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By diagonalization, it is possible to construct a real number $r \in [0,1]$ such that for every rational $q \in [0,1]$, there exists an index $i \in \mathbb{N}$ such that $r_i \neq q_i$ (where $x_i$ is the $i$'ith digit in $x$).

Can we make a stronger claim, and construct a real number $r \in [0,1]$ such that for every rational $q \in [0,1]$ there exists an index $i \in \mathbb{N}$ such that for every $j \geq i$, $r_j \neq q_j$?

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  • $\begingroup$ This is not a research-level question, and as such it is off-topic for this site. It might be more appropriate for math.stackexchange.com . Anyway, a hint: think about $q=k/9$ (assuming decimal digits). $\endgroup$ – Emil Jeřábek supports Monica Feb 2 at 9:35
  • $\begingroup$ Sorry! I didn't know this website is for research level questions. Should I delete the question? $\endgroup$ – Larry Feb 2 at 9:40
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No. Consider the rational numbers 0, 0.111..., 0.222..., 0.333.., ..., 0.888... Any real number either shares infinitely many digits with one of these, or it has only finitely many digits which are not 9. But then it is after some point only 9s, so is equal to a terminating rational and has infinitely many 0s.

If you prefer not to allow this equivalence, add 0.999... to the above list.

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