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Here's a little bit of a strange question: In class, my math teacher decided to let us take a break from the usual class to solve a mathematical puzzle:

\begin{align} i & = i \\ i^2 &= i^2 \\ \sqrt{-1} &= \sqrt{-1} \\ \sqrt{\frac{-1}{1}} &= \sqrt{\frac{1}{-1}} \\ \frac{\sqrt{-1}}{\sqrt{1}} &= \frac{\sqrt{1}}{\sqrt{-1}} \\ i &= \frac{1}{i} \end{align}

I assume that the transition from line 4 to 5 is wrong, but I don't see how :p.

Does anyone care to explain?

Eric Lee
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    Every nonzero complex number has two square roots. – Angina Seng Oct 15 '17 at 06:15
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    This exact thing has been posted before. See the link above/below. If that specific post is not identical enough of a duplicate for you, consider going through the other linked questions/answers attached to that post. The short explanation is that $\sqrt{\frac{a}{b}}\neq \frac{\sqrt{a}}{\sqrt{b}}$ in general unless $a$ and $b$ are both positive. – JMoravitz Oct 15 '17 at 06:15

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The function $t \mapsto \sqrt{t}$ is defined on $\mathbb{R}_{\geq 0}$. You can extend this notion to the whole real line (actually, to $\mathbb{C}$), but you lose some properties. Particularly, you lose the product rule you're using to "prove" the statement (the fourth equality to be precise).

qualcuno
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