If you raise an irrational number to a rational power,
it is possible to get something rational. For instance,
raise Sqrt 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 AB is rational.
Proof. We know that Sqrt is irrational. So, if
A=Sqrt and B=Sqrt satisfy the conclusion of the
theorem, then we are done. If they do not, then
SqrtSqrt is irrational, so let A be
this number. Then, letting B=Sqrt, it is easy to
verify that AB=2 which is rational and
hence would satisfy the conclusion of the theorem.
This proof is non-constructive because it
actually tell us whether SqrtSqrt is
rational or irrational!
The Math Behind the Fact:
Actually, SqrtSqrt can be shown to
be irrational, using something called the
Gelfond-Schneider 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 AB must be irrational
(in fact, transcendental).
But you don't need Gelfond-Schneider to construct
an explicit example, assuming you know transcendental numbers exist
(numbers that are not roots of non-zero polynomials with
Let x be any transcendental and q be any positive rational.
so all we have to show is that log_x(q) is irrational.
If log_x(q)=a/b then q=xa/b,
implying that xa-qb=0,
contradicting the transcendentality of x.
How to Cite this Page:
Su, Francis E., et al. "Rational Irrational Power."
Math Fun Facts.