$\epsilon$-$\delta$ proof that $f$ is continuous for $x\notin\mathbb Q$ but isn't for $x\in\mathbb Q$ I'm trying to give an $\epsilon$-$\delta$ proof that the following function $f$ is continuous for $x\notin\mathbb Q$ but isn't for $x\in\mathbb Q$. 
Let $f:\mathbb{A\subset R\to R}, \mathbb{A=\{x\in R| x>0\}}$ be given by:
$$
  f(x) = \begin{cases}
1/n,&x=m/n\in\mathbb Q
\\
0,&x\notin\mathbb Q
\end{cases}
$$
where $m/n$ is in the lowest terms.
Can anyone help me with this proof (I'd prefer an answer with an $\epsilon$-$\delta$ proof).
Thank you very much!

HINTS: To show that $f$ is continuous at an irrational $x$, show for any $n\in\Bbb Z^+$ you can choose $\delta>0$ small enough so that the interval $(x-\delta,x+\delta)$ contains no fraction with a denominator $\le n$ in lowest term. Use the fact that between two rationals with denominator $m$ there is a gap of at least $\frac1m$.
To show that $x$ is discontinuous at a rational $\frac{m}n$, let $\epsilon=\frac{m}n$ (or any smaller positive real), and just observe that every non-empty open interval in $\Bbb R$ contains an irrational number.
