Prove that there is a real number $a$ such that $\frac{1}{3} \leq \{ a^n \} \leq \frac{2}{3}$ for all $n=1,2,3,...$ Prove that there is a real number $a$ such that $\frac{1}{3} \leq \{ a^n \} \leq \frac{2}{3}$ for all $n=1,2,3,...$ Here, $\{ x \}$ denotes the fractional part of $x$.

My attempt: Clearly $a$ cannot be an integer because $\{ a^n \}=0$ for all $n \in \mathbb{N}$. Also $a$ cannot be a rational number because $\{ x \}$ is the same as $a$ modulo $1$. So the fractional part of $a$, at some point, will be less than $\frac{1}{3}$. 

I try to find irrational $a$ which satisfies the inequality but I got no luck. 

Can anyone give some hint to this question?
There exists $\alpha \in\left[\dfrac{16}{3},\dfrac{17}{3}\right]$ with the required property. To see this, we will construct an interval sequence
$$\left[\dfrac{16}{3},\dfrac{17}{3}\right]=[\alpha_{1},\beta_{1}]\supset [\alpha_{2},\beta_{2}]\supset\cdots\supset[\alpha_{n},\beta_{n}],$$
where $\alpha_{n}$ and $\beta_{n}$ are such that
$$\alpha^n_{n}-\dfrac{1}{3}=\beta^n_{n}-\dfrac{2}{3}=m_{n}\in \Bbb N^{+},$$
so that, for any $x\in [\alpha_{n},\beta_{n}]$, we have
$$\dfrac{1}{3}\le\{x^n\}\le\dfrac{2}{3}.$$

We construct the interval sequence by induction. Assume that we have $[\alpha_{n},\beta_{n}]$. Let
$$a=\alpha^{n+1}_{n},\quad\quad b=\beta^{n+1}_{n}.$$It follows that $$ 
b-a=(m_{n}+\dfrac{2}{3})\beta_{n}-(m_{n}+\dfrac{1}{3})\alpha_{n}&gt;\dfrac{\alpha_{n}}{3}&gt;\dfrac{5}{3}.$$
Then there exists $m_{n+1}\in \Bbb N^{+}$ such that
$$\left[m_{n+1}+\dfrac{1}{3},m_{n+1}+\dfrac{2}{3}\right]\subset[a,b].$$
We take
$$\alpha_{n+1}=\sqrt[n+1]{m_{n+1}+\dfrac{1}{3}},\qquad\beta_{n+1}=\sqrt[n+1]{m_{n+1}+\frac{2}{3}}.$$
Now $$\alpha^{n+1}_{n}=a&lt;\alpha^{n+1}_{n+1}=m_{n+1}+\dfrac{1}{3}&lt;\beta^{n+1}_{n+1}=m_{n+1}+\dfrac{2}{3}&lt;b=\beta^{n+1}_{n},$$and hence $\alpha_{n}\le\alpha_{n+1}&lt;\beta_{n+1}&lt;\beta_{n},$ or
$$[\alpha_{n},\beta_{n}]\supset[\alpha_{n+1},\beta_{n+1}].$$
Obviously it depends how approximately a≈5.62144865272, but a quick check with windows calculator shows that a^14 ≈ 31468458796.31579.. or a^15 ≈ 176898325303.72424... for instance