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Meaning of the square root and its contradiction

  1. Jun 20, 2010 #1
    Is square root a function in this way?

    [tex]f:\mathbb C\rightarrow\mathbb R^+[/tex]

    However contradiction can be drawn:

    [tex]\sqrt{x^2}=|x|\text{ and } i^2=-1[/tex]

    [tex]\sqrt{-1}=1[/tex] ??

    What is the problem in the definition or the deduction?
    Last edited: Jun 20, 2010
  2. jcsd
  3. Jun 20, 2010 #2


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    No, it's not- the square root function is from [itex]\mathbb C[/itex] to [itex]\mathbb C[/itex]. And functions, on the complex numbers, are not in general "single valued".

    The problem is with your definition, which is wrong.
  4. Jun 20, 2010 #3
    That's a quite narrow view. It might not be the most mathematically useful way, but you might as well define Sqrt to be the principal value. Just as some people use Ln() instead of ln(). This makes writing out equations much easier because you have actual functions.

    In this case the answer to the question is
    [tex]\sqrt{x^2}\neq |x|[/tex]
    for complex numbers!
  5. Jun 20, 2010 #4
    What does it mean for function not being single-valued? Does it violate the definition of function?
    Is it due to the definition of absolute value on complex number?
  6. Jun 20, 2010 #5
    I suppose you could say that. A very important identity for complex numbers is [tex]x \bar{ x } = | x |^2[/tex], so the correct identity relating square roots and absolute value (usually called modulus when extended to complex numbers) is [tex]| x | = \sqrt{ x \bar{ x } }.[/tex] The reason that you're familiar with the identity [tex]| x | = \sqrt{ x^2 }[/tex] for real numbers is that for a real number x, [tex]x=\bar{x}[/tex], so [tex]x\bar{x}=x^2[/tex].
  7. Jun 20, 2010 #6


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    The definition of "function" for real numbers requires that f(a) be a unique number. To use Gerunuk's phrase, that is too "narrow" a view for complex numbers.
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