How to prove rational number is a commutative field

[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?

 

[micromass]

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??

 

[colstat]

Hey, micromass. How do I check field axioms?
Do I say a and b belongs to rational numbers and go through the axiom?

 

[micromass]

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?

 

[colstat]

Thanks so much, I think I got it, that was very helpful! :)