# Problem 2 Solution

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 Solutions
Exercise 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 aA
In this case, for every aR, 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 aR
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_2A
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_3A
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.