Proof of the twin prime conjecture!
- From: eestath <stathopoulosee@xxxxxxxxx>
- Date: Wed, 22 Jul 2009 21:10:58 -0700 (PDT)
In the text n>2 and k>2 and we must use all the primes that are
required by eatch formula!
p_i<p_(i+1)
and
q_i<q_(i+1)
Suppose that the primes that satisfy the condition are finite (p and p
+2 prime) are denoted by p with index i (1<=i<=n).We create the
number
A_n =[(p_1)^a_1]*[(p_2)^a_2]...[(p_n)^a_n].
(a_n>=2 and a_(i-1)>a_i and a1<n+1)
The possible outcomes of composite numbers for A_n is given by the
function K and is equal with:
K(A_n)=2^n-2-n + {Π(a_i) from i=1 to i=n)}
Suppose that the primes that do not sutisfy the condition are denoted
by q and are infinite with index i (1<=i<=k).
We create the number
B_k =[(q_1)^b_1]*[(q_2)^b_2]...[(q_k)^b_k]. (b_k>=2 and b_(i-1)>a_i
and b1<k+1).
The possible outcomes of composite numbers for B_k is given by the
function K and is equal with:
K(B_k)=2^k-2-k + {Π(b_i) from i=1 to i=k)}
We know create the number C_(n+k)=(A_n)*(B_k). The product of A_n and
B_k.
The possible outcomes of C_(n+k) is given by the formula:
K(C_(n+k))=2^(n+k)-2-(n+k) + K(A_n)=2^n-2-n + {Π(a_n) from i=1 to i=n)}
*{Π(b_i) form 1 to k)
We create the following inequality that holds for some constant δ. δ
depents and must be bigger than δ >{Π(a_n) from i=1 to i=n)}/{Κ(Α_n)}
For that δ we have:
K(C_(n+k))<{K(A_n)}*{K(B_k)}*δ.
witch holds!
This inequality must hold in the form for every k>2:
{K(C_(n+k))} / {{K(B_k)}}* δ < K(A_n).
But the wright part of the equation diverges as k gets bigger!
[{K(C_(n+k+1))} / {{K(B_(k+1))}}* δ]-1> {K(C_(n+k))} / {{K(B_k)}}* δ
as k gets bigger the term gets bigger from the previous one and not
only that diverges also!
So K(A_n) mast get bigger too!
But K(A_n) is only a number!
Reductio ad absurdum!
If we suppose that there is a finite number of primes that sutisfy the
condition then we fall in to a contradiction!
So we prove that if there are 2 or more primes that sutisfy a condtion
then there are infinite!
Dimitris Sakkos and Dimitris Stathopoulos!
For everything we say we have a proof.
.
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