Commuting floor functions Positive irrational numbers $a,b>0$ are such that $\lfloor a\lfloor bx\rfloor\rfloor = \lfloor b\lfloor ax\rfloor\rfloor$ for all $x>0$. Must it be that $a=b$?
If $a$ and $b$ are allowed to be rational, this is not always true. For example, we can take $a=1$ and $b=\frac{1}{2}$. It is true that $\lfloor x/2\rfloor = \lfloor\lfloor x\rfloor/2\rfloor$ for any real number $x>0$, as when $2n\leq x<2n+2$ for some nonnegative integer $n$, both sides evaluate to $n$.
 A: Yes. Take irrational $a,b$ with $a \not = b$. 
First assume $a,b$ are linearly independent over $\mathbb{Q}$. Take $\epsilon > 0$ small (to be determined later). Let's look at $x$ of the form $x = \frac{k+\epsilon}{ab}$ for $k \in \mathbb{N}$. Then $\lfloor k+\epsilon-a\{\frac{k+\epsilon}{a}\}\rfloor = \lfloor a\lfloor \frac{k+\epsilon}{a}\rfloor\rfloor = \lfloor b\lfloor\frac{k+\epsilon}{b}\rfloor\rfloor = \lfloor k+\epsilon-b\{\frac{k+\epsilon}{b}\}\rfloor$, so, to get a contradiction, it suffices to show that there is some $k$ with $a\{\frac{k+\epsilon}{a}\} < \epsilon$ and $b\{\frac{k+\epsilon}{b}\} > \epsilon$. Let $\overline{a} = \frac{1}{a}$ and $\overline{b} = \frac{1}{b}$ for psychological ease, so that we wish to show there is some $k$ with $\{\overline{a}(k+\epsilon)\} < \overline{a}\epsilon$ and $\{\overline{b}(k+\epsilon)\} > \overline{b}\epsilon$. It's well known (and easy to prove with fourier) that $((\overline{a}k,\overline{b}k))_{k \ge 1}$ is uniformly distributed over the torus $\mathbb{T}^2$ (which is a fancy name for $(\mathbb{R}/\mathbb{Z})^2$), so $((\overline{a}(k+\epsilon),\overline{b}(k+\epsilon))_{k \ge 1}$ is as well, so we're done if we take $\epsilon = \frac{1}{2}\min(\frac{1}{\overline{a}},\frac{1}{\overline{b}})$, say.
Now assume $a,b$ are not linearly independent over $\mathbb{Q}$. Say $b = ar$ for some $r \in \mathbb{Q}$. By dividing by $r$ if need be, we may assume $r \not \in \mathbb{Z}$. Letting $x = \frac{k}{a}$, we have $\lfloor ark-\{kr\}a\rfloor = \lfloor ark\rfloor$, so, to get a contradiction, it suffices to find a $k$ with $\{kr\}a > \{ark\}$. But since $r \in \mathbb{Q}\setminus\mathbb{Z}$, if we restrict $k$ to be, for example, $1$ modulo the denominator of $r$, then the value $\{kr\}a$ if fixed, while $\{ark\}$ can be arbitrarily small, due to the irrationality of $a$ and thus equidistribution.
