[thaumasiotes]

Shoring up places where I thought my proof was weaker:

- I rely on the assumption that an irrational number with this property exists. (If it didn't, then the property would imply normality.) This is easy to fix; Champernowne's constant has this property and so the assumption is valid.

- I assert without proving that z' is irrational. We can prove this using the definition of a rational number as one whose decimal expansion repeats after some index. Since z is irrational, somewhere in its decimal expansion there is a digit not equal to 2. (Otherwise, every digit of z would be 2, and z would be the rational number 2/9.) Since z' successively repeats larger and larger stretches of z, this suffices to show that, for any index i into z', there is a higher index j > i such that the jth digit of z' is not 2.

- But we also know that a sequence of n "2"s in a row can be found within z' for any positive n. Assume that the jth digit of z' is not 2. Since we know that a sequence of 2j "2"s occurs later within z' -- it can't occur earlier because not enough digits have yet occurred -- any cycle in the digits of z' cannot yet have begun by index j.

- But since there is no maximum index into z' beyond which all digits are not 2, a cycle in the digits of z' cannot have begun at any index into z'. This shows that z' is irrational.

[thaumasiotes]

> Since we know that a sequence of 2j "2"s occurs later within z' -- it can't occur earlier because not enough digits have yet occurred -- any cycle in the digits of z' cannot yet have begun by index j.

This is wrong -- the cycle might begin at j and continue into a huge series of 2s.

But we cannot yet have completed one cycle by index j, and this property can be extended -- there is no index into z' at which one cycle could have been completed, and hence the digits of z' never cycle.

A much simpler proof that there is no cycle goes like this:

Suppose there is a cycle of length n. We know that a stretch of 2n consecutive 2s will appear after the beginning of the cycle. This implies that the cycle consists entirely of 2s. We also know that a non-2 digit will appear after the beginning of the cycle. This is a contradiction; there cannot be a cycle of any length.