Bertrand's postulate

Theorem

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Bertrand's postulate

Theorem

In number theory, Bertrand's postulate is the theorem that for any integer n > 3 {\displaystyle n>3} , there exists at least one prime number p {\displaystyle p} with n < p < 2 n − 2. {\displaystyle n<p<2n-2.} A less restrictive formulation is: for every n > 1 {\displaystyle n>1} , there is always at least one prime p {\displaystyle p} such that n < p < 2 n . {\displaystyle n<p<2n.} Another formulation, where p n {\displaystyle p_{n}} is the n {\displaystyle n} -th prime, is: for n ≥ 1 {\displaystyle n\geq 1} p n + 1 < 2 p n . {\displaystyle p_...

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From Wikipedia

In number theory, Bertrand's postulate is the theorem that for any integer n > 3 {\displaystyle n>3} , there exists at least one prime number p {\displaystyle p} with n < p < 2 n − 2. {\displaystyle n<p<2n-2.} A less restrictive formulation is: for every n > 1 {\displaystyle n>1} , there is always at least one prime p {\displaystyle p} such that n < p < 2 n . {\displaystyle n<p<2n.} Another formulation, where p n {\displaystyle p_{n}} is the n {\displaystyle n} -th prime, is: for n ≥ 1 {\displaystyle n\geq 1} p n + 1 < 2 p n . {\displaystyle p_{n+1}<2p_{n}.} This hypothesis was first conjectured in 1845 by Joseph Bertrand, who verified it for all integers up to 3,000,000. Chebyshev proved it in 1852 and so it is also called the Bertrand–Chebyshev theorem or Chebyshev's theorem. Chebyshev's theorem can also be stated as a relationship with π ( x ) {\displaystyle \pi (x)} , the prime-counting function (number of primes less than or equal to x {\displaystyle x} ): π ( x ) − π ( x 2 ) ≥ 1 , for all x ≥ 2. {\displaystyle \pi (x)-\pi {\bigl (}{\tfrac {x}{2}}{\bigr )}\geq 1,{\text{ for all }}x\geq 2.}

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