RD Sharma Solutions for Class 12 Maths Exercise 1.1 Chapter 1 Relation (Updated for 2021-22)

RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1: The RD Sharma Class 12 Solution PDF of Chapter 1 Exercise 1.1 Relations can be downloaded from the link given. Solutions for correct answer practice are prepared by experts in the best possible way and are easily understood by the students. In this practice, students will gain knowledge about relations and types. RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1 are detailed and detailed to make learning easier for students. In this exercise, there are two levels according to the increasing order of difficulty.

Download RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1 PDF

RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1

Access answers to RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1

1. Let A be the set of all human beings in a town at a particular time. Determine whether the following relation is reflexive, symmetric, and transitive:

(i) R = {(x, y): x and y work at the same place}

(ii) R = {(x, y): x and y live in the same locality}

(iii) R = {(x, y): x is wife of y}

(iv) R = {(x, y): x is father of y}

Solution:

(i) Given R = {(x, y): x and y work at the same place}

Now we have to check whether the relation is reflexive:

Let x be an arbitrary element of R. 

Then, x ∈R   

⇒ x and x work at the same place is true since they are the same.

⇒(x, x) ∈R [condition for reflexive relation]

So, R is a reflexive relation.

Now let us check the Symmetric relation:

Let (x, y) ∈R

⇒x and y work at the same place [given]

⇒y and x work at the same place

⇒(y, x) ∈R

So, R is a symmetric relation.

Transitive relation:

Let (x, y) ∈R and (y, z) ∈R. 

Then, x and y work at the same place. [Given]

y and z also work at the same place. [(y, z) ∈R]

⇒ x, y, and z all work at the same place.

⇒x and z work at the same place.

⇒ (x, z) ∈R

So, R is a transitive relation.

Hence R is reflexive, symmetric, and transitive.

(ii) Given R = {(x, y): x and y live in the same locality}

Now we have to check whether the relation R is reflexive, symmetric, and transitive.

Let x be an arbitrary element of R. 

Then, x ∈R

It is given that x and x live in the same locality is true since they are the same.

So, R is a reflexive relation.

Symmetry:

Let (x, y) ∈ R

⇒ x and y live in the same locality [given]

⇒ y and x live in the same locality

⇒ (y, x) ∈ R 

So, R is a symmetric relation.

Transitivity:

Let (x, y) ∈R and (y, z) ∈R. 

Then,

x and y live in the same locality and y and z live in the same locality

⇒ x, y, and z all live in the same locality

⇒ x and z live in the same locality 

⇒ (x, z) ∈ R

So, R is a transitive relation.

Hence R is reflexive, symmetric, and transitive.

(iii) Given R = {(x, y): x is wife of y}

Now we have to check whether the relation R is reflexive, symmetric, and transitive.

First, let us check whether the relation is reflexive:

Let x be an element of R. 

Then, x is the wife of x cannot be true.

⇒ (x, x) ∉R

So, R is not a reflexive relation.

Symmetric relation:

Let (x, y) ∈R

⇒ x is the wife of y 

⇒ x is female and y is male

⇒ y cannot be the wife of x as y is the husband of x

⇒ (y, x) ∉R  

So, R is not a symmetric relation.

Transitive relation:

Let (x, y) ∈R, but (y, z) ∉R

Since x is the wife of y, but y cannot be the wife of z, y is the husband of x.

⇒ x is not the wife of z

⇒(x, z) ∈R

So, R is a transitive relation.

(iv) Given R = {(x, y): x is father of y}

Now we have to check whether the relation R is reflexive, symmetric, and transitive.

Reflexivity:

Let x be an arbitrary element of R. 

Then, x is the father of x cannot be true since no one can be the father of himself.

So, R is not a reflexive relation.

Symmetry: 

Let (x, y) ∈R

⇒ x is a father of y

⇒ y is the son/daughter of x

⇒ (y, x) ∉R 

So, R is not a symmetric relation.

Transitivity:

Let (x, y) ∈R and (y, z) ∈R. 

Then, x is a father of y and y is a father of z

⇒ x is the grandfather of z

⇒ (x, z) ∉R

So, R is not a transitive relation.

2. Three relations R₁, R_2, and R₃ are defined on a set A = {a, b, c} as follows:
R₁ = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}
R₂ = {(a, a)}
R₃ = {(b, c)}
R₄ = {(a, b), (b, c), (c, a)}.

Find whether or not each of the relations R₁, R₂, R₃, R₄ on A is (i) reflexive (ii) symmetric, and (iii) transitive.

Solution:

(i) Consider R₁

Given R₁ = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}

Now we have check R₁ is reflexive, symmetric and transitive

Reflexive:

Given (a, a), (b, b) and (c, c) ∈ R₁

So, R₁ is reflexive.

Symmetric:

We see that the ordered pairs obtained by interchanging the components of R₁ are also in R₁.

So, R₁ is symmetric.

Transitive:

Here, (a, b) ∈R₁, (b, c) ∈R_1, and also (a, c) ∈R₁

So, R₁ is transitive.

(ii) Consider R₂

Given R₂ = {(a, a)}

Reflexive: 

Clearly (a, a) ∈R₂.

So, R₂ is reflexive.

Symmetric: 

Clearly (a, a) ∈R 

⇒ (a, a) ∈R.

So, R₂ is symmetric.

Transitive: 

R₂ is clearly a transitive relation since there is only one element in it.

(iii) Consider R₃

Given R₃ = {(b, c)}

Reflexive:

Here,(b, b)∉ R₃ neither (c, c) ∉ R₃

So, R₃ is not reflexive.

Symmetric:

Here, (b, c) ∈R₃, but (c, b) ∉R₃

So, R₃ is not symmetric.

Transitive:

Here, R₃ has only two elements.

Hence, R₃ is transitive.

(iv) Consider R₄

Given R₄ = {(a, b), (b, c), (c, a)}.

Reflexive:

Here, (a, a) ∉ R₄, (b, b) ∉ R₄ (c, c) ∉ R₄

So, R₄ is not reflexive.

Symmetric:

Here, (a, b) ∈ R₄, but (b, a) ∉ R₄.

So, R₄ is not symmetric

Transitive:

Here, (a, b) ∈R₄, (b, c) ∈R₄, but (a, c) ∉R₄

So, R₄ is not transitive.

3.  Test whether the following relation R₁, R₂, and R₃ are (i) reflexive (ii) symmetric, and (iii) transitive:

(i) R₁ on Q₀ defined by (a, b) ∈ R₁ ⇔ a = 1/b.

(ii) R₂ on Z defined by (a, b) ∈ R₂ ⇔ |a – b| ≤ 5

(iii) R₃ on R defined by (a, b) ∈ R₃ ⇔ a² – 4ab + 3b² = 0.

Solution:

(i) Given R₁ on Q₀ defined by (a, b) ∈ R₁ ⇔ a = 1/b.

Reflexivity:

Let a be an arbitrary element of R₁.

Then, a ∈ R₁

⇒ a ≠1/a for all a ∈ Q₀

So, R₁ is not reflexive.

Symmetry:

Let (a, b) ∈ R₁ 

Then, (a, b) ∈ R₁

Therefore we can write ‘a’ as a =1/b

⇒ b = 1/a

⇒ (b, a) ∈ R₁

So, R₁ is symmetric.

Transitivity:

Here, (a, b) ∈R₁ and (b, c) ∈R₂

⇒ a = 1/b and b = 1/c

⇒ a = 1/ (1/c) = c

⇒ a ≠ 1/c

⇒ (a, c) ∉ R₁

So, R₁ is not transitive.

 

(ii) Given R₂ on Z defined by (a, b) ∈ R₂ ⇔ |a – b| ≤ 5

Now we have to check whether R₂ is reflexive, symmetric, and transitive.

Reflexivity:

Let a be an arbitrary element of R₂.

Then, a ∈ R₂

On applying the given condition we get,

⇒ | a−a | = 0 ≤ 5

So, R₁ is reflexive.

Symmetry:

Let (a, b) ∈ R₂

⇒ |a−b| ≤ 5                    [Since, |a−b| = |b−a|]

⇒ |b−a| ≤ 5

⇒ (b, a) ∈ R₂

So, R2 is symmetric.

Transitivity:

Let (1, 3) ∈ R₂ and (3, 7) ∈R₂

⇒|1−3|≤5 and |3−7|≤5

But |1−7| ≰5 

⇒ (1, 7) ∉ R₂

So, R₂ is not transitive.

(iii) Given R₃ on R defined by (a, b) ∈ R₃ ⇔ a² – 4ab + 3b² = 0.

Now we have check whether R₂ is reflexive, symmetric and transitive.

Reflexivity:

Let a be an arbitrary element of R₃.

Then, a ∈ R₃

⇒ a² − 4a × a+ 3a²= 0 

So, R₃ is reflexive

Symmetry:

Let (a, b) ∈ R₃

⇒ a²−4ab+3b²=0

But b²−4ba+3a²≠0 for all a, b ∈ R

So, R₃ is not symmetric.

Transitivity:

Let (1, 2) ∈ R₃ and (2, 3) ∈ R₃

⇒ 1 − 8 + 6 = 0 and 4 – 24 + 27 = 0

But 1 – 12 + 9 ≠ 0

So, R₃ is not transitive.

4. Let A = {1, 2, 3}, and let R₁ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}, R₂ = {(2, 2), (3, 1), (1, 3)}, R₃ = {(1, 3), (3, 3)}. Find whether or not each of the relations R₁, R₂, R₃ on A is (i) reflexive (ii) symmetric (iii) transitive.

Solution:

Consider R₁

Given R₁ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}

Reflexivity:

Here, (1, 1), (2, 2), (3, 3) ∈R

So, R₁ is reflexive.

Symmetry:

Here, (2, 1) ∈ R₁,

But (1, 2) ∉ R₁

So, R₁ is not symmetric.

Transitivity:

Here, (2, 1) ∈R₁ and (1, 3) ∈R₁, 

But (2, 3) ∉R₁ 

So, R₁ is not transitive.

Now consider R₂

Given R₂ = {(2, 2), (3, 1), (1, 3)}

Reflexivity:

Clearly, (1, 1) and (3, 3) ∉R₂ 

So, R₂ is not reflexive.

Symmetry:

Here, (1, 3) ∈ R₂ and (3, 1) ∈ R₂

So, R₂ is symmetric.

Transitivity:

Here, (1, 3) ∈ R₂ and (3, 1) ∈ R₂ 

But (3, 3) ∉R₂

So, R₂ is not transitive.

Consider R₃

Given R₃ = {(1, 3), (3, 3)}

Reflexivity:

Clearly, (1, 1) ∉ R₃

So, R₃ is not reflexive.

Symmetry:

Here, (1, 3) ∈ R₃, but (3, 1) ∉ R₃

So, R₃ is not symmetric.

Transitivity:

Here, (1, 3) ∈ R₃ and (3, 3) ∈ R₃ 

Also, (1, 3) ∈ R₃

So, R₃ is transitive.

5. The following relation is defined on the set of real numbers.
(i) aRb if a – b > 0

(ii) aRb iff 1 + a b > 0

(iii) aRb if |a| ≤ b.

Find whether a relation is reflexive, symmetric, or transitive.

Solution:

(i) Consider aRb if a – b > 0

Now for this relation, we have to check whether it is reflexive, transitive, and symmetric.

Reflexivity:

Let a be an arbitrary element of R.

Then, a ∈ R

But a − a = 0 ≯ 0

So, this relation is not reflexive.

Symmetry:

Let (a, b) ∈ R

⇒ a − b > 0

⇒ − (b − a) >0

⇒ b − a < 0

So, the given relation is not symmetric.

Transitivity:

Let (a, b) ∈R and (b, c) ∈R. 

Then, a − b > 0 and b − c > 0

Adding the two, we get

a – b + b − c > 0

⇒ a – c > 0 

⇒ (a, c) ∈ R.

So, the given relation is transitive.

(ii) Consider aRb iff 1 + a b > 0

Now for this relation, we have to check whether it is reflexive, transitive, and symmetric.

Reflexivity:

Let a be an arbitrary element of R.

Then, a ∈ R

⇒ 1 + a × a > 0

i.e. 1 + a² > 0             [Since, square of any number is positive]

So, the given relation is reflexive.

Symmetry:

Let (a, b) ∈ R

⇒ 1 + a b > 0

⇒ 1 + b a > 0

⇒ (b, a) ∈ R

So, the given relation is symmetric.

Transitivity:

Let (a, b) ∈R and (b, c) ∈R

⇒1 + a b > 0 and 1 + b c >0

But 1+ ac ≯ 0

⇒ (a, c) ∉ R

So, the given relation is not transitive.

(iii) Consider aRb if |a| ≤ b.

Now for this relation, we have to check whether it is reflexive, transitive, and symmetric.

Reflexivity:

Let a be an arbitrary element of R.

Then, a ∈ R                  [Since, |a|=a]

⇒ |a|≮ a

So, R is not reflexive.

Symmetry:

Let (a, b) ∈ R

⇒ |a| ≤ b 

⇒ |b| ≰ a for all a, b ∈ R

⇒ (b, a) ∉ R 

So, R is not symmetric.

Transitivity:

Let (a, b) ∈ R and (b, c) ∈ R

⇒ |a| ≤ b and |b| ≤ c

Multiplying the corresponding sides, we get

|a| × |b| ≤ b c

⇒ |a| ≤ c

⇒ (a, c) ∈ R

Thus, R is transitive. 

6. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.

 

Solution:

Given R = {(a, b): b = a + 1}

Now for this relation, we have to check whether it is reflexive, transitive, and symmetric Reflexivity:

Let a be an arbitrary element of R.

Then, a = a + 1 cannot be true for all a ∈ A.

⇒ (a, a) ∉ R 

So, R is not reflexive on A.

Symmetry:

Let (a, b) ∈ R

⇒ b = a + 1

⇒ −a = −b + 1

⇒ a = b − 1

Thus, (b, a) ∉ R

So, R is not symmetric on A.

Transitivity: 

Let (1, 2) and (2, 3) ∈ R

⇒ 2 = 1 + 1 and 3 

2 + 1  is true.

But 3 ≠ 1+1

⇒ (1, 3) ∉ R

So, R is not transitive on A.

7. Check whether the relation R on R defined as R = {(a, b): a ≤ b³} is reflexive, symmetric, or transitive.

Solution:

Given R = {(a, b): a ≤ b³}

It is observed that (1/2, 1/2) in R as 1/2 > (1/2)³ = 1/8

∴ R is not reflexive.

Now,

(1, 2) ∈ R (as 1 < 2³ = 8)

But,

(2, 1) ∉ R (as 2 > 1³ = 1)

∴ R is not symmetric.

We have (3, 3/2), (3/2, 6/5) in “R as” 3 < (3/2)³ and 3/2 < (6/5)³

But (3, 6/5) ∉ R as 3 > (6/5)³

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

 

8. Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.

Solution:

Let A be a set.

Then, Identity relation IA=I_A is reflexive, since (a, a) ∈ A ∀a

The converse of it need not be necessarily true.

Consider the set A = {1, 2, 3}

Here,

Relation R = {(1, 1), (2, 2) , (3, 3), (2, 1), (1, 3)} is reflexive on A.

However, R is not an identity relation.

9. If A = {1, 2, 3, 4} define relations on A which have properties of being

(i) Reflexive, transitive but not symmetric

(ii) Symmetric but neither reflexive nor transitive.

(iii) Reflexive, symmetric and transitive.

Solution:

(i) The relation on A having properties of being reflexive, transitive, but not symmetric is

R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)}

Relation R satisfies reflexivity and transitivity.

⇒ (1, 1), (2, 2), (3, 3) ∈ R 

And (1, 1), (2, 1) ∈ R ⇒ (1, 1) ∈ R

However, (2, 1) ∈ R, but (1, 2) ∉ R

(ii)  The relation on A having properties of being reflexive, transitive, but not symmetric is

R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)}

Relation R satisfies reflexivity and transitivity.

⇒ (1, 1), (2, 2), (3, 3) ∈ R 

And (1, 1), (2, 1) ∈ R ⇒ (1, 1) ∈ R

However, (2, 1) ∈ R, but (1, 2) ∉ R

(iii) The relation on A having properties of being symmetric, reflexive, and transitive is

R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1)}

The relation R is an equivalence relation on A.

RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1: Important Topics

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

FAQs on RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1