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

We have $R = \{ (a,\;b);\;a \le {b^2}\} ,$ where $a,\;b \in R$

(i) Reflexivity

We observe that, $\cfrac{1}{3} \le {\left( {\cfrac{1}{3}} \right)^2}$ is not true.

Therefore, $\left( {\cfrac{1}{3},\;\cfrac{1}{3}} \right) \notin R.$ So, R is not reflexive.

(ii) Symmetry

We observe that, $1 \le {(2)^2}\;\;but\;\;2$ ${1^2}$

i.e., $(1,\;2) \in R$ but $(2,\;1) \notin R$

So, R is not symmetric.

(iii) Transitivity

We observe that, $10 \le {4^2}\;\;and\;\;4 \le {3^2}$ but $10$ ${(3)^2}$

i.e., $(10,\;4) \in R\;\;and\;\;(4,\;3) \in R\;\;but\;\;(10,\;3) \notin R$

So, R is not transitive.