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Irrational to the irrational=rational

Posted: Thu Aug 28, 2014 1:40 am UTC
by Farabor
I don't know if this is a well known result or not, but someone just showed me a neat, simple proof of the existence of two irrational real numbers x and y such that [math]x^y[/math] is rational. Thought I'd share the question here for those who might not have seen it to try it out.

Spoiler:
As an existence proof, it of course relies on the law of the excluded middle. Anyone who manages to do this without that, I'd love to know!

Re: Irrational to the irrational=rational

Posted: Thu Aug 28, 2014 3:20 am UTC
by Moole
Are we restricted to the reals? Because e and iπ are irrational, but e is rational. That's the nicest example I can think of. Of course, one still has to prove things like that e is irrational (not so hard) and so is π (kind of hard).

Though, a more conventional-but-not-classic answer might be...
Spoiler:
Let x be a positive transcendental number and n be a positive rational number. Then let y = log(n)/log(x). Clearly, xy=n is a rational. Further, if y were equal to a ratio of integers, a/b, then xa=nb, meaning that x would be an algebraic, contrary to definition. Thus, y is irrational and xy is an example of a power of two irrationals that equals a rational. We could make an even stronger statement like:

Suppose x is algebraic and has some integer power equal to a rational number. Choose the smallest positive integer k such that xk=p is rational. Then, let q be the smallest rational above 1 that p is a power of (or largest rational below one - they'll be reciprocal, so it doesn't matter). Then, as long as n is not an integer power of p, the logarithm log(n)/log(x) will not be rational. For instance, if x = 1/9 * cube root of 4, then k = 3 and xk=p=4/729=2^2*3^-6. Then, q=2*3^-3=2/27, by eliminating common factors in the exponents in the prime factorization (though 2^-1*3^3=27/2 would be equally admissible). So, anything which is not a power of 2/27 would have an irrational logarithm base 1/9 cube root of 4.

This provides us with the machinery to find a lot of examples. If we just take the en is irrational for any integer n (though this is not a trivial fact; but it could likely be easier to show that en is never a power of two or some statement like that, which would suffice), it follows that. Thus, if ey is rational, y is irrational. This provides us with lots of examples of irrational^irrational=rational, of the form elog(2)=2.


Spoiler:
Is the non-constructive proof you're thinking of the one involving the square root of two? It's a much more beautiful proof than what I wrote here. But the problem doesn't really need that kind of elegance because, well, there's a lot more irrationals than rationals, so it'd be

Re: Irrational to the irrational=rational

Posted: Thu Aug 28, 2014 3:33 am UTC
by Qaanol
Yeah, that’s a classic proof.

However, there is a lot more depth to the matter than is at first apparent. Notably, what does it even mean to raise a number to an irrational power?

One option is to take limits of rational exponents, which also means you have to show that doing so is well-defined.

You could instead define the exponential function exp() by its power series, which then requires a proof that it coincides with the standard definition for rational exponents.

Another way is to prove a bunch of things about the integral of 1/x, and then takes its inverse function.

Re: Irrational to the irrational=rational

Posted: Thu Aug 28, 2014 5:58 pm UTC
by Tirian
Qaanol wrote:One option is to take limits of rational exponents, which also means you have to show that doing so is well-defined.


If a > 1, then x -> a^x is isotone over the rationals. So it doesn't seem so hard to extend that to the reals by a^b = sup {a^q | q rational and q<=b}.

Re: Irrational to the irrational=rational

Posted: Thu Aug 28, 2014 6:20 pm UTC
by Moole
Tirian wrote:
Qaanol wrote:One option is to take limits of rational exponents, which also means you have to show that doing so is well-defined.


If a > 1, then x -> a^x is isotone over the rationals. So it doesn't seem so hard to extend that to the reals by a^b = sup {a^q | q rational and q<=b}.


One would probably also want to ensure that sup {a^q | q rational and q<=b} = inf {a^q | q rational and q<=b} - otherwise you might end up with problems, but that's easy to show: Let f(x) = a^x, mapping rationals to reals. It holds that f(x+n)=f(x)f(n) - so shifts of f are equivalent to scalings. Let f'(x)= inf {a^q | q rational and q<=b} - sup {a^q | q rational and q<=b}. Clearly, f'(x+n)=f'(x)f(n). Suppose there were any rational y where f'(y)!=0. If so, then by the above, f'(x)=ca^x for a constant c. This is obviously a problem, since f(1)-f(0) has to be at least the sum of the lengths of the discontinuities in (0,1) by the fact that f is isotone, but there are countably many discontinuities with length of at least c in that interval, so f(1)-f(0) would have to be infinite, which it is not. Therefore, f'(y)=0 everywhere and f is continuos.

Re: Irrational to the irrational=rational

Posted: Fri Aug 29, 2014 3:17 pm UTC
by Farabor
I'm glad I posted this, as the discussion above has proved useful and interesting!
Spoiler:
Yes, the one I was thinking of was the square root of 2 one.