Define f(x) by 
if 
is a rational number expressed in lowest terms, and f(x)=0 for irrational x.
(I've sometimes heard this called the ``ruler'' function, since its graph
vaguely resembles the markings on a ruler.) Then f has the surprising property
that it is continuous at all irrationals and discontinuous at all rationals.
It's easy to believe that f is discontinuous at all rationals, since for a
rational number , there are irrational numbers x arbitrarily
close to , but f(x)=0 is not getting close to .
It's a bit harder to see that f is continuous at any irrational x. Roughly
speaking, there's no way that rational numbers can approach an irrational number
x without their denominators going to infinity, so that f approaches 0. More
formally, take any 
. There is an integer q with 
. Look at all the rational numbers of the form 
.
Since x is irrational, it is not one of these numbers. Because of the way the
numbers , p=0, 
, 
, 
appear on the
number line, there is a closest number in this set to x (a careful proof of
this fact uses properties of the integers). Take 
to be smaller than
the distance from x to the closest number of the form . Then
no rational number within 
of x may be written as a fraction with
denominator less than or equal to q, so all numbers with of x must
have their function values within 
of f(x), so f is continuous at
any irrational x.