A more general question than in "A question about decimal representation of irrational numbers.": Since there is an infinite amount of irrational numbers could you always find one that contains a given finite decimal sequence?
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Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. – Community Feb 19 '22 at 10:23
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Of course you can, simply define it to be that way. Take your sequence of digits $a_1a_2...a_n$ and consider the number $0.a_1a_2...a_nb_1b_2...$ where $b_1, b_2...$ are the decimal digits of $\sqrt{2}$. This number will certainly be irrational, because if it was not that would imply $\sqrt{2}$ is rational.
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Ah, yeah that is indeed a simple solution. Overlooked the fact that there is a trivial solution to this generalization. – Mount-Chiliad Feb 19 '22 at 11:06