I am having some difficulty with the following question regarding a proof for continuity in Real Analysis. I have a function defined by two cases and I am being asked to prove that f is continuous at $x$ = $b$ if and only if $b$ is irrational.
f(x) = \begin{cases} 0, & \text{if $x$ is irrational } \\ \frac{1}{q}, & \text{if $x$ = $\frac{p}{q}$, in reduced form} \end{cases}
I began my proof using the epsilon delta definition for continuity but got stuck trying to show |f(x) - f(b| < $\epsilon$ , when x is rational. I then tried to show the contrapositive but ended up confusing myself further.
Hope someone can help ! :)
Let $b$ be an irrational number, and let $\varepsilon > 0$. Note that there are only finitely many rationals with denominator less than or equal to $1/\varepsilon$ in the interval $(b-1,b+1)$; it follows that such rationals with small denominator do not get "arbitrarily close" to $b$, i.e. there exists $\delta>0$ such that if $p/q$ is rational with $\vert b-p/q\vert < \delta$, then $q>1/\varepsilon$.
Now suppose $x\in \mathbb{R}$, and that $\vert b-x \vert < \delta$. Then $f(b) = 0$. If $x$ is irrational, then $f(x) =0$. As shown above, if $x$ is rational, then $f(x) < 1/(1/\varepsilon) = \varepsilon$. In either case, we have that $\vert f(b)-f(x) \vert < \varepsilon$, and so $f$ is continuous at $b$.