Is there a sense in which $4^{2} = -16$? Here's my train of thought:
I. $4^2=4^{(4/2)}$
II. $4^{(4/2)}=(4^4)^{(1/2)}$
III. $(4^4)^{(1/2)}=256^{(1/2)}$
IV. $256^{(1/2)}=\sqrt{256}$
square root of 256 has 2 solutions: 16 and -16
Can you explain me where I made the mistake and why am I not allowed to do whatever I did?
I thought that maybe (II) is wrong and only $4^{(4/2)}=(4^{(1/2)})^4$ is true,
but if $(x^a)^b=x^{(a\cdot b)}$ and numbers can switch places during multiplication and this does not change the outcome then:
$x^{(a\cdot b)}=x^{(b\cdot a)}$, and $(x^b)^a=x^{(b\cdot a)}$
Also, please don't insult me in the replies. I feel stupid enough already for not getting this.

OVERALL THEORY
If $x$ is a positive real number and $m$ and $n$ are integers, then
$$x^{\frac mn} = (x^m)^{\frac 1n} = (x^\frac 1n)^m = \sqrt[n]{x^m} = (\sqrt[n]x)^m$$
When $x$ is negative, all heck breaks loose. We can say that
$$\text{$x^m$ is positive when $m$ is even and is negative when $m$ is odd.}$$
So there is no sense in which $4^2 = -16$.
It is OK to work with negative numbers and integer powers. For example,
$$((-2)^3)^4 = (-2)^{12} = 4096$$
But, weird things can happen with a negative $x$ and fractional exponents. For example
$$
\begin{array}{c}
   (-8)^\frac 13 = \sqrt[3]{-8}= -2 \\
   (-8)^\frac 26 = \sqrt[6]{((-8)^2)}= \sqrt[6]{64} = 2 \\
\end{array}
$$
Since $\frac 13 = \frac 26$, we see that we have a problem.
There is no way to fix this problem. There is no way around it. Raising a negative real number to a fractional power is not well-defined.
PRINCIPAL SQUARE ROOTS
The roots of an expression, $f(x)-n$, is the set of all $x$ for which $f(x)=0$
The solutions to the equation $x^2 - n = 0$ are called the square roots of $n$.

*

*Negative numbers have no (real-valued) square roots.

*$0$ has one square root.

*Positive numbers have two square roots.

The principal square root of a number is defined to be the non negative root of that number. There is a really cool equation that encapsulates this.
$$\sqrt{x^2} = |x|$$
Hence, for example,
$$\sqrt{(-3)^2} = \sqrt{3^2} = |3| = |-3| = 3$$
So, if $x^2 = y$, then the principal value of $\sqrt y$ is $|x|$.
ORIGIN OF THE WORD ROOT
