31 AEMAN SPECTRUM DISCRETE MATHE 30 SETS, RELATION AND FUNCTION Eumple 6. Show that AN(B-C) = (A N B)-(A NC) R.H.S. (A OB)- (A nC) = (A NB)n(ANC= (AN B)n (A° uc) SECTION-II Sol. - [(A B)n A"]u[(AN B) nc]=[A NA')OB] u[A n (B nc RELATIONS (p n B)U [A N(B- C)]= ¢U[An (B-)] An(B-C) 1.12. Ordered Pair Let A and B be two given non-empty sets. If a E A and bEB, then (a, b) denotes an ordered pair whose first component is a and the second component is b. - L.H.S. The ordered pair (a, b) and (b, a) are different unless a =b. Also two ordered pairs (a, b) and (c, d) are equal if and only if a=c and b=d. EXERCISE 1.3 1 Verify the following identities: () AnBUC)= (A N B) U (A NC) Note 1. Ordered pairs (2, 5) and (5, 2) are not equal whereas the sets {5, 2} and {2, 5} are equal. AUBnC)=(A UB) N (A UC) Note 2. In the definition of an ordered pair, a and b may not be distinct i.e., (a, a) and (b, b) are also ordered pairs. where, A. B, C are three sets defined by A {1.2,4,5 }, B = {2,3, 5, 6 }, C={4, 5, 6, 7 1.13. Cartesian Product of Sets Cartesian Product of Two Sets. If A and B are two non-empty sets, then the set of all ordered pairs (a, b) where a E A and b E B is called the cartesian product of sets A and B and is denoted by A x B.. 2. If X and Y are two sets, then find X n (XU Y). 3. Show that )ACAUB (i) ANBCA. Prove the following: In symbols, A xB= {(a b): a E A, bE B} 4. ( BCAUB (i) ANBCB (i) BCAANB=B Note 1. A x B and B x A are different sets if A * B. () AC C and B C D AUBC CUD (v)BCC>ANB CAnC 2. A xB= when one or both 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 We know that ) An(A'UB) = ANB in) A -(A - B) = A OB A=A XA = {«,y):x,y E A} 7. Prove that A\ B' = B\A. A=AxA XA={x, y, z):x, y, zE A} 8. Prove the following : 1.14. Graphical Representation of A xB ((A-B)nB=o (G) An(B- A) = ¢ ii) AU (B- A) = AUB Draw two perpendicular lines X'OX and Y'OY intersecting at (i) A-B = A- (A NB) ) (A-C)U(B-C)=(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 - (B-A/" horizontal line X'OX represent the elements of A and on vertical 9. Prove that A n (B\C) = (A N B)MAN C). line Y'OY, represent the elements of B. Now if a E A, b E B, draw a vertical line through a and a ANSWERSs horizontal line through b. The point where they meet represents the ordered pair (a, b), The set of all such points obtained graphically 2. represents A x B. SPECTRUM DISCRETE MATHEM, 33 32 Note 1. If A is the set of all numbers, then A consists of all points in a line. A x A will consist of al SETS, RELATION AND FUNC tn the plane. Example 2. IfG = {7, 8} and H = {5, 4, 2}, find G x H and H xG. GxH (7,8} x {5,4,2 ={(7, 5), (7, 4), (7, 2). (8, 5), (8, 4), (8, 2)} HxG (5, 4, 2} x {7, 8}- {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8) Note 2. The ordered pair (a, b} represents a point whose co-ordinates are (a, 6). Sol. G {7, 8), H = {5,4, 2} Note 3. Let m (A) denote the number of elements of A. Then n (A x B) = n (A) X n (B). 1.15. Prove tha Example 3. IfA X B = {(a, x), (a. y). (6, x), (6, )}, find A and B. Ax (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 A and y E(B Uc)} = {a,):x*E A and (y E B ory E C)} Example 4. The cartesian product A x A has a elements among which are found (-1, 0) and (0, 1). Find the set A and the remaining elements of A x A. and B set of second elements = {r, y} = {r.1): («E A and y E B) or (r E A andyE C)} = {a. ): ay) E (A x B) or (z, ) E (A x C)} Sol. Since (-1, 0) E A X A and (0, 1) EA XA = {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 Now AX A has 9 elementsA has 3 elements Ax (BUC)= (A x B) U(A x C) A={-1, 0, 1} () L.H.S. A x (BnC) Remaining elements of A X A and (-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1). (1,0), (1, 1) Example 5. Let A = {1,2, 3}, B {3,4} and C = {4, 5,6}. Find ()A x(Bnc) (in) (A x B)n(A x C) Gi) A x (BUC) a. y):a E A and y E(B nc)} = {«,y):r¬A and (oEB and y EC} (iv) (A x B) U (A x C). a, y): (x EA and y E B) and (r E A andy E C)} Sol. Here A= {1,2,3, B= {3,4}, C={4, 5,6} BUC (3, 4} U {4, 5, 6} = {3,4,5, 6} =r. y): (a ) E (A X B) and (x. y) E (A x C)} (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 Ax (BnC)= (A x B) n(A x C) AXB {1,2, 3} x {3,4) ((1, 3), (1,4), (2, 3), (2,4), (3,3), (3,4)} 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} Ax (Bnc)={1,2,3} x {4) - {(1, 4),. (2, 4), (3, 4)} (in (A x B) n(A x C) = {(1, 3), (1, 4), (2, 3), (2,4), (3, 3), (3, 4)} ILLUSTRATIVE EXAMPLES Example 1. If (r+ 1,y-2)=(3, 1). find the values of x and y. n1,4), (0,5), (1, 6), (2, 4), 2,5),(2, 6), (3,4), (3. 5), (3, 6)} Sol. Here «+ 1,y-2) = (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 {(1,3), (1,4), (1, 5), (1.6), (2, 3), (2,4),(2, 5), (2,6), (3, 3), 3, 4), (3,5), (3,6) y-2-1 () (AxB) U(A x C)= {(1, 3), (1, 4), (2, 3), (2,4), (3, 3), (3, 4)} From (1). r=3-1-2 U1, 4), (1, 5), (1, 6). (2, 4), (2, 5), (2, 6), (3,4), (3, 5), (3,6)} From (2), y=1+2=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)} x2, y=3. SPECTRUM DISCRETE MATHEN EMAN Example 8. Let A be a non-empty set such that A xB=A x C. Show that B C. Sol. Here A xB =A x C Let b be any element of B. SETS, RELATION AND FUNCTION ..(1) Eumple 6. Let A =2.B (2.3,5).C - (-1,-21, then verify that A (BUC) = (A * B) U (A * C). ere A B(2.3.5), C- t--2 : of(1] (a, b) EA xB V a EA (a, b) E A xC BUC 2 3, 5} U{- 1, -2) = {2,3. 5, - 1, -2} bEC ...(2) AB- .)- (2.2),(2, 3), (2,5) bEB bEC BCC AC 1-- -2la-pa-3 Let cbe any element ofC (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 RH.S. (A B) U (A C) B C. anasa.s|u|-}-2Ja-n.a-a Example 9. For any three sets A, B, C prove that (A - B) x C= (A x C)- (B x C) L.H.S. (A B) xC Sol. -):x E (A - B) and yEC} = {«): («EA and x £ B) and yEC} {a, :rEA and y E C) and (x £ B and y E C) {):«, y) E (A x C) and (x, y) £ (B x C)} , ): (, )E (A x C)\(B x C)} = (A x C)\ (B x C) = (A x C)-(B x C) 2.2),(2.3),2, 5,2-0,2,-2 L.H.S. RH.S. Example 7. Let A = {1,2,3, B {2,3,4} and C {4,5). Verify that A x (Bnc)=(A x B) n(Ax¢ Sol. A= {1,2,3}, B {2, 3,4), C= {4, 5) R.H.S. BnC 12,3,4} 14,5)= 4 LH.S. A Bnc)= {1,2, 3} x {4) = {(1, 4), (2, 4), (3, 4)} Example 10. Let A -2.B-(2,3,5}, C -1,-2), then verify that A x (B-C)-(A x B)-(A x C) 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) Here A2B- 12,3.51, C-t-1,-2) Sol. AC-{1,2,3} * 14,5) 0, 4), (1, 5), (2, 4), 2, 5), 3, 4), (3, 5)} R.H.S. (A B) N(A* C) B-C 12,3, 5} --1,-2} - {2,3,5) ={(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. Ac- --(-2Ja-a-a {(1,4), (2, 4), (3, 4)} L.H.S. - R.H.S. 32 SPECTRUM DISCRETE MATHEMAT. 6 SETS, RELATION AND FUNCTION If R is the set of all real numbers, what do the cartesian products R x R and R x R x R 5. LHS. A (B C) 4. IfP={1, 2}, form the set P x P x P. RH.S. (4 B)- (A C) represent ? .a.auca s,es 2a-n.c.s Let A = {1, 2}) and B {3, 4). Write A xB. How many subsets will A x B have ? List them. 7. 6. IFA XB = {(p. 9), (p. r), (m, q), (m, r)}, find A and B. 8. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly. (2.2).(2.3),(2. 5) (i) If A and B are non-empty sets, then A x B is a non-empty set of ordered pairs (x, y) such that x E A and y E B. () IfP= {m, n} and Q= {n, m}, then PxQ= {(m, n), (n, m)}. LH.S. RH.S Example 11. If A and B be non-empty subsets, then show that A x B = B XA iff A=B Sol () Assume that A xB=B x A Letr be any element of A. (ii) IfA = {1, 2), B = {3, 4), then A x (Bn$) =Ù. If (a1, 2), (a2, 3), (as, 2), (a, 3), (a, 2) are in A x B and a, a2, a3, as and as are distinct. Find A and B. 9. Let A and B be two sets such that n (A) = 5 and n (B) =2. Now r E A y) EA <BVy ¬B , ) EB xA :of (l) Let A and B be two sets such that n (A) = 3 and n (B) = 2. If (r, 1), (, 2), (z, ) are in A x B, 10. r EB rE A x EB find A and B, where , y, z are distinct elements. 11. Let A = {1,2), B = {1,2, 3,4), C = {5, 6} and D {5, 6, 7, 8}. Verify that ( A x (BnC)=(A x B) n(A xKC) ACB .. Let: be any element of B Now: E B (.:) EA XBVrEA i) A x C is a subset of B x D. 12. Let A= {1,2,3}, B {4} and C = {5}. Verify that A x (BUC)= (A x B) U (A x C) (i) A x (B-C)= (A x B)- (A x C). 13. Let A = {1,2,3, 4} and S= {(a, b) : a ¬ A, b E A, a divides b}. Write S explicitly. 14. A, B, C are any three sets, then prove that (A n B) x C= (A x C)n (B x C). 15. IfA, B, C are any three sets, then prove that (A U B) x C =(A x C)U (B x C) For any three sets A, B, C, prove that A x (B-C)= (A x B) - (A x C). 17. Prove that (A x B) n(C x D) = (A NC) x (BN D). .)EB XA :of (l) :EA EB :¬A BCA From () and (I), we get, A= B i) Assume that A = B (2) 6. AxB=A xA of(2)) and BX A =A x A we have, A x B =B x A Hence A x B B xA iff A = B [: of (2)] 18. Let A and B be two non-empty sets having n elements in common. Prove that A x B and BXA have n elements in common. ANSWERS EXERCISE 14 1. x2, y=1| 2. 9 3. No find the values of r and y. 4. (1, 1, 1), (0, 1,2), (1, 2, 1), (1,2, 2), (2, 1, 1), (2, 1,2), (2, 2, ). (2,2, 2)} 5. Coordinates of all the points in two dimensional space 2. If the set A has 3 elements and the set B = (3, 4, 5), then find the number of elements in (A X B) Coordinates of all the points in three dimensional space 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} WEMAN SPECTRUM DISCRETE MATHE IfR is a relation from a set A to the set A, then R is calleda relation on A. Thus a relation on a set A is defined as any subset of A XA. Example: Let A = {1,2, 3 7. Suhsets of A x B are SETS, RELATION AND FUNCTION 1.3)). (1, 4)1. (2. 3)}. H(2, 4)}. 0, 3), (1, 4)}. t1, 3), 2.3)1. {(1. 3), (2, 4)}. 1, 4).(2, 3). (1, 4), (2. 4)), 2. 3). (2.4)):(0.3). (1. 4). (2. 3)} (1. 3), (1, 4), (2. 4, (1.4). (2. 3). (2, 4)}. {(1, 3). (2, 3), (2, 4)}. A X B Then A x A ={(1, ), (1,2), (1, 3), (2, 1), (2, 2), (2, 3), (3, ), 3,2), (3, 3)}. Let R {(1, 2), (2, 2), (3, 2), (3, 3)}. AX B has 16 subsets. Then RS A XA. Therefore R is a relation on the set A. 8. ( False Since (1, 2) ER, therefore I R2 ie, l is R related to 2. Here P {m, n}. Q = {n, m} Again, since (1,1) £ R so 1 R1 ie, 1 is not R related to 1. Domain of R = {1,2,3. Range of R={2, 3}. PxQ= {(m, n). (m, m). (n, n), (n, m)} Example: For any a, b E N, the set of natural numbers, define a relation R by a Rb ifa divides b. Then R-(1, 1, (1,2), (1, 3),.. 2,2),(2, 4)..3, 3), (3, 6),. Then R is clearly a subset of N x N and hence a relation on N. is the correct statement. (i) True (iin) True 9. Afa, a, as, d, as }, B= {2,3} 10. A x, y,z}, B = {1,2} (1,2) E R since 1 divides 2 (2, 1) ER since 2 does not divide 1. 13. 1. ), (1.2), (1, 3), (1, 4), (2,2), (2, 4). G, 3), (4, 4)} Example: Let A = {1, 2} and B = {3, 4} Then A x B={(1,3), (1, 4), (2, 3), (2, 4)} R {(1,3), 2, 4)} 1.16. Relation A relation R from a non-empty set A to a non-empty set B is a subset of the cartesian product A x The subset is derived by describing a relationship between the first element and the second element ofs ordered pairs in A x B. The second element is called the image of the first element. 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 Let Then RC A XB and hence R is a relation from A to B. IR3 since (1,3) E R. 1R 4 since (1,4) E R. to b. Domain ofR = {1,2}. Range of R ={3,4}. 1.18. Representation of Relations If(a, b) E R, then we write it an a R band read it, a is not related tob by the relation R. Example: Let A= {1,2, 3}, B {4,5} AXB={1,2, 3} x 14, 5) {(1,4), (1, 5), (2, 4), (2, 5), (3,4), 3,5)} A relation may be represented algebraically either by the Roster method or by the Set-builder method. 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= (2,4), 2, 5),3,5)} from A into B is represented by arrow diagram as shown in the figure given below : R (1, 2), (3,5)} 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 of the first components of the elements of K called the domain of R and the set of the second components of the elements of R is called the range ofk 3- 5 Thus, domain of R = { a: (a, b) E R}, and range of R = {b: (a, b) ER}. The whole set B is called the codomain of the relation R. 40 SPECTRUM DISCRETE MATHEN AEMAT 41 Above relation R can be represented in tabular form as follows SETS, RELATION AND FUNCTION 3. Inverse of a Relation 0 Let A, B be two sets and let R be a relation from a set A to a set B. Then, the inverse of R, denoted by R, is a relation from B to A and is defined by R {(b, a): (a, b) E 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 write 0 in the row containing 1 and the column containing 4. Also (2, 4) E R, so we write 1 in th containing 2 and the column containing 4. (a, b) E R (b, a) E R- the n and Domain of R = Range of Rand Range of R = Domain of R 4. Identity Relation Note: Total Number of Relations Let A be a set. Then, the relation la = {(a, a): a E A} on A is called the identity relation on A. Let A and B be two non-empt has finite sets having m and n elements respectively. Then A x Bh elements. Therefore number of subsets of A x B is 2mn. ILLUSTRATIVE EXAMPLES 1.19. Particular Types of Relations Example 1. Let A = {1, 2, 3, 4, 5} and B {2, 4, 6, 8, 10}. 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 divides b} be a relation from A into B. Find R. Show that domain of R Is A into B. These relations also include o and A xB. A and range of R is B. Sol. Here A = {1,2, 3, 4, 5}, B = {2, 4, 6, 8, 10} A XB={1, 2, 3, 4, 5} x {2, 4, 6, 8, 10) {(1, 2), (1, 4). (1, 6), (1, 8), (1, 10). (2, 2). (2, 4), (2, 6). (2, 8) (2, 10), (3, 2), (3, 4), (3, 6), (3, 8), (3, 10), (4, 2), (4, 4). (4, 6), (4. 8). (4, 10), (5, 2), (5, 4), (5, 6), (5, 8), (5, 10)} 1. Empty Relation IfR = 9, then R is called the empty relation. 2. Universal Relation Now R {(a, b): a E A, b E B, a divides b} IfR A x B, then R is called the universai relation. Note: Both the empty relation and the universal relation are called trivial relations. Consider the relation R on the set A = {1,2, 3, 4, 5} defined by 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)) Domain of R {1,2, 3, 4, 5} = A Range of R = {2, 4, 6, 8, 10} = B Example 2. Let R be a relation from Q into Q defined by R {(a, b):a-b= 16} R {(a, b): a, b E Q and a - b E Z}. Show that ( (a, a) ER for all E Q. () (a, b) E R implies (b, a) ER Now a-b* 16 for any two elements of A. (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 (a, b) : a, b E Q and a - bE Z} Ris the empty relation on A. () Since a- a = 0 E Z 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 >a-bEZ b-a EZ R {(a, b) ER:|a-b|2 0} (b, a) ER (ii) (a, b) ER, (b, c) ER a- b E Z, b-cEZ a-c=(a- b) + (b- c) EZ Now a-b|20 for all a, b E A (a, b) E R for all (a, b) E A xA * each element of set A is related to every element of set R A X A > R is universal relation on set A. '(a, c) E R SPECTRUM DISCRETE MATHEMAT. 4 42 Exampe3. Let A = 3. 5} andB = {7, 11}. Let R t(a. b): a E A. bEB, a- b is odd}. Show that R is an empty relation from A into B. SETS, RELATION AND FUNCTION Example 7. Determine the domain and range of the relation R defined by R {(x, v):rEN, yEN and x+y= 10} ( (in R{(a, y):x¬N, x <5,y=3. x+y= 10 y= 10-x Sol. Here A = {3. S}. B {7, 11} R (a. b): a E A. bEB, a - b is odd } Sol. () Now 3 -75- 4. 3- 11 = 8, 5 - 7=-2. 5 -11=6 are not odd numbers. XI =10-I =9 x =2 y= 10-2 =8 R is an empty relation. Eample4. Let A = {1.2.3, 4. 6}. Let R be the relation on A defined by (a b): a.bE A, b is exactly divisible by a} x =3 y= 10-3 7 x=4 y= 10-4-6 x =5 y= 10-5 =5 Write R in roster form (i) Find the domain of R x =6 y= 10- 6 = 41 (in Find range of R. 7 y=10-7 3 Sol Here A= {1, 2, 3, 4, 6} r=8 Y=10-8 =2 x9 y=10-9=1| R {(a, b): a.bEA, b is exactly divisible by a} () 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 of R = {1, 2, 3, , 8, 9} Domain ofR = {1. 2, 3. 4, 6} and range of R = {1,2, 3, ......, 8, 9} (ui) Range of R = {1,2. 3, 4, 6} 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, (1. :). 3, r). (4. y)}. Represent R in the tabular formn. Sol Here A = {1,2. 3, 4}. B= {. y. (in x+y = 10 When y3, then x +3 = 10 >x=7 there is no x E N, x <5, y= 3 which satisfies x +y = 10 domain ofR = ¢, range of R = Example 8. Let A = {1, 2, 3, 4, 5, 6}. Define a relation R on set A by R= {(«, y) :y=x + 1} () Depict this relation using an arrow diagram. R is a relation from A into B defined by R (1. x). (1. z), (3, r). (4. y)} domain of R = {1, 3, 4} and range of R = {x. y, z3. (i) Write down the domain, co-domain and range of R. Tabular form of relation R is Sol. y=x+1 () Putting x = 1,2, 3, 4, 5, 6, we get, y-2,3, 4, 5, 6, 7. Forx =6, we get y =7 which does not belong to set A. R {(1, 2), (2, 3), (3,4), (4, 5), (5, 6)} The arrow diagram representing R is as follows : 0 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 KAEMA SPECTRUM DISCRETE MAT 44 SETS, RELATION AND FUNCTION (in Domain of R = { I.2, 3, 4, 5} EXERCISE 1.5 Let A = {1, 2, 3, 4} and B {x, y, z). Let R be a relation from A into B defined by R {(1,), (1, z), (3, x), (4, y)}. Find the domain and range of R. and Range of R= {2, 3, 4, 5, 6. Example 9. The figure given below shows a relation R between the sets A and B. Write this rel ( Set builder form (i) Roster form. this relation R 2. Let R be the relation on Z defined by R {(a, b): a, b EZ,a-b is an integer. Find the domain and range of R. Let R be a relation from N into N defined by R {(a, b) : a, b E N and a = b*}. Are the following true ? ) (a, a) E R, for all a E N. 251 () (a, b) ER implies (6, a) E R (in (a, b) ER, (6, c) E R implies (a, e) ER 5 What is its domain and range? Justify your answer in each case. Sol. Given figure is 4. Let R be the relation on Z defined by R = {(a, b): a E Z, b E Z, a* =b°}. Find () R (i) domain of R (ii) Range of R. 5. Let R be the relation on the set N of naturals defined by a + 3 b= 12. Find (ii) Range of R ()R (ii) domain of R 6. Let A ={1,2,3, .., 14}. Define a relation R from A to A by R= {(r, y): 3x-y=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 of elements (x, y). where x is the square of y ie. x = y*. Therefore, relati R r. ):y=x +5,x is a natural number less than 4 ; x, yEN} in Roster from is Depict this relationship using roster form. Write down the domain and range. 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 and y is odd ;rE A,y E B. R {r, ):x= y', xEA, yE 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} The range of R is (-5,-3,-2,2,3, 5}. 9. Let A = {1, 2} and B = {3,4}. Find the number of relations from A into B. 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 R {(a, b) : a is sister ofb} is the empty relation and R' = {(a, b): the difference between heights of as b is less than 3 meters) is the universal relation. 11. Let A fx, y. z} and B = {1,2}. Find the number of relations from A into B. 12. The figure given below shows a relationship between the sets P and Q. Write this relation in () set-builder form (i) roaster form. 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 DISCRETE MATHEMAT 47 46 SETS, RELATION AND FUNCTION ANSWERS Example. We define a relation S on the set of real numbers R by a Sb ifa is less than b where a, b E R. It IS not a not a reflexive relation since for any aER, a is not less then a and hence (a, a) £ S. I. Domain = {1,3, 4). Range = { ax, y, z} 2. Domain = Z, Range =Z Example: The relation R defined on set of lines by R1, if l, is parallel to 1 is reflexive, since every line is parallel to itself. Example: The relation R defined on set of natural numbers a Rb ifa> b is not reflexive 3. ( No; (a, a) ER is true for a=1 (i) No: (4,2) E R but (2, 4) £ R (iin) No: (16, 4) E R, (4,2) E R but (16, 2) ¬ R a>a is not true. Irreflexive Relation 4. R = {(a, a): a EZ}U {(a, - a): a EZ} A relation R on a set A is ireflexive if a R a, i.e. (a, a) £ R for every a E A. () Domain Z (ii) Range = Z Ris irreflexive itf no element is related to itself. i) Domain = {9,6, 3} 5. ( R {(9. 1), (6, 2). (3, 3)} (üi) Range = {1,2,3} Example. Let A = {1,2} and let R= {(1, 2), (2, 1)} 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)} Ris not ireflexive as (2, 2) E R. 7. R {(1, 6).(2, 7). (3, 8)}, Domain ={1,2,3}, Range = {6, 7, 8} 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 12. () R={(x, y):y =x-2, xE A, y E B} (in R= {(5, 3). (6, 4), (7, 5)}, Domain= {5, 6,7}, Range = {3,4,5 Then AXA= {. D. (1,2),(0,3), 2, ),2, 2), 2, 3),0, ), G,2) 6.3} 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 Let R-{(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 a relation on N. (1, 2) ER since l is less than 2. Since (x, x) ERVx¬A, so R is a reflexive relation on A. But (2, 1) E R since 2 is not less than 1. Example. For a, bEN, the set of natural 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 R defined 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 number x, x divides x, so x Rx VXE Therefore R is a reflexive relation. if then 2 SPECTRUM DISCRETE MATHE MATN 49 48 Asymmetric Relations A relation R on a set A is asymmetric if whenever a R b, thenb R a. SETS, RELATION AND FUNCTION Example. 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)} Example. Let A = R, the set of real numbers and let R be the relation <'. R {(1, 1), (2, 2),(1, 3), (2, 3), (2, 1)} Ifa <b. then b a (b is not less than a), so *<* is asymmetric. Example. Let A = {1.23} and let R = {(1.2, (2, 1), (2, 3)}. Is R symmetnc, asymmetric or antisymmetrien Let Let R (1, 2), (2, 3), 2, 1)} etric? Sol. Here A ={1. 2,3},R ={(1,2). (2, 1), (2, 3)}. Then R and R, are both subsets of A x A. Therefore, R and R, are both relations on A. Symmetry: R is not symmetric either since (2, 3) ER but (3, 2) £R Also (a, b) ER and (b, c) ER> (a, c) ER. Thus R is a transitive relation. Asymmetry: R is not asymmetric since both (1, 2) and (2, 1) ER Further,(1, 2) and (2, 3) E R, but (1, 3)ER|. Antisymmetry: R is not antisymeetric since (1, 2) and (2, 1) ER. Thus R is not a transitive relation Example. For a, b EN, the set of natural numbers, define a R b if2a+b= 10. The natural numbers a andb satisfying the relation 2 a +b= 10 are given by: a 1,b 8, a=2, b=6, a=3, b= 4, a =4, b =2 Anti-Symmetric Relation A relation R on a set A is called an anti-symmetric relation if a Rb and b R a implies that a = b ie, if (a b) E R and (b, a) E R *a=b. OR R {(1,8), 2,6), (3, 4), (4, 2)} A relation R on a set A is called anti-symmetric if a, bEA (a b) Since (3, 4) ER and (4, 2) ER but (3, 2) £ R. Therefore R is not a transitive relation. and (a. b)ER ° (6, a) ER. Circular Relation A relation R is called circular if (a, b) E R, (b, c) ER (c, a) E R. Example. Let A be the set of all lines in a plane. Let Li, L, E A. We define a relation R on A by L RL L I| L ie, if L, is paralel to L2. Since in any plane there exist different lines Lj and Lz such that L,|U and La l| L but L, * L; i.e, Ly R L, and L, R L, but L, # L2, therefore R is not an anti-symmetric relatin Sol. Let R be retlexIve and circular Example. Show that a relation is reflexive and circular iff it is equivalence relation. (a, a) E R and (a, b) E R, (6, c) ER » (c, a) ¬R. Example. LetA= { 1,2,3} Since (c, a) E R and (a, a)ER and R is circular, we have (a, c) ER (c, a) ER » (4, c) E R, which shows that R is symmetric. Again (a, b) ER, (6, c) ER (¢, a) E R, since R is circular Then R {(1,1), (1.2), (2,1)} is a relation on the set A. Since (1.2) ER and (2,1) ER but 1 2, therefore R is not anti-symmetric relation. But R 3,3) is an anti-symmetric relation on A. (a, c) E R as R is proved to be symmetric Example. For a, bEN the setof natural numbers define a R bif asb R is transitive. Let a, bE N such that a Rband b R a. Ris reflexive, symmetric and transitive and so is an equivalence relation. as band b s a. a=b. Conversely, let R be reflexive, symmetric and transitive ( of transitivity) (a, b) E R, (b, c) ER * (a, c) ER (c, a) ER R is an anti-symmetric relation. Note: Compatible Relation : A relation R in A is said to be compatible relation if it is retlexive ( of symmetry) symmetric. R is circular and so R is reflexive and circular. Transitive Relation Equivalence Relation A relation R on a set A is called an equivalence relation if R is reflexive, symmetric and transitive. Example. Let X be the set of all triangles in a plane. A relation R on a set A is called a transitive relation if aRb,bRc aRcVa, b, c E A ie, if (a, b)ER For any two triangles A and A, in X define A R A2, ifA, and Az are congruent triangles. Then () Ris Reflexive. Since each triangle is congruent to itself, so ARA for each A in X. and (b, c) ¬R (a, c) ER where a, b, c EA. 51 EMATN SPECTRUM DISCRETE MATHEM 50 (in) Ris Symmetric. Let A and A: E X such that A, R A. Then A and A are congruent Hence A R A triang SETS, RELATION AND FUNCTION Proof: Let cEC and a EA (c, a) E (SoR)" iff(a, c) E SOR (in R is Transitive. Let A. A: A: E X such that A R A; and A; R Az. congruent Then A. triang, then congruent triangles and so are A: and A;. This implies that the Aj and Az are also congruent Hence , R A Now (a, c) E SoR which means there exist b E B such that (a, b) E R and (b, c) ES (b, a)ER and (c, 6) ES or (c, b) ES' and (6, a) ER' (c, a) E R"oS" (SoR)= Ro s' So, R is reflenive, symmetric and transitive. Therefore, R is an equivalence relation on X. So 1.23. Let A, B, C, D be sets. Suppose R is a relation from A to B, S is relation from B toC and T is a relation from C to D. Then show that Partial Order Relation ditions A relation R on a set X is said to be a partial order relation if it satisfies the following three conditione. ()rRx, for every a E X (reflexivity) i) xRy and y Rr r=y(anti-symmetry) (iu) x Ry and y R: x R: (ransitivity). x, y, : EX Remark: The only equivalence relation on a set X which is also a partial order relation on X is the i relation Iy. that is, the relation defined by xRY iff x =y. (RoS)oT Ro(SoT). Proof: Let (a, d)E (RoS)oT Then there exists somecEC such that identi (a, e)ERoS and (c, d) ET Since (a, c) E RoS, so there exist b in B such that 1.21. Composition of Relations (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 SoR {(a. e):3bEB for which (a, b) E R. (h. c)E S} (b, d) E SoT (a, b) E R and (b, d) E SOT Also if A be any non-empty set and R. S be any two relations on A. Then composition of R and denoted by SoR is defined as Again (a, d)E Ro(SoT) SoR {(a. c):3bEA for which (a, b) ER. (b. c) E S} Example. Let A = {1,2, 3, 4, 5, 6, 7 (Ros)oT C Ro(SoT) Similarly, RofSOT) C (RoS)T .(2) R {(1,2). (2, 5). (3, 6), (7, 4} From equations (1) and (2) and S- {(1, 4). (7. 5), (3. 7). (4. 3)} Ro(SoT) = (RoS)oT. 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 XSX R(X,) S R(X;) (i) R(X, U x) =R(X,) U R(X) (Gi) R(X, nXi) E R(X,) n R(X;) Proof: () Let bE R(X1) Then SoR {(7,3)} ( RoS {(3,4), (4. 6)} {(2, ).6, 2). (6, 3), (4. 7) S={(4, 1). (5, 7). (7. 3), (3, 4)} R 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, aE X Now we discuss some theorems on composition of relations. aEX bER(X;) ds 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 (SoR) = R oS R(X,)C RX,) 52 SPECTRUM DisCRETE MATH (in) Let b E R(X, U X,) 53 TA there exist some a E X, UX; SETS, RELATION AND FUNCTION (a, b) E R(X, UX) SuDpose that R is an equivalence relation on a set X. Then S.t. 1.26. aEla]Va EX. ( aE[b} ifand only if[ a] = [6] Va, bEX. n al [b] or [a] n [6]=¢Va,bEXie, any two equivalence classes are disjoint or identical. Now a E X, U X; a E Xi or a E X; If aEX bER(X}) Similarly if a EX; bE RX:) Proof: () Since R is an equivalence relation on X. Ris reflexive. aRaVa EX. > aE[a]VaEEX. so bE R(X) U R(X;) RX, UX) S R(X)) U RX) (1) Also we know X, C X, UX2 and X, C X, U X (i) Let a, b EX such that a E [b] By part () R(X,)SRX, U X) aRb ..(2) bRa, since R is equivalence relation R(X) CROX, U X) Now we show that [ a ] = [b]. Let pE[al1 R(X,) U ROX)E RX, U X) ..3) From (1) and (2) pRa R(X, UX)= R(X,) U R(X;). From, (1) and (3), p R b, since R is an equivalence relation (ii) Let b E R(X, nX; there exist some a E X, N X2 SopE[a) p¬[b]. Therefore [ a]S[b]. Now let g E [b] 5.t. (a. b)E R(OX, n X) qRb. Also b R a (From (2). qRa and hence q E [ al Now a EX, NX2 aE X1 and a EX2 bER(XI) and bE R(X,) b E R(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. From (),a E[a] aE[b]. since [ a) = [ b] Cii) Let a, bEX. If[ a]n[b]=p, then we have nothing to prove. Ifa]tb]* 9, then there exists p E X such that p E [a] n[b}. pE[a] and p E [ b] 1.25. Equivalence Class Let R be an equivalence relation on a non-empty set X. let a E X. Then the equivalence class of a denoted by [a), is defined as follows al- {bEX:bRa}. Example. Let A ={ 1,2, 3. R (1, 1), (2, 1), (1, 2), 2,2). (3, 3)} []={1,2} since only I and 2 are related tol Ipl-la] 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]. A = [1]U[3] and [1] O[3]=¢ R {(1,1), 2.2), (3,3), (1,2), (2,1)} Proof: Let R be an equivalence relation on a set X. Therefore R is also a reflexive relation. aRava EX. aE[al Va EX. Also Then ..) is an equivalence relation on A. where [a) denotes the equivalence class of a. SPECTRUM DISCRETE MATHEM 55 We prove that X = U[a). a EX 3ut t this is against our supposition that [a) and [b] are distinct equivalence classes. So, our supposition SETS, RELATION AND FUNCTION Let a EX. that (oln [o]#p is wrong. aln b]=. Then a E[a] Thus From ( a EX Therefore, X is union of distinct equivalence classes and any two distinct equivalence classes are XC l] disjoint. Lence the set of distinct equivalence classes of R forms a partition of X. a EX Since 1 For any partition of A, there is an equivalence relation on X whose equivalence classes are the sets in ll={bEX:bRa), therefore lalCXVa EX the partition. Proof: Let {A}iEn be a partition of X. Therefore we have aEX 0 X A (i) A NA Ý ifA *u where A and u E A. For a, bEX, define a R b if and only ifa, b are in the same A. From (2) and (3), we get X = U [a]. lf we delete the repetitions from this union, we get X a EX union of distinct equivalence of X under R. Now we prove that any two distinct equivalence classes are disjoint. Then for any a, b, c, E A, we have: (0Ris reflexive Let a EX. by () aE ,U A so that aE A, for some 2EA. Let la J and [ b] be any two distinct equivalence classes where a, bEX We want to prove that [a ] n[b]=¢. f possible, let [a] n [b]* o. 3rEX such thatx E [ a] n[b] r¬[a ] andx E [b]. xRa andx Rb 4EA Therefore aR a. (i) R is symmetric Let a Rb. a and b belong to A, for some a EA. band a belong to some Ag. bRa. a Rrand b Rx Now we prove that [ a ] = [b]. Let pEla (i) Let a R b and b R c. pRa. AlsoaRr a and b belong to same A for some a E A and b, c belong to same Ag for some ß E A. From (5 (From(4 pRx. Alsox Rb bE A, and Ag both i.e., b E A, N A. PRb (Since R is an equivalence relation a-B, For if a * ß, then by (i) A, N Ap pE [b] a and c belong to same A [alSb] aRc. Now let q E [6] Therefore R is reflexive, symmetric as well as transitive relation on X. Hence R is an equivalence relation on X. 9Rb. Also b Rx gRx.(Since R is equivalencerelation) (From (5 Now we prove that each equivalence class of X is equal to A. Let [ a] denote the equivalence of a for any a EX. Then a]{bEX:bRa}. Also x R a From(4 qRa gE la [bEX:b and a are in the same A, for some 2 E A} = A Hence (a]= [6]. Equivalence class of X is equal to Ai. 57 S6 SPECTRUM DISCRETE MATHEMA Conversely we prove that each A is equal to some equivalence of X. SETS, RELATION AND FUNCTION Consider any A 1.32. Let A be a set with partition tA,,A2,.., A,and let R be the relation induced by the partition. Take any a E A Then R is an equivalence relation. Such an a exist, since A, is non-empty. Proof: ( Suppose that a E A. Since {A, A2.A is a partition of A, So a E A, for some i. We prove that A = [ a] Let bE A. Then a and b belong to same A there is a set A, of the partition such that a E A, and a E A, * aRa bRa and hence bE [ a bEA r¬A^ * AS[a] Ris reflexive (i) Let a, bEA such that a R 6. there is a subset A, of the partition such that both a and b are in A Now, let x ¬[a] xRa and hence x, a belong to same A ie both band a belong to some A, bRa, But aE A r¬ A Ris symmetric. (i) Let a, b, c E A such that a R b and b R c. Hence A= [a] there are subsets A, and A, of the partition such that a, b, E A, and b, cE A, AlE is the set of all equivalence classes under the relation R. We claim that A, =A, 1.29. Quotient of A by R If possible, suppose that A, *A, Then A, nA, = as all sets of the partition are Let R be an equivalence relation on A. Then the collection of equivalence classes of the elements of is called Quotient of A by R and is denoted by A | R. AR [a:aEA}. disjoint. But b is in both A, and A, Hence A, n A, *p. we 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 union of that a R c and so R is transitive. disjoint equivalence classes. ILLUSTRATIVE EXAMPLES Let P-U [a), a E A PSA 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 symmetric but not transitive. ASP (v) Give an example of a relation which is reflexive and transitive but not symmetric. From (1) and (2), we get () Give an example of a relation which is reflexive and anti-symmetric but not transitive. A=P. (vi) Give an example of a relation which is symmetric and transitive but not reflexive. 1.31. Relation Induced by the Partition vi) Give an example of a relation which is reflexive and anti-symmetric but neither symmetric nor transitive. 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 induced by the partition if for all a, bEA, Vu) Give an example of a relation which is neither reflexive, nor symmetric, nor transitive nor anti- symmetric. aRb there is a subset 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 anti-symmetric. MAN SPECTRUM DISCRETE MATHEN. 58 SETS, RELATION AND FUNCTION Let A= {1, 2, 3). Sol. () Let A {2,3,4}. () Then A x A- {2.2).(2. 3). (2, 4), (3, 2). (3, 3). (3, 4). (4, 2). (4, 3), (4, 4)} AXA {(1, 1), (1, 2), (1, 3). (2, 1), (2, 2), (2, 3), (3, 1),. (3, 2), (3, 3)}. Then Let R {2.2). (3. 3), 4, 4). (2, 3), (4, 3), (3, 4)} R- {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}. Let Since RC A x A, therefore R is a relation on A. Ris a relation on A as R SAXA. R is reflexive since (a, a) ER Va E A. Ris a reflexive relation on A since (a, a) E RVa¬A. R is not symmetric since (2, 3) E R but (3, 2) £ R. Ris also a anti-symmetric relation on A since (a, b) ER and (b, a) E R. R is not transitive since (2, 3) and (3, 4) ER but (2, 4) R. a=b. Ris not transitive since (1, 2) and (2, 3) ER but (1, 3) £ R. (viLet A={1,2,3} Further R is not anti-symmetric since (3,4) and (4, 3) E R but 3 4. (nLet A = {1,2 Then Ax A= {(1, 1),(1, 2), (2, 1), (2. 2)}. AxA={(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3,1). (3, 2), (3, 3)} Then Let R {(1, 2). 2. 1)} R-{(1, 1), (1,2), (2, 1), (2, 2)}. Let Ris not