Integer parts of multiples of irrationals

Solution 1:

Using the known fact, we see that $S(\alpha)\cap S(\beta) = \varnothing $ If we have $$\frac{k}{\alpha}+\frac{l}{\beta} =1 $$ for some positive integers $k, l$

Conversely, we use Kronecker's theorem ( If $1$, $\alpha$ and $\beta$ are linearly independent over $\mathbb{Q}$ then the set $\{((n\alpha), (n\beta)) : n \in \mathbb{Z}^{+}\}$ is dense in $[0,1]^2$) , and detailed analysis in case when they are linearly dependent over rationals, we obtain that the condition above for $\alpha$ and $\beta$ is indeed holds when $S(\alpha)\cap S(\beta)=\varnothing$.

On the other hand, from the link above we can also deduce that if $\alpha$ rational, then $S(\alpha)\cap S(\beta)\neq \varnothing$ regardless $\beta$.

Hence, we have

(Theorem) $S(\alpha)\cap S(\beta) = \varnothing $ if and only if $\alpha$ and $\beta$ are irrational numbers satisfying $$\frac{k}{\alpha}+\frac{l}{\beta} =1 $$ for some positive integers $k, l$

If $S(\alpha)\cap S( \beta)=\varnothing$, and $S(\beta)\cap S(\gamma)=\varnothing$, then there are positive integers $k_1, k_2, k_3, k_4$ such that $$\frac{k_1}{\alpha}+\frac{k_2}{\beta}=\frac{k_3}{\beta}+\frac{k_4}{\gamma}=1$$

If $k_2\neq k_3$, then we have $$\frac{k_1k_3}{\alpha}-\frac{k_2k_4}{\gamma}=k_3-k_2$$ In this case, $S(\alpha)\cap S(\gamma)\neq \varnothing$.

If $k_2=k_3$, then we have $$\frac{k_1}{\alpha}=\frac{k_4}{\gamma}$$ This case also, $S(\alpha)\cap S(\gamma)\neq \varnothing$.

Hence, we have proved that for real numbers $\alpha, \beta$ and $\gamma$, $$ S(\alpha)\cap S(\beta)= S(\beta) \cap S(\gamma) =S(\alpha)\cap S(\gamma) = \varnothing $$ is impossible.