Solution:
(i) Reflexivity:
Suppose $\mathrm{p}$ is an arbitrary element of $\mathrm{S}$.
So now, $\mathrm{p} \leq \mathrm{p}$
$\Rightarrow(p, p) \in R$
Therefore, $\mathrm{R}$ is reflexive.
(ii) Transitivity:
Suppose $\mathrm{p}, \mathrm{q}$ and $\mathrm{r} \in \mathrm{S}$, such that $(\mathrm{p}, \mathrm{q}) \in \mathrm{R}$ and $(\mathrm{q}, \mathrm{r}) \in \mathrm{R}$
$\begin{array}{l}
\Rightarrow \mathrm{p} \leq \mathrm{q} \text { and } \mathrm{q} \leq \mathrm{r} \\
\Rightarrow \mathrm{p} \leq \mathrm{r} \\
\Rightarrow(\mathrm{p}, \mathrm{r}) \in \mathrm{R}
\end{array}$
Therefore, $\mathrm{R}$ is transitive.