Partitioning of transcendentals

1. Dec 1, 2012

RobertCairone

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 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$^{5}$ 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, $\pi$ = √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 $\pi$ 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.

2. Dec 1, 2012

Number Nine

I'm really not sure what you're trying to do here, or what the significance of "width" would possibly be. Why wouldn't the width of pi be equal to 1, since it's value is completely denoted by the symbol $\pi$? Your definition is completely dependent on arbitrary notational choices.

You're almost touching on something vaguely related to computability, but you mess it up by introducing concepts that aren't really necessary.

The probability of picking one of any countable number of real numbers is zero, yes.

You're negating your own premise here. Either we're picking a real number at random, or we're picking exclusively from the "enfranchised numbers". You can't have both.

3. Dec 1, 2012

Robert1986

I might be very wrong here, but this just seems to be very superficial. Basing mathematical definitions on our way of expressing them seems kind of silly. You give a base 10 example, but if these definitions were going to make any sense, it should work for any base, right?

$\lim_{n \to \infty}(1+1/n)^{1/n}$. What is the width of this? What if instead of that, I write $\lim_{n}(1+1/n)^{1/n}$? The point is that you have tied something that is a very deep property about numbers and have made it about as superficial as possible, and so it doesn't seem like you could be able to do a lot with it.

Every real number is the limit of a sequence of rational numbers. Now, if you can write out the nth term, then you have a finite width number. If not, then I guess you have an infinite width number. But, again, we are trying to make a serious mathematical definition based on whether or not, in our notation, we can make a formula for something.

I *think* this is one of the biggest problems early mathematicians had. They wanted to define a continuous function to be something like "a function which has a one line formula" (ie no piecewise stuff.) (Note, I have not actually researched this myself, I have just read about it in some popular math books.) But this wasn't rigorous enough, and that lead to the epsilon-delta stuff we do now.

4. Dec 1, 2012

RobertCairone

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 unless it is defined in terms of other numbers and/or operators. But given a clear definition for the value of this special number, I've no problem with creating a special symbol for it and assigning it to width 1. Just like we can define a one character symbol for the factorial operator. What doesn't make sense is to say "Here is a number I call '$\vartheta$'" and not be able to describe what its value is. That $\vartheta$ is not a number of width 1.

The point is that there are still only finitely many numbers of width 1, Exactly how many numbers will fall into each width is dependent on notation and doesn't matter much. What matters is that it's a finite set.

Sorry, I was being a bit tongue in cheek. I didn't mean pick any particular number, I meant pick any number of that class. The Enfranchised numbers are all well defined and readily expressible, but there are only countably many of them. Even though they make up every number you know, and every number you can know, among the reals there are almost none of them. I find that ironic and amusing.

If you pick a disenfranchised number, as you probably would if you could, you cannot tell me what it is with any finite combination of numbers and operators. If I ask you what number you picked, you can't say and I can't know what it is. This is a curious property for what are almost all the numbers.

5. Dec 1, 2012

Number Nine

So is 5 (and 3, and 100). So what?

It only a finite set because of our notational conventions.

True, but that's not what you said, and it has nothing to do with the probability of drawing a "disenfranchised number".

6. Dec 1, 2012

Michael Redei

I think you've touched upon a deeper problem here than you may realise.
How can I actually take up your offer to "pick" a number for myself? This whole offer seems like a conjuror offering me a pack of cards and saying, "Pick any card, don't show me what it is, and you mustn't look at it yourself." In what sense would I then have "picked" the card I hold in my hand, rather than any other one you still have in the pack? What difference is there between mine and all of yours?

If there's no way of detecting any difference between my number and all the (equally disenfrachised) numbers I didn't pick, we might as well consider them all to be equal -- just like the set of unicorns is equal to the set of mermaids: they are both the same empty set.

PS: What system of symbols we use to describe numbers isn't important, as long as we decide on one. Thereafter some numbers become "enfranchised" or "disenfranchised" within this system.

7. Dec 1, 2012

RobertCairone

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 very definition of irrational numbers.

8. Dec 1, 2012

Michael Redei

You don't need the rational numbers to define the reals. Try this:

Let R be a field with a total order ≥ such that, for all real numbers x, y and z:
• if x ≥ y then x + z ≥ y + z;
• if x ≥ 0 and y ≥ 0 then xy ≥ 0;
and every non-empty subset S of R that has an upper bound in R also has a least upper bound in R.

Any such set R is isomorphis to the set of reals.

9. Dec 1, 2012

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 infinite number of points without affecting the integral.

I agree the systems of symbols isn't important. But I rather suspect that enfranchised will map to enfranchised and the disenfranchised will remain disenfranchised. Simply changing the base of the number system won't do it, so it would have to be some fundamental changes to the arithmetic operators. I don't know what that might be.

I've been struggling with the idea that the indistinguishability of the disenfranchised means they could all be treated as equals. I don't buy that. Rather, I think they allow us to consider them as continuous.

10. Dec 1, 2012

Hurkyl

Staff Emeritus
If nothing else, when the conjurer draws another card from the deck and asks "Is this your card?", you can confidently answer "no".

An interesting point to note is that if you define the complex numbers as the algebraic closure of the real numbers, the two square roots of -1 are indistinguishable in the same sense you describe.

11. Dec 1, 2012

Michael Redei

I didn't say the set of disenfranchised numbers was empty. What I did say was that each disenfranchised number resembles the empty set in having no property that distinguishes it from other disnfranchised numbers.

Let's just assume we decide on some system for describing real numbers, and some real numbers now are disenfranchised. Can one say anything worthwhile about these numbers, as a whole, even if not about any one specifically? The set of all disnfranchised numbers is uncountable, sure, but that's nothing new -- there are other uncountable subsets of the reals. What makes the disenfranchised numbers interesting after we've defined them? What use are they, what insights do they offer?

12. Dec 1, 2012

Norwegian

Hi Robert,

Your notion of "width" is of course not well defined.

Your idea of distinguishing between numbers with finite and infinite "width" is not a new one. For example, the notion of a definable real number, does perhaps capture your idea, and more.
It is an elementary fact that every real number is a limit of a sequence of rationals.

13. Dec 1, 2012

Michael Redei

If I can pick a card, it automatically becomes "enfranchised" since it now has gained a defining property: it's the one in my hand. And if I can't pick even a first card, the conjurer can't pick a second one.

14. Dec 1, 2012

Robert1986

It is absolutely true. Every irrational number can be approximated by a rational number to any degree of accuracy. Surely you can see how this means every irrational number is a limit of a sequence of rationals. For example, if r is a number, for each N pick some rational number that is in the interval (r-(1/N),(r+(1/N)) but is not in (r-(1/(N-1),r-1/N) or (r+1/N,r+1/(N-1))

You completely missed my point. My example was to show you that non-rigorous definitions don't work in math.