Simplify $\sqrt{-3}$ I was reading about this known fallacy
$$
-1 = i^2 = i \cdot i = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = 1
$$
and according to Wikipedia "The fallacy is that the rule $\sqrt{xy} = \sqrt{x}\sqrt{y} $  is generally valid only if both x and y are positive"
So  my question is, how come we can say that $\sqrt{-3} = \sqrt{3}i$ ?. Aren't we applying the same mistake as the fallacy? Like $\sqrt{-3} = \sqrt{(-1)(3)} = \sqrt{-1}\sqrt{3} = \sqrt{3}i$ cannot be since -1 is negative.
Thanks for reading.
 A: First: the fact that the identity $\sqrt a\sqrt b=\sqrt{ab}$ is only valid for nonnegative reals $a,b{}$ does not mean that it is always false for other numbers.
But let's see the proof of the identity to see where it's essencial that $a$ and $b$ are nonnegative:

$$(\sqrt a\sqrt b)^2=\left(\sqrt a\right)^2\left(\sqrt b\right)^2=ab$$
  Therefore, $\sqrt{ab}=\sqrt a\sqrt b$.

Now, let's do this for $a=b=-1$:

$$(\sqrt {-1}\sqrt {-1})^2=\left(\sqrt {-1}\right)^2\left(\sqrt {-1}\right)^2=(-1)(-1)$$
  Therefore, $\sqrt{(-1)(-1)}=\sqrt {-1}\sqrt {-1}$.

The problem arises in the conclusion. The definition of $\sqrt{\:}$ says:
$$\sqrt{x}=y\iff y^2=x\text{ and }y\ge 0$$
The last condition is essential. Many books for high school student say that $\sqrt 4$ has "two solutions" but this is totally wrong. What has two solutions is the equation $x^2=4$, but $\sqrt 4$ is $2$ and not, ever, $-2$.
In the fallacy we are saying in essence that $\sqrt{(-1)(-1)}=\sqrt {-1}\sqrt {-1}$, but this is wrong because $\sqrt {-1}\sqrt {-1}<0$.
