Sums, differences, products and quotients of an irrational and a rational are irrational.

Prove that if $x \in \mathbb{Q}$ with $x \neq 0$ , and $y \in \mathbb{R} \smallsetminus \mathbb{Q}$ , then

$x+y,\quad x-y, \quad xy, \quad \frac{x}{y}, \quad \frac{y}{x}$

are all irrational, i.e., are all in $\mathbb{R} \smallsetminus \mathbb{Q}$ .

Proof. Since $x \in \mathbb{Q}$ , we know there are integers $m$ and $n$ such that $x = \frac{m}{n}$ . Now we consider each of the elements we wish to show are irrational.

Suppose otherwise, that $x+y \in \mathbb{Q}$ . Then, there exist $r,s \in \mathbb{Z}$ such that
$x+y = \frac{r}{s} \ \implies \ \frac{m}{n} +y = \frac{r}{s} \ \implies \ y = \frac{r}{s} - \frac{m}{n} \implies \ y = \frac{rn - ms}{sn} \ \implies \ y \in \mathbb{Q}.$

This contradicts our assumption that $y$ is irrational.
Since $y$ is irrational, so is $-y$ (by part (a), the sum of a rational and an irrational cannot be rational and since $y+(-y) = 0$ is rational, cannot have $-y$ rational). But then, $x-y = x+(-y)$ and by part (a) this sum must be irrational.
Suppose otherwise, that $xy \in \mathbb{Q}$ . Then there exist $r,s \in \mathbb{Z}$ such that $xy = \frac{r}{s}$ . Further, since $x \neq 0$ , we know $x^{-1} = \frac{n/m}$ exists.
$xy = \frac{r}{s} \ \implies \ \frac{m}{n} y = \frac{r}{s} \ \implies \ y = \frac{r}{s} \frac{n}{m} \ \implies \ y = \frac{rn}{sm} \in \mathbb{Q}.$

Contradicting our assumption that $y$ is irrational.
First, we since $y$ is irrational, we have $y \neq 0$ , and thus $y^{-1}$ exists. Further, since $yy^{-1} = 1$ is rational, and $y$ is irrational, by (c) we must have $y^{-1}$ irrational as well. Then by (c), since $\frac{x}{y} = xy^{-1}$ , we must have $xy^{-1}$ irrational.
By (d) we know $\frac{x}{y}$ is irrational, and since $\frac{x}{y} \frac{y}{x} = 1$ is rational, by (c), we must have $\frac{y}{x}$ irrational.