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How can I construct a rational sequence explicitly that converges to $ x $?

I tried to consider $x\in [0,1]$ with \begin{equation*} \bigcup_{i = 0}^{n-1} \left[ \frac{i}{n},\frac{i+1}{n} \right] = [0,1] , \end{equation*} then exists $0\leq i\leq n-1$ such that $x\in[i/n,(i+1)/n]$. And with $\lfloor x\rfloor$ y can traslate this interval such that for all $x\in\mathbb{R}$ \begin{equation*} x\in \bigcup_{i = 0}^{n-1} \left[\lfloor x\rfloor + \frac{i}{n},\frac{i+1}{n} + \lfloor x\rfloor\right] = [\lfloor x\rfloor,\lfloor x\rfloor+1] , \end{equation*} Then, reasoning as before, there exists $1\leq i\leq n-1$ such that $x\in[\lfloor x\rfloor+i/n , \lfloor x\rfloor + (i+1)/n]$.

But, taking limit I would like to decrease the interval and obtain a rational sequence that goes to $ x $ in terms of its integer part, $ i $ and $ n $.

I get a sequence $(q_{n})_{n\in\mathbb{N}}$ $$ q_{n} = \frac{\lfloor x\rfloor n+i}{n}, $$ but this limit is $\lfloor x\rfloor$. Could someone help me with this? (it is to prepare an assistantship class)

Regards!!

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  • $\begingroup$ Try $\dfrac{\lfloor nx\rfloor}{n}$ or $\dfrac{\lfloor ix\rfloor}{i}$ - I am not clear why you use both $\endgroup$ Commented Sep 3, 2021 at 1:02

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Here's an explicit sequence that uses infinite decimal expansions as its motivation.

$$ q_n = \frac{\lfloor 10^n x \rfloor}{10^n} $$

This sequence has the advantage that, for every term $q_j$ after $q_i$, $q_j$ is closer to $x$ than $q_i$ is or $q_j$ is the same distance away from $x$ as $q_i$ is.

This is not true of $p_n = n \mapsto \frac{\lfloor n x \rfloor}{x}$, which works, but can bounce around a bit.

For example, consider $x = \frac{2}{3} - 0.001 $. Then $p_2$ is $\frac{1}{2}$, but $p_3$ is $\frac{1}{3}$. $p_3$ is further away from $x$ than $p_2$ is, so we need to work slightly harder to prove convergence since we aren't monotonically approaching our target.

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