PROOF OF GOLDBACH THEOREM
by
Dr. S. K. Kapoor

"Every Even number greater than two can be written as a sum of a pair of primes."

PROOF

Step 1.

Let Even E = M+M.
We can construct a set D(1) =

We can designate the members of D(1) as duplexes and can have abbreviation for each member as d_r where r = 0 to M and as such, cardinality of D(1) = M+1 > M = E/2.

D(1) is the largest possible set from which we have to sort out at least a single d_r such that both r and 2M-r are primes to satisfy the conjecture.

For this, impliedly, we have to rule out all values of r and 2M-r where r or 2M-r are composites. Let us formally define d_r composite when either r or 2M-r is composite. Therefore, the whole exercise as such reduces to sorting out composites from the range of 0 to M as well as from M to 2M and finally to see if there exists any d_r which is not a composite.

For this, we have to construct a chain of subsets of D(1) such that D(2) contains only those d_r's where 2 is not a factor of r or 2M-r and then to have subset D(3) of D(2) such that for no member d_r of D(3), 3 is a factor of r or 2M-r, and so on.

Let us first of all construct a subset D(2) of D(1) such that D(2) = {d_r | r = odd M}.

Therefore, cardinality of D(2) shall be E/4.

Step 2.

Before, we construct further subsets and count their cardinalities, let us first of all see the following properties of number which may be availed for the purpose:

a. Let E = E E = AB.
    If A<E, then B>E.
    Therefore, in terms of primes E, we can sort out composites uptil E.

b. Let P={p₁, p₂, p₃, ......, p_u} be the set of all odd primes E such that p_k < p_k+1 for k=1 to u.
    Let N is largest odd natural number E. Therefore, 1/E < 1/N.
    Further, as:
    1/N = 1/33/55/77/9......(N-2)/N.
    Therefore, if we restrict the denominators (2m+1)'s of above fractions of the form (2m-1)/(2m+1) only to (2m+1) as a prime, then
    1/N < 1/33/55/79/11......(p_u-2)/p_u, as p_u is the largest prime E.
    Therefore,
    1/E
           < 1/N
           < 1/p_u = 1/33/55/79/11......(p_u-2)/p_u.

Step 2A.

Rule "one more than the previous one" gives us a natural number array which takes E steps uptil E as 1, 2, 3, ......, E.

Rule "proportionately" permits us to cover as a table of p uptil E only as E/p steps as p, 2p, 3p, ...... pp.

Taking p_r as (r)th prime and p_r+1 as (r+1)th prime, r>2, as such, both odds, shall be expressible as p_r+1 = p_r+2s for suitable whole number s and p_r+1 E.

Now, taking A(1) as array of whole numbers; A(2) as A(1) except multiples of 2; A(3) as A(2) except multiples of 3; and so on, A(p_r+1) as A(p_r) except multiples of p_r+1 shall be permitting us to reach at the cardinality of A(p_r+1) with the help of the cardinality of A(p_r) as:

1. Total multiples of p_r+1 uptil E are E/p_r+1.

2. Multiples of p_r+1 = odd multiples of p_r+1 + even multiples of p_r+1.

3. Even multiples of p_r+1 would be multiples of 2 and as such would be ruled out from A(1).

4. Therefore, at the most, only odd multiples of p_r+1 i.e. 1/2E/p_r+1 would remain to be ruled out from A(p_r).

5. Taking cardinality of A(p_r) E/p_r, the cardinality after accounting for 1/2E/p_r+1 i.e. odd multiples of p_r+1,

we shall be left with E/p_r - 1/2E/p_r+1 1/2E/p_r+1

6. The number 1/2E/p_r+1 gives us 1/4E/p_r+1 pairs (duplexes).

7. Now, from here 1/4E/p_r+1 pairs (duplexes), to be back to the array of whole numbers, first we have to straighten two arrays of duplexes from (1, 2M-1), (3, 2M-3), ..........., (M, M) to a single array amongst (1, 3, 5, 7, ...........2M-7, 2M-5, 2M-3, 2M-1) and then as a second step we shall be going from above odds' array to the whole numbers (1, 2, 3, 4, 5,...........,2M). This as such, in first step would take us from 1/4E/p_r+1 pairs to 1/2E/p_r+1 odds (singles) to E/p_r+1 to complete sequence of whole numbers (odds as well as evens).

8. With this, we can say we have reached from A(p_r) of cardinality E/p_r to A(p_r+1) with E/p_r+1 as cardinality.

9. The odds pairs of E/p_r = 1/4E/p_r to odds pairs of E/p_r+1 = 1/4E/p_r+1 gives us a multiplier (p_r+1-2)/p_r+1 which takes us from 1/4E/p_r to 1/4E/p_r+1 as that E/4(p_r+1) = (p_r+1-2)/p_r+1E/4p_r.

Hence we can apply the above multiplier i.e. (p_r+1-2)/p_r+1 and sequentially can cover uptil the largest prime E.

Step 3.

Let us construct a subset D(3) = D₁ (a subset of set D(2) where we are sorting out and cancelling d_r's of D(2) for which either r or 2M-r is composite with p₁ (first odd prime i.e. 3) as a factor).

We can write M = 3Q+R where Q is quotient and R is remainder obtained of division of M by 3. Therefore, the composites of the range 0 to M with 3 as factor would be Q-1. In fact, there are Q numbers which would be having 3 as a factor. But as 3 is divisible by 3 but is not a composite, therefore, the balance numbers would be Q-1. On the other hand 2M=32Q+2R would imply that 2R may be greater than 3 and as such the total numbers of the entire range 0 to 2M which may have 3 as a factor may be 2Q+1 but as this quotient 2Q+1 includes 3 also, therefore, the composite numbers would remain (2Q+1)-1 = 2Q.

Therefore, the maximum numbers uptil 2M which may have 3 as a factor and as such would be composites would be 2/3 of 2M. As such, the cardinality of the set which would remain after cancelling composites with 3 as a factor would be (3-2)/3 i.e. 1/3.

As such, the cardinality of subset D(3) = D₁ would be not less than (3-2)/3 of cardinality of D(2) i.e. 1/3E/4.

At a next step, the cardinality of subset D(4) = D₂ after cancelling out the composites with second odd prime i.e. 5 as a factor would be (5-2)/51/3E/4 i.e. 3/51/3E/4.

Sequentially, we may proceed uptil the largest odd prime p_u E and shall be having cardinality of D_u, after cancelling out the composites with all odd primes as

(p_u-2)/p_u(p_u-1-2)/p_u-1.... 3/51/3E/4

1/33/55/77/9......(N-2)/NE/4

1/NE/4

1/EE/4

E/4 2 for E 64

E=p+q for E Even 64 can be physically tested on the lines:

With this, the proof of the truth of the conjecture is complete.
                                                                                               Q.E.D.