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B Why is i^2 equal to negative one and not positive one?

  1. Oct 28, 2016 #1
    Someone please tell me what is wrong with this logic:

    i = √-1

    i2= √-1√-1 = √(-1)(-1) = √+1 = 1

    But also i2 = (√-1)1/2= -1
  2. jcsd
  3. Oct 28, 2016 #2


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    The square root is not unique any more in the complex numbers. The laws that work for positive real numbers don't work for complex numbers in general.
  4. Oct 28, 2016 #3


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    ... and the property of radicals that you used -- ##\sqrt{a}\sqrt{b} = \sqrt{ab}## -- is applicable only if both a and b are nonnegative.
  5. Oct 28, 2016 #4


    Staff: Mentor

  6. Oct 29, 2016 #5
    Thank you all for the response, much appreciated!
  7. Oct 29, 2016 #6


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    The symbol, i, was defined as the imaginary number which, when squared, gives -1. That definition should not be violated. It worked out very well with several other things. It allows every polynomial to be factored completely. It provides a good way to represent rotations in the two-dimensional complex plane with multiplication by complex numbers.
  8. Oct 29, 2016 #7


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  9. Oct 29, 2016 #8


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    How does it come you know all the exotic corners out there? I'm flabbergasted every single time.
    But that's not what I wanted to say. I like to take the chance to thank you for your Insight on the subject. I've linked it now for the third or forth time (hoping it will be read). Very useful.
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