**Spoiler:**

## Irrational to the irrational=rational

**Moderators:** gmalivuk, Moderators General, Prelates

### Irrational to the irrational=rational

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.

### Re: Irrational to the irrational=rational

Are we restricted to the reals? Because e and iπ are irrational, but e

Though, a more conventional-but-not-classic answer might be...

^{iπ}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:**

**Spoiler:**

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.

### Re: Irrational to the irrational=rational

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.

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.

wee free kings

### Re: Irrational to the irrational=rational

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

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.

### Re: Irrational to the irrational=rational

I'm glad I posted this, as the discussion above has proved useful and interesting!

**Spoiler:**

### Who is online

Users browsing this forum: No registered users and 11 guests