TITLE: Prove that $r+x$ is irrational
QUESTION [6 upvotes]: If $r$ is rational ($r$$\ne$$0$) and $x$ is irrational, prove that $r+x$ is irrational.
Assume that $r+x$ is rational. Then $r+x=(\frac{p}{q})$, where $p$ , $q$ are $\in$ $\mathbb{Z}$, and $p$ and $q$ are in lowest terms. Then we have $x=(\frac{p}{q})-r$= $(\frac{p-rq}{q})$. Since $p-rq$ is $\in$ $\mathbb{Z}$, $x$ is rational. This is a contradiction since $x$ was assumed to be irrational. Therefore, $r+x$ is irrational. QED
I wanted to try proving by contradiction. Just wanted to know if everything looked okay.

REPLY [4 votes]: Lemma: the difference of two rationals is rational.
Indeed, $$\frac ab-\frac cd=\frac{ad-bc}{bd}\in\mathbb Q.$$
Main theorem:
If $x+r$ is rational, so is $(x+r)-r=x+(r-r)=x$.

Note that this proof requires associativity of the addition of reals.