ProofOfBertrandsConjecture 2002-12-24 08:44:13 2007-02-14 09:41:13 Proof Bertrand's conjecture 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newtheorem*{lemma*}{Lemma} \PMlinkescapeword{term} \PMlinkescapeword{terms} \PMlinkescapeword{conjecture} \PMlinkescapeword{minimal} \PMlinkescapeword{formula} \PMlinkescapeword{information} This is a version of Erd\H{o}s's proof as it appears in Hardy and Wright. We begin with the following lemma. \begin{lemma*} Let $n$ be a positive integer and $p$ be a prime. The highest power of $p$ dividing $n!$ is $\displaystyle\sum_j\left\lfloor\frac{n}{p^j}\right\rfloor$, where $\lfloor x\rfloor$ is the floor of $x$. \end{lemma*} \begin{proof} Let $p^k$ divide $n!$ with $k$ as large as possible. Every $p$th term of the sequence $1,\dots,n$ is divisible by $p$, so contributes a factor to $p^k$. There are $\displaystyle\left\lfloor\frac{n}{p}\right\rfloor$ such factors. Every $p^2$th term contributes an extra factor above that, providing $\displaystyle\left\lfloor\frac{n}{p^2}\right\rfloor$ new factors. In general, the $p^j$th terms contribute $\displaystyle\left\lfloor\frac{n}{p^j}\right\rfloor$ extra factors to $p^k$. So the highest power of $p$ dividing $n!$ is $\displaystyle\sum_j\left\lfloor\frac{n}{p^j}\right\rfloor$. \end{proof} We now prove the theorem. \begin{proof}[Bertrand's conjecture] Let $n>2$ be a minimal counterexample to the claim. Thus there is no prime $p$ such that $n<p<2n$. In the sequence of primes \[ 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, \] each succeeding term is smaller than the double of its predecessor. This shows that $n\ge 2503$. The \PMlinkname{binomial expansion}{BinomialTheorem} of $(1+1)^{2n}$ has $2n+1$ terms and has largest term $\binom{2n}{n}$. Hence \[ \binom{2n}{n}\ge\frac{4^n}{2n+1}\ge\frac{4^n}{(2n)^2}. \] For a prime $p$ define $r(p,n)$ to be the highest power of $p$ dividing $\binom{2n}{n}$. To compute $r(p,n)$, we apply the lemma to $(2n)!$ and $n!$. We get that \[ r(p,n)=\sum_{j\ge 1}\left\lfloor\frac{2n}{p^j}\right\rfloor -2\sum_{j\ge 1}\left\lfloor\frac{n}{p^j}\right\rfloor =\sum_{j\ge 1}\left(\left\lfloor\frac{2n}{p^j}\right\rfloor -2\left\lfloor\frac{n}{p^j}\right\rfloor\right). \] The terms of the sum are all 0 or 1 and vanish when $\displaystyle j>\left\lfloor\frac{\log(2n)}{\log p}\right\rfloor$, so $\displaystyle r(p,n)\le\left\lfloor\frac{\log(2n)}{\log p}\right\rfloor$, that is, $p^{r(p,n)}\le 2n$. Now $\displaystyle\binom{2n}{n}=\prod_p p^{r(p,n)}$. By the inequality just proved, primes larger than $2n$ do not contribute to this product, and by assumption there are no primes between $n$ and $2n$. So \[ \binom{2n}{n}=\prod_{\substack{1\le p \le n\\ p\text{\ prime}}}p^{r(p,n)}. \] For $n>p>\frac{2n}{3}$, $\frac{3}{2}>\frac{n}{p}>1$ and so for $p>\frac{2n}{3}>\sqrt{2n}$ we can apply the previous formula for $r(p,n)$ and find that it is zero. So for all $n>4$, the contribution of the primes larger than $\frac{2n}{3}$ is zero. If $p>\sqrt{2n}$, all the terms for higher powers of $p$ vanish and $\displaystyle r(p,n)=\left\lfloor\frac{2n}{p}\right\rfloor-2\left\lfloor\frac{n}{p}\right\rfloor$. Since $r(p,n)$ is at most 1, an upper bound for the contribution for the primes between $\sqrt{2n}$ and $\frac{2n}{3}$ is the product of all primes smaller than $\frac{2n}{3}$. This product is $\exp\left(\vartheta(\frac{2n}{3})\right)$, where $\vartheta(n)$ is the Chebyshev function \[ \vartheta(n)=\sum_{\substack{p\le n\\p\text{\ prime}}}\log p. \] There are at most $\sqrt{2n}$ primes smaller than $\sqrt{2n}$ and by the inequality $p^{r(p,n)}\le 2n$ their product is less than $(2n)^{\sqrt{2n}}$. Combining this information, we get the inequality \[ \frac{4^n}{(2n)^2}\le\binom{2n}{n}\le(2n)^{\sqrt{2n}}\exp\left({\textstyle\vartheta(\frac{2n}{3})}\right). \] Taking logarithms and applying the \PMlinkname{upper bound of $n\log 4$ for $\vartheta(n)$}{UpperBoundOnVarthetan}, we obtain the inequality $\frac{n}{3}\log 4\le(\sqrt{2n}+2)\log(2n)$, which is false for sufficiently large $n$, say $n=2^{11}$. This shows that $n<2^{11}$. Since the conditions $n\ge 2503$ and $n<2^{11}$ are incompatible, there are no counterexamples to the claim. \end{proof} \begin{thebibliography}{9} \bibitem{HW} G.H. Hardy, E.M. Wright, \emph{An Introduction to the Theory of Numbers}, Oxford University Press, 1938. \end{thebibliography}