An Elementary proof for Bertrand's Postulate

TL;DR

Bertrand's postulate asks whether for every integer $n>3$ there exists a prime in $(n,2n)$. The paper offers an elementary, self-contained proof by constructing the index set $S=\{x:\, g_x\ge p_x\}$ and introducing the notions of Maximal gap and Max gap, along with a careful analysis of floor-function properties. It employs an irrationality-based contradiction involving $A=\frac{p_n-1}{p_n}\sqrt{p_{n+1}p_n}$ and $B=\sqrt{(p_{n+1}-2)(p_n-1)}$ to establish $p_n>g_n$ for $n>3$, which then yields $p_{n+1}<2p_n$ for all $n$. From these steps, the paper deduces that no $x$ can satisfy $g_x\ge p_x$, and consequently there must be a prime in every interval $(n,2n)$, thereby proving Bertrand's postulate. The approach provides an accessible, calculus-free path through the structure of prime gaps and offers a rigorous, elementary alternative to classical proofs.

Abstract

In this paper we give an elementary proof for Bertrand's postulate also known as Bertrand-Chebyshev theorem.

Theorems & Definitions (12)

• Lemma 3.1
• proof
• Lemma 3.2
• proof
• Lemma 3.3
• proof
• Lemma 3.4
• proof
• Theorem 4.1
• proof