In the first line of Chapter Three of my graduate level text from which I am working from, the author declares the following divisibility relation to be the basis of Chebyshev theorem, but has also been referred to as Bertrand's postulate:

$$S_0: n \lt p_j \lt 2n \Rightarrow p_j \quad\Bigg|\quad {2n\choose n}$$


Because I struggled with the first few exercises I attempted in my understanding of the author's justifications for steps taken, I decided that I must rigorously prove this introductory lemma before moving forward. The first step I took, was an attempt at minimization of the factors in the divisiblilty relation, instead of working with the factorials of the binomial coefficient in our original statement, obtaining a lemma that involves the lowest common multiples for the first $2n$ and $n$ natural numbers,which directly follows from our initial declaration, and so I found the proceding statement must also be true if and only if the statement $S_0$ is true:

$$S_1: n \lt p_j \lt 2n \Rightarrow p_j \quad\Bigg|\quad \frac{\operatorname{lcm}(1,2,3,...,2n)}{(\operatorname{lcm}(1,2,3,...,n))^2} $$


<Important Additional Note Enclosed>

As one user has already commented, in the strictest sense that one states $a | b$ one requires $\frac{b}{a} \in \mathbb Z$, in what I have stated here I at no point specified that $(\operatorname{lcm}(1,2,3,...,n))^2 | \operatorname{lcm}(1,2,3,...,2n)$ must be true.

When we speak of a divisibility relation $a | b$ in $a,b\in \mathbb Q$ as I have above, we no longer retain the assumptions as one makes in the conventional use of this type of lemma for regarding $a,b\in \mathbb Z$, rather than the requisite of $\frac{b}{a} \in \mathbb Z$ we instead have the requisite of the denominator of $b$ remaining unchanged or reduced by the division by $a$, but never is there an increase in it's total number of factors.

<Important Additional Note Enclosed>

We then consider the unique prime factorization product for the lowest common multiple expressions in $S_1$ from the consideration of the arithmetic law and the exponents of the terms in these unique factorization products implied by the identity for the lowest common multiple in terms of the second Chebyshev function:

$$\psi ( x ) =\sum _{j=1}^{\pi ( x ) }\Bigl\lfloor {\frac {\ln ( x ) }{\ln ( p_{{j}} ) }} \Bigr\rfloor \ln ( p_{{j}} ) \land \ln(\operatorname{lcm}(1,2,3,...,n))=\psi (n) $$ $$\Leftarrow \Rightarrow$$ $$\operatorname{lcm}(1,2,3,...,n)=\prod_{j=1}^{\pi(n)}p_j^{\Bigl\lfloor \frac{\ln(n)}{\ln(p_j)} \Bigr\rfloor} $$


This allows one to restate $S_1$ purely in terms of prime factorization products, in mutual consideration over which range our divisor product must be taken as per the inequality stated in both $S_0$ & $S_1$:

$$S_{3a}:\prod_{j=\pi(n)+1}^{\pi(2n)-1}p_j \quad\Bigg|\quad \frac{ \prod^{\pi(2n)}_{j=1}p_j^{\bigl\lfloor \frac{\ln(2n)}{\ln(p_j)} \bigr\rfloor}}{\prod_{j=1}^{\pi(n)}p_j^{2\bigl\lfloor \frac{\ln(n)}{\ln(p_j)} \bigr\rfloor}}$$


Let the numerator of $\frac{\operatorname{lcm}(1,2,3,...,2n)}{(\operatorname{lcm}(1,2,3,...,n))^2}$ be denoted as $\mathcal N_n$.

We can always be assured that:

$$S_{3b}: p_{\pi(2n)}=\frac{\mathcal N_n}{\prod_{j=\pi(n)+1}^{{\pi(2n)-1}}p_j} $$


From $S_3$ we can easily then construct an assertion regarding a finite set that dictates that a necessary requisite for $S_1$ (hence also our original statment $S_0$) to be true, $S_1$ is true if and only if it is true that for any $n$ greater than $1$:

$$S_4: \min\Biggl({\Bigl\{\Bigl\lfloor \frac{\ln(2n)}{p_j} \Bigr\rfloor-2\Bigl\lfloor \frac{\ln(n)}{p_j} \Bigr\rfloor}\Bigr\}_{j=\pi(n)+1..\pi(2n)-1}\Biggr)=1$$


We finally note the following to be true for all $n$ greater than $1$:

$$S_5: {\Bigl\{\Bigl\lfloor \frac{\ln(2n)}{p_j} \Bigr\rfloor-2\Bigl\lfloor \frac{\ln(n)}{p_j} \Bigr\rfloor}\Bigr\}_{j=\pi(n)+1..\pi(2n)-1}={\Bigl\{1}\Bigr\} $$


$S_5$ assuring that $S_4$ is true for all $n$ greater than 1, hence also $S_1$ and our original statement $S_0$ must also be true for all $n$ greater than 1 and our proof is complete.

  • 1
    $\begingroup$ $\frac{\operatorname{lcm}(1,2,3,...,2n)}{(\operatorname{lcm}(1,2,3,...,n))^2}$ isn't an integer (take $n=3$) while $\frac{(2n)!}{(n!)^2}$ is. $\endgroup$ – reuns Nov 23 '18 at 4:13
  • $\begingroup$ yeah but $5$ still divides $\frac{5}{3}$ right? $\endgroup$ – Adam Nov 23 '18 at 9:13
  • 3
    $\begingroup$ You have chosen a ridiculously complicated way to try to prove a very simple result. $2n\choose n$ is the quotient of the product of all the integers from $n+1$ to $2n$, inclusive, by the product of all the integers from $1$ to $n$, inclusive. The prime $p_j$ being strictly between $n$ and $2n$, it appears among the numbers whose product gives the numerator, so it divides the numerator. Being strictly greater than $n$, it does not divide any of the numbers whose product gives the denominator; since it is prime, it doesn't divide the denominator. Done! $\endgroup$ – Gerry Myerson Nov 27 '18 at 3:33
  • 1
    $\begingroup$ Reciprocity is a deep and important result, and worthy of a large number of proofs. That every prime between $n$ and $2n$ divides $2n\choose n$ is a triviality, and just what "insight worth consideration" does your proof provide? (and who will read all the way through it to find any such insight?). By the way, if you want to be certain I see a comment here, you have to include @Gerry in it. $\endgroup$ – Gerry Myerson Nov 28 '18 at 6:06
  • 1
    $\begingroup$ Look I think what it boils down to here is that I haven't looked at the problem for anywhere near the length of time I should have, and it is just wishful thinking on my behalf hoping that I have proven more than I really have, so I'll will step back from this idea and study the approach taken by Larry Freeman in this post math.stackexchange.com/questions/356912/… $\endgroup$ – Adam Dec 4 '18 at 22:04

Your Answer

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

Browse other questions tagged or ask your own question.