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As long as I've been working with mathematical notation, it seems like I should be embarrassed about being confused about this, but...

Does $\sqrt[n]{a}$ with $n \in \mathbb{N}$ in general stand for the set of all $n$ roots, or just one of them? When $a \in \mathbb{Z}$ and it's clear that the surd is also meant to represent real numbers only, the question becomes whether both the positive and negative roots are included when $n$ is even.

If it stands for all $n$ roots, then I would think $\sqrt{4}\sqrt{4}=\{-2,2\}\times\{-2,2\}=\{-4,4\}$. On the other hand, at least in most contexts it's obvious that $(\sqrt{4})^{2}$ is meant to equal positive 4 only - I suppose the idea is that in this case there is only one instance of $\sqrt{4}$, so one is not free to construct a product from one root from one instance and a different root from another instance. Does this mean that $\sqrt[n]{a}*...*\sqrt[n]{a}$ and $(\sqrt[n]{a})^{n}$ are strictly not interchangeable?

If it stands for only one root, it seems like in some cases it would be clumsy to define which one is meant, depending on what kind of object $a$ is.

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In the context of $\Bbb R $, if $n\in \Bbb N,$ it means the real $n$th root when $n$ is odd. And the non-negative real $n$th root when $n$ is even and $a\geq 0.$

In the context of $\Bbb C,$ if $n \in \Bbb N,$ it does not mean the $set$ of all $n$ roots, but an unspecified member of that set.

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  • $\begingroup$ How does it work in the context of doing arithmetic that a surd with a complex number inside stands neither for a set of numbers nor any one specific number? Can the whole expression not be said to "equal" anything? Especially with something like 1/((2i)^(1/2)+1+i); for one choice of root, it's undefined, but not the other. $\endgroup$ Commented Dec 5, 2017 at 10:51
  • $\begingroup$ $x=\sqrt {2i}$ is ambiguous. It's analogous to the $\pm$ symbol. $x=\sqrt {2i}\iff x^2=2i.$ $\endgroup$ Commented Dec 7, 2017 at 1:59
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If $a \in \mathbb R$, then $\sqrt[n]{a}$ always stands for a single value:

  • When $n$ is odd, there is no ambiguity because $x^n=a$ has only one real solution.

  • When $n$ is even, $\sqrt[n]{a}$ is the positive solution of $x^n=a$. This is only defined when $a\ge 0$.

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The standard meaning of $\sqrt[n]a$, when $a\in[0,+\infty)$, is the only real number greater than or equal to $0$ such that its $n$th power is equal to $a$, nor the set of all $n$th roots of $a$. With this convention, both$$\overbrace{\sqrt[n]a\times\cdots\times\sqrt[n]a}^{n\text{ times}}\text{ and }\left(\sqrt[n]a\right)^n$$have the same meaning: $a$.

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