# Class 12-science RD SHARMA Solutions Maths Chapter 1 - Relations

## Relations Exercise MCQ

### Solution 35

Given: R = {(1, 2)}

R is not reflexive as (1, 1), (2, 2), (3, 3) ∉ R.

Since (1, 2) ∈ R but (2, 1) ∉ R

Therefore, R is not symmetric.

Also, R is not transitive.

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15

### Solution 16

### Solution 17

### Solution 18

### Solution 19

### Solution 20

### Solution 21

### Solution 22

### Solution 23

### Solution 24

### Solution 25

### Solution 26

### Solution 27

### Solution 28

### Solution 29

### Solution 30

### Solution 31

### Solution 32

### Solution 33

### Solution 34

## Relations Exercise Ex. 1.1

### Solution 1(i)

### Solution 1(ii)

### Solution 1(iii)

### Solution 1(iv)

### Solution 2

### Solution 3(i)

### Solution 3(ii)

### Solution 3(iii)

### Solution 4

### Solution 5(i)

### Solution 5(ii)

### Solution 5(iii)

### Solution 6

### Solution 7(i)

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 16

### Solution 17

A relation R in A is said to be reflexive if aRa for all a∈A

R is said to be transitive if aRb and bRc ⇒ aRc

for all a, b, c ∈ A.

Hence for R to be reflexive (b, b) and (c, c) must be there in the set R.

Also for R to be transitive (a, c) must be in R because (a, b) ∈ R and (b, c) ∈ R so (a, c) must be in R.

So at least 3 ordered pairs must be added for R to be reflexive and transitive.

### Solution 18

A relation R in A is said to be reflexive if aRa for all a∈A, R is symmetric if aRb ⇒ bRa, for all a, b ∈ A and it is said to be transitive if aRb and bRc ⇒ aRc for all a, b, c ∈ A.

• x > y, x, y ϵ N

(x, y) ϵ {(2, 1), (3, 1).......(3, 2), (4, 2)....}

This is not reflexive as (1, 1), (2, 2)....are absent.

This is not symmetric as (2,1) is present but (1,2) is absent.

This is transitive as (3, 2) ϵ R and (2,1) ϵ R also (3,1) ϵ R ,similarly this property satisfies all cases.

• x + y = 10, x, y ϵ N

(x, y)ϵ {(1, 9), (9, 1), (2, 8), (8, 2), (3, 7), (7, 3), (4, 6), (6, 4), (5, 5)}

This is not reflexive as (1, 1),(2, 2)..... are absent.

This only follows the condition of symmetric set as (1, 9)ϵR also (9, 1)ϵR similarly other cases are also satisfy the condition.

This is not transitive because {(1, 9),(9, 1)}ϵR but (1, 1) is absent.

• xy is square of an integer, x, y ϵ N

(x, y) ϵ {(1, 1), (2, 2), (4, 1), (1, 4), (3, 3), (9, 1), (1, 9), (4, 4), (2, 8), (8, 2), (16, 1), (1, 16)...........}

This is reflexive as (1,1),(2,2)..... are present.

This is also symmetric because if aRb ⇒ bRa, for all a,bϵN.

This is transitive also because if aRb and bRc ⇒ aRc for all a, b, c ϵ N.

• x + 4y = 10, x, y ϵ N

(x, y) ϵ {(6, 1), (2, 2)}

This is not reflexive as (1, 1), (2, 2).....are absent.

This is not symmetric because (6,1) ϵ R but (1,6) is absent.

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

### Solution 7(ii)

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

Clearly, 1 < 2

That is, (1, 2) ∈ R

But 2 ≮ 1

So, (2, 1) ∉ R

Therefore, R is not symmetric.

Let (a, b), (b, c) ∈ R

∴ a < b and b < c

∴ a < c

Therefore, (a, c) ∈ R

Thus, R is transitive.

### Solution 15

Given: R = {(x, y) ∈ W × W such that x and y have at least one letter in common}

Clearly, (x, x) ∈ R as x and x have all the letters common.

Therefore, R is reflexive.

Now suppose, x and y have at least one letter in common.

So, y and x will also have at least one letter in common.

Therefore, R is symmetric.

Now take, x = Sun, y = Moon and z = Mars

Here, (x, y) ∈ R and (y, z) ∈ R as x and y have the letter "n" common whereas y and z have the letter "m" common.

But, x and z do not have any letter common.

Therefore, (x, z) ∉ R.

Thus, R is not transitive.

## Relations Exercise Ex. 1.2

### Solution 13

Given: R = {(a, b): a is divisor of b}

Any natural number "a" divides itself.

So, a is divisor of a.

Therefore, R is reflexive.

We know that 2 is divisor of 4.

So, (2, 4) ∈ R

But 4 does not divide 2, then (4, 2) ∉ R

Therefore, R is not symmetric.

As R is not symmetric.

Thus, R is not an equivalence relation.

### Solution 18

To check reflexive:

Now, which is real.

So, z_{1}
R z_{1}.

Therefore, R is reflexive.

To check symmetric:

Suppose z_{1}
R z_{2}

is real

Let where k is real.

Now,

Therefore, is real as -k is real.

Therefore, R is symmetric.

To check transitive:

Suppose z_{1}
R z_{2} and z_{2} R z_{3}

are real

Let

… (i)

Now, which is real.

Thus, R is transitive.

Hence, R is an equivalence relation.