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?

This is the special case $\,\rm G = \Bbb Q^{\times}\,$ and $\rm\,H = R^{\times}\,$ in the below complementary subgroup law. Here the composition law arises via the  following complementary view of the Subgroup Test ($\rm\color{#c00}{ST} $)
Theorem $ $ Let $\rm\,G\,$ be a nonempty subset of an abelian group $\rm\,H,\,$ with complement set $\rm\,\bar G = H\backslash G.\,$
Then $\rm\,G\,$ is a subgroup of $\rm\,H\iff G + \bar G\, =\, \bar G. $ 
Proof  $\ $ $\rm\,G\,$ is a subgroup of $\rm\,H\!\overset{\ \large \color{#c00}{\rm ST}}\iff\! G\,$ is closed under subtraction, so, complementing
$\begin{eqnarray} & &\ \ \rm G\text{ is a subgroup of }\, H\  fails\\
&\iff&\ \rm\ G\ -\ G\ \subseteq\, G\,\ \ fails\\
&\iff&\ \rm\ g_1\, -\ g_2 =\,\ \bar g\ \ \ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\
&\iff&\ \rm\ g_2\, +\ \bar g\ \ =\,\ g_1\  for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\
&\iff&\ \rm\ G\ +\ \bar G\ \subseteq\ \bar G\ \ fails\qquad\ {\bf QED}
\end{eqnarray}$ 
Instances of this law are ubiquitous in concrete number systems, e.g. below. For many further examples see some of my prior posts here (and also on sci.math).

