Do their exist irrational numbers α and β with α^β rational?

Question

A proof for your answer, constructive or otherwise, must be offered to merit the 10 points.

Answer 1

Yes, such numbers do exist and here is a proof:

sqrt(2) is well known to be irrational. Consider sqrt(2)^sqrt(2). If sqrt(2)^sqrt(2) is rational we let alpha=beta=sqrt(2) and we are done. Otherwise, sqrt(2)^sqrt(2) is irrational. In this case we see that

(sqrt(2)^sqrt(2))^sqrt(2) = sqrt(2)^(sqrt(2)sqrt(2)) = sqrt(2)^2=2

Thus we are done if we let alpha=sqrt(2)^sqrt(2) and beta=sqrt(2).

Since sqrt(2)^sqrt(2) is either rational or irrational and both of these cases imply the existence of the desired alpha and beta, we conclude that there do exist irrational numbers alpha and beta with alpha^beta rational.

Answer 2

It has been proven (by Galfondi) that e^pi is irrational. But pi^e and pi ^sqrt(2) have never been proven to be irrational.

No one as of yet has been able to prove or disprove that their exist irrational numbers α and β with α^β rational.

I do believe that you will not get a proof of your conjecture.

Answer 3

131 + n = 234

Answer 4

Assume j and k are irrational.

Assume j^k = (a/b) where a and b are integers
j = (a/b)^(1/k)

ln j = ln(a/b^(1/k))
lnj = ln(a/b) / k

1/lnj = k / ln(a/b)
(1/lnj)(ln(a/b)) = k
-ln(lnj)(ln(a/b)) = lnk
-ln(lnj^(a/b)) = lnk

That's about as far as I got... There's gotta be a way to get an expression for j being some ratio of 1 to b or 1 to a or a to b or something...
Interesting question.

ln (j^k) = ln(a/b)

ln(j^k) = lna - lnb

k = (lna - lnb) / lnj