
#1
Jan412, 06:03 AM

P: 12

It is known that for every n > 1 there is always at least one prime p such that n < p < 2n.
The Bertrand's theorem's proof I read is quite involved. I've thought of another way to prove it but I think it's too simple to be correct. Can anyone please point out my mistakes? below, Fundamental Theorem means the Fundamental Theorem of Arithmetic Proof: # 1) for any n > 1, n can be expressed as a product of prime factors (Fundamental Theorem) 2) Let PF[n] be a set containing all prime numbers that appear in the prime factorization of n. e.g. PF[6] = {2,3} 3) obviously, if x is in PF[n] then x ≤ n. 4) for all prime numbers that are smaller than n, we look for one that is not in PF[n], if it exists, let it be M. for M exists: 5a) consider K = n + M, K ≤ 2n (from 4) 6a) now, K is not divisible by any prime smaller than n, since for any prime that divides n, it does not divide M, and M does not divide n. (from 4) 7a) but K has a prime factorization (Fundamental Theorem) 8a) hence there must be a prime in the range [n, 2n] for M not exists: 5b) consider K' = n+1, K ≤ 2n (assumption, n>1) 6b) now, K is not divisible by any prime smaller than n, since for any prime that divides n, it does not divide 1. 7b) but K has a prime factorization (Fundamental Theorem) 8b) hence there must be a prime in the range [n, 2n] This proof seems alright to me, but after all I'm just an amateur. I would be grateful if anyone could point out the insufficiency of this proof so that I can learn from my mistakes. Thanks. 



#2
Jan412, 06:46 AM

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#3
Jan412, 06:53 AM

P: 12

ok, I understand now, thanks!




#4
Jan612, 08:34 AM

P: 42

Proving (a weaker form of) Bertrand's postulate 



#5
Jan612, 08:49 AM

P: 42





#6
Jan612, 10:18 AM

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#7
Jan612, 10:56 AM

P: 42




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