Pseudointegers - Revision history

SuneJ at 08:17, 23 February 2010

2010-02-23T08:17:24Z

← Older revision		Revision as of 01:17, 23 February 2010
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	<math>d=\lim_{n\to\infty}\frac{\|\varphi(X)\cap (0,n]\|}{n}</math>		<math>d=\lim_{n\to\infty}\frac{\|\varphi(X)\cap (0,n]\|}{n}</math>

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

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

SuneJ at 19:44, 21 February 2010

2010-02-21T19:44:31Z

← Older revision		Revision as of 12:44, 21 February 2010
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	2: If <math>a/b>log_x(y)</math>, a/b can't be in the set we take sup of. So <math>x^a> y^b</math>.		2: If <math>a/b>log_x(y)</math>, a/b can't be in the set we take sup of. So <math>x^a> y^b</math>.

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

	4: If <math>\log_x(y)\leq 0</math> (2) tells us that <math>x^1>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>\log_x(y)=\infty</math> (3) tells us that <math>x^n<y^1</math> for all n. Again we get a contradiction.		4: If <math>\log_x(y)\leq 0</math> (2) tells us that <math>x^1>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>\log_x(y)=\infty</math> (3) tells us that <math>x^n<y^1</math> for all n. Again we get a contradiction.

SuneJ at 19:34, 21 February 2010

2010-02-21T19:34:11Z

@@ Line 68: / Line 68: @@
 (notice that log are taken on numbers in (0,1) so they are negative) where <math>log</math> is the usual logarithm on the reals. Similarly we get <math>\frac{a}{b}<\log_x(y)\Rightarrow \frac{a}{b}\leq \frac{\log(d_y)}{\log(d_x)}</math>. This tells us that <math>\log_x(y)=\frac{\log(d_y)}{\log(d_x)}</math>. From this we get that <math>-\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>\log: X\to \mathbb{R}</math> defined by <math>\log(z)=-\log(d_x)\log_x(z)</math> for some <math>x\in X</math> where <math>d_x</math> exists and <math>d_x\in (0,1)</math>, doesn't depend on the x we choose. From this function we define <math>\varphi(z)=e^{\log(z)}</math>. If <math>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>\varphi(z)=e^{\log(z)}=e^{-\log(d_z)\log_z(z)}=(e^{\log(d_z)})^{-1}=\frac{1}{d_z}</math>.
 If there exists a <math>x\in X: d_x\in (0,1)</math>, we define
@@ Line 73: / Line 75: @@
 <math>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 to many pseudointegers" compared to the integers.
+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.

SuneJ at 17:44, 21 February 2010

2010-02-21T17:44:58Z

@@ Line 6: / Line 6: @@
 Here multiplication is defined by termwise addition: <math>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>p_1=(1,0,0,\dots)<p_2=(0,1,0,\dots)<...</math> is an increasing sequence. These pseudointegers are called (pseudo)primes.
 We would like to find a function <math>\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>x\in X, x\neq (0,0,0,\dots)</math>. We now define the logarithm in base x by
@@ Line 17: / Line 19: @@
 # <math>\log_x(yz)=\log_x(y)+\log_x(z)</math>
 # <math>y\leq z \Rightarrow\log_x(y)\leq \log_x(z)</math>
 Proof:
@@ Line 30: / Line 33: @@
 The proof of the opposite inequality is similar.
-: This is equvivalent to <math>\log_x(y)>\log_x(z)\Rightarrow y>z</math>, so assume that <math>\log_x(y)>\log_x(z)</math>. Now we can find <math>\log_x(y)>\frac{a}{b}>\log_x(z)</math>. This implies <math>z^b<x^a<y^b</math> and then <math>z<y</math>.
+: This is equivalent to <math>\log_x(y)>\log_x(z)\Rightarrow y>z</math>, so assume that <math>\log_x(y)>\log_x(z)</math>. Now we can find <math>\log_x(y)>\frac{a}{b}>\log_x(z)</math>. This implies <math>z^b<x^a<y^b</math> and then <math>z<y</math>.
 We can now define a function <math>\varphi_x(y)=e^{\log_x(y)}</math>. For this function we have:
@@ Line 36: / Line 41: @@
 # <math>y\leq z\Rightarrow\varphi_x(y)\leq \varphi_x(z)</math>
-To be continued...
+==Densities of the pseudointegers divisible by y==

SuneJ: New page: 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 pseudoin...

2010-02-20T16:42:45Z

New page: 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 pseudoin...

New page

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>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>a\geq b, c\geq d \Rightarrow ac\geq bd</math> and
* <math>\forall a\in X: |\{b\in X|b<a\}|<\infty </math>

Here multiplication is defined by termwise addition: <math>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>p_1=(1,0,0,\dots)<p_2=(0,1,0,\dots)<...</math> is an increasing sequence. These pseudointegers are called (pseudo)primes.

We would like to find a function <math>\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>x\in X, x\neq (0,0,0,\dots)</math>. We now define the logarithm in base x by
<math>\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

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

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

2: If <math>a/b>log_x(y)</math>, a/b can't be in the set we take sup of. So <math>x^a> y^b</math>.

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

4: If <math>\log_x(y)\leq 0</math> (2) tells us that <math>x^1>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>\log_x(y)=\infty</math> (3) tells us that <math>x^n<y^1</math> for all n. Again we get a contradiction.

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

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

We can now define a function <math>\varphi_x(y)=e^{\log_x(y)}</math>. For this function we have:
# <math>\varphi_x(yz)=\varphi_x(y)\varphi_x(z)</math>
# <math>y\leq z\Rightarrow\varphi_x(y)\leq \varphi_x(z)</math>

To be continued...