@14:34 We're assuming that p and 2p | (2n)! (and 3p does not divide (2n)!) implies that p²||(2n)!, but that's not true when p=2. Indeed, with p=2, p|2n C n when ⅔n

wow! that was so cool

Thank you for giving a CORRECT proof!

An amazing video for an amazing result! Best wishes for #some1 😉

@irrazionalex226

2 жыл бұрын

Thank you guys! 😎

@leif1075

2 жыл бұрын

@@irrazionalex226 why do you say this is elementary?? IF someone didn't know this beforehand I don't see why would anyone would.think of this tonsolve this postulate..even a math whiz like Ramanujan or Russell or me..ornwhybwould anyone think of the postulate or these tools in the first place?

Wow! Good luck ;)

@irrazionalex226

2 жыл бұрын

Thank you :)

Nice! Thank you.

@irrazionalex226

2 жыл бұрын

Thank you!

Great video, was struggling to understand and this helped me, thanks!! (I think the 'i' you write at 8:01 is maybe meant to be a '2'? Just a small typo though :) I was a bit confused for a while.)

An Italian wearing a colorful headband. A book on the shelf that says "Dio".

Beautiful proof!

@irrazionalex226

2 жыл бұрын

Thank you very much!

@leif1075

2 жыл бұрын

@@irrazionalex226 Omg this is ALL RIDICULOUS CRAZY TRICKS HOW I don't see why Anyone would think any of this AT ALL..can you please respond when you can??

Erdos consiguió demostrar este resultado importante de la teoría de números ,que abre la posibilidad de demostrar otras cosas aún más importantes en teoría de números.

overall it’s excellent. though it would be a minor problem, but i never learned that product over an empty set is 1. (or said that if statement was false then there is no p inside and then erase the term) also where can i find some introductions of landaus trick

One more question please: your first key observation: P > (2n)^.5 even if P > [2n chose n], is that still P|| [2n chose n] by some definition?

@irrazionalex226

2 жыл бұрын

In this case if P>[2n chose n] then P automatically does not divide the binomial coefficient. Still in general there are some primes strictly larger than sqrt(2n) and less than [2n chose n].

I like it more and more! Is q=2m+1 prime or not prime. Couldn't get it from the audio ...

@irrazionalex226

2 жыл бұрын

Yes sorry for the induction it is sufficient to ask for q=2m+1 being prime

@eggtimer2

2 жыл бұрын

@@irrazionalex226 no need to apologise! I have to thank you for your help and patience!

Can anybody please tell me how P^(alpha) is at most 2n ? How that formula concludes this result?

How did the application of the formula got that resul at 8:13?

molto bene

Hmmm, there are some inconsistencies. I like it overall..

@irrazionalex226

2 жыл бұрын

Thank you! Yes maybe I've said something incorrect, if you want you can write here the inconsistencies and I'll be glad to explain them better

@eggtimer2

2 жыл бұрын

@@irrazionalex226 you did a great job. Wondering why you got the 2+ out in the first lemma Just to pitnit back in?

@irrazionalex226

2 жыл бұрын

@@eggtimer2 yes in order to have in the final formula 2n and not (2n+1). In fact an easier computation show that (by a mean-inequality type of argument) (2n,n)>4^n/(2n+1)

@eggtimer2

2 жыл бұрын

@@irrazionalex226 Tha ks for your reply. But now I am confused: 2n was your starting point? Can you explain a bit more why that is. Appreciate that I take a lot of your time but really appreciate it.

@irrazionalex226

2 жыл бұрын

@@eggtimer2 yes the main idea behind this proof is to reach a contradiction by bounding the binomial coefficient C(2n,n) in two ways, showing that for every big enough n these two bounds are incompatible (assuming that Bertrand's postulate is false). In this way the first lemma gives exactly the first bound for the binomial coefficient we are looking for.

I believe it's the proof of Paul erdos( It's one of elementary proofs)

@irrazionalex226

2 жыл бұрын

Yes It Is exactly his proof 😎

Thank you for a clear and complete treatment of this very interesting theorem.

saved for later

@irrazionalex226

2 жыл бұрын

Thank you Juan!

*Taking notes for use in math Olympiad

the music is annoying thanks for the vid!

are you not counting 1 as a natural number?

@irrazionalex226

2 жыл бұрын

Sorry could you point the minute in which I've said that?

I don't see why on Earth ANYONE WOHLD EVER think of ANY of this if you were told this postulate and asked to.prove it..the first step especially comes out of nowhere..and I'm very good at math..

@ahoj7720

Жыл бұрын

As for many proofs this is presented in reverse. The starting point is the remark that binomial(2*n, n) is an integer, then exploring its prime factors. According to Erdös himself the key remark is that it cannot have factors between 2/3n and n. And I agree with you: Erdös was not anyone!

@DarinBrownSJDCMath

Жыл бұрын

Erdös likely discovered this proof not by seeking to find a new proof of Bertrand's postulate, but just by playing around with the prime factorizations of C(2n, n) and trying to see what he could get out of it.

Neck beard and bandana. Are we related?

Very bad quality of video