Is a number like .12112111211112111112... irrational?

Question

This number repeats forever in this "pattern", but is it irrational, and is there a way to prove whether it is or not?
(For clarity, between every successive number '2', we have an additional number '1' - so a single one, then a 2, then two ones, then a two, then three ones, etc for infinity)

Also, would this number be transendental?

What exactly is "this pattern"? As in the numbers of $1$ between $2$ is growing? This is not a periodic representation, hence irrational. — freakish, Mar 14 '25 at 12:46
This question is similar to: Is $0.1010010001000010000010000001 \ldots$ transcendental?. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — K B Dave, Mar 14 '25 at 13:21
@KBDave the questions are obviously different. Proving irrationality is fairly trivial in general, proving transcendentality (is that a word?) is rather hard usually. The proof in the linked question is what I would call quite hard, using advanced concepts. Whereas proving this number is not rational is pretty basic late high school early college stuff. — DRF, Mar 14 '25 at 13:27
Note that if $x= 0.1010010001000010000010000001\ldots$ and $y=0.12112111211112111112\ldots$ with the intended extensions then $y=10x-\frac89$. — Henry, Mar 14 '25 at 13:47
And now I see that the lowest voted (and non-accepted) answer to the link provided by @KBDave has the same answer that I wrote. — Lee Mosher, Mar 14 '25 at 13:50
This question is similar to: Prove that 2.101001000100001... is an irrational number.. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — Henry, Mar 14 '25 at 13:52
In particular, if you subtract $1/9=0.1111\dots$ this becomes whether $0.01001000100001\dots$ is rational. — Thomas Andrews, Mar 14 '25 at 14:05

score 1 · Accepted Answer · answered Mar 14 '25 at 13:22

I'll answer only your question about rationality --- the question about transcendental is very different, and our policy is to ask one question per post.

The pattern that a number satisfies if and only if it is rational is a very particular pattern with very carefully formulated requirements. Namely, writing the sequence of digits to the right of the decimal point in the form $.a_1a_2a_3\ldots$, the pattern says that there exists $N \ge 1$ and $K \ge 1$ such that for all $n \ge N$, $a_{n+K}=a_n$. In other words, as long as you go far enough (as long as $n \ge N$), the digits repeat after every $K$ steps ($a_{n+K}=a_n$).

Your pattern does not satisfy these requirements: no matter what $N$ and $K$ are, I can find a counterexample, i.e. I can find a value of $n \ge N$ such that $a_{n+K} \ne a_n$. Here's how. There are infinitely many occurrences of $a_n=2$ in your pattern. So even amongst $n \ge N$ there are still infinitely many occurrences of $a_n=2$. As we go further out along each of those occurrences, the block of $1$'s that follows gets longer and longer. So, I can find $n \ge N$ such that $a_n=2$ and such that the following block of $1$'s is longer than $K$. Therefore, $a_{n+K}=2 \ne 1 = a_n$.

Two fun corollaries about this, that neither diminish nor enhance the answer. First, this works for any positive integer base: 10, 2, 17, 42. Base 10 isn't special. Second, you can use this to show that for any integer, there's a number composed of $K$ digits equal to radix minus one (i.e., $99 \cdots 9$ in base 10) that's divisible by that integer. — TimWescott, Mar 14 '25 at 14:21
could you post a name for this theorem so folks can search on it? — TimWescott, Mar 14 '25 at 14:22

Is a number like .12112111211112111112... irrational?

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