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

• $a\geq b, c\geq d \Rightarrow ac\geq bd$ and
• $\forall a\in X: |\{b\in X|b\lta\}|\lt\infty$

Here multiplication is defined by termwise addition: $ab=(a_1,a_2,\dots)(b_1,b_2,\dots)=(a_1+b_1,a_2+b_2)$. For simplicity we furthermore assume that $p_1=(1,0,0,\dots)\ltp_2=(0,1,0,\dots)\lt...$ is an increasing sequence. These pseudointegers are called (pseudo)primes.

## Logarithm and $\varphi$

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

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

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

2: If $a/b\gtlog_x(y)$, a/b can't be in the set we take sup of. So $x^a\gt y^b$.

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

4: If $\log_x(y)\leq 0$ (2) tells us that $x^1\gty^n$ for all n. But that we have a infinite bounded sequence which contradicts an axiom of the ordering on the pseudointegers. If $\log_x(y)=\infty$ (3) tells us that $x^n\lty^1$ for all n. Again we get a contradiction.

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

6: This is equivalent to $\log_x(y)\gt\log_x(z)\Rightarrow y\gtz$, so assume that $\log_x(y)\gt\log_x(z)$. Now we can find $\log_x(y)\gt\frac{a}{b}\gt\log_x(z)$. This implies $z^b\ltx^a\lty^b$ and then $z\lty$.

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

We can now define a function $\varphi_x(y)=e^{\log_x(y)}$. For this function we have:

1. $\varphi_x(yz)=\varphi_x(y)\varphi_x(z)$
2. $y\leq z\Rightarrow\varphi_x(y)\leq \varphi_x(z)$

## Densities of the pseudointegers divisible by y

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

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

if the limit exist. Clearly $d_y\in [0,1]$ if the limit exist. More generally we can always define

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

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

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

$\{z\in X|xyz\leq n_p\}=\{z\in X|x(yz)\leq n_p\}=\{z\in X|yz\in\{a\in X|xa\leq n_p\}\}=\{z\in X|yz\leq|\{a\in X|xa\leq n_p\}|_p\}.$

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

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

If $d_x=0$ for some x then for any $y\neq (0,0,\dots)$ we know that $d_y$ exists and $d_y=0$: Assume for contradiction that $s_y\gt0$. Then for some n, $y^n\geq x$. This implies that $s_y^n=s_{y^n}\leq s_x=0$. Contradiction.

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

$\frac{a}{b}\gt\log_x(y)\Rightarrow x^a\gty^b\Rightarrow d_{x^a}\leq d_{y^b}\Rightarrow d_x^a\leq d_y^b\Rightarrow a\log(d_x)\leq b\log(d_y)\Rightarrow \frac{a}{b}\geq \frac{\log(d_y)}{\log(d_x)}$

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

## Density of the set of pseudointegers

If there exists a $x\in X: d_x\in (0,1)$, we define

$d=\lim_{n\to\infty}\frac{|\varphi(X)\cap (0,n]|}{n}$

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

## From $d$ to $d_x$

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

$d=\lim_{n\to\infty}\frac{|\varphi(X)\cap (0,n]|}{n}$

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

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

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