Irrational to the irrational=rational

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Farabor
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Irrational to the irrational=rational

Postby Farabor » Thu Aug 28, 2014 1:40 am UTC

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!

Moole
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Re: Irrational to the irrational=rational

Postby Moole » Thu Aug 28, 2014 3:20 am UTC

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
Mathematical hangover (n.): The feeling one gets in the morning when they realize that that short, elementary proof of the Riemann hypothesis that they came up with at midnight the night before is, in fact, nonsense.

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Qaanol
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Re: Irrational to the irrational=rational

Postby Qaanol » Thu Aug 28, 2014 3:33 am UTC

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.
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Tirian
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Re: Irrational to the irrational=rational

Postby Tirian » Thu Aug 28, 2014 5:58 pm UTC

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}.

Moole
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Re: Irrational to the irrational=rational

Postby Moole » Thu Aug 28, 2014 6:20 pm UTC

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.
Mathematical hangover (n.): The feeling one gets in the morning when they realize that that short, elementary proof of the Riemann hypothesis that they came up with at midnight the night before is, in fact, nonsense.

Farabor
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Re: Irrational to the irrational=rational

Postby Farabor » Fri Aug 29, 2014 3:17 pm UTC

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.


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