# Proving $y$ is irrational given $x$ is irrational

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 $$f(x) = x$$.

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

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)).

• 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 $\pi$?
– user239203
Commented Nov 3, 2019 at 23:32
• Hold on $-$ it's a trivial result that the odd digits of $\pi$ form an irrational number? Do you have a proof of that? Commented Nov 3, 2019 at 23:33
• 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? Commented Nov 3, 2019 at 23:34
• 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) Commented Nov 3, 2019 at 23:35

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.