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.
Best Answer
The comments essentially yield the full solution but I'll provide a quick proof for you.
Assume that $x$ is irrational and that $r$ is a rational number such that $x+r=s$ is rational. Then we have a contradiction when we observe $x= s-r$ since $x$ is assumed irrational and $s-r$ is rational (difference between rational numbers is rational).