Prove Goldbach's "Weak" Conjecture

I came up with a way to prove the Goldbach Conjecture using the fact that
all composites are eliminated using the prime factors up to the square root
of n. Here it is:

It was proved in 1989 by Chen and Wang that all even integers past 3.33*10^43000 are the sums of two primes. Based on the fact that all composite integers are divisible by at least one prime, I have devised a special process to eliminate enough composites. The formula is 1/pi-[(1/pi)(1/pi-1)]<[7/8(1/(ln n)-1)], where pi is the ith prime and pi-1 is the prime immediately preceding the ith prime. First, all even integers except 2 are composite. So, about half of all the integers up to any n are composite. Next, every third integer is divisible by 3. This means that ½-(1/2)(1/3)=1/3 of the integers are left after all integers divisible by 3 are eliminated. And every fifth integer is divisible by 5. The same formula is used 1/3-(1/3)(1/5)=5/15-1/15=4/15. Each succeeding prime is plugged in the equation in an analogous manner. So, 4/15-(4/15)(1/7)=28/105-4/105=24/105; 24/105-(24/105)(1/11)=275/1155-24/1155=251/1155; 251/1155-(251/1155)(1/13)=

3263/15015-251/15015=3012/15015; 3012/15015-(3012/15015)(1/17)=51204/255255-(51204/255255)(1/19)=972876/4849845-(972876/4849845)(1/23)=22376148/111546435-972876/111546435=21403272/111546435; 21403272/111546435-(21403272/111546435)(1/29)=

620694888/3234846615-21403272/3234846615=599291616/3234846615; 599291616/3234846615-

(599291616/3234846615)(1/31)=18578040096/100280245065-(599291616/100280245065)=

17978748480/100280245065;

Chebyshev proved that the minimum number of primes for any n is (7/8) (n/ln n). Using 100, 1000, 10000, and 100000 as examples for n will demonstrate that the Chebyshev minimum for primes will not decrease below the formula for composites above. The natural logarithm of 100 is approximately equal to 4.605. So, (7/8)(n/(ln n -1))=(7/8)(100/3.605)=(7/8)(0.277392)=24.27. The composite formula is less than the Chebyshev minimum for 100 (0.277392) at 4/15=0.266667. The prime numbers up to 5

generated 4/15, but 100 has the possible prime factors 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97. Eliminating all the composites up to the square root will

evidently eliminate all of the composites. This appears to prove the Goldbach Conjecture. For example,

for n equal to the even integer 102, the possible 6x-1 and 6x+1 primes are as follows:

5 97

7 95

11 91

13 89

17 85

19 83

23 79

25 77

29 73

31 71

35 67

37 65

41 61

43 59

47 55

49 53

About 1/5 of the remaining integers will be divisible by 5 and about 1/7 will be divisible by 7. Since the

Square root of 100 is 10 and the last prime before 10 is 7, all the integers left after the pairs with one integer or both integers divisible by 5 and 7 are eliminated will be prime pairs. After 25, 35, 49, 55, 65, 77, 85, 91, and 95 are eliminated, below is what is left:

5 97

13 89

19 83

23 79

29 73

31 71

41 61

43 59

All remaining are prime pairs as expected.

Estermann in 1938 proved that almost all even numbers are the sums of two primes. Vinogradov in 1954 proved that all sufficiently large odd integers are the sums of three odd primes. The sufficiently large
N has been reduced to 3.33x10^43000. All this information is available
on the website mathworld.wolfram.com.
Arranging the 6x+1 and 6x-1 integers that add to any even integer can be used to evidently prove the Goldbach Conjecture as follows: All the integers on the left side can be reduced to prime numbers. Since Bertrand's Postulate states that there are always at least two primes between n and 2n, there are always at least two primes between (n-1)/2 and 2n (49 and 98). Since 49 and 98 is approximately the right hand side of the list, there are always two primes to prove the Goldbach Conjecture by making two prime pairs adding up to any even integer.