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.

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