Let $a,b \in \mathbb N$ be coprime. Prove that for all $n\in \mathbb N$ such that $n>ab$ there are $r,s\in \mathbb N$ such that $n=ra+sb$.
I'm really stuck on this problem. I know that since $(a,b)=1$, $ja+kb=1$ for some $j,k\in \mathbb Z$. Also, $n=j'a+k'b$ has an assured solution if $(a,b)|n$.
What am I missing?
Assume you have found integers $r$ and $s$ such that $n = ra + sb$. If they are positive we are done. Otherwise one of them, say $r$, has to be positive. Also $r > b$, since $ra \ge n > ab$. But then $n = (r-b)a + (a+s)b$. Repeat this process until $s$ becomes positive.