- #1

- 56

- 0

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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- #1

- 56

- 0

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

Do I simply prove it is commutative under multiplication?

- #2

- 22,178

- 3,301

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??

- #3

- 56

- 0

Do I say a and b belongs to rational numbers and go through the axiom?

- #4

- 22,178

- 3,301

[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?

- #5

- 56

- 0

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

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