Theorem
Let’s think about the sequence of $[\alpha k]$ for $k \in \mathcal{Z}^+$ and $\alpha \in \mathcal{R}^+/\mathcal{Q}^+$ and $\alpha > 1$. For example, $[\sqrt{2}] \approxeq 1.414 = 1, [2 \sqrt{2}] \approxeq 2.828 = 2, [3 \sqrt{2}] \approxeq 4.242 = 4, [4 \sqrt{2}] \approxeq 5.656 = 5, [5 \sqrt{2}] \approxeq 7.071 = 7$ for $\alpha = \sqrt{2}$. Then, $[\alpha k]$ and $[\beta k]$ partitions an integer sequence if $\alpha^{-1} + \beta^{-1} = 1$. For example, $\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}/(\sqrt{2} - 1)} = \frac{1}{\sqrt{2}} + \frac{\sqrt{2} - 1}{\sqrt{2}} = 1$. Which means $[\sqrt{2}/(\sqrt{2} - 1)] \approxeq 3.414 = 3, [2\sqrt{2}/(\sqrt{2} - 1)] \approxeq 6.828 = 6$.
Proof
Proof will be done with contradictions for two parts.
No collision
There is something that exists in both $[\alpha k]$ and $[\beta k]$.
Then, there is positive integer $v, r, k \in \mathcal{Z}^+$ such that $v = [\alpha r] = [\beta k]$.
Which means $v \le \alpha r < v + 1$ and $ v \le \beta k < v + 1$.
By dividing each by $\alpha, \beta$, $\frac{v}{\alpha} \le r < \frac{v + 1}{\alpha}$ and $\frac{v}{\beta} \le k < \frac{v + 1}{\beta}$.
However, equality can’t be happen because $\alpha, \beta$ is irrational.
Now adding two equation results in $\frac{v}{\alpha} + \frac{v}{\beta} = v(\frac{1}{\alpha} + \frac{1}{\beta}) = v < r + k < \frac{v + 1}{\alpha} + \frac{v + 1}{\beta} < (v + 1)(\frac{1}{\alpha} + \frac{1}{\beta}) = v + 1$.
Which result in $v < r + k < v + 1$ and it’s a contradiction to $r + k$ is an integer.
Coverage
There is something that doesn’t exists in both $[\alpha k]$ and $[\beta k]$.
Then, there is positive integer $v, r, k \in \mathcal{Z}^+$ such that $[\alpha r] = v - 1, [\alpha (r + 1)] = v + 1, [\beta k] = v - 1, [\beta (k + 1)] = v + 1$.
Which means, $\alpha r < v, \alpha (r + 1) \ge v + 1, \beta k < v , \beta (k + 1) \ge v + 1$.
Notice that equality can’t happen in here either because $\alpha, \beta$ is irrational.
By dividing $\alpha, \beta$, it results in $r < \frac{v}{\alpha}, r + 1 > \frac{v + 1}{\alpha}, k < \frac{v}{\beta} , k + 1 > \frac{v + 1}{\beta}$.
Adding 1st and 3rd inequality, $r + k < \frac{v}{\alpha} + \frac{v}{\beta} = v(\frac{1}{\alpha} + \frac{1}{\beta}) = v$.
Similarly adding 2nd and 4th inequality, $r + 1 + k + 1 > \frac{v + 1}{\alpha} + \frac{v + 1}{\beta} = (v + 1)(\frac{1}{\alpha} + \frac{1}{\beta}) = v + 1$.
Which means $r + k > v + 1 - 2 = v - 1$.
By using both inequality, $v - 1 < r + k < v$ and it’s a contradiction to $r + k$ is an integer.
As a result, claim holds.