Question. Find, with proof, the possible values of a rational number $q$ for which $q+\sqrt{2}$ is a reduced quadratic irrational.
So, by definition a quadratic irrational is one of the form $u+v\sqrt{d}$ where $u,v\in\Bbb Q, v\neq 0$ and $d$ being square-free. Then, it is said to be reduced if it exceeds $1$ and has conjugate in the interval $(-1, 0)$.
For the question at hand; this corresponds to having $-1<q-\sqrt{2}<0$ and $q+\sqrt{2}>1$.
But I'm not quite sure how to proceed from here?
Since we have $-1 < q-\sqrt{2} < 0$ and $q+\sqrt{2} > 1$, we can simply move the square roots to get the inequalities $$-1 + \sqrt{2} < q < \sqrt{2} \text{ and } q > 1 - \sqrt{2}$$ Because $-1 + \sqrt{2} > 1 - \sqrt{2}$, our first inequality is all that is necessary. This means we can define a set $S$ of all rational numbers satisfying the inequality as $$S = \left\{q \in \mathbb{Q} \mid q \in (-1+\sqrt{2}, \sqrt{2})\right\}$$ This is the set of all rational numbers between $-1+\sqrt{2}$ and $\sqrt{2}$.