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Homework Help: 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

    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


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  4. Oct 12, 2012 #3

    Ray Vickson

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    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]

  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.
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