Is this complete partial order?

Question

Is $(\mathbb{N} , \#)$ complete partial order, where $m\#n$ iff $(\exists k \in \mathbb{N})m=kn$. I proved it's partial order. For completeness I take directed subset and I know there is upper bond $1$ but does it mean there is supremum?

Answer 1

HINT: Note that if $m\#n$ then $n\leq m$. Conclude that if $D$ is a directed subset then above every $m\in D$ there are only finitely many points; and recall that a finite partial order always have a maximal element.