A relation $R$ is defined on $\mathbb{N}$ by $aRb$ if $a|b$ $\in \mathbb{N}$. For $c, d \in \mathbb{N}$, under what condition is $c R^{-1} d$?
My Solution
By definition of $R$, $aRb$ if $a|b \in \mathbb{N}$. Thus, let $x \in \mathbb{N}$, then $b = ax$. Lets now consider $R^{-1}$, the inverse relation of $R$. For an ordered pair $\left(a, b\right) \in R^{-1}$, the ordered pair $\left(b, a\right) \in R$. Thus $\left(a,b\right)\in R^{-1}$ implies that $bRa$. Let $y\in\mathbb{N}$, then $bRa$ means that $a = by$ for $\left(a,b\right)\in R$. Since $b = ax$, it follows that $a = by = \left(ax\right)y = axy$. Dividing through by $a$, we have that $xy = 1$. Since $x, y \in \mathbb{N}, xy = 1$ if and only if $x=y=1$.
Thus, for $\left(a, b\right) = \left(a, axy\right) \in R{-1}, \left(b,a\right)\in R$ if and only if $\left(b,a\right) = \left(ax, a\right)$ and $x = y = 1$. Let $R_{1} \subset R$ defined $R_{1} = \left\{\left(a, a\right) : a \in \mathbb{N}\right\}$, then $R_{1}$ is precisely the relation that has the inverse $R^{-1}$ and $R_{1} = R^{-1}$.
Therefore, in particular, for $c, d \in \mathbb{N}, c R^{-1} d$ if and only if $c = d$.
Is my reasoning sound here; thus I am correct? I am self-learning all of these and a beginner as such so please keep your comments as simple as possible. Thank you.