# If $x$ is irrational, is $2x - \frac{1}{x}$ irrational?

... 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...).

• How about $$x^2=\sqrt3$$ Dec 7, 2019 at 1:34
• Take $x=1/\sqrt 2$ Dec 7, 2019 at 1:35
• The answer is no. Consider $x=\frac1{\sqrt2}$. Note that $x$ is clearly irrational, but $2x-\frac1x=0$. Dec 7, 2019 at 1:35
• Ha, I forgot to consider the case where $2x^2 - 1 = 0$, oh gosh... Well that was even easier than I expected. Dec 7, 2019 at 1:41
• Actually, after thinking about it, $\frac{1}{\sqrt{2}}$ is a special case trigonometrically ($\frac{\pi}{4}$), so it's included in the "first few cases" category. Are there more obvious counterexamples? Dec 7, 2019 at 1:51

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.
Consider a rational number $$q$$ such that $$\sqrt{q^2+8}\notin\mathbb Q$$ – not a very hard condition to fulfill (any integral $$q>1$$ makes this happen, for example). Then, $$x=\frac{q\pm\sqrt{q^2+8}}{4}$$, for both choices of the sign, will be irrational, while $$2x-\frac 1x=q$$ will be rational. In this manner, we can generate infinitely many counterexamples.
• @MrArsGravis The $2$ was actually a mistake, thanks for pointing that out. Dec 7, 2019 at 2:34
You can take $$x=a+\sqrt{a^2+1/2}$$ then $$2x-\frac1x=4a$$ in general $$\frac1{a+\sqrt b} =\frac{\sqrt b - a}{b-a^2}$$ in this case we just take $$b-a^2=1/2$$ in general you can make $$px+\frac qx$$ rational with such choice for $$x$$