Recent content by RobertCairone

But I don't think the disenfranchised is an empty set. Rather, compared to the disenfranchised, everything else is an empty set. Consider the function f(x) where f(x) = 0 if x is Enfranchised and f(x) = 1 if x is Disenfranchised. Then ∫^{0}_{1}f(x) = 1, since you can remove up to a countably...

I'm comfortable saying every Enfranchised number is the limit of a sequence of rational numbers. I'm not so sure about all the the Disenfranchised. Of course, the deltas and epsilons used must themselves be enfranchised, and I think they even have to be rational, since they are used in the...

True to an extent, just as the designation of a particular base is arbitrary. Numbers in binary necessarily take more characters to write than numbers in decimal. But \pi itself isn't a number, it is a symbol for a number. Unlike the digits 0 thru 9, the symbol \pi tells nothing about it's value...

Is this interesting? I have a way to partition the transcendentals into two disjoint groups. I begin by defining a property of the real numbers I call "width". Given a base b, say 10 for convenience, I write down all the numbers I can with a single digit. There are ten of them, ten...

So if the first corresponds to the integers, and the second to the primes, is there a naive interpretation for the third of these? Does this say there is a definable subset of the primes that still diverges?

Yes, the natural numbers would work as well, but I chose the primes as a smaller set of basis elements. I was looking for some kind of "minimalist" way of approximating the reals. The slowness by which the reciprocals of the primes diverges made them seem natural. Does anything diverge more slowly?

Is this interesting? I have a way to represent all of the real numbers with a subset of the open interval (0,1). I write as a binary decimal x = .a_{0}a_{1}a_{2}a_{3}... where x = \sum -1^{a_{0}+1}/p_{i} where p_{i} is the i^{}th prime. Since the reciprocals of the primes diverges, in this...