Is $\sqrt[n]{x^n} = |x|$ for all n?

Question

We know that $\sqrt{x^2} = |x|$. My question is:

Does this property apply to $\sqrt[n]{x^n}$, for all $n \in \mathbb{N}$?

I can't think on how i could check this. It is not very clear to me why $\sqrt{x^2} = |x|$ either. Thank you.

Answer 1

No. For a counterexample $\sqrt[3]{-8}=-2$

For even $n$, we define $\sqrt[n]{x}$ to be $\ge 0$ because we want $\sqrt[n]{x}$ to work like a function. That is, if we said "$\sqrt[n]{x}$ is that $y$ such that $y^n =x$", then we would get 2 solutions for $y$ for even $n$, but for it to work like a function we want just one. This is why $\sqrt[2]{x^2} = |x|$, rather than $\sqrt[2]{x^2}=\pm x$

Answer 2

No, for $n$ is odd is not true.

Answer 3

No. For instance, $\sqrt[3]{(-1)^3} = \sqrt[3]{-1} = -1$ .

Answer 4

Big picture:

For any real number $y$, there are $n $ complex numbers , $k $so that $k^n=y$. These are the $n $ complex roots.

If $n $ is odd then exactly one of the roots is real. We call that one "the" $n-th $ root of $y$. If $y $ is negative/positive then $\sqrt [n]{y}$ is negative/positive.

So if $n$ is odd $\sqrt {x^n}=x$.

If $n$ is even and $y $ is positive, two of these roots will be real. One will be positive real and one will be negative real. By convention we call the positive one "the" $n-th $ root of $y $.

So if $n $ is even then $x^n $ will always be positive (even if $x $ is negative) and will have two real $n-th $ roots; a positive one equal to $|x|$, and a negative one equal to $-|x|$. By convention, only the positive one is considered "the" $n-th $ root.

So if $n $ is even $\sqrt [n]{x^n}=|x|$.

Finally, if $n$ is even but $y $ is negative then none of the roots are real. There are no real even roots of a negative number.