Finding the R-minimal and R-maximal

Question

For any two integers $a,b$, we say that $a$ divides $b$ (written as $a\mid b$) if $b=ak$ for some $k∈\Bbb Z$. Let $A=\{3,6,7,9,12,14,21,42,252\}$. Consider the partial order relation $R$ on $A$ given by $aRb↔a∣b$. Find all the $R$-minimal and $R$-maximal elements of A. How would I go about finding the $R$-minimal and $R$-maximal elements of $A$ in this question? Would it just be the smallest number (which is $3$) for the $R$-minimal and the biggest number (which is $252$) for the $R$-maximal? If not, how would I go about solving this question?

Answer 1

Yes, $3$ is a $R$-minimal element, but not the only one. Another $R$-minimal element is $7$, since $a\mathrel R7$ never occurs, unless $a=7$. Actually, $3$ and $7$ are the only $R$-minimal elements of $A$: if $n\in A\setminus\{3,7\}$, then $3\mathrel Ra$ or $7\mathrel Ra$.

Can you find the $R$-maximal elements now?