Intro proofs problem3 Prove the quotient of an integer and a nonzero rational number is a rational number. 
Wondering if i did this correct
$m \in \mathbb{Z} $
$\frac{m}{n} \in \mathbb{Q} $
By the definition of nonzero rational,   $ n = \frac {a}{b} $ for some $ a,b\in \mathbb{Z}$ and  $b \neq 0$.
Consider : 
$ \frac {m}{n} = \frac {mb}{a}$. Let $x = mb, x \in \mathbb{Z} $
so:
$ \frac mn = \frac xa$
Therefore, by definition of rational numbers, the quotient of an integer and a nonzero rational number is a rational number. 
 A: 
$m \in \mathbb{Z} $
$\frac{m}{n} \in \mathbb{Q} $


*

*that second definition doesn't really make sense, if what you are trying to do is specify $n$. Better to introduce your rational variable separately:

$ r\in \mathbb{Q} $

By the definition of nonzero rational,   $ n = \frac {a}{b} $ for some $ a,b\in \mathbb{Z}$ and  $b \neq 0$


*

*this neglects the "non=zero" part of the definition:

By the definition of rational,   $ r = \frac {a}{b} $ for some $ a,b\in \mathbb{Z}$ and  $b \neq 0$, and since $r$ is non-zero, we also have $a\ne 0$

Consider :
$ \frac {m}{n} = \frac {mb}{a}$

Probably better to break this down a little more, and (at least briefly) use the operator version of division:
$ \frac mr = m \div r = m\div \frac ab = m \times \frac ba = \frac{mb}a$
($\frac ba$ is well-defined since we know that $a\ne 0$)

let $x = mb, x \in \mathbb{Z} $

$ x \in \mathbb{Z} $ should be an inference:
let $x = mb$; then since $m,b  \in \mathbb{Z}$ we also have $x \in \mathbb{Z} $

so:
$ \frac mn = \frac xa$


*

*Could add

$x,a\in \mathbb{Z}$ and $a \ne 0 \implies \frac xa \in \mathbb{Q}$

Therefore, by definition of rational numbers, the quotient of an integer and a nonzero rational number is a rational number.

$\square$
