Let p and q be distinct primes.
Divide the interval [0,1] into p equal subintervals by red dots, into q equal subintervals by blue dots,
and by p + q equal subintervals by (p+q) green dots.

Show that each interval between two green dots can contain at most one red or blue dot.

In the illustration below, p = 3 and q = 5, with 8 green intervals.

First of all, we note that all the dots inside the interval are distinct; this is easy to show using the fact that p and q are prime.

There can't be two red dots in a green interval, since the distance between two consecutive red dots (1/p) is greater
than the width of the interval (1/(p+q)). Similarly, there can't be two blue dots in a green interval, either.

Finally, there can't be a red dot and a blue dot in a green interval, since given any red dot (say, at a/p), and
any blue dot (say, at b/q), it's straightforward to verify that there's a green dot between them, at (a+b)/(p+q).