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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.

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  • $\begingroup$ Be careful. It is only possible to multiply $ab$ and $ac$ by $1/a$ if we know that $a \neq 0$. $\endgroup$ – N. F. Taussig Nov 17 '15 at 11:56
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The simplest way, IMO, would be the following:

$-ac$ 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

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  • $\begingroup$ 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. $\endgroup$ – Ami Gold Nov 16 '15 at 21:07
  • $\begingroup$ 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 $b=0$ $\endgroup$ – Alan Nov 16 '15 at 22:01
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a is either equal to 0 or not.if a is 0 then your are done. if a is not 0 then you can devide by a and get b=c

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