A reply to: A Short Formal Proof of Goldbach’s Conjecture

A Short Formal Proof of Goldbach’s Conjecture

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 $x$ 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 $b$, Miles correctly calculates the number of odd natural numbers less than $x$ as $\frac{x}{2}$. 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.

mm@milesmathis.com 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 $b$ 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.

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16 Responses to A reply to: A Short Formal Proof of Goldbach’s Conjecture

1. JH says:

I would be interested in understanding where Mr. Mathis made his error. The shorter version of his proof (http://milesmathis.com/gold4.html) contains only a few equations that seem to be logically sound.

You basically proved that Mr. Mathis made an error in his proof by looking at the end result – the ability to prove absurd theorems using his very proof. But where is the error? Mathematically, I am interested. Is it division of zero? Something else? It should be easy to find in the shorter version, it contains only a few paragraphs and many can be left out, because sometimes he only explains how he visualized the problem before coming up with the equations.

• Cee says:

Dear JH,

(First, allow me to ask one thing: You say that the equations in the article “seem to be logically sound” and I was wondering what you mean by that? Since the conclusion is invalid, the proof cannot be logically sound.)

I think the error is somewhere around this equation:
[1 – (4y/x)] /[1 – (1/x)] > 1 – (4y/x)

Mathis basically writes down all possible sums of two integers for a certain integer x. Let us only look at the case where x is even. For example, let’s take x = 8. This can be written as
1 + 7 (NPO+P)
2 + 6
3 + 5 (P+P)
4 + 4
5 + 3 (P+P)
6 + 2
7 + 1 (P+NPO)

where I have indicated Mathis’ terminology for the type of sums that may occur. These are clearly symmetric, so it suffices to look only at half of the sums. Since all primes are odd integers (except for 2, but Mathis does not count 2 as a prime) we only need to think about sums of odd integers.

Mathis then proceeds to calculate (assuming the conjecture is false) the fraction of sums where non-prime odd integers are summed with another non-prime odd integers relative to the number of sums where two odd integers (prime or non-prime) are summed. In his formula, he uses the symmetry mentioned above to avoid counting similar sums (e.g. 7+1 and 1+7) twice.

He compares this to the ratio of non-prime odd integers minus the number primes to the number of integers smaller than x and multiplies with 2 for some reason or another. This time however, he does not use the symmetry.
He then compares the two ratios and concludes that something is “disallowed”.

I don’t really see why the two ratios should be compared.

Cheers,
Cee

• janos havas says:

Hi I am a mathematician and have to admit that I feel a bit ashamed because I didnt see immediately the very smart refusal made by you that Mathis simply didnt use primality. I agree with you that this is enough and you wouldnt have to investigate the “scene of the crime”. What I am interested to know if you told Mathis his fundamental flaud. Because you mention bugging him about the other little details. If you did so what was his reply????? This is vital to me because he is very bright in some respect and I agree with a lot of his critisism against the “establishment”.
In his recently published book he repeats his “proof”. Will he ever admit being wrong? At least in one point? Then I think everything collapses in his mind so it shouldnt happen. Or ???

• Gus Mueller says:

I’ll go Mathis one better. I don’t consider 2 to be an even number. 2 is only divisible by itself and 1. 1 is also only divisible by itself and 1. Therefore it would absurdly follow that 1 is an even number.

• None says:

I think what you are saying is correct but it’s not very clear to the reader, or at least it was not to me. Having looked at the “condensed proof” here:
http://milesmathis.com/gold4.html
the error occurs very early:
“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) ”
Here the x/4 entry represents the unique “odd to odd” sums of x. So for example if x was 12 it would be 11-1, 9-3, 7-5. Cleary in this case if you wish to remove the y prime entries (of which there are “y primes less than x”) from the list, you would (naively) remove y/2 entries – not y as he has used in his equation. In the example of 12, since y = 4 (giving him the benefit of the doubt and not counting 2 as a prime leaving 4 primes (11, 7, 5, 3)), you end up with (12/4-4)/(12/4) which is -1/4 – clearly ludicrous.

I also said “naively” because you could not even take y/2 as the number of primes because their proportion is not evenly distributed. You’d presumably have to have a ln(x/2) somewhere in there along with other terms to qualify the primes count.

• None says:

After thinking about it a bit more, I think you are making a different point and my point is invalid (ie it is correct to subtract y entries rather than y/2 because the assumption, which he’s attempting to disprove by contradiction, is that at this point no prime is matched with a prime).

2. JH says:

Dear Cee,

To answer your first question: In a mathematical proof, you have a set of axioms in a mathematical framework (like algebra) that you use to prove your conjecture. What I meant by “logically sound” is that I went through Mathis’s proof line by line and put a little check mark next to each line consisting of a correct algebraic operation (or some logical definition, declaration etc.). At the end I had a check mark next to each line, which would mean his argument is solid. And because the conclusion is false as you said, what’s disproven is the integrity of the framework.
Of course much more probable answer is that I overlooked something, because last time I was in a math class was 5 years ago in college (engineering). But to me this is unfinished work, because if his conclusion is invalid, it must mean that he made a very specific mistake somewhere in the proof, which I am unable to find.

You can now disregard what I said in the previous paragraph, because I now understand the error in Mathis’s thinking (I wrote this up a day later):

In your next-to-last paragraph you mention his formula for “number of NPO-NPO sums as a fraction of odd sums” (Mathis-speak), which is 1 – (4y/x) and it uses the symmetry, yes.

Then in your last paragraph you mention his other formula, where he takes the number of NPO’s, pairs them up with P’s in sums for a given x and then takes the remaining NPO’s and pairs them up in NPO-NPO pairs. And you are right, he doesn’t use the symmetry this time, which means he counts e.g. 15+9 and 9+15 as two separate sums. Then he can’t compare those two formulas of course.

And I found a second mistake when looking into it: he multiplies by 2 to calculate the number of NPO-NPO pairs from the remaining NPO’s. But what he should actually do is divide by 2, multiplying is nonsensical. You don’t calculate the number of pairs of things by multiplying the number of those things by two, you divide.

Ok, I now understand where the error is.

Thanks Cee,

JH

3. Bassam. says:

I could not find a paper from the author proving Riemann’s Hypothesis… Please if someone knows where I could find such a paper let me knoow. If there are not any, I strongly urge the author to write one.

1. It is blasphemous to say that any proof of Goldbach’s Conjecture is simple, for Euler, not having published a proof for this conjecture said he could not prove it. What Euler could not proof could not be simple, and I strongly urge the author to define in his Riemann Hypothesis paper what a difficult proof would be like.

2. If we consider in this “proof” as it is, but we replace the word prime with odd square numbers, the reasoning would still hold, and yet a new theorem would be accepted, and published, saying that every even number could be written as the sum of odd squares.

3. If the author still thinks it is an easy proof, I would care to help him and show him some complications that should arise mathematicians while proving the conjectures.

• Gus Mueller says:

”2. If we consider in this “proof” as it is, but we replace the word prime with odd square numbers, the reasoning would still hold, and yet a new theorem would be accepted, and published, saying that every even number could be written as the sum of odd squares.”

Here you’re simply saying that any even number x could be written as the sum of x 1s. That is hardly ”publishable” except in the JMILC, the Journal of Mommy I Learned to Count.

• Snake4eva says:

“saying that every even number could be written as the sum of odd squares.” That statement is incorrect. The conjecture would read “Every even integer greater than 2 can be expressed as the sum of squares two primes.” The operate word is two primes, so no “Here you’re simply saying that any even number x could be written as the sum of x 1s” x could not be written as as sum of 1s.

4. Tom Arnall says:

Certainly I am missing something, but at the moment it seems to me that in any interval between L and H which contains no primes, you can easily prove that there cannot be an even number in the interval between L+(H-L)/3 and H which is the sum of two primes. In fact, you don’t need a logical proof for it but can instead simply:

pick an interval between L and H which is empty of primes, e.g., 370262-370361

pick an even number in the interval between L+(H-L)/3 and H, e.g., 370350

derive all the possible pairs of addends for the even number

As far as I can see, for every pair of addends, at least one addend falls within the interval between L and H, i.e., in an interval in which there are no primes, i.e., you have just found an even number number which is not the sum of two primes.

I strongly suspect that it is not goodbye-Goldbach time, since lots of folks have empirically verified that Goldbach holds for the interval i’ve chosen, but I can’t see the problem in my logic. Help!

• Cee says:

I don’t see why it should be the case that at least one of the addends must fall in the interval between L and H? Could you explain?

5. ManWhoStareAtScientistsWhoAreNotCrackpots says:

You just proved yourself to be a big fraud Cee, Miles Mathis, Field Prize year 2014.

• You have to be under 40 to get a Fields Medal (not Prize). Miles is too old.

• Gus Mueller says:

You also have to know at least some math,,,,,

6. Chickenmales says:

I think it’s good that Miles Mathis is trying to solve some of these problems, and although I haven’t read a lot of his stuff, I think his problem is his ego. He spends a fair bit of time writing about how good he is that the truth sometimes seems to take a back seat. It would be easier to read if he stuck to what he was writing about and submitted his proofs for review instead of declaring their absolute correctness and blindly defending against all criticisms.

Having said that, I appreciate his efforts to solve the big problems. There is a quest for truth in what he’s doing, but it seems to me that sometimes his mind takes a bit of a tangent and his soul isn’t willing to get back on course.

My 2 cents, I could be wrong 😉