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Main Question

If we look at the sequence $$x^n \bmod 1$$ for increasing values of $n$ where x is a real number, are we guaranteed that we can find some value in the sequence arbitrarily close to 0?

Another way of asking this might be to define a recursive function $$f(x, y) = \min(x^y \bmod 1,\,f(x, y-1))$$ where $f(x, 1) = x^y \bmod 1$ and $y \ge 1$. Then, my question becomes: for all real $x$, is it true that

$$\lim_{y \to \infty} f(x, y) = 0?$$

Why I'm Interested

I was coming up with a way to simulate any probability with just a series of single coin tosses.

I'm sure there are better ways to do it, but my first pass attempt was to take a fraction $\frac{i}{j}$, and convert it to $\frac{1}{\frac{j}{i}}$ and then raise that to the $n$th power until the denominator becomes an integer. And then since it's now in the form $\frac{1}{k}$, where $k$ is an integer, I figured it'd be easier to deal with.

But then I realized I didn't even know if it was possible to get it into that form with arbitrary accuracy. Thus, my question. (Of course, my question expands this and asks if this is always possible to do for reals, not just rational numbers.)

Pro Q
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    As far as I know, this is open. For $nx$ , $n$ running over the natural numbers, it is well known that the sequence is equidistributed mod 1 if and only if x is irrational. – Peter Jun 25 '20 at 07:34
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    If you take, for example $x=e$, then essentially you're looking at the decimal parts of $e^n$ ... I don't see any reason why this would approach zero. – Matti P. Jun 25 '20 at 07:36
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    The OP isn't looking at the limit of the fractional parts of $e^n$, but rather at the lim inf of those fractional parts. That seems much more likely to approach $0$. – Greg Martin Jun 25 '20 at 07:37
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    This is not known for $x=3/2$, see https://math.stackexchange.com/a/2160372/87023 for some citations – Chris Culter Jun 25 '20 at 08:19

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