Proof of Bertrand's postulate

Template:Short description In mathematics, Bertrand's postulate (now a theorem) states that, for each <math>n \ge 2</math>, there is a prime <math>p</math> such that <math>n<p<2n</math>. First conjectured in 1845 by Joseph Bertrand,<ref>Template:Citation.</ref> it was first proven by Chebyshev, and a shorter but also advanced proof was given by Ramanujan.<ref>Template:Citation</ref>

The following elementary proof was published by Paul Erdős in 1932, as one of his earliest mathematical publications.<ref>Template:Citation</ref> The basic idea is to show that the central binomial coefficients must have a prime factor within the interval <math>(n, 2n)</math> in order to be large enough. This is achieved through analysis of their factorizations.

The main steps of the proof are as follows. First, one shows that the contribution of every prime power factor <math>p^r</math> in the prime decomposition of the central binomial coefficient <math>\textstyle\binom{2n}{n}=(2n)!/(n!)^2</math> is at most <math>2n</math>; then, one shows that every prime larger than <math>\sqrt{2n}</math> appears at most once.

The next step is to prove that <math>\tbinom{2n}{n}</math> has no prime factors in the interval <math>(\tfrac{2n}{3}, n)</math>. As a consequence of these bounds, the contribution to the size of <math>\tbinom{2n}{n}</math> coming from the prime factors that are at most <math>n</math> grows asymptotically as <math>\theta^{\!\;n}</math> for some <math>\theta<4</math>. Since the asymptotic growth of the central binomial coefficient is at least <math>4^n\!/2n</math>, the conclusion is that, by contradiction and for large enough <math>n</math>, the binomial coefficient must have another prime factor, which can only lie between <math>n</math> and <math>2n</math>.

The argument given is valid for all <math>n\ge427</math>. The remaining values of <math>n</math> are verified by direct inspection, which completes the proof.

Lemmas in the proofEdit

The proof uses the following four lemmas to establish facts about the primes present in the central binomial coefficients.

Lemma 1Edit

For any integer <math>n>0</math>, we have

<math>\frac{4^n}{2n} \le \binom{2n}{n}.</math>

Proof: Applying the binomial theorem,

<math>4^n = (1 + 1)^{2n} = \sum_{k = 0}^{2n} \binom{2n}{k}=2+\sum_{k = 1}^{2n-1} \binom{2n}{k} \le 2n\binom{2n}{n},</math>

since <math>\tbinom{2n}{n}</math> is the largest term in the sum in the right-hand side, and the sum has <math>2n</math> terms (including the initial <math>2</math> outside the summation).

Lemma 2Edit

For a fixed prime <math>p</math>, define <math>R = R(n,p)</math> to be the p-adic order of <math>\tbinom{2n}{n}</math>, that is, the largest natural number <math>r</math> such that <math>p^r</math> divides <math>\tbinom{2n}{n}</math>.

For any prime <math>p</math>, <math>p^{R }\le 2n</math>.

Proof: The exponent of <math>p</math> in <math>n!</math> is given by Legendre's formula

<math>\sum_{j = 1}^\infty \left\lfloor \frac{n}{p^j} \right\rfloor\!,</math>

so

<math>R=\sum_{j = 1}^\infty \left\lfloor \frac{2n}{p^j} \right\rfloor - 2\sum_{j = 1}^\infty \left\lfloor \frac{n}{p^j} \right\rfloor=\sum_{j = 1}^\infty \left(\left\lfloor \frac{2n}{p^j} \right\rfloor - 2\!\left\lfloor \frac{n}{p^j} \right\rfloor\right)</math>

But each term of the last summation must be either zero (if <math>n/p^j \bmod 1<1/2</math>) or one (if <math>n/p^j\bmod 1\ge1/2</math>), and all terms with <math>j>\log_p(2n)</math> are zero. Therefore,

and

Lemma 3Edit

If <math>p</math> is an odd prime and <math>\frac{2n}{3} < p \leq n</math>, then <math>R(n,p) = 0.</math>

Proof: There are exactly two factors of <math>p</math> in the numerator of the expression <math>\tbinom{2n}{n}=(2n)!/(n!)^2</math>, coming from the two terms <math>p</math> and <math>2p</math> in <math>(2n)!</math>, and also two factors of <math>p</math> in the denominator from one copy of the term <math>p</math> in each of the two factors of <math>n!</math>. These factors all cancel, leaving no factors of <math>p</math> in <math>\tbinom{2n}{n}</math>. (The bound on <math>p</math> in the preconditions of the lemma ensures that <math>3p</math> is too large to be a term of the numerator, and the assumption that <math>p</math> is odd is needed to ensure that <math>2p</math> contributes only one factor of <math>p</math> to the numerator.)

Lemma 4Edit

An upper bound is supplied for the primorial function,

where the product is taken over all prime numbers <math>p</math> less than or equal to <math>n</math>.

For all <math>n\ge1</math>, <math>n\#<4^n</math>.

Proof: We use complete induction.

For <math>n=1,2</math> we have <math>1\#=1<4</math> and <math>2\#=2<4^2=16</math>.

Let us assume that the inequality holds for all <math>1\le n\le2k-1</math>. Since <math>n=2k>2</math> is composite, we have

Now let us assume that the inequality holds for all <math>1\le n\le2k</math>. Since <math>\binom{2k+1}{k}=\frac{(2k+1)!}{k!(k+1)!}</math> is an integer and all the primes <math>k+2\le p\le2k+1</math> appear only in the numerator, we have

<math>\frac{(2k+1)\#}{(k+1)\#}\le\binom{2k+1}{k}=\frac12\!\left[\binom{2k+1}{k}+\binom{2k+1}{k+1}\right]<\frac12(1+1)^{2k+1}=4^k .</math>

Therefore,

<math>(2k+1)\#=(k+1)\#\cdot\frac{(2k+1)\#}{(k+1)\#}\le4^{k+1}\binom{2k+1}{k}<4^{k+1}\cdot4^k=4^{2k+1}.</math>

Proof of Bertrand's PostulateEdit

Assume that there is a counterexample: an integer n ≥ 2 such that there is no prime p with n < p < 2n.

If 2 ≤ n < 630, then p can be chosen from among the prime numbers 3, 5, 7, 13, 23, 43, 83, 163, 317, 631 (each being the largest prime less than twice its predecessor) such that n < p < 2n. Therefore, n ≥ 630.

There are no prime factors p of <math>\textstyle\binom{2n}{n}</math> such that:

2n < p, because every factor must divide (2n)!;
p = 2n, because 2n is not prime;
n < p < 2n, because we assumed there is no such prime number;
2n/3 < p ≤ n: by Lemma 3.

Therefore, every prime factor p satisfies p ≤ 2n/3.

When <math>p > \sqrt{2n},</math> the number <math>\textstyle {2n \choose n} </math> has at most one factor of p. By Lemma 2, for any prime p we have p^R(p,n) ≤ 2n, and <math>\pi(x)\le x-1</math> since 1 is neither prime nor composite. Then, starting with Lemma 1 and decomposing the Template:Nowrap side into its prime factorization, and finally using Lemma 4, these bounds give:

<math>\frac{4^n}{2n}\le\binom{2n}{n}=\left(\,\prod_{p\,\le\,\sqrt{2n}}p^{R(p,n)}\right)\!\!\left(\prod_{\sqrt{2n}\,<\,p\,\le\,2n/3}\!\!\!\!\!\!\!p^{R(p,n)}\right)<\left(\,\prod_{p\,\le\,\sqrt{2n}}\!\!2n\right)\!\!\left(\prod_{p\,\le\,2n/3}\!\!p\right)\le(2n)^{\sqrt{2n}-1}4^{2n/3}.</math>

Therefore

<math>4^{n/3}<(2n)^\sqrt{2n}</math>, which simplifies to <math>2^\sqrt{2n}<(2n)^3.</math>

Taking the base-2 logarithm of both sides yields

By concavity of the right-hand side as a function of n, the last inequality is necessarily verified on an interval. Since it holds true for <math>n=426</math> and it does not for <math>n=427</math>, we obtain

But these cases have already been settled, and we conclude that no counterexample to the postulate is possible.

Addendum to proofEdit

It is possible to reduce the bound to <math>n=50</math>.

For <math>n\ge17,</math> we get <math>\pi(n)<\frac{n}{2}-1</math>, so we can say that the product <math>p^R</math> is at most <math>(2n)^{0.5\sqrt{2n}-1}</math>, which gives

<math>\begin{align}&\frac{4^n}{2n}\le\binom{2n}{n}\le(2n)^{0.5\sqrt{2n}-1}4^{2n/3}\\&4^{\sqrt{2n}}\le(2n)^3\\&2\sqrt{2n}\le3\log_2(2n)\end{align}</math>

which is true for <math>n=49</math> and false for <math>n=50</math>.

ReferencesEdit

Template:Reflist

External linksEdit

Chebyshev's Theorem and Bertrand's Postulate (Leo Goldmakher): https://web.williams.edu/Mathematics/lg5/Chebyshev.pdf
Proof of Bertrand's Postulate (UW Math Circle): https://sites.math.washington.edu/~mathcircle/circle/2013-14/advanced/mc-13a-w10.pdf
Proof in the Mizar system: http://mizar.org/version/current/html/nat_4.html#T56
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