Let's say I have an irrational number. For clarity, I'll use $\pi$ since it's pretty well known that $\pi \not \in \mathbb{Q}$. Regardless, let's assume I have a function $f(x)$ that takes an irrational number $x$, and outputs a number $y$.
Let $y$ be the concatenation of a number of digits of $x$ according to some rule. For the sake of this question, we'll define it as the first, third, fifth, to the $n$th odd decimal place. So,
$f(\pi) = 0.1196....$
Since $\pi$ is irrational and does not repeat, it should hold that $f(\pi)$, similarly, is irrational.
Does this hold for all irrational numbers? It does not. Does it hold for any? (
Major edit: I should note that $f(x)$, evidently, isn't a very well-defined function. The odd number example was just an example! I'm looking for a more general answer about these rules in general. Obviously, there must be some outputs of an arbitrarily-defined $f$ such that the output always produces an irrational number when $x$ is irrational. Take the incredibly trivially case in which $f(x) = x$.
It definitely doesn't hold for all irrational numbers. There is no reason why all odd decimals of some irrational couldn't be $0$, for example. In fact, take any arbitrary irrational number between $0$ and $1$ and wedge zeroes between all of its decimals; the result is still irrational (hint: try multiplying by $11$).
It is similarly trivial that there are some irrational numbers for which the rule does work (I strongly suspect it works $\pi$, for example, as you stated yourself, and it works for the above example by taking the even numbers (or shifting the decimals)).