$$\mathcal{N}= \sqrt{1\cdot 2 \cdot 3 \cdot 4 \cdots (N-1)\cdot N}$$

is irrational.

- Tolaso J Kos
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Let $N \in \mathbb{N} \mid N>1$. Prove that the number:

$$\mathcal{N}= \sqrt{1\cdot 2 \cdot 3 \cdot 4 \cdots (N-1)\cdot N}$$

is irrational.

$$\mathcal{N}= \sqrt{1\cdot 2 \cdot 3 \cdot 4 \cdots (N-1)\cdot N}$$

is irrational.

Hidden message

Let $p \leq N$ be the last prime. If we prove that between $p$ and $N$ does not exist a number that has $p$ as a factor we are done. So, we need to prove that $2p>N$. But this is exactly what Bertrand's postulate says.

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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