Let R , be a relation on the set N , defined by { ( x , y ) ∣ x , y ∈ N , 2 x + y = 41 } . Then prove that R , is neither reflexive nor symmetric and nor transitive.

Given: Let R , be a relation on the set N , defined by { ( x , y ) ∣ x , y ∈ N , 2 x + y = 41 } We have to prove that R , is neither reflexive nor symmetric and nor transitive. Checking 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 , on a set A , is not reflexive if there exists an element a ∈ A , such that ( a , a ) ∉ R Given relation is 2 x + y = 41 . . . i , In relation, we substitute x = a and y = a , we get 3 a = 41 ⇒ a = 41 3 , a ∉ N , so ( a , a ) ∉ R . ∴ Given relation is not reflexive. Checking Symmetric relation : A relation R , on a set A , is said to be a symmetric relation if ( a , b ) ∈ R ⇒ ( b , a ) ∈ R , for all a , b ∈ A . i.e. a R b ⇒ b R a , for all a , b ∈ A . In i substitute x = 1 , we get y = 39 , so ( 1 , 39 ) ∈ R , but, if we put x = 39 then y = - 37 ⇒ y ∉ N ⇒ ( 39 , 1 ) ∉ R . So, given relation is not symmetric. Checking Transitive relation: Let A , be any set. A relation R on A , is said to be a transitive relation if ( a , b ) ∈ R and ( b , c ) ∈ R ⇒ ( a , c ) ∈ R , for all a , b , c ∈ A i.e. a R b and b R c ⇒ a R c , for all a , b , c ∈ A In relation i put x = 20 , we get y = 1 , and putting x = 1 , we get y = 39 , but when we put x = 20 , we get y = 1 . So, ( 20 , 1 ) ∈ R , 1 , 39 ∈ R but ( 20 , 39 ) ∉ R . ∴ Given relation, R , is not transitive. Hence, proved.

Let n , be a fixed positive integer. Define a relation R , on the set of integers Z , a R b ⇔ n ∣ ( a - b ) . Then prove that R , is an equivalence.

Given: n , be a fixed positive integer. Define a relation R , on the set of integers Z , a R b ⇔ n ∣ a - b . We have to prove that the relation R , is equivalence. Equivalence relation: A relation R , on a set A , is said to be an equivalence relation on A if (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 ∈ A Checking Reflexive relation: Let a ∈ N , we have a - a = 0 = 0 × n a - b , is divisible by n ⇒ a , a ∈ R So, the relation is reflexive. Checking Symmetric relation: Let a , b ∈ R . Then, a - b , is divisible by n a - b = n k , for some k ∈ Z b - a = n - k b - a , is divisible by n . Thus, a , b ∈ R ⇒ b , a ∈ R , for all a , b ∈ Z So, given relation is symmetric. Checking Transitive relation: Let a , b ∈ R ⇒ a - b , is divisible by n ⇒ a - b = n k , for some k ∈ Z b , c ∈ R ⇒ b - c , is divisible by n ⇒ b - c = n l , for some l ∈ Z Therefore, a , b ∈ R and b - c ∈ R . a - b = n k and b - c = n l ⇒ a - b + b - c = n k + n l ⇒ a - c = n k + l a - c , is divisible by n ⇒ a - c ∈ R Thus, a , b ∈ R and b , c ∈ R a , c ∈ R , for all a , b , c ∈ Z So given relation is transitive. ∴ Given relation, R , is an equivalence relation.

Let S be a set of all square matrices of order 2 . If a relation R is defined on set S such that ARB ⇒ AB = BA , then identify the type of relation of R A , B ∈ S among reflexive, symmetric and transitive.

Given: S be a set of all square matrices of order 2 . If a relation R defined on set S such that ARB ⇒ AB = BA , We have to identify the type of relation of R A , B ∈ S among reflexive, symmetric, and transitive. Checking 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 on a set A is not reflexive if there exists an element a ∈ A such that a , a ∉ R . Given a relation R is ARB ⇒ AB = BA . . . 1 , Put B = A in 1 , we get A 2 = A 2 , So A , A ∈ R , So given relation R is reflexive. Checking symmetric relation : A relation R on a set A is said to be a symmetric relation if a , b ∈ R ⇒ b , a ∈ R for all a , b ∈ A . i.e. a R b ⇒ b R a for all a , b ∈ A Let A , B ∈ R we have to show that B , A ∈ R in relation 1 , we put B = A and A = B , We get B A = A B ⇒ B , A ∈ R , So, given relation R is symmetric. Checking transitive relation: Let A be any set. A relation R on A is said to be a transitive relation if a , b ∈ R and b , c ∈ R ⇒ a , c ∈ R for all a , b , c ∈ A i.e. a Rb and bRc ⇒ aRc for all a , b , c ∈ A Let A , B ∈ R and B , C ∈ R , we have to show that A , C ∈ R . From the relation 1 A B = B A and B C = C B , we have to show that A , C ∈ R , i.e. A C = C A , ⇒ A C B = C A B ⇒ A B C ≠ C B A ∴ A . C ∉ R So, given relation is R is not transitive, ∴ Given relation, R is reflexive and symmetric but not transitive.

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 2 - Set Theory and Relations>Exercise-1>Q 11

HARD

JEE Main/Advance

IMPORTANT

Earn 100

The Cartesian product $A \times A$ has $16$ , elements, $S = {(a, b) \in A \times A ∣ a < b}$ . $(- 1, 2)$ and $(0, 1)$ , are two elements belonging to $S$ . Find the set containing the remaining elements of $S$ .

Important Questions on Set Theory and Relations

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 2 - Set Theory and Relations>Exercise-1>Q 12

Identify the type of relation among reflexive, symmetric and transitive.

(i) $R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}$ , on set $A = {1, 2, 3}$

(ii) $P = \{(x, y) ∣ x^{2} + y^{2} = 1, x, y \in R\}$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 2 - Set Theory and Relations>Exercise-1>Q 13

Prove that the relation “less than” in the set of natural number is transitive but not reflexive and symmetric.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 2 - Set Theory and Relations>Exercise-1>Q 14

Let $A = {p, q, r}$ . Then which of the following is an equivalence relation on $A$ ?

(i) $R = {(p, q), (q, r), (p, r) (p, p)}$
(ii) $R = {(p, p), (q, q), (r, r) (q, p)}$
(iii) $R = {(p, p), (q, q), (r, r)}$
$(iv) R = {(p, p), (q, q), (r, r), (p, q), (q, r), (p, r)}$
$(v) R = {(p, p), (q, q), (r, r), (p, q), (q, p)}$