Problem 1 Solution

This page contains the NCERT mathematics class 12 chapter Relations and Functions Exercise 1.1 Problem 1 Solution. Solutions for other problems are available at Exercise 1.1 Solutions

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Exercise 1.1 Problem 1 Solution

1. Determine whether each of the following relations are reflexive, symmetric and transitive:

Relation R in the set A = {1, 2, 3, …, 13, 14} defined as R = {

(x, y) : 3x - y = 0

}

ii.

Relation R in the set N of natural numbers defined as R = {

(x, y) : y = x + 5

} and

x \lt 4

iii.

Relation R in the set A = {1, 2, 3, 4, 5, 6} as R =

\{(x, y) : y

is divisible by

x

}

iv.

Relation R in the set Z of all integers defined as R = {

(x, y) : x - y

is an integer}

Relation R in the set A of human beings in a town at a particular time given by

R = {

(x, y) : x

and

y

work at the same place}

R = {

(x, y) : x

and

y

live in the same locality}

R = {

(x, y) : x

is exactly 7 cm taller than

y

}

R = {

(x, y) : x

is wife of

y

}

R = {

(x, y) : x

is father of

y

}

i. Relation R in the set A = {1, 2, 3, …, 13, 14} defined as R = {

(x, y) : 3x - y = 0

}

⇒

3x - y = 0

⇒

3x = y

⇒

y = 3x

When

{x = 1}

{y = 3 \times x = 3 \times 1 = 3}

. As 3 ∈ A, (1, 3) ∈ R

When

{x = 2}

{y = 3 \times x = 3 \times 2 = 6}

. As 6 ∈ A, (2, 6) ∈ R

When

{x = 3}

{y = 3 \times x = 3 \times 3 = 9}

. As 9 ∈ A, (3, 9) ∈ R

When

{x = 4}

{y = 3 \times x = 3 \times 4 = 12}

. As 12 ∈ A, (3, 12) ∈ R

When

{x = 5}

{y = 3 \times x = 3 \times 5 = 15}

. As 15 ∉ A, (3, 15) ∉ R

These calculations can also be represented in the tabular form as follows:

x

y = 3x

(x, y)

∈ R?

3 × 1 = 3

✔

3 × 2 = 6

✔

3 × 3 = 9

✔

3 × 4 = 12

✔

3 × 5 = 15

❌

For higher values of

x

, the value of

y

will be more than 14. As all the elements in A are less than 14, we can stop here.

∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}

To Check whether R is Reflexive: The relation R in the set A is reflexive if

(a, a)

∈ R, for every

a

∈ A.

In this case, (1, 1) ∉ R, (2, 2) ∉ R, (3, 3) ∉ R,…. (13, 13) ∉ R, (14, 14) ∉ R.

∴ R is not Reflexive

Note: Ideally, we don’t have to check for all the elements. Even if one element for example (1,1) ∉ R, then we can say that R is not reflexive)

To Check whether R is Symmetric: The relation R in the set A is symmetric if

(a_1, a_2)

∈ R implies that

(a_2, a_1)

∈ R, for all

a_1, a_2

∈ A

Here we see that (1, 3) ∈ R, but (3, 1) ∉ R

∴ R is not Symmetric

To Check whether R is Transitive: The relation R in the set A is transitive if

(a_1, a_2)

∈ R and

(a_2, a_3)

∈ R implies that

(a_1, a_3)

∈ R, for all

a_1, a_2, a_3

∈ A

Here we see that (1, 3) ∈ R and also (3, 9) ∈ R but (1, 9) ∉ R

∴ R is not Transitive

∴ R is neither reflexive nor symmetric nor transitive.

ii. Relation R in the set N of natural numbers defined as R = {

(x, y) : y = x + 5

and

x \lt 4

}

As it is given that

{x \lt 4}

x

takes the values 1, 2 and 3.

When

{x = 1}

{y = x + 5 = 1 + 5 = 6}

. So, (1, 6) ∈ R

When

{x = 2}

{y = x + 5 = 2 + 5 = 7}

. So, (2, 7) ∈ R

When

{x = 3}

{y = x + 5 = 3 + 5 = 8}

. So, (3, 8) ∈ R

This can also be put in the tabular form as follows:

x

y = x + 5

(x, y)

∈ R?

1 + 5 = 6

✔

2 + 5 = 7

✔

3 + 5 = 8

✔

∴ R = {(1, 6), (2, 7), (3, 8)}

To Check whether R is Reflexive: The relation R in the set A is reflexive if

(a, a)

∈ R, for every

a

∈ A

In this case, (1, 1) ∉ R, (2, 2) ∉ R and (3, 3) ∉ R.

∴ R is not Reflexive

To Check whether R is Symmetric: The relation R in the set A is symmetric if

(a_1, a_2)

∈ R implies that

(a_2, a_1)

∈ R, for all

a_1, a_2

∈ A

In this case, we see that (1, 6) ∈ R, but (6, 1) ∉ R

∴ R is not Symmetric

To Check whether R is Transitive: The relation R in the set A is transitive if

(a_1, a_2)

∈ R and

(a_2, a_3)

∈ R implies that

(a_1, a_3)

∈ R, for all

a_1, a_2, a_3

∈ A

In this case, we see that (1, 3) ∈ R and also (3, 8) ∈ R but (1, 8) ∉ R

∴ R is not Transitive

∴ R is neither reflexive nor symmetric nor transitive.

iii. Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {

(x, y) : y

is divisible by

x

}

To Check whether R is Reflexive: The relation R in the set A is reflexive if

(a, a)

∈ R, for every

a

∈ A.

As we know, every element

x

is divisible by itself i.e.

x

. In otherwords,

(x, x)

∈ R

∴ R is Reflexive

To Check whether R is Symmetric: The relation R in the set A is symmetric if

(a_1, a_2)

∈ R implies that

(a_2, a_1)

∈ R, for all

a_1, a_2

∈ A

In this case, we can see that 6 is divisible by 3. However, 3 is not divisible by 6. In otherwords, (3, 6) ∈ R but (6, 3) ∉ R.

∴ R is not Symmetric

To Check whether R is Transitive: The relation R in the set A is transitive if

(a_1, a_2)

∈ R and

(a_2, a_3)

∈ R implies that

(a_1, a_3)

∈ R, for all

a_1, a_2, a_3

∈ A

Here we see that (1, 3) ∈ R as 3 is divisible by 1. Also, (3, 6) ∈ R as 6 is divisible by 3. Also we see that 6 is divisible by 1 i.e. (1, 6) ∈ R

∴ R is Transitive

∴ R is both reflexive and transitive but not symmetric.

iv. Relation R in the set Z of all integers defined as R = {

(x, y) : x - y

is an integer}

To Check whether R is Reflexive: The relation R in the set A is reflexive if

(a, a)

∈ R, for every

a

∈ A.

In this case, for any element

x

{x - x = 0}

. As 0 is also an integer, this property is true for every element

x

∈ Z. In otherwords,

(x, x)

∈ R

∴ R is Reflexive

To Check whether R is Symmetric: The relation R in the set A is symmetric if

(a_1, a_2)

∈ R implies that

(a_2, a_1)

∈ R, for all

a_1, a_2

∈ A

{x - y = k}

, where

k

is an integer.

Now,

{y - x = -(x - y) = -k}

. If

k

is an integer, then

-k

is also an integer. So, we see that if

(x, y)

∈ R then

(y, x)

∈ R.

∴ R is Symmetric

To Check whether R is Transitive: The relation R in the set A is transitive if

(a_1, a_2)

∈ R and

(a_2, a_3)

∈ R implies that

(a_1, a_3)

∈ R, for all

a_1, a_2, a_3

∈ A

Here we see that if

{(x - y)}

is an integer and

{(y - z)}

is another integer, then

{(x - z) = (x - y) + (y - z)}

is also an integer, as sum of two integers is integer. So, we can say that if

(x, y)

∈ R and

(y, z)

∈ R then

(x, z)

∈ R.

∴ R is Transitive

∴ R is reflexive, symmetric as well as transitive.

v (a). R = {

(x, y) : x

and

y

work at the same place}

To Check whether R is Reflexive: The relation R in the set A is reflexive if

(a, a)

∈ R, for every

a

∈ A.

In this case, if

x

is working at a given place, then the same person

x

is also working at the same place. So, clearly

(x, x)

∈ R.

∴ R is Reflexive

To Check whether R is Symmetric: The relation R in the set A is symmetric if

(a_1, a_2)

∈ R implies that

(a_2, a_1)

∈ R, for all

a_1, a_2

∈ A

x

and

y

are working at a given place, then it implies that

y

and

x

are also working at the same place. So, we can say that if

(x, y)

∈ R then

(y, x)

∈ R

∴ R is Symmetric

To Check whether R is Transitive: The relation R in the set A is transitive if

(a_1, a_2)

∈ R and

(a_2, a_3)

∈ R implies that

(a_1, a_3)

∈ R, for all

a_1, a_2, a_3

∈ A

x

and

y

are working at the given place and also

y

and

z

are working at that place, then it implies that both

x

and

z

are also working at the same place. In otherwords, if

(x, y)

∈ R,

(y, z)

∈ R then

(x, y)

∈ R.

∴ R is Transitive

∴ R is reflexive, symmetric as well as transitive.

v (b). R = {

(x, y) : x

and

y

live in the same locality}

To Check whether R is Reflexive: The relation R in the set A is reflexive if

(a, a)

∈ R, for every

a

∈ A.

In this case, if

x

is living at a given locality, then the same person

x

is also living at the same locality. So, clearly

(x, x)

∈ R.

∴ R is Reflexive

To Check whether R is Symmetric: The relation R in the set A is symmetric if

(a_1, a_2)

∈ R implies that

(a_2, a_1)

∈ R, for all

a_1, a_2

∈ A

x

and

y

are living at a given locality, then it implies that

y

and

x

are also living at the same locality. This implies that if

(x, y)

∈ R then

(y, x)

∈ R

∴ R is Symmetric

To Check whether R is Transitive: The relation R in the set A is transitive if

(a_1, a_2)

∈ R and

(a_2, a_3)

∈ R implies that

(a_1, a_3)

∈ R, for all

a_1, a_2, a_3

∈ A

x

and

y

are living at a given locality and also

y

and

z

are living at that locality, then it implies that both

x

and

z

are also living at the same locality.

In otherwords, if

(x, y)

∈ R,

(y, z)

∈ R then

(x, z)

∈ R.

∴ R is Transitive

∴ R is reflexive, symmetric as well as transitive.

v (c). R =

(x, y) : x

is exactly 7 cm taller than

y

To Check whether R is Reflexive: The relation R in the set A is reflexive if

(a, a)

∈ R, for every

a

∈ A.

As a person

x

can never be taller than oneself,

(x, x)

∉ R.

∴ R is not Reflexive

To Check whether R is Symmetric: The relation R in the set A is symmetric if

(a_1, a_2)

∈ R implies that

(a_2, a_1)

∈ R, for all

a_1, a_2

∈ A

x

is 7 cm taller than

y

, then we can not say that

y

is also taller than

x

. In fact,

y

will be shorter than

x

. So, in this case, if

(x, y)

∈ R then

(y, x)

∉ R

∴ R is not Symmetric

To Check whether R is Transitive: The relation R in the set A is transitive if

(a_1, a_2)

∈ R and

(a_2, a_3)

∈ R implies that

(a_1, a_3)

∈ R, for all

a_1, a_2, a_3

∈ A

x

is exactly 7 cm taller than

x

and

y

is exactly 7 cm taller than

z

, then

x

is (7 cm + 7 cm) = 14 cm taller than

z

. So,

x

is not exactly 7 cm taller than

z

. (Though

x

is taller than

z

but not exactly by 7 cm). In otherwords, if

(x, y)

∈ R,

(y, z)

∈ R then

(x, z)

∉ R.

∴ R is not Transitive

∴ R is neither reflexive nor symmetric nor transitive.

v (d). R = {

(x, y) : x

is wife of

y

}

To Check whether R is Reflexive: The relation R in the set A is reflexive if

(a, a)

∈ R, for every

a

∈ A.

As a person

x

can not be wife of oneself. So,

(x, x)

∉ R.

∴ R is not Reflexive

To Check whether R is Symmetric: The relation R in the set A is symmetric if

(a_1, a_2)

∈ R implies that

(a_2, a_1)

∈ R, for all

a_1, a_2

∈ A

x

is wife of

y

then

y

can not be wife of

x

(

y

will be husband of

x

). So, if

(x, y)

∈ R then

(y, x)

∉ R.

∴ R is not symmetric

To Check whether R is Transitive: The relation R in the set A is transitive if

(a_1, a_2)

∈ R and

(a_2, a_3)

∈ R implies that

(a_1, a_3)

∈ R, for all

a_1, a_2, a_3

∈ A

x

is wife of

y

, then

y

can not be wife of any other person

z

(as

y

is husband). So, when

(x, y)

∈ R,

(y, z)

∉ R.

∴ R is not Transitive

∴ R is neither reflexive nor symmetric nor transitive.

v (d). R = {

(x, y) : x

is father of

y

}

To Check whether R is Reflexive: The relation R in the set A is reflexive if

(a, a)

∈ R, for every

a

∈ A.

As a person

x

can not be father of oneself. So,

(x, x)

∉ R.

∴ R is not Reflexive

To Check whether R is Symmetric: The relation R in the set A is symmetric if

(a_1, a_2)

∈ R implies that

(a_2, a_1)

∈ R, for all

a_1, a_2

∈ A

x

is father of

y

then

y

can not be father of

x

(

y

will be child of

x

). So, if

(x, y)

∈ R then

(y, x)

∉ R.

∴ R is not symmetric

To Check whether R is Transitive: The relation R in the set A is transitive if

(a_1, a_2)

∈ R and

(a_2, a_3)

∈ R implies that

(a_1, a_3)

∈ R, for all

a_1, a_2, a_3

∈ A

x

is father of

y

and

y

is father of

z

, then

x

is not father of

z

(

x

will be grandfather of

z

). So, if

(x, y)

∈ R and

(y, z)

∈ R then

(x, z)

∉ R.

∴ R is not Transitive

∴ R is neither reflexive nor symmetric nor transitive.

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