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

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- terenzreignz

Am I allowed to assume that pi is transcendental?

- anonymous

Any transcendental number disproves this.

- terenzreignz

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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## More answers

- anonymous

I'm guessing the point is to prove the existence transcendental numbers.

- anonymous

Or rather, to do the proof without assuming they exist.

- anonymous

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"

- anonymous

But this also means "Every irrational number q, when taken to some integer number n, results in a rational number"

- anonymous

Probably should have noted earlier that if \(y\) is negative, then we can take the reciprocal of \(x\) which is still rational.

- anonymous

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.

- anonymous

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...

- anonymous

Oh wait! I got an idea, how about \[
\sqrt[\sqrt{2}]{2}
\]

- anonymous

This is obviously an irrational number

- anonymous

I don't see how that is relevant.

- anonymous

Which thing?

- anonymous

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

- anonymous

@oldrin.bataku What do you think then?

- anonymous

Without the for sight of transcendental numbers, how would you disprove it?

- anonymous

Hmmm, I think maybe a good way is to do a proof by contradiction.
Suppose \[
\sqrt{2}^\sqrt{2} = x^y
\]

- anonymous

We know the inverse of \(y\) is also rational, let's say \(z = 1/y\)

- anonymous

\[ \Large
x = \sqrt{2}^{ \sqrt{2}z}
\]

- anonymous

\[ \Large
x = 2^{\sqrt{2}z/2}
\]

- anonymous

\[ \Large
x^2 = 2^{\sqrt{2}z}
\]

- anonymous

Now if only I can show that \[ \Large
2^{\sqrt{2}z}
\]is not rational....

- anonymous

I know that a rational number times an irrational number is irrational, but an integer to an irrational power...?

- anonymous

@wio the question states \(x,y\) are *rational*, so \(\sqrt2\) is no good here.

- anonymous

'proof by contradiction'

- anonymous

@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.

- anonymous

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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