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),…..}.

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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 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.

 

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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

[0] = {x : x   0 (mod 5)} = {5p : 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 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.

 

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