2
$\begingroup$

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 $$n<p\leqslant2n.\tag{1.13}$$ $\text{P}\scriptstyle{\text{ROOF}}$. 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 $$k(p)\leqslant\left\lfloor\dfrac{\log2n}{\log p}\right\rfloor.\tag{1.15}$$

enter image description here

enter image description here

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

The idea is to notice that

$$\log(N) \leq \sum_{p|N}{\log(p)} + \sum_{k(p) \geq 2}{k(p)\log(p)}.$$

The first term is dealt with by $(1.16)$, the second one by the last estimate of the image before last.

$\endgroup$
  • 1
    $\begingroup$ Could you talk it detail ? I can't understand you. Thanks. $\endgroup$ – lanse7pty Jan 9 '19 at 11:44
  • $\begingroup$ Have you read your page $21$? $\endgroup$ – Mindlack Jan 9 '19 at 11:45
  • $\begingroup$ Yes, I have read it, but seemly, there is not anything about the case $k(p)=1$. $\endgroup$ – lanse7pty Jan 9 '19 at 11:58
  • $\begingroup$ 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. $\endgroup$ – Mindlack Jan 9 '19 at 12:08
  • $\begingroup$ I understand you, thanks. $\endgroup$ – lanse7pty Jan 9 '19 at 12:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.