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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 :)

fygesser
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    Because $\sqrt a \cdot \sqrt b = \sqrt{ab}$ may not hold for negative $a$ or negative $b$. – peterwhy Sep 24 '15 at 20:59
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    @fygesser Are you sure that $\sqrt{1} = 1$? Why not $\sqrt{1} = -1$ (because $(-1)^2 = 1$)? Do you see the flaw? As always, $\sqrt{-1}$ is very dangerous, especially when dealing with complex numbers. – flawr Sep 24 '15 at 21:01
  • @flawr But $\sqrt x$ for non-negative $x$ is the principal square root of $x$, and $\sqrt 1 = 1$. – peterwhy Sep 24 '15 at 21:02
  • Certain mathematical rules do not apply to imaginary numbers. You can actually prove contradictions if you naively apply all normal mathematical rules to imaginary numbers- which is one of the reasons they were abandoned for a good few decades after they were first used, because mathematicians saw them as useless. – Parthian Shot Sep 24 '15 at 21:03

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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

  • $f(1)=1$.

Noah Schweber
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    You can see this issue in the comments - peterwhy is speaking of $\sqrt{x}$ as a function, and flawr is speaking of $\sqrt{x}$ as a relation. – Noah Schweber Sep 24 '15 at 21:06
  • oh ok. I would blame people who call $\sqrt\cdot$ function for non-negative real input as "square root" instead of principal square root, hiding the fact that there are up to two square roots of a number, and at the same time leading some other people to think that $\sqrt 4 = \pm 2$ is true. – peterwhy Sep 24 '15 at 21:16
  • @peterwhy I see no reason to blame people who call the $\sqrt{\ }$ function defined on the non-negative real halfline, the square root function, and no reason either to think that $\sqrt{4}=\pm2$ since $\sqrt{\ }$ is real valued and $\pm2$ is not a real number. – Did Sep 24 '15 at 21:30