, 2022

First note that $\frac{\beta}{\alpha}$ is an irrational number (otherwise there are positive integers $k_1, k_2$ such that $k_1\alpha = k_2\beta$, which leads to a contradiction).
If $\alpha = \frac{q}{p}$ is a rational number, then there is a positive integer $k_2$ such that the decimal part of $\frac{k_2\beta}{q}$ is less than $\frac{1}{q}$. Taking $k_1 = p\lfloor\frac{k_2\beta}{q}\rfloor$ gives that
$$k_1\alpha = q \cdot \lfloor\frac{k_2\beta}{q}\rfloor < k_2\beta < q\lfloor\frac{k_2\beta}{q}\rfloor + 1,$$
which contradicts to the assumption. Thus $\alpha$ must be irrational. Similarly, $\beta$ must be irrational as well.

A pair of positive integers $(a, b)$ is called distinguished if $0 < b\beta - a\alpha < 1$. By our assumption, there is a unique positive integer $t$ such that $a\alpha < t < b\beta$. Denoting $u = t - a\alpha, v = b\beta - t$, we say a distinguished pair $(a, b)$ corresponds to the intermediate number $t$ together with a distant pair $(u, v)$. We will prove the following.

Lemma. Assume two distinguished pairs $(a_1, b_1)$, $(a_2, b_2)$ correspond to distant pairs $(u_1, v_1)$, $(u_2, v_2)$, respectively. Then $\frac{u_1}{v_1} = \frac{u_2}{v_2}$.

Proof of Lemma. Suppose that the intermediate numbers of these two distinguished pairs are $t_1, t_2$, respectively. Assume for the sake of contradiction that $\frac{u_1}{v_1} > \frac{u_2}{v_2}$. We also take $\epsilon = \frac{u_1v_2 - u_2v_1}{2} > 0$.
Since $\frac{\beta}{\alpha}$ is an irrational number, there are positive integers $a_0, b_0$ such that $0 < a_0\alpha - b_0\beta < \epsilon$. On the other hand, our assumption implies that there is another integer $t_0$ satisfying $b_0\beta < t_0 < a_0\alpha$. Denote $u_0 = a_0\alpha - t_0$ and $v_0 = t_0 - b_0\beta$. If $\frac{u_1}{u_0} - \frac{v_1}{v_0} > 1$, then one can take $L = \lfloor\frac{v_1}{v_0}\rfloor + 1$ such that $\frac{u_1}{u_0} > L > \frac{v_1}{v_0}$, i.e.,
$$\begin{aligned} u_1 - Lu_0 &= (t_1 + Lt_0) - (a_1 + La_0)\alpha > 0, \\ v_1 - Lv_0 &= (t_1 + Lt_0) - (b_1 + Lb_0)\beta < 0. \end{aligned}$$
Set $k_1 = a_1 + La_0$ and $k_2 = b_1 + Lb_0$. Then $\lfloor k_1\alpha \rfloor = \lfloor k_2\beta \rfloor = t_1 + Lt_0 - 1$, which leads to a contradiction. This further implies that $\frac{u_1}{u_0} - \frac{v_1}{v_0} \le 1$. Using a similar argument, we can show that $\frac{u_2}{u_0} - \frac{v_2}{v_0} \ge -1$. Therefore,
$$u_1 - \frac{u_0}{v_0}v_1 \le u_0, \quad u_2 - \frac{u_0}{v_0}v_2 \ge -u_0, \quad \Rightarrow \quad u_1v_2 - u_2v_1 \le u_0(v_1 + v_2) < 2\epsilon.$$
However, this violates our choice of $\epsilon$. The lemma is proved.

Now we see all distinguished pairs $(a, b)$ share the same ratio $\frac{u}{v} = \frac{t - a\alpha}{b\beta - t}$. Denote this common $\lambda = \frac{v}{u+v} \in (0, 1)$; then we have an equality $\lambda \cdot a\alpha + (1 - \lambda) \cdot b\beta = t \in \mathbb{Z}$. We call linear combinations of distinguished pairs with integer coefficients nice pairs. For each nice pair $(c, d)$, it is not hard to see that
$$\lambda \cdot c\alpha + (1 - \lambda) \cdot d\beta = c \cdot \lambda\alpha + d \cdot (1 - \lambda)\beta \in \mathbb{Z}.$$
Take a nice pair $(a, b)$ and set $\delta = b\beta - a\alpha \in (0, 1)$. We claim that for all $M > 0$, any pair $(c, d)$ of integers satisfying $0 < d\beta - c\alpha < M$ and $c > \frac{a}{\delta}M$, is nice. Indeed, it suffices to take $L = \lfloor \frac{d\beta - c\alpha}{\delta} \rfloor < \frac{M}{\delta} < \frac{c}{a}$ so that $(d - Lb)\beta - (c - La)\alpha = (d\beta - c\alpha) - L\delta \in (0, \delta) \subset (0, 1)$, and therefore the pair $(c - La, d - Lb)$ must be distinguished and the pair $(c, d) = (c - La, d - Lb) + L \cdot (a, b)$ must be a nice pair.

Now take $M = \alpha + 2\beta$, and integers $c_0 > \frac{a}{\delta}M + 1$ and $d_0 = \lfloor \frac{c_0\alpha}{\beta} \rfloor$ satisfying the condition $0 < d_0\beta - c_0\alpha < \beta$. Using the analysis above, $(c_0, d_0)$, $(c_0, d_0 + 1)$, and $(c_0 - 1, d_0)$ are nice pairs, and their linear combinations $(0, 1)$ and $(1, 0)$ are nice pairs as well. This implies in particular that $\lambda\alpha$ and $(1 - \lambda)\beta$ are (positive) integers. Taking positive integers $m_2 = \lambda\alpha$ and $m_1 = (1 - \lambda)\beta$, we get $\frac{m_2}{\alpha} + \frac{m_1}{\beta} = \lambda + (1 - \lambda) = 1$, or equivalently $m_1\alpha + m_2\beta = \alpha\beta$. We conclude the proof.