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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.

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    Note that 1+1=0 mod 2, in which case every equality in the post can be interpreted as a true statement (we can take $i=1$ in this context). This shows that the error in the argument is not intrinsic to the algebra, but depends on properties of the ground field/context in which the equations are set.2012-06-14

3 Answers 3

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

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    @Derek: because you get a contradiction otherwise? In the complex numbers it's only valid when $a,b$ have arguments in $[0,\pi)$ due to the standard branch cut. (Technically I think exactly one of $a,b$ may also have argument $\pi$, i.e. be on the negative real axis).2012-06-14
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You cannot split the square root term since that is only valid for non-negative numbers.

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

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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.