[tex]\displaystyle Sa~presupunem~ca~exista~a,b \in \mathbb{Z}~astfel~incat~\\ \\ \sqrt{6}+ 4\sqrt{2}=a+b \sqrt{2}. \\ \\ Atunci~\sqrt{6}+ \underbrace{(4-b)}_{k} \sqrt{2}=a. \\ \\ Deci~\sqrt{6}+k \sqrt{2} \in \mathbb{Z}.~Atunci~si~(\sqrt{6}+k \sqrt{2})^2 \in \mathbb{Z} \Leftrightarrow \\ \\ \Leftrightarrow 6+4k \sqrt{3}+2k^2 \in \mathbb{Z} \Rightarrow 4k \sqrt{3} \in \mathbb{Z} \Rightarrow k=0~(caci~k=4-b \in \mathbb{Z}). \\ \\ Rezulta~b=4,~si~deci~a= \sqrt{6} \notin \mathbb{Z}. [/tex]
[tex]\displaystyle Prin~urmare~ \sqrt{6}+4 \sqrt{2} \notin \{a+b \sqrt{2}~|~a,b \in \mathbb{Z} \}.[/tex]
