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Math Help - I've never understood why this is the case

  1. #1
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    I've never understood why this is the case

    \sqrt{4} \bullet \sqrt{9} = 2 \bullet 3 = 6

    When I do it this way --- \sqrt{4 \bullet 9} = \sqrt{36} = 6

    I get the same answer either way I write it. However if i do this:

    \sqrt{-1} \bullet \sqrt{-1} = 1

    or if I write:

    \sqrt{-1 \bullet -1} Either way I get answer of 1 because

    -1 \bullet -1 = 1 and \sqrt{1} is just one.

    However complex numbers say that i^2 is equal to -1

    How can i^2 be equal to -1 if
    \sqrt{-1} \bullet \sqrt{-1} equals \sqrt{1} or just 1?

    By my logic i^2 should be just 1. So there is obviously a rule I am missing here that would lead me to believe this. If someone could explain to me why \sqrt{4 \bullet 9} = \sqrt{36} = 6 but \sqrt{-1 \bullet -1} doesnt not equal \sqrt{1} or just 1 . It would be much appreciated!

    Sorry in advance if my code got messed up, I havent used it before.
    Last edited by mitchelljk; February 12th 2009 at 09:03 PM.
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  2. #2
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    Quote Originally Posted by mitchelljk View Post
    How can \iota^2 be equal to -1 if
    \sqrt{-1} \bullet \sqrt{-1} equals \sqrt{1} or just 1?
    Simple. \sqrt{-1}\sqrt{-1} does not equal 1.

    Here's where you're going wrong: the property that you are using does not necessarily work for negative values. \sqrt a\sqrt b = \sqrt{ab}\text{ for }a, b\geq0, but not necessarily if a or b is negative.
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  3. #3
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    Awesome thanks. I knew it would be something simple like that.
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  4. #4
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    No problem. Check out this page for an interesting "proof" that 2=1, using this same property incorrectly. They give a nice explanation of the error.
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  5. #5
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    Talking

    Quote Originally Posted by mitchelljk View Post
    ...there is obviously a rule I am missing here....
    The rule is that, by adding the ability to take the square root of negative values, you lose the ability to swap the values around. If you allow for negatives inside the radical, then you must deal with that negative sign (that is, you must do the "i" part) first.
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