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

[QUOTE=Crook]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[/QUOTE]

If what is true?

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.

[QUOTE=Wacky]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.
[/QUOTE]

This is Bertrand's Postulate and is actually true for all n > 1 (or for all positive integers if you replace n < p < 2n by n < p <= 2n).

Dave

[QUOTE=Wacky]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.[/QUOTE]

If we can exclude the unity (as is usually done with lists of primes).

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.

([B]edit:[/B] Dave beat me to the post, I hadn't seen his before my post)

[QUOTE=Uncwilly]If we can exclude the unity (as is usually done with lists of primes).[/QUOTE]

I specifically failed to exclude unity because I wanted to make the point that one MUST be very precise when they state a postulate. A slight variation in the specification can greatly change the answer.

[QUOTE=Wacky]I specifically failed to exclude unity because I wanted to make the point that one MUST be very precise when they state a postulate. A slight variation in the specification can greatly change the answer.[/QUOTE]

of course the details are crucial, but they aren't the meat of the subject; belittling newcomers for fumbling on the secret handshake does nothing but alienate possible future contributors.

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 [i]assume[/i] you meant cases where n is a positive integer greater than 1, and in that case..."

So,from what I understood, there is no proof that for any integer n, there is a prime p, such as n<p<2n ?

[QUOTE=Crook]So,from what I understood, there is no proof that for any integer n, there is a prime p, such as n<p<2n ?[/QUOTE]

As far as I know, there is such a proof, but it is very difficult.

If there is a proof, could you tell me who proved it, and where i can find at least an outline of it? thanks

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

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: [url]http://matholymp.com/TUTORIALS/Bertrand.pdf[/url]

Regards