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Basic Error?

  1. Mar 7, 2008 #1
    Could someone please sort out this contradiction which must come from some very basic error - but where and which error? If you raise -3 to the power of 1/2, this gives the square root of -3 which has no real value, but if you raise it to the power of 2/4, you are finding the fourth root of -3 squared, which is the fourth root of +9 which is real. What is wrong here?

    Thanks in anticipation.
     
    Last edited: Mar 7, 2008
  2. jcsd
  3. Mar 7, 2008 #2
  4. Mar 7, 2008 #3

    HallsofIvy

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    Roughly speaking, the "laws of exponents" do not apply to complex numbers in the same way they apply to real numbers. But certainly look at the link Diffy mentioned.
     
  5. Mar 7, 2008 #4
    Thanks for that, but I do no fully follow the replies in the link. One reply says that when you write p[tex]^{a/b}[/tex], a and b must be mutually prime. The demonstration that p[tex]^{a/b}[/tex]= [tex]\sqrt{p^{a}}[/tex]seems to work whether they are or not.

    e.g. p[tex]^{a_{1}/b}[/tex] X p[tex]^{a_{2}/b}[/tex] ....X p[tex]^{a_{b}/b}[/tex] = p[tex]^{ab/b}= p^{a}[/tex].

    Thus p[tex]^{a/b}[/tex] = [tex]\sqrt{p^{a}}[/tex]. Where is the requirement there that they should be mutually prime? Or is it that the requirement is created by the need not to get into the contradiction?
     
    Last edited: Mar 7, 2008
  6. Mar 7, 2008 #5
    I am not sure where that person was going with that reply. Certainly one can compute an answer for [tex]64^{2/6}[/tex].
    I think that what he was getting at is that you are in tricky waters when you start using equalities. For example [tex]64^{2/6}[/tex] and [tex]64^{1/3}[/tex] aren't necessarily equal. Consider the polynomials that these two expressions are solutions to, Sqrt(x^6) = 64 and x^3 = 64. For the first wouldn't you say the answer is 4 or -4? and for the second there is only one answer 4. Therefore how could you say the two statements are equal?
     
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