I have to proove in the most 'primitive' way the following:
if $ab=ac$ then either $a=0$ or $b=c$.
I could think only about the following solution: considering the given $ab=ac$, let's subtract from both sides of the equation $ac$. Thus we get $ab-ac=0$. It means that:
- either both $ab$ and $ac$ equal $0$, so the difference equals $0$,
- or $ab$ and $ac$ represent a real number, so it is possible to multiply $ab$ and $ac$ both by $1/a$. This way we get that $b-c=0$, so $b=c$ as asked,
- or $b\neq c$, so $a$ must be zero.
I'm not very sure about step 3.