A property of the floor function $x\mapsto \lfloor x\rfloor$ This is problem is from Vinogradov's elementary number theory book.
For any real number $a>0$, define $f_a:\mathbb{N}\rightarrow\mathbb{Z_+}$ by $x\mapsto \lfloor ax\rfloor$.
Given $\alpha,\beta>0$, show that $f_\alpha$, $f_\beta$ are injective,  $f_\alpha(\mathbb {N})\cap f_\beta(\mathbb{N})=\emptyset$, and $\mathbb{N}=f_\alpha(\mathbb{N})\cup f_\beta(\mathbb{N})$ if and only if $\alpha,\beta\in\mathbb{R}\setminus\mathbb{Q}$ and $\frac1\alpha + \frac1\beta =1$
Sufficiency is not difficult to prove. The part I am struggling with is necessity, where I don't seem able to control the gaps between values of $f_\alpha$ (or $f_\beta$ for that matter.) Any hints will be appreciated.
 A: Start with $N\in\mathbb{N}$ fixed. Let $x_N$ be the number of natural numbers such that $1\leq [\alpha x]\leq N$. There are two cases:
(a) $\alpha x_N < N < N+1\leq \alpha (x_N+1)$
(b) $N\leq \alpha x_N < N+1 \leq \alpha(x_N+1)$
In case (a), $x_N<\frac{N}{\alpha}<x_N +1$ which means that $x_N=\frac{N}{\alpha}-\delta\,$ for some $0<\delta<1$.
In case (b), $\frac{N}{\alpha}\leq x_N<\frac{N}{\alpha}+\frac{1}{\alpha}$ which means that $x_N=\frac{N}{\alpha}+\delta\,$ for some $0\leq \delta<\frac{1}{\alpha}$.
All in all, one has that
$$x_N=\frac{N}{\alpha}+\delta_N, \quad\text{for some}\quad |\delta_N|<\max\big\{1,\tfrac{1}{\alpha}\big\}$$
Similarly, if $y_N$ is the number of natural numbers such that $1\leq [\beta y]\leq N$, then
$$y_N=\frac{N}{\beta}+\varepsilon_N, \quad\text{for some}\quad |\varepsilon_N|<\max\big\{1,\tfrac{1}{\beta}\big\}$$
By assumption, if $1\leq k\leq N$, then either $k=[\alpha x]$ for some $x\in\mathbb{N}$, or $k=[\beta y]$ for some $y\in\mathbb{N}$, but not both. Consequently
$$
N=x_N+y_N=\frac{N}{\alpha}+\frac{N}{\beta}+\delta_N+\varepsilon_N
$$
Dividing by $N$ on both sides of the equation above leads to
$$
1=\frac{1}{\alpha}+\frac{1}{\beta}+\frac{\delta_N+\beta_N}{N}
$$
Clearly $\frac{\delta_N+\beta_N}{N}\xrightarrow{N\rightarrow\infty}0$ from where one part of the conclusion follows. This also shows that $\alpha,\beta>1$.
To check that $\alpha$ (similarly $\beta$) are irrational, suppose $\alpha=\frac{p}{q}$ where $p,q>0$ are irreducible integers, i.e. $\operatorname{g.m.d}(p,q)=1$. Then, $q<p$ and
$$[\alpha q]=[\beta(p-q)]$$
in contradiction to $f_\alpha(\mathbb{N})\cap f_\beta(\mathbb{N})=\emptyset$.
