Fermat's Last theorem

HallsofIvy

Here is a proof that comes awfully close to that:

Prove the there exist irrational a, b such that a[sup]b[/sub] is rational.

Look at [tex]\sqrt{2}^\sqrt{2}[/itex]. It is not known (last time I checked) if that number is rational or irrational. However:

If it is rational, then we are done.
If it is irrational, then [tex]\(\sqrt{2}^\sqrt{2}\)^\sqrt{2}\)= \sqrt{2}^2= 2[/tex] is rational.

We have proved that there exist irrationals a, b such that a^b is rational but are unable to say what a and b are!

matticus

yeah that proof was in my discrete math book, it is cool. the example i was going to use was that given any number we know there is a bigger prime, despite the fact that at some point we won't be able to construct it. neither of these are the same thing, but i think one of the things that attracts me to math is that you can know things exist without having to see them exist. where in science statistical evidence is very important, in math it's almost irrelevant.