Given $x\in\langle 1,\infty\rangle\cap\mathbb Q$ Prove: $$\exists\lim_{n\to\infty}\{x^n\}\implies x\in\mathbb N$$ $\{\cdot\}$ is a fractional part of the number.

So far, even I couldn't prove it generally, I've used a calculator to notice that $f(x_n)>f(x_{n-1})$ and $f(x_n)>f(x_{n+1})$ if $x$ is not a natural number, where $f(x_n)=\{x^n\}$. That could mean $f(x_n)$ is not monotonous (maybe divergent) sequence, and maybe that leads to a contradiction. Yet, I couldn't come to any ideal solution though.

Your help is greatly appreciated. Thanks in advance.

  • 1
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    – saulspatz
    Commented Feb 18, 2020 at 16:56
  • $\begingroup$ If $x=k+r$ where $k$ is a positive integer and $0< r< 1$ then for any $\epsilon > 0$ there are infinite $n,m$ so that there is a integer $M < (k+r)^n < M+\epsilon$ and there is an integer $N$ so that $N-\epsilon < (k+r)^m < N$. Use archimedean principal. $\endgroup$
    – fleablood
    Commented Feb 19, 2020 at 1:30

1 Answer 1


Assume $x=\frac{q}{p}, q>p>2, (q,p)=1$ and let the limit of {$x^n$} be $l$. Pick $\epsilon >0$ very small such that $100p(x+1)\epsilon < 1$ say.

Then since {$ml$}, $m\ge 1$ integer is uniformly distributed modulo $1$ if $l$ irrational (edit later - here ud is overkill as noted in the comments, dense or accumulation at zero will do and that follows easily from Dirichlet-box arguments), or there is {$ml$}$=0$ if $l$ rational, and since obviously {$mx^n$} $\to${$ml$} for $m$ positive integer fixed, we can find $m,N$, s.t {$mx^n$} $\le \epsilon, n \ge N$.

But this means there are integers $r_n \le mx^n, n \ge N$, $mx^n-r_n \le \epsilon, n \ge N$. Multiplying the relation by $x$ and subtracting the relation for $n+1$ gives:

(edit per question in comments) - $mx^{n+1}-r_nx \le x\epsilon$ and $mx^{n+1}-r_{n+1} \le \epsilon$, so

$|xr_n-r_{n+1}|=|(mx^{n+1}-r_nx)-(mx^{n+1}-r_{n+1})| \le |mx^{n+1}-r_nx|+|mx^{n+1}-r_{n+1}|$, so

$|xr_n-r_{n+1}| \le (x+1)\epsilon$. However if $xr_n-r_{n+1} \ne 0$, we have $|xr_n-r_{n+1}|=\frac{|qr_n-pr_{n+1}|}{p} \ge \frac{1}{p}$, so $p(x+1)\epsilon \ge 1$ which contradicts the choice of $\epsilon$ above. Hence we must have $xr_n=r_{n+1}, n \ge N$.

This means $qr_n=pr_{n+1}, n \ge N$ so $r_{N+1}=Aq$ for some fixed integer $A>0$. But using $qr_{N+1}=pr_{N+2}$ we get that $p$ divides $A$ and $r_{N+2}=\frac{q^2A}{p}$

But now using $qr_{N+2}=pr_{N+3}$ we get $p^2$ divides $A$ and $r_{N+3}=\frac{q^3A}{p^2}$ so it is obvious that continuing this we eventually run out of powers of $p$ dividing $A$ so we get a contradiction! Done!

Note that the same proof applies when there are potentially finitely many limit points as we have joint uniform distribution (edit later - see comment above - dense will do) of the fractional parts of $ml_1,..., ml_k$ for rationally independent irrationals $l_1,..l_k$ and by appropriate integral multiplications we can reduce all the limit points to such and $0$ and find $m$ as before.

  • $\begingroup$ I think you don't need uniformly distributed - dense will do. $\endgroup$ Commented Feb 19, 2020 at 5:07
  • $\begingroup$ @marty - completely agree; ud is overkill, just accumulation at zero is needed and that is very easy with the usual Dirichlet-box type arguments $\endgroup$
    – Conrad
    Commented Feb 19, 2020 at 12:06
  • $\begingroup$ @Conrad - it seems I didn't quite get how did you subtract the relation for n+1 $\endgroup$
    – VIVID
    Commented Feb 19, 2020 at 14:42
  • $\begingroup$ edited to show it explicitly $\endgroup$
    – Conrad
    Commented Feb 19, 2020 at 14:50

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