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-1 is not 1, so where is the mistake?
$i^2$ why is it $-1$ when you can show it is $1$?

Can someone please point out what I'm doing wrong here?

$$ \frac{1}{-1} = \frac{-1}{1} \implies \sqrt{\frac{1}{-1}} = \sqrt{\frac{-1}{1}} $$ which should imply that $$ \frac{\sqrt{1}}{\sqrt{-1}} = \frac{\sqrt{-1}}{\sqrt{1}} $$

But the last equality is not true. Is there something really obvious that I'm overlooking here?

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marked as duplicate by J. M., GEdgar, Jennifer Dylan, Ross Millikan, M Turgeon Aug 22 '12 at 17:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Square roots behave slightly differently on complex numbers. This has been covered on the site before. – Asaf Karagila Aug 22 '12 at 16:42
The issue is that $\sqrt{ab}\neq \sqrt{a}\sqrt{b}$. This question has been asked here before in numerous variations. – Alex Becker Aug 22 '12 at 16:42
$\sqrt{ab} = \sqrt{a}\sqrt{b}$ only holds when $a, b \ge 0.$ Hence $\sqrt{\frac{1}{-1}} \neq \frac{\sqrt{1}}{\sqrt{-1}}.$ – user2468 Aug 22 '12 at 16:56

1 Answer 1

Assume that $f \colon A \to B$ is a given function. From $x=y$ you can get $f(x)=f(y)$, but only when $x$ and $y$ belong to $A$. In your example, $f=\sqrt{\cdot}$, $A=[0,+\infty)$ and $B=\mathbb{R}$. But $x=-1/1 = -1 \notin A$, and you can't conclude.

Of course, you may say that $A =\mathbb{C}$, but then $f$ becomes multi-valued.

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