$1 +1$ is $0$ ?​ 
Possible Duplicate:
-1 is not 1, so where is the mistake?
$i^2$ why is it $-1$ when you can show it is $1$? 

So:
$$
\begin{align}
    1+1 &= 1 + \sqrt{1} \\
    &= 1 + \sqrt{1 \times 1} \\
    &= 1 + \sqrt{-1 \times -1} \\
    &= 1 + \sqrt{-1} \times \sqrt{-1} \\
    &= 1 + i \times i \\
    &= 1 + (-1) \\
    &= 1 - 1\\
    &= 0
  \end{align}
$$
I can't see anything wrong there, and I can't see anything wrong in $1+1=2$ too. Clearly, $1+1$ is $2$, but I really want to know where is the incorrect part in the above.

$$\sqrt{ab} = \sqrt{a} \times \sqrt{b}$$ is valid only for non-negative real numbers $a$ and $b$. Hence, the error is in the step $$\sqrt{(-1) \times (-1)} = \sqrt{-1} \times \sqrt{-1}$$
