Discontinuous at rationals and differentiable at irrationals? We know that there exist real functions which are continuous at each irrational and dis-
continuous at each rational number. 

But does there exist a function $f: \mathbb{R} \to \mathbb{R}$ that is differentiable at every irrational and discontinuous at every rational?

 A: I think that there can be no such function.
Suppose that $f$ is discontinuous at some point $q$. There must be constants $A,B$ so that $f(s) < A < B < f(t)$ holds for points $s,t$ arbitrarily near to $q$. 
Suppose that $x_0<q$ satisfies $q-x_0 < B-A$. Choose sequences $s_n, t_n > x_0$ converging to $q$ and satisfying $f(s_n) < A < B < f(t_n)$. Then we have 
$$\frac{f(t_n) - f(x_0)}{t_n - x_0} - \frac{f(s_n) - f(x_0)}{s_n - x_0} > \frac{B - f(x_0)}{t_n - x_0} - \frac{A - f(x_0)}{s_n - x_0} \to \frac{B-A}{q - x_0} > 1$$
and may conclude that there exist $s,t$ as close to $q$ as we like satisfying
$$\frac{f(t) - f(x_0)}{t - x_0} - \frac{f(s) - f(x_0)}{s - x_0} > 1$$
Arguing similarly when $x_0 > q$ and trivially (since the Newton quotients will be unbounded as a result of the discontinuity) when $x_0 = q$ we get the same conclusion with only $|x_0 - q| < B-A$.
For $n$ a positive integer let $X_n$ denote all of the points $x_0$ in $\mathbf{R}$ for which there exist $s,t$ at distance less than $1/n$ from $x_0$ satisfying the preceding inequality. Our argument implies that, for every $n$, $X_n$ is a neighbourhood of every point $q$ at which $f$ is discontinuous (consider the points $x_0$ whose distance from $q$ is less than $1/n$ and $B-A = B_q - A_q$). If it happens that the points of discontinuity are dense (as in the case of $\mathbf{Q}$) then this implies that every $X_n$ contains an open dense set and thus $\bigcap_n X_n$ is second category in $\mathbf{R}$.
The point is of course that no point at which $f$ is differentiable can be in all the $X_n$ so if $f$ is discontinuous on the rationals then $f$ is differentiable on at most a set of 1st category in $\mathbf{R}$ (which the irrationals are not).
A: It came up in an answer that has been deleted that a solution to this problem can be found as "solution 2" in the following file: http://www.isibang.ac.in/~statmath/problems/soljan09.pdf.  Another reference is "A theorem concerning functions discontinuous on a dense set" by Fort.
