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Fie \( a_1,a_2,...,a_n \in \mathbb{Q}_+^* ,\ n \in \mathbb{N}^* \).
Sa se arate ca \( \sqrt{a_1}+\sqrt{a_2}+...+\sqrt{a_n}\in \mathbb{Q} \Leftrightarrow \sqrt{a_1} , \sqrt{a_2} , ... , \sqrt{a_n} \in \mathbb{Q} \).

Incercati sa folositi faptul ca daca \( p_1<\dots<p_m \) sunt numere prime, atunci \( \sqrt{p_m}\notin\mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_{m-1}}) \).

Notam cu \( \alpha =\sqrt{a_1}+...+\sqrt{a_n} \) si aratam ca \( Q(\alpha) =Q(\sqrt{a_1},...,sqrt{a_n}) \).
Aratam ca \( \sqrt{a_1} \in Q(\alpha) \). Fie \( A=\{\sqrt{a_1} \pm \sqrt{a_2} \pm ... \pm \sqrt{a_n}\} \) toate cele \( 2^{n-1} \) combinatii si \( B=\{-\sqrt{a_1} \pm \sqrt{a_2} \pm ... \pm \sqrt{a_n}\} \).

Fie \( f=\prod_{a \in A} (X-a) \) si \( g=\prod_{b \in B} (X-b) \).

Avem ca \( f,g \in Q(\sqrt{a_1})[X] \). Atunci \( f=f_1 +\sqrt{a_1} f_2 \) si \( g=f_1-\sqrt{a_1} f_2 \), \( f_1,f_2 \in Q[X] \) (cine are o demonstratie la acest lucru il rog sa o posteze).
Se observa ca \( f(\alpha)=0 \Rightarrow f_1(\alpha) + \sqrt{a_1}f_2(\alpha)=0 \)
Daca \( f_1(\alpha)=f_2(\alpha)=0 \Rightarrow g(\alpha)=0 \) fals din constructie
Daca \( f_1(\alpha),f_2(\alpha)\neq 0 \Rightarrow \sqrt{a_1} \in Q(\alpha) \)

Analog \( \sqrt{a_i} \in Q(\alpha) \Rightarrow Q(\sqrt{a_1},...,sqrt{a_n}) \subset Q(\alpha) \) cealalta incluziune fiind evidenta.