Relations

1.2.1 Definition.
Let A and B be two non-empty sets, then every subset of A × B defines a relation from A to B and every
relation from A to B is a subset of A × B.
Let R  A  B and (a, b)  R. Then we say that a is related to b by the relation R and write it as a R b . If
(a, b)  R , we write it as a R b .
Example: Let A = {1, 2, 5, 8, 9}, B = {1, 3} we set a relation from A to B as: a R b iff a  b; a  A, b  B .
Then R = {(1, 1)}, (1, 3), (2, 3)}  A × B
(1) Total number of relations : Let A and B be two non-empty finite sets consisting of m and n elements
respectively. Then A × B consists of mn ordered pairs. So, total number of subset of A × B is 2mn. Since each
ps
subset of A × B defines relation from A to B, so total number of relations from A to B is 2mn. Among these 2mn
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relations the void relation  and the universal relation A × B are trivial relations from A to B.
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(2) Domain and range of a relation : Let R be a relation from a set A to a set B. Then the set of all first
components or coordinates of the ordered pairs belonging to R is called the domain of R, while the set of all
ud

second components or coordinates of the ordered pairs in R is called the range of R.

St

Thus, Dom (R) = {a : (a, b)  R} and Range (R) = {b : (a, b)  R}.

It is evident from the definition that the domain of a relation from A to B is a subset of A and its range is a
subset of B.
(3) Relation on a set : Let A be a non-void set. Then, a relation from A to itself i.e. a subset of A × A is
called a relation on set A.

Example: 1 Let A = {1, 2, 3}. The total number of distinct relations that can be defined over A is
(a) 29 (b) 6 (c) 8 (d) None of these
Solution: (a) n( A  A)  n( A).n( A)  3  92

So, the total number of subsets of A  A is 29 and a subset of A  A is a relation over the set A.
Example: 2 Let X  {1, 2, 3, 4, 5} and Y  {1, 3, 5, 7, 9} . Which of the following is/are relations from X to Y
(a) R1  {( x, y)| y  2  x, x  X, y  Y } (b) R2  {(1,1), (2,1), (3, 3), (4, 3), (5, 5)}
(c) R3  {(1,1), (1, 3)(3, 5), (3, 7), (5, 7)} (d) R4  {(1, 3), (2, 5), (2, 4), (7, 9)}
Solution: (a,b,c) R4 is not a relation from X to Y, because (7, 9)  R4 but (7, 9)  X  Y .
Example: 3 Given two finite sets A and B such that n(A) = 2, n(B) = 3. Then total number of relations from
A to B is
(a) 4 (b) 8 (c) 64 (d) None of these
Solution: (c) Here n( A  B) = 2 × 3 = 6
Since every subset of A × B defines a relation from A to B, number of relation from A to B is
equal to number of subsets of A  B  26  64, which is given in (c).

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Example: 4 The relation R defined on the set of natural numbers as {(a, b) : a differs from b by 3}, is given
by
(a) {(1, 4, (2, 5), (3, 6),.....} (b) {(4, 1), (5, 2), (6, 3),.....} (c){(1, 3), (2, 6), (3, 9),..}
Solution: (b) R  {(a, b) : a, b  N , a  b  3} = {((n  3), n) : n  N }  {(4, 1), (5, 2), (6, 3).....}

1.2.2 Inverse Relation.

–1
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 1  {(b, a) : (a, b)  R}

Clearly (a, b)  R  (b, a)  R–1 . Also, Dom (R) = Range (R 1 ) and Range (R) = Dom (R 1 )
Example : Let A = {a, b, c}, B = {1, 2, 3} and R = {(a, 1), (a, 3), (b, 3), (c, 3)}.
–1
Then, (i) R = {(1, a), (3, a), (3, b), (3, c)}
(ii) Dom (R) = {a, b, c} = Range (R 1 )

(iii) Range (R) = {1, 3} = Dom (R 1 )

Example: 5 Let A = {1, 2, 3}, B = {1, 3, 5}. A relation R : A  B is defined by R = {(1, 3), (1, 5), (2, 1)}. Then R 1 is
defined by
(a) {(1,2), (3,1), (1,3), (1,5)} (b)
ps {(1, 2), (3, 1), (2, 1)} (c) {(1, 2), (5, 1), (3, 1)}(d)
te
1 1
Solution: (c) ( x, y)  R  (y, x)  R ,  R  {(3,1), (5,1), (1, 2)} .
ys

Example: 6 The relation R is defined on the set of natural numbers as {(a, b) : a = 2b}. Then R 1 is given by
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(a) {(2, 1), (4, 2), (6, 3).....} (b) {(1, 2), (2, 4), (3, 6)....} (c) R 1 is not defined (d)
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Solution: (b) R = {(2, 1), (4, 2), (6, 3),......} So, R 1 = {(1, 2), (2, 4), (3, 6),.....}.

1.2.3 Types of Relations.

(1) Reflexive relation : A relation R on a set A is said to be reflexive if every element of A is related to itself.
Thus, R is reflexive  (a, a)  R for all a  A.
A relation R on a set A is not reflexive if there exists an element a  A such that (a, a)  R.
Example: Let A = {1, 2, 3} and R = {(1, 1); (1, 3)}
Then R is not reflexive since 3  A but (3, 3)  R
:  The identity relation on a non-void set A is always reflexive relation on A. However, a reflexive relation
on A is not necessarily the identity relation on A.
 The universal relation on a non-void set A is reflexive.
(2) Symmetric relation : A relation R on a set A is said to be a symmetric relation iff
(a, b)  R  (b, a)  R for all a, b  A
i.e. aRb  bRa for all a, b  A.
it should be noted that R is symmetric iff R 1  R
 The identity and the universal relations on a non-void set are symmetric relations.

 A relation R on a set A is not a symmetric relation if there are at least two elements a, b  A such that
(a, b)  R but (b, a)  R.
 A reflexive relation on a set A is not necessarily symmetric.

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 (3) Anti-symmetric relation : Let A be any set. A relation R on set A is said to be an anti-symmetric relation
iff (a, b)  R and (b, a)  R  a = b for all a, b  A.
Thus, if a  b then a may be related to b or b may be related to a, but never both.
Example: Let N be the set of natural numbers. A relation R  N  N is defined by xRy iff x divides y(i.e., x/y).

Then x R y, y R x  x divides y, y divides x  x  y

 The universal relation on a set A containing at least two elements is not anti-symmetric, because if
a  b are in A, then a is related to b and b is related to a under the universal relation will imply that
a = b but a  b.
 The set {(a, a) : a  A}  D is called the diagonal line of A  A . Then “the relation R in A is
antisymmetric iff R  R 1  D ”.
(4) Transitive relation : Let A be any set. A relation R on set A is said to be a transitive relation iff
(a, b)  R and (b, c)  R  (a, c)  R for all a, b, c  A i.e., aRb and bRc  aRc for all a, b, c  A.
In other words, if a is related to b, b is related to c, then a is related to c.
ps
Transitivity fails only when there exists a, b, c such that a R b, b R c but a R c.
te

Example: Consider the set A = {1, 2, 3} and the relations

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R1  {(1, 2),(1, 3)} ; R 2 = {(1, 2)}; R3 = {(1, 1)}; R 4 = {(1, 2), (2, 1), (1, 1)}
ud
St

Then R1 , R 2 , R3 are transitive while R4 is not transitive since in R 4 ,(2, 1) R 4 ;(1, 2) R 4 but (2, 2)  R4 .

 The relation ‘is congruent to’ on the set T of all triangles in a plane is a transitive relation.
(5) Identity relation : Let A be a set. Then the relation IA = {(a, a) : a  A} on A is called the identity relation on A.
In other words, a relation IA on A is called the identity relation if every element of A is related to itself only. Every
identity relation will be reflexive, symmetric and transitive.
Example: On the set = {1, 2, 3}, R = {(1, 1), (2, 2), (3, 3)} is the identity relation on A .

Note :  It is interesting to note that every identity relation is reflexive but every reflexive relation need not be
an identity relation.
Also, identity relation is reflexive, symmetric and transitive.
(6) Equivalence relation : A relation R on a set A is said to be an equivalence relation on A iff
(i) It is reflexive i.e. (a, a)  R for all a  A
(ii) It is symmetric i.e. (a, b)  R  (b, a)  R, for all a, b  A
(iii) It is transitive i.e. (a, b)  R and (b, c)  R  (a, c)  R for all a, b, c  A.

Note :  Congruence modulo (m) : Let m be an arbitrary but fixed integer. Two integers a and b are said to
be congruence modulo m if a  b is divisible by m and we write a  b (mod m).

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 Thus a  b (mod m)  a  b is divisible by m. For example, 18  3 (mod 5) because 18 – 3 = 15
which is divisible by 5. Similarly, 3  13 (mod 2) because 3 – 13 = –10 which is divisible by 2. But 25
 2 (mod 4) because 4 is not a divisor of 25 – 3 = 22.
The relation “Congruence modulo m” is an equivalence relation.

Important Tips
 If R and S are two equivalence relations on a set A , then R  S is also an equivalence relation
on A.
 The union of two equivalence relations on a set is not necessarily an equivalence relation on the
set.
 The inverse of an equivalence relation is an equivalence relation.

1.2.4 Equivalence Classes of an Equivalence Relation.

Let R be equivalence relation in A(  ) . Let a  A . Then the equivalence class of a, denoted by [a] or {a } is defined
as the set of all those points of A which are related to a under the relation R. Thus [a] = {x  A : x R a}.
It is easy to see that
ps
(1) b  [a]  a  [b] (2) b  [a]  [a]  [b] (3) Two equivalence classes are either disjoint or identical.
te
ys

As an example we consider a very important equivalence relation x  y(mod n) iff n divides ( x  y), n is a fixed
positive integer. Consider n  5. Then
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[0] = {x : x  0 (mod 5)} = {5p : p  Z} = { 0,  5,  10,  15,.....}

[1] = {x : x  1(mod 5)}  {x : x  1  5k, k  Z}  {5k  1 : k  Z}  {1, 6,11, .......,  4,  9, .....} .

One can easily see that there are only 5 distinct equivalence classes viz. [0], [1], [2], [3] and [4], when n = 5.

Example: 7 Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, the minimum number of ordered pairs which
when added to R make it an equivalence relation is
(a) 5 (b) 6 (c) 7 (d) 8
Solution: (c) R is reflexive if it contains (1, 1), (2, 2), (3, 3)
 (1, 2)  R, (2, 3)  R

 R is symmetric if (2, 1), (3, 2)  R. Now, R  {(1, 1), (2, 2), (3, 3), (2, 1), (3, 2), (2, 3), (1, 2)}

R will be transitive if (3, 1); (1, 3)  R. Thus, R becomes an equivalence relation by adding (1, 1) (2, 2) (3,
3) (2, 1) (3,2) (1, 3) (3, 1). Hence, the total number of ordered pairs is 7.
Example: 8 The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on set A = {1, 2, 3} is
(a) Reflexive but not symmetric (b) Reflexive but not
transitive
(c) Symmetric and Transitive (d) Neither symmetric nor
transitive
Solution: (a) Since (1, 1); (2, 2); (3, 3)  R therefore R is reflexive. (1, 2)  R but (2, 1)  R, therefore R is not
symmetric. It can be easily seen that R is transitive.
Example: 9 Let R be the relation on the set R of all real numbers defined by a R b iff | a  b | 1 . Then R is

04
 (a) Reflexive and Symmetric (b) Symmetric only (c) Transitive only (d)
Solution: (a) | a  a | 0  1 a R a  a  R

 R is reflexive, Again a R b  | a  b | 1 | b  a | 1  bRa

1 1 1
 R is symmetric, Again 1R and R1 but  1
2 2 2
 R is not anti-symmetric
Further, 1 R 2 and 2 R 3 but 1 R 3
[ |1  3| 2  1 ]

 R is not transitive.
Example: 10 The relation "less than" in the set of natural numbers is [UPSEAT 1994, 98; AMU
1999]
(a) Only symmetric (b) Only transitive (c) Only reflexive (d) Equivalence relation
Solution: (b) Since x  y, y  z  x  z  x, y, z  N

 x R y, yR z  x R z ,  Relation is transitive ,  x  y does not give y  x ,  Relation is not symmetric.

Since x  x does not hold, hence relation is not reflexive.

Example: 11 With reference to a universal set, the inclusion of a subset in another, is relation, which is
(a) Symmetric only (b) Equivalence relation ps (c) Reflexive only (d) None of these
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Solution: (d) Since A  A  relation ' ' is reflexive.
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Since A  B, B  C  A  C
ud

 relation ' ' is transitive.

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But A  B,  B  A ,  Relation is not symmetric.

Example: 12 Let A  {2, 4, 6, 8} . A relation R on A is defined by R  {(2, 4), (4, 2), (4, 6), (6, 4)} . Then R is

(a) Anti-symmetric (b) Reflexive (c) Symmetric (d) Transitive

Solution: (c) Given A = {2, 4, 6, 8}
R = {(2, 4)(4, 2) (4, 6) (6, 4)}
(a, b)  R  (b, a)  R and also R 1  R . Hence R is symmetric.
Example: 13 Let P  {( x, y)| x 2  y 2  1, x, y  R} . Then P is

(a) Reflexive (b) Symmetric (c) Transitive (d) Anti-symmetric

Solution: (b) Obviously, the relation is not reflexive and transitive but it is symmetric, because x 2  y 2  1  y 2  x 2  1 .
Example: 14 Let R be a relation on the set N of natural numbers defined by nRm  n is a factor of m (i.e., n|m). Then
R is
(a) Reflexive and symmetric (b) Transitive and symmetric
(c) Equivalence (d) Reflexive, transitive but not symmetric
Solution: (d) Since n | n for all n N , therefore R is reflexive. Since 2 | 6 but 6 | 2 , therefore R is not symmetric.

Let n R m and m R p  n|m and m|p  n|p  nRp. So R is transitive.

Example: 15 Let R be an equivalence relation on a finite set A having n elements. Then the number of ordered pairs
in R is
(a) Less than n (b) Greater than or equal to n (c) Less than or equal to n (d)
Solution: (b) Since R is an equivalence relation on set A, therefore (a, a)  R for all a  A . Hence, R has at least n
ordered pairs.

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Example: 16 Let N denote the set of all natural numbers and R be the relation on N  N defined by (a, b) R (c, d) if
ad(b  c)  bc(a  d), then R is [Roorkee 1995]
(a) Symmetric only (b) Reflexive only (c) Transitive only (d) An equivalence
relation
Solution: (d) For (a, b), (c, d)  N × N
(a, b)R(c, d)  ad(b  c)  bc(a  d)

Reflexive: Since ab(b  a) = ba(a  b)ab  N ,

 (a, b)R(a, b) ,  R is reflexive.

Symmetric: For (a, b), (c, d)  N  N , let (a, b)R(c, d)

 ad(b  c)  bc(a  d)  bc(a  d)  ad(b  c)  cb(d  a)  da(c  b)  (c, d)R(a, b)

 R is symmetric
Transitive: For (a, b), (c, d), (e, f )  N  N , Let (a, b)R(c, d), (c, d)R(e, f )

 ad(b  c)  bc(a  d) , cf (d  e)  de(c  f )

 adb  adc  bca  bcd .....(i) and cfd  cfe  dec  def .......(ii)
(i) × ef  (ii) × ab gives, adbef  adcef  cfdab  cfeab = bcaef  bcdef  decab  defab

ps
 adcf (b  e)  bcde(a  f )  af (b  e)  be(a  f )  (a, b)R (e, f ) .  R is transitive. Hence R is an equivalence
te
relation.
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Example: 17 For real numbers x and y, we write x Ry  x  y  2 is an irrational number. Then the relation R is
ud

(a) Reflexive (b) Symmetric (c) Transitive (d) None of these

St

Solution: (a) For any x  R, we have x  x  2  2 an irrational number.

 xRx for all x. So, R is reflexive.

R is not symmetric, because 2R1 but 1 R 2 , R is not transitive also because 2 R 1 and 1R 2 2

but 2 R 2 2 .

Example: 18 Let X be a family of sets and R be a relation on X defined by ‘A is disjoint from B’. Then R is
(a) Reflexive (b) Symmetric (c) Anti-symmetric (d) Transitive
Solution: (b) Clearly, the relation is symmetric but it is neither reflexive nor transitive.
Example: 19 Let R and S be two non-void relations on a set A. Which of the following statements is false
(a) R and S are transitive  R  S is transitive (b) R and S are transitive  R  S is transitive
(c) R and S are symmetric  R  S is symmetric (d) R and S are reflexive  R  S is reflexive
Solution: (a) Let A  {1, 2, 3} and R = {(1, 1), (1, 2)}, S = {(2, 2) (2, 3)} be transitive relations on A.

Then R  S = {(1, 1); (1, 2); (2, 2); (2, 3)}

Obviously, R  S is not transitive. Since (1, 2)  R  S and (2, 3)  R  S but (1, 3)  R  S .

Example: 20 The solution set of 8 x  6(mod 14), x  Z , are

(a) [8]  [6] (b) [8]  [14] (c) [6]  [13] (d) [8]  [6]  [13]
1
Solution: (c) 8 x  6  14 P (P  Z )  x  [14 P  6] , x  Z
8
1
x = (7 P  3)  x = 6, 13, 20, 27, 34, 41, 48,.......
4

06
  Solution set = {6, 20, 34, 48,.....}  {13, 27, 41, ......} = [6]  [13].
Where [6], [13] are equivalence classes of 6 and 13 respectively.

1.2.5 Composition of Relations.

Let R and S be two relations from sets A to B and B to C respectively. Then we can define a relation SoR from A to C
such that (a, c)  SoR   b  B such that (a, b)  R and (b, c)  S.
This relation is called the composition of R and S.
For example, if A = {1, 2, 3}, B = {a, b, c, d}, C = {p, q, r, s} be three sets such that R = {(1, a), (2, c), (1, c), (2, d)} is a
relation from A to B and S = {(a, s), (b, q), (c, r)} is a relation from B to C. Then SoR is a relation from A to C given by SoR
= {(1, s) (2, r) (1, r)}
In this case RoS does not exist.
In general RoS  SoR. Also (SoR)–1 = R–1oS–1.

Example: 21 If R is a relation from a set A to a set B and S is a relation from B to a set C, then the relation SoR
(a) Is from A to C (b) Is from C to A (c) Does not exist (d) None of these
Solution: (a) It is obvious.
Example: 22 If R  A  B and S  B  C be two relations, then (SoR)1  ps
te
(a) S 1oR 1 (b) R 1oS 1 (c) SoR (d) RoS
ys

Solution: (b) It is obvious.

ud

Example: 23 If R be a relation < from A = {1,2, 3, 4} to B = {1, 3, 5} i.e., (a, b)  R  a  b, then RoR 1 is
St

(a) {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
(b) {(3, 1) (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}
(c) {(3, 3), (3, 5), (5, 3), (5, 5)}
(d) {(3, 3) (3, 4), (4, 5)}
Solution: (c) We have, R = {(1, 3); (1, 5); (2, 3); (2, 5); (3, 5); (4, 5)}
R 1  {(3, 1), (5, 1), (3, 2), (5, 2); (5, 3); (5, 4)}

Hence RoR 1 = {(3, 3); (3, 5); (5, 3); (5, 5)}
Example: 24 Let a relation R be defined by R = {(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)} then R 1oR is
(a) {(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)} (b) {(1, 1), (4, 4), (7, 7), (3, 3)}
(c) {(1, 5), (1, 6), (3, 6)} (d) None of these
1 1
Solution: (a) We first find R , we have R  {(5, 4); (4, 1); (6, 4); (6, 7); (7, 3)} we now obtain the elements of R 1 oR we first

pick the element of R and then of R 1 . Since (4, 5)  R and (5, 4)  R 1 , we have (4, 4)  R 1oR

Similarly, (1, 4)  R, (4, 1)  R 1  (1, 1)  R 1oR

(4, 6)  R, (6, 4)  R 1  (4, 4)  R 1oR, (4, 6)  R, (6, 7)  R 1  (4, 7)  R 1oR

(7, 6)  R, (6, 4)  R 1  (7, 4)  R 1oR, (7, 6)  R, (6, 7)  R 1  (7, 7)  R 1oR

(3, 7)  R, (7, 3)  R 1  (3, 3)  R 1oR,

Hence R 1oR  {(1, 1); (4, 4); (4, 7); (7, 4), (7, 7); (3, 3)}.

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1.2.6 Axiomatic Definitions of the Set of Natural Numbers (Peano's Axioms).
The set N of natural numbers (N = {1, 2, 3, 4......}) is a set satisfying the following axioms (known as peano's axioms)
(1) N is not empty.

(2) There exist an injective (one-one) map S : N  N given by S(n)  n  , where n is the immediate successor of

n in N i.e., n  1  n .
(3) The successor mapping S is not surjective (onto).
(4) If M  N such that,

(i) M contains an element which is not the successor of any element in N, and

(ii) m  M  m   M , then M  N
This is called the axiom of induction. We denote the unique element which is not the successor of any element is 1.
Also, we get 1  2, 2  3 .

Note :  Addition in N is defined as,

n  1  n

n  m   (n  m)
 Multiplication in N is defined by,
ps
te

n .1  n
ys

n.m   n.m  n
ud
St

08
 Assignment
Level-1
1. A relation from P to Q is [AMU 1998]
(a) A universal set of P × Q (b) P × Q (c) An equivalent set of P × Q (d) A subset of P × Q
2. Let R be a relation from a set A to set B, then
(a) R = A  B (b) R = A  B (c) R  A × B (d) R  B × A
3. Let A = {a, b, c} and B = {1, 2}. Consider a relation R defined from set A to set B. Then R is equal to set[Kurukshetra CEE 1995
(a) A (b) B (c) A × B (d) B × A
4. Let n(A) = n. Then the number of all relations on A is
2
(a) 2n (b) 2(n)! (c) 2n (d) None of these
5.
relations from A to B is
ps
If R is a relation from a finite set A having m elements to a finite set B having n elements, then the number of
te

(a) 2mn (b) 2mn  1 (c) 2mn (d) mn

ys

6. Let R be a reflexive relation on a finite set A having n-elements, and let there be m ordered pairs in R. Then
ud

(a) m  n (b) m  n (c) m  n (d) None of these

St

7. The relation R defined on the set A = {1, 2, 3, 4, 5} by R = {(x, y) : | x 2  y 2 | 16} is given by

(a) {(1, 1), (2, 1), (3, 1), (4, 1), (2, 3)} (b) {(2, 2), (3, 2), (4, 2), (2, 4)}
(c) {(3, 3), (3, 4), (5, 4), (4, 3), (3, 1)} (d) None of these
8. A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by; xRy  x is relatively prime to y. Then domain of R is
(a) {2, 3, 5} (b) {3, 5} (c) {2, 3, 4} (d) {2, 3, 4, 5}
9. Let R be a relation on N defined by x  2y  8 . The domain of R is
(a) {2, 4, 8} (b) {2, 4, 6, 8} (c) {2, 4, 6} (d) {1, 2, 3, 4}
10. If R  {( x, y)| x, y  Z, x 2  y 2  4} is a relation in Z, then domain of R is
(a) {0, 1, 2} (b) {0, – 1, – 2} (c) {– 2, – 1, 0, 1, 2} (d) None of these
11. If A = {1, 2, 3} , B = {1, 4, 6, 9} and R is a relation from A to B defined by ‘x is greater than y’. The range of R is
(a) {1, 4, 6, 9} (b) {4, 6, 9} (c) {1} (d) None of these
12. R is a relation from {11, 12, 13} to {8, 10, 12} defined by y  x  3 . Then R 1 is
(a) {(8, 11), (10, 13)} (b) {(11, 18), (13, 10)} (c) {(10, 13), (8, 11)} (d) None of these
13. Let A = {1, 2, 3}, B = {1, 3, 5}. If relation R from A to B is given by R ={(1, 3), (2, 5), (3, 3)}. Then R 1 is
(a) {(3, 3), (3, 1), (5, 2)} (b) {(1, 3), (2, 5), (3, 3)} (c) {(1, 3), (5, 2)} (d) None of these
14. Let R be a reflexive relation on a set A and I be the identity relation on A. Then
(a) R  I (b) I  R (c) R  I (d) None of these
15. Let A = {1, 2, 3, 4} and R be a relation in A given by R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1), (3, 1), (1, 3)}. Then
R is
(a) Reflexive (b) Symmetric (c) Transitive (d) An equivalence relation

09
16. An integer m is said to be related to another integer n if m is a multiple of n. Then the relation is
(a) Reflexive and symmetric (b) Reflexive and transitive (c) Symmetric and
transitive (d) Equivalence relation

17. The relation R defined in N as aRb  b is divisible by a is

(a) Reflexive but not symmetric (b) Symmetric but not transitive (c) Symmetric and
transitive (d) None of these
18. Let R be a relation on a set A such that R  R 1 , then R is
(a) Reflexive (b) Symmetric (c) Transitive (d) None of these
19. Let R = {(a, a)} be a relation on a set A. Then R is
(a) Symmetric (b) Antisymmetric
(c) Symmetric and antisymmetric (d) Neither symmetric nor anti-
symmetric
20. The relation "is subset of" on the power set P(A) of a set A is
(a) Symmetric (b) Anti-symmetric (c) Equivalency relation (d) None of these
21. The relation R defined on a set A is antisymmetric if (a, b)  R  (b, a)  R for
(a) Every (a, b)  R (b) No (a, b)  R (c) No (a, b), a  b, R (d) None of these
22. ps
In the set A = {1, 2, 3, 4, 5}, a relation R is defined by R = {(x, y)| x, y  A and x < y}. Then R is
te
(a) Reflexive (b) Symmetric (c) Transitive (d) None of these
Let A be the non-void set of the children in a family. The relation x is a brother of y on A is
ys

23.

(a) Reflexive (b) Symmetric (c) Transitive (d) None of these

ud

24. Let A = {1, 2, 3, 4} and let R= {(2, 2), (3, 3), (4, 4), (1, 2)} be a relation on A. Then R is
St

(a) Reflexive (b) Symmetric (c) Transitive (d) None of these

25. The void relation on a set A is
(a) Reflexive (b) Symmetric and transitive (c) Reflexive and symmetric (d)Reflexive and transitive
26. Let R1 be a relation defined by R1  {(a, b)| a  b, a, b  R} . Then R1 is
(a) An equivalence relation on R (b) Reflexive, transitive but not
symmetric
(c) Symmetric, Transitive but not reflexive (d) Neither transitive not reflexive but symmetric
27. Let A = {p, q, r}. Which of the following is an equivalence relation on A
(a) R1 = {(p, q), (q, r), (p, r), (p, p)} (b) R2 = {(r, q), (r, p), (r, r), (q, q)}
(c) R3 = {(p, p), (q, q), (r, r), (p, q)} (d) None of these
28. Which one of the following relations on R is an equivalence relation
(a) a R1 b | a || b | (b) aR2b  a  b (c) aR3b  a divides b (d) aR4 b  a  b

29. If R is an equivalence relation on a set A, then R 1 is

(a) Reflexive only (b) Symmetric but not transitive (c) Equivalence (d)
30. R is a relation over the set of real numbers and it is given by nm  0 . Then R is
(a) Symmetric and transitive (b) Reflexive and symmetric (c) A partial order relation(d)
31. In order that a relation R defined on a non-empty set A is an equivalence relation, it is sufficient, if R
(a) Is reflextive (b) Is symmetric
(c) Is transitive (d) Possesses all the above three properties
32. The relation "congruence modulo m" is
(a) Reflexive only (b) Transitive only (c) Symmetric only (d) An equivalence relation

10
33. Solution set of x  3 (mod 7), x  Z, is given by
(a) {3} (b) {7 p  3 : p  Z} (c) {7 p  3 : p  Z} (d) None of these
34. Let R and S be two equivalence relations on a set A. Then
(a) R  S is an equivalence relation on A (b) R  S is an equivalence relation on A
(c) R  S is an equivalence relation on A (d) None of these
35. Let R and S be two relations on a set A. Then
(a) R and S are transitive, then R  S is also transitive (b) R and S are transitive, then R  S is also transitive
(c) R and S are reflexive, then R  S is also reflexive (d) R and S are symmetric then R  S is also symmetric
36. Let R = {(1, 3), (2, 2), (3, 2)} and S = {(2, 1), (3, 2), (2, 3)} be two relations on set A = {1, 2, 3}. Then RoS =
(a) {(1, 3), (2, 2), (3, 2), (2, 1), (2, 3)} (b) {(3, 2), (1, 3)}
(c) {(2, 3), (3, 2), (2, 2)} (d) {(2, 3), (3, 2)}
37. In problem 36, RoS 1 
(a) {(2, 2), (3, 2) (b) {(1, 2), (2, 2), (3, 2)} (c) {(1, 2), (2, 2)} (d) {(1, 2), (2, 2), (3, 2), (2,
3)}

Level-2
38. Let R be a relation on the set N be defined by {(x, y)| x, y  N, 2x + y = 41}. Then R is
(a) Reflexive (b) Symmetric ps
(c) Transitive (d) None of these
te
39. Let L denote the set of all straight lines in a plane. Let a relation R be defined by R   , ,   L . Then R is
ys

(a) Reflexive (b) Symmetric (c) Transitive (d) None of these

ud

40. Let T be the set of all triangles in the Euclidean plane, and let a relation R be defined on T by aRb iff a  b, a, b  T . Then R is
St

(a) Reflexive but not transitive (b) Transitive but not symmetric (c) Equivalence (d) None of these
41. Two points P and Q in a plane are related if OP = OQ, where O is a fixed point. This relation is
(a) Partial order relation (b) Equivalence relation (c) Reflexive but not symmetric (d)Reflexive but not transitive
42. Let r be a relation over the set N × N and it is defined by (a, b)r(c, d)  a  d  b  c. Then r is

(a) Reflexive only (b) Symmetric only (c) Transitive only (d) An equivalence relation
43. Let L be the set of all straight lines in the Euclidean plane. Two lines l1 and l 2 are said to be related by the relation R iff l1 is
parallel to l 2 . Then the relation R is

(a) Reflexive (b) Symmetric (c) Transitive (d) Equivalence

44. Let n be a fixed positive integer. Define a relation R on the set Z of integers by, aRb  n | a  b |. Then R is

(a) Reflexive (b) Symmetric (c) Transitive (d) Equivalence

11
 Answersheet

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
d c c c a a d d c c c a a b a,b b a b c b
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
c c b,c c b b d a c d d d c b b,c,d c b d b c
41 42 43 44 ps
te
b d a,b a,b
ys

,c, ,c,
d d
ud
St

12