sqrt(2)^sqrt(2) : rational or irrational?

By Jeff Davies on December 12, 1997 :

$begin{displaymath}sqrt{2}^{sqrt{2}}end{displaymath}$

rational or irrational?

By Anonymous :

[The following brief reply is followed by some clarifications of the concepts in darker blue. - The Editor]

It's a highly non-trivial question, but the answer is simple: it's irrational, even transcendental . This is a consequence of the Gelfond-Schneider Theorem: a^b is transcendental provided a,b are algebraic, b is irrational, and a is neither 0 nor 1.
(The problem of proving this was one of the ones Hilbert famously proposed in 1900 , and it was solved independently by G. and S. in, I think, 1936.)

Did this come up in the context of the excellent proof that one can raise an irrational number to an irrational power and get a rational number? In case the answer is "no" and you haven't seen this: let a=sqrt(2), b=a^a , c=b^a . Note that c=2 since it's (a^a )^a = a^a*a = a² = 2. So either a^a =b or b^a =c is an example of the phenomenon.

g

By Anonymous : That was actually a reply I made to a personal e-mail message asking `can you help?'; I mention this merely to excuse the perhaps excessive terseness of what I said. So, a few words of clarification.

transcendental A number is transcendental if it isn't a root of any algebraic equation with integer coefficients. So, for instance, the square root of 2 is not transcendental (it's a root of $X^{2} - 2 = 0$ ).
In a sense that can be made perfectly precise, there are enormously more transcendental numbers than algebraic ones (`algebraic' = `not transcendental').
Proving that a given number is transcendental is usually rather hard. Some well-known numbers that are transcendental:
$e$ and $π$ . The proofs that they are transcendental are each about a page long, if you write them very tersely.
The Gelfond-Schneider Theorem, which as I mentioned implies that $\sqrt{2}^{\sqrt{2}}$ is transcendental, is much harder.
One other thing: a consequence of the fact that $π$ is transcendental is that the old problem of `squaring the circle' is impossible: that is, it's not possible to construct, using the usual methods with ruler and compass alone, a square with area equal to the area of a given circle. This problem goes back to the ancient Greeks.
Hilbert's problems
At the International Congress of Mathematicians in Paris in 1900, David Hilbert (possibly the greatest mathematician then living) proposed a list of 23 problems that he suggested mathematicians would do well to work on. They have all proved extremely fertile; some have been solved and some have not.
The seventh of these problems was: determine whether, if $a$ and $b$ are algebraic, $\log_{b} (a)$ (the logarithm of $a$ to base $b$ ) is always either rational or transcendental.
This is equivalent to: is $a^{b}$ always transcendental when $a$ and $b$ are algebraic, $b$ is irrational, and $a$ is neither 0 nor 1?
Gelfond and Schneider each (independently) proved that the answer is `yes', in 1934 (not 1936, as I wrongly said). Some very far-reaching generalisations of this theorem have since been proved, but there are still lots of unresolved questions. As a random example: I don't think anyone has any idea how to prove that $e + π$ is transcendental (though it presumably is).
Irrational number to irrational power
What I wrote was perhaps a bit too abbreviated, so I'll go over it a little more slowly.
FACT: there are irrational numbers $x$ , $y$ such that $x^{y}$ is rational.
PROOF: Let $a = \sqrt{2}$ ; $b = a^{a}$ ; $c = b^{a}$ .
If $b$ is rational then we are already done: $x = a$ , $y = a$ .
If $b$ is irrational then we can take $x = b$ , $y = a$ .
For (1) these $x$ , $y$ are irrational,
and (2) $x^{y} = b^{a} = (a^{a}) a = a^{a \times a} = a^{2} = 2$ ,
which is rational. QED.
As it happens, $b$ is irrational; but, as I said, that's not at all easy to prove. In fact I don't know of any numbers $a$ , $b$ , $c$ for which it's easy to prove
- $a$ , $b$ are irrational
- $c$ is rational
- $a^{b} = c$ .

Hmm. That's more than the `few words' I predicted at the start. Never mind.