In the book Rings that are Nearly Associative, has the follow definitions:
$1)$ An algebra $A$ is called an $\textit{algebra with division}$ if, for any elements $a, b \in A$, with $a \neq 0$, each of the equations is solveble in $A$:
$$ax=b \qquad ya=b$$
$2)$ If each of this equations has one and only one solution, and $A$ has an identity element, then $A$ is called a $\textit{division ring}$.
I want to prove that every alternative algebra with division is a division ring.
My attempt:
Let $x_1,x_2$ solutions of $ax=b$, then $ax_1=b$ and $ax_2=b$
$ax_1-ax_2=0 \Rightarrow a(x_1 - x_2)=0$
$a(a(x_1-x_2))=a \cdot 0 \Rightarrow a^2(x_1-x_2)=0$
By the left Moufang identity, we have $a^n(x_1-x_2) = 0$, for all $n\geq 3$.
Since $a \neq 0 $, then $x_1-x_2=0$, therefore $x_1=x_2$
The same for $ya=b$.
My attempt is correct?, something is wrong? Thank you!