One of the most amusing instances of indirect proof - proof by contradiction - is the establishment of the existence of two irrational numbers and such that is rational. Indeed, is irrational. Then if is rational the problem is solved with . Otherwise, it is solved with and because then .
Stan Dolan has recently proved that practically all rational numbers in the interval are in the form , where is irrational.
The claim is really elementary if we consider a smaller interval . On this interval function is clearly monotone increasing. To prove that this is also true for requires a little calculus: and is positive for .
Function maps continuously interval onto interval . Let be rational. There is such that . Assume is also rational. More accurately, assume and , both fractions in lowest terms. Then . This implies
If , . We shall show that this is an exceptional case. In other words, an assumption that leads to a contradiction. Since and are mutually prime, every factor of divides . The converse is also true because and are also mutually prime. It follows that
Assume is a common prime factor of and . We'll write
where both and are prime to . Substitution gives
In particular, . Further, so that . But and are assumed to be mutually prime, implying . We thus led to
But this is a contradiction, for, for all positive integer , .
To sum up, the identity , with a rational is possible in two cases only: when is an integer or irrational. Put differently, every rational number in the interval , except for those in the form , with integer , are in the form with irrational.