{
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  "body": "\n## Statement\n\nLet $n$ be a positive integer.\nThen there is a [[prime number|0SN0MAEKYX]] $p$ such that $n \u003c p \\le 2 n$.\n\n## Remarks\n\nFor example, for $n = 10$ the primes $11, 13, 17, 19$ are in the range $10 \u003c p \\le 20$.\nAn equivalent form: if $p_k$ is the $k$-th prime, then $p_{k+1} \u003c 2 p_k$ for each $k \\ge 1$.\nThe [[prime number theorem|0EHBY5Y84N]] gives a stronger result for large $n$, but Bertrand's postulate holds for each $n \\ge 1$.\n\n## History\n\nJoseph Bertrand stated the result as a conjecture in 1845. Chebyshev proved it in 1852.\n\n## Explanations\n\n### L1 human\n\nBetween each number and its double, there is always at least one prime.\nSo the gap after a prime $p$ to the next prime is always smaller than $p$.\n",
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  "name": "Bertrand's postulate",
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