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Is there difference between $\sqrt [q]{x^p}$ and $x^{p/q}$ ?

There seems to be no difference when just dealing with positive reals, but what about when $x$ is negative (or even complex)?

p.s. Maybe there's in fact no difference but I'm too suspicious, I just want to be sure...

Git Gud
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JSCB
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3 Answers3

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Some people prefer to consider $\sqrt[n]{\cdot}$ as the inverse of $(\cdot)^n$. In particular, they write freely $\sqrt[3]{-1}$. On the other hand they prefer not to write $(-1)^{\frac{1}{3}}$ because this would be different than $(-1)^{\frac{2}{6}}$. I have seen this distinction mainly in high-school teachers, because at that level it might be hard to explain that power rules may be false for negative numbers.

Another remark is that, for $x>0$ and $\alpha \in \mathbb{R}$, $x^\alpha = e^{\alpha \log x}$, and this should be true when $\alpha = \frac{p}{q} \in \mathbb{Q}$. There are people who read $x^{\frac{p}{q}}$ as a special case of $x^\alpha$, and they want $x > 0$. It s therefore useful to distinguish between $\sqrt[q]{x}$ and $x^{\frac{1}{q}}$, because the first may be defined also for $x<0$.

Siminore
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There is already the potential for a difference in meaning without recourse to complex numbers: compare $\sqrt[2]{(-1)^2} = \sqrt{1}$ to $(-1)^{2/2} = (-1)^1$.

The point is that there isn't really a way to define a (singly-valued) function $x^y$ across negative/complex values of $x,y$ so that $(x^a)^b$ is always equal to $x^{ab}$. This makes order of operations matter, so there can be subtle distinctions between "raise to power $p$, then power $1/q$" and "raise to power $p/q$".

Erick Wong
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There is no difference between $\sqrt[p]{x^q}$ and $x^{\frac{q}{p}}$, they are just different forms of notation for the same thing.

naslundx
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