Can I assume that "dividing on both sides" as a step in this proof is a well-defined operation?

Theorem: Suppose $x in mathbb{Q}$ with $x neq 0$ and $y in mathbb{R – Q}$. Then $xy$ is irrational.

Proof. We proceed by contradiction. Suppose $xy in mathbb{Q}$. Then $xy = frac{a}{b}$, where $a, b in mathbb{Z}, b neq 0$. By hypothesis $x$ is rational, so we can write $x = frac{c}{d}$, $c, d in mathbb{Z}$ and $d neq 0$. Then
$$y = frac{a}{bx} = frac{ad}{bc}$$

By closure of integer multiplication, $ad = m in mathbb{Z}$ and $bc = n in mathbb{Z}$. Then $y = frac{m}{n}$, so $y in mathbb{Q}$.

But this is a contradiction, since by hypothesis $y in mathbb{R – Q}$.

Then our assumption that $xy$ is rational must be false. Then $xy$ is irrational. $square$.

Discussion

The proof as written hinges on the fact that $xy = frac{a}{b} leftrightarrow y = frac{a}{bx}$.

It feels a bit silly to ask, but is it reasonable to assume that (given "basic facts" about the rationals and integers such that one would have at the beginning of an introductory analysis course) I can "divide/multiply both sides" to obtain this result, and also the next result that $frac{a}{bfrac{c}{d}} = frac{ad}{bc}$?