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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:Supposed proof of lemma.

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 Second lemma

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

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    $\begingroup$ 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 $n=3$. $\endgroup$ – mathlove Oct 30 '18 at 12:15
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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.

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  • $\begingroup$ I thought so (I meant to mention that this was what I believed they meant). The rest of the above proof works if you assume they meant the floor. $\endgroup$ – Isky Mathews Oct 30 '18 at 21:22

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