Show that $F=0.0110001111…$ is irrational, with $F$‘s decimal expansion being $1,3,5,7,9,…$ consecutive $0$s interspersed with $2,4,6,8,…$ consecutive $1$s.
A friend sent me this recently. The first thing I tried was to multiply by a power of $10$ and subtract to get some sort of repeating form. This failed. Then I realized that technique would only have been successful if I had been trying to prove $F$’s rationality, not irrationality. Then I realized that the only real technique I have to show irrationality is the famous one used for $\sqrt{2}$, which I don’t think works for this number. Heck, I’m even a little bit lost in trying to find a closed algebraic form for $F$. I’m aware of the concept of that an irrational number has no “repetition” in its expansion in some sense, but I don’t know that that is sufficient for a rigorous proof in this context. Any hints would be greatly welcome.
Best Answer
Not a very creative proof, but let's try this: we say that a decimal of the form $$0.a_1a_2a_3a_4a_5\cdots$$ ($a_i\in\{0,1,...,9\}$) is eventually periodic if there exists $N\in\mathbb{N}$ and $k\in\mathbb{N}$ such that $$(a_N,a_{N+1},\cdots ,a_{N+k})=(a_{(N+lk)+1},a_{(N+lk)+2},\cdots a_{(N+lk)+(k+1)})$$ for any $l\geq 1$. Now suppose for contradiction that $F$ is eventually periodic, and let the $(k+1)$-tuple $(a_N,a_{N+1},\cdots ,a_{N+k})$ denote the numbers in one period of $F$. By the definition of $F$, it is clear that $$(a_N,a_{N+1},\cdots ,a_{N+k})\neq (0,\cdots,0)\text{ and }(a_N,a_{N+1},\cdots ,a_{N+k})\neq (1,\cdots,1).$$ (Indeed, if the above were to hold, then that would imply that the digits of $F$ are eventually constant, which is absurd.) Hence, $$(a_N,a_{N+1},\cdots ,a_{N+k})$$ would have to consist of both $1$'s and $0$'s. Then take $l$ large enough so that the length of repeated $1$'s is at least, say, $10(k+1)$. This would imply that for some $l\geq 1$, the string $(a_{(N+lk)+1},a_{(N+lk)+2},\cdots, a_{(N+lk)+(k+1)})$ is completely contained in the string of $1$'s, so that $$(a_N,a_{N+1},\cdots ,a_{N+k})=(a_{(N+lk)+1},a_{(N+lk)+2},\cdots, a_{(N+lk)+(k+1)})=(1,\cdots,1)$$ but this contradicts our earlier assumption. Hence $F$ is not eventually periodic. It is also clear that $F$ is not eventually terminating. We conclude $F$ is irrational.