## sauravshakya 2 years ago Prove OR Disprove Every irrational number can be expressed as (x)^y where x and y are rational numbers.

1. terenzreignz

Am I allowed to assume that pi is transcendental?

2. oldrin.bataku

Any transcendental number disproves this.

3. 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

4. wio

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

5. wio

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

6. wio

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"

7. wio

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

8. wio

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

9. wio

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.

10. wio

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...

11. wio

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

12. wio

This is obviously an irrational number

13. oldrin.bataku

I don't see how that is relevant.

14. wio

Which thing?

15. wio

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

16. wio

@oldrin.bataku What do you think then?

17. wio

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

18. wio

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

19. wio

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

20. wio

$\Large x = \sqrt{2}^{ \sqrt{2}z}$

21. wio

$\Large x = 2^{\sqrt{2}z/2}$

22. wio

$\Large x^2 = 2^{\sqrt{2}z}$

23. wio

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

24. wio

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

25. oldrin.bataku

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

26. wio

27. oldrin.bataku

@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.

28. wio

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.