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/
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