An inequality involving irrational numbers Let $x > 1$ be an irrational number , then how to show that there exist $y \in (1,2)$ such that $0<yx^n-[yx^n]<\dfrac 1{x-1} , \forall n\in \mathbb N$ ? , here $[.]$ denotes the greatest integer function .  
 A: For $x\leqslant2$, it is easy to find $y\in(1,2)$ such that $yx^n$ is not integral for any $n\in\mathbb{N}$. Then
$$0<yx^n-\lfloor yx^n\rfloor<1\leqslant\frac{1}{x-1}$$ for all $n\in\mathbb{N}$ as desired. So we may assume $x>2$.
Define an integer sequence $\{a_n\}$ recursiely by
$$a_{n+1}:=\lceil xa_n\rceil,~a_0:=1.$$
Then we define
$$b_n:=x^{-n}a_n,$$
and
$$c_n:=x^{-n}\biggl(a_n+\frac{1}{x-1}\biggr).$$
Since $x$ is an irrational number, we have $xa_n\not\in\mathbb{Z}$. Hence $a_{n+1}=\lceil xa_n\rceil>xa_n$, which implies $b_{n+1}>b_n$, i.e. $\{b_n\}$ is strictly increasing.
On the other hand, we have $a_{n+1}=\lceil xa_n\rceil<xa_n+1$. In other words,
$$a_{n+1}+\frac{1}{x-1}<x\biggl(a_n+\frac{1}{x-1}\biggr),$$
which implies $c_{n+1}<c_n$, i.e. $\{c_n\}$ is strictly decreasing.
Because $b_n<c_n$, the monotonic sequences $\{b_n\}$ and $\{c_n\}$ are both bounded, therefore convergent. Moreover, it is obvious that $c_n-b_n\to0$ as $n\to\infty$. So we can take
$$y:=\lim_{n\to\infty}b_n=\lim_{n\to\infty}c_n.$$
As the limit of strictly monotonic sequences, we get $b_n<y<c_n$, or equivalently,
$$a_n<yx^n<a_n+\frac{1}{x-1}$$
for all $n\in\mathbb{N}$. Since $x>2$, the right hand side is less than $a_n+1$. Thus $a_n\in\mathbb{Z}$ derives $\lfloor yx^n\rfloor=a_n$, then
$$0<yx^n-\lfloor yx^n\rfloor<\frac{1}{x-1}$$ for all $n\in\mathbb{N}$. Besides, $b_0<y<c_0$ implies $y\in(1,2)$. This completes the proof.
