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. ^_^

  • 1
    $\begingroup$ Regarding your nickname: it's time. $\endgroup$ Commented Apr 10, 2013 at 1:14
  • $\begingroup$ Duly noted. ^_^ $\endgroup$ Commented Apr 10, 2013 at 1:54
  • $\begingroup$ (was previously ICanChangeThisLaterRight) $\endgroup$ Commented Apr 10, 2013 at 1:56

2 Answers 2


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 $n$.

  • $\begingroup$ Nifty...thanks! Somewhere between 7.3769347535 and 7.3769347577 lurks a number whose powers never stray within 1/3 of an integer... $\endgroup$ Commented Apr 10, 2013 at 2:35

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 $0.543...$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.