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This is inspired by the comment in Minimum values of the sequence $\{n\sqrt{2}\}$ that, by Kroneckers Approximation Theorem, the fractional part of $n\sqrt{2}$ is dense in $[0, 1]$.

My question is that, given an $x \in [0,1]$, is there an explicit construction of a sequence $(n_k)$ such that $\{n_k\sqrt{2}\} \to x$? In particular, an explicit function of the form $n_k = f(k)$ would be nice.

Replace $\sqrt{2}$ by other irrationals for extra credit.

Is this easier for some particular class of reals?

The continued fraction for $\sqrt{2}$ probably comes into play.

marty cohen
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1 Answers1

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You can simply find a sequence $b_n$ that converges to $\sqrt 2$, such as $$b_{k+1}=1+\frac 1{1+b_k},\quad b_1=1,\tag 1$$

or $$b_k=\frac{\lfloor 10^k \sqrt 2\rfloor}{10^k}.\tag 2$$

By replacing $n_k=\dfrac 1{b_k/2}$, $(n_k)$ converges to $\dfrac 1{2\sqrt 2}$.

Therefore $\{n_k\sqrt 2\}$ converges to $x=1/2$.

Generally, for $r>0$ and $x\in[0,1)$, we find a sequence $b_n$ that converges to $\sqrt r$. Set $n_k=\dfrac {x}{b_k}$, then $\{n_k\sqrt r\}$ converges to $x$.

Tianlalu
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