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*.

Mon May 5 12:53:33 CDT 1997