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I think that {$a^n$} (where {x} is x (mod 1)) , where $a$ is fixed rational greater than 1 and $n$ is positive integer, is dense in $[0,1]$ is unsolved. However what about {$a^n$} is arbitrary small for some n (a is fixed rational as well)

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If ${a^n}$ is shown to take on arbitrarily small values, for every possible rational (nonintegral) $a>1$, then that would mean that $0$ had been shown to be a limit point of the fractional parts of all the possible choices of base $a$ for the powers $a^n$. If this were known it likely would have appeared on the Wolfram page but I didn't see it there. The subject seems complex in general. – coffeemath Jul 28 '13 at 4:09
seems like you are saying that n is fixed as we vary a. i will try and rephrase the problem – 61plus Jul 29 '13 at 4:37
Actually I see that it means for a fixed $a$ one computes the fractional parts of $a,a^2,a^3,\cdots$ and checks whether that sequence has $0$ as a limit point. This idea of limit points of these sequences (for fixed $a$) is what is discussed on the Wolfram page I linked in my previous comment. – coffeemath Jul 29 '13 at 8:50

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