Warren D Smith <warren.wds@gmail.com> wrote:

--Andy, your random number algorithm works, but only with probability 1. But the FACT that it works with prob=1 proves EXISTENCE of many normal numbers, ...

I'd like to return to what I said two days ago: This also works for defining what's meant by the normality of an infinite list of real numbers that are between 0 and 1. Phrased another way, any large excerpt from such a sequence would have approximately equal distribution across the real number line between 0 and 1, and taken pairwise and graphed against each other would have an approximately equal distribution across the unit square, and taken as triples and graphed against each other would have an approximately equal distribution across the unit cube, etc. Such a sequence would generate a normal number in base B if you choose digits by binning the numbers in the sequence into bins of size 1/B. For instance if your sequence was the non-integer parts of (3/2)^N, whose first few terms are: 0.500000, 0.250000, 0.375000, 0.062500, 0.593750, 0.390625, 0.085938, ... in base 10 you'd simply take the first digit of each, to generate the number: 0.5230530... This would also work for base factorial. You'd just use a bin size of 1/N for the Nth digit. But is any infinite sequence of reals known to be normal in that sense? I used (3/2)^N only for illustrative purposes. The non-integer part of successive powers of X is never a normal sequence, at least not if X is real. Can you see why not? Anyhow, it still wouldn't get you a number that's known to be normal in every base, since this binning technique would get you a *different* normal number in each base.