I've only recently started learning about imaginary numbers, and there is one thing I cannot really wrap my head around:
$i^2 = i*i = {\sqrt{-1}} * {\sqrt{-1}} = {\sqrt{(-1) * (-1)}} = {\sqrt{1}} = 1$
I'm aware that by definition $i^2 = -1$, but as far as I know, even definitions have to obey mathematical rules.
So where is the mistake? :)
EDIT:
Thank you all, seems like the wrong bit is ${\sqrt{-1}}*{\sqrt{-1}}$
Cheers :)
$\endgroup$ 41 Answer
$\begingroup$There's a confusion about how to treat $\sqrt{x}$, here.
If we want a function called "square root," then this function will not be very well-behaved - in general, we will not have $\sqrt{ab}=\sqrt{a}\sqrt{b}$ for $a, b$ not both positive. This will break the third equality in your argument.
If we want to talk about "square root" as a relation, then we will always have "$\sqrt{a}\sqrt{b}=\sqrt{ab}$," in the sense that, if $x$ and $y$ are square roots of $a$ and $b$, respectively, then $xy$ is a square root of $ab$. But we won't have statements like "$\sqrt{1}=1$" - the last equality in your argument will break.
Generally, when we speak of square roots, we are talking about the function version, so "$\sqrt{1}=1$" is correct, but $\sqrt{ab}=\sqrt{a}\sqrt{b}$ need not be.
EDIT: Note that what your argument has done is show - perfectly rigorously! - that there is no function $f(x)$ defined on all of $\mathbb{C}$, satisfying
$f(x)^2=x$,
$f(xy)=f(x)f(y)$, and
$f(1)=1$.
