if a and b are irrational, does it follow that a^b is irrational?

Given a and b are irrational, a^b could be rational. The interesting thing about your second expression to consider is, given (sqrt 2 ^ sqrt 2) ^ sqrt 2,

we know it is equal to (sqrt 2) ^2 by multiplying exponents. This gives us a value of 2.

What we still don't know in the example is whether or not (squrt 2 ^ sqrt 2) is rational or irrational.

Hiker, you've just about got it. We know that expression (2) is rational, right? We don't really know if sqrt(2)^sqrt(2) is rational, as you mentioned. If it's irrational, then we have (2) is irrational^irrational=rational, which contradicts the hypothesis. But if it's rational, then expression (1) is a contradiction. So the answer to the original question would be no.

EDIT: steiner1745 answers the question of whether (1) is rational...nicely done. I remembered that it was indeed irrational, but had never seen a reason behind it. For the purposes of showing that a^b need not be irrational even when a and b are irrational though, we don't need to know this bit.

Oh, and steiner1745, when I was first presented with this proposition a few months ago, e^(i*pi) was my first response too. My instructor, a graph theorist, didn't like my use of complex numbers :)

Consider

e^(iπ) = -1.

e and iπ are certainly irrational, so the answer to

your question is no.

BTW. A poster asked whether √2^√2 is rational.

This number is not only irrational, it is transcendental.

This follows from the theorem of Gelfond-Schneider,

which states that a^b is transcendental if

1). a is algebraic, a .ne. 0 or 1

and

2. b is algebraic and irrational.

Since √2 is a root of x²-2 = 0 and √2 is irrational,

the result follows.

No. The first example proves that 2#s that are irrational can be multiplied to be a rational #. Therefore, it does not logically follow.

sqrt(2)^sqrt(2) = 1.6325....

(sqrt(2)^sqrt(2))^sqrt(2) = 2 - that's what my calculator says, I don't know how to prove it.