The relation R on the set of real numbers defined by $R= \{(a,b): a≤ b^2\}$ is

neither reflexive nor symmetric nor transitive

The correct answer is Option (1) → neither reflexive nor symmetric nor transitive

Given relation: $R=\{(a,b):a\le b^{2}\}$ on $\mathbb{R}$

Reflexive: For reflexivity, $(a,a)\in R\ \forall a$ $\Rightarrow a\le a^{2}$ must hold for all real $a$. For $a=\frac{1}{2}$, $a^{2}=\frac{1}{4}$, and $a\le a^{2}$ is false. Hence, not reflexive.

Symmetric: If $(a,b)\in R$, then $a\le b^{2}$. For symmetry, need $(b,a)\in R \Rightarrow b\le a^{2}$. Take $a=1,\ b=2$: $1\le4$ (true), but $2\le1$ (false). Hence, not symmetric.

Transitive: If $(a,b)\in R$ and $(b,c)\in R$, then $a\le b^{2}$ and $b\le c^{2}\Rightarrow a\le (c^{2})^{2}=c^{4}$. But need $a\le c^{2}$ for transitivity, which does not always hold. Example: $a=3,\ b=2,\ c=1$ ⇒ $3\le4$ (true), $2\le1$ (false). So not transitive.

Answer: neither reflexive nor symmetric nor transitive