Possible fake proof of $1= -1$

Question

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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?

Actually this one is more like Simple Complex Number Problem: 1 = -1 — MJD, Sep 09 '12 at 00:39
@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? — MJD, Sep 09 '12 at 00:41
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 09 '12 at 00:41
Could someone explain why $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$ is only true for positive numbers? — Rivasa, Sep 09 '12 at 00:51
@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 $\operatorname{Log} (z_1 z_2) \neq \operatorname{Log} (z_1) + \operatorname{Log} (z_2)$ — Argon, Sep 09 '12 at 00:57
@binn Strictly speaking, yes. It is still a used word, however; see Wikipedia. — Argon, Sep 09 '12 at 01:07
$\sqrt{-1\cdot-1}\neq\sqrt{-1}\sqrt{-1}$. 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. — Tortoise, Nov 05 '12 at 00:51

score 4 · Accepted Answer · answered Sep 09 '12 at 00:41

4

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

answered Sep 09 '12 at 00:41

binn

1 Answers1