Generating conditions on $b$ $\in$ $(0,1)$ s.t. $a^{1/b}$ $>$ $0$ $\forall$ $a$ $\in$ $\mathbb{R}$ \ {$0$}? Consider the expression $a^{1/b}$ .
I want to ensure that this expression results in a positive number by choosing the value $b$ myself, i.e. $a^{1/b}$ > $0$. I assume that $a$ $\in$ $\mathbb{R}$ \ {$0$} and $b$ $\in$ $(0,1)$, i.e.
$a$ is a non-zero real and $b$ lies on the open interval between $0$ and $1$. I want to generate conditions of $b$ under which the expression $a^{1/b}$ > $0$.
Obviously if $a$ $>$ $0$ then $a^{1/b}$ > $0$ $\forall$ $b\in(0,1)$. But $a$ doesn't have to be strictly positive. If $a$ is strictly negative then $a^{1/b}$ > $0$ if $1/b$ $\in $ even $\mathbb{N}$. To my knowledge the only values of $b$ that will satisfy this are $b$ = $1/2^n $ $\forall$ $n\in\mathbb{N}$ $>=$$1$.
So my question is, can I generate conditions on $b$ such that the property $a^{1/b}$ > $0$ will hold always, or will the condition that $b$ = $1/2^n $ $\forall$ $n\in\mathbb{N}$ $>=$$1$ have to suffice to show $a^{1/b}$> $0$? Certainly $1/b$ doesn't have to be an even natural for $a^{1/b}$ > $0$ to hold, right? After all if $b=3/4$ then $1/b=4/3$ and $a^{1/b}$ $>$ $0$ $\forall$ $a$ $\in$ $\mathbb{R}$ \ {$0$}  By conditions I'm referring to properties of $b$ like which sets it belongs to. Since $b$ $\in$ $(0,1)$ , $b$ can't be an integer which was my first thought, so then I started researching if properties of evenness and oddness can be extended to the rational numbers and if there might be some way that I could exploit such 'even' or 'odd' rational numbers to show that the reciprocal will result in an 'even-like' number that isn't a natural but none-the-less evaluates the entire expression to a positive.
Is the answer that $b$ must satisfy $b$ = $c/d$ $\in$ $\mathbb{Q}$ where $d$ must be even and $c$ must be odd? Is this the least restrictive condition necessary to show $a^{1/b}$ > $0$ or just sufficient?
 A: The answer to this question makes use of complex numbers. Any negative real can be written as $a=re^{i\pi}$ for some $r>0$.
Then $a^{1/b} = r^{1/b} e^{i\pi/b}$. If $a^{1/b}>0$ then $r^{1/b} e^{i\pi/b}>0$ which is true if and only if $\frac{\pi}{b} = 2 \pi k$ for some integer $k$. This means that $b = \frac{1}{m}$ for any positive even integer $m$, since you also require $b\in(0,1)$.
So any positive, even denominator works. Not only a power of $2$. But for example $${(-1)}^{4/3} = e^{\frac{4}{3}i\pi} = \cos(\frac{4\pi}{3}) + i \sin(\frac{4\pi}{3}) = -\frac12 - i\frac{\sqrt{3}}{2},$$ which is not a real number.

Edit: Replying to a comment
There is an issue here that stems from the fact that complex roots are multivalued. Any negative number can be written as $a=re^{(2l+1)\pi i}$ and thus, if $b=\frac{2l+1}{2k}$ for some integers $k,l$, then $a^{1/b} = r^{1/b}e^{2k\pi i} = r^{1/b}>0$. However, this is one of the many ($2l + 1$ in this case) values of $a^{1/b}$.
So if you want to answer the question "For what values of $b$ it is true that for all $a<0$, at least one of the values of $a^{1/b}$ is positive?" then the answer is any number $b\in (0,1)$ such that $b=x/y$ with $x$ odd and $y$ even. Indeed, one of the 3 values of $(-1)^{4/3}$ is $1$.
If $x=1$, as in my original answer above, then we don't have a root anymore, and the only value of $a^{1/b}$ is positive. I believe this answers your original question more accurately, which is why I answered as I did.
