On continuous extension of Thomae function.

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Let $f: \mathbb{Q}_0 \to \mathbb{R}: \frac {p}{q} \mapsto \frac{1}{q}, (p,q) = 1, q > 0$

Does $f$ have a continuous extension to $\mathbb{R}$?

My attempt:

I showed that $\lim_{x \to a, x \in \mathbb{Q_0}} f(x) =0$ for any $a \in \mathbb{Q}$.

Hence, $f$ is continuous nowhere.

It follows that $f$ cannot have a continuous extension, since the restriction of such an extension to $\mathbb{Q}_0$ would be continuous.

Is this correct?