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December 22nd, 2016, 11:53 PM   #1
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Bidimensional Ordinal

Can anoyone let me write an example of a Bidimensional Ordinal ?

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Stefano
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December 23rd, 2016, 01:14 AM   #2
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Sorry to be more clear: bidimensional not means just

{1,1},{1,2}.....{1,n}; {2,1},{1,2}.....{2,n}...; {m,1},{1,2}.....{m,n}

means that there is a relation (so a function or another ordinal) between n and m, so we can produce $n X m$ by one "relation" / function, only.

I think I discover what I call a Tridimensional Ordinal, that solve Riemann hypo problem, of course using transfinite induction.

As I dislike Abstract Algebra I discover Set Theory was very powerfull and extremly easy to be understood.

I will not flood the forum again since I understood is not the right place... pls just remember, one day, who discover what...

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Stefano
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December 23rd, 2016, 01:31 AM   #3
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I think you are just ordering the Cartesian product of two ordinals by lexicographic order. Yes? In which case (don't quote me on this) I think that's just the product of the ordinals. But I always get the order backwards for ordinal multiplication.

Last edited by Maschke; December 23rd, 2016 at 01:33 AM.
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December 23rd, 2016, 02:51 AM   #4
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Let me say it's more deep than the concept of lexicographic since involves a function, that can be also f.ex. the Rarional Derivate or the Rational Integral, or else...

For Rational or integral derivate I mean f.ex. the Rational couple of values $x=1/K ; y'$ of the derivate $y'=nX^{n-1}$ (as example of one continous function we can use as a analitical extention of our ordinal $x=1/K ; y'$.

So we can make a well ordered set of Apple's trees, a well ordered set of Apples, a well ordered set of Apple's seeds

in a well ordered set of "Fruit's Trees with fruits and seeds".

We can say this is a well ordered set in case we can prove 2 on 3 sets only are well ordered, and the 3-th follows since for any apple there are for example (we have to know that by a known math relation) 3 seeds ?

Here in the example of the Rational derivate if you start from $Y=X^n$ considering just $X\in\mathbb{Q}$ you have $n-1$ following derivate already full of well ordered sets... (till of course the last Flat non significative one).

And, of course you can also create a serie of well ordered set betweens the lines of different $n-th$ derivate / integral since it's clear how to move up or down...

Last edited by complicatemodulus; December 23rd, 2016 at 03:00 AM.
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December 23rd, 2016, 03:26 AM   #5
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To be clear my proof for Riemann Hypo follow as:

Since Naturals are well ordered,

Since Powers of Numbers (Integer, Rational or Complex under the conditions are made just of Integers or Rationals parts) are well ordered too,

Primes are well ordered too since:

$z= n!/n^2$

can be taken as ordinal who stick $n$ whos:

$z\in\mathbb{N}$ in the Non Prime Set

and

$z\in\mathbb{Q-N}$ in the Prime Set

Since we know how to build Rh function and who is the first zero for the first prime and who is the first zero for the first non prime, the same for the 2th and the 3th and the n-th, and the one of the n-th+1 for both...

Riemann's 1/2 integer part of S power, follows by the transfinite law of induction
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