Question about irrational numbers and finite subsequences of their decimal places.

Question

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?

Answer 1

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.