# A different version of Thomae's function and differentiability

So, you have a function

$f(x) = x^2$ if $x \in \mathbb{Q}$

and $f(x) = 0$ if $x \in \mathbb{R} \setminus \mathbb{Q}$

I'm almost certain it is differentiable at $0$ and nowhere else and I was thinking that the proof for this involved that $\mathbb{Q}$ is dense in $\mathbb{R}$ and so are the irrationals, but I'm not sure how to approach it.

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Show that $f$ discontinuous at any point $x\ne0$. – David Mitra Dec 16 '12 at 2:12