let $\alpha=\sqrt{2}+\sqrt{3}.$ then $\frac{1}{\alpha}=\sqrt{3}-\sqrt{2}$ and thus $\sqrt{3}=\frac{\alpha^2+1}{2\alpha} \in \mathbb{Q}(\alpha)$ and $\sqrt{2}=\alpha - \sqrt{3} \in \mathbb{Q}(\alpha).$