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I wish to prove that $(-x)^r\neq y$ for $x>0$ and $x,y,r\in \mathbb{R}$ and $r\notin\mathbb{Q}$.

I tried to prove in this way;$$(-x)^r=(-1)^r\cdot x^r=x^r\cdot e^{\pi i r}=x^r(\cos(\pi r)+i\sin(\pi r))$$ For that to be real, $\sin(\pi r)$ has to be $0$;$$\sin(\pi r)=0\\\pi r=0+\pi k , \ \ \,\,\, k\in\mathbb{Z}\\r=k$$ But something must be wrong with this proof, because it proves that $r$ must be an integer, but really it doesn't have to; $(-1)^{\frac{1}{3}}=-1$.

So what is wrong with this proof, and how do I prove correctly?

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    @spaceisdarkgreen Why not? one of the third order roots of -1 is -1. – 2017-01-02
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    I dont know the context of the proof but observe that, in a complex environment, $(-x)^r:=e^{r\ln (-x)}=e^{r (\ln |x|+i\alpha)}$, then the reality of this number depend on $r\alpha$. – 2017-01-02

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Well, there are 2 different ways to define the power function (in complex world). If you wish to have only one solution for any power, than you can use the Euler formula like you used it, and in this way $(-1)^{\frac{1}{3}}$ won't be $-1$, it will be $(-1)^{\frac{1}{3}}=e^{\frac{i\pi}{3}}=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}=0.5+0.5\sqrt{3}i$.

In the second way, you have to write the Euler formula like that; $e^{i\pi r}=\cos(r(\pi+2\pi k))+i\sin(r(\pi+2\pi k))$ for any integer $k$.

So now you can prove your original question in this way;$$(-x)^r=(-1)^r\cdot x^r=x^r\cdot e^{i\pi r}=x^r(\cos(\pi r+2\pi rk)+i\sin(\pi r+2\pi rk))\\\sin(\pi r+2\pi rk)=0\\\pi r(1+2k)=0+\pi n\\r=\frac{n}{1+2k}\in\mathbb{Q}$$. EDIT: You can see more about the 2 ways here.

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    Thank you for this important answer. I liked it. – 2017-01-02