The post A question about decimal representation of irrational numbers. asked: "Is this true that any finite word of the alphabet $\mathcal{A_9}=\{0,1,2, \ldots,8,9\}$ appears somewhere in the decimal representation of $\sqrt{2}$ ?" (A)
The given answers say it's a very similar proposition to saying "$\sqrt{2}$ is a normal number." (B)
The phrase "similar" confuses me. It is not equal (A $\iff$ B). So can A be true if $\sqrt{2}$ is an irrational number?
(The start of the sentance in A "Is this true..." suggest that there is probably a source to this question, but I couldn't find anything.)
No, numbers in which every finite sequence of digits occurs are called disjunctive, disjunctivity is weaker then normality, in the sense being normal implies being disjunctive but the converse is not true.
Take a look at this https://oeis.org/wiki/Disjunctive_numbers for more info.
As explained in the link $0.10200300000040000000000005...$ is disjunctive in Base $10$ but not normal.