Given in a set of real numbers, the relation $R=\{(a,b):a\leq b^2\}$:

Let $a = \frac{1}{3} \; \rightarrow \ a^2 = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Clearly in this case, since $\frac{1}{3} \not \leq (\frac{1}{3})^2$, i.e., $\frac{1}{3} \not \leq \frac{1}{9}$,

Therefore since $(\frac{1}{3}.\frac{1}{3}) \not \in R$, $R$ is not reflexive.

Let $a = 1$ and $b = 3 \rightarrow \; a \leq b^2 \rightarrow 1 \leq 3 \times 3 \rightarrow 1 \leq 9$.

However, $a^2 = 1 \times 1 = 1 \rightarrow 3 \not \leq 1$ i.e., $ b \not \leq a^2 $.

Therefore while $(x,y) \in R$, $(y,x) \not \in R$. Hence is $R$ is not symmetric.

Let $a = 2$ and $b = -3 \; \rightarrow \ b^2 = -3 \times -3 = 9 \rightarrow 2 \leq 9 \rightarrow a \leq b^2$.

Let $b = -3$ and $c =1 \; \rightarrow \ c^2 = 1 \times 1 = 1 \rightarrow -3 \leq 1 \rightarrow b \leq c^2$.

However $a =2$ and $c =1 \; \rightarrow \ c^2 = 1 \times 1 = 1 \rightarrow 2 \not \leq 1 \rightarrow a \not \leq c^2$.

Hence $R$ is not transitive.