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Is this Notation correct?

For example:

$$\sqrt[-100]{100}$$

I think this is wrong, because

$$100^{\frac{1}{-100}}=100^{\frac{-1}{100}}=\sqrt[100]{100^{-1}}=\sqrt[100]{\frac1{100}}$$

Am I correct?

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    $\begingroup$ I think this notation is correct but should not be used to avoid confusion. $\endgroup$ Commented Dec 27, 2017 at 11:49

3 Answers 3

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The notation $\sqrt[-100]{100}$ is correct, albeit not commonly used. In fact the whole equality chain \begin{align*} \color{blue}{\sqrt[-100]{100}}=100^{\frac{1}{-100}}=100^{\frac{-1}{100}}=\sqrt[100]{100^{-1}}=\sqrt[100]{\frac1{100}}\tag{1} \end{align*} is correct.

Sometimes we can read in analysis books a definition of rational powers which goes like: Let $a>0$ and $r=\frac{p}{q}$ where $p,q$ are integers, $q>0$, then we define \begin{align*} a^r\equiv a^{\frac{p}{q}}:=\sqrt[q]{a^p}\tag{2} \end{align*}

Note that the definition above justifies only the following representations in (1) \begin{align*} 100^{\frac{-1}{100}}=\sqrt[100]{100^{-1}}=\sqrt[100]{\frac1{100}} \end{align*}

But since we are allowed to use the notation \begin{align*} \frac{-p}{q}=-\frac{p}{q}=\frac{p}{-q} \end{align*} an extension of the notation (2) to each representation in (1) is admissible.

Note: As a plausibility check note that Wolfram Alpha accepts $\sqrt[-100]{100}$ and suggests the following simplified representation \begin{align*} \sqrt[-100]{100}=\frac{1}{\sqrt[50]{10}} \end{align*}

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it can be simplified to $$\frac{1}{(10^2)^{1/100}}=\frac{1}{10^{1/50}}$$

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    $\begingroup$ Sir, I don’t think the OP has invoked a square root notation. $\endgroup$
    – user371838
    Commented Dec 27, 2017 at 11:54
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    $\begingroup$ but i can see a square root notation above? or i will need glases $\endgroup$ Commented Dec 27, 2017 at 11:55
  • $\begingroup$ I took it to be the $-100$th root. An extension of $\sqrt[3]{2}$, $\sqrt[4]{2}$ etc but with an unusual negative. $\endgroup$
    – badjohn
    Commented Dec 27, 2017 at 12:02
  • $\begingroup$ It is in similar sense to $\sqrt[3] x, \sqrt[-5] x$, sir. $\endgroup$
    – user371838
    Commented Dec 27, 2017 at 12:04
  • $\begingroup$ i saw it also here themathpage.com/Alg/rational-exponents.htm $\endgroup$ Commented Dec 27, 2017 at 12:09
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It's unusual and hence clarification would be advisable to avoid misinterpretation. With appropriate clarification then it would be acceptable. You can even invent your own notation provided that you define it. However, when clearer alternatives are easily available, what's the point?

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  • $\begingroup$ I saw this notation new, today $\endgroup$
    – Newuser
    Commented Dec 27, 2017 at 11:53
  • $\begingroup$ Can you give us a link? It is possible that it may have made sense in the context. $\endgroup$
    – badjohn
    Commented Dec 27, 2017 at 11:58

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