Let $x \in \mathbb Q \setminus \{0\}$ and $y \in \mathbb R\setminus \mathbb Q$. Prove that $\frac{x}{y} \in \mathbb R \setminus \mathbb Q$ Let $x \in \mathbb Q\setminus  \{0\}$ and $y \in \mathbb R\setminus \mathbb Q.$ Prove that $\frac{x}{y} \in \mathbb R \setminus \mathbb Q$
I saw this question in a basic analysis test but it confuses me because intuitively it makes sense but how do you show mathematically that the set of rationals is not in the solution space?

From the hypothesis we have that
$$\exists n,m \in \mathbb{Z} \setminus \{ 0 \} : x = \frac{m}{n}$$
and that
$$ y \in \mathbb{R} \setminus \mathbb{Q}$$
Now assume that
$$ \frac{x}{y} \in \mathbb{Q}$$
Now we
$$ \exists p,q \in \mathbb{Z} : \frac{x}{y} = \frac{p}{q}$$
Well we can write this as
$$ \frac{1}{y} \frac{m}{n} = \frac{p}{q} $$
With some further algebraic manipulations can you see why this would be a contradiction?
