# A question regarding a possible misprint on showing why a part bertrands lemma leads to a certain result

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$$ does not divide $$\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 $$\pmatrix{2n\\n}$$= $$\pmatrix{6\\3}=\tfrac{6.5.4.3.2.1}{3.2.1.3.2.1.}=\tfrac{6.5.4}{3.2}=5.2^2$$ 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.
• 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 – excalibirr Mar 24 '19 at 23:44