Assume $r,s \in\mathbb{Q}$. Prove $\frac{r}{s},r-s \in\mathbb{Q}$ I have attempted this proof by contradiction. Beginning with assuming to the contrary that each a and b are irrational but was not sure if I did it correctly. Any help would be greatly appreciated.
Assume $r,s \in\mathbb{Q}$.
a) Prove $\frac{r}{s}\in\mathbb{Q}$
b) Prove $r-s \in\mathbb{Q}$

$x\in\mathbb Q\iff \exists a,b\in\mathbb Z: b\ne0,\  x=\frac ab$
Let $r=\frac pq$,$s=\frac tu$, where $p,q,t,u\in\mathbb Z$ and $u,q\ne0$. Then,
If $s=0$, $\frac rs$ is not defined. Assuming, $s\ne0$,thus, $t\ne0$. $\frac rs=\frac {pu}{qt}$, where $pu,qt\in\mathbb Z$, as $t,q\ne0$; $qt\ne0$. Thus, $\frac rs\in\mathbb Q$.
$r-s=\frac {pu-tq}{qu}$, where $pu-tq,qu\in\mathbb Z$, and $qu\ne$ as $q,n\ne0$. Thus, $r-s\in\mathbb Q$
