Proof that $\sqrt{x}=-\sqrt{x}$

Question

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

The idea popped into my head while I was evaluating an integral. I have a feeling that I made some obvious mistake because the "proof" is so simple, but I don't see any flaw. Of course, there must be a flaw somewhere. What is it?

Answer 1

Most fake proofs involving square roots rest, at some point, on the false identity $$\sqrt{a^2} = a$$ This identity seems natural and true, which is why it fools us. The correct identity is $$\sqrt{a^2} = |a|$$

Answer 2

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

Perhaps you are making the mistake of thinking $(\sqrt{(-1)^2} = (-1)^{2/2} = (-1)^{1} = -1$, but this too is mistaken because $(-1)^{1/2}$ does not exist in the real numbers, so $\Big((-1)^{1/2}\Big)^2$ does not exist, whereas $\Big((-1)^{2}\Big)^{1/2} = 1$ does indeed exist.

Answer 3

$\sqrt{(-1)^2 \cdot x}=\sqrt{(-1)^2}\cdot \sqrt{x}$and $\sqrt{(-1)^2}=-1$ are wrong.