Solution:

$R=\{(a, b): a>b\}$ on $N$ (given)

Non-Reflexivity:
Assume $a$ be an arbitrary element of $\mathrm{N}$
a cannot be greater than $a$
$\Rightarrow(\mathrm{a}, \mathrm{a}) \notin \mathrm{R}$
Therefore, $R$ is not reflexive.

Symmetric:
Assume $a$ and $b \in N$, such that $(a, b) \in R$
$\Rightarrow a>b$
$\Rightarrow$ $b$ cannot be greater than $a$
$\Rightarrow(\mathrm{b}, \mathrm{a}) \notin \mathrm{R}$
Therefore, $\mathrm{R}$ is not symmetric.

Transitivity:
Suppose $a, b$ and $c \in N$, such that $(a, b) \in R$ and $(b, c) \in R$
$\Rightarrow \mathrm{a}>\mathrm{b}$ and $\mathrm{b}>\mathrm{c}$
$\begin{array}{l}
\Rightarrow \mathrm{a}>\mathrm{c} \\
\Rightarrow(\mathrm{a}, \mathrm{c}) \in \mathrm{R}
\end{array}$
Therefore, $\mathrm{R}$ is transitive.
As a result, $\mathrm{R}$ is transitive but neither symmetric nor reflexive.