Is there a rational number #x# such that #sqrt(x)# is irrational, but #sqrt(x)^sqrt(x)# is rational?

The motivation for this question is to consider: #sqrt(2)^sqrt(2)# and #(sqrt(2)^sqrt(2))^sqrt(2)#.

Note that #sqrt(2)# is irrational, and #(sqrt(2)^sqrt(2))^sqrt(2) = sqrt(2)^2 = 2#.

So either #sqrt(2)^sqrt(2)# is rational and we find that this rational number raised to an irrational power #sqrt(2)# is also rational, or #sqrt(2)^sqrt(2)# is irrational and we find that this irrational number raised to an irrational power is rational.

Going back to the question asked, I think there is no such rational number #x#, but I also suspect that it is difficult to prove.

1 Answer
George C.
Apr 30, 2017

No. If #sqrt(x)# is irrational, then #sqrt(x)^sqrt(x)# is transcendental by the Gelfond-Schneider theorem.

Explanation:

The Gelfond-Schneider theorem says that if #a, b# are algebraic numbers with #a != 0#, #a != 1# and #b# irrational, then #a^b# is transcendental (which implies that it is irrational).

A proof of the Gelfond-Schneider theorem can be found at http://people.math.sc.edu/filaseta/gradcourses/Math785/Math785Notes8.pdf

In our example, #sqrt(x)# is irrational, so it will not be equal to #0# or #1# and is algebraic, since it's a root of #t^2-x = 0# with #x# rational.

Hence with #a=b=sqrt(x)# we can deduce that #sqrt(x)^sqrt(x) = a^b# is transcendental, so irrational.

Related questions
See all questions in Square Roots and Irrational Numbers
Impact of this question
6459 views around the world
You can reuse this answer
Creative Commons License