# Simplest Proof for an Elementary Number Theory Condition

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:

1. either both $ab$ and $ac$ equal $0$, so the difference equals $0$,
2. 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,
3. or $b\neq c$, so $a$ must be zero.

I'm not very sure about step 3.

• Be careful. It is only possible to multiply $ab$ and $ac$ by $1/a$ if we know that $a \neq 0$. – N. F. Taussig Nov 17 '15 at 11:56

$-ac$ from both sides gives you $ab-ac=0$ Left distibutitivity gives you $a(b-c)=0$.
The fact that you're in an integral domain thus gives you either $a=0$ or $b-c=0$, in the latter case, add $c$ to either side to get $b=c • first of all it's the real numbers domain. secondly, I'm not sure if I can use the distributivity here. tried to avoid this use. – Ami Gold Nov 16 '15 at 21:07 • The real numbers are an example of an integral domain, an integral domain is any number system in which$ab=0$implies$a=0$or$b=0\$ – Alan Nov 16 '15 at 22:01