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Is root over a^2=modolus of a?

  1. Jun 9, 2007 #1
    Is "root over a^2=modolus of a?"

    mod of a is always positive, but root over a^2 can both be positive or negative. So how these two can be equal to each other?
    I have found this in a math textbook. But I can't convince myself about it.

    It will be very helpful if you give a proof.

    Thnx in anticipation.
     
    Last edited: Jun 9, 2007
  2. jcsd
  3. Jun 9, 2007 #2

    matt grime

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    There are two square roots of any number, agreed, but the symbol x^{1/2} is explicitly chosen to be one of them, and for positive real numbers, the positive one is always chosen.
     
  4. Jun 9, 2007 #3
    so root over (-5)^2=5
    RIGHT?
     
  5. Jun 9, 2007 #4
    Q:When x^{1/2} is a negative real number?
    A: Never!
    So the negative square root of x is expressed by the symbol -x^{1/2}.
    Am I right? :grumpy:
     
  6. Jun 9, 2007 #5
    Why not...
     
  7. Jun 9, 2007 #6

    uart

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    Yes that's correct. This way [tex]\sqrt{.}[/tex] is a function (single valued) and we can always refer to the positive root of [tex]x^2=1[/tex] as [tex]\sqrt{x}[/tex], or the negative root, [tex]-\sqrt{x}[/tex], or both roots, [tex]\pm \, \sqrt{x}[/tex] as we wish.
     
    Last edited: Jun 9, 2007
  8. Jun 10, 2007 #7
    This is a dilemma (I think I spelt it wrong...) I had when calculating ranges of certain functions. Because of the way computer have worked since the 70's functions have been defined (especially the SQRT function), to take only one value - the positive value...
     
  9. Jun 10, 2007 #8

    matt grime

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    Really? You think that functions are defined to be single valued owing to theinvention of computers in the 70s?
     
  10. Jun 10, 2007 #9
    yes, and I have got a few people that agree strongly with this...
     
  11. Jun 10, 2007 #10

    matt grime

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    I doubt we can pin down the first time someone wrote down the formal definition of a fucntion, but it was many decades before the 1970s. One need only look at the notion of branch cuts and Riemann surfaces (c. 1900) to notice that.
     
  12. Jun 10, 2007 #11
    :rofl:
    It is great to get rid of every piece of misconceptions.
    I am a happy man now. :approve:[tex]^2=1[/tex]
     
    Last edited: Jun 10, 2007
  13. Jun 10, 2007 #12

    Office_Shredder

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    Next thing you know, arcsin(x) will only return a value between 0 and 2Pi!
    :grumpy:
     
  14. Jun 10, 2007 #13
    Well yes. This is one thing that really irritates me.
     
  15. Jun 10, 2007 #14

    Hurkyl

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    You mean -pi/2 and pi/2. :wink:
     
  16. Jun 11, 2007 #15

    uart

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    Yes that was a typo :blushing:, I meant to say :
    ... we can always refer to the positive root of [tex]x^2=a[/tex] as [tex]\sqrt{a}[/tex] ...
     
    Last edited: Jun 11, 2007
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