## Yes, only integers

Reading my post ‘only integers behave nicely’, I came up with another proof (and another variation of it) of the following fact:

Let $(\alpha, \beta)$ be a pair of real numbers having the property that: $\alpha [\beta n] = \beta [\alpha n]$ for all natural numbers $n$, where $[x]$ denotes the largest integer $\le x$. Then $\alpha$ and $\beta$ are either the same, one of them equals $0$, or they are both integral. So let’s assume that the last 2 possibilites do not hold, and dive right into it.

Dividing by $\beta[\beta n]$ (which is possible for large enough $n$), we see that the ratio of $\alpha$ and $\beta$ is rational, say $\dfrac{p}{q}$, with $p$ and $q$ coprime.  Substitute $\dfrac{\beta p}{q}$ for $\alpha$, dividing by $\beta$ and multiplying by $q$ we obtain:

$p [\beta n] = q [\beta n \dfrac{p}{q}]$

Since the RHS is divisible by $q$, LHS must be a multiple of $q$ too. In particular this is true for $n = q$. Since we may assume that the fractional part of $\beta$ is between $0$ and $1$ (duh), this implies that  the fractional part of $\beta$ is actually contained in the interval $(0, \dfrac{1}{q})$. Otherwise we would have that $n$ times the integer part of $\beta$ is divisible by $q$ (since we assumed for the moment that $n = q$), while $n$ times the fractional part of $\beta$ is (rounded down) NOT divisible by $q$, while their sum has to be a multiple of $q$, because $p$ and $q$ are coprime. Let $n = mq$ now, with $m$ integral. By induction we now see that the fractional part of $\beta$ is contained in the interval $(0, \dfrac{1}{mq})$, for all $m$; contradiction.

Another (admittedly only slightly different) route to take, after we showed that $\beta - [\beta] \in (0, \dfrac{1}{q})$, is the following:

Let $m$ (as it definitely resembles the $m$ in the inductionproof) be the smallest integer such that the fractional part of $m\beta$ is larger than $\dfrac{1}{q}$. Then $m\beta - [m\beta] < \dfrac{2}{q}$ (otherwise $[\dfrac{m+1}{2}]$ would have been a smaller integer such that the fractional part of $m\beta$ is larger $\dfrac{1}{q}$, which contradicts our definition of $m$. But now, letting $\delta$ be the fractional part of $\beta$, setting $n = mq$ and keeping in mind that $[\beta n]$ is a multiple of $q$, we have the following:

$[\beta n] = [(b + \delta)n] = bmq + [\delta mq] = bmq + 1$

Which shows $q = 1$. By symmetry, $p =1$ and once again, Q.E.D.