Associated Topics || Dr. Math Home || Search Dr. Math

### 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/
```
Associated Topics:
High School Number Theory

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search