nLab Bertrand's postulate

Introduction

Bertrand’s postulate, first proven by analytic methods by Chebyshev in 1852, is one of the few simple facts providing precise bounds on the occurrence of prime numbers.

Bertrand’s postulate

Given an integer $i \geq 1$, let $p_{i}$ denote the $i^{th}$ prime. Bertrand’s postulate is the following.

Theorem

For any integer $n \geq 1$, we have that $p_{n+1} \lt 2p_{n}$.

Remark

There are a number of equivalent formulations of Theorem . One is that, for any integer $n \gt 3$, there is a prime $p$ such that $n \lt p \lt 2n$.

Proof given Goldbach’s conjecture

Around 2005, it was noticed by Henry J. Ricardo and Yoshihiro Tanaka (Ricardo 05) that Bertrand’s postulate follows from the Goldbach conjecture. We record the proof (of the formulation in Remark ).

Proof

If the Goldbach conjecture holds, there are prime numbers $p_{1}$ and $p_{2}$ such that $2n = p_{1} + p_{2}$. We must have that at least one of $p_{1}$ and $p_{2}$ is greater than or equal to $n$. Without loss of generality, suppose that $p_{1}$ has this property.

If $n$ is not prime, then it is immediate that $n \lt p_{1} \lt 2n$, as required. Suppose instead that $n$ is prime. Then $n+1$ is composite, since it must be divisible by $2$.

By Goldbach’s conjecture once more, there are prime numbers $p_{1}'$ and $p_{2}'$ such that $2(n+1) = p_{1}' + p_{2}'$. As above, at least one of $p_{1}'$ and $p_{2}'$ must be greater than or equal to $n+1$. Without loss of generality, suppose that $p_{1}'$ has this property.

Then since $n+1$ is not prime, we have that $n+1 \lt p_{1}' \lt 2(n+1)$, and thus that $n \lt p_{1}' \lt 2(n+1)$.

Now, $p_{1}'$ is not equal to $2n+1$, for this would imply that $p_{2}'=1$, which is impossible. Moreover, $p_{1}'$ is not equal to $2n$, since $2n$ is not prime. We deduce that $p_{1}' \lt 2n$, as required.

References

• Henry J. Ricardo, Goldbach’s conjecture implies Bertrand’s postulate, Amer. Math. Monthly, 112, pg. 492, 2005