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Re: Who gonna prove Goldbach conjecture?
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Arrow Re: Who gonna prove Goldbach conjecture? - 01-27-2007, 03:43 AM

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Originally Posted by AntonioLao View Post
Recently, I came across or rather I realized that the union and intersection operators of sets could not have any inverse even thought complementation exist for them. I'm wondering could this non-inverse property be used to prove the conjecture?
are we comfortable using the expression 10r, as infinity, so that any infinitesimal number can be represented to the right of 10r.10, and any number infitesimally smaller than the first number will be placed further to the right 10r.10.10, and that we truly wanted to we can multiply this by 2 to get 20r.20.20, hence we have developed a system where the infinity to the left of the leading sequence is assumed to be true, so that we could preserve the structure of 10r.10.10 if we wanted to introduce it to another number infinitely bigger, 10r.00., we would get in addtion 10r.10r.10.10, noted quite diligently that the decimal places do not denote value but rather denote relative value on an infinite and infinitesimal scale.
  
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Re: Who gonna prove Goldbach conjecture?
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Re: Who gonna prove Goldbach conjecture? - 02-21-2007, 03:40 PM

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Originally Posted by theunify
are we comfortable using the expression 10r, as infinity
Thanks for your mathematical descriptions. I still would prefer the binary system of numbers.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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Re: Who gonna prove Goldbach conjecture? - 07-14-2008, 05:41 PM

There is now a simple algorithmic proof of this conjecture. Maybe a conference is needed to finally announce it to the public.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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Smile Re: Who gonna prove Goldbach conjecture? - 07-14-2008, 05:49 PM

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There is now a simple algorithmic proof of this conjecture. Maybe a conference is needed to finally announce it to the public.
We will announce it tomorrow,as it never really comes!





regards michael.


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Re: Who gonna prove Goldbach conjecture? - 07-14-2008, 06:00 PM

Then we definitely announce it on your birthday.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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Smile Re: Who gonna prove Goldbach conjecture? - 07-14-2008, 06:06 PM

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Then we definitely announce it on your birthday.

Okay that sounds fair!



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Re: Who gonna prove Goldbach conjecture? - 07-14-2008, 06:08 PM

So let me know ahead of time when the day is near.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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Re: Who gonna prove Goldbach conjecture? - 07-19-2008, 02:07 PM

The proof of this conjecture is as easy as adding 0 plus 1 or 1 plus 1 or 1 plus 2. What is needed is a Diophantine sieve without unity.


Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²
  
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