Misbehaving Imaginary Fractions

[a.powell]

Homework Statement

Why is
[itex]\sqrt{\frac{1}{-1}} \neq \sqrt{\frac{-1}{1}}[/itex]
when quite obviously
[itex]\frac{1}{-1} = \frac{-1}{1}[/itex]

Homework Equations

N/A

The Attempt at a Solution

By the above inequality, I mean when one calculates [itex]\sqrt{\frac{1}{-1}}[/itex] as [itex]\frac{\sqrt{1}}{\sqrt{-1}}[/itex], and [itex]\sqrt{\frac{-1}{1}}[/itex] as [itex]\frac{\sqrt{-1}}{\sqrt{1}}[/itex]. 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?

 

[vela]

The usual rules you're using only apply for positive real numbers.

http://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities

 

[Ray Vickson]

Homework Statement

Why is
[itex]\sqrt{\frac{1}{-1}} \neq \sqrt{\frac{-1}{1}}[/itex]
when quite obviously
[itex]\frac{1}{-1} = \frac{-1}{1}[/itex]

Homework Equations

N/A

The Attempt at a Solution

By the above inequality, I mean when one calculates [itex]\sqrt{\frac{1}{-1}}[/itex] as [itex]\frac{\sqrt{1}}{\sqrt{-1}}[/itex], and [itex]\sqrt{\frac{-1}{1}}[/itex] as [itex]\frac{\sqrt{-1}}{\sqrt{1}}[/itex]. 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?

The rule you learned is not universally true: the identity
[tex] \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}[/tex]
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
[tex] \sqrt{ab} = \sqrt{a} \sqrt{b}[/tex]
and for
[tex] \log(ab) = \log(a) + \log(b),[/tex]
etc.

RGV

 

[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.