“Proof” that $\frac{1}{7}$ is irrational

Question

Sketched this out this morning and it’s been bugging me:

1. If we enumerate ℚ with each element as its decimal expansion, we know that any new number we make a la the old diagonalization trick must be irrational, else we’d have a reductio proving ℚ‘s uncountability
2. This is true irrespective of which enumeration/method of “tweaking” the diagonal we pick
3. So make the following map for the digits of the diagonal:

• $0 \rightarrow 1$
• $1,2 \rightarrow 4$
• $3,4 \rightarrow 2$
• $5,6 \rightarrow 8$
• $7 \rightarrow 5$
• $9 \rightarrow 7$

4. Then pick an enumeration which cycles through ℚ in a way that guarantees the above digits repeat when we apply the map

Clearly this is me being dumb somewhere but I’m not sure where exactly?

Answer 1

At some point, there will be $\frac 1 7 = 0.\overline{142857}$ in your enumeration of $\mathbb Q$. At that point you have to pick a digit in your new number that makes sure it is not equal to $\frac 1 7$, rendering step $4$ impossible.

Tweaking your enumeration on the go to avoid hitting $0.\overline{142857}$ indefinitely you will end up with an enumeration of (a subset of) $\mathbb Q\setminus\{\tfrac 1 7\}$. Not being on that list doesn't prove irrationality. It just proves your new number is irrational or equal to a rational number not on the list, like $\frac 1 7$.