Fractional part of rational power arbitrary small

Question

I think that $\{a^n\}$ (where $\{x\}$ is $x \pmod 1$), where $a$ is fixed rational greater than 1 and $n$ is positive integer, is dense in $[0,1]$ is unsolved. However what about $\{a^n\}$ is arbitrary small for some $n$ ($a$ is fixed rational as well).

Answer 1

Edit: sorry, I now realize that the question asks whether 0 is a limit point of $\{a^n\}$. My answer below is not a paraphrase of the question but is something weaker. But I still hope it helps.

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Let me rephrase the question: Let $a$ be a rational greater than 1, is it true that $\lim_{n \to \infty}\{a^n\} = 0$?

When $a$ is an integer, yes, $\lim_{n \to \infty}\{a^n\} = \lim_{n \to \infty} 0 = 0$.

However, When $a$ is a noninteger rational, $\lim_{n \to \infty}\{a^n\} \ne 0$.

Proof by contradiction, assume $\lim_{n \to \infty}\{a^n\} = 0$. let $$ be the denominator of $\{a\}$, we can find an $$ such that $∀ n > $, $\{x^n\} < \frac{1}{aq}$.

Let $ > $, we have

$$\{a^m\} < \frac{1}{aq} \tag{1}\label{1}$$ and

$$\{a^{m+1}\} < \frac{1}{aq} < \frac{1}{q} \tag{2}\label{2} $$ where

\begin{align} \{a^{m+1}\} &= \{ (⌊a^m⌋ + \{a^m\}) (⌊a⌋ + \{a\}) \} \\ & = \{ ⌊a^m⌋⌊a⌋ + ⌊a^m⌋\{a\} + \{a^m\}(⌊a⌋ + \{a\}) \} \\ &= \{⌊a^m⌋\{a\} + \{a^m\}(⌊a⌋ + \{a\})\} \\ &= \{⌊a^m⌋\{a\} + \{a^m\}a\} \\ & = \{\{⌊a^m⌋\{a\}\} + \{a^m\}a\} \end{align}

where the last step comes from the observation that { + } = {{} + }.

Note $\{⌊a^m⌋\{a\}\}$ is one of $1/, 2/, 3/... (-1) / $, in particular

$$ \frac{1}{q} \le \{⌊a^m⌋\{a\}\} \le \frac{q-1}{q} \tag{3}\label{3}$$

We have $1/ < \{⌊a^m⌋\{a\}\} + \{a^m\}a < 1$ by \eqref{1} \eqref{3}, which implies $\{a^{m+1}\} = \{\{⌊a^m⌋\{a\}\} + \{a^m\}a\} > 1/$ which contradicts \eqref{2}.