Proof:

Let $r$ be a rational number and $x$ be an irrational number. We need to prove that $r + x$ is irrational using proof by contradiction.

Assume that $r + x$ is rational. This is the negation of what we need to prove.

Since $r + x$ is rational, we can write

$$r + x = \frac{p}{q}$$

where $p$ and $q$ are integers and $q \neq 0$.

Since $r$ is rational, we can write

$$r = \frac{m}{n}$$

where $m$ and $n$ are integers and $n \neq 0$.

Now,

$$x = (r + x) - r = \frac{p}{q} - \frac{m}{n} = \frac{np - mq}{nq}$$

Since $np - mq$ and $nq$ are integers and $nq \neq 0$, $x$ is a rational number.

This is a contradiction, because we have shown $x$ to be rational, but by our hypothesis, $x$ is irrational.

The contradiction has arisen because of the faulty assumption that $r + x$ is rational. Therefore, $r + x$ is irrational.