False Proof of 1=-1

Question

I found the following proof online and am confused as to where the error occurs.

$1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=(\sqrt{-1})^2=-1$

My guess is that the error occurs here: $\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}$, but I'm not sure how to show that.

Is my guess correct? How might I prove this?

Thank you!

Answer 1

Well,

$$\sqrt{ab} = \sqrt a \sqrt b$$

is only true when $a$ and $b$ are non-negative.

Answer 2

Indeed what you are proving is that in the complex numbers you don't have (in general) $$\sqrt{xy}=\sqrt{x}\sqrt{y}$$ Because you find a counterexample.

Answer 3

For complex numbers $a$ and $b$, it is not in general true that $(ab)^{1/2} = a^{1/2}b^{1/2}$, so when they write $$ \sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)}, $$ this is incorrect. The issue is that $a$ and $b$ must each by nonnegative.