I could not realize yet if this yields a rider where the branches don't interesect each other.
Isn't that obvious? To intersect they would need to have the same ratio of n and m. But there is only a single irreducible representation of n/m. (In other words, if n/m = n'/m' then nm' = n'm, and have the same factors. But since none of the factors in n is in m, they must all be in n', and none of the factors in n' is in m', and must be in n. So n = n'. This relies on the factorization of a number to be unique.)
Isn't that obvious? To intersect they would need to have the same ratio of n and m. But there is only a single irreducible representation of n/m. (In other words, if n/m = n'/m' then nm' = n'm, and have the same factors. But since none of the factors in n is in m, they must all be in n', and none of the factors in n' is in m', and must be in n. So n = n'. This relies on the factorization of a number to be unique.)