Mutually disjoint sets of all integer powers?

[Ventrella]

I identified what appears to be a partitioning of all integers > 1 into mutually disjoint sets. Each set consists of an infinite series of integers that are all the powers of what I am calling a "root" r (r is an integer that has no integer roots of its own, meaning: there is no number x^n that equals r, where x > 1 and n > 0).

For example: here are the first few integers of the first 5 sets:

2^n = 2, 4, 8, 16, 32...
3^n = 3, 9, 27, 81...
5^n = 5, 25, 125...
6^n = 6, 36...
7^n = 7, 49...

These roots include all the prime numbers, but they also include some composites. Analogous to how the primes are fundamental to multiplication, these roots are fundamental to exponentiation.

I am curious if there is an official name for these sets. Have I used proper definitions and terms? Is my assumption correct that these are mutually disjoint sets, the union of which are all the positive integers?

Thank you!

 

[andrewkirk]

The terminology is that each of the sets is the orbit of its root (which would be called a 'generator' in group theory), where we consider the multiplicative group of positive integers as acting on itself.

Yes your assumption is correct. The union of all the orbits is all the integers because any integer that is not in another orbit is the root of its own orbit.

To see that the orbits are disjoint, we proceed as follows.
If two orbits intersect then, considering the prime factorisation of an element in the intersection, and using the uniqueness of prime factorisations, we see that the roots of the two orbits must have the same set of prime factors
Let the prime factorisations of the roots be ##p_1^{a_1}...p_m^{a_n}## and ##p_1^{b_1}...p_n^{b_n}##, where all ##a_i## and ##b_i## are positive integers. The set of all ##a_i## must be coprime (GCF=1), because if the GCF is ##k>1## then the item we thought was the root is the ##k##th element in the orbit of the lower integer ##p_1^{a_1/k}...p_m^{a_n/k}##. the same goes for the set of ##b_i##.

The ##k##th elements of the two orbits are ####p_1^{ka_1}...p_m^{ka_n}## and ##p_1^{kb_1}...p_n^{kb_n}##. Let the first element in the intersection of the orbits be the ##j##th element of the first orbit and the ##k##th element of the second orbit and assume WLOG that ##k>j##. Then we must have ##ja_i=kb_i## for all ##i##, so that ##b_i=a_i\frac jk##.

Let ##c/d## be the form of the fraction ##j/k## that has all possible cancellations made, so that ##c,d## are coprime. Then we have ##b_i=a_i\frac cd## for all ##i##. So it must be the case that ##d## divides ##a_i## for all ##i##. Hence the ##a_i## are not coprime, contrary to assumption. Hence by contradiction, the intersection of the two orbits must be empty.

 

[WWGD]

I identified what appears to be a partitioning of all integers > 1 into mutually disjoint sets. Each set consists of an infinite series of integers that are all the powers of what I am calling a "root" r (r is an integer that has no integer roots of its own, meaning: there is no number x^n that equals r, where x > 1 and n > 0).

For example: here are the first few integers of the first 5 sets:

2^n = 2, 4, 8, 16, 32...
3^n = 3, 9, 27, 81...
5^n = 5, 25, 125...
6^n = 6, 36...
7^n = 7, 49...

These roots include all the prime numbers, but they also include some composites. Analogous to how the primes are fundamental to multiplication, these roots are fundamental to exponentiation.

I am curious if there is an official name for these sets. Have I used proper definitions and terms? Is my assumption correct that these are mutually disjoint sets, the union of which are all the positive integers?

Thank you!

It seems you just need for your generators to not be powers of other generators ( obvious) but I don't know if this is the only way. Primes by themselves will not be enough, e.g., 6 will not be generated by primes.