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I am a high school student and we just learned about radical and radical notation. Our teacher says index of radical must be integer and greater than 2 by definition. But I can’t understand why we can’t have radical with negative or rational indexes?

For example why can’t we have either of these?

$$\sqrt[\frac32]8=8^{\frac1{\left(\frac32\right)}}=8^{\frac23}=\sqrt[3]{8^2}=\sqrt[3]{64}=4$$

$$\sqrt[-2]4=4^{\frac1{-2}}=4^{-\frac12}=\sqrt[2]{4^{-1}}=\sqrt[2]{\frac14}=\frac12$$

Our teacher says it’s because negative and rational indexes are not defined for radical notations but why they are not defined? They certainly have answers.

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  • $\begingroup$ Compare radical with exponents. $\endgroup$ – Mauro ALLEGRANZA Feb 9 '17 at 14:54
  • $\begingroup$ Your teacher says true. $\endgroup$ – Nosrati Feb 9 '17 at 14:54
  • $\begingroup$ There isn't really any reason other than tradition. If $r \not= 0$ and you have a way to determine what $x^{1/r}$ means you could just as well define $\sqrt[r]{x} = x^{1/r}$. $\endgroup$ – Umberto P. Feb 9 '17 at 15:00
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Nothing prevents you from choosing to define a meaning for something like $\sqrt[-2/3]{x}$.

It's just not something that is usually done, because the only "reasonable" choice of definition would to be to make it mean the same as what we already have the notation $x^{-3/2}$ for -- and since the latter notation is simpler and easier to read, there is no particular demand for also writing it $\sqrt[-2/3]{\cdots}$.

In short, the fact that we usually don't define this is not out of any kind of mathematical necessity, but simply because there doesn't seem to be any need to.

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  • $\begingroup$ I have a vague recollection that Isaac Asimov defined fractional roots in Realm of Numbers. On the other hand, I was 8 years old when I read that book, so I may merely be imagining that. $\endgroup$ – David K Feb 9 '17 at 15:15

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