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Negative numbers to non-interger powers

  1. Jul 17, 2006 #1
    Hello,

    How would you determine the value of such numbers as -3^1.5?

    Thanks,
    -scott
     
  2. jcsd
  3. Jul 17, 2006 #2

    Integral

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    [tex] - 3 ^{ 1.5} = - 3^ {\frac 3 2} [/tex]

    Do either the square root or the cube first, either way results in the square root of a negative number. So the solution lies in the imaginary plane.
     
    Last edited: Jul 17, 2006
  4. Jul 17, 2006 #3
    Have you ever worked with complex numbers before? Define i = (-1)^0.5
    -3^1.5 = (3)^1.5 * (-1)^1.5 = 3^1.5 * (-1)^1 * (-1)^0.5 =
    [tex] -i3\sqrt{3}[/tex]. By extension of this procedure one can define negative numbers to any power.
     
  5. Jul 17, 2006 #4
    hmm... idk i would just put it in the caculator :D but i do know that a negative number squared by an odd nuber is X i idk if this helps
     
  6. Jul 18, 2006 #5
    As you wrote it, you do not have a square root of a negative number:
    [tex]-3^{1.5} = -3^{\frac{3}{2}} = - \sqrt{27} = - 3\sqrt{3}[/tex]

    On the other hand,
    [tex](-3)^{1.5} = (-3)^{\frac{3}{2}} = \sqrt{-27} = 3i \sqrt{3}[/tex]

    [tex](-3)^{1.5} = (-3)^{\frac{3}{2}} = \sqrt{-27} = \sqrt{27} \sqrt{-1} = \boxed{3i \sqrt{3}}[/tex]
    not
    [tex]-3i \sqrt{3}[/tex]

    *Also, note that:
    [tex]\forall x < 0, x^k \notin \mathbb{R} \; \text{ if } \, k \notin \mathbb{Q}[/tex]
     
    Last edited: Jul 18, 2006
  7. Jul 18, 2006 #6

    HallsofIvy

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    You posted this 9 times?! I deleted the other 8.

    A good point!


    While "the" square root of a positive real number is define to be the positive root, that is not true for complex numbers, where most functions are "multi-valued". (-3)1.5 has two values,
    [itex]3i\sqrt{3}[/itex] and [itex]-3i\sqrt{3}[/itex]

    ?? or even if k is rational: 1.5 is certainly rational! Did you mean
    [tex]k \notin \mathbb {I}[/tex]?
     
    Last edited: Jul 18, 2006
  8. Jul 18, 2006 #7
    Not really ~~

    The problem was that I couldn't edit the [tex]\LaTeX[/tex] just by using the "Edit" or "Go Advanced". I had to post, delete, and repost :frown: (for some time, so it "appeared" as though I posted nine times)

    I see; so the answer is then
    [tex]\pm 3i\sqrt{3}[/tex]

    I didn't intend to be an 'if and only if' statement...

    Not really//
    my statement there wasn't necessarily for scott_alesk's question;
    it's just a property I like, that's related to complex numbers (~which this thread involves~).

    What I meant to say was that any negative real raised to an irrational power cannot be a real number (the imaginary part of all solutions for [itex]x^k[/itex] is always nonzero if [itex]x<0[/itex] and [itex]k \notin \mathbb{Q}[/itex]).

    (However, is this also the case if [itex]k\notin \mathbb{Z}[/itex] ? But doesn't [itex] (-1)^{1 / 3} [/itex] have one real solution equal to [itex]-1[/itex] ?)
     
    Last edited: Jul 18, 2006
  9. Jul 18, 2006 #8

    0rthodontist

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    Sometimes after you've changed the latex by editing and committed the changes, they don't show until you refresh the page.
     
    Last edited: Jul 18, 2006
  10. Jul 18, 2006 #9

    HallsofIvy

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    My reader is so messed up it doesn't always show the changes even after "refresh"! Sometimes what I do is make the changes, copy it, then delete the post (clicking "physically remove") and paste into a new post.
     
  11. Jul 18, 2006 #10
    ~That's exactly what I did! (..."nine times" :frown:)
     
  12. Jul 19, 2006 #11

    HallsofIvy

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    Note the part about clicking "physically remove"!
     
  13. Jul 19, 2006 #12

    Hurkyl

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    I often open the image itself in a new window, and reload that. It's easy to do in opera, but I don't know about the other major browsers.
     
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