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GOLDBACH CONJECTURE
Goldbach conjecture: ”Each even number greater than six, could be written as sum of two primes”
This conjecture is almost experimentally proven on computer , but we still miss theoretical proof. I’ll try to reach it by the method showed below.
Let define set S: the S is set of primes a1,a2,a3,a4,a5…an ,such that a1<a2<a3<a4<a5<…<an. Now we need interval of numbers L (0,a1*a2*a3*a4*a5*…*an). We have to prove that in interval L exists one more prime number a_(n+1):
If we take number m=a1*a2*a3*a4*a5…*an, and increase it for 1,we got the number
n=m+1=a1*a2*a3*a4*a5…an+1.Number n is odd and it’s obvious that it cannot be combination of the elements of the set S.
(a2)^p(a3)^q(a4)^r(a5)^s(a6)^t=a1a2a3a4a5a6a7+1
(a2)^p(a3)^q(a4)^r(a5)^s(a6)^ta1a2a3a4a5a6a7=1
a2* ((a2)^(p1)(a3)^q(a4)^r(a5)^s(a6)^ta1a3a4a5a6a7)=1
but it’s obvious that difference between is >=1 so we have the proof that in interval L has to be one more prime at least to form the number n,so we can write that:
a1*a2*a3*a4*a5*…*a_(n1)>a_n ........................................(i)
That means
a1*a2>a3
a1*a2*a3>a4
so the inequality i could be wriiten as
a1*a2*(a1*a2)*(a1*a2*a1*a2.....=(2*3)^(2^(n2))>a_n
or
6^(2^(n2))>an
if we approximate 6 with 2^3
then
2^(3*2^(n2))>an
if we approximate 3 with 2^2
we go that
2^(2^(n))>an.......(D)
according to
2^k>k^2 for k>5
the largest value an can have to satisfy inequality (D) is
an<k^2.....(U)
because (D) is satisfied for all k
so that means
2^(2n)>an
and again repeat (U)
(2n)^2>an
4n^2>an
n^2>an/4
and it is easy to prove that
a1*a2*a3*a4*a5*...a_(n1)>6an:
or
a3*a4*a5*...*a_(n1)>an...........(G):
Proof is the same as to prove (A):
t=a3*a4*a5*a6*a7+1 isn't divisible by the elements of the set B=
(a3,a4,a5,a6,a7) because if it is divisible then t could be written as
(a3)^q(a4)^r(a5)^s(a6)^t=a3a4a5a6a7+1
(a3)^q(a4)^r(a5)^s(a6)^ta3a4a5a6a7=1
a3* ((a3)^(q1)(a4)^r(a5)^s(a6)^ta4a5a6a7)=1
it's obvious that expression in the brackets is higher or equal 1.
end of proof.
But that doesn't mean that t isn't divisible by 6.
And it's obvious that the smallest number which is divisible by set B
is a3*a4*a5*a6*a7+a3 because
a3* ((a3)^(q1)(a4)^r(a5)^s(a6)^ta4a5a6a7)=a3
or
which is possible
((a3)^(q1)(a4)^r(a5)^s(a6)^ta4a5a6a7)=1
a3=5
odd number isn't divisible by 2
so the number
t+1(or +3) isn't divisible by 2
because number t1 is odd.
So if t+1 is divisible by 3 then it could be written as 3^k and t+3
is 3^k+2 and it cannot be divisible by 3 and as it is odd it cannot
been divisible by 2 so a3*a4*a5*a6>a7
or for all n
a3*a4*a5*...*a_(n1)>an..............(G)
End of proof.
according to (G)
4n^2>6a_n
or
n^2>3*a_n/2
to simplify inequality we it is allowed to say:
n^2>a_n....................(Z)
now
We have to determine number of numbers that can be sum of k pairs
of different primes. For example 24=13+11=17+7=23+1.the number of
that combinations is n*(n+1)/2
when we sub it from n^2 we get that of n primes could be at least
formed n(n1)/2
now let sub of an, n (because there are n prime numbers in
interval a_n) and then divide it by 2 to eliminate odd numbers and
using (Z)
(n^2n)/2>(a_nn)/2
(n^2n)/2=n(n1)/2>(a_nn)/2..............(R)
where (a_nn)/2 is actually number od even numbers in interval
(0,an] so the GoldBach conjecture is proven.
Goldbach conjecture: ”Each even number greater than six, could be written as sum of two primes”
This conjecture is almost experimentally proven on computer , but we still miss theoretical proof. I’ll try to reach it by the method showed below.
Let define set S: the S is set of primes a1,a2,a3,a4,a5…an ,such that a1<a2<a3<a4<a5<…<an. Now we need interval of numbers L (0,a1*a2*a3*a4*a5*…*an). We have to prove that in interval L exists one more prime number a_(n+1):
If we take number m=a1*a2*a3*a4*a5…*an, and increase it for 1,we got the number
n=m+1=a1*a2*a3*a4*a5…an+1.Number n is odd and it’s obvious that it cannot be combination of the elements of the set S.
(a2)^p(a3)^q(a4)^r(a5)^s(a6)^t=a1a2a3a4a5a6a7+1
(a2)^p(a3)^q(a4)^r(a5)^s(a6)^ta1a2a3a4a5a6a7=1
a2* ((a2)^(p1)(a3)^q(a4)^r(a5)^s(a6)^ta1a3a4a5a6a7)=1
but it’s obvious that difference between is >=1 so we have the proof that in interval L has to be one more prime at least to form the number n,so we can write that:
a1*a2*a3*a4*a5*…*a_(n1)>a_n ........................................(i)
That means
a1*a2>a3
a1*a2*a3>a4
so the inequality i could be wriiten as
a1*a2*(a1*a2)*(a1*a2*a1*a2.....=(2*3)^(2^(n2))>a_n
or
6^(2^(n2))>an
if we approximate 6 with 2^3
then
2^(3*2^(n2))>an
if we approximate 3 with 2^2
we go that
2^(2^(n))>an.......(D)
according to
2^k>k^2 for k>5
the largest value an can have to satisfy inequality (D) is
an<k^2.....(U)
because (D) is satisfied for all k
so that means
2^(2n)>an
and again repeat (U)
(2n)^2>an
4n^2>an
n^2>an/4
and it is easy to prove that
a1*a2*a3*a4*a5*...a_(n1)>6an:
or
a3*a4*a5*...*a_(n1)>an...........(G):
Proof is the same as to prove (A):
t=a3*a4*a5*a6*a7+1 isn't divisible by the elements of the set B=
(a3,a4,a5,a6,a7) because if it is divisible then t could be written as
(a3)^q(a4)^r(a5)^s(a6)^t=a3a4a5a6a7+1
(a3)^q(a4)^r(a5)^s(a6)^ta3a4a5a6a7=1
a3* ((a3)^(q1)(a4)^r(a5)^s(a6)^ta4a5a6a7)=1
it's obvious that expression in the brackets is higher or equal 1.
end of proof.
But that doesn't mean that t isn't divisible by 6.
And it's obvious that the smallest number which is divisible by set B
is a3*a4*a5*a6*a7+a3 because
a3* ((a3)^(q1)(a4)^r(a5)^s(a6)^ta4a5a6a7)=a3
or
which is possible
((a3)^(q1)(a4)^r(a5)^s(a6)^ta4a5a6a7)=1
a3=5
odd number isn't divisible by 2
so the number
t+1(or +3) isn't divisible by 2
because number t1 is odd.
So if t+1 is divisible by 3 then it could be written as 3^k and t+3
is 3^k+2 and it cannot be divisible by 3 and as it is odd it cannot
been divisible by 2 so a3*a4*a5*a6>a7
or for all n
a3*a4*a5*...*a_(n1)>an..............(G)
End of proof.
according to (G)
4n^2>6a_n
or
n^2>3*a_n/2
to simplify inequality we it is allowed to say:
n^2>a_n....................(Z)
now
We have to determine number of numbers that can be sum of k pairs
of different primes. For example 24=13+11=17+7=23+1.the number of
that combinations is n*(n+1)/2
when we sub it from n^2 we get that of n primes could be at least
formed n(n1)/2
now let sub of an, n (because there are n prime numbers in
interval a_n) and then divide it by 2 to eliminate odd numbers and
using (Z)
(n^2n)/2>(a_nn)/2
(n^2n)/2=n(n1)/2>(a_nn)/2..............(R)
where (a_nn)/2 is actually number od even numbers in interval
(0,an] so the GoldBach conjecture is proven.
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