*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 2* ^{mn}*. Since each subset of

*A*×

*B*defines relation from

*A*to

*B*, so total number of relations from

*A*to

*B*is 2

*. Among these 2*

^{mn}*relations the void relation*

^{mn}*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* = 2*b}*. 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),…..}.

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**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 *I _{A}* = {(

*a*,

*a*) :

*a*Î

*A*} on

*A*is called the identity relation on

*A*.

In other words, a relation *I _{A}* 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.*

** **

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** ****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)} = {5

*p*:

*p*Î

*Z*} = { [1] = .

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)

*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) .

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**Example: 20** The solution set of , are

(a) [8] È [6] (b) [8] È [14] (c) [6] È [13] (d) [8] È [6] È [13]

**Solution:** (c) Þ

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

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

Where [6], [13] 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 *S*o*R* 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 *R*o*S* ¹ *SoR*. Also (*SoR*)^{–1} = *R*^{–1}*oS*^{–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.

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