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I just came across this annotation in my school's maths compendium:

Screenshot

The compendium is very brief and doesn't explain what this means.

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This is the inverse function of $a^n$. Hence $\sqrt[n]a$ means, you look for a number $b$, which when multiplied $n$ times with itself results in $a$.

For instance: We know that $2^3 = 8$, so $\sqrt[3]8 = 2$, $\sqrt[5]{-1}=-1$ because $(-1)^5 = -1$. $\sqrt[4]3 \approx 1.31607$ because $1.31607^4 \approx 3$.

If there is no number at the top of the root symbol, it means $n=2$, so $\sqrt[2]a = \sqrt a$.

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    $\begingroup$ Very well explained, thank you $\endgroup$ – Hubro Aug 13 '13 at 9:41
  • $\begingroup$ I would not recommend to use this notation with negative numbers as you did. $\endgroup$ – Tom-Tom Sep 23 '15 at 11:48
  • $\begingroup$ Could you explain why $(-1)^5=-1\sqrt[4]{3}$? I'm really confused on that. $\endgroup$ – Travis Apr 6 '17 at 11:24
  • $\begingroup$ @Travis It is not. These are two examples. The first example is: $(-1)^5 = -1$ and the second one is $\sqrt[4]3 \approx 1.31607$. $\endgroup$ – Jakube Apr 6 '17 at 12:17
  • $\begingroup$ Oh, didn't see the comma. Thanks. $\endgroup$ – Travis Apr 6 '17 at 12:19
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It means that instead of the "square root of a" you are now considering the "nth root of a". This is the same as writing $a^{1 \over n}$. And just like the square root is "undone" by applying a squared term, i.e., $(\sqrt a)^2 = a$, so the nth root is "undone" by applying the nth power, i.e., $(\sqrt[n] a)^n = a$.

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$\sqrt[n] a$=$a^ \frac 1 n$

Also,if $\sqrt[n] a$=$x$ then $x^n=a$

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    $\begingroup$ @Olorun-Didn't notice that it was 2 year old...may be it was recently modified....saw it in the newest questions tab... $\endgroup$ – tatan Sep 23 '15 at 6:29
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    $\begingroup$ Different answers teach different and sometimes new things, and I see no problem with new answers, given to old posts. $\endgroup$ – codezombie Dec 26 '16 at 11:20
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    $\begingroup$ Minor point: I notice quite a few elementary algebra books as well as some writers here taking the view that the n-th root of x is defined as x to the power 1/n. I disagree strongly. For an elementary student, the idea of saying "what raised to the n-th power gives x?" is more straightforward than saying "just raise x to the power 1/n" particularly since the way one does this without a calculator is the former, not the latter. That is, x to the power 1/n should be the entity being defined, with it defined as I've described - and not the reverse. $\endgroup$ – Dr. Michael W. Ecker Dec 11 '19 at 7:16

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