Talk:Proof of Bertrand's postulate

message: I fail to see why assuming there is no prime p such that n < p < 2n-2 won't work. This allows the possibility of 2n-1 being prime which would make the right side 
(2n-1)(2n)^{\sqrt{2n}}(2n/3)\# but still maintains the asymptotic behavior that will lead to a contradiction.
Likewise assuming that there is no prime p such that n < p < 2n-4 would allow 2n-1 and 2n-3 to be prime and give a right side bound of
(2n-3)(2n-1)(2n)^{\sqrt{2n}}(2n/3)\# still maintaining the asymptotic contradiction.
I must be missing something obvious here.  —Preceding unsigned comment added by 71.112.88.200 (talk) 00:40, 24 September 2010 (UTC)
thread_title:  Why can't the proof be strengthened? 

message: Ramanujan's proof is needed as to explain the prime numbers which bare his name.
http://www.imsc.res.in/~rao/ramanujan/CamUnivCpapers/Cpaper24/page1.htm
The beauty of this proof is not in the simplicity, but in the foundation which can be used for other proofs. Namely, with the help of a new paper by Hashimoto (July, 2008), we can prove Legendre's conjecture.
Can we make another section which explains Ramanujan proof? I am not a good writer, but do understand the general concept of why it is important. Namely, the increase in the number of Ramanujan primes, Ri, increases the number of primes between n and 2*n with n > Ri. This is the foundation of other proofs.
Hashimoto (July, 2008) http://arxiv.org/abs/0807.3690
Reddwarf2956 (talk) 11:35, 29 September 2009 (UTC)
thread_title:  Another Proof 

message: I really enjoyed this very clear exposition! I have a couple of remarks.
* Although this is written in form of a proof by contradiction, the argument by contradiction is not really needed. One could put it more naturally (IMO) as an  a priori bound  on the numbers n that fail to satisfy the thesis, plus a direct examination of the remaining small cases, a kind of customary procedure in mathematics. Thus, if there is no prime between n and 2n then n verifies an inequality yielding to n < 468, and then one checks the remaining cases (so I'd suggest to move the first lines of the proof, about the small values of n, to the end.)
* The slightly stronger lower bound for the central binomial coefficient, 
:\frac{4^n}{2n} \le {2n \choose n} 
is also available (we may just put together \scriptstyle{2n \choose 0} and \scriptstyle{2n \choose 2n} in the sum, so that the number of terms in the sum is now 2n). The latter gives a somehow nicer final inequality   \scriptstyle 2^{\frac{2n}{3}}\le (2n)^{\sqrt{2n}+1}\; , that yields quickly to n < 468 (the conclusion, although standard, may be not immediate, and of interest to some reader). --pma 15:46, 19 September 2012 (UTC)
replies: text: Let me add a link to a very nice explanation about why we like more direct proofs (even in form of contrapositive) than a proof by contradiction: http://mathoverflow.net/questions/12342/reductio-ad-absurdum-or-the-contrapositive/12400#12400 (BTW, another famous theorem about prime numbers, the Euclid proof of infinitude, is also sometimes written in form of reductio ad absurdum, although  the original proof was constructive!) --pma 16:21, 19 September 2012 (UTC)
thread_title: Two remarks

message: This is a really nice article...the only thing is, the in-line math mode is visually distracting, esp. if the HTML font size is small.  Unfortunately, I don't see any way around this without completely disrupting the readability of the proof.  Maybe this is an issue that will be solved as the wiki evolves. Revolver 18:42, 31 Jan 2004 (UTC)
replies: text: This doesn't help the people who don't have Wikipedia logins, but you could try turning on Mathjax (in preferences ⇒ appearance ⇒ math). It will probably look better that way. —David Eppstein (talk) 00:17, 25 November 2012 (UTC)
thread_title: Font

message: Please see math.SE for a review of Lemma 4. Mark Hurd (talk) 00:25, 3 February 2015 (UTC)
replies: text: Lemma 4 appears to have a factual error. At the time of this writing it claims that \binom{2n}{n} <2^{2n-2} when n≥3. But simply plugging in n=3 gives 20<16, and plugging in n=4 gives 70<64. Furthermore, no justification is given for the claimed (but incorrect) inequality. Qaanol (talk) 17:47, 5 April 2015 (UTC)
replies: text: In fact, the bound in Lemma 4,  x\#  \le 2^{2x-3} is a bit stronger that needed, which is  x\#  \le 4^x. 
pma 07:50, 20 December 2018 (UTC)
thread_title:  Lemma 4 being reviewed 

message: 1. "When  p > \sqrt{2n}, the number \textstyle {2n \choose n}  has at most one factor of p."
:This needs to be explained.
2. This whole section goes way too fast with multiple steps in a single line:
"Then, starting with Lemma 1 and decomposing the right-hand side into its prime factorization, and finally using Lemma 4, these bounds give:
:\frac{4^n}{2n }
       \le \binom{2n}{n}
       = \left(\prod_{p \le \sqrt{2n}} p^{R(p,n)}\right) \left(\prod_{\sqrt{2n} < p \le \frac{2n}{3}} p^{R(p,n)}\right)
       < (2n)^{\sqrt{2n}}  \prod_{1 < p \leq \frac{2n}{3} } p
       = (2n)^{\sqrt{2n}} \left( \frac{2n}{3}\right)\# \le (2n)^{\sqrt{2n}} 4^{2n/3}.
Taking logarithms yields to
:{\frac{\log 4}{3}}n  \le (\sqrt{2n}+1)\log 2n\; .
By concavity of the right-hand side as a function of n, the last inequality is necessarily verified on an interval."
:I know that whoever posted the proof as it currently appears did a lot of work, and I am grateful for that.
:Nevertheless, the last few steps need to be teased apart so that only one step occurs per line, so that it becomes possible to follow the reasoning. 
:Also:   Replace "yields to"   with just   "yields". 2601:200:C000:1A0:306B:A689:CCD1:440 (talk) 09:46, 20 September 2021 (UTC)
thread_title:  Things that need explaning 

message: The 3rd point • is false ! 
Any prime p such that n < p < 2n appears exactly once in the decomposition of 2n choose n. So all the proof shoul'd be reviewed. 176.177.54.9 (talk) 14:22, 7 December 2022 (UTC)
thread_title:  Section : Proof of the Bretrand's postulate