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April 13th, 2012, 10:00 PM  #1 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  A proof of Bertrand's postulate
Bertrand's postulate also called the BertrandChebyshev theorem or Chebyshev's theorem, states that if n>3, there is always at least one prime p between n and 2n2. Equivalently, if n>1, then there is always at least one prime p such that n<p<2n. Here is my proof : Now we create a set A={n,n+1,n+2,n+3……………..2n1,2n} with n+1 consecutive positive integers and all n+1,n+2,2n1 are (n1) composite numbers We have n+1=n! +2, n+2=n!+3,2n1=n! +n ? n=n!+1 (1) and 2n=n!+n+1=2n!+2(2) from (2) n!=n1 there is no such solution that is the creation of the set A is impossible ? There must exit at least one prime number p such that n<p<2n.and the theorem is finished . If there is anything wrong in this pooof please let me know My greatest appreciation to you . Albert.Teng from Taiwan 
April 13th, 2012, 11:37 PM  #2 
Senior Member Joined: Nov 2011 Posts: 595 Thanks: 16  Re: A proof of Bertrand's postulate
Sorry but can you explain me why you say n+1=n! +2 ??? I guess I might misunderstand something here...I assumed n! is "factorial n", but this equation has of course no solution, so I am wondering what you mean. Thanks

April 14th, 2012, 12:04 AM  #3 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Re: A proof of Bertrand's postulate
we suppose n+1,n+2,n+3,2n1 are all consecutive composite numbers (all together we have n1 elements) so we set : n+1=n! +2, n+2=n!+3,2n1=n! +n (for example if we want to creat 9 consecutive composite numbers we set 10! +2 ,10!+3,,10!+10) Albert 
April 14th, 2012, 10:07 AM  #4 
Senior Member Joined: Apr 2010 Posts: 215 Thanks: 0  Re: A proof of Bertrand's postulate
The sequence {n .. 2n} is not the same as the sequence {n! .. n!+n}.

April 14th, 2012, 07:07 PM  #5  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: A proof of Bertrand's postulate Quote:
 
April 15th, 2012, 03:34 AM  #6 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Re: A proof of Bertrand's postulate
I will say in this way: Step 1:creat a set B with n1 consecutive composite numbers a1=n!+2 .a2=n!+3,,a3=n!+4,an1=n!+n B={a1,a2,a3,a4an1}={n!+2,n!+3,,n!+n} Set B is possible Step 2:create a set A={n,a1,a2,a3,an1,2n} that is a1=n+1 2n=(an1)+1=n!+n+1 n=(a1)1=n!+1 we get 2n=2n! +2 = n!+n+1 again n! = n1 there is no solution for this equation it means the creation of set A is failed ,so the proof is completed 
April 15th, 2012, 12:32 PM  #7 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: A proof of Bertrand's postulate
What you wrote doesn't even resemble a proof of Bertrand's postulate. 
April 15th, 2012, 04:09 PM  #8 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: A proof of Bertrand's postulate
Your step 1, ie getting a sequence of n1 consecutive composites of the form n!+2, ..., n!+n is sound and well known. But what does it even have to do with your step 2? The only conceivable relevance I can figure out is if you had a proof that the sequence from n!+2 to n!+n was always the SMALLEST sequence of n1 consecutive composites. But there are counterexamples galore right from the start. Consider n = 4. While 4!+2, 4!+3 and 4!+4 are indeed three consecutive composites, so are 8, 9 and 10. Perhaps you could express in words what linkage you are trying to set up between the two sequences n!+2, ..., n!+n and n+1, ..., 2n1?

April 16th, 2012, 01:22 AM  #9  
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Re: A proof of Bertrand's postulate Quote:
B=(8,9,10} A must be : A={7,8,9,10,11,12,13,14} or A={6,7,8,9,10,11,12} For every element p in B must satisfy : n<p<2n at this time we should creat 5 or even 6 consecutive composites instead of 3  
April 16th, 2012, 05:42 AM  #10  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: A proof of Bertrand's postulate Quote:
 

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