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$i^2$ 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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marked as duplicate by Brian M. Scott, Benjamin Lim, Marvis, Rahul, Austin Mohr May 15 '12 at 8:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
The incorrect step is the assumption that $\sqrt{ab} = \sqrt{a} \sqrt{b}$ While this holds for nonnegative numbers, it does not hold for negative numbers. This has to do with the convention that $\sqrt 4 = 2$ instead of $-2$. I should also note that this is a duplicate-post-in-idea, so it will probably be closed. –  mixedmath May 15 '12 at 8:07
    
This question is almost an exact duplicate; others whose answers you may find helpful include this one and this one. –  Brian M. Scott May 15 '12 at 8:10
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1 Answer 1

up vote 2 down vote accepted

$\sqrt{-1 \cdot -1}$ 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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Oke I did not no that rule. Thanks :) –  Lotte Laat May 15 '12 at 8:10
    
*know​​​​​​​​​​ –  Derek 朕會功夫 Jun 14 '12 at 20:32
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