Solution
(a) We first prove the following statement: For each positive integer $m$ there exist positive integers $n \leq m$ and $x_{m}$ such that $\left|n \alpha-x_{m}\right|2$ while $n_{2}>n_{1}$ we conclude that $x_{m}=\left[n_{2} \alpha\right]-\left[n_{1} \alpha\right] \geq[\alpha] \geq 2$ is also a positive integer.
As proved above, for each integer $m \geq 3$, there exist positive integers $n \leq m$ and $x_{m}$ such that $\left|n \alpha-x_{m}\right|2$ we conclude that for each $i \in \mathbb{N}$ we have $n_{i+2}-2 n_{i+1}+n_{i}=0$. So, $n_{1} \alpha, n_{2} \alpha, \cdots, n_{i} \alpha, \cdots$ is itself an arithmetic progression. Therefore, the difference of the two arithmetic progressions is another infinite arithmetic progression: $\left\{n_{1} \alpha\right\},\left\{n_{2} \alpha\right\}, \cdots,\left\{n_{i} \alpha\right\}, \cdots$.

However, the arithmetic progressions cannot be bounded, unless their ratio is 0 . Hence $\left\{n_{2} \alpha\right\}=\left\{n_{1} \alpha\right\}$, which yields that $n_{2} \alpha-n_{1} \alpha=\left[n_{2} \alpha\right]-\left[n_{1} \alpha\right] \in \mathbb{Z}$ and therefore $\alpha \in \mathbb{Q}$, which is a contradiction with our assumption (also note that $n_{2} \neq n_{1}$ since they belong to an infinite arithmetic progression).