# Does every irrational number contain "$666$" in its decimal expansion?

Let $$x\in\mathbb{R}$$. Say $$x$$ is satanic if $$x$$ contains "$$666$$" somewhere in its decimal expansions. Let $$S$$ be the set of such numbers. Is $$\mathbb{R} \setminus \mathbb{Q} \subseteq S$$ More generally, for any finite string of digits, say $$y = y_1y_2y_3\ldots y_{n-1}y_n$$, does $$x\in \mathbb{R}\setminus\mathbb{Q}$$ necessarily contain $$y$$ in its decimal expansion.

It seems intuitive that, at least, for any $$y$$, $$y$$ should be contained in $$\textit{almost every }$$ irrational number, since the irrational numbers have an infinite decimal expansion. But, I could envision there exists some irrational number that follows some pattern such that it doesn't contain some $$y$$.

• For an example that came up in a recent Question, see here. It is pretty well known that a decimal expansion which neither terminates nor (eventually) repeats periodically is thus representative of an irrational number. Jan 12, 2022 at 2:59

No. There are even transcendental numbers all of whose digits are $$0$$ and $$1$$.
On the other hand, in terms of cardinality there are as many non-satanic numbers as satanic numbers, because there are $$2^{\aleph_0}$$ of each: just consider numbers whose decimal expansion contains nothing but 0s and 1s.
For a concrete example, take $$x=0.101001000100001000001\ldots1\underbrace{00\dots0}_{n\text{ zeros}}1\underbrace{00\dots0}_{n+1\text{ zeros}}1\dots.$$ This decimal expansion isn't periodic (since it contains arbitrarily long runs of zeros), and so $$x$$ is irrational. You can do something similar to avoid any string of digits you want, not just $$666$$.