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I came across this interesting problem in my book :

What kind of quantities are $\frac{2}{0}$ and $\sqrt{-2}$?

All I know is that $\frac{2}{0}$ is undefined while the second one is a complex quantity $i \sqrt2$.

So the second one's a complex quantity but what about the first one?

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    $\begingroup$ $\sqrt{-2}=i\sqrt 2$ $\endgroup$ – pmichel31415 Apr 16 '16 at 9:31
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    $\begingroup$ @Mandrathax typo fixed. $\endgroup$ – Hisenberg Apr 16 '16 at 9:34
  • $\begingroup$ $\frac{2}{0}$ is never defined, because it's not useful. Defining $\sqrt{-2}$ is useful. See math.stackexchange.com/questions/259584/…. $\endgroup$ – user236182 Apr 16 '16 at 9:39
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    $\begingroup$ To quote Humpty-Dumpty, they mean whatever you want them to mean in a given context. $\endgroup$ – franz lemmermeyer Apr 16 '16 at 10:34
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Well as you said $\sqrt {-2}=i\sqrt 2$. You could also say $\frac 2 0 =\infty$.

In fact, there's a set called the extended real number line which is $\mathbb R\cup\{-\infty,\infty\}$. It's interesting because it gives $<$ a lower and upper bound on $\mathbb R$.

However it's much more tricky than $i$. You have a well defined algebra on complex number, which is not the case on the extended real number line. For instance, what is $\infty-\infty$? $\frac \infty \infty$? $0\times \infty$? Not only are these quantities undefined but they can have different values, sort of. Like $\frac{n^2} n \longrightarrow \infty$ although $n^2\longrightarrow \infty$ and $n\longrightarrow \infty$, but also $\frac n {n^2} \longrightarrow 0$.

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    $\begingroup$ Problematic is also, that if your arithmetic works as expected then $\infty = \frac{2}{0} = \frac{2}{-0} = -\frac{2}{0} = -\infty$ (?) $\endgroup$ – Stefan Perko Apr 16 '16 at 11:22
  • $\begingroup$ Even in using the extended real numbers, $\frac{2}{0}$ is undefined. $\endgroup$ – Morgan Rodgers Apr 16 '16 at 11:33

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