Misbehaving Imaginary Fractions

1. Oct 12, 2012

a.powell

1. The problem statement, all variables and given/known data
Why is
$\sqrt{\frac{1}{-1}} \neq \sqrt{\frac{-1}{1}}$
when quite obviously
$\frac{1}{-1} = \frac{-1}{1}$

2. Relevant equations
N/A

3. The attempt at a solution
By the above inequality, I mean when one calculates $\sqrt{\frac{1}{-1}}$ as $\frac{\sqrt{1}}{\sqrt{-1}}$, and $\sqrt{\frac{-1}{1}}$ as $\frac{\sqrt{-1}}{\sqrt{1}}$. Is it just that the "rule" which is taught at school for taking roots of fractions just doesn't always apply? That'd seem a little arbitrary.

This isn't a homework question, it's just something that popped into my mind while lying in bed, interspersed among much more interesting thoughts on isospin. I'm a final-year mathematician at university and I can't believe I'm asking such a basic question, but my migraine-addled brain won't let me work out why the above is true. It seems so trivial and pathetically simple that I must be missing something really obvious. Could someone shed some light and save me from my shame in asking such a question?

2. Oct 12, 2012

vela

Staff Emeritus
3. Oct 12, 2012

Ray Vickson

The rule you learned is not universally true: the identity
$$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$$
is true for b > 0 and a ≥ 0. As you have seen, it is not true when a < 0 and/or b < 0. The same can be said for the identity
$$\sqrt{ab} = \sqrt{a} \sqrt{b}$$
and for
$$\log(ab) = \log(a) + \log(b),$$
etc.

RGV

4. Oct 12, 2012

a.powell

So it really was a case of the taught rule not applying everywhere, I'm a little disappointed it was that simple to be honest. Oh well, thank you both for saving me, back to bed for more thoughts on isospin.

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