It is known that for every n > 1 there is always at least one prime p such that n < p < 2n.(adsbygoogle = window.adsbygoogle || []).push({});

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:

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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.

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# Proving (a weaker form of) Bertrand's postulate

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