations on the set A. Ris notreflexive as (a, a) ER [ (a, b) ER » b=a+l a b11 We now show that R UR' is reflexive relation on A. (in (a, b) ER > (b, a) E R Let a E A. Risnot symmetric. R'. ( R and R' are reflexive on A) (iui) (a, b) E R, (b, c) ER (a, c)ER ( a , a) E R and (a, a) E [ (a, b), (b, c) E R b = a t 1, c=b+l > c=a+2 (a, a)E R UR' and RNR'Va E A. RUR and RNR' arereflexiverelations on A. Ris not ransitive. Example 3. Let R be the relation defined on the set of natural numbers N as Esample 5. For any relation bRa. Prove that R in a set A, we can define the inverse relation R by a R'bif and only if R={a v):xEN, yEN, 2x+y=41} A s a subset of A x A, R ' = {(6, a): (a, b) E R verify whether R is Find the domain and range of this relation R Also ) reflexive (i) symmetric (ii) transitive (in Ris symmetric if and onlyif R R Sol. 2x *) y=412x Sol. () Risdefined by a R'b iffb Ra y=41 2(1)=41 239 Ris defined by (a, b) E R' iff (b, a) E R Ris defined by (b, a) E R' iff (a, b) E R y412 (2)=41437 3 y=41 2(3) =41  6 = 35 as a subsetofA X A,R' {(6, a): (a, b) E R} 4 y=41  2 (4) = 4 1  8 = 33 () Ris symmetric if and only if bRa=aRb ********** ******* ****************' But b Ra aRT'b x19 y=412 (19) =41  38 =3 aR' b=aRb 20 y=41 2 (20)=41  40= 1 Hence R is symmetric iff R = R . x=21 y=41 2(21)= 41 4 2  1 N EXample 6. ILet Z be the set of all integers andR be the relation on Z defined as R = {(1, 39), (2, 37). (3, 35), (4, 33), . . . , (20, 1) R {(a, b): a, bE Z and (a  b) is divisible by 5). 20 Prove that R is domain of R ={1,2,3,4,. an equivalence relation. 39 and range of R { 1,3, 5, 7, 63 62 SPECTRUM DISCRETE MATHEM EMATK RELATION AND FUNCTION multipleof 4 Sol. R {(a. b): 5 divides a  b}, where R is in the set Z of integers. SETS, is a multiple of 4 acisa ) a  a  0  50 (ab)+ (bc) (a, c)ER 5 divides a a (a, a) E R Ris reflexive. laels amultiple of 4 R is transitive. ) Let (a. b) ER equivalence relation. R is an S divides ab which are related to I elements Set of {a E A: a1  is a multiple of 4} {a EA: (a, 1) ER} = ab 5 nforsome n E Z »ba=5(n) » 5 divides ba (b,)ER = (:111=0, 51=4 and 91=8 are multiples of 4]  {1,5,9} (a. b)ER (6a) ER R IS symmetric. (i) R= {(a, b) : a =b} aava EA, uin Let (a, b) and (b, c) E R Ris reflexive. 5 divides a b and bcboth ER > a=b > b=a (b, a) ER Again, (a, b) ab= S and bc= 5 n, for some n.n2 EZ Ris symmetric. and (6, c) ER (ab)+(bc) = 5n + 5n2 Next, (a,b) ER a =b and b=c a =c (a, c) ER ac=5 (n +n2) 5 divides a  c (a, c) ER Ris transitive. (a, b), (b, c) E R (4, c) ER Ris an equivalence relation. Set ofelements of Awhich are related to {a ¬ A: (a, 1) E R} ={a ¬ A:a= 1} {1} l = Ris transitive = From (), (i), (ii) it follows that R is anequivalence relation. Example &. Prove that the following defines an equivalence relation on the xyplane: 12}, given by (,y)R(6,9 ifxs and yt are both integers. ofthe relation R in the set A {x EZ:0 sxs = Example 7. Show thateach Sol () Since xx and yy are integers R(a, b): abisa multiple of4} )R .) (n R {(a,b):a=b} relation is reflexive. 1 in each equivalence relation. Find the set of all elements related to case. is an (n Let (r,) R (6,) A= EZ:0 SrS 12} = {0,1,2,3,4, 5,6, 7,8, 9, 10, 11, 12} Sol I  s and ytare integers R (a, b):abis amultiple of4} S  X and y are integers As aa=0 is divisible by 4 6,0) R(x,) (a, a) ER V aE A. relation is symmetric. R is reflexive. Gii) Let (x, )R (6, ) and (6, ) R (u, v) Next, let (a, b) ER Xs,y1,su, tv are integers Now xu= (xs) + lab is divisible by 4 ~ (ba) is divisible by 4 bal is divisible by 4 (su) integer = and yv= (b, a) ER (y)+((v) integer = X  u and yv are R is symmetric. integers ,y)R (4, v) Again, (a, b) E Rand (b, c)ER if(x,y) R(s, ) and (5, ) R (u, v), then (x, y) R (4, v)  a  b  isa multiple of 4 and bclis a multiple of 4 relation is symmetric. abis a multiple of4 and bcisa multiple of 4 Hence the result. 65 64 SPECTRUM DISCRETE MATHEM. FUNCTION shouMA AND Example 9. IfR is the relation in N x N defined by (a, b) R (c, d) if and only if a+d=b+ SETS, RELATION c. b= c (mod m) m) and is equivalence relation. tha (in Let a E b(mod Sol. Here(a, b)R(¢,d) a+d=b+c. .a=btkh m and b =ctkzm m +k; m ) Now (a, 6) R (a. b) if a + b =b+ a, which is true. a=ctk2 (k +ki) m relation R is reflexive. a=c+ > a =c(mod m) (i) Now (a, 5) R (c, d) ac +km a+d=b+c d+a=c+b c+b=dt relationis transitive a (c, d) R (a, b) b are real ifa s b where a, Hence the result relation R is symmetric. defined on the set of real numbers by aRb be a relation (in) Now (a, b) R (c. d) and (c, d) R (e,f) Crample 12. Let R order relation. Then R is partial a a+d=b+ cand c+f=d+e (a+ d)+ (c+f) =(b+ c) +(d+ e) number. number a and henceaRa. * a+f=b+e reflexive since asafor any real R is R b and bRa. (a, b)R(e.fD Sol. that a b be two real numbers such Ris antisymmetric. an Let aand relation R is transitive. antisymmetric bsa. *a=b Ris * a s b and Rc. Now R is reflexive. symmetric and transitive numbers such that aRband b transitive. Let a, b, c be any real (i) Ris relation R is an equivalence relation. asband b s c. a s c i e , aRc * R is transitive. Example 10. In Nx N, show that the relation defined by (a, b) R (c, d if ad=bc is an equivale and transitive. Ris reflexive, antisymmetric relation. relation on the set of real numbers. R i s a partial order of Sol. Here (a. b) R (c, d) « ad=be () Now (a, b) R (a, b) ifab=ba, which is true order relation. But union 13. Show that intersection of two partial order relations is a partial Example relation R is reflexive. two partial order relations need not be partial order relation. Give suitable example. () Now (a, b) R (c, d and R, be two partial order relation on a nonempty set X Sol. Suppose that R, ad=bc da=cb cb=da (c, d)R(a, b) We show that R, n R, is partial order relation on X relation R is symmetric. R,ORis reflexive :Leta E Xbe arbitrary (ii) Now (a, b) R (c, d) and (c, d) R (e,f) Then (a, a) ER, and (a, a) E R2, since R1, R2 both being partial order relations are reflexive ( a d) ef)= (6 e) (de) adef=bcde ad=bcand cf=de So (a,a) E R, OR (af) dc) =(b e) (d e) af=be (a, b)R (e,) RRis reflexive R, OR; relation R is transitive ) R, nR is Antisymmetric: Let a, bE X such that(a, b) ER, OR and (6, a) E Now R is reflexive, symmetric and transitive {(a, b) E Rand (a, b) E R,} and {(b, a) ER, and (b, a) ER} .'. relation R is an equivalence relation. {la, b) E R, and (6, a) ERI} and {(a, bER, and(b, a) E Ra} Example 11. Prove that mod m relation is an equivalence relation. a b and a=b (: both R, & R, are antisymmetric] Sol. () Since a E a (mod m) a=b relation is reflexive Thus R, n R, is antisymmetric in Let a E b (mod m) ti) R, nRistransitive: Let a, b, c E X suchthat a =b+km (a, b) E R, n R, and (6, c) ER, NR; ba=(k) m bE a (mod m) relation is symmetric tla, b) E R, and (a, b) E R,} and {(b, c) ER, and (b, c)ER,} SPECTRUM DISCRETE MATH b6 {(a. b) E R, and (b, c) E R,} and i(a, b) E R; and(b, c) E R,} MATHEMAN RELATION SETS, NCTION ANDFUNCT following five relation on set A = {1,2, 3} Consider the (a. c) E R and (a. c) ER [: both Ri and R are 7. 3), (3, 3)} 2), (1, (a. c) ER, NR: transili R (1, 1), (1, 1), (1, 2), (2, 1), (2, 2), (3, 3)} S {(1. Thus R, N R; is transitive. T (1, )(2, 2) (1, 2) (2, 3)} Hence R, OR is partial order relation. Empty relation But R UR, need not be partial order relation AXA=Universal relation A is Forexample: Let X {1,2,3} then relations = on whether or not each of above Determine xX= {0. 1). a.2), 0. 3).(2, 1). (2. 2),(2, 3), (6, ), (3.2),(3.3)} (ii) Transitive () Antisymmetric Reflexive (i) Symmetric Let R i(1. ). (2.2). (3, 3). (1, 2). (3. 1)} are defined on the set of natural numbers three relations 8. The folowing R (1, 1). (2, 2). (3, 3). (1, 3), (2, 1)} R {( ) :*<y, x¬N, y¬N} Then R, and R; are subsets of X x X. Therefore R, and R, are both relations on X. 10, xEN, yEN} S { ) : x+y= R, is reflexive since (a, a) ER,V aEX ory=1, xEN, y¬N} T{( ):*=y R is antisymmetric since for no (a, b) E R, with a * b, Explain clearly which of the above relations are Transitive. we have (b, a) E R ( Reflexive (i) Symmetric (i) R istransitivesince (a, b) E R, and (6, c) E R, > (4, c)ERI and y share common factor other than 1. Let R be the relation N defined by R, if on x a 9. X Determine the reflexivity and transitivity of R. R is partial order relations on A~ B matrices with real entries given by Similarly, R is also partial order relation on X 10. Show that the relation'~ in the set of 2 2 invertible x But R UR{(1. ). (2,2), 3, 3). (1, 2), (3, 1), (1, 3), (2, 1)} iffB A is symmetric but not reflexive. Is it transitive ? 11. Suppose R and S are symmetric relations on a set A. Show that R n S is also symmetric. with (1.2) ER, U R, and(2, 1) ER, U R,but 1 2 12. IfR and R' are symmetric relations on a set A, then R nR' is also a symmetric relation on A. RUR is not antisymmetric set is again a symmetric relation on that set. 13. Show that the union of two symmetric relations on a Hence R UR, is not a partial order relation. 14. For any relation R in a set A, we can define the inverse relation R by a R' biff b R a. Prove EXERCISE 1.6 that R is symmetric iff R '=R. 1. Give an example of relation which is transitive but neither reflexive nor symmetric n I5. Give an example ofrelation R on A= {1. 2, 3} havingthe stated property: antisymmetric. R is neither symmetric nor antisymmetric, and R is transitive but R UR' is not transitive. 2. Give an example of a relation which is antisymmetric and transitive but neither reflexive m 16. Ris arelation relation ? on set ofpositive integers s.t. R (a, b): ab is odd integer)} 1s R an equivalence symmetriC. symmetric and antisymmetricand 17. 3. Give example of relation R on A {1, 2, = 3} which is both For two lines ,, in a plane o, the relation R defined by R if and only if t, is perpendicular is neither symmetric nor antisymmetric. to is neither reflexive nor transitive but symmetric. Hence R is not an equivalence relation. 4. Show that the relation R in R defined as R = {(a, b): a s b}, is reflexive and transitive bur 18. Let Rbe symmetric and transitive relation on a set A. If for each x E A, 3 y¬A such that (r, y) symmetric. ER. Then R is an equivalence relation. . Show that the relation R in the set R of real numbers, defined as 19. Prove that the intersection of two equivalence relations on a nonempty set is again an R= {(a, b): sb} equivalence relation on that set. a isneitherreflexive nor symmetric nortransitive. 20. Show that 6. Check whether the relation R in R defined by R, U R2 may not be an equivalence relation set X if on a Ri, R2 are equivalence R relations on X. {(a, b) : asb} is reflexive, symmetric or transitive. 69 SPECTRUM DISCRETE MATH 21. Is inclusion of a subset in another, in the context of a universal set, an FUNCTION equivalence rel AND class of subsets of the sets ? Justify your answer. relation ing iSETS, RELATION 22 Show that the relation R in the set Z of integers given by R {(a, b):2 divides a SECTIONII ab} b} is an relation cquiva 23. The relation RC N FUNCTIONS x Nis defined by (a, b) ER if and only if 5 divides b a. Show that R equivalencerelation. of Function from X to Y if for every 24. In the set N of all natural numbers. Let relation R be defined by Definition a function 1.33. nonempty sets. subsetfof A Xx Y is called Y be two Let X and R r.yr EN.y ¬N:xyis divisible by m} a unique E Y such that y Ef or we sayy =f(r). (x, y) c). (4, b)} isa f= {(1, a), (2, b), (3, exists Show that R is an equivalence relation. EX. there then subset {a. b, c, d} {1, 2, 3, 4}, Y = X = 25. For a. bER the set of real numbers, defined a Sb if a=b then S is an Forexample: E X, we have y E Y such that y=f(r). equivalence r from X to Y :: for each x nR. Telal function is also represented by following figure Above function 26. Let R be a relation on the set of A of ordered pairs of positive integers defined by (x. y) R( X and only if xv =yu. Show that R is an equivalence relation. 27. For EQthe set ofrelational mumbers, defineRi and onlyifad=bes b that R is an equivalence relation on Q. 28. Let N denote the set of all natural numbers and R be the relation NxN defined by (a, b) R (c, d) a d(b+ c)=be(a +d). Check whether R is an equivale relation on N xN. Another Definition of Function with unique rulefwhich associates each element of X a 29. IfR is an sets. Then a equivalence relation on a set A, then so is R Let X and Y be two nonempty is called a function from X to Y. 30. For any a, b E N, the set of natural numbers, define a Rb if and only ifa divides b. Then D. eclement of Y partial order relation. The other terms used for functions are mappings or transformations. We denote this mapping by X Y. 31. In power set P (X) ofX, show that the relation 'C is a partial order. Is it an equivalence relatia f:XYor The set X is called the domain of fand is written as D= X. The set Y is called codomain off. ANSWERS If an element y E Y is associated with an element x of X under the rule, then y is called the image of 6. Not reflexive, not symmetric, not transitive. r under the rule f, denoted by f(r). 7. Ris not reflexive, not symmetric, antisymmetric, transitive. The set consisting of images of all the elements of X under/is called Image set or Range off and is S is reflexive, symmetric, not antisymmetric, transitive written as R or ran f. T is not reflexive, not symmetric, antisymmetric not transitive, Ry Uf;xEX} =f(X) or Ry b:yft) wherexEX} =S(X) is not reflexive, symmetric, antisymmetric, transitive Clearly fCX) CY A XA is reflexive, symmetric, not antisymmetric, transit In above example D, {1, 2, 3, 4} = 8. ( R S, T are not reflexive. Ry=a, b, c} andcodomain is {a, b, c, d. Note. Difference between Relation and Function (i) R, Tare not symmetric and S is symmetric (ii) R is transitive but S, T are not transitive. Function is special of that of relation. A relation may relate each element of the domain to a case a 9. Ris neither reflexive nor transitive. nore thanofone element the element of the range, but function relates each element of the domain to and only one codomain. a one 10. No. 16. No 21. No. 28. Yes 31. Yes It should be noted that every function is a relation but every relation is not a function. Consider X= {1,2,3, 4,), Y {5,6, 7}
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