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Possible Duplicate:
-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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    See [this wikipedia page](http://en.wikipedia.org/wiki/Mathematical_fallacy#Power_and_root).2012-09-09
  • 2
    Actually this one is more like [Simple Complex Number Problem: 1 = -1](http://math.stackexchange.com/questions/3210/simple-complex-number-problem-1-1?lq=1)2012-09-09
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    @MJD, I saw that, but I found this different, so I asked.2012-09-09
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    @link which "that"? And they seem very similar to me; have you tried transferring the explanation in the other questions to see if it makes sense here too?2012-09-09
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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 $1=-1$?2012-09-09
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    Duplicate: [Square root of 1 is (not) -1](http://math.stackexchange.com/questions/145364/square-root-of-1-is-not-1)2012-09-09
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    Duplicate: [i^2 why is it −1 when you can show it is 1?](http://math.stackexchange.com/questions/49169/i2-why-is-it-1-when-you-can-show-it-is-1)2012-09-09
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    Duplicate: [-1 is not 1, so where is the mistake?](http://math.stackexchange.com/questions/438/1-is-not-1-so-where-is-the-mistake)2012-09-09
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    @GerryMyerson, sure? Sometimes simplicity is key.2012-09-09
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    Duplicate: [paradoxical answers using 'i'](http://math.stackexchange.com/questions/83379/paradoxical-answers-using-i)2012-09-09
  • 1
    Duplicate:[A contradiction involving exponents](http://math.stackexchange.com/questions/72000/a-contradiction-involving-exponents)2012-09-09
  • 0
    Could someone explain why $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$ is only true for positive numbers?2012-09-09
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    @Link If $x \in \mathbb {C}$ then $\sqrt{x}=\exp\left( \frac{1}{2} \operatorname{Log} x\right)$. However, one must recall that $\operatorname{Log} x = \log |x|+i(\theta+2\pi k)$ for $k=0, \pm 1, \cdots$, i.e. it is a multivalued function. This means that $\operatorname{Log} (z_1 z_2) \neq \operatorname{Log} (z_1) + \operatorname{Log} (z_2)$2012-09-09
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    @argon: is "multivalued function" not an oxymoron2012-09-09
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    @binn Strictly speaking, yes. It is still a used word, however; see [Wikipedia](http://en.wikipedia.org/wiki/Multivalued_function).2012-09-09
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    $\sqrt{-1\cdot-1}\neq\sqrt{-1}\sqrt{-1}$. If we replace $-1$ with $-2$ (just for good measure) and simplify, we arrive at $2\neq\sqrt{-2}\sqrt{-2}$, which seems reasonable enough.2012-11-05

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