Proof by Contradiction relating to rational and irrational numbers

Question

I've been given the question: given $x,y\in\mathbb{R}\setminus\mathbb{Q}$ and $x+y =\frac{m}{n}$, prove $x-y$ is irrational. I tried solving this using a proof by contradiction but I feel like I got a bit off base and I feel like I've screwed up somewhere.

Proof by Contradiction
$x,y \in \mathbb{R} \backslash \mathbb{Q}$
$x + y = \frac{m}{n}$, $m,n \in \mathbb{Z}$, $n \neq 0 $
$y = \frac{m}{n} - x$
Assume $x-y$ is rational

$x-y = \frac{p}{q}$, $p, q \in \mathbb{Z}$, $q \neq 0$
$x - (\frac{m}{n} - x) = \frac{p}{q}$, using $y = \frac{m}{n} - x$
$2x - \frac{m}{n} = \frac{p}{q}$
$\frac{2xn-m}{n} = \frac{p}{q}$
$q(2xn-m) = pn$
$2xn-m = \frac{pn}{q}$
$2xn = \frac{pn}{q} + m$
$x = \frac{(\frac{pn}{q} + m)}{2n}$

$\therefore$ Contradiction as $x$ is irrational

Here is my working. Have I made any mistakes and is it okay to assume that since all the variables in the final fraction are integers that it is rational?

Answer 1

It's correct. A bit too lengthy, though. From $$ 2x-\frac{m}{n}=\frac{p}{q} $$ you can derive $$ 2x=\frac{m}{n}+\frac{p}{q}=\frac{mq+np}{nq} $$ so $$ x=\frac{mq+np}{2nq} $$ would be rational, because you assumed $m,n,p,q\in\mathbb{Z}$ with $n\ne0$ and $q\ne0$, so $mq+np\in\mathbb{Z}$ and $2nq\ne0$.

Answer 2

Since $x+y$ is rational, if $x - y$ is rational, then so are $(x + y) - (x - y)$ and $(x + y) + (x - y)$

Answer 3

From your "$2x -\frac{m}{n} = \frac{p}{q}$, you should realize you have "twice an irrational is the sum of two rationals": "$2x = \frac{p}{q} + \frac{m}{n}$. The sum of two rationals is rational, twice an irrational is irrational, and you have your contradiction at your third line.