roots: square roots, cube roots, x^1/4, x^1/5

Question

I was having confusion over the concepts of roots. My confusion begins as: I know that if $x^{1/n}=y$ then $y^n=x$ must be true.

Case 1: With square roots, let's take $y=4^{1/2}$

We know $y=2$ is a solution as $2^2=4$, but $y=-2$ must also be a solution as $(-2)^2=4$ is also true. But I browsing here I read that $(x^2)^{1/2}$ is actually equivalent to absolute value of $x$, hence only $y=2$ is solution of my above case.

Case 2: With cubic root, let's restrict ourselves to the real no.s's domain, let's take $y=(-8)^{1/3}$, then $y=-2$ seems feasible solution as $(-2)^3=-8$ but why in this case absolute value after cube root is not required. Also, why $(-x)^{1/3}=-(x)^{1/3}$, but $(-x)^{1/2}$ is not equal to $-(x)^{1/2}$

The same apparent ambiguity goes with $x^{1/4}$ and $x^{1/5}$

So, in general what is the general function definition of $x^{1/n}$?

Compare the phrases "a n'th root" vs "the n'th root." If we refer to the root, it implies there is only one that we are interested in, and we generally take it to be the "principal root." — JMoravitz, Jul 22 '19 at 18:53
note that $(-x)^2=x^2$ but $(-x)^3=-x^3$, and $2$ and $3$ can be replaced with other even and odd numbers, respectively — J. W. Tanner, Jul 22 '19 at 19:05

score 0 · Answer 1 · answered Jul 22 '19 at 19:15

In general a real or complex number has $n$, $n^{th}$ roots in complex numbers.

In case that you are interested in real roots if $n$ is odd there is only one $n^{th} $ root of the real number $x$ and it has the same sign as $x$

On the other hand if $n$ is even the real number $x$ has two real $n^{th} $ roots one positive and one negative.

For example we have $\sqrt {25}=5$ and its opposite $-\sqrt {25}=-5$ as two real square roots of $25$

For $ x=-64$ there is only one real cube root and it is $-4$

roots: square roots, cube roots, x^1/4, x^1/5

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