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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 $f(x) = x$.

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    $\begingroup$ Famously, $\sum_{k=2}^\infty 10^{-k!}$ is irrational, and all its non-zero digits are in even positions. $\endgroup$
    – user239203
    Commented Nov 3, 2019 at 23:30

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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)).

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  • $\begingroup$ Has it actually been proved that $\sum_{k=0}^\infty \pi_{2k+1}10^{-k}$ is irrational, where $\pi_k$ is the $k$-th decimal digit of $\pi$? $\endgroup$
    – user239203
    Commented Nov 3, 2019 at 23:32
  • $\begingroup$ Hold on $-$ it's a trivial result that the odd digits of $\pi$ form an irrational number? Do you have a proof of that? $\endgroup$
    – TonyK
    Commented Nov 3, 2019 at 23:33
  • $\begingroup$ Yeah, right after I posted this I answered my own question with essentially exactly the same trivial case, too. As a less general question: what makes it work for $\pi$? is it because it is known to not be...structured? $\endgroup$
    – Genie Wiley
    Commented Nov 3, 2019 at 23:34
  • $\begingroup$ I might have misphrased that - I strongly suspect it holds for $\pi$, and it is trivial such numbers exist (the number I constructed above, but now take the even numbers instead) $\endgroup$
    – Alexander Geldhof
    Commented Nov 3, 2019 at 23:35
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We could define an irrational I, that alternates 0's with the digits of pi.

$f(I)$ as defined above will return $0$ which is rational.

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