# Does $lcm\{1,2,…,n\} = \prod_{p\leq n, p\in\mathbb{P}}p^{\lceil \frac{log(n)}{log(p)}\rceil}$?

I am trying to understand Apery's proof of the irrationality of $$\zeta(3)$$ from start to end, with this document. I apologise for having 2 questions in one, but both are relatively simple (I just need to be sure I completely understand each part of the proof).

It says in the preliminaries that $$lcm\{1,2,...,n\} = \prod_{p\leq n, p\in\mathbb{P}}p^{\lceil \frac{log(n)}{log(p)}\rceil}$$ and gives the following proof: But it seems to have a mistake - when it says $$\lceil x \rceil$$ is the highest integer power of $$p$$ such that it's smaller than $$n$$, as surely $$p^{\lceil x\rceil}\geq n$$?

Following this and relatedly, it says that Why is this true? I'm satisfied that it proves the second statement, but why are the lemma and the proved statement equivalent?

• For the first, if it is the ceiling function, then I think that it is a mistake. For the second, it seems that the equality in Lemma 3.1.2. does not hold for $n=3$. – mathlove Oct 30 '18 at 12:15

You need to replace the ceiling function with the floor function. Let $$N=lcm(1,\dots,n)$$. If $$p$$ is a prime and $$k$$ and integer bigger or equal than $$1$$, then you have:
1) if $$p^k\le n$$, then $$p^k|N$$ by definition of $$N$$;
2) if $$p^k|N$$, then there must be an integer $$1\le m\le n$$ such that $$p^k|m$$, by definition of $$lcm$$ (and prime numbers). Therefore $$p^k\le n$$.
It follows that for any prime $$p$$, $$p^k|N$$ if and only if $$p^k\le n$$, which is equivalent to $$k\le \lfloor \log_p n \rfloor$$. Therefore the exponent of the power of $$p$$ in the prime factorization of $$N$$ is $$\lfloor \log_p n \rfloor$$ as claimed.