RD Sharma Class 12 Exercise 1.1 Relation Solutions Maths

RD Sharma Class 12 Exercise 1.1 relations Solutions Maths - Download PDF Free Online

Updated on Jan 18, 2022 01:21 PM IST

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RD Sharma Class 12 Solutions Chapter 1 Relations - Other Exercise

Chapter 1 Relations Ex 1.2

Relations Excercise: 1.1

Relations Exercise 1.1 Question 1 (i)

Answer : Reflexive, symmetric, transitive.
Hint :
If

R

is reflexive

\Rightarrow (a, a) \in R for all a \in A

R

is symmetric

\Rightarrow (a, b) \in R \Rightarrow (b, a) \in R_{for all} a, b \in A

R

is transitive

\Rightarrow (a, b) \in R and (b, c) \in R \Rightarrow (a, c) \in R for all a, b, c \in A

Given :

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

Solution : A relation

R

on set A is said to be reflexive if every element of A is related to itself.
Thus,

R

is reflexive

\Rightarrow (a, a) \in R for all a \in A

A relation

R

on set A is said to be symmetric relation if

(a, b) \in R \Rightarrow (b, a) \in R

for all

a, b \in A

i.e.

a R b \Rightarrow b R a for all a, b \in A

A relation

R

on set A is said to be transitive relation
If

(a, b) \in R and (b, a) \in R \Rightarrow (c, a) \in R for all a, b, c \in A

i.e

a R b

and

b R c \Rightarrow a R c

for all

a, b, c \in A

Now, let's get back to the actual problem
For Reflexive :

x

and

x

work at same place
Similarly,

y

and

y

work at the same place.
So,

R

is reflexive.
For Symmetric :
It is given that

x

and

y

work at the same place.
So, we can say that

y

and

x

work at the same place.
So,

R

is symmetric.
For Transitive:
Let

z

be a person such that

y

and

z

work at the same place.
And we know that

x

and

y

work at the same place.
So, we can say that

x

and

z

work at the same place.
So,

R

is Transitive.

Relations Exercise 1.1 Question 1 (ii)

Answer : Reflexive, symmetric, transitive.
Hint :
If

R

is reflexive

\Rightarrow (a, a) \in R for all a \in A

R

is symmetric

\Rightarrow (a, b) \in R \Rightarrow (b, a) \in R_{for all} a, b \in A

R

is transitive

\Rightarrow (a, b) \in R and (b, c) \in R \Rightarrow (a, c) \in R for all a, b, c \in A

Given :

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

Solution :
A relation

R

on set A is said to be reflexive if every element of A is related to itself.
Thus,

R

is reflexive

\Leftrightarrow (a, a) \in R for all a \in A

A relation

R

on set A is said to be symmetric relation if

(a, b) \in R \Rightarrow (b, a) \in R

for all

a, b \in A

I.e.

a R b \Rightarrow b R a

for all

a, b \in A

A relation

R

on set A is said to be transitive relation if

(a, b) \in R and (b, c) \in R \Rightarrow (c, a) \in R

for all

a, b, c \in A

i.e

a R b and b R c \Rightarrow a R c for all a, b, c \in A

For Reflexive:

x

and

x

live in the same locality.
Similarly,

y

and

y

live in same locality
So,

R

is reflexive.
For Symmetric:

x

and

y

live in same locality.
So, we can easily say that

y

and

x

live in same locality.
So,

R

is symmetric.
For Transitive:
Let

z

be a person;

z \in A

and

z

and

y

live in same locality
And it is given that

x

and

y

live in same locality
So, we can say that

x

and

z

live in the same locality.
So,

R

is Transitive.

Relations Exercise 1.1 Question 1 (iii)

Answer:
Neither reflexive, nor symmetric, not transitive.
Hints :
If

R

is reflexive

\Rightarrow (a, a) \in R for all a \in A

R

is symmetric

\Rightarrow (a, b) \in R \Rightarrow (b, a) \in R for all a, b \in A

R

is transitive

\Rightarrow (a, b) \in R, (b, c) \in R \Rightarrow (a, c) \in R for all a, b, c \in A

Given :
$R = {(x, y) : x is wife of y}$
Solution :
A relation

R

on set

A

is said to be reflexive if every element of

A

is related to itself.
Thus,

R

is reflexive

\Leftrightarrow (a, a) \in R for all a \in A

.
A relation

R

on set

A

is said to be symmetric relation if

(a, b) \in R \Rightarrow (b, a) \in R

for all

a, b \in A

i.e;

a R b \Rightarrow b R a for all a, b \in A

A relation

R

on set

A

is said to be transitive relation if

(a, b) \in R

and

(b, c) \in R \Rightarrow

(c, a) \in R

for all

a, b, c \in A

i.e;

a R b

and

b R c \Rightarrow a R c

for all

a, b, c \in A

For Reflexive:

x

is not wife of

x

and

y

is not wife of

y

So,

R

is not reflexive.
For Symmetric:

x

is the wife of

y

but

y

is not the wife of

x

.
So,

R

is not symmetric.
For Transitive:
Let

z

be a person;

z \in A

such that

y

is the wife of

z

.
And it is given that

x

is the wife of

y

but this case is not possible. Also, here we can’t show x is the wife of z.
So, R is not transitive.

Relations Exercise 1.1 Question 1 (iv)

Answer:
Neither reflexive, nor symmetric nor transitive.
Hints :
If

R

is reflexive

\Rightarrow (a, a) \in R for all a \in A

R

is symmetric

\Rightarrow (a, b) \in R \Rightarrow (b, a) \in R for all a, b \in A

R

is transitive

\Rightarrow (a, b) \in R, (b, c) \in R \Rightarrow (a, c) \in R for all a, b, c \in A

Given :
$R = {(x, y) : x is father of y}$
Solution :
A relation

R

on set

A

is said to be reflexive if every element of

A

is related to itself.
Thus,

R

is reflexive

\Leftrightarrow (a, a) \in R_{for all} a \in A

A relation

R

on set

A

is said to be symmetric relation if

(a, b) \in R \Rightarrow (b, a) \in R

for all

a, b \in A

i.e

a R b \Rightarrow b R a

for all

a, b \in A

A relation

R

on set

A

is said to be transitive relation if

(a, b) \in R

and

(b, c) \in R \Rightarrow

(c, a) \in R

for all

a, b, c \in A

i.e

a R b

and

b R c \Rightarrow a R c

for all

a, b, c \in A

For Reflexive:

x

is not father of

x

and

y

is not father of

y

So,

R

is not reflexive
For Symmetric:
It is given that

x

is the father of

y

.
But we can say that

y

is not the father of

x

.
So

R

is not symmetric.
For Transitive:
Let

z

be a person;

z \in A

such that

y

is father of

z

and it is given that x is a father of y.
Then

x

is grandfather of

z

So,

R

is not Transitive.

Relations Exercise 1.1 Question 2

Answer:

R_{1}

is reflexive and transitive but not symmetric

R_{2}

is reflexive, symmetric, and transitive.

R_{3}

is transitive but neither reflexive nor symmetric

R_{4}

is neither reflexive nor symmetric nor transitive
Hint :
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If ^{$(a, b)$} is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If ^{$(a, b)$}and

(b, c)

\in R

, then ^{$(a, c)$}

\in R

for every

a, b \in A

Given :
Set

A = {a, b, c}

\begin{aligned} R_{1} = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)} \\ R_{2} = {(a, a)} \\ R_{3} = {(b, c)} \\ R_{4} = {(a, b), (b, c), (c, a)} \end{aligned}

Consider

R_{1} = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}

Reflexive :
Given

(a, a), (b, b)

and

(c, c) \in R_{1}

So,

R_{1}

is reflexive.
For Symmetric:
We see that the ordered pairs obtained by interchanging the components of

R_{1}

are not in

R_{1}

.
For ex :

(a, b) \in R_{1} but (b, a) \notin R_{1}

So,

R_{1}

is not symmetric.
For Transitive:
Here,

(a, b) \in R_{1} and (b, c) \in R_{1}

but

(a, c) \in R_{1}

So,

R_{1}

is transitive
(ii) Consider

R_{2}

R_{2} = {(a, a)}

Reflexive:
clearly

(a, a) \in R_{2}

So,

R_{2}

is reflexive.
Symmetric:
Clearly

(a, a) \in R_{2}

So,

R_{2}

is symmetric.
Transitive:

R_{2}

is a transitive relation, since there is only one element in it.
(iii) Consider

R_{3}

R_{3} = {(b, c)}

Reflexive:
Here neither

(b, b) \notin R_{3}

nor

(c, c) \notin R_{3}

So,

R_{3}

is not reflexive
Symmetric:
Here neither

(b, c) \in R_{3}

nor

(c, b) \notin R_{3}

So,

R_{3}

is not symmetric.
Transitive:

R_{3}

has only one element
Hence

R_{3}

is transitive.
(iv) Consider

R_{4} = {(a, b), (b, c), (c, a)}

Reflexive:
Here

(a, b) \in R_{4} but (b, a) \notin R_{4}

So,

R_{4}

is not reflexive
Symmetric:
Here

(a, b) \in R_{4} but (b, a) \notin R_{4}

So,

R_{4}

is not symmetric
Transitive:
Here

(a, b) \in R_{4}, (b, c) \in R_{4}

but

(a, c) \notin R_{4}

Hence

R_{4}

is not transitive.

Relations Exercise 1.1 Question 3 (i)

Answer: $R_{1}$ is symmetric but neither reflexive nor transitive.
Hint :
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b)

and

(b, c) \in R

, then

(a, c) \in R

for every

a, b, c \in A

Given :

R_{1} on Q_{0}

defined by

(a, b) \in R_{1} \Leftrightarrow a = \frac{1}{b}

Solution :
Reflexivity:
Let

a

be an arbitrary element of

R_{1}

Then,

a \in R_{1}

a \neq \frac{1}{a} for all a \in Q_{0}

So,

R_{1}

is not reflexive
Symmetry:
Let

(a, b) \in R_{1}

Therefore, we can write 'a' as

a = \frac{1}{b}

b = \frac{1}{a}

Then

(b, a) \in R_{1}

So,

R_{1}

is symmetric.
For Transitive:
Let

(a, b) \in R_{1} and (b, c) \in R_{1}

\begin{aligned} a = \frac{1}{b} and b = \frac{1}{c} \\ a = \frac{1}{(\frac{1}{c})} \Rightarrow c \\ a \neq \frac{1}{c} \\ (a, c) \notin R_{1} \end{aligned}

So,

R_{1}

is not transitive.

Relations Exercise 1.1 Question 3 (ii)

Answer:

R_{2}

is reflexive and symmetric but not transitive.
Hint :
A relation R on set A is
Reflexive relation:

If $(a, a) \in R$ for every $a \in A$
Symmetric relation:
if $(a, b)$ is true then $(b, a)$ is also true for every $a, b \in A$
Transitive relation:
If $(a, b)$ and $(b, c) \in R$ , then $(a, c) \in R$ for every $a, b, c \in A$
Given:
$R_{2} on Z$ defined by $(a, b) \in R_{2} \Leftrightarrow | a - b | \leq 5$
Solution :
Reflexivity:
Let a be an arbitrary element of $R_{2}$
Then, $a \in R_{2}$
On applying the given condition, we get
$| a - a | = 0 \leq 5$
So, $R_{2}$ is reflexive
Symmetry :
Let $(a, b) \in R_{2} | a - b | \leq 5$
[Since $| a - b | = | b - a |$ ]
Then $| b - a | \leq 5$
$(b, a) \in R_{2}$
So, $R_{2}$ is symmetric.
Transitivity :
Let $(1, 3) \in R_{2} and (3, 7) \in R_{2}$
$| 1 - 3 | \leq 5 and | 3 - 7 | \leq 5$
But, $| 1 - 7 | \leq 5$
$(1, 7) \neq R_{2}$
So, $R_{2}$ is not transitive.

Relations Exercise 1.1 Question 3 (iii)

Answer:

R_{3}

is reflexive but neither symmetric nor transitive
Hint :
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in R

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
if

(a, b)

and

(b, c) \in R

, then

(a, c) \in R

for every

a, b, c \in A

Given :

R_{3}

on R defined by

(a, b) \in R_{3} \Leftrightarrow a^{2} - 4 a b + 3 b^{2} = 0

Solution :
Reflexivity:
Let a be an arbitrary element of

R_{3}

Then,

a \in R_{3}

a^{2} - 4 a \times a + 3 a^{2} = 0

So,

R_{3}

is reflexive
Symmetry :
Let :

(a, b) \in R_{3}

a^{2} - 4 a b + 3 b^{2} = 0

But

b^{2} - 4 b a + 3 a^{2} \neq 0

for all

a, b \in R

So,

R_{3}

is not symmetric.
Transitivity:
Let

(a, b) \in R_{3} and (b, c) \in R_{3}

\begin{aligned} a^{2} - 4 a b + 3 b^{2} = 0 . . . (i) \\ and b^{2} - 4 b c + 3 c^{2} = 0 . . . . (i i) \end{aligned}

Adding (i) and (ii) we get,

\begin{aligned} a^{2} - 4 a b + 3 b^{2} + b^{2} - 4 b c + 3 c^{2} = 0 \\ \Rightarrow a^{2} - 4 a b + 4 b^{2} - 4 b c + 3 c^{2} = 0 \\ But a^{2} - 4 a c + 3 c^{2} = - 4 a c - 4 a b + 4 b^{2} - 4 b c \neq 0 \\ \Rightarrow a^{2} - 4 a c + 3 c^{2} \neq 0 \end{aligned}

(a, c) \notin R_{3}

So,

R_{3}

is not transitive.

Relations Exercise 1.1 Question 4

Answer:

R_{1}

is reflexive but neither symmetric nor transitive

R_{2}

is symmetric but neither reflexive nor transitive

R_{3}

is transitive but neither reflexive nor symmetric
Hint :
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in R

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
if

(a, b)

and

(b, c) \in R

, then

(a, c) \in R

for every

a, b, c \in A

Given :

\begin{aligned} R_{1} = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)} \\ R_{2} = {(2, 2), (3, 1), (1, 3)} \\ R_{3} = {(1, 3), (3, 3)} \end{aligned}

Solution :
Consider

R_{1}

R_{1} = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}

Reflexivity:
Here

(1, 1), (2, 2), (3, 3) \in R

So,

R_{1}

is Reflexive
Symmetric :
Here

(2, 1) \in R_{1}

But,

(1, 2) \notin R_{1}

So,

R_{1}

is not symmetric
Transitivity:

\begin{aligned} Here (2, 1) \in R_{1} and (1, 3) \in R_{1} \\ But (2, 3) \notin R_{1} \end{aligned}

So,

R_{1}

is not transitive.
Now consider

R_{2}

R_{2} = {(2, 2), (3, 1), (1, 3)}

Reflexivity :
Clearly,

(1, 1) and (3, 3) \notin R_{2}

So,

R_{2}

is not reflexive.
Symmetric :

Here (1, 3) \in R_{2} and (3, 1) \in R_{2}

So,

R_{2}

is symmetric.
Transitivity:

\begin{aligned} Here \\ (1, 3) \in R_{2} and (3, 1) \in R_{2} \\ But (3, 3) \notin R_{2} \end{aligned}

So,

R_{2}

is not transitive.
Now consider

R_{3}

R_{3} = {(1, 3), (3, 3)}

Reflexivity:
Clearly,

(1, 1) \notin R_{3}

So,

R_{3}

is not reflexive.
Symmetry:
Here

(1, 3) \in R_{3} but (3, 1) \notin R_{3}

So,

R_{3}

is not symmetric.
Transitivity:
Here

(1, 3) \in R_{3} and (3, 3) \in R_{3}

But

(1, 3) \in R_{3}

So,

R_{3}

is transitive.

Relation Exercise 1.1 Question 5 (i)

Answer: Given relation is transitive
Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b)

and

(b, c) \in R

, then

(a, c) \in R

for every

a, b, c \in A

Given :

a R b if a - b > 0

Solution :
Reflexivity:
Let

a

be an arbitrary element of

R

Then,

a \in A

But

a - a = 0 ≯ 0

So, this relation is not reflexive.
Symmetry :
Let

\begin{aligned} (a, b) \in R \\ a - b > 0 \\ - (b - a) > 0 \\ b - a < 0 \end{aligned}

So, the given relation is not symmetric.
Transitivity :

\begin{aligned} Let (a, b) \in R and (b, c) \in R \\ Then, a - b > 0 . . . (i) \\ b - c > 0 . . . (i i) \end{aligned}

Adding eq (i) & (ii), we get

\begin{aligned} a - b + b - c > 0 \\ a - c > 0 \\ (a, c) \in R \end{aligned}

So, the given relation is transitive.

Relation Exercise 1.1 Question 5 (ii)

Answer:
Reflexive and symmetric but not transitive.
Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in R

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b)

and

(b, c) \in R

, then

(a, c) \in R

for every

a, b, c \in A

Given :

a R b if 1 + a b > 0

Solution :
Reflexivity:
Let

a

be an arbitrary element of

R

Then,

a \in R

\begin{aligned} 1 + a \times a > 0 \\ 1 + a^{2} > 0 \end{aligned}

Since, Square of any number is positive.
So, the given relation is reflexive.
Symmetry:
Let

(a, b) \in R

\begin{aligned} 1 + a b > 0 \\ 1 + b a > 0 \\ (b, a) \in R \end{aligned}

So, the given relation is symmetric.
Transitivity:

\begin{aligned} Let (a, b) \in R and (b, c) \in R \\ Let a = - 8, b = - 2, c = \frac{1}{4} \\ Then 1 + a b > 0 i.e; 1 + (- 8) (- 2) = 17 > 0 \\ and 1 + b c > 0 i.e; 1 + (- 2) \frac{1}{4} = \frac{1}{2} > 0 But 1 + a c \neq 0 i.e; 1 + (- 8) \frac{1}{4} = - 1 ≯ 0 \\ (a, c) ϵ R \end{aligned}

So, the given relation is not transitive.

Relation Exercise 1.1 Question 5 (iii)

Answer: Transitive neither reflexive nor symmetric.
Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in R

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b)

and

(b, c) \in R

, then

(a, c) \in R

for every

a, b, c \in A

Given :

a R b if | a | \leq b

Solution :
Reflexivity:
Let -a be an arbitrary element of R
Then

- a \in R

\Rightarrow | - a | \neq - a

So,

R

is not reflexive
Symmetry:
Let

(a, b) \in R

\begin{aligned} | a | \leq b \\ | b | ≶ a for all a, b \in R \\ (b, a) \notin R \end{aligned}

So,

R

is not symmetric.
Transitivity:

\begin{aligned} Let (a, b) \in R_{and} (b, c) \in R \\ | a | \leq b and | b | \leq c for a, b, c \in R \end{aligned}

Multiplying the corresponding sides, we get

\begin{aligned} | a | \times | b | \leq b c \\ | a | \leq c \\ (a, c) \in R \end{aligned}

Thus,

R

is transitive

Relation Exercise 1.1 Question 6

Answer:

R

is neither reflexive nor symmetric nor transitive.
Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in R

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b)

and

(b, c) \in R

, then

(a, c) \in R

for every

a, b, c \in A

Given :

R = {(a, b) : b = a + 1}

Solution :
Let

a

be an arbitrary element of set A.
Then,

a = a + 1

cannot be true for all

a \in A

(a, a) \notin R

So,

R

is not reflexive on

A

Symmetry :

\begin{aligned} Let (a, b) \in R \\ b = a + 1 \\ a = b - 1 \\ - a = - b + 1 \\ Thus \\ (b, a) \notin R \end{aligned}

So,

R

is not symmetric on

A

Transitivity :
Let

(1, 2) and (2, 3) \in R

\begin{aligned} 2 = 1 + 1 and \\ 3 = 2 + 1 is true \\ But 3 \neq 1 + 1 \\ (1, 3) \notin R \end{aligned}

So,

R

is not transitive on

A

Relation Exercise 1.1 Question 7

Answer:

R

is neither reflexive nor symmetric nor transitive.
Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in R

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b)

and

(b, c) \in R

, then

(a, c) \in R

for every

a, b, c \in A

Solution :
Reflexive :
It is observed that

(1 / 2, 1 / 2) is R as 1 / 2 > (1 / 2)^{3} = \frac{1}{8}

∴ R is not reflexive.

Symmetric :
Now

(1, 2) \in R (as 1 < 2^{3} = 8)

But

(2, 1) \notin R (as 2 > 1^{3} = 1)

R

is not symmetric
Transitive:
We have

(3, \frac{3}{2}), (\frac{3}{2}, \frac{6}{5}) in R as 3 < {(\frac{3}{2})}^{3}

and

(\frac{3}{2}) < {(\frac{6}{5})}^{3}

but

(3, \frac{6}{5}) \notin R as 3 > {(\frac{6}{5})}^{3}

R

is not transitive
Hence,

R

is neither reflexive, nor symmetric nor transitive

Relations Exercise 1.1 Question 8

Answer:
Hence, prove every identity relation on a set is reflexive but the converse is not necessarily true.
Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:

If (a, b) and (b, c) \in R, then (a, c) \in R for every a, b, c \in A

Given: Every Identity relation is reflexive.
Solution:
Let

A

be a set

A = {1, 2, 3}

Then,

I_{A} = {(1, 1), (2, 2), (3, 3)}

Identity relation is reflexive, since

(a, a) \in A \forall a

The converse of it need not be necessarily true.
Counter example:
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.

Relations Exercise 1.1 Question 9(i)

Answer:
$R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)}$
Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:

If (a, b) and (b, c) \in R, then (a, c) \in R for every a, b, c \in A

Given:
$A = {1, 2, 3, 4}$
Solution:
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) \in R

And

(2, 2), (2, 1) \in R \Rightarrow (2, 1) \in R

However,

(2, 1) \in R_{but} (1, 2) \notin R

Relations Exercise 1.1 Question 9 (ii)

Answer: $R = {(1, 2), (2, 1)}$
Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:

If (a, b) and (b, c) \in R, then (a, c) \in R for every a, b, c \in A

Given:
$A = {1, 2, 3}$
Solution:
The relation on A having properties of being symmetric but neither transitive nor reflexive.
Let

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

Now,

(1, 2) \in R, (2, 1) \in R

So, it is symmetric.
Clearly

R

is not transitive

(1, 2) \in R, (2, 1) \in R but (1, 1) \notin R

Relations Exercise 1.1 Question 9(iii)

Answer:

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

The relation

R

is an equivalence relation on

A

Hint:
A relation

R

on set

A

is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:

If (a, b) and (b, c) \in R, then (a, c) \in R for every a, b, c \in A

Given:
$A = {1, 2, 3, 4}$
Solution:
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

Relations Exercise 1.1 Question 10

Answer:
Domain of

R

is x ∈ N where x∈

{1, 2, 3. . . . . . . . . . . .20}

and
range of

R

is y ∈ N where

{39, 37, 35, . . . . . .3, 1}

The relation having properties of being neither symmetric nor transitive nor reflexive.
Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a \in A

Transitive relation:

If (a, b) and (b, c) \in R, then (a, c) \in R for every a, b, c \in A

Given:
$R = {(x, y) : x, y \in N, 2 x + y = 41}$
Solution:
The domain of

R is x \in N, such that 2 x + y = 41

x = (41 - y) / 2

Since

y \in N

, largest value that

x

can take corresponds to the smallest value that

y

can take.

∴ x = {1, 2, 3 \dots .20}

Range

R

of is

y \in N

such that
Since

\begin{matrix} 2 x + y = 41 \\ y = 41 - 2 x \\ x = {1, 2, 3 \dots 20} \end{matrix}

y = {39, 37, 35, \dots . .3, 1}

Since

(2, 2) \notin R, R

,is not reflexive.
Also, since

(1, 39) \in R (39, 1) \notin R, R

is not symmetric.
Finally, since

(15, 11) \in R and (11, 19) \in R but (15, 19) \notin R, R

is not transitive.

Relations Exercise 1.1 Question 11

Answer:
No, it is not true that every relation which is symmetric and transitive is also reflexive.
Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b)

and

(b, a)

\in R

, then

(a, c) \in R

for every

a, b \in A

Given:
Answer that, whether every relation which is symmetric and transitive is also reflexive.
Solution:
No, it is not necessary that a relation that is symmetric and transitive is reflexive as well.
For example:

R = {(1, 1), (1, 2), (2, 1)} on A = {1, 2}

is symmetric and transitive but not reflexive.
Because

(2, 2) \notin R

Relations Exercise 1.1 Question 13

Answer:
Hence proved, the relation “

\geq

” on the set

R

of all real numbers is reflexive and transitive but not symmetric.
Hint:
A relation

R

on set

A

is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b) and (b, c) \in R

then

(a, c) \in R

for every

a, b, c \in A

Given:
Relation is ”

\geq

”on the

R

of all real numbers.
Solution:
Reflexivity:
Let

a \in R

a \geq a

\geq

"is reflexive.
Transitive:
Let

a, b, c \in R

Such that

a \geq b and b \geq c

Then

a \geq c

" \geq "

is transitive.
Symmetric: Let

a, b \in R

Such that

a \geq b but b ⪈ a

" \geq "

not symmetric

Relations Exercise 1.1 Question 14(ii)

Answer: $R = {(a, b) : a^{3} \geq b^{3}}$
Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:

If (a, b) and (b, c) \in R, then (a, c) \in R for every a, b, c \in A

Given:
We have to give the example of a relation which is reflexive and transitive but not symmetric.
Solution:
The relation having properties of being reflexive and transitive but not symmetric.
Define a relation

R

R

as:

R = {(a, b) : a^{3} \geq b^{3}}

Clearly

(a, a) \in R as a^{3} = a^{3}

Therefore

R

is reflexive.
Now

(2, 1) \in R (as 2^{3} \geq 1^{3})

But

(1, 2) \notin R (as 1^{3} < 2^{3})

Therefore

R

is not symmetric.
Now let

\begin{aligned} (a, b), (b, c) \in R \\ a^{3} \geq b^{3} and b^{3} \geq c^{3} then a^{3} \geq c^{3} (a, c) \in R \end{aligned}

$R$ is transitive.
Hence, relation $R$ is reflexive and transitive but not symmetric.

Relations Exercise 1.1 Question 14(iii)

Answer: $R = {(- 5, - 6), (- 6, - 5), (- 5, - 5)}$
Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:

If (a, b) and (b, c) \in R, then (a, c) \in R for every a, b, c \in A

Given:
We have to give the example of a relation which is symmetric and transitive but not reflexive.
Solution:
The relation having properties of being symmetric and transitive but not reflexive.
Let

A = {- 5, - 6}

Define a relation

R

A

R = {(- 5, - 6), (- 6, - 5), (- 5, - 5)}

Relation

R

is not reflexive as

(- 6, - 6) \notin R

Relation

R

is symmetric as

(- 5, - 6), (- 6, - 5) \in R

It is seen that

(- 5, - 6), (- 6, - 5) \in R

Also

(- 5, - 5) \in R

The relation

R

is transitive.
Hence relation

R

is symmetric and transitive but not reflexive.

Relations Exercise 1.1 Question 14(iv)

Answer: $R = {(5, 6), (6, 5)}$
Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b) and (b, c) \in R

, then

(a, c) \in R

for every

a, b, c \in A

Given:
$Let A = {5, 6, 7} .$
Solution:
$Define a relation R on A as R = {(5, 6), (6, 5)} . Relation R is not reflexive as (5, 5), (6, 6), (7, 7) \notin R .$
$\begin{aligned} Now, as (5, 6) \in R and also (6, 5) \in R, R is symmetric. \\ \Rightarrow (5, 6), (6, 5) \in R, but (5, 5) \notin R \end{aligned}$
$Therefore, R is not transitive. Hence, relation R is symmetric but not reflexive or transitive.$

Relations Exercise 1.1 Question 14(v)

Answer:

R = {(a, b) : a < b}

Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b) and (b, c) \in R, then (a, c) \in R for every a, b, c \in A

Given:
We have to give the example of a relation which is transitive but neither symmetric nor reflexive.
Solution:
The relation having properties of being transitive but neither symmetric nor reflexive.
Consider a relation R in

R

defined as:

R = {(a, b) : a < b}

For any

a \in R

we have

(a, a) \notin R

, since

a

can’t be strictly less than

a

itself.
Infact

a = a

∴ Relation

R

is not reflexive.
Now,

(1, 2) \in R (as 1 < 2)

But 2 is not less than 1
∴

(2, 1) \notin R

∴

R

is not symmetric
Now let

(a, b), (b, c) \in R

\begin{aligned} a < b and b < c \\ a < c \\ (a, c) \in R \end{aligned}

R

is transitive.

Relation Exercise 1.1 Question 15

Answer :

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

Hint :
A relation R on set A is
Reflexive relation:

a, b, c \in A

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b)

and , then

(b, c) \in R

for every

(a, c) \in R

Given :

\begin{aligned} Relation R = {(1, 2), (2, 3)} on the set \\ A = {1, 2, 3} \end{aligned}

Solution :
To make

R

reflexive we will add

(1, 1), (2, 2), (3, 3)

to get

R^{'} = {(1, 2), (2, 3), (1, 1), (2, 2), (3, 3)}

is reflexive.
Again, to make

R

symmetric we will add

(3, 2)

and

(2, 1)

R^{''} = {(1, 2), (2, 3), (1, 1), (2, 2), (3, 3), (3, 2), (2, 1)}

is reflexive and symmetric .
To make

R

transitive we will add

(1, 3)

and

(3, 1)

R^{'''} = {(1, 2), (2, 3), (1, 1), (2, 2), (3, 3), (3, 2), (2, 1), (1, 3), (3, 1)}

is reflexive and symmetric and transitive.

Relation Exercise 1.1 Question 16

Answer:
only 1 ordered pair maybe added to

R

so that it may become a transitive relation on

A

Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b)

and

(b, c) \in R

, then

(a, c) \in R

for every

a, b, c \in A

Given :

A = {1, 2, 3}

Solution :

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

be a relation on

A

To make

R

transitive we shall add

(1, 3)

only

R^{'} = {(1, 2), (1, 1), (2, 3), (1, 3)}

As we know,
Transitive relation

x=y a n d y=z Then x=z

Note: for

R

to be transitive

(a, c)

must be in

R

because

(a, b) \in R and (b, c) \in R

So,

(a, c)

must be in

R

Relation Exercise 1.1 Question 17

Answer:
At least 3 ordered pairs must be added for

R

to be reflexive and transitive
Hint:
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b)

and

(b, c) \in R

, then

(a, c) \in R

for every

a, b, c \in A

Given: Minimum number of order pair that makes R reflexive and transitive.
Solution:
A relation

R

A

is said to be reflexive if

a R a

for all

a \in A

R

is said to be transitive if

a R b

and bRc then aRc for all

a, b, c \in A

Hence for

R

to be reflexive

(b, b)

and

(c, c)

must be there in set

R

Also for

R

to be transitive

(a, c)

must be in

R

because

(a, b) \in R (b, c) \in R

So,

(a, c)

must be in

R

So, at least 3 ordered pairs must be added for

R

to be reflexive and transitive.

Relation Exercise 1.1 Question 18 (i)

Answer : Transitive
Hint :
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b)

and

(b, c) \in R

then

(a, c) \in R

for every

a, b, c \in A

Given :

x > y, x, y \in N (x, y) \in {(2, 1), (3, 1) \dots \dots (3, 2), (4, 2) \dots}

Solution :
This is not reflexive as

(1, 1), (2, 2) \dots \dots

are absents.
This is not symmetric as

(2, 1)

is present but

(1, 2)

is absent
This is transitive as

(3, 2) \in R and (2, 1) \in R also (3, 1) \in R

Hence, this relation is not satisfying reflexivity and symmetricity.

Relation Exercise 1.1 Question 18 (ii)

Answer : Symmetric
Hint :
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b)

and

(b, c) \in R

then

(a, c) \in R

for every

a, b, c \in A

Given :

x + y = 10, x, y \in N (x, y) \in {(9, 1), (1, 9), (2, 8), (8, 2), (3, 7), (7, 3), (4, 6), (6, 4), (5, 5)}

Solution :
This is not reflexive as

(1, 1), (2, 2) \dots \dots

are absent.
This only follows the condition of symmetry as

(1, 9) \in R also (9, 1) \in R

This is not transitive because

{(1, 9), (9, 1)} \in R but (1, 1)

is absent.
Hence, this relation is not satisfying reflexivity and transitivity.

Relation Exercise 1.1 Question 18 (iii)

Answer : Symmetric, Reflexive and transitive
Hint :
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b)

and

(b, c) \in R

then

(a, c) \in R

for every

a, b, c \in A

Given :

x y is square of an integer x, y \in N

(x, y) \in {(1, 1), (2, 2), (4, 1), (1, 4) (3, 3), (9, 1), (1, 9), (4, 4), (2, 8), (8, 2), (16, 1), (1, 16) \dots}

Solution :
This is reflexive as

(1, 1), (2, 2) \dots \dots

are present.
This is also symmetric because

a R b \Leftrightarrow b R a for all a, b \in N

This is transitive because if

a R b and b R c \Leftrightarrow a R c a, b, c \in N

This relation is reflexive, symmetric, and transitive.

Relation Exercise 1.1 Question 18 (iv)

Answer : This relation is neither symmetric nor reflexive nor transitive.
Hint :
A relation R on set A is
Reflexive relation:
If

(a, a) \in R

for every

a \in A

Symmetric relation:
If

(a, b)

is true then

(b, a)

is also true for every

a, b \in A

Transitive relation:
If

(a, b)

and

(b, c) \in R

then

(a, c) \in R

for every

a, b, c \in A

Given :

x + 4 y = 10, x, y \in N (x, y) \in {(6, 1), (2, 2)}

Solution :

This is not reflexive as $(1, 1), (6, 6)$ are absent.

This is not symmetric as $(6, 1)$ is present but $(1, 6)$ is absent.

This is not transitive as there are only two elements in the set having no element in common.

This relation is neither symmetric nor reflexive nor transitive.

Void relation
Universal relation
Identity relation
Reflexive relation
Symmetric relation
Transitive relation
Antisymmetric relation
Equivalence relation
Theorems based on relations

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RD Sharma Class 12 Solutions Chapter 1 Relations - Other Exercise
Relations Excercise: 1.1
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