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 April 13th, 2012, 09:00 PM #1 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 A proof of Bertrand's postulate Bertrand's postulate also called the Bertrand-Chebyshev theorem or Chebyshev's theorem, states that if n>3, there is always at least one prime p between n and 2n-2. Equivalently, if n>1, then there is always at least one prime p such that n
 April 13th, 2012, 10:37 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 13th, 2012, 11:04 PM #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,----------------------2n-1 are all consecutive composite numbers (all together we have n-1 elements) so we set : n+1=n! +2, n+2=n!+3,----------------------2n-1=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, 09:07 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, 06:07 PM   #5
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Re: A proof of Bertrand's postulate

Quote:
 Originally Posted by Albert.Teng If there is anything wrong in this pooof please let me know My greatest appreciation to you .
The first mistake I saw was the assumption that n+1=n! +2.

 April 15th, 2012, 02:34 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 n-1 consecutive composite numbers a1=n!+2 .a2=n!+3,,a3=n!+4,-----------------------------an-1=n!+n B={a1,a2,a3,a4------------------------------------------an-1}={n!+2,n!+3,------------------------------,n!+n} Set B is possible Step 2:create a set A={n,a1,a2,a3,---------------------------an-1,2n} that is a1=n+1 2n=(an-1)+1=n!+n+1 n=(a1)-1=n!+1 we get 2n=2n! +2 = n!+n+1 again n! = n-1 there is no solution for this equation it means the creation of set A is failed ,so the proof is completed
 April 15th, 2012, 11:32 AM #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, 03:09 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 n-1 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 n-1 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, ..., 2n-1?
April 16th, 2012, 12:22 AM   #9
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Re: A proof of Bertrand's postulate

Quote:
 johnr said :Consider n = 4. While 4!+2, 4!+3 and 4!+4 are indeed three consecutive composites, so are 8, 9 and 10.
If we use 8,9,10 as 3 consecutive composites then
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, 04:42 AM   #10
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Re: A proof of Bertrand's postulate

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 Originally Posted by Albert.Teng For every element p in B must satisfy : n
But nothing that you've written implies this. You just made a bunch of unrelated statements and concluded what you wanted out of the blue.

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