Ordered integral domain 
If $a>0$ and $b>0$, both $a$ and $b$ are integers, and $a|b$. Use ordered integral domain to prove $a<b$.

I wrote: We can write that $b=an$, where $b$ is some positive integer and
we get $b\left(\frac1n\right)=a$;
$\frac1n < 1$;
that proves $b>a$.
Is this correct?
 A: The problem as stated cannot be proved, e.g. for $a=3$ and $b=3$ we have $a\mid b$ but $a \not < b$.
Assuming the problem was to prove $a\le b$, you should be careful how you conclude that $\frac 1 n \le 1$.  When you write $b=an$ for some integer $n$, you should explain why $n$ cannot be negative or $0$.
With rings and integral domains, you should generally avoid division.  In your case, $\frac 1 n$ is not an integer unless $n=\pm 1$. Writing $\frac 1 n$ means we have to use the rationals or reals.
A more careful proof would be:


*

*$a \mid b$ means $b=an$ for some integer $n$.

*Argue that $n$ is positive, e.g. we know $n\ne0$, otherwise $b=an=a(0)=0$; assuming we've already proven positive $\times$ negative $=$ negative then $n$ cannot be negative.

*Write $b-a = an-a = a(n-1)$.

*Case 1: if $n=1$ then $b=a(1)=a$ so $a\le b$ and we're done.

*Case 2: if $n\ne1$ argue (since $n$ positive) that $n-1 > 0$, so $b-a = a(n-1) =$ positive $\times$ positive $=$ positive, i.e. $b-a>0$, i.e. $a<b$.


To be really careful, using just the axioms of an ordered integral domain, you'd need/prove extra properties of the integers, e.g. that there are no integers between $0$ and $1$.
