text: # How to prove rational number is a commutative field
  
  1. Sep 10, 2011
  
  ### colstat
  
  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. Sep 10, 2011
  
  ### micromass
  
  Staff Emeritus
  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??
  
  3. Sep 10, 2011
  
  ### colstat
  
  Hey, micromass. How do I check field axioms?
  Do I say a and b belongs to rational numbers and go through the axiom?
  
  4. Sep 10, 2011
  
  ### micromass
  
  Staff Emeritus
  OK, let me do an example. Let me check commutativity of addition. Take two fraction a/b and c/d. We must prove that
  
  $$\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}$$
  
  Indeed:
  
  $$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$$
  
  and
  
  $$\frac{c}{d}+\frac{a}{b}=\frac{bc+ad}{db}$$
  
  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?
  
  5. Sep 10, 2011
  
  ### colstat
  
  Thanks so much, I think I got it, that was very helpful! :)