Pseudointegers

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 [math]\displaystyle{ X=l_0(\mathbb{N}_0) }[/math] of sequences over the non-negative integers that are 0 from some point, with a total ordering that fulfills:

• [math]\displaystyle{ a\geq b, c\geq d \Rightarrow ac\geq bd }[/math] and
• [math]\displaystyle{ \forall a\in X: |\{b\in X|b\lt a\}|\lt \infty }[/math]

Here multiplication is defined by termwise addition: [math]\displaystyle{ ab=(a_1,a_2,\dots)(b_1,b_2,\dots)=(a_1+b_1,a_2+b_2) }[/math]. For simplicity we furthermore assume that [math]\displaystyle{ p_1=(1,0,0,\dots)\lt p_2=(0,1,0,\dots)\lt ... }[/math] is an increasing sequence. These pseudointegers are called (pseudo)primes.

Logarithm and [math]\displaystyle{ \varphi }[/math]

We would like to find a function [math]\displaystyle{ \varphi: X\to \mathbb{R} }[/math] that preserves multiplication and order. In general, this function will not be injective. Look at a fixed element [math]\displaystyle{ x\in X, x\neq (0,0,0,\dots) }[/math]. We now define the logarithm in base x by [math]\displaystyle{ \log_x(y)=\sup\{a/b|a\in\mathbb{N}_0,b\in \mathbb{N},x^a\leq y^b\}. }[/math] This function fulfills some basis formulas

1. [math]\displaystyle{ \log_x(x)=1 }[/math]
2. [math]\displaystyle{ a/b\gt log_x(y)\Rightarrow x^a\gt y^b }[/math]
3. [math]\displaystyle{ a/b\lt log_x(y)\Rightarrow x^a\lt y^b }[/math]
4. [math]\displaystyle{ \forall y\in X\setminus \{(0,0,\dots)\}: \log_x(y)\in (0,\infty) }[/math]
5. [math]\displaystyle{ \log_x(yz)=\log_x(y)+\log_x(z) }[/math]
6. [math]\displaystyle{ y\leq z \Rightarrow\log_x(y)\leq \log_x(z) }[/math]
7. [math]\displaystyle{ \log_x(z)=\log_y(z)\log_x(y) }[/math]

Proof: 1: We have [math]\displaystyle{ x^1\leq x^1 }[/math], so [math]\displaystyle{ \log_x(x)\geq 1/1=1 }[/math]. If [math]\displaystyle{ a\gt b }[/math] then [math]\displaystyle{ x^a\gt x^b }[/math], since otherwise [math]\displaystyle{ x^{b+n(a-b)} }[/math] would be an infinite decreasing sequence. This shows [math]\displaystyle{ \log_x(x)\leq 1 }[/math].

2: If [math]\displaystyle{ a/b\gt log_x(y) }[/math], a/b can't be in the set we take sup of. So [math]\displaystyle{ x^a\gt y^b }[/math].

3: For [math]\displaystyle{ y=(0,0,\dots) }[/math] we have [math]\displaystyle{ \log_x(y)=0 }[/math] and the theorem is true, so we may assume [math]\displaystyle{ y\neq (0,0,\dots) }[/math]. If [math]\displaystyle{ a/b \lt \log_x(y) }[/math] we can find c and d so that [math]\displaystyle{ a/b\lt c/d }[/math] and [math]\displaystyle{ x^c\leq y^d }[/math]. Now we have [math]\displaystyle{ (x^a)^{bc}=x^{abc}\leq y^{abd}=(y^b)^{ad} }[/math] and using [math]\displaystyle{ ad\lt bd }[/math] and that the power function is increasing for [math]\displaystyle{ y\neq (0,0,\dots) }[/math] (see the proof of 1) we get [math]\displaystyle{ x^a\lt y^b }[/math]

4: If [math]\displaystyle{ \log_x(y)\leq 0 }[/math] (2) tells us that [math]\displaystyle{ x^1\gt y^n }[/math] for all n. But that we have a infinite bounded sequence which contradicts an axiom of the ordering on the pseudointegers. If [math]\displaystyle{ \log_x(y)=\infty }[/math] (3) tells us that [math]\displaystyle{ x^n\lt y^1 }[/math] for all n. Again we get a contradiction.

5: First I show [math]\displaystyle{ \log_x(yz)\leq \log_x(y)+\log_x(y) }[/math]. Assume for contradiction that [math]\displaystyle{ \log_x(yz)\gt \log_x(y)+\log_x(y) }[/math]. Now we can find [math]\displaystyle{ a_1,a_2,b }[/math] so that [math]\displaystyle{ \frac{a_1}{b}\gt \log_x(y) }[/math], [math]\displaystyle{ \frac{a_2}{b}\gt \log_x(z) }[/math], and [math]\displaystyle{ \log_x(yz)\gt \frac{a_1+a_2}{b} }[/math]. Form (2) we know that [math]\displaystyle{ x^{a_1}\gt y^b }[/math] and [math]\displaystyle{ x^{a_2}\gt z^b }[/math]. And thus [math]\displaystyle{ x^{a_1+a_2}\gt y^bz^b=(yz)^b }[/math]. Using (3) we get a contradiction with [math]\displaystyle{ \log_x(yz)\gt \frac{a_1+a_2}{b} }[/math]. The proof of the opposite inequality is similar.

6: This is equivalent to [math]\displaystyle{ \log_x(y)\gt \log_x(z)\Rightarrow y\gt z }[/math], so assume that [math]\displaystyle{ \log_x(y)\gt \log_x(z) }[/math]. Now we can find [math]\displaystyle{ \log_x(y)\gt \frac{a}{b}\gt \log_x(z) }[/math]. This implies [math]\displaystyle{ z^b\lt x^a\lt y^b }[/math] and then [math]\displaystyle{ z\lt y }[/math].

7: First I show that [math]\displaystyle{ \log_x(z)\leq \log_y(z)\log_x(y) }[/math]. Assume for contradiction that [math]\displaystyle{ \log_x(z)\gt \log_y(z)\log_x(y) }[/math]. Now we can find [math]\displaystyle{ a,b,c\in\mathbb{N} }[/math] with [math]\displaystyle{ \log_x(z)\gt \frac{a}{b} }[/math], [math]\displaystyle{ \frac{a}{c}\gt \log_y(z) }[/math], and [math]\displaystyle{ \frac{c}{b}\gt \log_x(y) }[/math]. Using (2) we get [math]\displaystyle{ y^a\gt z^c }[/math] and [math]\displaystyle{ x^c\gt y^b }[/math] and from this we get [math]\displaystyle{ x^{ac}\gt y^{ab}\gt z^{bc} }[/math] and then [math]\displaystyle{ x^a\gt z^b }[/math] contradicting [math]\displaystyle{ \log_x(z)\gt \frac{a}{b} }[/math]. The proof of the opposite inequality is similar.

We can now define a function [math]\displaystyle{ \varphi_x(y)=e^{\log_x(y)} }[/math]. For this function we have:

1. [math]\displaystyle{ \varphi_x(yz)=\varphi_x(y)\varphi_x(z) }[/math]
2. [math]\displaystyle{ y\leq z\Rightarrow\varphi_x(y)\leq \varphi_x(z) }[/math]

Densities of the pseudointegers divisible by y

Let [math]\displaystyle{ n_p }[/math] where [math]\displaystyle{ n\in\mathbb{N} }[/math] denote the nth pseudointeger (here p stands for pseudo, it is not a variable) and let [math]\displaystyle{ [n_p] }[/math] denote the set of the n smallest pseudointegers. We define the density of the pseudointegers divisible by y to be

[math]\displaystyle{ d_y=\lim_{n\to\infty}\frac{|yX\cap [n_p]|}{n}=\lim_{n\to\infty}\frac{|\{z\in X|yz\leq n_p\}|}{n} }[/math]

if the limit exist. Clearly [math]\displaystyle{ d_y\in [0,1] }[/math] if the limit exist. More generally we can always define

[math]\displaystyle{ i_y=\liminf_{n\to\infty}\frac{|yX\cap [n_p]|}{n}=\liminf_{n\to\infty}\frac{|\{z\in X|yz\leq n_p\}|}{n} }[/math]

[math]\displaystyle{ s_y=\limsup_{n\to\infty}\frac{|yX\cap [n_p]|}{n}=\limsup_{n\to\infty}\frac{|\{z\in X|yz\leq n_p\}|}{n} }[/math]

Using that the set [math]\displaystyle{ \{z\in X|yz\leq n_p\} }[/math] is a set of the form [math]\displaystyle{ [m_p] }[/math] for some m, we get:

[math]\displaystyle{ \{z\in X|xyz\leq n_p\}=\{z\in X|x(yz)\leq n_p\}=\{z\in X|yz\in\{a\in X|xa\leq n_p\}\}=\{z\in X|yz\leq|\{a\in X|xa\leq n_p\}|_p\}. }[/math]

From this we get the inequalities: [math]\displaystyle{ i_xi_y\leq i_{xy} }[/math] and [math]\displaystyle{ s_{xy}\leq s_xs_y }[/math]. In particular if [math]\displaystyle{ d_x }[/math] and [math]\displaystyle{ d_y }[/math] exists, then [math]\displaystyle{ d_{xy} }[/math] exists and [math]\displaystyle{ d_{xy}=d_xd_y }[/math]. By induction we get that [math]\displaystyle{ d_{x^n} }[/math] exists and [math]\displaystyle{ d_{x^n}=d_x^n }[/math]. If [math]\displaystyle{ x\leq y }[/math] we have [math]\displaystyle{ \{z\in X|yz\leq n_p\}\subset \{z\in X|yz\leq n_p\} }[/math] so in general [math]\displaystyle{ i_y\leq i_x }[/math] and [math]\displaystyle{ s_y\leq s_x }[/math] and if both densities exists [math]\displaystyle{ d_y\leq d_x }[/math].

If [math]\displaystyle{ d_x=1 }[/math] for some [math]\displaystyle{ x\neq (0,0,\dots) }[/math] we have [math]\displaystyle{ d_{x^n}=1 }[/math] of all n, and for any y there is a n so that [math]\displaystyle{ y\geq x^n }[/math]. This gives us [math]\displaystyle{ 1=i_{x^n}\leq i_y }[/math] which implies that [math]\displaystyle{ d_y }[/math] exists for any y and [math]\displaystyle{ d_y=1 }[/math].

If [math]\displaystyle{ d_x=0 }[/math] for some x then for any [math]\displaystyle{ y\neq (0,0,\dots) }[/math] we know that [math]\displaystyle{ d_y }[/math] exists and [math]\displaystyle{ d_y=0 }[/math]: Assume for contradiction that [math]\displaystyle{ s_y\gt 0 }[/math]. Then for some n, [math]\displaystyle{ y^n\geq x }[/math]. This implies that [math]\displaystyle{ s_y^n=s_{y^n}\leq s_x=0 }[/math]. Contradiction.

Assume that [math]\displaystyle{ d_x }[/math] and [math]\displaystyle{ d_y }[/math] exists and [math]\displaystyle{ d_x\in (0,1) }[/math]. Now we know that [math]\displaystyle{ d_y\in (0,1) }[/math] (I haven't been able to show this without the assumption that [math]\displaystyle{ d_y }[/math] exists). We know that for [math]\displaystyle{ a,b\in \mathbb{N} }[/math]:

[math]\displaystyle{ \frac{a}{b}\gt \log_x(y)\Rightarrow x^a\gt y^b\Rightarrow d_{x^a}\leq d_{y^b}\Rightarrow d_x^a\leq d_y^b\Rightarrow a\log(d_x)\leq b\log(d_y)\Rightarrow \frac{a}{b}\geq \frac{\log(d_y)}{\log(d_x)} }[/math]

(notice that log are taken on numbers in (0,1) so they are negative) where [math]\displaystyle{ log }[/math] is the usual logarithm on the reals. Similarly we get [math]\displaystyle{ \frac{a}{b}\lt \log_x(y)\Rightarrow \frac{a}{b}\leq \frac{\log(d_y)}{\log(d_x)} }[/math]. This tells us that [math]\displaystyle{ \log_x(y)=\frac{\log(d_y)}{\log(d_x)} }[/math]. From this we get that [math]\displaystyle{ -\log(d_x)\log_x(z)=-\log(d_x)\log_y(z)\log_x(y)=-\log(d_y)\log_y(z) }[/math], so the function [math]\displaystyle{ \log: X\to \mathbb{R} }[/math] defined by [math]\displaystyle{ \log(z)=-\log(d_x)\log_x(z) }[/math] for some [math]\displaystyle{ x\in X }[/math] where [math]\displaystyle{ d_x }[/math] exists and [math]\displaystyle{ d_x\in (0,1) }[/math], doesn't depend on the x we choose. From this function we define [math]\displaystyle{ \varphi(z)=e^{\log(z)} }[/math]. If [math]\displaystyle{ d_z }[/math] exists (again, it would be nice to find a proof that it does or to prove that it does have to) we know that [math]\displaystyle{ \varphi(z)=e^{\log(z)}=e^{-\log(d_z)\log_z(z)}=(e^{\log(d_z)})^{-1}=\frac{1}{d_z} }[/math].

Density of the set of pseudointegers

If there exists a [math]\displaystyle{ x\in X: d_x\in (0,1) }[/math], we define

[math]\displaystyle{ d=\lim_{n\to\infty}\frac{|\varphi(X)\cap (0,n]|}{n} }[/math]

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 [math]\displaystyle{ d_x }[/math] exists for all [math]\displaystyle{ x\in X }[/math]. I don't think it is enough to assume that [math]\displaystyle{ d_x\in (0,1) }[/math] for one x, but it might be enough to assume that [math]\displaystyle{ d_x,d_y \in (0,1) }[/math] for some x,y such that [math]\displaystyle{ \log_x(y) }[/math] is irrational.

From [math]\displaystyle{ d }[/math] to [math]\displaystyle{ d_x }[/math]

Let [math]\displaystyle{ \varphi: X \to\mathbb{R} }[/math] be any function such that [math]\displaystyle{ \varphi(x)\varphi(y)=\varphi(xy) }[/math] and [math]\displaystyle{ x\leq y \Rightarrow \varphi(x)\leq \varphi(y) }[/math], and such that the density

[math]\displaystyle{ d=\lim_{n\to\infty}\frac{|\varphi(X)\cap (0,n]|}{n} }[/math]

exists and [math]\displaystyle{ d\in (0,\infty) }[/math] (in particular this implies that [math]\displaystyle{ \lim_{n\to\infty}\varphi(n_p)=\infty }[/math]). We can now find [math]\displaystyle{ d_x }[/math]:

[math]\displaystyle{ d_x=\lim_{n\to\infty}\frac{|\{z\in X|xz\leq n_p\}|}{n}=\lim_{n\to\infty}\frac{|\{z\in X|\varphi(x)\varphi(z)\leq \varphi(n_p)\}|}{|\{z\in X|\varphi(z)\leq \varphi(n_p)\}|}=\lim_{n\to\infty}\frac{|\varphi(X)\cap (0,\frac{\varphi(n_p)}{\varphi(x)}]|}{\varphi(n_p)}\frac{\varphi(n_p)}{|\varphi(X)\cap (0,\varphi(n_p)]|}=\frac{1}{\varphi(x)}. }[/math]

Notice that in general we don't have [math]\displaystyle{ \{z\in X|xz\leq n_p\}=\{z\in X|\varphi(x)\varphi(z)\leq \varphi(n_p)\} }[/math] but [math]\displaystyle{ \{z\in X|\varphi(x)\varphi(z)\lt \varphi(n_p)\}\subset\{z\in X|xz\leq n_p\}\subset\{z\in X|\varphi(x)\varphi(z)\leq \varphi(n_p)\} }[/math]. The above works because we are taking limits.