Let Z be the set of all integers and Z0 be the set of all non-zero integers. Let a relation R on Z × Z0 be defined as

Solution:

We observe the following properties of R.

Reflexivity:

Let $(a, b)$ be an arbitrary element of $\mathrm{Z} \times \mathrm{Z}_{0}$. Then,

$(a, b) \in \mathrm{Z} \times \mathrm{Z}_{0}$

$\Rightarrow a, b \in \mathrm{Z}, \mathrm{Z}_{0}$

$\Rightarrow a b=b a$

$\Rightarrow(a, b) \in R$ for all $(a, b) \in Z \times Z_{0}$

So, $R$ is reflexive on $Z \times Z_{0}$.

Symmetry:

Let $(a, b),(c, d) \in Z \times Z_{0}$ such that $(a, b) R(c, d)$. Then,

$(a, b) R(c, d)$

$\Rightarrow a d=b c$

$\Rightarrow c b=d a$

$\Rightarrow(c, d) R(a, b)$

Thus,

$(a, b) R(c, d) \Rightarrow(c, d) R(a, b)$ for all $(a, b),(c, d) \in Z \times Z_{0}$

So, $R$ is symmetric on $Z \times Z_{0}$.

Transitivity:

Let $(a, b),(c, d),(e, f) \in N \times N_{0}$ such that $(a, b) R(c, d)$ and $(c, d) R(e, f) .$ Then,

$\left.\begin{array}{l}(a, b) R(c, d) \Rightarrow a d=b c \\ (c, d) R(e, f) \Rightarrow c f=d e\end{array}\right\} \Rightarrow(a d)(c f)=(b c)(d e)$

$\Rightarrow a f=b e$

$\Rightarrow(a, b) R(e, f)$

Thus,

$(a, b) R(c, d)$ and $(c, d) R(e, f) \Rightarrow(a, b) R(e, f)$

$\Rightarrow(a, b) R(e, f)$ for all values $(a, b),(c, d),(e, f) \in N \times N_{0}$

So, $R$ is transitive on $N \times N_{0}$.