# JEE Notes: Relations and Functions notes for JEE Mains and Advanced – Part 1 Important notes of Relations and Functions for JEE Mains and Advanced.

## Definition: Relation

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  and (a, b) Î R. Then we say that a is related to b by the relation R and write it as . If ,  we write it as.

Example: Let A = {1, 2, 5, 8, 9}, B = {1, 3} we set a relation from A to B as: a R b iff  . 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 subset of A × B defines relation from A to B, so total number of relations from A to B is 2mn. Among these 2mn relations the void relation f and the universal relation A × B are trivial relations from A to B.

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

Thus, Dom (R) = {a : (a, b) Î R} and Range (R) = {b : (a, b) Î 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)                              (b) 6                                  (c) 8                            (d) None of these

Solution: (a)

So, the total number of subsets of  is  and a subset of  is a relation over the set A.

Example: 2   Let  and . Which of the following is/are relations from X to Y

(a)                               (b)

(c)                                 (d)

Solution: (a,b,c)  is not a relation from X to Y, because (7, 9)  but (7, 9) .

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  = 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  which is given in (c).

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),..}                                   (d) None of these

Solution: (b) =

## Inverse Relation

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–1, is a relation from B to A and is defined by

Clearly (a, b) Î R Û (b, a) Î R–1 .      Also, Dom (R) = Range  and Range (R) = Dom

Example :  Let A = {a, b, c}, B = {1, 2, 3} and R = {(a, 1), (a, 3), (b, 3), (c, 3)}.

Then,  (i)  R–1 = {(1, a), (3, a), (3, b), (3, c)}

(ii)  Dom (R) = {a, b, c} = Range

(iii)  Range (R) = {1, 3} = Dom

Example: 5   Let A = {1, 2, 3}, B  = {1, 3, 5}. A relation  is defined by R = {(1, 3), (1, 5), (2, 1)}. Then  is defined by

(a) {(1,2), (3,1), (1,3), (1,5)}                               (b) {(1, 2), (3, 1), (2, 1)}      (c)                            {(1, 2), (5, 1), (3, 1)}     (d) None of these

Solution: (c) ,   .

Example: 6   The relation R is defined on the set of natural numbers as {(a, b) : a = 2b}. Then  is given by

(a) {(2, 1), (4, 2), (6, 3)…..}                                 (b) {(1, 2), (2, 4), (3, 6)….}      (c)                             is not defined           (d) None of these

Solution: (b) R = {(2, 1), (4, 2), (6, 3),……} So,  = {(1, 2), (2, 4), (3, 6),…..}.

#### Free Coaching of IIT JEE by IITians

Get the free and most effective coaching from IIT Graduates and score less than 5000 rank in JEE Mains and Advanced. To join the program, follow the steps:

2. Join the learning group code: 517329

### 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 Û (a, a) Î R for all a Î A.

A relation R on a set A is not reflexive if there exists an element a Î A such that (a, a) Ï R.

Example: Let A = {1, 2, 3} and R = {(1, 1); (1, 3)}

Then R  is not reflexive since  but (3, 3) Ï R

Note : q 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.

q 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) Î R Þ (b, a) Î R for all a, b Î A

i.e.                                              aRb Þ bRa for all a, b Î A.

it should be noted that R  is symmetric iff

Note :  q  The identity and the universal relations on a non-void set are symmetric relations.

q A relation R on a set A is not a symmetric relation if there are at least two elements a, b Î A such that (a, b) Î R but (b, a) Ï R.

q  A reflexive relation on a set A  is not necessarily symmetric.

(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) Î R and (b, a) Î R Þ a = b for all a, b Î A.

Thus, if a ¹ 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  is defined by  iff x divides y(i.e., x/y).

Then  divides y, y divides

Note : q  The identity relation on a set A is an anti-symmetric relation.

q  The universal relation on a set A containing at least two elements is not anti-symmetric, because if       a ¹ 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 ¹ b.

q  The set  is called the diagonal line of . Then “the relation R in A is antisymmetric iff ”.

(4) Transitive relation : Let A be any set. A relation R on set A is said to be a transitive relation iff

(a, b) Î R and (b, c) Î R Þ (a, c) Î R for all a, b, c Î A i.e.,  aRb and bRc Þ aRc for all a, b, c Î A.

In other words, if a is related to b, b is related to c, then a is related to c.

Transitivity fails only when there exists a, b, c such that a R b, b R c but a R c.

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

; = {(1, 2)}; = {(1, 1)};  = {(1, 2), (2, 1), (1, 1)}

Then , ,  are transitive while  is not transitive since in  but .

Note : q  The identity and the universal relations on a non-void sets are transitive.

q  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 Î 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 : q  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) Î R for all a Î A

(ii) It is symmetric i.e. (a, b) Î R Þ (b, a) Î R, for all a, b Î A

(iii) It is transitive i.e. (a, b) Î R and (b, c) Î R Þ (a, c) Î R for all a, b, c Î A.

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

Thus  (mod m)  is divisible by m. For example,  (mod 5) because 18 – 3 = 15 which is divisible by 5. Similarly,  (mod 2) because 3 – 13 = –10 which is divisible by 2. But 25 ¹ 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 Ç 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.

#### Free Coaching of IIT JEE by IITians

Get the free and most effective coaching from IIT Graduates and score less than 5000 rank in JEE Mains and Advanced. To join the program, follow the steps:

2. Join the learning group code: 517329

### Equivalence Classes of an Equivalence Relation.

Let R be equivalence relation in . Let . Then the equivalence class of a, denoted by [a] or  is defined as the set of all those points of A which are related to a under the relation R. Thus [a] = {x Î A : x R a}.

It is easy to see that

(1)           (2)        (3) Two equivalence classes are either disjoint or identical.

As an example we consider a very important equivalence relation  iff n divides  is a fixed positive integer. Consider  Then

 = {x : x   0 (mod 5)} = {5p : p Î Z} = {

 = .

One can easily see that there are only 5 distinct equivalence classes viz. , , ,  and , 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)

R is symmetric if (2, 1), (3, 2) Î R. Now,

R will be transitive if (3, 1); (1, 3) Î 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) Î R therefore R is reflexive. (1, 2) Î R but (2, 1) Ï 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 . Then R is                                      ]

(a) Reflexive and Symmetric (b)Symmetric only

(c)Transitive only                   (d)Anti-symmetric only

Solution: (a)

R is reflexive,     Again a R b Þ

R is symmetric,   Again  and  but

R is not anti-symmetric

Further, 1 R 2 and 2 R 3 but 1 R 3

[]

R is not transitive.

Example: 10 The relation “less than” in the set of natural numbers is

(a) Only symmetric     (b) Only transitive

(c) Only reflexive       (d) Equivalence relation

Solution: (b) Since

,  Relation is transitive ,    does not give ,  Relation is not symmetric.

Since  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                                                            [Karnataka CET 1995]

(a) Symmetric only     (b) Equivalence relation   (c)Reflexive only   (d) None of these

Solution: (d) Since  relation  is reflexive.

Since

relation  is transitive.

But Þ,   Relation is not symmetric.

Example: 12 Let . A relation R on A is defined by . Then R is                                            [Karnataka CET 1995]

(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) Î R Þ (b, a) Î R and also .  Hence R is symmetric.

Example: 13 Let . 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 .

Example: 14 Let R be a relation on the set N of natural numbers defined by nRm Û 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 , therefore R is reflexive. Since 2 | 6 but , therefore R is not symmetric.

Let n R m and m R p Þ n|m and m|p Þ n|p Þ 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)None of these

Solution: (b) Since R  is an equivalence relation on set A, therefore (a, a) Î R for all . Hence, R has at least n ordered pairs.

Example: 16 Let N denote the set of all natural numbers and R be the relation on  defined by (a, b) R (c, d) if  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) Î N × N

Reflexive: Since  = ,

,  R is reflexive.

Symmetric: For , let

Þ  Þ  Þ

R  is symmetric

Transitive:  For  Let

,

Þ           …..(i)          and                     …….(ii)

(i) ×  (ii) ×  gives,   =

Þ  Þ   Þ .  R is transitive. Hence R  is an equivalence relation.

Example: 17 For real numbers x  and y, we write x Ry Û  is an irrational number. Then the relation R  is

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

Solution: (a) For any  we have  an irrational number.

Þ  for all x. So, R is reflexive.

R is not symmetric, because  but , R is not transitive also because R 1 and  but.

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 Þ R È S is transitive

(b) R and S are transitive Þ R Ç S is transitive

(c) R and S are symmetric Þ R È S is symmetric

(d)R and S are reflexive Þ R Ç S is reflexive

Solution: (a) Let  and R = {(1, 1), (1, 2)}, S = {(2, 2) (2, 3)} be transitive relations on A.

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

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

#### Free Coaching of IIT JEE by IITians

Get the free and most effective coaching from IIT Graduates and score less than 5000 rank in JEE Mains and Advanced. To join the program, follow the steps:

2. Join the learning group code: 517329

Example: 20 The solution set of , are

(a)  È                  (b)  È                   (c)  È             (d)  È  È 

Solution: (c)   Þ

Þ x  =    Þ x = 6, 13, 20, 27, 34, 41, 48,…….

Solution set = {6, 20, 34, 48,…..} È {13, 27, 41, ……} =  È .

Where ,  are equivalence classes of 6 and 13 respectively.

### 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) Î SoR Û \$ b Î B such that (a, b) Î R and (b, c) Î 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 ¹ 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  and  be two relations, then

(a)                      (b)                         (c)                        (d)

Solution: (b) It is obvious.

Example: 23 If R be a relation < from A = {1,2, 3, 4} to B = {1, 3, 5} i.e.,  then  is

(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)}

{(3, 1), (5, 1), (3, 2), (5, 2); (5, 3); (5, 4)}

Hence = {(3, 3); (3, 5); (5, 3); (5, 5)}

### 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  given by , where  is the immediate successor of  in N i.e., .

(3) The successor mapping S is not surjective (onto).

(4) If  such that,

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

(ii) , then

This is called the axiom of induction. We denote the unique element which is not the successor of any element is 1.

#### Free Coaching of IIT JEE by IITians

Get the free and most effective coaching from IIT Graduates and score less than 5000 rank in JEE Mains and Advanced. To join the program, follow the steps: