How to prove that if $R$ is the divisibility partial order over $S=\mathbb N-\{0,1\}$ then $RR^{-1}=S\times S$?

Question

Let $S=\mathbb N-\{0,1\}$ and let $R$ be a partial order relation over $S$. $(a,b)\in R$ iff $b$ is divisible by $a$ without remainder. Prove that $RR^{-1}=S\times S$.

Intuitively it makes sense, but I don't know how to prove this formally. I know that for $RR^{-1}=\{(a,c)|\exists b \in R((a,b)\in R\land (b,c)\in R^{-1})$. But how do I prove that there's always $b$ for any $a,c$?

Answer 1

According to your definition of $RR^{-1}$, if $(x,y)\in S\times S$, then $x|xy$ and $y|xy$ so that taking $a=x, c=y, b=xy$ we get $(x,y)\in RR^{-1}$. Hence, $S\times S= RR^{-1}$.

Answer 2

Well, what does it mean that $(a,b)\in R$? What does it mean that $(b,c)\in R^{-1}$?

This won't hold for every relation $R$ on $S,$ but it holds for this one because of how $R$ is defined. For example, it is not true that $R^{-1}R=S\times S.$