<s>[INST] $1 +1$ 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.
 [/INST] We have that $\sqrt{-1} \times \sqrt{-1} = (\sqrt{-1})^{2} = -1$ but $\sqrt{-1 \times -1} = \sqrt{1} = 1$. So $\sqrt{-1 \times -1} \neq \sqrt{-1} \times \sqrt{-1}$ which is the error.
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