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Is this too Obvious? (additive inverse)

  1. Jul 1, 2009 #1
    1. The problem statement, all variables and given/known data

    In my new Linear Algebra book, it is discussing the basic properties of complex numbers

    It says that "you should verify, using the familiar properties of Real numbers, that addition and multiplication on C satisfy the following properties..."

    ...
    ....
    ....
    additive inverse: for every z[itex]\epsilon[/itex]C, there exists a unique number w[itex]\epsilon[/itex]C such that z+w=0

    I have never been asked to show this before and have always just taken it as a given.
     
  2. jcsd
  3. Jul 1, 2009 #2

    Dick

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    I think you can assume real numbers have additive inverses. Write z=a+bi and say what w must is.
     
  4. Jul 1, 2009 #3
    Right. So let z = (a + bi) then w = -(a + bi)

    That just seems silly to me. Did I really verify anything?

    Anyway, thanks! :smile:
     
  5. Jul 1, 2009 #4

    Dick

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    It's bit better to write that as w=(-a)+(-b)i rather than -(a+bi). That way you've shown w is a real number plus a real number times i. I'll certainly admit it's not a difficult proof.
     
  6. Jul 2, 2009 #5
    I like to think of the additive inverse of a complex number visually. If a complex number is represented as a vector on the Argand diagram, then its additive inverse is the "opposite" of that vector.
     
  7. Jul 2, 2009 #6

    HallsofIvy

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    Yes, but that is irrelevant to this particular problem.
     
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