Is e Normal?

Wikipedia does this, adding the extra term "simply normal" for the case where we are talking about relative to a single base:

    We say that x is simply normal in base b if
    the sequence S(x,b) is simply normal, and
    that x is normal in base b if the sequence
    S(x,b) is normal.  The number x is called a
    normal number (or sometimes an absolutely
    normal number) if it is normal in base b
    for every integer b greater than 1.

https://en.wikipedia.org/wiki/Normal_number

  <span style="font-family:Symbol">p</span>

    <span style='font-family:Symbol'>p</span>

It's generally expected that non-integer square roots are normal, and yet they always have repeating CFs.

All that combines to mean that I would be surprised if e turns out not to be normal.

Note that for a base 10 normal irrational by the definition of "convergence" in this case means || (number of digits of value a) / (number of digits of value b) ||_\infty < ε (sup norm over all a, b from 0 to 9, a ≠ b) for some ε > 0 such that for some fixed n (the number of digits in the decimal expansion, halted), we have that whenever N > n this inequality holds.

In the case where there are a fixed number m of leading "a"s (whatever digit a is), we have ||(m + dig_a)/(dig_b)||_\infty = ||m/dig_b + dig_a/dig_b||_\infty -- we only need this to be less than SOME fixed ε (not necessarily the same ε as above) for every N > n for some n we choose.

By assumption ||dig_a / dig_b||_\infty < ε whenever n > N. Choose n' > n such that dig_b satisfies m/dig_b < ε

(Note: this is possible as dig_b increases as n' grows: since m is constant and if dig_b were to be constant for n > N' then ||dig_a/dig_b||_\infty would be larger than ε for some n).

Then we have ||m/dig_b||_\infty < ε and ||dig_a/dig_b||_\infty < ε so ||m/dig_b + dig_a/dig_b||_\infty < 2ε.

Thus it converges -- satisfying the same definition as above chooseing n = n' and ε = 2ε.

If bits were copyrighted, you could just transform them into different bits. Change a single bit and then it's no longer copyrighted. Or xor them all. Or only every other bit. Or xor the whole thing with a different file, like a one time pad.

You could also change the content itself. Say it's a video. Shift all the pixels to the right. Or tint it slightly red. Or play it on a monitor and record it with a camera, then invert the resulting video.

There are an infinite number of possible transforms that result an infinite number of possible bits. What matters is information, not bits. They are not the same thing! Information is the bits, and the method you use to transform them together.

Often it's proposed that you can compress a file by storing the index where it occurs in pi. The problem is the index where your file starts will be bigger than the actual file! So you no longer need the file, but you still need this other set of bits that represents the index. The bits have changed, but the information has not.

If you don't have the index, then the whole procedure is entirely useless. Yes, pi theoretically contains season 5 of breaking bad - and an infinite number of variations and derivative works of it. But you will never ever find it unless you already know where to look.

And no computer could possibly calculate pi to that many digits in millions of years, so the whole thing is impractical anyway.