I am struggling to prove that if $a$ is a real number, then
a) $(a^{-1})^{-1} = a$, and
b) $(-a)^{-1} = -a^{-1}$.
I have done the rest of the theorem but it is just these two that are difficult. To prove them I can only use the axioms of multiplication: multiplication is associative and commutative, "one" is a real number, and every non-zero real number has a multiplicative inverse.
a) $a^{-1} \cdot a = 1 \Rightarrow \left(a^{-1}\right)^{-1} = a$. This follows from the definition of the multiplicative inverse.
b) Let $a^{-1} = b$.
Then $ab = 1$, so $(-a)(-b) = ab = 1$.
This implies $(-a)^{-1} = -b = -(a^{-1})$.