Inequality involving the distance to the nearest integer

Question

i've been searching for several days trying to prove this inequality below. Let $\alpha$ an irrationnal number.

First let's write for $n,j>0$ some integers, $\sigma_j^{+}(n)=1$ if $0<\{n\alpha\}<\frac{1}{2}2^{-2^j}$ and $0$ otherwise, where $\{\}$ means the fractional part.

Now let's write $q(j)\geq 0$, the smallest integer $q$ such that $0<||q\alpha||< \frac{1}{2}2^{-2^j}$, where $||x||=\min_{k\in\mathbb{Z}}|x-k|=\min(\{x\},1-\{x\})$.

I don't find a way to prove that, for $n$ and $n'$ two distincts integers such that $\sigma_j^{+}(n)=\sigma_j^{+}(n')=1$, then $q(j)\leq |n-n'|$.

You can find this assertion in this proof: Brujo Theorem

Thank you for your help !

Answer 1

A friend of mine proved it (thanks to him !):

Let $\epsilon>0$. For $n > n'$ two distincts integers such that $0<\{n\alpha\}< \epsilon$ and $0<\{n'\alpha\}< \epsilon$, we will prove that $||(n-n')\alpha||<\epsilon$ which if enought to conclude the proof by definition of $q(j)$.

First, we have for some integers $k,k'\in\mathbb{Z}$, $k<n\alpha<k+\epsilon$ and $k'<n'\alpha<k'+\epsilon$.

So we have $k-k'-\epsilon<(n-n')\alpha<k-k'+\epsilon$.

• If we have $k-k <(n-n')\alpha<k-k'+\epsilon$, then $\{(n-n')\alpha\}<\epsilon$;
• Else, if $k-k'-\epsilon<(n-n')\alpha<k-k'$, then we have $k'-k <-(n-n')\alpha< k'-k+\epsilon$ and so $1-\{(n-n')\alpha\}=\{-(n-n')\alpha\}<\epsilon$.

Finaly, we get $||(n-n')\alpha||=\min\{\{(n-n')\alpha\},1-\{(n-n')\alpha\}\}<\epsilon$.