Why isn't $\sqrt[12]{\left(-1\right)^6}$ equal to $\sqrt{-1}$? Why isn't $\sqrt[12]{\left(-1\right)^6}$ equal to $\sqrt{-1}$? Clearly, the square root isn't defined. We have divided the index and the exponent by $6$. The theorem says that $$\sqrt[nk]{a^{mk}}=\sqrt[n]{a^m}$$ where $a\ge0$. How does it work when $a<0$, or it doesn't?
PP. I see that we can write $\sqrt[12]{(-1)^6}=\sqrt[12]{1^6}=1,$ but my question still holds.
 A: Because it is a compositon of two functions $f(x)= \sqrt[12]{x}$ and $g(x)=x^6$. You need $$(f\circ g)(-1) = f(g(-1)) = f(1) = 1$$
So, you act on $-1$ with $g$ and then on result with $f$.
A: If you call $a=\sqrt[12]{(-1)^6}$, this means by definition that $a^{12}=(-1)^6$. So $a^{12}=1$. There are two real roots to this equation, namely $a=1$ and $a=-1$ (there are ten other complex roots, including $i$ and $-i$).
Notwithstanding the above, the usual convention for the symbol $\sqrt{\ \ 
 \ }$ is to mean the positive root.
A: A related property of complex square roots that sometimes shows up in these 1=-1 proofs is that the $\sqrt{ ab}=\sqrt{a}\sqrt{b}$ property of positive real numers $a,b$ does not hold for complex numbers, i.e.  $\sqrt{ ab}=-\sqrt{a}\sqrt{b}$, if $a,b<0$. That’s the exact point where these proofs are false.
A: If $\ x\in\mathbb{R}\ $ and $\ x>0,\ $ then $\ x^y\ $ is well-defined and unique for any $\ y\in\mathbb{R}.\ $ This is, for example, an exercise left to the reader in Rudin's PMA chapter $1.$
If $\ x\in\mathbb{R}\ $ and $\ x<0,\ $ then $\ x^y\ $ is well-defined if and only if $\ y\ $ is an integer. Basically, attempts to define $\ x^y\ $ if $\ x<0\ $ always runs into problems, one of which you encountered in your question.
However, in $\ \mathbb{C}\ $ we do not get analogous issues: if $\ x,y\in\mathbb{C}\ $ then $\ x^y\in\mathbb{C}.$
A: $\sqrt[n]{a^m}$ says do calculations in this order:

*

*Find $a^m$.

*Find the greatest real number you can raise to the $n$th power to get that previous result.

$\sqrt[nk]{a^{mk}}$ says do calculations in this order:

*

*Find $mk$ and separately $nk$.

*Find $a$ raised to the first result from 1.

*Find the greatest real number you can raise to (1)'s second result,such that you get the result from 2.

My point is that the sequence of things that you do in the two expressions are very different according to purely the order of operations. So in some sense it should be surprising that they would ever end up giving you the same result.
And that's right: they don't give the same result. Except in certain conditions, it can be proved that they will give the same result. One of those conditions is when $a>0$, in which case it can be proved these two things lead to the same end. That proof relies on $a$ being positive. In another situation where $a<0$, maybe the two things will end up the same, maybe not.
