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-1 is not 1, so where is the mistake?
Simple Complex Number Problem: 1 = -1

Well, I remembered this after having Algebra II a year ago, is it possible that this is a valid proof that $1 = -1$?

$$ 1 = \sqrt{1} = \sqrt{-1\cdot-1} = \sqrt{-1} \cdot \sqrt{-1} = i \cdot i = i^2 = -1 $$

$$ \therefore 1 = -1 $$

So is this actually fully valid? Or can it be disproved?

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marked as duplicate by MJD, Argon, Douglas S. Stones, Brian M. Scott, Gerry Myerson Sep 9 '12 at 0:41

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See this wikipedia page. –  Envious Page Sep 9 '12 at 0:34
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Actually this one is more like Simple Complex Number Problem: 1 = -1 –  MJD Sep 9 '12 at 0:39
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Do you really think it is even remotely possible that there is a valid proof - and a proof using nothing more than two lines of high school algebra, at that - a valid proof that $1=-1$? –  Gerry Myerson Sep 9 '12 at 0:41
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Duplicate: Square root of 1 is (not) -1 –  Argon Sep 9 '12 at 0:41
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Duplicate: i^2 why is it −1 when you can show it is 1? –  Argon Sep 9 '12 at 0:42

1 Answer 1

I think the problem is between $\sqrt{ -1 \dot{} -1 }$ and $\sqrt{-1} \dot{} \sqrt{-1}$.

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