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

(N/A) $(i)$ Reflexive: $A$ relation $R$ is reflexive if $(a, a) \in R$ for all $a \in \mathbb{R}$.
Consider $a = \frac{1}{2}$. Since $\frac{1}{2} > (\frac{1}{2})^2 = \frac{1}{4}$,it follows that $(\frac{1}{2}, \frac{1}{2}) \notin R$.
Thus,$R$ is not reflexive.
$(ii)$ Symmetric: $A$ relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$.
Consider $(1, 4) \in R$ because $1 \leq 4^2 = 16$. However,$4 \not\leq 1^2 = 1$,so $(4, 1) \notin R$.
Thus,$R$ is not symmetric.
$(iii)$ Transitive: $A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$.
Consider $(3, 2) \in R$ (since $3 \leq 2^2 = 4$) and $(2, 1.5) \in R$ (since $2 \leq 1.5^2 = 2.25$).
However,$3 \not\leq 1.5^2 = 2.25$,so $(3, 1.5) \notin R$.
Thus,$R$ is not transitive.
Conclusion: The relation $R$ is neither reflexive,nor symmetric,nor transitive.