Difference between revisions of "Bertrand's Postulate"
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==Proof== | ==Proof== | ||
− | It is similar to the proof of Chebyshev's estimates in the [[Prime Number Theorem|prime number theorem]] article but requires a closer look at the [[combinations|binomial coefficient]] <math>2n | + | It is similar to the proof of Chebyshev's estimates in the [[Prime Number Theorem|prime number theorem]] article but requires a closer look at the [[combinations|binomial coefficient]] <math>\binom{2n}{n}</math>. Assuming that the reader is familiar with that proof, the Bertrand postulate can be proved as follows. |
− | Note that the power with which a prime <math>p</math> satisfying <math>\frac{2n}3<p\le n</math> appears in the prime factorization of <math>2n | + | Note that the power with which a prime <math>p</math> satisfying <math>\frac{2n}3<p\le n</math> appears in the prime factorization of <math>\binom{2n}{n}</math> is <math>\left\lfloor\frac{2n}{p}\right\rfloor- |
2\left\lfloor\frac{n}{p}\right\rfloor=2-2=0</math>. Thus, | 2\left\lfloor\frac{n}{p}\right\rfloor=2-2=0</math>. Thus, | ||
− | <math>\frac{2^{2n}}{(2n+1)}\le{2n | + | <math>\frac{2^{2n}}{(2n+1)}\le{\binom{2n}{n}}\le |
\left(\prod_{p\le\sqrt{2n}}p^{\lfloor\log_p (2n)\rfloor}\right)\cdot | \left(\prod_{p\le\sqrt{2n}}p^{\lfloor\log_p (2n)\rfloor}\right)\cdot | ||
\left(\prod_{\sqrt{2n}<p\le\frac{2n}3}p\right)\cdot | \left(\prod_{\sqrt{2n}<p\le\frac{2n}3}p\right)\cdot | ||
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The right hand side is strictly greater than <math>1</math> for <math>n\ge 500</math>, so it remains to prove the Bertrand postulate for <math>n<500</math>. In order to do it, it suffices to present a sequence of primes starting with <math>2</math> in which each prime does not exceed twice the previous one, and the last prime is above <math>500</math>. One such possible sequence is | The right hand side is strictly greater than <math>1</math> for <math>n\ge 500</math>, so it remains to prove the Bertrand postulate for <math>n<500</math>. In order to do it, it suffices to present a sequence of primes starting with <math>2</math> in which each prime does not exceed twice the previous one, and the last prime is above <math>500</math>. One such possible sequence is | ||
− | <math>2,3,5,7,13,23,43,83,163,317,631</math>. | + | <math>2,3,5,7,13,23,43,83,163,317,631</math>. Another possible sequence is |
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+ | [[category:Axioms]] | ||
{{stub}} | {{stub}} | ||
+ | [[category:Mathematics]] |
Latest revision as of 07:44, 18 October 2024
Formulation
Bertrand's postulate states that for any positive integer , there is a prime between and . Despite its name, it is, in fact, a theorem. A more widely known version states that there is a prime between and .
Proof
It is similar to the proof of Chebyshev's estimates in the prime number theorem article but requires a closer look at the binomial coefficient . Assuming that the reader is familiar with that proof, the Bertrand postulate can be proved as follows.
Note that the power with which a prime satisfying appears in the prime factorization of is . Thus,
.
The first product does not exceed and the second one does not exceed . Thus,
The right hand side is strictly greater than for , so it remains to prove the Bertrand postulate for . In order to do it, it suffices to present a sequence of primes starting with in which each prime does not exceed twice the previous one, and the last prime is above . One such possible sequence is . Another possible sequence is This article is a stub. Help us out by expanding it.