Here in the second comment I do not understand why $\omega^\omega$ corresponds to irrational numbers? : In my experience one typically identifies $ω^ω$ with the irrational elements of R; and then we call them "reals" because they are equinumerous, and in particular "isomorphic up to a countable set".
QUESTION: What can we say about irrationality of $f=(2,2,2,2,...)$ goes to $(0,0,1,0,0,1,0,0,1,0,0,...)$ which is not irrational since it repeats the $0,0,1$ pattern forever,right? And yet, it doesn't end wiht eventually all $1$'s.
As noted above, you're conflating two different homeomorphisms between $\omega^\omega$ and $\mathbb{R}\setminus\mathbb{Q}$. It seems you're most interested in the continued fraction map, so let me say a bit about that one.
Given an infinite sequence $A=(a_i)_{i\in\omega}$ of natural numbers - which in logic include $0$, so you'll see some "$1+$"s here which you won't in other contexts - the corresponding continued fraction is $$F_A=(1+a_0)+{1\over (1+a_1)+{1\over (1+a_2)+{1\over ...}}}.$$ The following is a basic fact about continued fractions (and not a result from logic):
See here for a proof. This implies immediately that the map $A\mapsto F_A$ is a bijection from $\omega^\omega$ to $\mathbb{R}\setminus \mathbb{Q}$, and with a bit of thought we can also see that it's continuous in both directions.
Note that the proof of the basic fact above is nontrivial. This is the main virtue of the bijection mentioned in my answer to your previous question: it's less natural but much easier to see that it is indeed a homeomorphism.