Spectrum of irrational number (exercise 3.13) I'm trying to come up with a solution for exercise 3.13 from Concrete math. The exercise asks to prove that $Spec(\alpha)$ and $Spec(\beta)$ partition positive integers if and only if $\alpha$ and $\beta$ are irrational and $$1/\alpha + 1/\beta = 1$$
This mean that if condition is true, than following should hold:
$$
\lfloor{n \alpha}\rfloor \neq \lfloor{m \frac{\alpha}{\alpha - 1}\rfloor},\quad \mid n, m \in \mathbb{N}
$$
for the case when $\alpha > 2$. This means that there exist no such combination of $m$ and $n$, such that previous condition is true.
Let's replace floor functions with sums:
$$
\sum_i [ i < n \alpha ] \neq \sum_i [i < m \frac{\alpha}{\alpha - 1}]
$$
Consider the case when $n\alpha > m \frac{\alpha}{\alpha - 1}$. Then left sum can be split as follows:
$$
\sum_i [ i < n \alpha ] = \sum_i [i < m \frac{\alpha}{\alpha - 1}] + \sum_i [  m \frac{\alpha}{\alpha - 1} \leq i < n \alpha ]
$$
A sum with $m$ cancels out in each side of inequality, and the result is:
$$
\sum_i [  m \frac{\alpha}{\alpha - 1} \leq i < n \alpha ] \neq 0
$$
In other words there exist no such combination of $n$ and $m$ such that the last sum is equal to 0.
And I'm in stuck in here, because I do not see why. And if there exist such combination, then the same number can pop up in the spectra of $\alpha$ and $\beta$.
I would appreciate, if you either explain me why the last sum can't be 0, or if you show me a mistake in my reasoning.
 A: Suppose that $\alpha$ and $\beta$ are irrational numbers such that $\frac1\alpha+\frac1\beta=1$. Suppose further that there are $m,n\in\Bbb Z^+$ such that $\lfloor m\alpha\rfloor=\lfloor n\beta\rfloor=\ell$, say, so that
$$\ell\le m\alpha,n\beta<\ell+1\;.$$
Then 
$$\frac{\ell}m\le\alpha<\frac{\ell+1}m\qquad\text{and}\qquad\frac{\ell}n\le\beta<\frac{\ell+1}n\;.$$
Moreover, $\frac{\ell}m$ and $\frac{\ell}n$ are rational while $\alpha$ and $\beta$ are not, so we can strengthen this to
$$\frac{\ell}m<\alpha<\frac{\ell+1}m\qquad\text{and}\qquad\frac{\ell}n<\beta<\frac{\ell+1}n\;.$$
Then
$$\frac{m}\ell>\frac1\alpha>\frac{m}{\ell+1}\qquad\text{and}\qquad\frac{n}\ell>\frac1\beta>\frac{n}{\ell+1}\;,$$
so
$$\frac{m+n}\ell>\frac1\alpha+\frac1\beta>\frac{m+n}{\ell+1}\;,$$
i.e.,
$$\frac{m+n}\ell>1>\frac{m+n}{\ell+1}\;.$$
But then $\ell<m+n<\ell+1$, which is impossible, since $\ell,m$, and $n$ are integers. This shows that $\operatorname{Spec}(\alpha)\cap\operatorname{Spec}(\beta)=\varnothing$.
As you’ve probably already realized, the proof that $\operatorname{Spec}(\alpha)\cup\operatorname{Spec}(\beta)=\Bbb Z^+$ is essentially the same as that given on pages $77$ and $78$ for the special case $\alpha=\sqrt2$ and $\beta=2+\sqrt2$.
