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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 integers of width 1. There are more numbers of width 2, the ninety from 10 to 99, plus the first nine negative integers, plus nine decimals, .1, .2, .3, etc. That's 118 of width 2. On width 3 we get a lot more. More integers, more decimals, and some numbers like 1/3 and 1/7. We have a finite number of understandable mathematical symbols, and we have a finite number of digits, so we have a finite number of elements of each width (we could allow for factorials and add five numbers to width 2, being 5!, 6!, etc. We wouldn't include 9[itex]^{5}[/itex] in width 2, as the exponent is a binary operator needing explicit specification. But it is width 3). The width of any number is defined as the minimum width that gives its value.

It's not always easy to know the minimum width. For example, [itex]\pi[/itex] = √6*Ʃi=1,∞1/i^2. For that typography, that's a width of 14. In other conventions or by using other expansions the width could be different. In any case, the value of [itex]\pi[/itex] is completely specified in a finite number of defined symbols.

There are two types of numbers - those with finite width and those without. I call the first set the "Enfranchised" numbers and the second set "Disenfranchised" since they have no clear representation. Note that the Enfranchised numbers contain all of the integers, all of the rationals, all of the algebraics, and many of the transcendentals.

The Disenfranchised therefore contain only transcendentals.

As there are no more than a countably infinite number of Enfranchised numbers, the Disenfranchised must be uncountable.

So if I were to ask you to pick a number at random, the probability that you would pick an Enfranchised number is exactly zero. And yet you have no way to pick anything but an Enfranchised number, so the probability is 1. That'd be my bet.

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# Partitioning of transcendentals

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