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Amine Ben hammou
webNeat
I am a software engineer with focus on web development and performance optimization.
We all remember from our fundamental mathematics classes the emphasis on the irrationality of $\sqrt{2}$ using a simple proof by contradiction.
Well, it turns out that we can extend this theorem more broadly to the square roots of prime numbers with the same kind of demonstration, and that's what we're going to demonstrate here in this article. But first, let's have some reminders about what prime numbers are.
Let $p$ be a natural number, we say that $p$ is prime if and only if $p$ is different from 1 and it has only 1 and itself as divisors. Let's denote this set of prime numbers as $\mathbb{P}$.
$$\forall p \in \mathbb{N}, \ p \in \mathbb{P} \iff \begin{cases} p \neq 1 \newline \nexists(a, b) \in \mathbb{N}^2/ \ a \neq 1 \And b \neq p \ \Longrightarrow \ p = ab \end{cases}$$
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