Timeline for Awfully sophisticated proof for simple facts
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 31 at 9:04 | comment | added | Fedor Petrov | @KConrad "large and relatively prime to a couple of specific integers" - that's how Euclid proved infinitude of primes, right? | |
| Dec 1, 2016 at 5:36 | comment | added | KConrad | @OstapChervak it is not true that irrationality of $\pi$ depends on infinitude of the primes. Niven's proof at en.wikipedia.org/wiki/… does not use primes. I suspect you are thinking of proofs of transcendence of $\pi$. Often the most accessible such proofs use a large auxiliary prime, but there is no need for that number to be prime (only to be large and relatively prime to a couple of specific integers). See my answer at mathoverflow.net/questions/21367/…. | |
| May 12, 2015 at 11:52 | comment | added | Vim | I believe I saw this one in R. Courant's famous book What Is Mathematics. | |
| Jan 4, 2013 at 21:03 | comment | added | Ostap Chervak | Irrartionality of $\pi$ uses infinitude of primes, so proofs by zeta function are all circular – | |
| Nov 5, 2010 at 11:59 | comment | added | Boris Bukh | @Vince: Yes, we could. Apery proved in 1979 that $\zeta(3)$ is irrational. | |
| Nov 4, 2010 at 18:56 | comment | added | Vince | Could we instead look at $\zeta(2)$, which is $\sum_n \frac{1}{n^2}$ due to Euler, which in turn is $\frac{\pi^2}{6}$ which is irrational? I ask because I don't immediately see why $\zeta(3)$ is irrational. | |
| Oct 18, 2010 at 13:13 | comment | added | J.C. Ottem | @Todd: You're right, I meant the prime number theorem (and you compare it to $\sum \frac1{n \log n}$. | |
| Oct 18, 2010 at 5:06 | comment | added | Thomas Bloom | @Qiaochu,Mark: It does (they need to embed [1,N] in $Z_p$ for some prime bigger than N to get a nice field structure for some arguments to work). | |
| Oct 18, 2010 at 2:33 | comment | added | sigoldberg1 | As an aside, Titchmarsh argued the infinitude of the primes because zeta(1) is infinite on page 1 of his book. | |
| Oct 17, 2010 at 22:04 | comment | added | Qiaochu Yuan | @Mark: surely the proof of the Green-Tao theorem uses at some point the infinitude of the primes... | |
| Oct 17, 2010 at 19:39 | comment | added | Mark | Even better, there are infinitely many primes because there are arbitrarily long arithmetic progressions in them (the Green-Tao theorem). | |
| Oct 17, 2010 at 17:17 | comment | added | Barry | It sounds like he is trying to bound the partial sums from below by half the harmonic series since for each n, there exists a prime p with 1/2n < 1/p < 1/n. Of course the problem is that multiple n's can lead to the same p so this won't work. In any case, the "Proof's from the Book" proof of Bertrand's postulate given by Erdos is simple enough that I don't think the above proof, even if valid, would be nuking a mosquito. | |
| Oct 17, 2010 at 16:51 | comment | added | Andrés E. Caicedo | @J.C. Ottem, How does Bertrand's postulate give us divergence? | |
| Oct 17, 2010 at 15:47 | comment | added | J.C. Ottem | Or since $\frac12+\frac13+\frac15+\frac17+\ldots$ diverges (by Bertrand's postulate). | |
| Oct 17, 2010 at 15:41 | history | answered | Boris Bukh | CC BY-SA 2.5 |