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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.

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    I have changed the formatting of the title so as to [make it take up less vertical space](https://math.meta.stackexchange.com/a/9686/290189) -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See [here](https://math.meta.stackexchange.com/a/9730) for more information. Please take this into consideration for future questions. Thanks in advance.2018-03-12

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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 given you the value $2^2$: both $2$ and $-2$. There are two number 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.

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    I presume that's a typo in your first expression... :)2010-12-05
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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.2010-12-05
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    I remain envious of you, Arturo; I'm not that eloquent at 4 A.M. ;P2010-12-05
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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. :P2010-12-05
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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 $x$."2010-12-05
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    @Jonas: yes. I'll rewrite. Good point (he says 11 hours later, after some sleep).2010-12-05
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    Does that mean that $i\neq\sqrt{-1}$, since $(\sqrt{-1})^2=1$?2018-02-08
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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 $1$.2018-02-09
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The comments for this answer have more value than the answer I posted. I am removing the contents of my answer but leaving the responses in tact ( that is why I am not deleting this post, if there is an alternative method please let me know). Regarads

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    We can however speak of *principal* roots...2010-12-05
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    @Arjang: "it is meaningless and has no place in mathemtics": I disagree. For each even $n$, one can define $f:[0,\infty)\to[0,\infty)$ by $f(x)=$ the unique positive $n^\text{th}$ root of $x$. A standard notation for this function is $f(x)=\sqrt[n]{x}$. Although this definition is not given in the question, I believe it is clear that this is what is being used.2010-12-05
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    Where does the restriction of domain comes from? yes, if the domain is restricted to positive real numbers then of course, but if we are going to be working with positive real numbers then $ x + 1 = 0 $ has no solution. In mathematics anything that is not stated makes the statement unclear and ambiguous. It has no place in mathematics.2010-12-05
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    The logarithm is restricted for reasons of convenience, and that carries over to roots.2010-12-05
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    Are you asking for reasons why one might consider the positive square root function? There are many applications. For example, it is useful in defining a norm in an inner product space by $\|x\|=\sqrt{\langle x,x\rangle}$. Whether one restricts to a particular domain or prefers a particular root function will depend on context. In a given context, it can be very meaningful to do so, and have a good place in mathematics.2010-12-05
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    Down vote without explaining Why not $\sqrt{(-1)^2} = {(-1)}^{2 ( ^1/_2 )} = {-1}^{( ^1/_1 )} =-1$ instead?2010-12-05
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    Because $\sqrt{(-1)^2}$ is, by the definition which is clear from the context of the question, the positive square root of $(-1)^2$, namely $1$, and not what you wrote.2010-12-05
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    @Jonas Meyer , no I am not asking why in specific places one would want to define and restrict the domain. But I am asking why in elementary mathematics where students know about negative numbers all of a sudden the domain is magically restricted to positive numbers? The formula did not specify anything about it's domain and it is wants us to believe that the only number that can squared to be equal to 4 is 2 and not -2 ! How comes one can not find this fragmented formula in any analysis books? Has anyone seen this formula in any reputable analysis book?2010-12-05
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    @Jonas Meyer: $\sqrt{(-1)^2} = {(-1)}^{2 ( ^1/_2 )} = {-1}^{( ^1/_1 )} =-1 $ is elementary, $(x^a)^b = x^{ab}$ not because I say so, but I say so because it is so. so not unless one can show that $(x^a)^b \neq x^{ab}$ then $\sqrt{(-1)^2} = {(-1)}^{2 ( ^1/_2 )} = {-1}^{( ^1/_1 )} =-1 $ , other wise show that $(x^a)^b \neq x^{ab}$.2010-12-05
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    @Arjang: I understand where you're coming from (second to last comment). Perhaps I'm just overly familiar with elementary math :). However, it is just a matter of notation/definitions. Once one has chosen a convention when working with real roots of real numbers what $\sqrt[n]{x}$ means, then one will still have to worry, in the even case, about the other $n^{th}$ root of a positive number. So for instance, $\sqrt{7}$ may be defined to be positive, by convention, and then to solve the equation $x^2-7=0$, one can write $x=\sqrt{7}$ and $x=-\sqrt{7}$, with the meaning of each unambiguous.2010-12-05
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    "magically" - I think that's where your problem is coming from. We restrict domains and ranges because it's *convenient*. Certainly you are free to construct your system where the principal square root is negative...2010-12-05
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    My last comment. If $(x^a)^b=x^{ab}$ as you are claiming (without a restriction such as $x\geq0$), then you are not working with well-defined functions, but at best multi-valued functions. These have their place, but it is not the content of this question. Finally, I have nothing against your points about nonuniqueness of roots, I just disagree with your assessment of the validity of the math in question.2010-12-05
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    "I think that's where your problem is coming from. We restrict domains and ranges because it's convenient." 2 questions, 1 who is "We" and 2. What is convenient? The restriction of domain was not stated. If the domain is restricted in order to have a function with only principal roots as the range then it must be explicitly stated. Other wise "we" can just keep at it that what was meant and/or intended or not intended and, how is one is meant to interpret/not interpret it without getting anywhere. Mathematics that is not explicit is not mathematics.2010-12-05
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    @ Jonas Meyer : I fully agree with your last comment, and I know that you have nothing against my points, if anything I am grateful for your time and effort. Just because we argue about a point it doesn't mean we argue against one another. I see it as we argue together to learn together from one another. Plus nothing makes me happier to be shown my lack of reasoning or not having considered some aspects. In this case the implicit assumption that restriction of domain is to work with single values function. Actually that is the reason/motivation/explanation for the formula.2010-12-05
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    It is well known that $\sqrt{x^2}=\pm x$. This is how we were tought in school.2010-12-05
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    @Mathsfacts: Unfortunately, you are right about what is taught in a lot of schools and probably that is why it dawns on many people relatively late into their mathematical studies that $\sqrt{x}$ is defined to be the non-negative square root. There is a hint of this use even at school level: the teacher writes the roots of a quadratic equation on the board as $(-b \pm \sqrt{b^2 - 4ac})/2a$ but if we are free to assume $\sqrt{x}$ implies the negative root as well then we could equally write the roots of our quadratic as $(-b + \sqrt{b^2 - 4ac})/2a.$2010-12-05
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    When I first saw the standard formula, I recall asking myself why the minus sign was present and decided it must to ensure some people didn't forget the negative root! I think it would have been great if someone had posed a question such as “What is the definition of $\sqrt{x}$? as much of the discussion has centred around this important point, which trips up many people, and unfortunately it's a bit hidden here.2010-12-05
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    @Derek Jennings, Derek, you are right. The motivation for the formula is to make a multivalued function into a single valued function (thanks to Jonas Meyer post), but the only way to say that a number $a$ is an $n^{th}$ root of $x$ is to see if $a^n=x$. At the elementary level they never mention this and never explicitly state the motivation for using the formula of this discussion. This formula is only taught at elementary level and it is never seen again in proper math.2010-12-05
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    This is an interesting forum, while the discussion in this post is being used to edit the up voted answer, this post itself is keep getting down voted. The original answer posted has borrowed from the discussion that followed from my post without citing any attribution. The original answer is nothing like what I replied to at the beginning of this discussion. The statement in the original answer "Every number has unique cubic root" is only valid when the domain is restricted. There is way too much implicit assumptions going on here.2010-12-05
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    @Arjang: I only just now noticed your answer and the multitude of comments that followed it, as it happens, so I resent a bit the claim that I "borrowed from the discussion that followed from my post without citing any attribution." I did not do so; I was following up on the comment Jonas Meyer left in my response, not borrowing from you, and I apologize if you felt this was the case.2010-12-05
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    @Arjang: Pretty much every single comment you can make in mathematics is only valid in a particular context, with a particular set of conventions in place. The context of the question makes it abundantly clear that we are dealing with real valued functions of real variable; so the inputs are always assumed to be real numbers, as are the outputs, and the operations are assumed to be functions (that is, *single-valued* functions). The decision that "the" square root will be the positive root is a pretty universal convention, though of course I agree it is merely a convention. (cont)2010-12-05
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    @Arjang: (cont) While certainly other contexts will make the answer invalid, it seems unlikely to be helpful to add to what is already something confusing the OP by mentioning that in fact *neither* of the two equations are "right" in the sense that if you change the setting then the answers are wrong. Sure: if we go to complex numbers, then *neither* square nor cubic roots are unique, and we have to get into multi-valued functions and principal branches. But I think someone who is having trouble grasping the issue of the parity in $R$ is not going to benefit from being told *that.*2010-12-05
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    @Arjang: Now, to answer the question you rhetorically post in the first line of your answer: I disagree. One thing is "the solutions to $x^2-a=0$" (or to $x^2=a$), and another thing is **the** "square root of $a$". The latter (implicit in the use of the singular definite article) is understood to refer to a (single-valued) function; closely connected to the set of solutions of $x^2-a=0$, to be sure, but not identical to it. Also, "the only way to know if $x$ is the square root is to see if $x^2=a$" (perhaps true), is different from "$x$ is the square root **if and only if** $x^2=a$"2010-12-05
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    @ Arturo Magidin, My apologies for implying that you borrowed from the discussion that followed. What you realized on your own , I only realized by the posts that I received. I have/had no explicit indicator that the results that was posted was borrowed at all. And without having an explicit statement from you within the context that I was operating under it was implicitly assumed ( wrongly ).2010-12-05
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    @Arjang: No harm, no foul. (-:2010-12-06
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    @everyone (who tried to help me understand): The point is that by the definition one is trying to make a single valued function that defines principal roots as the values. This is done purely by definition and establishes the convention. There is no deeper reason than having a base foundation that everyone can agree on.2010-12-22
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    @Arjang: Well said.2010-12-27