Added union s.e.v proof and irrationality of sqrt

contents/number_theory.tex

@@ -333,3 +333,40 @@ $\rightarrow\leftarrow$
 $\implies |P| = \infty$
 
 Il existe une infinité de nombre premiers.
+
+\langsubsection{Irrationnalité}{Irrationality}
+
+\langsubsubsection{$\forall n \in \N, \sqrt{n}$ est soit un nombre premier ou un carré parfait}{$\sqrt{n}$ is either a prime number or a perfect square}
+
+\begin{theorem_sq} \label{theorem:sqrt_prime}
+$\Pn$ is the set of all prime numbers \ref{definition:prime_number}.
+$\forall p \in \Pn, \sqrt{p} \notin \Q$
+\end{theorem_sq}
+
+The classical proof of the irrationality of 2 is a specific case of \ref{theorem:sqrt_prime}.
+
+\begin{proof}
+
+By contradiction let's assume $\sqrt{p} \in \Q$
+
+$a \in \Z, b \in \N^*, \text{PGCD}(a,b) = 1, \sqrt{p} = \frac{a}{b}$
+
+$\Rightarrow p = (\frac{a}{b})^2 = \frac{a^2}{b^2}$
+
+$\Rightarrow b^2p = a^2$
+
+$\Rightarrow p|a$
+
+Let $c \in \N^*$, $a = pc$
+
+$\Rightarrow b^2 p = (pc)^2=p^2c^2$
+
+$\Rightarrow b^2 = pc^2$
+
+$\Rightarrow p|b$
+
+$\Rightarrow (p|b \land p|a \land \text{PGCD}(a,b)=1) \Rightarrow \bot$
+
+$\Rightarrow \sqrt{p} \notin \Q$
+
+\end{proof}