Hans Havermann <gladhobo@teksavvy.com> wrote:

MathWorld's article on 'base' cautions that "the representation of a given integer in an irrational base may be nonunique"

I believe that's true of the representation of any number in any non-integer base (e.g. 2.5 in base 1.5). I was speaking of *any* representation, not necessarily the canonical greedy-algorithm representation. Sorry if that wasn't clear. This is unfortunate, since it means I can't test bases for integer termination by calculating the representation of integers in the obvious way. The rule for integer representations that terminate may not be obvious. In base sqrt(2), the rule is to leave odd powers blank. In base phi, the rule is to never have two consecutive "1"s. Also, the tiniest amount of precision loss wrecks the algorithm. What's the square of the square root of 2? 1.9999999999999999? That's less than 2, so 2 in base sqrt(2) can't start with 100. Sigh.

(but this is true also of integer bases: 1=.999? in base ten).

Only of numbers that terminate. (In base 10, those are rationals whose denominator only contains 2 and 5 as prime factors.) I think that in non-integer bases, every number (integers, rationals, and irrationals) has multiple representations. The canonical form is to use the greedy algorithm, so 1.00000 is standard and 0.99999... is not.

More relevant perhaps is the issue of normalcy.

I think you mean "normality." "Normalcy" is a neologism invented by President Harding.

My understanding of the concept is that it cannot be applied to irrational bases. What then is the frequency distribution of the digits (0, 1, 2, 3) of seventeen in base pi?

1/pi, 1/pi, 1/pi, and (pi-3)/pi respectively, assuming 17 is normal in base pi. I'd be very surprised if it isn't. Of course multiple representations are possible, and perhaps you could go out of your way to skew the statistics. However, you can't get away with simply never using a "3" -- I checked. Weirdness: Sagan spoke of pi in base eleven. I just looked at eleven in base pi, and it doesn't appear to contain any "3"s at all! Just how far in do you have to go to find a "3"? If there are no "3"s at all, does that count as a message? (This is with the usual greedy algorithm, of course.) Your ellipses, including one of mine that you quoted, are turning into question marks. I suspect some Microsoft product at your end is "helpfully" turning each instance of three consecutive dots into the single Microsoft-proprietary character that represents an ellipsis (octal 205, decimal 133, hexadecimal 85), and the list software is then "helpfully" replacing any character whose high bit is "1" with a question mark. Please beat your software into submission. Thanks. And, moderator, please tweak the list software to allow ISO-8859-1 accented letters through. It's okay to nuke Microsoft droppings and UTF-8 mojibake. Thanks. You get extra credit for turning Microsoft droppings back into what they were supposed to have been, e.g. the above-mentioned character back into three dots.