20041229, 14:17  #1 
Nov 2004
2^{4} Posts 
n<p<2n ?
I know it's trivial for most of you but can you please tell me if this is true and if it's proved ? thanks

20041229, 14:22  #2  
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
Quote:
Do you mean: For any integer, n, there exists a prime, p, such that n<p<2n ? The answer is that, as stated, the premise is false. On the other hand, if you mean: For any prime, p, there exists an integer, n, such that n<p<2n. That premise is obviously true. Last fiddled with by Wacky on 20041229 at 14:25 

20041229, 14:45  #3  
May 2004
1010000_{2} Posts 
Quote:
Dave Last fiddled with by dave_dm on 20041229 at 14:45 

20041229, 14:46  #4  
6809 > 6502
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Aug 2003
101×103 Posts
22427_{8} Posts 
Quote:
2<3<2x2 3<5<2x3 4<5<7<2x4 5<7<2x5 6<7<11<2x6 7<11<13<2x7 8<11<13<2x8 9<11<13<17<2x9 10<11<13<17<19<2x10 11<13<17<19<2x11 12<13<17<19<23<2x12 13<17<19<23<2x13 14<17<19<23<2x14 15<17<19<23<29<2x15 I can only see that as numbers grow, the gap between the numbers grows. And there seems to be no gaps between primes so large below n=100 and n=1000 that there might not be a case were n<p<2n is not true. (edit: Dave beat me to the post, I hadn't seen his before my post) Last fiddled with by Uncwilly on 20041229 at 14:50 

20041229, 14:59  #5  
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
Quote:


20041229, 17:16  #6  
Cranksta Rap Ayatollah
Jul 2003
641_{10} Posts 
Quote:
Is your point important? Absolutely. However, why stifle interest on the case of a technicality? Perhaps you could include it as a caveat in a more satisfying answer. Something along the lines of "You should really be more careful when stating postulates, but I will assume you meant cases where n is a positive integer greater than 1, and in that case..." 

20041230, 14:56  #7 
Nov 2004
2^{4} Posts 
So,from what I understood, there is no proof that for any integer n, there is a prime p, such as n<p<2n ?

20041230, 15:03  #8  
Dec 2003
Hopefully Near M48
3336_{8} Posts 
Quote:


20041230, 15:19  #9 
Nov 2004
2^{4} Posts 
If there is a proof, could you tell me who proved it, and where i can find at least an outline of it? thanks

20041230, 15:38  #10 
May 2004
2^{4}×5 Posts 
I don't know how many different proofs there are. The one I have seen is easy in that it uses nothing advanced than binomial coefficients and could probably be understood by your average motivated maths undergrad.
Apparently Sylvester proved Bertrand's Postulate first. But I only know this from (literally) 5 seconds searching on Google. Not hard. Dave 
20041230, 15:50  #11 
Nov 2004
2^{4} Posts 
I found what I was looking for. Indeed, google helped me. Chebyshev proved it in the XIXth century. Here is the link of the proof for whom is interested: http://matholymp.com/TUTORIALS/Bertrand.pdf
Regards 