The set $M=\{r\in\mathbb Q:\ r>1\}$ is a commutative semigroup with respect to the multiplication. For any integers $a>b\ge1$, we clearly have $$\frac ab=\prod_{n=b}^{a-1}\frac{n+1}n.$$ So the set $\{(n+1)/n:\ n=1,2,3,\ldots\}$ generates the semigroup $M$.
A positive integer $n$ is called practical if each $m=1,\ldots,n$ can be written as the sum of some distinct divisors of $n$. All practical numbers greater than one are even. See https://oeis.org/A005153 for the sequence of practical numbers. Motivated by Questions 476578 and 476951, we pose the following new question involving practical numbers.
Question. Does the set $$S=\left\{\frac{n+1}n:\ n\in\mathbb Z^+,\ \text{and}\ 2n\ \text{is practical}\right\} =\left\{\frac{q+2}q:\ q>1\ \text{is practical}\right\}$$ generate the multiplicative semigroup $M$?
We believe that the answer is positive. Moreover, we conjecture that any $r\in M$ can be written as a product of finitely many distinct elements of $S$. For example, $6$ is the product of the six distinct numbers $$\frac21,\ \frac32,\ \frac 43,\ \frac 54,\ \frac 98,\ \frac{16}{15}$$ which belong to the set $S$ since $2,\,4,\,6,\,8,\,16,\,30$ are all practical.
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