This page contains the

**NCERT mathematics class 12 chapter Relations and Functions Exercise 1.1 Problem 2 Solution**. Solutions for other problems are available at Exercise 1.1 SolutionsExercise 1.1 Problem 2 Solution

2. Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a \le a^2} is neither reflexive nor symmetric nor transitive.

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 every a ∈ R, it should satisfy the relation, a \le a^2

When {a = 2}, {a^2 = 4}. So, 2 ≤ 2²

When {a = \dfrac{1}{5}}, {a^2 = \dfrac{1}{25}}. So, {\dfrac{1}{5}\nleq\left(\dfrac{1}{25}\right)^2}

Thus (a, a) ∈ R is not valid for all real numbers a ∈ 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

We have 3 ≤ 4² and also 4 ≤ 3². In this case, (a, b) ∈ R and also (b, a) ∈ R

However, 1 ≤ 2² but {2 \nleq 1^2}. In this case, (a, b) ∈ R but (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

This means that if {a \le b^2} and {b \le c^2} then {a \le c^2}

We have 3 ≤ (-5)² and -5 ≤ 1² but {3 \nleq 1^2}

∴ R is not Transitive

∴ R is neither reflexive nor symmetric nor transitive.