$1 = \sqrt{1} = \sqrt{(-1)^2} = \sqrt{-1}\sqrt{-1} = i\cdot i = -1$. I know the mistake is here $\sqrt{(-1)^2} = \sqrt{-1}\sqrt{-1}$ because everything else seems right to me, but I don't understand why?
I guess this is the correct way : $\sqrt{(-1)^2} = |\sqrt{-1}| |\sqrt{-1}|$ ?
On
The fact that $x^2 = y^2$ does not imply $x =y$ (as $x = -5$ and $y = 5$ show means that $\sqrt{k}$ means "the number $v$ so that $v^2 = k$" is not an actual well defined statement (because there isn't necessarily one number) and that it is not necessarily $\sqrt{x^2} =x$ nor that $\sqrt{x}^2 = x$.
And it doesn't follow that $\sqrt{ab} =\sqrt a \sqrt b$.
ANd just like $(-1)^2 = (1)^2 = 1$ we also have two numbers $(i)^2 = (-i)^2 = -1$.
If we go through the proof critically:
$1 = \sqrt 1$ sure that's notation. We define $\sqrt{x}$ to be one of the two values so that $v^2 = x$ and in the case of $x > 0$ we arbitrarily choose the positive one.
$\sqrt {1} = \sqrt {(-1)*(-1)}$. Sure. That's just an existential fact.
$\sqrt{(-1)}\sqrt{(-1)}$. Um. No. It is true that there is a $w$ so that $w^2 = -1$ and that therefore $(w*w)^2 = w^2w^2$ but is doesn't follow that $w$ is the one we designate as $\sqrt{-1}$ nor does it for that $w*w$ is the one we desiganta as $\sqrt{(-1)(-1)}$.
In general it is not true that $\sqrt{ab}=\sqrt a \sqrt b$. It is true if a and b are positive reals, but square root behaves strangely for negative numbers (and more generally, complex numbers). This is the error.
The more difficult part of the issue, if you're interested, is figuring out what square root means for complex numbers. And what people do, in the field, is call it a multi valued function. Basically there are two numbers that, when squared, give you your original number. And trying to do algebra objects like that isn't straightforward or predictable, for the purpose of this conversation.