… as stated in the title, a rather simple question:
Given $x \in \mathbb{R}\setminus\mathbb{Q}$, is
$$2x – \frac{1}{x} \in \mathbb{R}\setminus\mathbb{Q}?$$
If $x^2 \in \mathbb{Q},$ it is easy to show that this is true via $2x – \frac{1}{x} = \frac{1}{x} \left(2x^2 – 1\right)$ and subgroup properties. However, for irrational $x^2$, I cannot find a proof or counterexample either way, however my knowledge of number theory is more than limited.
Edit: I obviously forgot the case $x = \pm \frac{1}{\sqrt{2}}$, which also makes the stated proof for rational $x^2$ fail. Might it still be true for all $x \in \mathbb{R}\setminus\left(\mathbb{Q} \cup \left\{ -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\} \right)$?
For the interested: The context is from quantum mechanics; I want to determine whether a system is periodic, which depends on whether the energy levels (determined by a cosine dispersion relation in this case) have rational quotients. And by dividing $\cos(2x)$ by $\cos(x)$ using the double-angle formula, you get the given expression, which – after sprinkling in some Niven – is an elegant sufficient condition for non-periodicity for all but the very first few cases (of course, this could be proved on a case-by-case basis, but… I mean…).
Best Answer
Suppose $2x - \dfrac 1 x = a$ and $a$ is rational. Multiplying both sides by $x,$ we get $$ 2x^2 - 1 = ax. $$ This is a quadratic equation whose solution is $$ x = \frac{a\pm \sqrt{a^2 + 8}} 4 $$ Choose $a$ so that $\sqrt{a^2+8}$ is irrational, and then you have $x$ irrational and $2x-\dfrac 1 x$ rational.