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Field axioms

  1. Jul 8, 2009 #1
    hall,

    i need to prove by using the field axioms that:

    (-a)(-b)=ab, i think i know how to this but I'm very insecure with using those

    axioms cause i want to make sure I'm not using my intuition.

    i tried something like: (-a)(-b)=(-1)(a)(-1)b=(-1)(-1)(a)(b)=ab and i guess it's wrong (in the formal way).

    could someone show me how this mechanism works in this case?

    thanks,

    Omri
     
  2. jcsd
  3. Jul 8, 2009 #2
    Well the way you presented it, you actually used what you were trying to prove in the last equality.

    Can you prove that -(ab) = (-a)(b)? The hint is that you'll need the distributive property. If you can prove this, then (-a)(-b) = ab is the exact same proof.
     
  4. Jul 8, 2009 #3
    so if i undrstood you correctly, i can write
    -a(b+(-b))=0
    -ab+(-a)(-b)=0
    ab-ab+(-a)(-b)=0+ab
    (-a)(-b)=ab
    is that too much stepd for an answer?
    thanks
    Omri
     
  5. Jul 9, 2009 #4
    I think you are implicitly using the fact that (-a)b = -ab, but otherwise that looks fine.
     
  6. Jul 9, 2009 #5
    You will have to cite which axiom you are using at each step.
    You probably have them listed with some numbering, so use that.
     
  7. Jul 9, 2009 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You need to understand that, in the field axioms, "-a" does NOT mean (-1)(a). It means "the additive inverse of a". In order to prove that (-a)(-b)= ab, you need to show that "if x+ a= 0 and y+ b= 0, then xy= ab". You might start by looking at (x+a)(x+b)= 0(0)= 0.

    (Yes, you can then show that if x+a= 0, x= -1(a) where "-1" is defined as the additive inverse of the multiplicative identity but I was talking about using the axioms.)
     
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