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.

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]

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\ 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 equvivalent 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].

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]

To be continued...