What can we say about the value of $ [(n+1) \alpha ]- [ n \alpha]$, where $ \alpha$ is any irrational number?
Can this be further simplified? Here $[x]$ denotes the largest integer less or equal to $x$.
Any help would be appreciated. Thanks in advance.
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Let $x,y \in \mathbb R$. Then we can write
$$ x = m+\{x\} \:\:\text{and}\:\: y = n+\{y\}, $$
where $m,n \in \mathbb Z$ and $\{x\},\{y\} \in [0,1)$.
Thus,
$$ x+y = m+n+\{x\}+\{y\}, $$
where $m+n \in \mathbb Z$ and $\{x\}+\{y\} \in [0,2)$.
Therefore,
$$ \lfloor x+y \rfloor = \begin{cases} m+n & \:\text{if}\: \{x\}+\{y\}<1; \\ m+n+1 & \:\text{if}\: \{x\}+\{y\} \ge 1. \end{cases} $$
With $x=n\alpha$ and $y=\alpha$, this is
$$ \lfloor n\alpha+\alpha \rfloor = \begin{cases} \lfloor n\alpha \rfloor + \lfloor \alpha \rfloor & \:\text{if}\: \{n\alpha\}+\{\alpha\}<1; \\ \lfloor n\alpha \rfloor + \lfloor \alpha \rfloor + 1 & \:\text{if}\: \{n\alpha\}+\{\alpha\} \ge 1. \end{cases} $$
Therefore,
$$ \lfloor (n+1)\alpha \rfloor - \lfloor n\alpha \rfloor = \begin{cases} \lfloor \alpha \rfloor & \:\text{if}\: \{n\alpha\}+\{\alpha\}<1; \\ \lfloor \alpha \rfloor + 1 & \:\text{if}\: \{n\alpha\}+\{\alpha\} \ge 1. \end{cases} $$
If $a_n=[n\alpha]$ then $$a_n\le n\alpha<a_n+1,\quad a_{n+1}\le(n+1)\alpha<a_{n+1}+1$$
Hence $$\alpha-1=(n+1)\alpha-1-n\alpha< a_{n+1}-a_n<(n+1)\alpha-n\alpha+1=\alpha+1$$
Since the difference is an integer we conclude $$\lceil\alpha-1\rceil\le a_{n+1}-a_n\le\lfloor \alpha\rfloor$$
For example, $\pi$ has sequence $3, 6, 9, 12, 15, 18, 21, 25, 28, 31, 34$ with differences either 3 or 4.