How to prove rational number is a commutative field

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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?
 
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colstat said:
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?

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??
 
Hey, micromass. How do I check field axioms?
Do I say a and b belongs to rational numbers and go through the axiom?
 
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?
 
Thanks so much, I think I got it, that was very helpful! :)