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SImpe question about square root function

  1. May 6, 2010 #1
    Is it a right mathematical operation (a*b)^0.5 = a^0.5 *b^0.5 when a and b either both are negative or one of them is negative.
  2. jcsd
  3. May 6, 2010 #2


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  4. May 6, 2010 #3


    Staff: Mentor

    If a >= 0 and b >=0, [tex]\sqrt{ab} = \sqrt{a}\sqrt{b}[/tex]
  5. May 6, 2010 #4
    If a <= 0 and b <=0, [tex]\sqrt{ab} = -\sqrt{a}\sqrt{b}[/tex]
  6. May 6, 2010 #5


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    No, that's not true at all.
    [tex]\sqrt{(-2)(-8)} = \sqrt{16} = 4 = \sqrt{2}\sqrt{8}[/tex]
  7. May 7, 2010 #6


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    And finally for the last case, a<0 b[itex]\geq[/itex]0

  8. May 7, 2010 #7

    Char. Limit

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    Actually, if only one of a and b are positive, we don't know if the answer is multiplied by i or by -i, I believe. That non-property is used to show the fallacy in a certain 1=2 argument.
  9. May 7, 2010 #8
    Has this thread something to do with the recent release of Alice in Wonderland?

    Mark44: [itex]\sqrt{(-2)(-8)}=\sqrt{2}\sqrt{8}[/itex] (true) [itex]=(\sqrt{-2}\sqrt{-8})/i^2=-\sqrt{-2}\sqrt{-8}[/itex] (as in Xitami's formula)

    Or setting [itex]b=1,\sqrt{a}\neq 0[/itex]:

    Mentallic: [itex]1=i[/itex]

    Char Limit: [itex]1=\pm i[/itex]
  10. May 7, 2010 #9


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    Of course :yuck: I felt I can venture into this forbidden territory of complex numbers without any issues... I was wrong.

    I'll have some fun here to try and explain where Martin's second expression could've arisen from.

    a>0, b>0



    [tex]=\frac{-1}{i^2}\left( i\sqrt{a}\right)\left( i\sqrt{b}\right)[/tex]



    as required... I suppose :tongue2:

    Disclaimer: The first line, while mathematically sound, isn't allowed to be used in this square root function.

    I'm not sure what you're getting at here. We were assuming a<0 so [itex]\sqrt{a}\neq 0[/itex] is a given.
  11. May 8, 2010 #10
    If by second expression you mean [itex]\sqrt{2}\sqrt{8}[/itex] it was copied from Mark44's post (as was the first expression [itex]\sqrt{(-2)(-8)}[/itex], but for some reason itex doesn't extend the square root sign - it should have read [itex]\sqrt{(-2)(-8)}[/itex]).

    The third expression comes from [itex]\sqrt{-2}=i\sqrt{2},\sqrt{-8}=i\sqrt{8}[/itex] and substitution in the second expression.

    The fourth expression comes from replacing [itex]i^2[/itex] by [itex]-1[/itex] and looks very much like what Xitami said and Mark44's post is claiming to be wrong.

    What I'm getting at is you're claiming [itex]\sqrt{ab}=i\sqrt{a}\sqrt{b}[/itex] when [itex]a<0,b\geq 0[/itex]. Substituing [itex]b=1[/itex] (which is [itex]\geq 0[/itex]) gives, for [itex]a<0[/itex]


    or [itex]\sqrt{a}=i\sqrt{a}[/itex]

    and after division by [itex]\sqrt{a}[/itex] which is possible because, as you point out [itex]a<0[/itex] therefore [itex]\sqrt{a}\neq 0[/itex]

  12. May 8, 2010 #11


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    The "some reason" is that you switched from [ tex] tags to [ itex] tags.
  13. May 8, 2010 #12


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    but for a<0, [tex]\sqrt{a}=i\sqrt{-a}[/tex]
  14. May 8, 2010 #13


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    This isn't right. Neither equality is valid under the given conditions.

    Since the possiblility of b being zero doesn't add anything useful, let me assume that b > 0 for the rest of the discussion, and that a < 0 as before.

    [tex]\sqrt{ab} \neq \sqrt{-a}\sqrt{b}[/tex]
    In the expression on the left, ab < 0 so its square root is imaginary. In the expression on the right, both radicands are positive, so each square root is real and positive, making the product real and positive.

    [tex]\sqrt{-a}\sqrt{b} \neq i\sqrt{a}\sqrt{b}[/tex]
    For reasons already given, the left side above is real and positive. The right side above is also real, but negative. The reason for this is that
    [tex]i\sqrt{a}\sqrt{b} = i^2 \sqrt{-a}\sqrt{b}= -\sqrt{-a}\sqrt{b}[/tex]

    It might be helpful to show this with numbers instead of variables, replacing a with -2 and b with 8.
    [tex]i\sqrt{-2}\sqrt{8} = i^2 \sqrt{2}\sqrt{8}= -\sqrt{2}\sqrt{8}[/tex]

    I think we have succeeded in going well beyond what the OP asked, and might have further confused him/her with a lot of extraneous information. I hope we are done here.
  15. May 8, 2010 #14
    I might very well agree with that. You can divide by [itex]i\sqrt{-a}[/itex] and prove [itex]1=i[/itex] that way if you like.

    But if you want to leave the rabbit hole you'll need to change your original equation.
  16. May 8, 2010 #15
    I thought that was more or less what I said.

    I don't think there was any confusion up to and including Xitami's first (correct) post, but it went downhill from there. All that was required at that point was a correct formula for [itex]\sqrt{ab}[/itex] for real [itex]a,b[/itex] of opposite sign.
  17. May 8, 2010 #16

    Char. Limit

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    I think that what we have here is notation error...

    How about this then?

    For positive real a and b...


    Of course, a similar argument cannot be done for -a and -b because the square root function can't be "split" with two negative radicands. No, it can't.
  18. May 8, 2010 #17
    If you want to avoid confusion it's probably not a good idea to change the meaning of [itex]a[/itex] and [itex]b[/itex] from those in the original question (repeated under).

    For real [itex]a[/itex] and [itex]b[/itex], [itex](ab)^{0.5}=a^{0.5}b^{0.5}[/itex] unless both [itex]a[/itex] and [itex]b[/itex] are negative, in which case [itex](ab)^{0.5}=-a^{0.5}b^{0.5}[/itex].
  19. May 8, 2010 #18

    Char. Limit

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    No, if a and b are both negative, than the result is invalid. And I already proved the result if either a or b is negative.

    So if both a and b are negative, the result is invalid (actually multivalued).

    If either a or b is negative, my result follows above.
  20. May 8, 2010 #19
    From the original question:

    From your post:

    What I was suggesting is that this is likely to lead to confusion.

    By "the result" here, do you mean the result given in the original question or the result I gave in answer?

    If the former then I would agree with you. That is the import of the minus sign in the last expression in my answer.

    If the latter, can you give a specific pair of real values for a and b where you think the answer is incorrect?

    With the usual definition of [itex]z^w[/itex] for complex [itex]z,w[/itex] viz. [itex]z^w=e^{w Log(z)}[/itex] the Log is taken to be the principal log, so the function is single valued, so I think your remarks about the result being multivalued are mistaken if by "the result" in this case you mean one of (ab)0.5, a0.5 or b0.5.
    Last edited: May 8, 2010
  21. May 8, 2010 #20

    Char. Limit

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    I mean your result.


    I only used multiplication and the square root of a positive number. If a and b are negative, the answer is a real, positive number, and your result is invalid. Assuming we don't take the principal root (always possible)...

    [tex]\sqrt{(-2)(-8)}=\sqrt{16}=4 , -4[/tex]

    Which is multivalued. So if the principal root gives you a real, positive number, and the general root gives you multiple values, you won't receive a singlevalued negative real number from your operations.

    (Note: I did post the wrong Wikipedia section... They are talking about complex numbers, not negative numbers. Sorry for that mixup.)
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