Let $N>1$. Prove that the number:
$$\mathcal{N}= \sqrt{1\cdot 2 \cdot 3 \cdot 4 \cdots (N-1)\cdot N}$$
is irrational.
Solution
Let $p$ be the last prime that is less or equal to $N$. It suffices to prove that between $p$ and $N$ there does not exist a number that has $p$ as a factor. More specifically, we must prove that $2p>N$. However, this is a known , yet a very difficult theorem of Number Theory, better known as Bertrand's postulate. This theorem says that: anywhere in between $n$ and $2n$ there exists a prime.
If on the contrary was $p<2p<N$ then there would exist a prime $q$ such that:
$$p<q<2p\leq N$$
contradicting the hypothesis that $p$ is the last prime. Hence $\mathcal{N}$ is irrational.
$$\mathcal{N}= \sqrt{1\cdot 2 \cdot 3 \cdot 4 \cdots (N-1)\cdot N}$$
is irrational.
Solution
Let $p$ be the last prime that is less or equal to $N$. It suffices to prove that between $p$ and $N$ there does not exist a number that has $p$ as a factor. More specifically, we must prove that $2p>N$. However, this is a known , yet a very difficult theorem of Number Theory, better known as Bertrand's postulate. This theorem says that: anywhere in between $n$ and $2n$ there exists a prime.
If on the contrary was $p<2p<N$ then there would exist a prime $q$ such that:
$$p<q<2p\leq N$$
contradicting the hypothesis that $p$ is the last prime. Hence $\mathcal{N}$ is irrational.
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