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I am interested in an all-encompassing answer that explains under what conditions $a^{m/n}:=\sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m$, when $a\in \mathbb{R}$; $m,n \in \mathbb{Z}$; and $m/n$ is in lowest terms.

This is certainly true when $a\ge 0$, but what about negative values for $a$? If $m/n$ is in lowest terms, then is the only necessity that $n$ be odd?

BSplitter
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    How do you define $a^{m/n}$? – José Carlos Santos Aug 25 '24 at 18:27
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    Fair question. I suppose we can define it as $\sqrt[n]{a^m}$. So then the question asks, "when is $\sqrt[n]{a^m} = (\sqrt[n]{a})^m$?" – BSplitter Aug 25 '24 at 18:32
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    How do you define $\sqrt[n]{a}$? What if there are multiple numbers $x$ such that $x^n = a$? – Daniel P Aug 25 '24 at 18:33
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    Cf. this question – J. W. Tanner Aug 25 '24 at 18:34
  • @DanielP What is an example where that would occur? I've edited my question to clarify that I'm restricting my attention to real values of $a$ and integer values of $m$ and $n$. Is there any ambiguity there? – BSplitter Aug 25 '24 at 18:45
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    Math might be the only field where asking a question is just as difficult and require as much scrutiny as answering. What does $a^{\frac mn}$ mean? What does $\sqrt[n]a$ mean? When we defined this things (and we did systematically define them) what criteria did we watch out for. These mostly derive from the observation (by continuity) that for any positive $c$ and any natural number $k$ there is a unique positive $b$ so that $b^k=c$ and we define $\sqrt[k] c=c^{\frac 1k}=b$. That doesn't apply to neg numbers (which toggle and hop like fleas) so we only define for $a>0$.... at first. – fleablood Aug 25 '24 at 18:47
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    "What is an example where" $x^4 = 16$. Is $x = 2$ or is $x= -2$. – fleablood Aug 25 '24 at 18:48
  • $(-1)^2=1$ and $1^2=1$ – J. W. Tanner Aug 25 '24 at 18:49
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    It's irritating to down vote. The OP is genuinely asking a genuine question to the best of his knowledge. – fleablood Aug 25 '24 at 18:49
  • It's true that $(-2)^k$ is well-defined (so that $(-2)^2=4$ and $(-2)^3=-8$) but that doesn't mean $x^4=16$ can be solved by $(-2)^4=16$ and ergo $\sqrt[4]{16}=-2$ or and even though saying an umbrella statement of $\sqrt[4]{16}=\pm 2$ might seem like a good compromise, the bottom line is we only define $a^x$ where $x$ is not an integer for $a>0$. (that $a^x; x$ is an integer and $a< 0$ is okay is a bit of thorn. Technically these are two entirely different concepts with different definitions. Do you want more detail on this? We can give some even though we don't want to.) – fleablood Aug 25 '24 at 18:57
  • @fleablood Yeah asking this is genuinely quite tough. I suppose I knew that for $a>0$, $\sqrt[n]{a}$ can have multiple roots, but I was mostly concerned with negative values of $a$, since rules for rational exponents with positive values of $a$ seem fairly clear to me.

    My main concern when asking this question was to avoid making an incorrect statement about negative values of $a$. My takeaway from this discussion is that when we have negative $a$ values, great care should be taken, and rewriting it like I did in the question title may be unwise.

     – BSplitter Aug 25 '24 at 19:04
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    To define $a^x$ for a negative base $a$ requires complex analysis definition and with complex numbers $b^k = a\ne 0$ will have $k$ different root (whether $a$ is positive, neg. or complex) and we need a definition for a primary root. Turns out a primary root is not as important as you'd think. We well always have $\sqrt[k]{a^m}$ and $(\sqrt[k]{a})^m$ we'll both be $k$th roots of the same number. So really you question is one of "book-keeping" as to how we file and sort the multiple roots. The times $\sqrt[k]{a^m}\ne (\sqrt[k]a)^m$ will any be a jump from one arbitrary filing to another. – fleablood Aug 25 '24 at 19:32
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    Consider $\sqrt[4]{{-16}^5}$ vs $(\sqrt[4]{-16})^5$. $\sqrt[4]{-16}$ can be any of $2(\pm\cos \frac \pi 4\pm i\sin\frac \pi 4)$. Raise those to the fifth power you get any of $32(\pm\cos\frac\pi 4\pm i\sin\frac\pi 4)$. Likewise $\sqrt[4]{(-16)^5}=\sqrt[4]{-1048576}$ which can be any of the four $32(\pm\cos\frac\pi 4\pm i\sin\frac\pi 4)$ so you always get the same set of potential values. But which one you define as THE answer to one or THE answer to the other will be an arbitrary file choosing. – fleablood Aug 25 '24 at 19:45
  • I think case where where $\sqrt[n]{a^m}$ ($a$ complex) might not equal $\sqrt[n]{a}^m$ would be where $m\times \arg a > 2\pi$. We define $\sqrt[n]{a^m}$ to have an argument that is $\frac 1n$ of the argument of $a^m$ which would be $\frac{m\times \arg a - 2\pi}n$. But $(\sqrt[n]a)^m$ would have the argument $m\times \frac {\arg a}n=\frac {m\times \arg a}n$. But both when raised to the $n$ power will have the same angular argument $m\times \arg a \equiv m\times \arg a - 2\pi$. ($\arg a$ is the angular measurement of the complex number $a$ on the complex plane.) – fleablood Aug 25 '24 at 20:52

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