Apparently $\sqrt{ab}$ = $\sqrt{a}\sqrt{b}$ is only true if a and b are both positive or if a is negative and b is positive or if a is positive and b is negative. In other words, a and b can't both be negative.
Is it possible to algebraically prove this? Or is it just a result of the way the square root function is defined?
I know of 1 way to prove this radical property, but I'm still not sure why it won't work for negative numbers.
Let x = $\sqrt{ab}$.
Let y = $\sqrt{a}\sqrt{b}$
Square both sides for both equations.
$x^2 = (\sqrt{ab})^2 = ab$
$y^2 = (\sqrt{a}\sqrt{b})^2 = (\sqrt{a}\sqrt{b})(\sqrt{a}\sqrt{b}) = (\sqrt{a})^2(\sqrt{b})^2 = ab$
$\therefore x^2 = y^2$
$x^2-y^2=0$
$(x+y)(x-y)=0$
$\therefore x = y$ or $x = -y$
Or
$\therefore y = x$ or $y = -x$
A lot of people will go through this line of reasoning (shown below) in order to justify why a and b can't both be negative.
Considering that mathematicians define $i^2=-1$ or $i = \sqrt{-1}$
$1 = \sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1} = (i)(i) = i^2 = -1 $
But this is only a specific instance where this property fails us. This isn't a rigorous or at least satisfying proof of why $\sqrt{ab}$ = $\sqrt{a}\sqrt{b}$ can only be true if a and b are not both negative.
Note: I just started learning about complex and imaginary numbers and I am no means an expert in mathematical proofs, so if you do know the answer to this question please try (if possible) your best to answer the question without using too much complex or high-order math that I won't be able to understand.
This is perhaps rather a case of imprecise language than an issue with proof. The square root function is defined to have it's domain in positive real numbers (including zero). Whenever people say "square root of $x$", I must assume $x$ is a positive real number.
There is a completely different relation (and not a function) that I'll call $R(x)$, which relates any complex number with two other complex numbers. If $(x = r e^{\theta i})$ with $r$ a positive real number, then I have that $R(x) = \{\sqrt{r}e^{\theta/2 i}, \sqrt{r}e^{(\theta/2 +\pi)i}\}$.
Note that, in the particular case $\theta=0$, we have $R(x) = \{\sqrt{r}, -\sqrt{r}\}$.
That being said, if someone explicitly asks for the square root of $-1$, I'll forgive this person's abuse of language, and understand that he/she asked for $R(-1)$, which is $\{ i, -i \}$, and in some cases I might even consider that the person wanted to hear "$i$" only.
In your case, if both $a$ and $b$ are negative, at the left-hand side, I'll have a traditional square root written, whereas, on the right hand-side I should have a product of $R(a)$ and $R(b)$, which is undefined.