[kosievdmerwe]

Interesting, my first inclination is to be skeptical that the property would still hold, since we're "potentially" slicing up a necessary sequence and adding 2s in the middle.

However infinities are weird* and I think you could construct a proof by contradiction making use of the fact that a sequence of N digits is embedded in infinitely many longer sequences most of which that won't have been broken up by the inserted 2s.

* I'm always skeptical when dealing with infinities and probabilities. Human intuition doesn't gel well with either concept.

[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.

[thaumasiotes]

> my first inclination is to be skeptical that the property would still hold, since we're "potentially" slicing up a necessary sequence and adding 2s in the middle.

No. If the sequence you want occurs between places a and b of z, then it is a substring of the full sequence between places 1 and b of z, and all such sequences are included within the expansion of z'. (Going by example again, if you're interested in the sequence that occurs between decimal places 41,028 and 315,001 of z, then that sequence will occur within the part of z' that repeats places 1 through 315,001 of z.)