$1 +1$ is $0$ ?​

Question

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.

Answer 1

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

Answer 2

You cannot split the square root term since that is only valid for non-negative numbers.

$$\sqrt{-1 \cdot -1} \ne \sqrt{-1}\cdot \sqrt{-1}$$

Answer 3

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.