We have been considering more than one type of "pseudointegers", but here is the most general I think is useful in the EDP (I think we should create another article about EDP on pseudointegers):
A set of pseudointegers is the set of sequences over the non-negative integers that are 0 from some point, with a total ordering that fulfills:
Here multiplication is defined by termwise addition: . For simplicity we furthermore assume that is an increasing sequence. These pseudointegers are called (pseudo)primes.
We would like to find a function that preserves multiplication and order. In general, this function will not be injective. Look at a fixed element . We now define the logarithm in base x by This function fulfills some basis formulas
- logx(x) = 1
- logx(yz) = logx(y) + logx(z)
- logx(z) = logy(z)logx(y)
Proof: 1: We have , so . If a > b then xa > xb, since otherwise xb + n(a − b) would be an infinite decreasing sequence. This shows .
2: If a / b > logx(y), a/b can't be in the set we take sup of. So xa > yb.
3: For we have logx(y) = 0 and the theorem is true, so we may assume . If a / b < logx(y) we can find c and d so that a / b < c / d and . Now we have and using ad < bd and that the power function is increasing for (see the proof of 1) we get xa < yb
4: If (2) tells us that x1 > yn for all n. But that we have a infinite bounded sequence which contradicts an axiom of the ordering on the pseudointegers. If (3) tells us that xn < y1 for all n. Again we get a contradiction.
5: First I show . Assume for contradiction that logx(yz) > logx(y) + logx(y). Now we can find a1,a2,b so that , , and . Form (2) we know that and . And thus . Using (3) we get a contradiction with . The proof of the opposite inequality is similar.
6: This is equivalent to , so assume that logx(y) > logx(z). Now we can find . This implies zb < xa < yb and then z < y.
7: First I show that . Assume for contradiction that logx(z) > logy(z)logx(y). Now we can find with , , and . Using (2) we get ya > zc and xc > yb and from this we get xac > yab > zbc and then xa > zb contradicting . The proof of the opposite inequality is similar.
We can now define a function . For this function we have:
Densities of the pseudointegers divisible by y
Let np where denote the nth pseudointeger (here p stands for pseudo, it is not a variable) and let [np] denote the set of the n smallest pseudointegers. We define the density of the pseudointegers divisible by y to be
if the limit exist. Clearly if the limit exist. More generally we can always define
Using that the set is a set of the form [mp] for some m, we get:
From this we get the inequalities: and . In particular if dx and dy exists, then dxy exists and dxy = dxdy. By induction we get that exists and . If we have so in general and and if both densities exists .
If dx = 1 for some we have of all n, and for any y there is a n so that . This gives us which implies that dy exists for any y and dy = 1.
If dx = 0 for some x then for any we know that dy exists and dy = 0: Assume for contradiction that sy > 0. Then for some n, . This implies that . Contradiction.
Assume that dx and dy exists and . Now we know that (I haven't been able to show this without the assumption that dy exists). We know that for :
(notice that log are taken on numbers in (0,1) so they are negative) where log is the usual logarithm on the reals. Similarly we get . This tells us that . From this we get that − log(dx)logx(z) = − log(dx)logy(z)logx(y) = − log(dy)logy(z), so the function defined by log(z) = − log(dx)logx(z) for some where dx exists and , doesn't depend on the x we choose. From this function we define . If dz exists (again, it would be nice to find a proof that it does or to prove that it does have to) we know that .
Density of the set of pseudointegers
If there exists a , we define
if the limit exist. This limit can be greater than 1 and for the integers it is 1. This means that looking only at the structure of X, we can "see if there is too few/many pseudointegers" compared to the integers.
Here I would like to have a proof that d exists under some assumptions. Using the next paragraph this would imply that dx exists for all . I don't think it is enough to assume that for one x, but it might be enough to assume that for some x,y such that logx(y) is irrational.
From d to dx
Let be any function such that and , and such that the density
exists and (in particular this implies that ). We can now find dx:
Notice that in general we don't have but . The above works because we are taking limits.