Question

Let $R$ be a relation from $Q$ to $Q$ defined by $R=\left\{\right(a,b):a,b\in Q\text{and}a-b\in Z\}$.
Show that:
$\left(i\right)(a,a)\in R$ for all $a\in Q$
$\left(ii\right)(a,b)\in R$ implies that $(b,a)\in R$
$\left(iii\right)(a,b)\in R\text{and}(b,c)\in R$ implies that $(a,c)\in R$

Answer 1

$\left(i\right)$ Given:$R=\left\{\right(a,b):a,b\in Q\text{and}a-b\in Z\}$.
If $a,a\in Q$, i.e.,$a$ is in set $Q$
Also $a-a=0\in Z$
Hence, it proved that $(a,a)\in R$ for all $a\in Q$

$\left(ii\right)$ Given:$R=\left\{\right(a,b):a,b\in Q\text{and}a-b\in Z\}$.
$(a,b)\in R$ implies that $a-b\in Z$
So, $b-a\in Z(\because$negative of an integer is also an integer$)$
$\therefore (b,a)\in Z$
Hence, $(a,b)\in R$ implies that $(b,a)\in R$

$\left(iii\right)$ Given:$R=\left\{\right(a,b):a,b\in Q\text{and}a-b\in Z\}$.
$(a,b)\text{and}(b,c)\in R$
Implies that $a-b\in Z,b-c\in Z$
So, $a-c=(a-b)+(b-c)\in Z$
$\therefore (a,c)\in R$
Hence, it proved that $(a,b)\in R\text{and}(b,c)\in R$ implies that $(a,c)\in R$