Let R be a relation on \mathbb{Z} defined by R=\{(p,q)\in\mathbb{Z}\times\mat

tricotasu 2021-08-19 Answered
Let R be a relation on \(\displaystyle{\mathbb{{{Z}}}}\) defined by
\(\displaystyle{R}={\left\lbrace{\left({p},{q}\right)}\in{\mathbb{{{Z}}}}\times{\mathbb{{{Z}}}}{\mid}{p}-{q}\right.}\) is a multiple of \(\displaystyle{3}\rbrace\)
a) Show that R is reflexive.
b) Show that R is symmetric.
c) Show that R is transitive.

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Expert Answer

BleabyinfibiaG
Answered 2021-08-20 Author has 3342 answers

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.

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