a) For any element \(\displaystyle{p}\in{\mathbb{{{Z}}}}\)

\(\displaystyle{p}-{p}={0}\)

is a multiple of 0.

Thus, \(\displaystyle{\left({p},{p}\right)}\in{R}\Rightarrow{R}\) is reflexive.

b) Let \(\displaystyle{\left({p},{q}\right)}\in{R}\) Then

\(\displaystyle{p}-{q}\) is a multiple of 3, that is \(\displaystyle{p}-{q}={3}{k}\) for some \(\displaystyle{k}\in{\mathbb{{{Z}}}}\)

To show that \(\displaystyle{\left({q},{p}\right)}\in{R}\) Consider

\(\displaystyle{q}-{p}=-{\left({p}-{q}\right)}\)

\(\displaystyle=-{3}{k}={3}{m}\)

where \(\displaystyle{m}=-{k}\in{\mathbb{{{Z}}}}\)

So \(\displaystyle{q}-{p}\) is a multiple of 3

\(\displaystyle\Rightarrow{\left({q},{p}\right)}\in{R}\)

\(\displaystyle\Rightarrow{R}\) is symmetric

c) Let \(\displaystyle{\left({p},{q}\right)}\in{R}\) and \(\displaystyle{\left({q},{r}\right)}\in{R}\) Then both \(\displaystyle{p}-{q}\) and \(\displaystyle{q}-{r}\) are multiples of 3, that is,

\(\displaystyle{p}-{q}={3}{k},{k}\in{\mathbb{{{Z}}}}\) and \(\displaystyle{q}-{r}={3}{m},{m}\in{R}\)

To show \(\displaystyle{\left({p},{r}\right)}\in{R}.\) For that, consider

\(\displaystyle{p}-{r}={p}-{q}+{q}-{r}\)

\(\displaystyle={3}{k}+{3}{m}={3}{\left({k}+{m}\right)}\)

\(\displaystyle\Rightarrow{p}-{r}\) is a multiple of 3.

\(\displaystyle\Rightarrow{\left({p},{r}\right)}\in{R}\)

Thus, R is transitive.