1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Misbehaving Imaginary Fractions

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Why is
    [itex]\sqrt{\frac{1}{-1}} \neq \sqrt{\frac{-1}{1}}[/itex]
    when quite obviously
    [itex]\frac{1}{-1} = \frac{-1}{1}[/itex]

    2. Relevant equations
    N/A


    3. 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?
     
  2. jcsd
  3. Oct 12, 2012 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

  4. Oct 12, 2012 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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
     
  5. Oct 12, 2012 #4
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Misbehaving Imaginary Fractions
  1. Imaginary power? (Replies: 2)

  2. Imaginary roots (Replies: 7)

Loading...