Let $x$ be a negative real number and $0<alpha <1$, I would like to know why ${(x)}^{alpha }$ is not a negative real number.

For example wolfram alpha shows that

$$

{(-5)}^{0.33 }=0.856097+1.47919i

$$

but it’s not $-1.70906$.

My **question**:

When is it true that

$$

{(x)}^{alpha }=-{(x)}^{alpha},

$$ for $x<0$ and $0<alpha <1$ ?

## Answer

$(x)^alpha=-(x)^alpha$ does not make sense since this implies that $x^alpha=0$. You are probably asking if $(x)^alpha=-(-x)^alpha$ is true in general for $x<0$ and $0<alpha<1$.

The answer is NO.

For $x<0$ and $0<alpha<1$, one has to refers to complex analysis in order to make sense of $x^alpha$. If $alpha=frac{n}{m}$ for some integers $m$ and $n$, then

essentially you need to know the $m$-th root of unity. If otherwise $alpha$ is irrational, then

$$

x^alpha:=e^{alphalog x}

$$

and the right hand side is a long story.

[Added:] $x^alpha=-(-x)^alpha$ for $x<0$ and $0<alpha<1$ could be true if one stricts oneself to the *real* setting and further assumes that $alpha=dfrac{1}{m}$ where $m$ is an odd interger. For instance, you do have

$$

(-2)^{1/3}=-(2)^{1/3}

$$

where $y=(-2)^{1/3}$ is defined as a *real* number such that $y^3=-2$.