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Help requested understanding Erdos's proof of Sylvester's theorem that, if n >= 2k, n choose k has a prime divisor > k
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4
Last Post:
Apr 22, 2014 1:49 AM



Paul
Posts:
735
Registered:
7/12/10


Re: Help requested understanding Erdos's proof of Sylvester's theorem that, if n >= 2k, n choose k has a prime divisor > k
Posted:
Apr 21, 2014 4:49 PM


On Monday, April 21, 2014 12:26:06 PM UTC+1, Timothy Murphy wrote: > Paul wrote: > > > > > I'm stuck on a proof of Sylvester's theorem that, if n >= 2k, n choose k > > > has a prime divisor > k. > > > > > > The proof is available from the URL: > > > http://profs.sci.univr.it/~bellin/philsci/Erdos.pdf > > > > > > I don't follow why the claim (3) on page 285 holds, although I follow > > > everything up to that. > > > > You seem to specialize in extremely complicated problems! > > > > I haven't followed the argument here completely, > > but as far as I can see, at this point Erdos is saying that > > if \sqrt[r+1]{n} < p <= \sqrt[r]{n} > > then p occurs to power r on the left, > > and p divides each of the first r terms on the right. > > > > > Alternatively, if anyone can refer me to a free online proof of the same > > > result, that would also be of great interest. > > > > I rather doubt if anyone has simplified Erdos' proof. > > As far as I can see, Erdos is generalizing Chebyshev's theorem > > that there exists a prime between n and 2n. > > I think the proof of this result by Erdos is relatively simple, > > using the same kind of argument used here. > > But I'm sure you know that proof. > > Thanks a lot for your thoughts. I'm not sure "extremely complicated problems" is really what you mean. I think that I post for technical details about proofs, and that these details are messy and convoluted and don't lead clearly into the mainstream of mathematical thought. That is why many of my questions are unanswered. Basically, I read proofs, and then I post here when I get stuck on a proof. I'm not sure what I can do better.
I think that you are thinking along the right lines but your claim doesn't seem correct. If p is small then p does not divide [2 * ceil (n/2) choose ceil(n/2)]. So p does not divide "each of the first r terms on the right". Maybe you mean "each of the last r terms on the right"?
Paul Epstein



