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Negative times negative is positive?

  1. Mar 3, 2012 #1
    Probably the stupidest question I have ever asked, but is it possible to prove that the multiplication of two negatives yields a positive? Go easy on me I've asked better questions :D

    BiP
     
  2. jcsd
  3. Mar 3, 2012 #2

    Char. Limit

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    How rigorous a proof are you looking for?
     
  4. Mar 3, 2012 #3
    As rigorous as rigorous gets :D
     
  5. Mar 3, 2012 #4

    micromass

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    What axioms are you accepting??
     
  6. Mar 3, 2012 #5
    These are the usual axioms of real numbers system:

    http://www.gap-system.org/~john/analysis/Lectures/L5.html

    Specifically, look at the 2nd axiom of order.

    EDIT:
    Then, prove the following:
    [tex]
    a < 0 \Rightarrow -a > 0
    [/tex]
    By axiom II.c
    [tex]
    0 > a \Rightarrow 0 + (-a) > a + (-a) \Leftrightarrow -a > 0
    [/tex]
    Q.E.D.

    Then, look at the following:
    [tex]
    a, b < 0 \Rightarrow a \cdot b = (-a) \cdot (-b)
    [/tex]
    Then you have a product of two positive numbers, which by the quoted axiom is positive.
     
  7. Mar 3, 2012 #6
    Ahh, so this is only true for ordered fields then? Is there any number system where the field is not ordered (i.e. does not satisfy axiom II) ?

    BiP
     
  8. Mar 3, 2012 #7

    micromass

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    Sure, [itex]\mathbb{Z}_2[/itex] is not an ordered field.

    If you don't know what it is: it's just the set {0,1} with

    0+0=1+1=0, 1+0=0+1=1
    0*0=0*1=1*0=0, 1*1=1
     
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