I was playing with sequences and I thought of a really simple proof of this theorem. Bertrand's postulate has been proved by Chebyshev but I thought this was interesting.

Theorem: $2p_{n}>p_{n+1}$ where $p_{n}$ is the prime number sequence.


Given the difference between consecutive primes $p_{n+1}-p_n=g_n,$ define a sequence $a_n$ to be the number that when added to $g_n$ gives $p_n.$ So we have,


Solving for $a_n$ we have,


For $n>1$ the primes differ at the very least by $2$, so $a_n$ must be greater than $0$ and therefore $2p_n>p_{n+1}.$ $\square$

Am I missing something or is this proof sound?

  • 1
    $\begingroup$ primes differing by 2 means $g_n>0$, not $a_n>0$. A counterexample to Bertrand is exactly when $a_n<0$. $\endgroup$ – Matthew Towers May 31 '16 at 8:25
  • $\begingroup$ I see, but $a_n$ cannot be less than 0 because it is what must be added to $g_n$ to get $p_n$ and this is clearly greater than 0. Because $p_n$ is clearly greater than $g_n$ no? $\endgroup$ – e2theipi2026 May 31 '16 at 8:29

At first glance, your proof is probably wrong simply because it is too simple. It really does look like you are missing something. Sure, there exists simple proofs, but it should immediatelly raise an alarm as they aren't that common, especially as-yet-undiscovered very simple proofs of complex theorems.

A deeper inspection reveals your mistake:

You haven't shown why $a_n>0$, which is exactly what you need to show. There are two ways of looking at what you did and how it is wrong:

  1. You define $a_n$ as "what needs to be added to $g_n$ to get $p_n$. In this step, you assumed that $g_n$ is indeed smaller than $p_n$, but you don't know that yet. For all you know, $g_n$ could be greater than $p_n$, so then $a_n$ would be smaller.
  2. Equivalently, you defined $a_n$ as the solution to the equation $g_n+a_n=p_n$, meaning you defined $a_n$ to be $2p_n-p_{n+1}$. From this definition you can see that proving $a_n>0$ is equivalent to proving the entire theorem, so you haven't moved anywhere yet.
  • $\begingroup$ I see. But, $p_n$ is always greater than $g_n$ for $n>2$ no? The sequence $a_n$ must be greater than $0$ as a result. Am I wrong in this? $\endgroup$ – e2theipi2026 May 31 '16 at 8:33
  • $\begingroup$ @e2theipi2026 $p_n$ is greater than $g_n$ if and only if $a_n$ is greater than $2p_n>p_{n+1}$, which is what you still have to show. $\endgroup$ – 5xum May 31 '16 at 8:34
  • $\begingroup$ @e2theipi2026 Here's a general hint: if you think something is true, and even obviously true, then prove it. Try to draw a sketch of the proof. If you cannot, then the thing is (obviously) not obvioulsy true. $\endgroup$ – 5xum May 31 '16 at 8:35
  • $\begingroup$ Ok. Thank you for your help. $\endgroup$ – e2theipi2026 May 31 '16 at 8:36
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    $\begingroup$ Yah I get it. I'm beginning with the assumption that $p_n>g_n$ and I haven't proved that. O well, knew it was too good to be true. $\endgroup$ – e2theipi2026 May 31 '16 at 8:40

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