In one of his papers on Goldbach’s conjecture, Miles Mathis claims to prove the conjecture in a formalized way. While the lack of definitions and explanations of some terminology used by Miles makes the paper harder to follow, we can still find some inconsistencies and flaws to point out.
Even though the method of the proof is itself flawed, let us jump right at an inconsistency in one of the definitions. Let be an even number greater than 2. Miles writes
Let y = the number of primes less than x.
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)
When calculating , Miles correctly calculates the number of odd natural numbers less than as . But when substracting the number of primes less than x from this value, he wrongly assumes that all primes are odd numbers as well. Most of them are, but 2 is not. The formula can of course easily be corrected. However, upon sending him an email about this mistake Miles wrote
I don’t include 2 as a prime. The current definition of prime is muddled, in my opinion. If you include 2, you logically should include 1 and 0 as primes.
firstname.lastname@example.org to the author of this blog, Jun 12 2010
While it is perfectly reasonable to make up one’s own definitions, stating them is certainly necessary when giving a proof! Excluding 2 from the prime numbers, the given formula for is correct – But now we find that the conjecture Miles is about to prove does not hold any longer for the number 4! (Again, this is not a very big problem because conjecture could still be true for all even integers larger than 4 and the problem at hand would just be one of definition. However, the authors felt that it is worthwhile to point out these inconsistencies to the readers.)
But there is a more serious problem with Mathis’ proof. At no point does Mathis use any properties of the prime numbers, except that they are odd. If the proof were valid, replacing the word prime in the proof with any word that signifies a set of odd numbers would again give a valid proof and thus every even integer would be representable as the sum of two such numbers. That is certainly not the case for all sets of odd numbers, showing that Mathis’ proof is not correct.