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Prove OR Disprove Every irrational number can be expressed as (x)^y where x and y are rational numbers.

Mathematics
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Am I allowed to assume that pi is transcendental?
Any transcendental number disproves this.
Suppose pi (which is irrational) can be expressed as (x)^y where x and y are rational. Then (x)^y can be expressed as \[\huge b^{\frac{p}{q}}\] Where p and q are integers and b is rational Then pi is a solution of \[\huge x - b^{\frac{p}{q}}=0\] \[\huge x = b^{\frac{p}{q}}\] \[\huge x^{q} = b^{p}\] \[\huge x^{q} - b^{p} = 0\]effectively contradicting the transcendental-ness of pi XD

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Other answers:

I'm guessing the point is to prove the existence transcendental numbers.
Or rather, to do the proof without assuming they exist.
Let's say \(x = m/n\) and \(y=a/b\) \[ x^{y} = \left( \frac{m}{n}\right)^{a/b} = \left( \frac{m^a}{n^a}\right)^{1/b} = \sqrt[b]{x^a} \]We know \(x^a\) is still rational so the question then just becomes "every irrational number can be expressed as an integer root of a rational number"
But this also means "Every irrational number q, when taken to some integer number n, results in a rational number"
Probably should have noted earlier that if \(y\) is negative, then we can take the reciprocal of \(x\) which is still rational.
Anyway, I'm not sure where to continue from here, but I think: "Every irrational number q, when taken to some integer number n, results in a rational number" Is a bit easier of a thing to disprove. That's what my gut says.
We could note that: \[ \Large \sum_{n=1}^{\infty} \frac{1}{n^{2}} \]Converges, and thus is an irrational number...and try to show that there is no way putting it to any integer power would give you a rational number?? I think...
Oh wait! I got an idea, how about \[ \sqrt[\sqrt{2}]{2} \]
This is obviously an irrational number
I don't see how that is relevant.
Which thing?
Anyway, if we take it to the \(n\)th power: \[ \sqrt[\sqrt{2}]{2}^n= \sqrt{2}^{n/\sqrt{2}} \]No matter what we set \(n\) to (it must be an integer), we are going to get and irational number to an irrational power. hmmmm
@oldrin.bataku What do you think then?
Without the for sight of transcendental numbers, how would you disprove it?
Hmmm, I think maybe a good way is to do a proof by contradiction. Suppose \[ \sqrt{2}^\sqrt{2} = x^y \]
We know the inverse of \(y\) is also rational, let's say \(z = 1/y\)
\[ \Large x = \sqrt{2}^{ \sqrt{2}z} \]
\[ \Large x = 2^{\sqrt{2}z/2} \]
\[ \Large x^2 = 2^{\sqrt{2}z} \]
Now if only I can show that \[ \Large 2^{\sqrt{2}z} \]is not rational....
I know that a rational number times an irrational number is irrational, but an integer to an irrational power...?
@wio the question states \(x,y\) are *rational*, so \(\sqrt2\) is no good here.
'proof by contradiction'
@wio ... that doesn't disprove the claim however, at all. The only relevance \(\sqrt2\) has is that it's an algebraic irrational number that works for our claim, i.e. \(2^\frac12\) where \(2,\frac12\) are rational.
We know that \(x^2\) is rational. If \(2^{\sqrt{2}z}\) isn't rational then it's a contradiction, that's where I was going.

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