Bertrand's postulate

Statement

Let $n$ be a positive integer. Then there is a prime number $p$ such that $n < p \le 2 n$.

Remarks

For example, for $n = 10$ the primes $11, 13, 17, 19$ are in the range $10 < p \le 20$. An equivalent form: if $p_k$ is the $k$-th prime, then $p_{k+1} < 2 p_k$ for each $k \ge 1$. The prime number theorem gives a stronger result for large $n$, but Bertrand's postulate holds for each $n \ge 1$.

History

Joseph Bertrand stated the result as a conjecture in 1845. Chebyshev proved it in 1852.

Explanations

L1 human

Between each number and its double, there is always at least one prime. So the gap after a prime $p$ to the next prime is always smaller than $p$.