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    I know it is not really a study help but i couldn't find any sort of maths debate part.

    i = √-1 = √1/-1 = 1/i = i / i^2 = -i

    What is going on, anyone?
    (I think it might simply be a simple flaw rather like 1/0 is undefined in real numbers although I don't really think so.)
    I would appreciate any opinions and especially a hardcore proof of some sort...
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    How do you go from i/i^2 to -i?


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    (Original post by harribo12345)
    How do you go from i/i^2 to -i?
    \frac{i}{i^2} = \frac{i}{-1} = -i

    Can't see the mistake myself, but I think the mistake has gotten be in the transition between \sqrt{-1} = \frac{\sqrt{1}}{\sqrt{-1}}, since the rest of it checks out.

    Edit: a quick check on Wikipedia shows that the rule \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} only applies to non-negative values of a and b, so there's the mistake. :yes:
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    The square root function is multivalued. Both i and -i give -1 when squared.

    Perhaps writing the numbers in exponential form may give some insight:

     \sqrt{-1} = \sqrt{e^{i\pi}} = e^{i \frac{\pi}{2}} = i

    Or

     \sqrt{-1} = \sqrt{e^{-i\pi}} =  e^{-i\frac{\pi}{2}} = -i

    e^{i\pi} and e^{-i\pi} are both valid ways of writing -1, but they given different results when the square root is taken.
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    (Original post by Kerch)
    The square root function is multivalued. Both i and -i give -1 when squared.

    Perhaps writing the numbers in exponential form may give some insight:

     \sqrt{-1} = \sqrt{e^{i\pi}} = e^{i \frac{\pi}{2}} = i

    Or

     \sqrt{-1} = \sqrt{e^{-i\pi}} =  e^{-i\frac{\pi}{2}} = -i

    e^{i\pi} and e^{-i\pi} are both valid ways of writing -1, but they given different results when the square root is taken.
    It's true that i and -i are algebraically the same (hence why when solving complex roots, the 'mirror' complex number is always a solution, also). However, they are quantitatively different: hence why the i = -i statement, in the form it is above, is false. This is easily seeable if you divide the equation by i and seeing what comes out.
 
 
 
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