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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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  • $\begingroup$ 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. $\endgroup$ – davidlowryduda May 15 '12 at 8:07
  • $\begingroup$ This question is almost an exact duplicate; others whose answers you may find helpful include this one and this one. $\endgroup$ – Brian M. Scott May 15 '12 at 8:10
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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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  • $\begingroup$ Oke I did not no that rule. Thanks :) $\endgroup$ – Lotte Laat May 15 '12 at 8:10
  • $\begingroup$ *know​​​​​​​​​​ $\endgroup$ – Derek 朕會功夫 Jun 14 '12 at 20:32

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