# RD Sharma Solutions for Class 12 Maths Exercise 1.2 Chapter 1 Relation (Updated for 2021-22)

RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2: Students will mostly gain conceptual knowledge of equivalence relations through this practice. If a relation is reflexive, symmetric, and transitive, it is said to be an equivalence relationship. Class 12 is a crucial year for students in terms of shaping and achieving their long-term objectives. These RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2 problems’ solutions are created in a form of comprehensive manner, which develops students’ problem-solving abilities.

## Download RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2 PDF

RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2

### Access answers to Maths RD Sharma Solutions For Class 12 Chapter 1 – Relations Exercise 1.2 Question

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

Solution:

Given R = {(a, b): a – b is divisible by 3; a, b ∈ Z} is a relation

To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.

Let us check these properties on R.

Reflexivity:

Let a be an arbitrary element of R.

Then, a – a = 0 = 0 × 3

⇒ a − a is divisible by 3

⇒ (a, a) ∈ R for all a ∈ Z

So, R is reflexive on Z.

Symmetry:

Let (a, b) ∈ R

⇒ a − b is divisible by 3

⇒ a − b = 3p for some p ∈ Z

⇒ b − a = 3 (−p)

Here, −p ∈ Z

⇒ b − a is divisible by 3

⇒ (b, a) ∈ R for all a, b ∈ Z

So, R is symmetric on Z.

Transitivity:

Let (a, b) and (b, c) ∈ R

⇒ a − b and b − c is divisible by 3

⇒ a – b = 3p for some p ∈ Z

And b − c = 3q for some q ∈ Z

Adding the above two equations, we get

a − b + b – c = 3p + 3q

⇒ a − c = 3 (p + q)

Here, p + q ∈ Z

⇒ a − c is divisible by 3

⇒ (a, c) ∈ R for all a, c ∈ Z

So, R is transitive on Z.

Therefore R is reflexive, symmetric, and transitive.

Hence, R is an equivalence relation on Z.

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

Solution:

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

To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.

Let us check these properties on R.

Reflexivity:

Let a be an arbitrary element of the set Z.

Then, a ∈ R

⇒ a − a = 0 = 0 × 2

⇒ 2 divides a − a

⇒ (a, a) ∈ R for all a ∈ Z

So, R is reflexive on Z.

Symmetry:

Let (a, b) ∈ R

⇒ 2 divides a − b

⇒ (a-b)/2 = p for some p ∈ Z

⇒ (b-a)/2 = – p

Here, −p ∈ Z

⇒ 2 divides b − a

⇒ (b, a) ∈ R for all a, b ∈ Z

So, R is symmetric on Z

Transitivity:

Let (a, b) and (b, c) ∈ R

⇒ 2 divides a−b and 2 divides b−c

⇒ (a-b)/2 = p and (b-c)/2 = q for some p, q ∈ Z

Adding the above two equations, we get

(a – b)/2 + (b – c)/2 = p + q

⇒ (a – c)/2 = p +q

Here, p+ q ∈ Z

⇒ 2 divides a − c

⇒ (a, c) ∈ R for all a, c ∈ Z

So, R is transitive on Z.

Therefore R is reflexive, symmetric, and transitive.

Hence, R is an equivalence relation on Z.

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

To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.

Let us check these properties on R.

Reflexivity:

Let a be an arbitrary element of R. Then,

⇒ a − a = 0 = 0 × 5

⇒ a − a is divisible by 5

⇒ (a, a) ∈ R for all a ∈ Z

So, R is reflexive on Z.

Symmetry:

Let (a, b) ∈ R

⇒ a − b is divisible by 5

⇒ a − b = 5p for some p ∈ Z

⇒ b − a = 5 (−p)

Here, −p ∈ Z [Since p ∈ Z]

⇒ b − a is divisible by 5

⇒ (b, a) ∈ R for all a, b ∈ Z

So, R is symmetric on Z.

Transitivity:

Let (a, b) and (b, c) ∈ R

⇒ a − b is divisible by 5

⇒ a − b = 5p for some Z

Also, b − c is divisible by 5

⇒ b − c = 5q for some Z

Adding the above two equations, we get

a −b + b − c = 5p + 5q

⇒ a − c = 5 ( p + q )

⇒ a − c is divisible by 5

Here, p + q ∈ Z

⇒ (a, c) ∈ R for all a, c ∈ Z

So, R is transitive on Z.

Therefore R is reflexive, symmetric, and transitive.

Hence, R is an equivalence relation on Z.

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.
Show that R is an equivalence relation on Z.

Solution:

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

To prove equivalence relation the given relation must be reflexive, symmetric, and transitive.

Let us check these properties on R.

Reflexivity:

Let a ∈ N

Here, a − a = 0 = 0 × n

⇒ a − a is divisible by n

⇒ (a, a) ∈ R

⇒ (a, a) ∈ R for all a ∈ Z

So, R is reflexive on Z.

Symmetry:

Let (a, b) ∈ R

Here, a − b is divisible by n

⇒ a − b = n p for some p ∈ Z

⇒ b − a = n (−p)

⇒ b − a is divisible by n [ p ∈ Z⇒ − p ∈ Z]

⇒ (b, a) ∈ R

So, R is symmetric on Z.

Transitivity:

Let (a, b) and (b, c) ∈ R

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

⇒ a − b= n p for some p ∈ Z

And b−c = n q for some q ∈ Z

a – b + b − c = n p + n q

⇒ a − c = n (p + q)

⇒ (a, c) ∈ R for all a, c ∈ Z

So, R is transitive on Z.

Therefore R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on Z.

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

To prove equivalence relation the given relation must be reflexive, symmetric and transitive.

Let us check these properties on R.

Reflexivity:

Let a be an arbitrary element of Z.

Then, a ∈ R

Clearly, a + a = 2a is even for all a ∈ Z.

⇒ (a, a) ∈ R for all a ∈ Z

So, R is reflexive on Z.

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.

Transitivity:

Let (a, b) and (b, c) ∈ R

⇒ a + b and b + c are even

Now, let a + b = 2x for some x ∈ Z

And b + c = 2y for some y ∈ Z

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 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?

Solution:

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

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

To prove equivalence relation the given relation must be reflexive, symmetric and transitive.

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

Let us check these properties on R.

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.

Symmetry:

Let (m, n) ∈ R.

Then, m − n is divisible by 13

⇒ m − n = 13p

Here, p ∈ Z

⇒ n – m = 13 (−p)

Here, −p ∈ Z

⇒ 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

⇒ m – n = 13p and n − o = 13q for some p, q ∈ Z

Adding the above two equations, we get

m – n + n − o = 13p + 13q

⇒ m−o = 13 (p + q)

Here, p + q ∈ Z

⇒ 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.

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

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

To prove equivalence relation the given relation must be reflexive, symmetric and transitive.

Reflexivity:

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

Then, (a, b) ∈ A

⇒ a b = b a

⇒ (a, b) R (a, b)

Thus, R is reflexive on A.

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.

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:

Given set A = {x ∈ Z; 0 ≤ x ≤ 12}

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

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

To prove equivalence relation the given relation must be reflexive, symmetric and transitive.

Reflexivity:

Let a be an arbitrary element of A.

Then, a ∈ R

⇒ a = a [Since every element is equal to itself]

⇒ (a, a) ∈ R for all a ∈ A

So, R is reflexive on A.

Symmetry:

Let (a, b) ∈ R

⇒ a b

⇒ b = a

⇒ (b, a) ∈ R for all a, b ∈ A

So, R is symmetric on A.

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, R is an equivalence relation on A.

Therefore R is reflexive, symmetric and transitive.

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

## RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2: Important Topics From The Chapter

Let us have a look at important topics covered in this exercise.

• Types of relations
• Void relation
• Universal relation
• Identity relation
• Reflexive relation
• Symmetric relation
• Transitive relation
• Antisymmetric relation
We have provided complete details of RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.2. If you have any queries related to CBSE Class 12 Exam, feel free to ask us in the comment section below.

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