Help with proof involving field axioms of real numbers I need help with this proof.
If $a\ne0$ then $(a^{-1})^{-1}=a$
Which axiom do I use and what is $(a^{-1})^{-1}$ equal to.
I tried this but I'm stuck $(a^{-1})^{-1}=(\frac{1}{a})^{-1}$ What comes next?
Thanks a lot in advance!
 A: The notation $a^{-1}$ is used to denote the unique element $b\in F$ such that $a\cdot b=b\cdot a=1$ (you should have an "axiom of multiplicative inverse" to this effect). Note the symmetry here, $a$ plays thge same role for $b$ that $b$ plays for $a$. In other words, we immediately get that $a=b^{-1}=(a^{-1})^{-1}$.
A: To show that an element is the inverse of anouther, we need to show that when we multiply them together we get the identity.
First, observe $a^{-1}*a=e$, by the definition of $a^{-1}$. 
Similarly $a*a^{-1}=e$
And hence $a$ is the inverse element for $a^{-1}$, as required.
A: For any b |=0 , b^-1 = the unique c satisfying cb =1 ,the field axioms guarantee that there is exactly one such c . So if you locate a c satisfying cb = 1 then you may conclude that b^-1 = c . Also by the definition , b^-1 = this c, that is b (b^-1)b=1`. Also multiplication is commutative in a field so also b(b^-1)=1
Well now you are given that  a(a^-1)=1  .  So you have located for b =a^-1  the required c,namely c=a . So conclude that (a^-1^-1 =a .
A text or  an exam solution would just say tersely : Since we are given a(a^-1)=1 we have by definition of inverse that (a*-1)^-1 = a . Hope this helps smn 
