TITLE: Proof by Contradiction relating to rational and irrational numbers
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: 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.