1. ## Fields

I'm terrible at these proofs, so I was wondering if someone could please check this proof for me:

Question: Suppose a and b are non-zero elements of a field F. Using only the field axioms, prove that $a^{-1}b^{-1}$ is a multiplicative inverse of ab.

Proof:

$(ab)(a^{-1}b^{-1}) = (aba^{-1})(b^{-1}) =$ by associativity.

$(aba^{-1})(b^{-1}) = (aa^{-1}b)(b^{-1})$ by commutativity.

$(aa^{-1}b)(b^{-1}) = (aa^{-1})(bb^{-1})$ by associativity.

$(aa^{-1})(bb^{-1}) = 1*1$ By the axiom that every non-zero real number has a multiplicative inverse.

$1*1 = 1$ By the axiom that 1 is the multiplicative identity.

Since the multiplicative inverse of an element of a field is unique, this implies $a^{-1}b^{-1}$ is the multiplicative inverse of ab.

How does it look? Do I have to mention at the end that the multiplicative inverse of an element of a field is unique, or would it be good enough if I left that out?

2. Originally Posted by JG89
I'm terrible at these proofs, so I was wondering if someone could please check this proof for me:

Question: Suppose a and b are non-zero elements of a field F. Using only the field axioms, prove that $a^{-1}b^{-1}$ is a multiplicative inverse of ab.

Proof:

$(ab)(a^{-1}b^{-1}) = (aba^{-1})(b^{-1}) =$ by associativity.

$(aba^{-1})(b^{-1}) = (aa^{-1}b)(b^{-1})$ by commutativity.

$(aa^{-1}b)(b^{-1}) = (aa^{-1})(bb^{-1})$ by associativity.

$(aa^{-1})(bb^{-1}) = 1*1$ By the axiom that every non-zero real number has a multiplicative inverse.

$1*1 = 1$ By the axiom that 1 is the multiplicative identity.

Since the multiplicative inverse of an element of a field is unique, this implies $a^{-1}b^{-1}$ is the multiplicative inverse of ab.

How does it look? Do I have to mention at the end that the multiplicative inverse of an element of a field is unique, or would it be good enough if I left that out?
The proof looks fine. I don't think you need to include that the inverse is unique. It is of course unique, but you don't need it directly in your proof. For instance, you don't say that 1 is the unique multiplicative identity, even though it is. But I am just talking about presentation and your proof is exactly right.

3. Thanks for checking