If you raise an irrational number to a rational power,
it is possible to get something rational. For instance,
raise Sqrt[2] to the power 2 and you'll get 2.
But what happens if you raise an irrational number
to an irrational power? Can this ever be rational?
The answer is yes, and we'll prove it without
having to find specific numbers that do the trick!
Theorem. There exist irrational numbers A and B
so that A^{B} is rational.
Proof. We know that Sqrt[2] is irrational. So, if
A=Sqrt[2] and B=Sqrt[2] satisfy the conclusion of the
theorem, then we are done. If they do not, then
Sqrt[2]^{Sqrt[2]} is irrational, so let A be
this number. Then, letting B=Sqrt[2], it is easy to
verify that A^{B}=2 which is rational and
hence would satisfy the conclusion of the theorem.
QED.
This proof is nonconstructive because it
(amazingly) doesn't
actually tell us whether Sqrt[2]^{Sqrt[2]} is
rational or irrational!
The Math Behind the Fact:
Actually, Sqrt[2]^{Sqrt[2]} can be shown to
be irrational, using something called the
GelfondSchneider Theorem (1934), which says that if A and B
are roots of polynomials, and A is not 0 or 1 and B is
irrational, then A^{B} must be irrational
(in fact, transcendental).
But you don't need GelfondSchneider to construct
an explicit example, assuming you know transcendental numbers exist
(numbers that are not roots of nonzero polynomials with
integer coefficients).
Let x be any transcendental and q be any positive rational.
Then x^{log_x(q)}=q
so all we have to show is that log_x(q) is irrational.
If log_x(q)=a/b then q=x^{a/b},
implying that x^{a}q^{b}=0,
contradicting the transcendentality of x.
How to Cite this Page:
Su, Francis E., et al. "Rational Irrational Power."
Math Fun Facts.
<http://www.math.hmc.edu/funfacts>.

