Class 12 RD Sharma Solutions – Chapter 1 Relations – Exercise 1.2 | Set 1

Question 1. Show that the relation R = {(a,b): a-b is divisible by 3;, a, b ∈ Z} is an equivalence relation. 

Solution: 

According to question, relation R = {(a,b): a-b is divisible by 3;, a, b ∈ Z} 

We have to show that R is an equivalence relation. 

(i) reflexibity: 

let a = z 

=> a – a = 0 

=> 0 is divisible by 3 

:. a is divisible by 3 

(a, a) ∈ R, 

Hence, relation R is reflexive. 

(ii) symmetry: 

Let a, b∈ Z and (a, b) ∈ R 

=> a – b is divisible by 3 

=> a – b = 3x, for some x ∈ Z 

=> b -a = 3 (-x) 

here, -x ∈ Z 

:. b – a is divisible by 3 

Hence, (b- a) ∈ R for all a, b ∈ Z 

So, relation R is symmetric.

(iii) transitivity: 

let (a, b) and (b, c) ∈ R 

a – b and b – c is divisible by 3 

a – b = 3x, for some x ∈ Z & (eqn. 1)

b – c = 3y, for some y ∈ Z (eqn. 2) 

Adding the above equations 1 and 2, we get 

a – c = 3(x + y) 

here, x and y ∈ Z 

so, a – c is divisible by 3 

Hence, a – c ∈ R for all a, c ∈ Z 

thus relation R is transitive. 

As we know if a relation is reflexive, symmetric and transitive at the same time, then it is called an equivalence relation. 

Therefore, Relation R is an equivalence relation.

Question 2. Show that the relation R on the set of integers, given by R = {(a, b): 2 divides a – b}, is an equivalence relation. 

Solution: 

Given, R = {(a, b): 2 divides a – b} which is defined on Z 

We have to show that relation R is an equivalence relation

(i) reflexibility : 

let na be an arbitrary element on set Z 

then a ∈  R

a – a = 0 

=> 0 X 2 = 0 

2 divides a – a 

=> (a, a) ∈  R 

:. Relation R is a reflexive relation.

Now, (ii) symmetric : 

let (a,  b) ∈  R

 2 divides a – b 

a – b = 2x for some x ∈  Z 

b – a = 2 (- x) where – x ∈ Z 

2 divides b – a 

=> (b, a) ∈  R 

 and  (a, b) ∈  R 

therefore, relation R is symmetric. 

(iii) transitivity : 

let a, b, c ∈  Z such that (a, b) ∈  R and (b, c) ∈  R 

then, (a, b) ∈  R => 2 divides b – a 

b – a = 2x for some x ∈  Z  (eqn.1)

and(b – c) ∈  R 

2 divides c – b => c – b = 2y for some y ∈  Z  (eqn 2)

on solving equation 1 and 2, 

c – a = 2 (x + y) where x + y ∈  Z 

2 divides c – a 

(a, c) ∈ R 

Therefore, relation R is transitive. 

Hence, relation R is an equivalence relation. 

Question 3. Prove that the relation R on Z defined by (a, b) ∈ R <=> a – b is divisible by 5 is an equivalence relation on Z.

Solution:

Given, relation R on Z defined by (a, b) ∈ R <=> a – b is divisible by 5

as, we have to prove it a equivalence relation, the relation R must have to be reflexive, symmetric as well as transitive.

(i) Reflexibility : 

Let a be an arbitrary element of R 

=> a – a = 0 

=> 0 is divisible by 5 

=> a – a is divisible by 5 

=> (a, a) ∈ R for all a ∈ Z

:. relation R is reflexive.

Again, (ii) symmetry

Let (a, b) ∈ R

=> a − b is divisible by 5

=> a − b = 5x for some x ∈ Z  , b − a = 5 (−x)

since (−x) ∈ Z 

b − a is divisible by 5

 (b, a) ∈ R for all a, b ∈ Z

So, relation R is symmetric. 

(iii) transitivity ; 

Let (a, b) and (b, c) ∈ R

=> a − b is divisible by 5

=>a − b = 5x for some x ∈ Z (eqn. 1)

Also, b − c is divisible by 5

=> b − c = 5y for some y ∈ Z  (eqn. 2)

Adding the above two equations,

a −b + b − c = 5x + 5y

=> a − c = 5 (x + y)

=> a − c is divisible by 5

Since, x + y ∈ Z

=>(a, c) ∈ R for all a, c ∈ Z

So, R is transitive on Z.

As relation R is reflexive, symmetric and transitive,

Hence, R is an equivalence relation on Z.

Question 4. Let n be a fixed positive integer. Define a relation R on Z as follows : (a, b) ∈ R <=> a – b is divisible by n. 

Solution:

 Given (a, b) ∈ R ⇔ a − b is divisible by n is a relation R defined on Z.

 it is necessary that the relation R should be reflexive, symmetric and transitive.

(i) Reflexivity:

Let a ∈ N

 a − a = 0 

= 0 × n

=> a − a is divisible by n

=> (a, a) ∈ R

=> (a, a) ∈ R for all a ∈ Z

:.R is reflexive on Z.

(ii) Symmetry:

Let (a, b) ∈ R

 a − b is divisible by n

=> a − b = nx for some x ∈ Z

=> b − a = n (−x)

=> b − a is divisible by n

=> (b, a) ∈ R  

So, relation R is symmetric on Z.

Similarly (iii) Transitivity:

Let (a, b) and (b, c) ∈ R

 a − b is divisible by n and b − c is divisible by n.

=> a − b= n x for some x ∈ Z  (eqn . 1)

And b−c = ny for some y ∈ Z   (eqn. 2)

a – b + b − c = nx + n y

=> a − c = n (p + q)

=> (a, c) ∈ R for all a, c ∈ Z

So, relation R is transitive on Z.

Therefore, R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on Z.

Question 5. Let Z be the set of integers. Show that the relation R  = {(a, b): a, b ∈ Z and a + b is even} is an equivalence relation on Z.

Solution: 

Given R = {(a, b): a, b ∈ Z and a + b is even} is a relation defined on R.

Also given that Z be the set of integers

it is necessary that the given relation should be reflexive, symmetric and transitive.

Reflexivity:

Let a be an arbitrary element of Z.  

Then, a ∈ R

a + a = 2a is even for all a ∈ Z.

=> (a, a) ∈ R for all a ∈ Z

So, R is reflexive on Z.

(ii) Symmetry:

Let (a, b) ∈ R

=> a + b is even

=> b + a is even

=>(b, a) ∈ R for all a, b ∈ Z

So, R is symmetric on Z.

(iii) Transitivity:

Let (a, b) and (b, c) ∈ R

=> a + b and b + c are even

let a + b = 2x for some x ∈ Z  (eqn. 1)

And b + c = 2y for some y ∈ Z (eqn. 2)

Adding the above two equations, we get

A + 2b + c = 2x + 2y

=> a + c = 2 (x + y − b), which is even for all x, y, b ∈ Z

Thus, (a, c) ∈ R

So, R is transitive on Z.

Therefore, relation R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on Z

6. m is said to be related to n if m and n are integers and m − n is divisible by 13. Does this define an equivalence relation?

Question 6. m is said to be related to n if m and n are integers and m − n is divisible by 13. Does this define an equivalence relation?

Solution:

Given that m is said to be related to n if m and n are integers and m − n is divisible by 13

So, we have to check whether the given relation is equivalence or not.

Let R = {(m, n): m, n ∈ Z : m − n is divisible by 13}

Reflexivity: 

Let m be an arbitrary element of Z.  

Then, m ∈ R

=> m − m = 0 = 0 × 13

=> m − m is divisible by 13

=> (m, m) is reflexive on Z.

Now, Symmetry:  

Let (m, n) ∈ R.  

Then, m − n is divisible by 13

=> m − n = 13p

Here, p ∈ Z

=>n – m = 13 (−p)  

=> n − m is divisible by 13

=>(n, m) ∈ R for all m, n ∈ Z  

So, R is symmetric on Z.

Transitivity:  

Let (m, n) and (n, o) ∈ R

=>m − n and n − o are divisible by 13  (eqn. 1)

=>m – n = 13p and n − o = 13q for some p, q ∈ Z  (eqn.2)

Adding the above two equations, we get

=>m – n + n − o = 13p + 13q

=> m−o = 13 (p + q)

=> m − o is divisible by 13

=>(m, o) ∈ R for all m, o ∈ Z

So, R is transitive on Z.

Therefore, R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on Z.

Question 7. Let R be a relation on the set A of ordered pair of integers defined by (x, y) R (u, v) if xv = y u. Show that R is an equivalence relation.

Solution:

First let R be a relation on A

It is given that set A of ordered pair of integers defined by (x, y) R (u, v) if xv = y u

We have to check whether the given relation is equivalence or not.

Reflexivity:  

Let (a, b) be an arbitrary element of the set A.  

Then, (a, b) ∈ A

=> a b = b a  

=>(a, b) R (a, b)

Thus, R is reflexive on A.

Again, Symmetry:  

Let (x, y) and (u, v) ∈ A such that (x, y) R (u, v). Then,

x v = y u

=> v x = u y

=> u y = v x

=>(u, v) R (x, y)

So, R is symmetric on A.

Transitivity:  

Let (x, y), (u, v) and (p, q) ∈R such that (x, y) R (u, v) and (u, v) R (p, q)

=> x v = y u and u q = v p

Multiplying the corresponding sides, we get

x v × u q = y u × v p

=> x q = y p

=>(x, y) R (p, q)

So, R is transitive on A.

Therefore, R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on A.

Question 8. Show that the relation R on the set A = {x ∈ Z; 0 ≤ x ≤ 12}, given by R = {(a, b): a = b}, is an equivalence relation. Find the set of all elements related to 1.

Solution:

According to question, set A = {x ∈ Z; 0 ≤ x ≤ 12}

Also given that relation R = {(a, b): a = b} is defined on set A

We have to find whether the given relation is equivalence or not.

(i) Reflexivity:  

Let a be an arbitrary element of A.

Then, a ∈ R

=>a = a (because, every element is equal to itself)

=> (a, a) ∈ R for all a ∈ A

:. R is reflexive on A.

(ii) Symmetry:  

Let (a, b) ∈ R

=>b = a

=> (b, a) ∈ R for all a, b ∈ A

So, R is symmetric on A.

(iii) Transitivity: 

Let (a, b) and (b, c) ∈ R

=> a =b and b = c

=>a = b c

=>a = c

=> (a, c) ∈ R

So, R is transitive on A.

Hence, relation R is an equivalence relation on A.

Therefore, relation R is reflexive, symmetric and transitive.

The set of all elements related to 1 is {1}.

Question 9. Let L be the set of all lines in XY – plane and R be the relation in L defined as R = {(L_1, L₂) : L₁ is parallel to L₂}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Solution:

Given, L is the set of lines.

 R = {(L₁, L₂) : L₁ is parallel to L₂} be a relation on L 

Now, (i) reflexibility : 

as we know, a line is always parallel to itself,

So, L₁, L₂ ∈ R 

Therefore, R is reflexive. 

Again, (ii) symmetry : 

Assume, L₁, L₂ ∈  L and (L1, L2) ∈ R

Since, L₁ is parallel to L₂

:. L₂ is also parallel to L₁

Thus, relation R is symmetric. 

(iii) transitivity : 

let L₁, L₂, L₃ ∈ R in such a way that (L₁, L₂) ∈ R and (L₂, L₃) ∈ R

L₁ is parallel to L₂

L₂ is parallel to L₃ 

Therefore L₁ is parallel to L₃

Hence, relation R is transitive.

So, relation R is an equivalence relation.  

Now, the set of all lines related to line y = 2x + 4 is y = 2x + c for all c ∈ R.

Question 10. Show that the relation R, defined on the set A of all polygons as R = {(P1, P2):P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4, and 5?

Solution:

The relation R is defined as R = {(P₁, P₂):P₁ and P₂ have the same number of sides} 

(i) reflexibility: 

Let P be any polygon in A 

Then, P and P have same number of sides 

(P, P) ∈ R 

Thus, relation R is reflexive. 

(ii) symmetry: 

Let P₁ and P₂ be any polygon in A such that, (P₁, P₂ ) ∈ R 

(P₁, P₂) ∈ R

P₁ and P₂ have same number of sides

so, P₂ and P₁ will have also same number of sides

Hence (P₁, P₂) ∈ R 

so, relation R is symmetric.

at last (iii) transitivity : 

let P₁, P₂, P₃ be three polygons in A in such a way that (P₁ , P₂) ∈ R and (P₂, P₃) ∈ R 

then, (P₁, P₂₎ have same number of sides

(P₂, P₃) have same number of sides

:. P₁ and P₃ will also have same number of sides.

Hence, relation R is transitive 

so, relation R is an equivalence relation. 

Now, let P be a polygon in A such that (P, T) ∈ R, where T is a right angle triangle with sides 3, 4 and 5 

Then, (P, T) ∈ R

polygon P and triangle T have same number of sides 

Hence, the set of all elements in A related to T is the set of all triangles in A.