exponents of rational numbers close to integers I'm not sure how you can prove the base case for the answer to this question. The base case states:
Lemma. If $x \ge 1$ is a rational number and $\alpha\ne 0$ is a real number such that $\lim_{n\to\infty}\alpha x^n-m_n=0$ where $m_n\in\mathbb{N},$ then $x\in\mathbb{Z}$.
Does it also hold if $x$ can be real instead of just rational?
Some attemps: 
I've thought about raising $m_n$ to some power $p$ and then doing the reverse by raising to power $1/p$ so it gives a tighter bound around $m_n$, but this doesn't appear to work since there may be a lot of integers besides $m_n^p$ within the bound.
 A: Here's a proof assuming $x$ is rational.  Write $x=\frac{a}{b}$ for $a,b\in\mathbb{N}$ in lowest terms.  Choose $\epsilon>0$ such that $x\epsilon<\frac{1}{2b}$ and $\epsilon<\frac{1}{2b}$ (actually, the latter is automatic since $x\geq 1$).  Choose $N$ such that $|\alpha x^n-m_n|<\epsilon$ for all $n\geq N$ and such that $m_N\neq 0$ (the latter is possible since $x\geq 1$ and $\alpha\neq 0$).  In particular, for any $n\geq N$, we can write $$\alpha x^n=m_n+\delta$$ where $|\delta|<\epsilon$ ($\delta$ depends on $n$).  Multiplying by $x$, we get $$\alpha x^{n+1}=xm_n+\delta x.$$  Note that $xm_n$ is a rational number with denominator $b$, and $|\delta x|<\frac{1}{2b}$.  If $xm_n$ were not an integer, then it would be at least $\frac{1}{b}$ away from an integer, and so $xm_n+\delta x$ would be at least $\frac{1}{2b}$ away from an integer.  But by hypothesis, $\alpha x^{n+1}$ is within $\epsilon$ of $m_{n+1}$, and $\epsilon<\frac{1}{2b}$.  Thus $xm_n$ must be an integer, and clearly in fact we must have $xm_n=m_{n+1}$.
Thus $xm_n=m_{n+1}$ for all $n\geq N$.  But this means $x^km_N=m_{N+k}$ is an integer for all $k\in\mathbb{N}$.  This implies $m_N$ is divisible by $b^k$ for all $k$.  Since $m_N\neq 0$, this is only possible if $b=1$, i.e. if $x$ is an integer.
A: Here is a proof for more general statement: http://artofproblemsolving.com/community/c6h4556p334225
It's given as a lemma in the solution to problem 27 (section 4.1) in Russian 1987 book
 В.А.Садовничий, А.А.Григорьян, С.В.Конягин "Задачи студенческих математических олимпиад"
("Problems of mathematical olympiads for university students")
