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?
 A: Do you know the fact that the set  rational numbers is closed under all four arithmetic operations, $+,-,\times, /$ (avoiding division by zero)? Thats all to it.
A: We start saying that $y \in \mathbb{R} \setminus \mathbb{Q}.$
Since $x \in \mathbb{Q} \setminus \{0\}$, then there exists two coprime integers $a$ and $b$ ($a \neq 0, b \neq 0$), such that 
$$x = \frac{a}{b}.$$
Now, suppose that $$\frac{x}{y} \in \mathbb{Q}.$$
Then, there exists two coprime integers $c$ and $d$ ($c \neq 0, d \neq 0$), such that 
$$\frac{x}{y} = \frac{c}{d}$$
Clearly, $$ y = \frac{xd}{c} = \frac{ad}{bc} \in \mathbb{Q}.$$
We found a contradiction. Then $$\frac{x}{y} \not\in \mathbb{Q}$$ and the following must be true:
$$\frac{x}{y} \in \mathbb{R} \setminus \mathbb{Q}.$$
A: 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?
A: 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).

