Timeline for What is the simplest proof that the density of primes goes to zero?
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| Jan 17, 2021 at 20:08 | comment | added | GH from MO | @FedorPetrov: I agree that proving $\prod_p(1-1/p)=0$ via $\sum_p 1/p=\infty$ was a bit silly. | |
| Jan 17, 2021 at 18:44 | comment | added | GH from MO | @dodd: Well, it is not clear why you think my proof is incorrect. I (and others) tried to argue with you, there was no authority involved. Majority, yes, but this happens with most correct proofs (majority agrees). I end the discussion here. (BTW my argument also relied, ultimately, on the work of Erdős.) | |
| Jan 17, 2021 at 18:35 | comment | added | markvs | You are attempting a "proof by authority" and a "proof by majority". Next you will say that 97% of scientists consider the proof correct. The fact is that with all possible simpifications your "proof" if written completely is about 3 times longer than Edos' which the OP called "involved". | |
| Jan 17, 2021 at 18:26 | comment | added | GH from MO | @dodd: The proof is correct, as many of us tried to explain to you. Whether it is simpler than Erdős's proof of $\pi(x)\ll x/\log x$ is subjective. I would say it is not much simpler, but with Terry Tao's simplification (see his post), it is definitely simpler. | |
| Jan 17, 2021 at 18:19 | comment | added | markvs | @GHfromMO: assuming a version of this is correct, do you really think it is simpler than Erdos' proof? | |
| Jan 17, 2021 at 17:52 | comment | added | GH from MO | @dodd: At the end of my previous comment, I meant $\exp(-1/p)$ instead of $\exp(1/p)$. | |
| Jan 17, 2021 at 17:29 | comment | added | GH from MO | @dodd: For any positive integer $n$, the density of integers coprime to $n$ exists and equals to $\phi(n)/n$. So the upper density of the primes is at most the infimum of $\phi(n)/n$ over all positive integers $n$. This infimum is precisely $\prod_p(1-1/p)$, which is zero by the upper bound $1-1/p<\exp(1/p)$ and $\sum 1/p=\infty$. | |
| Jan 17, 2021 at 17:27 | comment | added | Fedor Petrov | I would say $\prod(1-1/p)=0$ follows from $\prod_{p\leqslant n}(1-1/p)\leqslant 1/(1+1/2+\ldots+1/n)$, you do not need the detour via $\sum 1/p=\infty$. | |
| Jan 17, 2021 at 17:26 | comment | added | Fedor Petrov | @dodd the implication $A\Rightarrow B$ is true whenever $A$ and $B$ are both true, and this is so. Note that we do not prove "if $\sum 1/n_k=\infty$ then $n_k$'s have zero density" for any sequence of positive integer numbers. Only for primes. It is bit counterintuitive (we deduce that are few primes from the fact that there are many primes), but it works. | |
| Jan 17, 2021 at 17:25 | comment | added | Terry Tao | @dodd Actually, the material implication is true (true implies true). The proof you have in mind is invalid, but that isn't the proof used here. Please reread all the discussion carefully. | |
| Jan 17, 2021 at 17:23 | comment | added | GH from MO | @WillSawin: By "at least" you meant "at most". | |
| Jan 17, 2021 at 17:23 | comment | added | markvs | @TerryTao: So as a result of the two steps he "proves" that if $\sum_p 1/p=\infty$, then primes have zero density. That implication is wrong. | |
| Jan 17, 2021 at 17:21 | comment | added | GH from MO | @TerryTao: Thanks for your insightful and enlightening comments! | |
| Jan 17, 2021 at 17:20 | comment | added | Terry Tao | In fact, what is really being proven here is that any sequence of mutually coprime natural numbers necessarily has zero natural density. | |
| Jan 17, 2021 at 17:18 | comment | added | Terry Tao | @dodd The argument is in two steps. The first is that $\sum_p 1/p=\infty$ implies $\prod_p (1-1/p)=0$. This step is general and would also apply to the natural numbers $n$. The second is to use the Chinese remainder theorem (and the sieve of Eratosthenes) to show that $\prod_p (1-1/p)=0$ implies that the primes have density zero. Here we use the fact, specific to the primes, that large primes are coprime to all small primes. | |
| Jan 17, 2021 at 17:16 | comment | added | Terry Tao | @FedorPetrov Ah, of course, you're right (I am so used to a normalising factor like $1/\log x$ when summing reciprocals that I didn't notice that it is absent in this case!). | |
| Jan 17, 2021 at 17:12 | comment | added | markvs | @FedorPetrov: I have read it carefully. The answer is wrong. He claims that $\sum 1/p =\infty$ implies $\pi(x)=o(x)$. That is not true since $\sum 1/n=\infty$. Moreover $\sum n^2= \infty$. | |
| Jan 17, 2021 at 17:11 | comment | added | Fedor Petrov | @TerryTao I think, we get a genuine zero density. If $\pi(n)\geqslant 2\kappa\cdot n$ for a constant $\kappa>0$ and all $n$ from a certain infinite set $S$, then we may choose a subsequence $n_1<n_2<\ldots$ in $S$ such that $\pi(n_k)-\pi(n_{k-1})>\kappa n_k$ that implies $\sum_{n_{k-1}<p\leqslant n_k} 1/p>\kappa$, and totally $\sum 1/p=\infty$. | |
| Jan 17, 2021 at 17:09 | comment | added | Terry Tao | Correction: one only gets zero logarithmic density (as opposed to zero natural density) of the primes with my version of the argument. | |
| Jan 17, 2021 at 17:09 | comment | added | Fedor Petrov | @dodd this answer is correct, read it carefully | |
| Jan 17, 2021 at 17:08 | comment | added | Will Sawin | @dodd It does, using the Chinese remainder theorem to deduce that for any finite set $p_1,\dots, p_n$ of primes, the density of primes is at least $\prod_{i=1}^n (1-\frac{1}{p_i})$. | |
| Jan 17, 2021 at 17:06 | comment | added | markvs | This answer is wrong. The equality $\sum 1/p=\infty$ does not imply inequality $\pi(x)=o(x)$. | |
| Jan 17, 2021 at 17:04 | comment | added | Terry Tao | If $\sum_p 1/p < \infty$ then one already has that the density of primes goes to zero by comparison with the harmonic series $\sum_n 1/n = \infty$. So one can give a self-contained proof by splitting into the two cases $\sum_p 1/p = \infty$, $\sum_p 1/p < \infty$ and treating the two cases separately. | |
| Jan 17, 2021 at 16:57 | history | answered | GH from MO | CC BY-SA 4.0 |