A Short Formal Proof
of Goldbach's Conjecture

by Miles Mathis

Abstract and Note: Since my "simple" proof published several years ago has failed to take hold of the public's imagination, I have returned to offer an even simpler, shorter proof. This one is also what one might call formal, as well as general. This proof is equivalent in all ways to my first proof, but it is mainly equations, and I hold all visualizations and most commentary. While I still find the first proof appealing for its visualization, I was embarrassed to find it contained a couple of important typos, even after all these years, typos that may well have caused confusion. I have re-written that first paper to conform better to this one, and to jettison some of the remaining disclarities. I hope that the two together may now find some measure of comprehension, since I still believe the solution to be a rather simple one.

In a slightly longer paper, I proved Goldbach's Conjecture with densities and with a visualization. Since densities are just fractions, this more formal proof for Goldbach can be stated with simple fractions.

Given: For any even number x , there are x/2 sums, x terms, and x – 1 numbers in the sums (the last sum is always a repeating number).

Depending on the given x, our number of sums may be even or odd, so I will develop separate proofs for each case. If our number of sums is even, x/4 of our sums will be odd-odd sums.

In the sums, it is not terms that meet each other, it is numbers that meet each other. Therefore we must express numbers of primes and non-primes relative to numbers, not to terms. This is important.

Let y = the number of primes less than x.
Let a = the number of primes less than x, as a fraction of numbers less than x.
a = y/(x – 1)
Let b = the number of non-prime odds less than x, as a fraction of numbers less than x.
b = [(x/2) – y] /(x – 1)

If Goldbach's Conjecture is false, then no prime will meet a prime. This means that each prime will meet a non-prime odd (NPO) in the sums. The remaining NPO's must meet each other, in which case our number of NPO-NPO sums as a fraction of odd sums would be represented by [(x/4) – y]/(x/4)
or 1 – (4y/x).

Notice that if we have no P-P sums, our NPO-NPO sums must be at a minimum. The greatest number possible of NPO's will be covering P's, you see (it helps to look at some real charts to get a feel for this). Since a and b are constant for each x, if we create another NPO-NPO sum, we must create a P-P sum. To create another NPO-NPO sum, we must uncover a P.

Now, our fraction of NPO's is in fact [(x/2) – y] /(x – 1), as I have shown. If we subtract away the fraction of primes (the ones covered by NPO's), like this

{[(x/2) – y] /(x – 1)} – [y/(x – 1)] = [(x/2) – 2y] /(x – 1)

Then multiply by 2 to indicate the meeting of two numbers in a sum, like this

2[(x/2) – 2y] /(x – 1) = (x – 4y) /(x – 1) = [1 – (4y/x)] /[1 – (1/x)]

We find that our fraction of NPO-NPO meetings should be [1 – (4y/x)] /[1 – (1/x)], instead of 1 – (4y/x)

But, [1 – (4y/x)] /[1 – (1/x)] > 1 – (4y/x)

Therefore, the term 1 – (4y/x) is disallowed. It is disallowed for this reason:

Again, the creation of a P-P sum will create more NPO-NPO sums; so when P-P is forbidden, the fraction of NPO-NPO sums will be at a minimum.

Therefore, the minimum fraction of NPO-NPO sums for any given values of x and y is [1 – (4y/x)] /[1 – (1/x)]. 1 – (4y/x) is less than the minimum, therefore it is mathematically disallowed.

Because the number of sums must be an integer value, the minimum fraction of NPO-NPO sums for any given value of x and y cannot be [(x/4) – y]/(x/4) . In order to satisfy our actual minimum, it must be [(x/4) – y + 1]/(x/4). Notice the plus 1. We must add an NPO-NPO sum to satisfy our calculated minimum.

Since a and b are constant, and cannot change for a given x, you cannot create another NPO-NPO sum without also creating a P-P sum. We have just been forced to create one P-P sum.

This proof works for all values of y and x, no matter how small the fraction y/x becomes. It works for all prime densities.

This proves Goldbach's Conjecture for even numbers with an even number of sums.

This simple solution is made possible by the fact that the last sum is always composed of a repeating number. As you now see, this means we have one fewer numbers in our charts than we have terms, skewing the fractions up. Because our number fractions are slightly higher than our term fractions, our minimum meetings are affected. The simple math shows that to meet this new minimum, we must add 1 to the term minimum. And if we do that, we must create a prime pair. If you get lost in the math, remember that it is all caused by the inequality between numbers and terms.

If our number of sums is odd, (x + 2)/4 of our sums will be odd-odd sums.

a = y/(x – 1)
b = {(x + 2)/2] – y} /(x – 1)

If Goldbach's Conjecture is false, then no prime will meet a prime. This means that each prime will meet a non-prime odd (NPO) in the sums, in which case our number of NPO-NPO sums as a fraction of odd sums is represented by

{[(x + 2)/4] – y}/[(x + 2)/4]
or (x + 2 – 4 y)/(x + 2) .

But our fraction of NPO's is {(x + 2)/2]– y}/(x – 1). If we subtract away the fraction of primes (the ones covered by NPO's), like this

{[(x + 2)/2] – y} /(x – 1)} – [y/(x – 1)] = {[(x + 2)/2] – 2y} /(x – 1)

Then multiply by 2 to indicate the meeting of two numbers in a sum, like this

2{[(x + 2)/2] – 2y} /(x – 1) = (x + 2 – 4y) /(x – 1)

We find that

(x + 2 – 4y) /(x – 1) > (x + 2 – 4 y)/(x + 2)

Therefore, the term (x + 2 – 4 y)/(x + 2) is disallowed.

This proves Goldbach's Conjecture for even numbers with an odd number of sums.

Go to longer paper.

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