[NameError]

Reminds me of a cool proof I saw recently that there are two numbers a and b such that a and b are both irrational, but a^b is rational:

Take sqrt(2)^sqrt(2), which is either rational or not. If it's rational, we're done. If not, consider sqrt(2) ^ (sqrt(2) ^ sqrt(2)). Since (a^b)^c = a^bc, we get sqrt(2) ^ (sqrt(2))^2 = sqrt(2)^2 = 2, which is rational!

It feels like a bit of a sleight of hand, since we don't actually have to know whether sqrt(2)^sqrt(2) is rational for the proof to work.

[tzs]

I wonder what the easiest to prove example of a, b irrational with a^b rational is?

The easiest I can think of offhand would be e^log(2). To prove that we need to prove that e is irrational and the log(2) is irrational.

To prove log(2) is irrational one approach is to prove that e^r is irrational for rational r != 0, which would imply that if log(2) is rational then e^log(2) would be irrational. To prove that e^r is irrational for irrational r it suffices to prove that e^n is irrational for all positive integers n.

We'd also get the e is irrational out of that by taking n = 1, and that would complete our proof that e^log(2) is an example of irrational a, b with a^b rational.

So, all we need now is a proof that e^n is irrational for integers n > 0.

The techniques used in Niven's simple proof that pi is irrational, which was discussed here [1], can be generalized to e^n. You can find that proof in Niven's book "Irrational Numbers" or in Aigner & Ziegler's "Proofs from THE BOOK".

That can also be proved by proving that e is transcendental. Normally proofs that specific numbers are transcendental (other than numbers specifically constructed to be transcendental) are fairly advanced but for e you can do it with first year undergraduate calculus. There's a chapter in Spivak's "Calculus" that does it, and there's a proof in the aforementioned "Irrational Numbers".

[1] https://news.ycombinator.com/item?id=41178560

[cruegge]

I think a = sqrt(2), b = log(9)/log(2) with a^b = 3 is easier. To show that b is irrational, assume b = n/m for integer n, m. Then 9^m = 2^n, which can't be the case since the lhs is odd and the rhs is even.

[tyilo]

Also see https://math.andrej.com/2009/12/28/constructive-gem-irration... for a similar proof using 2^(log_2 3)

[xanderlewis]

> To prove that e^r is irrational for irrational r

You mean for rational r, don’t you?

[tzs]

Yup!

[brianush1]

pretty sure you have a typo, should be "If not, consider (sqrt(2) ^ sqrt(2)) ^ sqrt(2)."

[seanhunter]

Well the proof I would use is let a = e and b = i(pi).

e^(i theta) = cos theta + i sin theta (Euler's identity) thus e^(i pi) = cos pi + i sin pi = -1 + i(0) = -1

We know that e and i pi are irrational (in fact i pi isn't even a real) and -1 is rational.

Therefore there exist two numbers a and b such that both a and b are irrational but a^b is rational.

In fact log of just about anything is irrational so e^(log x) works as well for just about all rational x, but Euler's identity is cool so I wanted to use that.