$a/b$ is just $a/\mathrm{gcd}(a,b)$. Here's a perversely highbrow explanation, just for fun.
Since the lattice of natural numbers under divisibility has meets of arbitrary nonempty sets (the greatest common divisor extends fine to infinite sets) any unbounded meet-preserving functor $f$ has a left adjoint $g$ given by $g(a)=\mathrm{inf}n:a \leq f(n)$. In the case that $f$ is multiplication by $b$, we have for $a/b$ the least $n$ such that $a$ divides $n\times b$, which is as described in the first paragraph.