Pseudointegers

From Polymath1Wiki

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
$\forall a\in X: |\{b\in X|b<a\}|<\infty$

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 $p_1=(1,0,0,\dots)<p_2=(0,1,0,\dots)<...$ 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

$log x (x) = 1$
$a/b> log_x(y)\Rightarrow x^a> y^b$
$a/b< log_x(y)\Rightarrow x^a< y^b$
$\forall y\in X\setminus \{(0,0,\dots)\}: \log_x(y)\in (0,\infty)$
$log x (y z) = log x (y) + log x (z)$
$y\leq z \Rightarrow\log_x(y)\leq \log_x(z)$
$log x (z) = log y (z)log x (y)$

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

2: If $a / b > l o g x (y)$ , a/b can't be in the set we take sup of. So $x a > y b$ .

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 $a / b < log x (y)$ we can find c and d so that $a / b < c / d$ and $x^c\leq y^d$ . Now we have $(x^a)^{bc}=x^{abc}\leq y^{abd}=(y^b)^{ad}$ and using $a d < b d$ and that the power function is increasing for $y\neq (0,0,\dots)$ (see the proof of 1) we get $x a < y b$

4: If $\log_x(y)\leq 0$ (2) tells us that $x 1 > y n$ 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 $x n < y 1$ 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 $log x (y z) > log x (y) + log x (y)$ . Now we can find $a 1, a 2, b$ so that $\frac{a_1}{b}>\log_x(y)$ , $\frac{a_2}{b}>\log_x(z)$ , and $\log_x(yz)>\frac{a_1+a_2}{b}$ . Form (2) we know that $x^{a_1}>y^b$ and $x^{a_2}>z^b$ . And thus $x^{a_1+a_2}>y^bz^b=(yz)^b$ . Using (3) we get a contradiction with $\log_x(yz)>\frac{a_1+a_2}{b}$ . The proof of the opposite inequality is similar.

6: This is equivalent to $\log_x(y)>\log_x(z)\Rightarrow y>z$ , so assume that $log x (y) > log x (z)$ . Now we can find $\log_x(y)>\frac{a}{b}>\log_x(z)$ . This implies $z b < x a < y b$ and then $z < y$ .

7: First I show that $\log_x(z)\leq \log_y(z)\log_x(y)$ . Assume for contradiction that $log x (z) > log y (z)log x (y)$ . Now we can find $a,b,c\in\mathbb{N}$ with $\log_x(z)>\frac{a}{b}$ , $\frac{a}{c}>\log_y(z)$ , and $\frac{c}{b}>\log_x(y)$ . Using (2) we get $y a > z c$ and $x c > y b$ and from this we get $x a c > y a b > z b c$ and then $x a > z b$ contradicting $\log_x(z)>\frac{a}{b}$ . 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:

$\varphi_x(yz)=\varphi_x(y)\varphi_x(z)$
$y\leq z\Rightarrow\varphi_x(y)\leq \varphi_x(z)$

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 $d x y$ exists and $d x y = d x d y$ . 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 $s y > 0$ . 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 $a,b\in \mathbb{N}$ :

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

(notice that log are taken on numbers in (0,1) so they are negative) where $l o g$ is the usual logarithm on the reals. Similarly we get $\frac{a}{b}<\log_x(y)\Rightarrow \frac{a}{b}\leq \frac{\log(d_y)}{\log(d_x)}$ . 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 $\{z\in X|\varphi(x)\varphi(z)< \varphi(n_p)\}\subset\{z\in X|xz\leq n_p\}\subset\{z\in X|\varphi(x)\varphi(z)\leq \varphi(n_p)\}$ . The above works because we are taking limits.

Pseudointegers

From Polymath1Wiki

Contents

Logarithm and $\varphi$

Densities of the pseudointegers divisible by y

Density of the set of pseudointegers

From $d$ to $d x$

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Pseudointegers

From Polymath1Wiki

Contents

Logarithm and

Densities of the pseudointegers divisible by y

Density of the set of pseudointegers

From d to dx

Views

Personal tools

Navigation

Search

Toolbox

Logarithm and $\varphi$

From $d$ to $d x$