Why does $\sqrt a\sqrt b =\sqrt {ab}$ only hold when at least one of $a$ and $b$ is a positive number? I've just been introduced to complex numbers, and I have found it surprising that the radical rule apparently holds even when one of $a$ and $b$ is a negative number. However, if both $a$ and $b$ are negative, then this rule doesn't work. Why is this?
Here is my attempt at proving the radical rule for positive $a,b$. I was wondering if this proof could be generalised for negative $a,b$, and whether this could form part of the explanation for when the radical rule holds. (Unfortunately, though, I am yet to learn about how the natural logarithm works when it can accept complex arguments.)
\begin{align}
\sqrt a\sqrt b &= a^{\frac{1}{2}}b^{\frac{1}{2}} \\
&=e^{\frac{1}{2}\ln a}\times e^{\frac{1}{2}\ln b} \\
&=e^{\frac{1}{2}\ln a+\frac{1}{2}\ln b} \\
&=e^{\frac{1}{2}\ln(ab)} \\
&=(ab)^{\frac{1}{2}} \\
&=\sqrt{ab}
\end{align}
I have also heard that part of the reason for why the radical rule only works in certain cases is because there is no way of ordering $i$ and $-i$. In other words, there is no way of saying that $i$ is 'greater than' $-i$, or vice versa. Taking this idea to the extreme, does this mean that we can't even say that $5$ is greater than $3$ when working with the complex plane?
 A: TLDR:
There's no good way of defining complex roots that allows us to have this property. But we do get close anyway.

Instead of going into your proof, let's first address the assertion that motivated his question, viz. $$\sqrt{a}\cdot \sqrt{b} = \sqrt{ab} \iff a \in \mathbb{R} \ \text{or} \ b \in \mathbb{R}$$.

Square roots in real numbers:
When we deal with the real number system, the square root operation $(\sqrt{})$, is usually defined to yield a positive number. The reason for this is that while we do have two (positive and negative) choices for the square root it is simply more convenient to fix one unique output for the operation. More precisely, fixing a single 'kind' of output allows us to make 'square-rooting' into a function, which can be advantageous when you're dealing with rigorous math. Thus, all non-negative real numbers (the set $\mathbb{R}^{+}$) are now to be associated with another unique real number.
How many ways can we do this? Infinitely many! We may choose to have all the square roots be negative, we may have just one of them be positive (say, $\sqrt{9} = +3$) and let all the others be negative. We can have all but $3$ numbers have negative roots. The choices are unlimited. But there is one choice that is more...natural than the others. Let each root be positive. This is simple (no exceptions for certain numbers that have different signs), and people are generally more familiar with positive numbers, so it sort-of makes more sense. Also, since we can only calculate square roots of non-negative numbers (unless we use complex numbers), there is a nice sense of symmetry with only having non-negative roots. Both the 'input' and the 'output' have the same sign.
However, this is not the complete story. It turns out that one of the simplest functions we can define is the modulus or mod function. It is represented by two vertical bars between which we keep a number, say $a$, like thus: $\vert a \vert$. The function basically turns any number 'inside' it positive. In other words, it returns its 'absolute value', or 'magnitude'. That is, if the number $a$ is positive it remains as such. If it is negative, we take its negative and get a positive output!
$$f(a) = \vert a \vert = \begin{cases}
a & a \geq 0 \\
-a & a \lt 0
\end{cases}$$
Foe eg. $\vert -3 \vert = -(-3) =3$ and $\vert 4 \vert = 4$.
Thus, we generally define things as follows:

*

*Every (non-negative) real number has two square roots- the positive one and the negative one. 'Square root' as such is an operation whose output is a pair of real numbers. For eg. $\sqrt{9} = (-3, +3)$. We can still define it as a function, but it would not map (associate) non-negative real number to positive real numbers ($\mathbb{R}^{*} \to \mathbb{R}^{*}, \mathbb{R}^{*} = \mathbb{R}^{+} \cup \{ 0 \} $), but instead from non-negatives to pairs of (one negative and one positive) real numbers ($\mathbb{R}^{*} \to (\mathbb{-R}^{*} \times \mathbb{R}^{*})$, where $\mathbb{-R}^{*}$ is the set of non-positive numbers). This is simply represented as $\pm \sqrt{a}$.

*However, we further define the positive square roots as $\vert \sqrt{a} \vert$, which selects (using the implied mod operation), the positive of the two square roots. Further, while we decide to have the first case be represented as $\pm \sqrt{a}$, and represent the the single positive square root by simply $\sqrt{a}$. Thus, when we write $\sqrt{a}$, it is actully shorthand for $\vert \pm \sqrt{a} \vert$ or $\vert \sqrt{a} \vert$.

A direct consequence of this is that $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$. In fact this is not all! We can prove a stronger result: $$\sqrt{a_{1}} \cdot \sqrt{a_{2}}  \dots \sqrt{a_{n}} = \sqrt{a_{1} \cdot a_{2} \ldots a_{n}}$$, which is true since all the roots are (by definition) non-negative.
Obviously, when we are still limited to real numbers, the $a, b, a_{1}, a_{2}, \ldots$ are all non-negative.

Square roots in complex numbers:
When we extend our number system to $\mathbb{C}$, the set of complex numbers $a + i \cdot b$, with $a$ and $b$ being real numbers; we would like to preserve the nice-looking properties from above. Thus, we want our definitions of things like 'square-root' and 'signs' for complex numbers to allow for that (with the rider that doing this doesn't break properties that are even more dear to us). And it turns out that to an extent, we can indeed do this!
Take any complex number, say $s = a+ i \cdot b$. How do we find its square root? You probably know the answer if you're studying complex numbers.
...
The answer is $\sqrt{(a + i \cdot b)} = (a^{2} + b^{2})^{1/4} \cdot (\cos (\arctan (b/a))/2 + i \cdot \sin (\arctan (b/a))/2)$.
You can do this easily using de Moiver's formula or the Euler form of complex numbers.
The key point here is that just like in the case of real numbers, we have two complex roots for any complex number. (In fact, the analogy is limited. We get $3$ cubic roots, $4$ fourth roots, etc in complex numbers. This is related to the Fundamental Theorem of Algebra. You can easily see these results using the fact that $tan^{-1}$ has multiple branches in its co-domain and doesn't become a function unless we choose one.) Note that one can justify this using considerations identical to that in the real case.
We can use these facts to reach an important result: The statement $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ holds for complex numbers to within a variation in sign.
To see this is not very difficult. Recall that complex numbers can be separated into their magnitude and argument. When you take the square root of a number, you take the (real-valued) square root of its magnitude and halve its argument. Thus, the product of magnitudes on both sides must be the same $\sqrt{ \text{mag}(c_{1})} \cdot \sqrt{ \text{mag}(c_{2})} = \sqrt{ \text{mag}(c_{1} \cdot c_{2})}$, as all quantities involved are real (in fact, also non-negative). Elementary trigonometry (made simple by a direct application of de Moivre) shows that the arguments must give the same result to within sign.
To make the next leap, you need to look at ordering of complex numbers.

Ordering in complex numbers:
Here's the deal. Complex numbers form what we call a non-ordered field. This basically means that given any two complex numbers, $c_{1}$ and $c_{2}$ (one or both of which may be purely real or imaginary), we cannot meaningfully say that one is larger or smaller than the other (they can, of course still be equal).
The term 'meaningful' above is key. You actually can still come up with an infinite number of 'sets of rules' that you can use to 'rank' or 'order' all complex numbers. But none of them would be as blindingly 'common-sensical', 'natural' or 'nice' as the usual 'greater than/ less then/ equal to' ordering rules of real numbers. More precisely, there is a certain nice set of properties that any candidate for ordering rules must possess before mathematicians will consider it an ordering. And it turns out that complex numbers simply don't have any such simple orderings.

Complex roots CONT.:
Knowing all this, we're now at the same stage as we were when we chose the signs for square-roots of real numbers. Suppose that we'd said that all real roots are positive except for that of $4$, which is instead taken to be $-2$. Can you then see how $\sqrt{4} \cdot \sqrt{9} = - (6) = - \sqrt{36} \neq \sqrt{36}$?
Thus, we are at an identical point. How to define square-rooting in a nice way, except that in this case, our criteria for niceness is that the $\sqrt{c_{1}} \cdot \sqrt{c_{2}} = \sqrt{c_{1} \cdot c_{2}}$ rule must be obeyed to obtain symmetry with the real case, while in the real case, we simply wanted to get a consistent way to determine the signs of square roots. Note that the criteria of simplicity applies in both cases, we don't want to use complicated rules to decide the sign in $\mathbb{R}$, and we don't our property to be at the cost of great complexity or sacrifice of other properties (say, commutativity) in $\mathbb{C}$.
So, here's the difference in both cases. The real numbers had the nice positive-negative division to make our assignment job easier. The complex numbers, in contrast, don't possess any good ordering that allows us to decide a convention of choosing the 'bigger' root or something by analogy with the reals. Certainly no nice rules that also give us the property. Again, you could probably come up with some complicated convention that gave you a pseudo-ordering and preserved the product property, but it'd feel put-on and unappealing, and mathematicians don't like such things.
So, there's your answer:
There's no good way of defining complex roots that allows us to have the property.

Note that if we take the modulus (magnitude) of each square root, like we did for the real square root, the property does hold fully. But then, we lose a lot of important information about the number (there are infinitely many real numbers with a given magnitude, so the sign is critical). In the reals, we only lost the sign (in fact, we choose just one, and know what the other is, so we lost nothing). Thus, this is not very satisfactory either.

A discussion on choosing properties and definitions:
Now, the thing to remember is this, mathematics doesn't involve following rules and operations that just are 'out-there' for some weird reason. Rather, we start off by choosing a set of definitions (which may be intuitive/ implicit instead of being explicitly written down - but only for the very basic concepts), and a set of axioms or rules that govern the 'behavior' of the stuff we've already defined. For example, using definitions (which we no longer take as seriously as he did) of some basic things such as points and lines and some rules governing them, Euclid formulated all of the basic geometry that people learn in schools. And he didn't just state things that he claimed were true, instead he took the basic definitions and axioms, and used them to systematically prove that all the results must be true as a logical consequence of the initial assumptions.
However, here's the catch, the initial rules that Euclid chose were not unique. Instead, he had the choice of taking any consistent set of axioms for his geometry. He simply made the choice that seemed to make the most sense to him based on how geometrical 'objects' such as lines tended to behave in real life - i.e. how things seemed to worked when you drew stuff on the ground, or took measurements for applications such as partitioning fields. And here's the clincher, when people several centuries later actually changed one of those rules, they came up with very different geometric systems! New geometries, that not only had unexpected properties that appeared to be at odds with common experience, but that later (loosely speaking) turned out to be better at describing the actual geometric structure of our world!
The takeaway is this:
We can define things as we want (as long as they are consistent). But we like to keep them in line with our intuitions and analogous to real world things. I don't know anyone, especially no mathematician, that wants to spend great time and  effort going over stuff that is totally alien and unrelateable. Maybe there are some post-modernists that can help you with that ;) (That was a joke. No offense intended to any post-modernists.) But who knows, maybe a formulation that obeys this property would be found useful some day. It's just that no one is likely to buy that stock right now. If you read this till the end and weren't bored out of your mind, than maybe this wasn't a waste :P
A: If $x=\sqrt{a}$ and $y=\sqrt{b}$, then $(xy)^2\stackrel{(1)}{=}x^2y^2=ab$ and $xy\stackrel{(2)}{=}\color{blue}{\pm}\sqrt{ab}$. This reasoning works for both $\sqrt{a},\,\sqrt{b}\in\Bbb R$ and $\sqrt{a},\,\sqrt{b}\in\Bbb C$, because (1) uses commutativity (also needed on the display line below) and (2) uses the nonexistence of zero divisors, so that$$(u-v)(u+v)\stackrel{(1)}{=}u^2-v^2=0\implies u\mp v=0.$$To lose the $\color{blue}{\pm}$ when $\sqrt{a},\,\sqrt{b}\in\Bbb R$, we use the fact that then $a,\,b\ge0$, and their square roots are defined as the non-negative choices for $x,\,y$; then $x,\,y,\,\sqrt{ab}$ are all $\ge0$, so $\sqrt{ab}$ is $xy$ as opposed to $-xy$. In particular, this uses the fact that non-negative reals are closed under multiplication. But there is no analogous half $H$ of $\Bbb C$ in which we can place square roots, so that (i) for each $z\in\Bbb C\setminus\{0\}$ either $z\in H$ or $z\in-H$ but not both, and (ii) $z,\,w\in H\implies zw\in H$.
A: You're proof looks good for $a$ and $b$ positive numbers!  Some weird things can happen when you're working with complex numbers and logarithms become a little harder, so I won't touch them.  Without loss of generality, assume that $a<0$ and $b \geq 0$.  Then we have that $$ \sqrt{a} \sqrt{b} = i \sqrt{|a|} \sqrt{b} = i \sqrt{|a|b} = \sqrt{ab}.$$  This proves the rule if exactly one of $a$ and $b$ is negative.  If they are both negative, we see that $$ \sqrt{a} \sqrt{b} = i \sqrt{a} i \sqrt{b} = i^2 \sqrt{a}\sqrt{b} = - \sqrt{ab}. $$  You are correct that in the context of the complex numbers, we don't think about ordering 3 and 5, for example, but they are still comparable as real numbers.  Hope this helps!
A: Not as comprehensive as the other answers, but it's the same reason as why we don't allow for division by zero, it causes rules to collapse: A classic "proof" that $1=-1$ assumes that you may factor square roots, regardless of them being well defined.

Basic sketch of "proof": $(-1)^2=1=\sqrt{1},$ $\sqrt{(-1)^2}=\sqrt{1},$ so $-1=1.$

This is obviously false, because this would mean $1+1=0,$ and although this is perfectly fine mod 2, we're not assuming that. This is why we require that the square root be multiplicative for only positive values $a,b$ as per your question.
