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There is a formula given in my module:

$$ \sqrt[n]{a^n} = \begin{cases} \, a &\text{ if $n$ is odd } \\ |a| &\text{ if $n$ is even } \end{cases} $$

I don't really understand the differences between them, kindly explain with an example.

F.A.
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Quixotic
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1 Answers1

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$\sqrt{(-1)^2} =\sqrt{1} = 1 = |-1|$ because in order to make the square root an unambiguous operation, we agree that the square root of a nonnegative number $x$ is always the (unique) nonnegative number $r$ such that $r^2=x$. But with cubic roots there is no problem: $\sqrt[3]{-1}=-1$, because every real number has a unique cubic root.

The same is true with 4th, 6th, 8th, 10th, etc. powers, since $a^n = (-a)^n$, and the 4th, 6th, 8th, 10th, etc. roots are defined to be the unique nonnegative real number that "works", so that they are unambiguous.

That is, there are two numbers which when squared will give you the value $2^2$: both $2$ and $-2$. There are two numbers that when taken to the fourth power will give you $(-6)^4$: both $-6$ and $6$. And so on. Generally, both $a$ and $-a$ will, when raised to an even $n$th power, give the same answer: $a^n = (-a)^n$. And we agree that a square root (fourth root, sixth root, etc.) will always be the nonnegative answer, so the $n$th root of $a^n$ will be $|a|$ when $n$ is even. (Don't let the big $-$ in "$-a$" fool you; that does not mean that $-a$ is negative, it just means the additive inverse of whatever $a$ is; if $a$ is positive, then $-a$ is negative, but if $a$ is negative, say $a=-3$, then $-a$ is positive, $-a = -(-3) = 3$. Repeat after me: the proper way to pronounce "-a" is not "negative a", the proper pronunciation is "minus a").

But if $n$ is odd, then every number has a unique $n$th root. In particular, the only number that when cubed gives $2^3$ is $2$; the only number which, when raised to the fifth power, gives $(-6)^5$, is $-6$. There is no longer the problem that both $6$ and $-6$ are possible answers, so we can simply say that the cubic root of $(-2)^3$ is $-2$, the fifth root of $7^5$ is $7$, etc.

J. W. Tanner
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Arturo Magidin
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    I presume that's a typo in your first expression... :) – J. M. ain't a mathematician Dec 05 '10 at 10:02
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    @Tretwick Marian: $i$ is the number one imagines at 4am in the morning... the time here. I should really go to bed now. – Arturo Magidin Dec 05 '10 at 10:07
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    I remain envious of you, Arturo; I'm not that eloquent at 4 A.M. ;P – J. M. ain't a mathematician Dec 05 '10 at 10:09
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    ...and if you catch yourself saying "negative a" again, the penance is to stand in the corner and keep repeating "minus a" until you are parched. :P – J. M. ain't a mathematician Dec 05 '10 at 10:18
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    @Arturo: Of course, by "square roots are always nonnegative" you mean something more like "$\sqrt{x}$ is defined to be the nonnegative square root of a nonnegative number $x$." – Jonas Meyer Dec 05 '10 at 13:16
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    @Jonas: yes. I'll rewrite. Good point (he says 11 hours later, after some sleep). – Arturo Magidin Dec 05 '10 at 21:03
  • Does that mean that $i\neq\sqrt{-1}$, since $(\sqrt{-1})^2=1$? – cansomeonehelpmeout Feb 08 '18 at 20:05
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    @cansomeonehelpmeout: No; $\sqrt{-1}$ does not make sense when you are working with the real function: $\sqrt$ is only defined on nonnegative reals. (Note that the first line is $\sqrt{(-1)^2}$, not merely $\sqrt{-1}^2$). The complex function is actually taken to be multi-valued (this makes more sense over the complex numbers), and so you must pick a branch of the square root ahead of time to determine whether you have $\sqrt{-1} = i$ or $\sqrt{-1}=-i$; either way, $(\sqrt{-1})^2 = -1$, not $1$. – Arturo Magidin Feb 09 '18 at 01:06