Probability of Relatively Prime Integers

Question

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?

Answer 1

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!