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A number theory paper I wrote was recently rejected from a journal due to "working under the untenable hypothesis that the natural density behaves like a probability measure (as it is not $\sigma$-additive and not stable under finite intersections)".

My paper was about providing a lattice-theoretic proof that the probability two Gaussian integers is relatively prime is $\frac{1}{\zeta_{\mathbb{Q}(i)}(2)}$, where $\zeta_{\mathbb{Q}(i)}(2)$ is the Dedekind zeta function over the Gaussian integers (as the known proof by Collins and Johnson from 1989 used difficult analytic number theory techniques). The main objection from the referee was to an argument of the following form:

Theorem: Let $x,y \in \mathbb{Z}$ be randomly chosen. Then, $$P(\gcd(x,y)=1) = \frac{1}{\zeta(2)}$$ Proof: We have that $P(\gcd(x,y)=1) = P(p \nmid x \cap p \nmid y, \forall p)$, where $p$ is prime. For a fixed prime $p$, the probability that $p \mid x$ and $p \mid y$ is $\frac{1}{p^2}$ and so we have that the probability this does not occur is $1 - \frac{1}{p^2}$. Therefore, taking the product over all primes we have that $$P(\gcd(x,y)=1) = \prod_{p \; \text{prime}} \left(1 - \frac{1}{p^2}\right) = \left(\prod_{p \; \text{prime}} \left(\frac{1}{1-p^{-2}}\right) \right)^{-1} = \frac{1}{\zeta(2)}$$

Why is this invalid reasoning?

Samuel Reid
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  • Have you read these answers? http://math.stackexchange.com/questions/37520/on-the-probability-that-two-positive-integers-are-relatively-prime?rq=1 – Matthew Conroy Feb 21 '14 at 20:46
  • This almost seems like a MathOverflow-level question. – John Feb 21 '14 at 20:47
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    The $\gcd$ stuff at the end can be rewritten as an asymptotic density argument. Perhaps so can your Gaussian integer calculation. – André Nicolas Feb 21 '14 at 20:53
  • @AndréNicolas: What is the necessity of the asymptotic density argument? If you are referring to, say Theorem 332 of Hardy and Wright (which proves $P(\text{gcd}(x,y)=1)=\frac{1}{\zeta(2)}$ using the Mobius function, inclusion-exclusion, and Big O notation on special numbers), then this is not an answer to my question. I realize you can give a proof that way, and Johnson and Collins gave a similar style proof for the case of the Gaussian integers. My question is what the problem is with my above proof over the integers (as my Gaussian integer proof is of a similar style)? – Samuel Reid Feb 21 '14 at 21:33
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    You are using properties of probability. A probability measure on the natural numbers is an assignment of specific non-negative reals $p_k$ such that $\sum_1^\infty p_k=1$. Obviously such a measure cannot assign equal weights to all natural numbers. So for what specific measure are your probabilities calculated? – André Nicolas Feb 21 '14 at 21:53
  • @AndréNicolas: The probability 1/p is calculated, and is to be understood as, the limit when N→∞ from the "probability of randomly selecting a prime p in the natural numbers 1 to N". That is, we assign only a probability measure on the first N numbers, find the probability, and then extend it. Does that work? – Samuel Reid Feb 21 '14 at 23:27
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    That's asymptotic density, don't call the limit, if it exists, probability. Informal arguments that treat it as if it were probability can lead to incorrect conclusions. – André Nicolas Feb 21 '14 at 23:47
  • @AndréNicolas: Thank you so much for the clarification! So, if I were to replace the word probability with asymptotic density and have an intermediary step with a limit, then the proof is correct? In the statement of the theorem can I use the word "probability" as in the Johnson and Collins paper (http://link.springer.com/chapter/10.1007%2F3-540-51084-2_23)? Or must we state the theorems with "asymptotic density" instead? I think that my argument is the same, but that I have been misusing the terminology. If so, you can add that answer and I will accept it! – Samuel Reid Feb 22 '14 at 00:35
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    That paper also misuses the terminology! I would not mind it much in the title, but if it is used in the body I would get upset. Yes, if density is used, and existence of asymptotic density (limit) is proved, everything is fine. A bit of history. It had been known for a while that if the limit of $\pi(x)/(x/\ln x)$ exists, that limit must be $1$. The hard part of the Prime Number Theorem was the proof of existence. – André Nicolas Feb 22 '14 at 00:40
  • @AndréNicolas: I see. So, I will re-write the paper with the use of asymptotic density and a limit and then re-submit the paper. Thank you so much for your help! – Samuel Reid Feb 22 '14 at 00:41
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    You are welcome. Good luck! Perhaps you ran into a referee in a bad mood. In a better mood, (s)he might have made suggestions about what changes to make. – André Nicolas Feb 22 '14 at 00:45

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Here is an example that shows how your reasoning can go wrong.

Say that a positive integer $n$ is $k$-special if $n \equiv k \bmod p_k^2$, where $p_k$ is the $k$th prime. The density of $k$-special positive integers is of course $1/p_k^2$.

Now say that $n$ is ordinary if $n$ is not $k$-special for any $n$. Applying your reasoning, we get that the density of ordinary positive integers is $$ \prod_k \left(1 - \frac{1}{p_k^2} \right) = \frac{1}{\zeta(2)}. $$ But note that $n$ is always $n$-special. Therefore, there are no ordinary positive integers!

Mirlan
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