# Significance of $\sqrt[n]{a^n}$?!

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.

$\sqrt{(-1)^2} =\sqrt{1} = 1 = |-1|$ 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: $\sqrt{-1}=-1$, 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.

• I presume that's a typo in your first expression... :) – J. M. isn't a mathematician Dec 5 '10 at 10:02
• @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 5 '10 at 10:07
• I remain envious of you, Arturo; I'm not that eloquent at 4 A.M. ;P – J. M. isn't a mathematician Dec 5 '10 at 10:09
• ...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. isn't a mathematician Dec 5 '10 at 10:18
• @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 5 '10 at 13:16

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

• @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. – Jonas Meyer Dec 5 '10 at 13:10
• 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. – Jonas Meyer Dec 5 '10 at 13:48
• @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 $(-b + \sqrt{b^2 - 4ac})/2a.$ – Derek Jennings Dec 5 '10 at 19:39
• @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) – Arturo Magidin Dec 5 '10 at 23:31
• @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. – Arturo Magidin Dec 5 '10 at 23:33