Rational + irrational = always irrational?

Suppose $x$ is irrational and $x+\dfrac pq=\dfrac mn$ then, $x=\dfrac mn-\dfrac pq=\dfrac{mq-np}{nq}$ so, $x$ would then be rational. :)

Look at the contrapositive: If $x$ is rational, then $x+n$ is rational. Clearly this is a true statement.

In fact, for any rational number $ r $ it is true that the irrationality of $ x+r $ implies the irrationality of $ x $. This is due to the fact that the rationals are closed under addition. Assume that $ x+r $ is irrational and (for contradiction) that $ x $ is rational, by the fact that the rationals are closed under addition ($\mathbb {Q}$ is a field) you get that $ x+r $ is rational. Contradiction.