Intuitive reason why $\sqrt a\cdot\sqrt b \neq \sqrt {ab}$ when $a$ and $b$ are negative As a rule of thumb I have learnt that $\sqrt a\cdot\sqrt b \neq \sqrt {ab}$ if $a$ and $b$ are negative.
I have not understood why this was a problem and took the word of my teacher.
Is this something so fundamental that it should be taken for granted or is there a reason why we can't
I'm sorry for this basic question but I would like to understand this concept a bit more .
 A: To disprove a statement, you need only provide one counterexample. We have
$$\sqrt{-1}\sqrt{-1} = -1 \neq 1 = \sqrt{(-1)(-1)}.$$
A: 
$T = \sqrt{a} \times \sqrt{b}$ where $a,b$ are each negative.

If the focus is on Real Analysis, then (for example) $\sqrt{a}$ is undefined, so the equation above is gibberish.
If the focus is on Complex Analysis, and $a < 0, i = \sqrt{-1}$, then $$\sqrt{a} = i\sqrt{|a|}, \tag1$$
so $$T = (i^2) \sqrt{|a|} \times \sqrt{|b|} = -1\sqrt{|ab|}.\tag2$$
It just so happens that when $a,b < 0$, then $(ab) = |ab|.$
In that event, within the Complex Analysis framework, you have
$$T = (-1)\sqrt{(ab)}.$$
Actually, the situation in Complex Analysis is somewhat more convoluted.  In Real Analysis, there is the generally accepted convention that (for example) $\sqrt{4}$ refers to $+2$ rather than $-2$.
In Complex Analysis, it is unclear (at least to me) how generally accepted the convention is that (for example) $\sqrt{-4} = 2i$, rather than $-2i$.
A: The reason for the subtlety is that the square root is a multivalued function, i.e. for any $y$ there is more than one (two, in fact) $x$ such that $x^2 = y$. For example, if $y = 4$, then $x=2$ or $x=-2$, if $y = -9$ then $x = 3i$ or $x = -3i$ where $i$ is the imaginary unit.
In order to see how the multivalued nature of the square root undermines the product rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$, recall that every complex number can be written in the polar form as $z = re^{i\theta}$ for a non-negative $r$ and an angle $\theta$ called the argument. This notation makes it clear that raising $z$ to a power $z^a$ corresponds to multiplying the angle $\theta$ by the exponent $a$. Thus, taking a square root corresponds to halving $\theta$. Note that $\theta$ is periodic
$$
e^{i\theta} = e^{i(\theta + 2\pi)}.
$$
Halving the angles on the left hand side and the right hand side allows us to arrive at two different square roots: both $e^{i\theta/2}$ and $e^{i(\theta/2 + \pi)}$ are valid square roots of $e^{i\theta}$. Because $e^{i\pi} = -1$ (see Euler's identity), the two square roots differ in sign (c.f. $2$ and $-2$, $3i$ and $-3i$ in the examples above).
The reason that the product rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ fails is that due to periodicity in argument the two sides can "wind around" a different number of times. Using the example from one of the other answers
$$
\sqrt{-1}\sqrt{-1} = \sqrt{e^{i\pi}}\sqrt{e^{i\pi}} = e^{i\pi/2}e^{i\pi/2} = e^{i\pi} = -1 \\
\sqrt{(-1)(-1)} = \sqrt{e^{i\pi}e^{i\pi}} = \color{green}{\sqrt{e^{i2\pi}} = \sqrt{e^0}} = \sqrt{1} = 1.
$$
Note how in the step marked in green we have reduced the angle $2\pi$ to zero. This corresponds to taking the principal value of a multivalued function. If we chose not to do this and instead halved the angle in that step then we would have obtained the same result as in the first equation
$$
\sqrt{(-1)(-1)} = \sqrt{e^{i\pi}e^{i\pi}} = \color{red}{\sqrt{e^{i2\pi}} = e^{i2\pi/2}} = e^{i\pi} = -1.
$$
Note however that the transformation in red is in fact incorrect, because it fails to choose the principal value of a multivalued function.
Thus, the product rule fails because to use square root in algebraic expressions (and to treat it as a function more generally) we force it to take the principal value and that arbitrary choice cannot be made consistently across expressions that wind the argument different number of times, such as $\sqrt{a}\sqrt{b}$ and $\sqrt{ab}$. More concretely, in $\sqrt{ab}$ the product of $a$ and $b$ is allowed to wind a full $2\pi$ argument before being forced to the principal value by square root, but in $\sqrt{a}\sqrt{b}$ each argument is halved before they can add up to a full $2\pi$.
In summary, in an expression like $\sqrt{a}\sqrt{b} = \sqrt{ab}$ the principal value is chosen from among the set of valid square roots at different stages of the calculation and it turns out that exactly when you take the principal value affects the result.

Remark: Note that the principal value of the square root of a non-negative real number does in fact satisfy the product rule, so if $a > 0$ and $b > 0$ then $\sqrt{a}\sqrt{b} = \sqrt{ab}$.
A: According to Varberg, Purcell & Rigdon, the symbol $\sqrt{ \quad}$ means the positive root for a number, so, the property doesn't hold for negative numbers because there's no positive root. That's a possible reason why your teacher gave the property as a default.
On the other hand I would start proving that $\sqrt{a}\sqrt{b} = \sqrt{ab}$ to get more knowledge of the property
Firstly I would say that, $\sqrt{a} = x $ and $\sqrt{b} = y$ and with some algebra we also have the following equalities $x^2 = a$ and $y^2 = b$ then to start the prove
$$
\begin{align*}
x^2    &= a \\
x^2y^2 & = ab \\
\sqrt{x^2y^2} & = \sqrt{ab} \\
\sqrt{x^2}\sqrt{y^2} & = \sqrt{ab} \\
xy & = \sqrt{ab} \\
\end{align*}
$$
On the other hand,
$$
\begin{align*}
x    &= \sqrt{a} \\
x \sqrt{b}  &= \sqrt{a}\sqrt{b} \\
xy  &= \sqrt{a}\sqrt{b} \\
\end{align*}
$$
So, we can conclude that
$$
\sqrt{a}\sqrt{b} = \sqrt{ab} \quad \dagger
$$
The proof shows that in order to the property to hold, squared numbers where used, so, those must be positive numbers.
Hope was useful
