Rashiminos
Fool Prophet
One of the twin prime conjectures is that there infinitely many pairs of primes that have a difference of 2.
2014.04.28 I have solved the twin prime conjecture.
- Thread starter RD-BH
- Start date
RD-BH
Human
The Basics:
... I is an integer
... (2I+-1) is an odd integer
... P[n-1] = nth Prime
... (x/n) is whole when integer n is a divisor of integer x
... (i^4) = 1
... (i^(4(x/n))) = 1 when integer n is a divisor of integer x
... Let n = an element of P[] and count the elements of P[] that are divisors of x
... ... if the count is 0, then integer x is not divisible by P[]
... ... the numerator is not divisible by any element of P[]
... ... ... ... P[x]=2 leaves r+-1, P[x]=3 leaves r+-1,P[x]>=5 leaves r+-2,+-4
... ... ... therefore (i^(4(x/P[n])) does not equal 1 for any element of P[]
... ... ... therefore f(I) = 0
... I is an integer
... (2I+-1) is an odd integer
... P[n-1] = nth Prime
... (x/n) is whole when integer n is a divisor of integer x
... (i^4) = 1
... (i^(4(x/n))) = 1 when integer n is a divisor of integer x
... Let n = an element of P[] and count the elements of P[] that are divisors of x
... ... if the count is 0, then integer x is not divisible by P[]
... ... the numerator is not divisible by any element of P[]
... ... ... ... P[x]=2 leaves r+-1, P[x]=3 leaves r+-1,P[x]>=5 leaves r+-2,+-4
... ... ... therefore (i^(4(x/P[n])) does not equal 1 for any element of P[]
... ... ... therefore f(I) = 0
Rashiminos
Fool Prophet
Then the product P1*P2*P3*...*Pn diverges as n -> infinity and the (edit) first equation in the OP is not well-defined.
RD-BH
Human
(6*(P[2])*(2I+-1)/2)+-3+-1)
(6*(P[3])*(2I+-1)/2)+-3+-1)
(6*(P[4])*(2I+-1)/2)+-3+-1)
if x[100] = (any set of 100 odd primes)
(P[0]*(prod_{n=0}^{99}{x[n]})*((2I+-1)/2)+-3+-1) is indivisible by any element of x[] for all I
Perfection
The Great Head.
the infinite product is in parentheses, the only term in parentheses is Pn. You're saying multiply all the prime numbers. You can't do that because it's divergent.
Also, please refrain from using plus-or-minus signs. The interpretation is often ambiguous.
Also, please refrain from using plus-or-minus signs. The interpretation is often ambiguous.
^Unless it is not a multiplication of all the primes, but one featuring some sort of sieve (which would be Irrational number-based). Not that i see that (beyond a guess) in the noted bits. Moreover i am not very familiar with modern sieve application (i know of its origin, regarding primes, and then some stuff from the 18th century).
warpus
Sommerswerd asked me to change this
Why can't we get a formal writeup of the proof? It makes me think that there really isn't a proof....
If one actually manages to complete a worthy paper on such problems, and:
a) they do not have access to a math journal, or a math uni department they can sent the work to
b) they feel that in the case the work is good it may be stolen
then it makes a lot of sense to try to post it in some places on the web. Of course CivFanatics is not the best choice for that sort of thing (afaik there are by now dedicated math forums where some sections are about members posting this sort of work, supposedly with the view that they are protected in case of a breakthrough, etc).
Anyway, i wish good luck on anyone trying their mind with those sorts of math problems.
Rashiminos
Fool Prophet
This is basically stating that the set of primes is not finite.
As for the numerator, it's not an integer, let alone a positive integer.
If it were, then it has a prime factorization. This would imply there exists a natural number x such that Px divides that numerator, or f(I)>=1.
As for the numerator, it's not an integer, let alone a positive integer.
If it were, then it has a prime factorization. This would imply there exists a natural number x such that Px divides that numerator, or f(I)>=1.
Unless I misunderstand something, the last line is false. Because you are taking the absolute of every number in the sum, the function can only be zero, when the i^4*y term is -1 for all x. For that to happen, y must be an odd multiple of 1/2. As 2 is prime, it cancels out the 1/2 in the y term. So for all Px, except for 2, the y term is not an odd multiple of 1/2 and thus the sum is not zero.
timtofly
One Day
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Once I had the glance on the mathematical structure it spelled "An infinity plus one and divided by two is zero ?" how so ? I am here , You are here unless You want to convice me that I do not exist maybe I don't but that would only mean that You do not exist as well and this is a paradox. Since You dont exist You cannot tell me that I don't You see
Unless you are a prime?
Rashiminos
Fool Prophet
The function has a floor operator. Each term in the sum is either 0 or 1 (as long as the exponentiation is defined in terms of complex numbers)*.
|(a+bi)| = (a^2 + b^2)^1/2,
0 <= |(0+i)^x + 1| <= 2, x is a positive rational number
Consider floor (|(0+i)^x + 1|/2), x is a positive rational number
*Which may be more of a problem than I initially thought when I originally made this post.
RD-BH
Human
This is number theory (indivisibility) and is indeed based on a sieve.
The sieve shows the ratio of indivisible twin pairs to the period of x primes to be:
Being based on Euler's work and needing this form for Riemann ... what is a man to do?
... efforting.
This.
The attached stuff seems vaguely number theory related, but I don't see it as showing anything about twin primes, and certainly there is no proof.
More damning, a google search for "Russ Carley jr math" brings up nothing.
Extraordinary claims require extraordinary evidence, and instead we are presented with no evidence
RD-BH
Human
This.
The attached stuff seems vaguely number theory related, but I don't see it as showing anything about twin primes, and certainly there is no proof.
More damning, a google search for "Russ Carley jr math" brings up nothing.
Extraordinary claims require extraordinary evidence, and instead we are presented with no evidence
Fair enough, I'm a nobody ...
Perfection
The Great Head.
Whatever we're all 99th percentilers here.
I can't get past my issues I outlined in my previous post.
I can't get past my issues I outlined in my previous post.
RD-BH
Human
If one actually manages to complete a worthy paper on such problems, and:
a) they do not have access to a math journal, or a math uni department they can sent the work to
b) they feel that in the case the work is good it may be stolen
then it makes a lot of sense to try to post it in some places on the web. Of course CivFanatics is not the best choice for that sort of thing (afaik there are by now dedicated math forums where some sections are about members posting this sort of work, supposedly with the view that they are protected in case of a breakthrough, etc).
Anyway, i wish good luck on anyone trying their mind with those sorts of math problems.
I need money, endorsements, or a professional affiliation.
I have been trying to get someone/anyone to look at my proof since 2006.
I don't have the correct 'political' affiliations mathwise.
What I have right now, is a promise of a peer review ... time will tell.
They are professional mathematicians and may help me clarify the proof for publication.
I've had people claim my work as their own before, I want the formula somewhere public (I've posted in more than one location) to have a 'paper' trail as it were.
Also, I like Civilization 8)
[note] Do you think feedback from <99%ile would be more helpful? 8)
Perfection
The Great Head.
Well from what I see your equation doesn't follow typical mathematical notation conventions. That makes me dubious about the correctness of your proof.
Rashiminos
Fool Prophet
Given the edit of the OP, it's supposed to be some kind of counting function.
AdamCrock
Polish Pirate
Mathematics is devised by man and that is why it is not perfect. Who am I to say that 1 is 1 or 2 is 2 maybe we are all terribly wrong.
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