User blog comment:Vel!/Happy Pi Day/@comment-5150073-20150315122531/@comment-5529393-20150316051550

Ikosarakt1, that's an intriguing question. What can be said about the set of strings that represent ordinals using the encoding in the ITTM article, or the corresponding real numbers in [0,1]? We do know that, if we reduce the string to correspond to a smaller ordinal, either by removing a leading 0 or taking every 2^nth binary digit in the case that the leading digit is one, and keep repeating the process, then we will reach 0 in a finite number of steps. It follows that the set of such reals has measure 0. Further, there must be postive integers n and m such that, for all k, the digit in position m + 2^n k is 0. So the real cannot be normal, as any binary string of length longer than n with no 0's of distance n apart cannot appear in the binary expansion except for a finite number of times. While we can't say for sure one way or another, generally when there is a "reasonable" definition of a set of measure 0 (like nonnormal reals), and there is no particular reason to believe that a particular real "should" be in the set, one would expect that the real is probably not in the set, as a randomly chosen real has probability 0 of being in the set. So people generally feel that it is more likely than not that pi is normal, and therefore does not represent an ordinal according to the given encoding.