Why doesn't $\sqrt{ab}=\sqrt{a}\sqrt{b}$ work for $a,b<0$? It seems that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is only defined for $a,b\geq 0$, because it doesn't work for $a,b<0$.
I can see that it doesn't work. I would like to know why it doesn't work. Is there a less circular reason than "by definition"?
This comes up in Khan Academy's i as the principal root of -1 and wikipedia's square root faulty proof:
$$
\begin{align*}
-1&=ii\\
&=\sqrt{-1}\sqrt{-1}\\
&=\sqrt{(-1)(-1)}\\
&=\sqrt{1}\\
&=1
\end{align*}
$$
They say $ii=\sqrt{-1}\sqrt{-1}$ is OK, and the faulty step is $\sqrt{-1}\sqrt{-1 }=\sqrt{(-1)(-1)}$, because $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is only defined for $a,b\geq 0$.
 A: Nice question. Think of it this way.
Let $A=\sqrt{ab}$
Let $B=(\sqrt{a})(\sqrt{b})$
$A^2=(\sqrt{ab})^2=ab$
$B^2=((\sqrt{a} )^2)((\sqrt{b} )^2)=ab$
Which means-
$A^2=B^2$
Which further means - 
$\pm{A}=\pm{B}$
Now you understood what happens? $A$ is not always equal to $B$. 
Now I'll prove why it's not equal only if $a,b\lt0$
Condition: $a$ and $b$ are negative. 
Let $\sqrt{a}=ix$ and $\sqrt{b}=iy$
Where $i=\sqrt{-1}$
$A=\sqrt{ab}=xy$ 
(note: $ab$ is positive so there roots would simply be $x$ and $y$) 
$B=(ix)(iy)=-xy$
Thus, when $a,b\lt0$ then $A=-B$
I hope I answered your doubts. 
A: Since $(\sqrt{a}\sqrt{b})^2=ab$, $\sqrt{a}\sqrt{b}=\pm\sqrt{ab}$. The only way to get rid of the $\pm$ sign for a more specific conclusion is by comparing the two sides of the equation. For $a,\,b\ge 0$, square roots are non-negative do we're fine. With negative numbers, the two $i$ factors on the LHS ruin it. If you want a fancier explanation than that, no inverse of $z^2$ on $\mathbb{C}$ is continuous, as can be seen by considering its phase.
