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Let $s \in [0,1] \cap \mathbb{Q}$. Take the decimal expansion of $s$ as $\sum_{n=1}^\infty \frac{x_n}{10^n}$ for a sequence $(x_n) \subset \{0,1,...,9\}$ that does not eventually end in a tail of 9's (implying uniqueness).

Suppose $(x_n)$ does not also end in a tail of 0's. Suppose $(x_n)$ has finitely many 0's. Let $P_n$ denote the $n$th prime number. Is $s_p=\sum_{n=1}^\infty \frac{x_{P_n}}{10^{P_n}}$ irrational?

It seems the decimal expansion of $s_p$ is supposedly 'random'. If true, it implies that pretty much every rational number "contains" the decimal expansion of an irrational number, which sorta makes sense considering the uncountability of $\mathbb{R}-\mathbb{Q}$. It also makes sense in that you are able to add two irrational numbers to produce a rational number.

I'll do an example. Let $s=\frac{1}{3}=0.\bar{3}$. Then $s_p=0.03303030003030...$, which is clearly irrational.

Edit: @DanielFischer showed $s=\frac{1}{99}=0.\overline{01}$ is a counterexample, as every odd-indexed digit is 0. Therefore $x_{P_n}=0$ since $P_n$ is odd for $n > 1$, implying $s_p=\frac{1}{100}$.

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    $\begingroup$ $(x_n)$ eventually settles down into a recurring sequence (perhaps you already know this). But I would be surprised if it was known one way or the other whether your sum is always irrational or not. $\endgroup$ – TonyK Dec 11 '19 at 20:38
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    $\begingroup$ Because if the decimal representation has finitely many non-zero digits, then $s_p$ will have finitely many terms. $\endgroup$ – egorovik Dec 11 '19 at 20:38
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    $\begingroup$ I see. I didn't read that part. Then you can provably show that $s_p$ is transcendental. They will probably be a Liouville number, since the sequence of primes has arbitrarily large gaps. But one would need to fiddle with the distribution of those gaps, how frequently they occur. $\endgroup$ – egorovik Dec 11 '19 at 20:41
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    $\begingroup$ Let $s = \frac{1}{99}$. Then $s_p = \frac{1}{100}$. You need a condition ensuring that $x_{p_n} \neq 0$ infinitely often. $\endgroup$ – Daniel Fischer Dec 11 '19 at 20:46
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    $\begingroup$ No, @SpencerKraisler, it isn't. $\frac{1}{99} = 0.\overline{01}$ doesn't end in a tail of $0$s, but all $x_{2m+1}$ are $0$, and there are only finitely many even primes. $\endgroup$ – Daniel Fischer Dec 11 '19 at 20:56
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Unless the $x_{p_n}$s are all eventually $0$, the maximum number of consecutive $0$s between nonzero terms in $0.0x_2x_30x_50x_7000x_{11}0x_{13}\ldots$ grows without limit, since the maximum number of consecutive composite numbers grows without limit (e.g., $n!+2,n!+3,n!+4,\ldots,n!+n$), so no number of the OP's given form, with the assumption that $x_n=0$ for only finitely many $n$, can be rational.

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  • $\begingroup$ Oh, I just saw Daniel Fischer's comment below the OP. This answer duplicates that comment. $\endgroup$ – Barry Cipra Dec 11 '19 at 22:15

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