Let us define the sequence $$a_{n}=\{a^n\}$$ with $a > 0$ where $\{a^n\}$ denotes the fractional part.
How could we show that there is no positive real numbers $a\in Q$ such that this sequence $a_{n}$ is strictly increasing.
I can find example such $a_{n}$ is increasing,such as $$a=1+\varepsilon,\varepsilon\to 0,$$
Now achille hui have find Nice a example when $a$ is irrational,
$$a=(2+\sqrt{3})$$
Now I think when $a\in Q$ is true,can someone prove it?
The statement is false. Let $a = 2+\sqrt{3}$ and $b = 2 - \sqrt{3}$. It is easy to see the sequence defined by
$$T_n = a^n + b^n, n \in \mathbb{N}$$
satisfies the recurrence relation $T_n = 4T_{n-1} - T_{n-2}$ for $n \ge 2$.
Since $T_0 = 2$ and $T_1 = 4$ are integers, so does all $T_n$.
Notice $b \in (0,1)$. This means for all $n > 1$, we have $\{\;a^n\;\} = 1-b^n$ strictly increasing and converges to $1$.
As achille hui shows, there are irrational $a$ for which $\{a^n\}$ is strictly increasing (any number of the form $a=\lceil \sqrt k\rceil +\sqrt k$ where $k$ is not a perfect square will do). For rational $a=\frac bc$ this is impossible: Assume otherwise and let $L=\lim\{a^n\}$. Then $\epsilon_n=L-\{a^n\}$ is a decreasing sequence of positive reals. We have $a^n=\{a^n\}+\lfloor a^n\rfloor = L-\epsilon_n+\lfloor a^n\rfloor$. Then $$ L-\epsilon_{n+1}+\lfloor a^{n+1}\rfloor= a^{n+1}=a(L-\epsilon_{n}+\lfloor a^n\rfloor),$$ hence (multiply with $c$ and rearrange) $$ (b-c)L+\underbrace{c\epsilon_{n+1}-b\epsilon_{n}}_{\to 0} = \underbrace{\lfloor a^{n+1}\rfloor c-b\lfloor a^n\rfloor}_{\in\mathbb Z}.$$ The right hand side is an integer for all $n$ and the left hand side converges to $(b-c)L$, hence $(b-c)L$ is an integer. But then $b\epsilon_n-c\epsilon_{n+1}$ is an integer and hence $=0$ for $n$ large enough. Using such $n$, we see $a=\frac{\epsilon_{n+1}}{\epsilon_n}<1$. But for $0<a<1$ the sequence $\{a^n\}=a^n$ is decreasing, contradiction!
