[bizaro86]

Originally Posted by psyang

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.

Spoiler!

Thanks! I agree that to consider the problem complete a proof that those are both irrational is necessary, but I didn't feel like it, and probably wouldn't be able to do that any more

A very slightly modified version of my solution works for your amended problem as well.

A=sqrt(2)
B= log-sqrt(2)(3) ie a logarithm of base sqrt(2) of 3

Which by the same rules of exponents/logarithms equals 3.

It is fairly trivial to prove that sqrt(2) is irrational - I haven't done that for 15 years either but recall the rough outline. If you assume it is rational in the form a/b and that a/b is fully reduced then a and b can't both be even, and by squaring both sides and moving the denominator you get 2b^2=a^2. That means a is even so b can't be even.

But if a^2 is even then a is even, so its a multiple of 2, and if you sub 2x for a (as its an even number) into the previous you get 2b^2=4x^2 which means that b must be even and is a contradiction.

I'm quite confident that logarithm is irrational as well but leave that as an exercise to the reader