Why do books say $\sqrt{xy}=\sqrt x\sqrt y$ My question is simple. I'm teaching in a college (pre-calculus course) and I'm asking myself why high school books say $\sqrt{xy}=\sqrt x\sqrt y$. This is false (counterexample is $x=y=-1$).
How can I state this rule in a more general and correct manner?
Thanks
 A: $\sqrt{x}:\mathbb{R}_{\ge0}\to\mathbb{R}_{\ge0}$ is defined as the unique $y\in\mathbb{R}_{\ge0}$ so that $y^2=x$ where $\mathbb{R}_{\ge0}=\left\{x\in\mathbb{R}:x\ge0\right\}$.
For this function, and this should be the function in your book, $\sqrt{xy}=\sqrt{x}\sqrt{y}$.

Unfortunately, after complex numbers are introduced, sometimes writers define an inverse of $z\to z^2$ on $\mathbb{C}$ minus some branch cut and call it $\sqrt{z}$. However, for this function, it is not the case that $\sqrt{zw}=\sqrt{z}\sqrt{w}$. 
A: Almost every mathematical statement may become wrong/absurd without its proper context. 
For instance, the apparently harmless
$$ (xy)^2 = x^2 y^2 $$
might not hold if $x$ and $y$ are two matrices or two differential operators. 
So the "right way" is just to be clear about the context, just like in everyday's life.
A: Corrected:
If $x$ and $y$ are allowed to be complex, your proposed counterexample is no such thing:
When $x = -1$ and $y = -1$,
$$\sqrt{xy} = \sqrt 1 = \pm 1$$
and
$$\sqrt x\sqrt y = \sqrt {-1} \sqrt {-1} = (\pm i) (\pm i) = \pm (-1) = \pm 1 $$ 
So the high-school text books are more correct than they seem to be. 

The above interprets $ \sqrt x $ as a multivalued function. If we take $\sqrt x$ as the principle value, the contention is, as you say, false. 
A: In general this assumes $\arg(x) + \arg(y) \leq \pi || \arg(x) + \arg(x) \geq 3\pi || \arg(x) \leq \pi {\;\rm and\;} \arg(y) > \pi || \arg(x) > \pi {\;\rm and\;} \arg(y) \leq \pi $
Hope I am not missing any particular case. I think someone should once and forever give the most general domail of the applicability of the $\sqrt{ x y}=\sqrt{x}\sqrt{y}$ formula! Please don't down-vote, that's not fare...
