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Math Help - Why is (-2)^(-2)^(-2) an imaginary number?

  1. #1
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    Why is (-2)^(-2)^(-2) an imaginary number?

    I'm kind of curious.. Shouldn't (-2)^(-2)^(-2)= (-2)^0,25..?
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  2. #2
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    Quote Originally Posted by narko View Post
    I'm kind of curious.. Shouldn't (-2)^(-2)^(-2)= (-2)^0,25..?
    It is. At least, that's how a calculator interprets it. And (-2)^(1/4) = 0.841 + 0.841i (correct to three decimal places).

    But another interprettation might well be [(-2)^(-2)]^(-2) = (1/4)^(-2), which will give a different answer.
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    hmm, so negative numbers cannot be exponentiated by fractions (like -10^0,5=imaginary number).. well OK, thanks. We haven't learned yet how to calculate things like 10^0,5 without calculators or programs yet. Thank you, again.

    PS. Heh, these are interesting numbers. So basically (-2)^(-2)^(-2) have multiple answers depending on how u interpret it (I always thought you had to solve these issues in "(-2)^(-2)^(-2)= (-2)^0,25" this way, and not this; "0,25^-2" way)..
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  4. #4
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    Quote Originally Posted by narko View Post
    hmm, so negative numbers cannot be exponentiated by fractions (like -10^0,5=imaginary number).. well OK, thanks. We haven't learned yet how to calculate things like 10^0,5 without calculators or programs yet.
    Wrong.

    If n is odd, then (-ve)^1/n = -ve. eg. (-8)^1/3 = -2. (There are two other answers that are both non-real).

    If n is even, then (-ve)^1/n is not real. eg. (-4)^1/2 = -2i. (There are two other answers that are also non-real).
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    Quote Originally Posted by mr fantastic View Post
    But another interprettation might well be [(-2)^(-2)]^(-2) = (1/4)^(-2), which will give a different answer.
    The standard interpretation is (-2)^[(-2)^(-2)].

    Reference: Wikipedia - Tetration

    Quote Originally Posted by mr fantastic View Post
    If n is even, then (-ve)^1/n is not real. eg. (-4)^1/2 = -2i. (There are two other answers that are also non-real).
    Is this a typo? I thought the only solutions were 2i and -2i.
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  6. #6
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    hmm, OK. I think I understand this allot better. for example:

    10^1/5 solving this would need:

    x^5=10
    and then u use the 5√ 10 to solve it.

    Thinking of it, we HAVE went through this. heh I just forgot it. 10^2/6= 3/1√10 and so on.

    And yeah, now that I can see this it is only logical that even exponent fractions are imaginary (square root of -1?) while odd ones are not (3√(-1)=-1*-1*-1=-1).
    Last edited by narko; August 13th 2010 at 08:03 AM. Reason: ahh I forgot
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    Hello, narko!

    \text{Why is }\,(-2)^{(-2)^{(-2)}}\:\text{ an imaginary number?}

    \text{Shouldn't }\:(-2)^{(-2)^{(-2)}} \:=\:(-2)^{0.25}\;?

    You are correct . . .

    Now how do you take the 4th root of a negative number?

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