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How to prove rational number is a commutative field

  1. Sep 10, 2011 #1
    I was wondering how to prove rational number is a commutative field.

    Personally, I didn't think the word "commutative" is necessary, how about others?

    Do I simply prove it is commutative under multiplication?
     
  2. jcsd
  3. Sep 10, 2011 #2

    micromass

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    You'll need to check the field axioms. See http://en.wikipedia.org/wiki/Field_(mathematics)

    That is, you need to check

    • Associativity of addition and multiplication
    • Commutativity of addition and multiplication
    • Existence of the additive and multiplicative identities
    • Existence of the additive and multiplicative inverses for each nonzero element
    • Distributivity

    So;, which one is troubling you??
     
  4. Sep 10, 2011 #3
    Hey, micromass. How do I check field axioms?
    Do I say a and b belongs to rational numbers and go through the axiom?
     
  5. Sep 10, 2011 #4

    micromass

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    OK, let me do an example. Let me check commutativity of addition. Take two fraction a/b and c/d. We must prove that

    [tex]\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}[/tex]

    Indeed:

    [tex]\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}[/tex]

    and

    [tex]\frac{c}{d}+\frac{a}{b}=\frac{bc+ad}{db}[/tex]

    because addition and multiplications in the integers is commutative, we got that the two right-hand sides above are equal. Thus the left-hand-sides are also equal. Thus addition is commutative.

    Can you prove all the other ones?
     
  6. Sep 10, 2011 #5
    Thanks so much, I think I got it, that was very helpful! :)
     
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