# Proof of Bertrand's postulate

The following proof is from the 19th page of
Everest, Graham; Ward, Thomas, An introduction to number theory, Graduate Texts in Mathematics 232. London: Springer (ISBN 1-85233-917-9/hbk). x, 294 p. (2005). ZBL1089.11001.

In fact, I think this proof is not finished. For the red line, only $$k(p)\ge 2$$ is disproved. For the case $$k(p)=1$$, there is not any talk. How to understand it ? Thanks.

$$\textbf{Theorem 1.9.}\ [\text{B}{\scriptstyle{\text{ERTRAND'S}}} \text{ P}\scriptstyle{\text{OSTULATE}}]\$$ If $$n\geqslant1$$, then there is at least one prime $$p$$ with the property that $$n $$\text{P}\scriptstyle{\text{ROOF}}$$. For any real number $$x$$, let $$\lfloor x\rfloor$$ denote the integer part of $$x$$. Thus $$\lfloor x\rfloor$$ is the greatest integer less than or equal to $$x$$. Let $$p$$ be any prime. Then $$\left\lfloor\dfrac np\right\rfloor+\left\lfloor\dfrac n{p^2}\right\rfloor+\left\lfloor\dfrac n{p^3}\right\rfloor+\cdots$$ is the largest power of $$p$$ dividing $$n!$$ (see Exercise $$8.7(a)$$ on p. $$162$$). Fix $$n\geqslant 1$$ and let $$N=\prod_{p\leqslant2n}p^{k(p)}$$ be the prime decomposition of $$N=(2n)!/(n!)^2$$. The number of times that a given prime $$p$$ divides $$N$$ is the difference between the number of times it divides $$(2n)!$$ and $$(n!)^2$$, so $$k(p)=\sum_{m=1}^\infty\left(\left\lfloor\dfrac{2n}{p^m}\right\rfloor-2\left\lfloor\dfrac n{p^m}\right\rfloor\right),\tag{1.14}$$ and each of the terms in the sum is either $$0$$ or $$1$$, depending on whether $$\left\lfloor\frac{2n}{p^m}\right\rfloor$$ is odd or even. If $$p^m>2n$$ the term is certainly $$0$$, so $$k(p)\leqslant\left\lfloor\dfrac{\log2n}{\log p}\right\rfloor.\tag{1.15}$$

• Already the definition of $k(p)$ is confusing. – Peter Jan 9 '19 at 10:06
• @Peter I think the definition of $k(p)$ is very ok – lanse7pty Jan 9 '19 at 10:11
• But hard to actually understand. We should find out what $k(p)=1$ means. Perhaps this case is trivial for the desired proof. – Peter Jan 9 '19 at 10:13
• $k(p)$ is the exponent of the prime $p$ in the prime decomposition of $N$. $k(p)$ is well-defined, unique and greater or equal to $0$. – Snake707 Jan 9 '19 at 10:30
• @Snake707 I did not claim that $k(p)$ is not well defined. – Peter Jan 9 '19 at 10:57

$$\log(N) \leq \sum_{p|N}{\log(p)} + \sum_{k(p) \geq 2}{k(p)\log(p)}.$$
The first term is dealt with by $$(1.16)$$, the second one by the last estimate of the image before last.
• Have you read your page $21$? – Mindlack Jan 9 '19 at 11:45
• Yes, I have read it, but seemly, there is not anything about the case $k(p)=1$. – lanse7pty Jan 9 '19 at 11:58
• In the derivation of $(1.17)$, you count every prime once, and then you count all primes with $k(p) \geq 2$ $k(p)$ times. So primes with $k(p)=1$ are counted once and primes with $k(p) > 1$ are counted $k(p)+1$ times. – Mindlack Jan 9 '19 at 12:08