When we prove the existence part of the division algorithm, namely "Given natural numbers $a, n$, there exist integers $q$ and $r$ such that $a = nq + r$ with $0\leq r < n$" , I constructed the set $S = \{i \in\mathbb{Z}: in > a \}$. Since $an > a$, $a \in S$ so $S$ is not empty. By Well-Ordering Principle, $S$ has a smallest element, call it $j$. Then I tried to say we know $(j-1)n \leq a$ because if $(j-1)n > a$ then $j$ would not be the smallest element of $S$. And then also used this set to prove the existence of the remainder, that $a - (j-1)n \geq 0$ and $a - (j-1)n < n$ since if $a - (j-1)n > n$ then $a > jn$, which is a contradiction of how we defined $j$. My professor wrote that my choice of element was wrong and that I need $jn$, not just $j$, so basically I need a different set. I don't understand why I can't just use $j$?
I tried to search for the answer but I couldn't find any discussing which set to use for the proof. Thank you in advance for your help!