Is this an acceptable way to do a proof? I'm going through the introduction to Apostol's Calculus and am trying to prove the theorems that he gives. 
Would this be an example of an acceptable proof?

THM 1.9 If $a\neq (0)$ then $b/a = b \cdot a^{-1}$

Proof:

=$\frac ba = b\cdot a^{-1}$

Multiplying each side by a we get:

$a(\frac ba)=(b \cdot a^{-1}) \cdot a $

For the left side we use Thm1.8 and obtain b.

For the right side we use the associative, reciprocal, and identity axioms to obtain:

$=^{1.8} b =(b \cdot a^{-1}) \cdot a =b \dot (a \cdot a^{-1})= b \cdot 1 = b$





1.8 Given a and b with a $\neq 0$, there is exactly one x such taht ax=b. This x is denoted by b/a and is called the quoteient of b and a. In particular, 1/a is also written $a^{-1}$ and is called the reciprocal of a.

 A: Three complete sentences in one paragraph.

By definition, we have $a\cdot a^{-1}=1$. If we multiply by $b$ both sides of this equation, we see that $(a\cdot a^{-1})\cdot b=1\cdot b=b$ and then, associating differently the first member of this last equality, that  $a\cdot (a^{-1}\cdot b)=b$. It follows that $b/a=a^{-1}\cdot b$, which equals $b\cdot a^{-1}$ simply because multiplication is commutative.

The worst problem with what your wrote is that you started with what you wanted to prove and modified it until you got something you knew, while in the immense majority of cases you should start from what you know and get to what you want to know. In many cases, this is just a matter of rewriting, but writing things in your direction is generlly harder and takes much more work to do correctly.
A different way to phrase it is:

As $a\cdot (b\cdot a^{-1})=a\cdot(a^{-1}\cdot b)=(a\cdot a^{-1})\cdot b=1\cdot b=b$, by definition we have $b\cdot a^{-1} = b/a$.

or, if you really must,


As \begin{align}a\cdot (b\cdot a^{-1})&=a\cdot(a^{-1}\cdot b) &&\text{by commutativity of $\cdot$}\\&=(a\cdot a^{-1})\cdot b &&\text{by associativity of $\cdot$}\\&=1\cdot b&&\text{by the definition of $a^{-1}$}\\&=b&&\text{because $1$ is a unit for $\cdot$,}\end{align} by definition we have $b\cdot a^{-1} = b/a$.


A: Once you have proved that $\rm\: a(ba^{-1}) = b,\:$ then you can conclude by saying: by the uniqueness theorem 1.8, for solutions of $\rm\:a x = b,\:$ we  deduce that $\rm\:ba^{-1}\! = b/a.\:$ The proof should explicitly say that it invokes a uniqueness theorem (a standard and powerful way of deducing equalities).
Thus you need only use your axioms to prove that $\rm\: a(ba^{-1}) = b,\:$ then conclude as above. 
As for the attempted proof, it seems to prove that $\rm\: b/a = ba^{-1}\Rightarrow\: b = b,\:$ by multiplying both sides by $\rm\:a,\:$ then applying field axioms. I cannot judge if this is correct proof that $\rm\,b/a = ba^{-1}$ from what little is written. It could possibly denote the following correct proof. Let $\rm\:x' = b/a\:$ and $\rm\:x = ba^{-1}.\:$ Multiplying both by $\rm\ a\ $ shows that $\rm\: ax' = b = ax,\:$ so, by the uniqueness theorem, we deduce that $\rm\:x' = x.\:$  But it could also denote a possibly incorrect proof that starts by already assuming what is to be proved, then attempting to deduce a true equality (a common error).
In any case, as I said at the start, you need only prove that $\rm\:ax =b\:$ for $\rm\:x = ba^{-1}.$ There's no need to prove also that $\rm\:x' = b/a\:$ is a solution, since that is true definition, i.e. by Theorem 1.8 there exists a unique solution, denoted by $\rm\:b/a.\:$ If you've proved that $\rm\:ba^{-1}$ is a solution, then it follows immediately that $\rm\:ba^{-1}\! = b/a\:$ by the uniqueness half of the Theorem 1.8. The proof is done. 
