Irrational Number^(Irrational Power)

[nycmathguy]

 

[MathLover1]

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 non-constructive because it (amazingly) doesn’t actually tell us whether sqrt(2) ^sqrt(2) is rational or irrational!

Actually, sqrt(2) ^sqrt(2) 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 A^B 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 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.

 

[nycmathguy]

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 non-constructive because it (amazingly) doesn’t actually tell us whether sqrt(2) ^sqrt(2) is rational or irrational!

Actually, sqrt(2) ^sqrt(2) 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 A^B 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 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.

This is truly a comprehensive reply. Thanks. I will post some word problems later this evening. Back to Calculus this weekend.

 

[Phrzby Phil]

For those unfamiliar with this most interesting question, I think it important to point out explicitly that the proof relies on the Law of the Excluded Middle (P or notP), which some mathematicians prefer to avoid, partly because it indeed allows non-constructive proofs, which, again, some mathematicians like to avoid. Not right or wrong IMO, just each person's philosophical approach to proofs.

When I show this, I first ask the student if they agree with "P or notP". Most agree (what could be more obvious) - but then feel somewhat unsettled at the result.