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This is probably a duplicate but I can't find, if you do let me know and I will delete.

Why is $x^{p/q}$ ill-defined for $x<0$.

I can see that it is, $(-1)^{1/3} \neq (-1)^{2/6}$, but why?

I define $x^{p/q}=\sqrt[q]{x^p}=(\sqrt[q]{x})^p$.

Also how does this affect calculus, in examples what about,

$$\int_{-1}^{0} \sqrt[3]{x} dx ~~~~?= ~~~~\int_{-1}^{0} x^{1/3} dx$$

$\frac{d}{dx}(\sqrt[3]{x})$ at $x=-1$.

OmG
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Ahmed S. Attaalla
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    If $x < 0$ and $r$ is rational, then $x^r$ is defined only if it is possible to write $r$ as $p/q$ with $q$ odd. In that case, $x^r$ is defined as $(\sqrt[q]{x})^p$, and this is independent of the choice of representation $r = p/q$, so long as $q$ is odd. (For example, $(\sqrt[25]{x})^{15}$ and $(\sqrt[15]{x})^9$ give the same result.) In this case, it is meaningless to try to write $(\sqrt[q]{x})^p$ when $q$ is even. For example, $(\sqrt[6]{x})^2$ is undefined. It is true that $\sqrt[6]{x^2}$ is defined, but it is not equal to $x^{2/6} = x^{1/3} = \sqrt[3]{x} = (\sqrt[9]{x})^3 = \dots$. – user49640 May 31 '17 at 05:43
  • A similar question, though not a duplicate, can be found here: https://math.stackexchange.com/questions/1628759/what-are-the-laws-of-rational-exponents The answers there might shed some light on your problem. – user49640 May 31 '17 at 05:59
  • You might also want to see https://math.stackexchange.com/questions/2303412/where-is-the-fallacy-i-1/2303454#2303454 for some similar answers to this – SEWillB May 31 '17 at 06:29

2 Answers2

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$$y=x^{p/q}$$ is understood as a solution of

$$y^q=x^p,$$ which makes sense when the sign are compatible. For negative $x$, this excludes odd $p$ with even $q$.

Ambiguity arises when the fraction $p/q$ can be simplified by $2$, like

$$\frac{2p}{2q}=\frac pq,$$ and though the first fraction is even/even (giving a positive power), the second can be odd/even (undefined) or even/odd (negative).

A better definition could be to enforce simplification of the fraction,

$$y^{q/\gcd(p,q)}=x^{p/gcd(p,q)}$$ which avoids the ambiguity.

All cases can be summarized by

  • $(-1)^{1/1}\to -1$

  • $(-1)^{1/2}\to \text{undefined}$

  • $(-1)^{2/1}\to +1$

  • $(-1)^{2/2}\to(-1)^{1/1}=-1$

Without this convention, the "natural" rule

$$x^{p/q}=\sqrt[q]{x^p}={\sqrt[q]x}^p$$ doesn't hold.

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There are some well-known ambiguities in defining say $(-1)^{1/2}$ because there are two candidates, $i$ and $-i$, just as there are two candidates for $2^{1/2}$ and by convention we choose the positive one. However, we can adopt a convention whereby the candidate with positive imaginary part is chosen, and so define $(-1)^{1/2}=i$. From there we can go to $(-1)^{n/2}$ for integers $n$.

It's harder to extend the convention further, but for even $m>2$, we can say that $(-1)^{1/m}=\exp(\pi i/m)$. If $m$ is odd, we can adopt the same convention, but it's kind of counterintuitive, since $(-1)^{1/m}$ will not be $-1$ as we expect, for example $(-1)^{1/3}$ will be $\exp(\pi i/3)$ and not $-1$. Alternatively, we can say that $(-1)^{1/m}=-1$ for all odd $m$. I think it is the difficulty making these conventions coexist that is the source of the difficulties you see.

When you write $(-1)^{1/3} \neq (-1)^{2/6}$, I assume you mean to use a definition like $(-1)^{2/6}=((-1)^{1/6})^2$. With my $\exp$ convention, I think you do get consistency, i.e., $\exp(\pi i/6)^2$ $=\exp(\pi i/3)$ $=(-1)^{1/3}$.

ForgotALot
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