$$\mathcal{N}= \sqrt{1\cdot 2 \cdot 3 \cdot 4 \cdots (N-1)\cdot N}$$
is irrational.
Hidden message
Welcome to mathimatikoi.org forum; Enjoy your visit here.
Irrational number
-
Tolaso J Kos
- Administration team
- Articles: 2
- Posts: 861
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
Irrational number
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.
Imagination is much more important than knowledge.
Re: Irrational number
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.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$