# Does the decimal expansion of a rational number contain the decimal expansion of an irrational number?

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 $$s_p=\frac{1}{100}$$.

• $(x_n)$ 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. – TonyK Dec 11 '19 at 20:38
• Because if the decimal representation has finitely many non-zero digits, then $s_p$ will have finitely many terms. – egorovik Dec 11 '19 at 20:38
• 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. – egorovik Dec 11 '19 at 20:41
• Let $s = \frac{1}{99}$. Then $s_p = \frac{1}{100}$. You need a condition ensuring that $x_{p_n} \neq 0$ infinitely often. – Daniel Fischer Dec 11 '19 at 20:46
• 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. – Daniel Fischer Dec 11 '19 at 20:56

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.