Yea, you're a genius! Now I understand what you tried to do. But I don't know if it will be really fruitfull, but at least sound interesting.

Tell me if I got it right:

You was right about the bijection (only if you first statements hold).

Anyways we have that the sets

and

are really differents but they have interesting links.

Lets re-start with a new version of your definitons: let

Ia)

for fixed

IIa)

for fixed

second step:

Ib)

IIb)

First:
maps

on a subset of the naturals that now we call

and like with the primes there are naturals numbers cutted off.

We want

to be injective (that is your condition

)

but is not invertible because is restricted to

: in fact

is bijective.

Your idea is to extend

to make it invertible adding more elements in the domain, in other words extending it to

(but you want do it adding real numbers)

Finally we have a bijection, in fact

holds!

for example using

for the addition we have:

then we build

inverting the naturals using

maps the naturals only if the domain is

then in our set we have numbers that are not naturals

To be honest I don't know if the extension of the naturals to

is a subset of the reals (as you defined)

but if we have that

preserves the order we can find this result

in our example this lead us to this

so the rank is not a natural number.

My question is: is it rational, irrational? Maybe it is trascendental, but this is beyond my limits.

Second:
About the relations with

now the differencese are clear.

We have that

because

in other words the function IIb) is multivalued, and the set

is the set of all possibles solution, in symbols:

that is equivalent to