I got a question in an assignment but I'm sure it has to be a misprint, any clarification on this would be much appreciated ( though I want to work the actual problem out myself)

It's about Bertrands postulate, particularly we've been asked to consider the lemma which states that for $n>2$ and $\tfrac{2n}{3}<p\leq n$, $p$ does not divide $\pmatrix{2n\\n}$, and then explain why this means that all prime factors of $\pmatrix{2n\\n}$ satisfy $p\leq \tfrac{2n}{3}$.

but then take for example $n=3$ then $\pmatrix{2n\\n}$= $\pmatrix{6\\3}=\tfrac{}{}=\tfrac{6.5.4}{3.2}=5.2^2$ and so 5 appears in the prime decomposition. But then this contradicts what we were trying to show because $\tfrac{2n}{3}=\tfrac{6}{3}=2$and 5 is certainly not less than 2. Surely the thing we should be trying to show is that $p\geq\tfrac{2n}{3}$ , no ?


First, it is correct that $\binom{2n}{n}$ has no prime factor $p$ such that $\frac{2n}3 \lt p \le n$. It follows that all prime factors $p$ are either $\le \frac{2n}3$ or greater than $n$. As you've seen with $n = 3$, both cases can occur.

Since the intent is to prove Bertrand's postulate, i.e., that there is some prime between $n$ and $2n$, I suspect the plan was to assume the contrary and eventually arrive at a contradiction. Under this assumption, all prime factors of $\binom{2n}{n}$ must be either $\le \frac{2n}3$ or greater than $2n$. If that is what's going on, your next step is simply to eliminate the possibility that $p \gt 2n$. That isn't hard. The difficult step is to show that the contribution of the primes $\lt \frac{2n}3$ is not actually large enough to produce $\binom{2n}{n}$, yielding the desired contradiction.

  • $\begingroup$ Sorry I should have been more clear, we don't actually have to prove bertrands postulate, just explain why the part of the lemma I stated means that $p\leq\tfrac{2n}{3}$. But I assume your answer is still what I would need to use to explain this $\endgroup$ – excalibirr Mar 24 '19 at 23:44

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