[psyang]

Originally Posted by bizaro86

Spoiler!

I was working on this as a laws of exponents problem and was messing around with 'e'. I suspect there is a general proof here but one example proves that it does exist, and e to the power of ln(2) equals 2, and both e and ln(2) are irrational.

I vaguely remember proving than ln(2) is irrational, but that was probably 15 years ago, so I'm going to go with, "it can be shown that ln(2) and e are irrational" and 2 is obviously rational. Neither have terminating or repeating decimals.

That wasn't what I was looking for but I think it's a great solution. Simple and immediately obvious. As you say, you do have to prove that the one term is irrational - and if we are being pedantic, you should have to prove that both terms are irrational. One isn't too hard, but the other is a bit more difficult.

Funny that I first heard this problem in my first year algebra course over 30 years ago, and I've seen it pop up every now and then, but never saw your solution.

Let me amend the problem a little. Prove that there exists two irrational, non-transcendental numbers (so no pi or e, for example) a and b such that a^b is rational.