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** $j \geq i$, $r\_j \neq q\_j$?



Real number which is different from all rationals
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.


