Need opinion on proof (if $ax=a$, then $x=1$) I haven't written a lot of proofs so I need the opinion of the experts on my proof of a simple proposition.
Here's the various properties I used:
(P10) (Trichotomy law) For every number $a$, one and only one of the
following holds:
(i) $a = 0$,
(ii) $a$ is in the collection $P$,
(iii) $—a$ is in the collection $P$.
(P7) For every number $a$ not equal to $0$, there is a number $a^{-1}$ such that
$a \cdot a^{-1} = a^{-1} \cdot a = 1$.
(P6) If $a$ is any number, then
$a \cdot 1 = 1 \cdot a = a$.
If $ax=a$ for some number $a$ different from $0$, then $x=1$.(Spivak's calculus.)
I consider two cases: $a>0$ or $a<0$.
By definition (Given in Spivak's calculus):
$a>b$ if $a-b$ is in the collection $P$ ($P$ being the collection of all positive numbers.)
$a>0$ because $a-0$ is in collection $P$ by trichotomy law (P10)
So $ax=a$
by P7 $aa^{-1}x= aa^{-1}$
$1x=1$
by P6 $x=1$
Second case:
By definition:
$a< b$ if $b>a$
$a<0$ because $0>a$
Now, we do the same thing as the previous case.
Proposition proven!
Here's Spivak's answer
$1=a^{-1}a=a^{-1}(ax)=(a^{-1}a)x=1x=x$
I guess he does his proof by "construction", beginning by $a=a$, then $aa^{-1}= a*a^{-1}$, etc.
Any opinions? Thank you!

To explain Spivak's proof rigorously, here it is step by step:
Hypothesis: Let $ax = a$, and let $1$ be the identity such that $1a = a = a1$ for all $a$.
Then,
$$\begin{align*}
1 &= a^{-1}a \tag{by definition of inverse}\\
  &= a^{-1}(ax) \tag{since $ax = a$, and symmetry of $=$} \\
  &= (a^{-1}a)x \tag{by associativity} \\
  &= 1x \tag{by definition of inverse} \\
  &= x.  \tag{by definition of identity}
\end{align*}
$$
Therefore, $x=1$.
