Doubt about Artin’s Algebra Theorem 2.3.3 Proof

I was able to follow most of the proof but I don’t understand how the author concludes that $r=0$ at the final part. I would appreciate it if someone could clarify. Thanks

If $r$ wasn't 0, it would be a positive integer, but the inequality $0\leq r < a$ tells us that it would be a positive integer smaller than $a$. However, since we chose $a$ to be the smallest positive integer in $S$ and $r\in S$, this yields a contradiction.

We know that $r = n-qa$ is an element of $S$ by the fact that $S$ is a group. If for a contradiction $r > 0$, then we know that $r$ will satisfy the compound inequality $0 < r < a$. However, this means that $r$ is an element of $S$ that is smaller than $a$, a contradiction to how we chose $a$ to begin with. Hence, it must be the case that $r = 0$.

The proof states that $n=qa+r$, where $n\in S$ and $a$ is the smallest positive integer in $S$. Then $qa\in S$, since $S$ is a subgroup.

Then $r=n-qa$ lies also in $S$ as $S$ is a subgroup. But $r$ is the remainder of dividing $a$ into $n$ and so by def. $0\leq r <a$ But $a$ is smallest with the properties $a>0$ and $s\in S$. This forces $r$ to be $0$.