... 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...).

  • $\begingroup$ How about $$x^2=\sqrt3$$ $\endgroup$ Dec 7, 2019 at 1:34
  • 1
    $\begingroup$ Take $x=1/\sqrt 2$ $\endgroup$
    – kingW3
    Dec 7, 2019 at 1:35
  • $\begingroup$ The answer is no. Consider $x=\frac1{\sqrt2}$. Note that $x$ is clearly irrational, but $2x-\frac1x=0$. $\endgroup$ Dec 7, 2019 at 1:35
  • $\begingroup$ Ha, I forgot to consider the case where $2x^2 - 1 = 0$, oh gosh... Well that was even easier than I expected. $\endgroup$ Dec 7, 2019 at 1:41
  • $\begingroup$ 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? $\endgroup$ Dec 7, 2019 at 1:51

3 Answers 3


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.

  • $\begingroup$ Absolutely. I guess I was too afraid of the word number theory to manipulate a polynomial. $\endgroup$ Dec 7, 2019 at 2:27

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.

  • $\begingroup$ The 2 instead of a 4 in the denominator threw me off, but I like the fact that you point out that this generates an infinite number of counterexamples. $\endgroup$ Dec 7, 2019 at 2:33
  • $\begingroup$ @MrArsGravis The $2$ was actually a mistake, thanks for pointing that out. $\endgroup$
    – ViHdzP
    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 $x$


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