Prime Number Proof: p_2n Greater Than 2*p_n

Date: 10/07/2001 at 11:42:33
From: Henry Dabrowski
Subject: Prime numbers

Hi, Dr. Math,

I have a problem. How to prove that for n > 1 we have:

p_2n > 2*p_n

where p_n is an n-th prime number.

Thank you very much.
Best regards.
Henry

Date: 10/07/2001 at 15:44:14
From: Doctor Roy
Subject: Re: Prime numbers

Hello,

This is certainly true.

Let's prove this statement by induction. 

For n = 2, the 4th prime number is 7, and two times the 2nd prime 
number is 6. So the statement holds for n = 2. 

For n > 1, we have that primes can only be of the form 6n + 1, or 6n-1 
(to be very careful, primes must be of this form, but not every number 
of this form is prime. This means that only two of every six numbers 
can be prime. So, fewer than half (actually at most a third but much 
less than that) can be prime.  

If p_2n = 2*p_n, then half the numbers from p_n to p_2n are prime (or 
the odd numbers are prime). However, if we get rid of multiples of 3 
(by saying that primes must be of the form 6n + 1 or 6n - 1, then we 
see that the original inequality must hold.

I hope this helps.

- Doctor Roy, The Math Forum
  http://mathforum.org/dr.math/