# 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
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 aA.
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_2A
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_3A
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.