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 $\displaystyle a^{-1}b^{-1} $ is a multiplicative inverse of ab.

Proof:

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

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

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

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

$\displaystyle 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 $\displaystyle 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?