Reals with integer powers bounded away from integers?

Question

Do there exist real numbers whose integer powers are bounded away from integers? More precisely, for an arbitrary constant 0 < $\epsilon$ < 1/2, does there exist real x such that for all positive integer n,
$ \epsilon < (x^n)$ mod 1 < $1-\epsilon $?

Pisot numbers more or less provide the opposite of this behavior, and at first I briefly thought that applying Hurwitz' Theorem to the logs might help...but ln(3)/ln(2) isn't rational, and $2^n$ and $3^n$ seem to keep a minimum unit distance apart. ^_^

Answer 1

We should be able to construct such a real number $x$. Let $\epsilon\in(0,1/2)$ be fixed. If $S\subset \mathbb{R}_{>0}$ and $r>0$, we write $S^r$ for the set of positive $r$-th powers of elements of $S$. Recursively define a sequence of intervals $I_n$ of the form $[N_n+\epsilon,N_n+1-\epsilon]$ where the $N_n$ are positive integers. Let $I_1=[N_1+\epsilon,N_1+1-\epsilon]$ where $N_1$ chosen large enough that $N_1(1-2\epsilon)>2$. Once $I_n$ has been defined, $I_n^{(n+1)/n}$ has length at least $2$, so we can find an integer $I_{n+1}$ so that $[I_{n+1}+\epsilon,I_{n+1}-\epsilon]\subset I_n^{(n+1)/n}$. Then the intervals $I_n^{1/n}$ are nested and have diameter going to 0, and we choose $x$ to be the unique element in the intersection of $I_n^{1/n}$. This $x$ is constructed in such a way that $x^n\in[I_n+\epsilon,I_n+1-\epsilon]$ for all $n$.

Answer 2

Such numbers exist. Here is a way to construct one of size close to $n$, for $n\ge 2$:

Let $a_1 = n + \frac{1}{2}$. Inductively, for each $k > 1$ let $a_k = (\lfloor a_{k-1}^k\rfloor+\frac{1}{2})^{1/k}$. Then $a = \lim_{k\rightarrow\infty} a_k$ exists and is between $n$ and $n+1$.

More precisely, we have $|a_k-a_{k-1}| = |a_k^k-a_{k-1}^k|/(\sum_{i=0}^{k-1}a_k^ia_{k-1}^{k-1-i}) < \frac{1}{2kn^{k-1}}$, so $|a-a_k| < \sum_{j>k} \frac{1}{2jn^{j-1}} < \frac{1}{kn^k}$. By the same argument, if we let $\alpha_k = \inf_{j\ge k} a_k$ then we have $|a-a_k| < \frac{1}{k\alpha_k^k}$, and since $\alpha_k > a-\frac{1}{kn^k}$ we have

$|a^k-a_k^k| < |a-a_k|k(a+\frac{1}{kn^k})^{k-1} < \frac{(a+\frac{1}{kn^k})^{k-1}}{(a-\frac{1}{kn^k})^{k}} < \frac{e^{1/n^k}}{n}.$

Thus the number $a$ we've constructed will satisfy your condition with $\epsilon = \frac{1}{2}-\frac{e^{1/n}}{n}$.

Example: For $n = 10$, we have $a = 10.51174467290...$, and the smallest fractional part of $a^k$ for $k$ from $1$ to $100$ is $0.452...$. The largest is $0.543...$.