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

1. terenzreignz Group Title

Am I allowed to assume that pi is transcendental?

2. oldrin.bataku Group Title

Any transcendental number disproves this.

3. terenzreignz Group Title

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 Group Title

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

5. wio Group Title

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

6. wio Group Title

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 Group Title

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

8. wio Group Title

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

9. wio Group Title

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 Group Title

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 Group Title

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

12. wio Group Title

This is obviously an irrational number

13. oldrin.bataku Group Title

I don't see how that is relevant.

14. wio Group Title

Which thing?

15. wio Group Title

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 Group Title

@oldrin.bataku What do you think then?

17. wio Group Title

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

18. wio Group Title

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

19. wio Group Title

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

20. wio Group Title

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

21. wio Group Title

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

22. wio Group Title

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

23. wio Group Title

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

24. wio Group Title

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

25. oldrin.bataku Group Title

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

26. wio Group Title

@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.