Reciprical of i

This is a discussion on Reciprical of i within the A Brief History of Cprogramming.com forums, part of the Community Boards category; i is defined as the square root of -1, (sqr(-1) or (-1)^(.5) ) A friend just asked me what 1/ ...

  1. #1
    Pursuing knowledge confuted's Avatar
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    Question Reciprical of i

    i is defined as the square root of -1, (sqr(-1) or (-1)^(.5) )

    A friend just asked me what 1/i is, so I punched it into the TI-89 and got -i. This would be great, except I can't get the algebra to do that.

    x=1/i
    x^2=1/(i^2)
    x^2=1/(-1)
    x^2=-1
    x=sqr(-1)
    x=i

    i!=-i, so I decided to try it with 2/i

    x=2/i
    x^2=4/(i^2)
    x^2=4/-1
    x^2=-4
    x=sqr(-4)
    x=sqr(-1*4)
    x=2 * sqr(-1)
    x=2i

    Same result...I tried again for 3 and ended up with 3i. I don't see any errors in my algebra, but as far as I know, the TI-89 is infallible. I once saw it described as "a massive beast of knowledge" on these boards, and I agree. Would someone please enlighten me as to the nature of my error?
    Last edited by confuted; 04-12-2003 at 04:36 PM.
    Away.

  2. #2
    The Earth is not flat. Clyde's Avatar
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    When you solve for a square root you get two answers plus and minus hence:

    x=1/i
    x^2=1/(i^2)
    x^2=1/(-1)
    x^2=-1
    x=sqr(-1)
    x=i
    becomes

    x = 1/i
    x^2 = 1/(i^2)
    x^2 = 1/(-1)
    x^2 = -1
    x = +/- sqr(-1)
    x = +/- i

    Since x cannot be +i:

    1/i = i
    1 = i^2
    1 = -1 ..... bzzzt

    it must be - i:

    1/i = -i
    1 = i * -i
    1 = -(i*i)
    1 = -(-1)
    1 = 1
    Last edited by Clyde; 04-12-2003 at 06:46 PM.

  3. #3
    Pursuing knowledge confuted's Avatar
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    Thank you, I feel dumb now As I said, the TI-89 knows all...
    Away.

  4. #4
    Programming Sex-God Polymorphic OOP's Avatar
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    no need to square at all, just rationalize

    1/i

    Multiply by i / i

    i/(-1)

    which is

    -i
    Last edited by Polymorphic OOP; 04-12-2003 at 08:23 PM.

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