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Possible Duplicate:
$i^2$ why is it $-1$ when you can show it is $1$?

I was thinking on the following line of thoughts: $1 = \sqrt{1} = \sqrt{-1 \cdot -1} = \sqrt{-1} \cdot \sqrt{-1} = i^2 = -1$

Of course this is not true, but I was wondering which step in this 'line of thoughts' is forbidden to make?

Thanks for the explanation.

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    [This question](http://math.stackexchange.com/questions/49169/i2-why-is-it-1-when-you-can-show-it-is-1) is almost an exact duplicate; others whose answers you may find helpful include [this one](http://math.stackexchange.com/questions/438/1-is-not-1-so-where-is-the-mistake) and [this one](http://math.stackexchange.com/questions/84436/which-step-in-this-process-allows-me-to-erroneously-conclude-that-i-1).2012-05-15

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$\sqrt{-1 \cdot -1}$ is not equal to $\sqrt{-1} \cdot \sqrt{-1}$. The formula $\sqrt{ab} = \sqrt{a}\sqrt{b}$ is only valid when both $a,b$ are nonnegative real numbers.

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    *know​​​​​​​​​​2012-06-14