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Math Help - X^(1/3 -1/3)

  1. #1
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    Question X^(1/3 -1/3)

    What would happen if

    X^(1/3 - 1/3)

    I know that X^(1/3 - 2/3) = X^(-1/3)

    What would happen then. It can't be 0/3.
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  2. #2
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    Quote Originally Posted by azurephantom View Post
    What would happen if

    X^(1/3 - 1/3)

    I know that X^(1/3 - 2/3) = X^(-1/3)

    What would happen then. It can't be 0/3.
    got news for you, it is ...

    x^{\frac{1}{3} - \frac{1}{3}} = x^0 = 1<br />
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  3. #3
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    Quote Originally Posted by azurephantom View Post
    What would happen if

    X^(1/3 - 1/3)

    I know that X^(1/3 - 2/3) = X^(-1/3)

    What would happen then. It can't be 0/3.
    Well, you probably shouldn't write 0/3's - but thats what it is, which is just 0.
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  4. #4
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    Awesome I can use this to kill off some parts of a problem.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by skeeter View Post
    got news for you, it is ...

    x^{\frac{1}{3} - \frac{1}{3}} = x^0 = 1<br />
    Yes!... of course!... but an interesting question: is that true for any real [or even complex...] x?...

    Kind regards

    \chi \sigma
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  6. #6
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    Quote Originally Posted by chisigma View Post
    Yes!... of course!... but an interesting question: is that true for any real [or even complex...] x?...

    Kind regards

    \chi \sigma
    not true for x = 0 ... what do you think about complex values of x?
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  7. #7
    MHF Contributor chisigma's Avatar
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    The problem I intend to propose is the following: for real or complex x and real or complex a is...

    x^{a - a} = \frac {x^{a}}{x^{a}}

    ... that means a number devided by himself. Well!... and whay is not...

    \frac {x^{a}}{x^{a}} = 1

    ... for all x?...

    Kind regards

    \chi \sigma
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