Problem 5 Solution

This page contains the NCERT mathematics class 12 chapter Relations and Functions Exercise 1.1 Problem 5 Solution. Solutions for other problems are available at Exercise 1.1 Solutions

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Exercise 1.1 Problem 5 Solution

5. Check whether the relation R in R defined by R = {

(a, b) : a \le b^3

} is reflexive, symmetric or transitive.

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

(a, a)

∈ R, for every

a

∈ A.

When

{a = -2}

, we’ve

{a^3 = -8}

. But

-2 \nleq (-2)^3

∴ 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

Consider

{a = 1}

and

{b = 2}

we’ve 1 ≤ 2³ but

2 \nleq 1^3

⇒ When

(a, b)

∈ R, we have

(b, a)

∉ 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

Consider the case,

{a = 2}

{b = 3}

and

{c = 4}

We’ve

{b^3 = 3^3 = 27}

and 4³ = 64

So, 2 ≤ 3³, 3 ≤ 4³ and 2 ≤ 4³.

However, when

{a = 4}

{b = \dfrac{7}{4}}

and

{c = \dfrac{5}{4}}

We’ve

{b^3 = \left(\dfrac{7}{4}\right)^{3} = 5.359}

and

{c^3 = \left(\dfrac{5}{4}\right)^{3} = 1.953}

In this case, 4 ≤

\left(\dfrac{7}{4}\right)^3

and

\dfrac{7}{4}

≤

\left(\dfrac{5}{4}\right)^3

, but

{4 \nleq \left(\dfrac{5}{4}\right)^3}

∴ R is not Transitive

Hence, in this case, R is neither reflexive nor symmetric nor transitive.

Back to Exercise 1.1

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