ATRACTOR

\(z=3+5\sqrt{2}\;\;\;\;\;\;\;\;\;\; w=2-\sqrt{2}\)

\[\begin{array}{l} \;\;\;3+5\sqrt{2}=\left(2-\sqrt{2}\right)\left(q_{1}+q_{2}\sqrt{2}\right)+r_{1}+r_{2}\sqrt{2}\Leftrightarrow\\ \Leftrightarrow\frac{r_{1}+r_{2}\sqrt{2}}{2-\sqrt{2}}=\frac{3+5\sqrt{2}}{2-\sqrt{2}}-\left(q_{1}+q_{2}\sqrt{2}\right)\Leftrightarrow\\ \Leftrightarrow\frac{r_{1}+r_{2}\sqrt{2}}{2-\sqrt{2}}=\frac{\left(3+5\sqrt{2}\right)\left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}-\left(q_{1}+q_{2}\sqrt{2}\right)\Leftrightarrow\\ \Leftrightarrow\frac{r_{1}+r_{2}\sqrt{2}}{2-\sqrt{2}}=8-q_{1}+\left(\frac{13}{2}-q_{2}\right)\sqrt{2} \end{array}\]

Basta escolher \(q_{1}=8\) e \(q_{2}=6\) ; vem então

\(r_{1}+r_{2}\sqrt{2}=3+5\sqrt{2}-\left(2-\sqrt{2}\right)\left(8+6\sqrt{2}\right)=-1+\sqrt{2}\)

e temos

\(3+5\sqrt{2}=\left(2-\sqrt{2}\right)\left(8+6\sqrt{2}\right)+\left(-1+\sqrt{2}\right)\), com

\(v\left(-1+\sqrt{2}\right)=1<2=v\left(2-\sqrt{2}\right)\)