Is this a simple and sound proof of Bertrand's Postulate in the form $2p_{n}>p_{n+1}$?

Question

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.

Proof:

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,

$p_{n+1}-p_n+a_n=p_n.$

Solving for $a_n$ we have,

$a_n=2p_n-p_{n+1}.$

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?

Answer 1

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.

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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.