The seventy-fifth question: For any positive real number $x$, define the set $A(x)=\left\{[n x] \mid n \in Z^{+}\right\}$. Find all irrational numbers $\alpha$ greater than 1 such that: if a positive real number $\beta$ satisfies $A(\beta) \subseteq A(\alpha)$, then $\frac{\beta}{\alpha}$ is an integer.
The 75th question,
Solution: First, prove a lemma: If $x, y \in R$, and for any positive integer $n$, we have $[n x]=[n y]$, then $x=y$. Proof of the lemma: In fact, according to the condition, for any positive integer $n$, we have $ny - 22$, and $A(\beta) \subseteq A(\alpha)$, then for any positive integer $n$, there exists a positive integer $m$ such that $[n \beta] = [m \alpha]$.
First, take $n=1$, then there exists $k \in Z^{+}$, such that $[\beta] = [k \alpha]$.
We will prove by induction on $n$: For any positive integer $n \in Z^{+}$, we have $[n \beta] = [n k \alpha]$.
When $n=1$, the conclusion is obviously true. Assume the conclusion holds for all positive integers less than $n (n \geq 2)$. Now consider the case of $n$.

Since $[(n k-1) \alpha] + [\alpha] \leq [n k \alpha] \leq [k \alpha] + [(n-1) k \alpha] + 1 = [\beta] + [(n-1) \beta] + 1 \leq [\mathrm{n} \beta] + 1$, hence $[\mathrm{n} \beta] \geq [(n k-1) \alpha] + [\alpha] - 1 > [(n k-1) \alpha]$.

Also, $[(n k+1) \alpha] \geq [k \alpha] + [(n-1) k \alpha] + [\alpha] = [\beta] + [(n-1) \beta] + [\alpha] \geq [n \beta] - 1 + [\alpha] > [n \beta]$, so there must be $[n \beta] = [n k \alpha]$.

According to mathematical induction, for any positive integer $n$, we have $[n \beta] = [n k \alpha]$. Therefore, according to the lemma, $\beta = k \alpha$, thus $\frac{\beta}{\alpha} = k \in Z$.
(2) Next, prove: An irrational number $\alpha$ greater than 1 and less than 2 does not satisfy the condition.
Let $\beta = \frac{\alpha}{2 - \alpha}$, then $\frac{\beta}{\alpha} = \frac{1}{2 - \alpha} \notin Z$. We will prove: $A(\beta) \subseteq A(\alpha)$. That is, we only need to prove: For any positive integer $n$, there exists a positive integer $m$ such that $[n \beta] = [m \alpha]$.

Let $[n \beta] = s$, then $s \leq n \beta = \frac{n \alpha}{2 - \alpha} < s + 1 \Rightarrow 2s - s \alpha \leq n \alpha < 2(s + 1) - (s + 1) \alpha$, rearranging gives $s \leq \frac{(n + s) \alpha}{2} < \frac{(n + s + 1) \alpha}{2} < s + 1$, thus $[n \beta] = s = \left[\frac{(n + s) \alpha}{2}\right] = \left[\frac{(n + s + 1) \alpha}{2}\right]$. Note that $\frac{n + s}{2}$ and $\frac{n + s + 1}{2}$ must include one integer, take this integer as $m$, then $[n \beta] = [m \alpha]$.
In summary, the $\alpha$ that satisfies the condition is all irrational numbers greater than 2.
Note: In fact, when $1 < \alpha < 2$, take $\gamma = \frac{\alpha}{\alpha - 1}$, then $\frac{1}{\alpha} + \frac{1}{\gamma} = 1$, according to the Beatty's theorem, $A(\alpha)$ and $A(\gamma)$ form a partition of $Z^{+}$. Also, $A(\gamma) \subseteq A\left(\frac{\gamma}{2}\right)$, hence $Z^{+} \backslash A\left(\frac{\gamma}{2}\right) \subseteq A(\alpha)$. Then take $\beta = \frac{1}{1 - \frac{2}{\gamma}} = \frac{\alpha}{2 - \alpha}$, according to Beatty's theorem, $A(\beta) = Z^{+} \backslash A\left(\frac{\gamma}{2}\right) \subseteq A(\alpha)$.