0
$\begingroup$

Possible Duplicate:
-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?

  • 0
    Square roots behave slightly differently on complex numbers. This has been covered on the site before.2012-08-22
  • 3
    The issue is that $\sqrt{ab}\neq \sqrt{a}\sqrt{b}$. This question has been asked here before in numerous variations.2012-08-22
  • 2
    $\sqrt{ab} = \sqrt{a}\sqrt{b}$ only holds when $a, b \ge 0.$ Hence $\sqrt{\frac{1}{-1}} \neq \frac{\sqrt{1}}{\sqrt{-1}}.$2012-08-22

1 Answers 1