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Question :

Explain this fallacy:

$$-1=\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt{1}=1$$

(Complex Variables Schaum c1:1.128)

Btw, it's not a homework, i'm curious why it can be false?

Well, i'm considering we're not allowed to multiply these two negative numbers under the roots.

But, what is the reason? Could it be proven? Or it's just an agreement?

Jyrki Lahtonen
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user516076
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    Who says that the property $\sqrt{xy}=\sqrt{x}\sqrt{y}$ extends to all real numbers? If you properly want to understand squareroot functions defined on complex numbers, you'll have to understand branch cuts and perhaps even some Riemann Surface theory. – Mathematician 42 Oct 30 '19 at 09:58
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    By evaluating each expression, it becomes clear that the step from the second to the third expression is wrong. As already mentioned , the reason is that the "root-law" does not hold for all real numbers. – Peter Oct 30 '19 at 10:00
  • Think about it this way. We assumed $\sqrt{-1\cdot -1} = \sqrt{-1} \sqrt{-1}$, but that leads to $1=-1$. What does that tell about our assumption? We can be sure that $-1$ and $1$ are different in $\mathbb C$, so it must be that the assumption about the square root 'distributivity' in all of $\mathbb R$ (or $\mathbb C$) is false. – AlvinL Oct 30 '19 at 10:15

1 Answers1

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$\sqrt{(-1)(-1)}$ is not equal to $\sqrt{-1}\cdot\sqrt{-1}$.

ajotatxe
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