"It was first suggested that a googolplex should be 1, followed by writing zeros until you got tired."
http://www-users.cs.york.ac.uk/~susan/cyc/g/googol.htm
http://www-users.cs.york.ac.uk/~susan/cyc/g/googol.htm
million, billion, trillion... then what?
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Fraggle Rocker
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But what is the basis for stating that a set of numbers that is infinitely countable is "smaller" than one that is not? Our system of defining numbers as the ratio of two integers does, after all, apply to irrational numbers. Rational numbers are merely a special case in which the numerator and denominator do not have to approach infinity. This strikes me as being an artifact of the particular kind of mathematics we have developed over the past 2,500 years. Square roots are just as easy and compact as ratios to define in our notational system. So are the solutions to any exponentiation which can be described with integers. And therefore, so are the solutions to any second-order exponentiation that can be described with the solutions to first-order exponentiations.
The set of all numbers of the form A ^ (B/C), where A, B, and C are integers, is just as countable as the set of all numbers of the form B/C. The mapping of a space of three or any arbitrary number of dimensions onto a two- or one-dimensional space is trivial.
By reiteration, the set of all numbers of the form D ^ (E/F), where D, E, and F are of the form set forth above, is just as countable.
It would seem that the solution to any formula that we can express using integers must be as countable as the integers. This includes trigonometric, logarithmic, and hyperbolic formulas.
We could end up mapping a space with an infinite number of dimensions onto a one-dimensional space, which is merely another way of saying that the space has an additional infinite dimension. Do two dimensions of infinity make a set of numbers "larger" than one? Is there a precedent for this?
Our ability to express irrational numbers compactly is only limited by the formulas we have discovered to date. We cannot assume that this set of formulas is not infinite. Will it, therefore, encompass all numbers so that there is only one infinity?
The set of all numbers of the form A ^ (B/C), where A, B, and C are integers, is just as countable as the set of all numbers of the form B/C. The mapping of a space of three or any arbitrary number of dimensions onto a two- or one-dimensional space is trivial.
By reiteration, the set of all numbers of the form D ^ (E/F), where D, E, and F are of the form set forth above, is just as countable.
It would seem that the solution to any formula that we can express using integers must be as countable as the integers. This includes trigonometric, logarithmic, and hyperbolic formulas.
We could end up mapping a space with an infinite number of dimensions onto a one-dimensional space, which is merely another way of saying that the space has an additional infinite dimension. Do two dimensions of infinity make a set of numbers "larger" than one? Is there a precedent for this?
Our ability to express irrational numbers compactly is only limited by the formulas we have discovered to date. We cannot assume that this set of formulas is not infinite. Will it, therefore, encompass all numbers so that there is only one infinity?
Fraggle...
The reason rational numbers are countable is because, roughly speaking, there exists a way to list them in a set that includes all rational numbers. Irration numbers are not countable because there does not exist a way to list them in a set (when one thinks they listed all the irrational numbers, you can always create a new one).
How is A^(B/C) countable if you let B/C = 1/2 and K = {A^(1/2) : A is in [0,1]}? It includes an infinite number of irrational numbers and A is not even countable.
Aleph-0 is any set having the same cardinality as the set of integers, like rational numbers... Let Aleph-0 = X
X<sup>X</sup> = c, which c is the nondenumerable set of real numbers, called the continuum. Also, X + c = c.
The reason rational numbers are countable is because, roughly speaking, there exists a way to list them in a set that includes all rational numbers. Irration numbers are not countable because there does not exist a way to list them in a set (when one thinks they listed all the irrational numbers, you can always create a new one).
How is A^(B/C) countable if you let B/C = 1/2 and K = {A^(1/2) : A is in [0,1]}? It includes an infinite number of irrational numbers and A is not even countable.
Aleph-0 is any set having the same cardinality as the set of integers, like rational numbers... Let Aleph-0 = X
X<sup>X</sup> = c, which c is the nondenumerable set of real numbers, called the continuum. Also, X + c = c.
abyssoft said:
Complete BS. I use these all the time. In fact, none of us would be here if it weren't for myriadaplex.
Fraggle Rocker
Staff member
I understand this argument but it seems unproveable. I have demonstrated that rationality is a sufficient but not necessary condition for countability. Based upon history, each time we expand mathematics we stand a good chance in the process of developing a way to list a new set of irrational numbers that includes one or more of the sets already covered. Therefore we cannot say that the possibility does not exist that we will discover a way to list all numbers.Absane said:
Isn't any set of numbers defined by a function of A, B, and C countable if A, B, and C are countable?
That certainly seems to be the thesis upon which the countability of rational numbers is based: All numbers of the form A/B are countable so long as A and B are integers.
So I count
1 ^ (1/1)/; 1 ^ (1/2); 1 ^ (2/1); 1 ^ (2/2); 2 ^ (1/1); 2 ^ (1/2); 2 ^ (2/1); 2 ^ (2/2);
1 ^ (1/3); 1 ^ (2/3); 2 ^ (1/3); 2 ^ (2/3); 1 ^ (3/2). . . .
or any other way of mapping three variables onto a one-dimensional vector. I get all their values and this particular mapping allows them all to approach infinity at the same decreasing rate.
A ^ (B/C) is indeed countable so long as A, B, and C are integers.
Fraggle Rocker said:
I refer you to a page that might convince you: here .
Yes. But then again, this IS the counting function! If A is a subset (may or may not be proper) of rational numbers, then given F:A->B that is onto, then B is countable.
But your list is only a proper subset of the positive reals. Tell me when A^(B/C) = Pi or e or Euler's Constant.
As I said before, this is true. Because you can list these numbers. Yes, irrationals are included in the list but not all of them, just a proper subset of them.
Edit: sorry.. what I actually said was "How is A^(B/C) countable if you let B/C = 1/2 and K = {A^(1/2) : A is in [0,1]}? It includes an infinite number of irrational numbers and A is not even countable."
I never did actually confirm your statement as true, but it is.
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Fraggle Rocker
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I understand that "to count" by definition is the ordination of integers. Therefore we can only count numbers that can be expressed as formulas that are combinations of cardinal numbers. I do indeed intend to play by those rules.Absane said:
Pi cannot be expressed by my formula or any other that we know of... yet. That does not mean that we won't discover it some day, which is my point.
I'm not familiar with Euler's constant but apparently it can be expressed as an infinite series of simple rational numbers. Unfortunately the people who named e did not foresee Google so it's a little difficult to research, but if memory serves me there is also an infinite series for that one. The ordination of infinite series is not so much a mathematical problem as a lexicographic one. If we can sort all infinite series by their alphanumeric text, that seems like a promising first step toward countability.
I realize that this leaves a lot of irrational numbers unaccounted for and therefore uncountable using today's math. But I still think we're being a little too proud of our accomplishments to date in mathematics when we state that we know that there will never be a way to list all irrational numbers.
Fraggle Rocker
Staff member
Is this an ordered list of all numbers, rational and irrational?
The integer part is trivial. The fractional parts, in binary notation:
.1, .01, .11, .001, .101, .011, .111, .0001, .1001, etc.
It's just counting from one to infinity, reversing the order of the bits, and putting a binary point in front of it. The leading zeros become trailing zeros and by the same convention are not written.
There cannot be any number which is not in that series or the integrity of our numbering system is called into question. Its infinitude is exactly the same as that of integers because they map one-to-one.
All numbers can be expressed as (A, B) where both are positive integers. B is the fractional part written backwards and its place in my list is defined by counting.
The integer part is trivial. The fractional parts, in binary notation:
.1, .01, .11, .001, .101, .011, .111, .0001, .1001, etc.
It's just counting from one to infinity, reversing the order of the bits, and putting a binary point in front of it. The leading zeros become trailing zeros and by the same convention are not written.
There cannot be any number which is not in that series or the integrity of our numbering system is called into question. Its infinitude is exactly the same as that of integers because they map one-to-one.
All numbers can be expressed as (A, B) where both are positive integers. B is the fractional part written backwards and its place in my list is defined by counting.
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Well, there are formulas for any number you want... e<sup>-1</sup> = 1 - 1/1! + 1/2! - 1/3! +..., for example. Euler's constant is lim(x->oo) (1 + 1/2 + 1/3 + 1/4 + ... x) - ln(x).
But essentially here is the deal. When you list all the rational numbers, and assume you have listed all of them, you naturally try to create a new one to add to the list. You cannot, as you will always find that one already in the list.
As for irrational number, you do the same thing. You assume you listed them all. Then you try to create a new number that is not in the list, and you do! Therefore, you assumption that you listed them all is false. How can a school count it's students if it does not have all the students present to be counted?
I think your arguement would (given enough time) come down to arguing the axioms of mathematics (as you seem to be suggestion "new math."). I can take that, as they are just assumed anyway. There is no way to prove them, as was proved by Godel's Incompleteness Theorem's.
But essentially here is the deal. When you list all the rational numbers, and assume you have listed all of them, you naturally try to create a new one to add to the list. You cannot, as you will always find that one already in the list.
As for irrational number, you do the same thing. You assume you listed them all. Then you try to create a new number that is not in the list, and you do! Therefore, you assumption that you listed them all is false. How can a school count it's students if it does not have all the students present to be counted?
I think your arguement would (given enough time) come down to arguing the axioms of mathematics (as you seem to be suggestion "new math."). I can take that, as they are just assumed anyway. There is no way to prove them, as was proved by Godel's Incompleteness Theorem's.
Fraggle Rocker said:
Are you asking if the set {.1, .01, .11, .001, .101, .011, .111, .0001, .1001, ...} Forms the interval (0,1) <font face="MS PMincho" size="4">⊂</font> R?
And also, if that is the case, since we found a way to line up the <b>all</b> the numbers in (0,1), we can expand on that for all R and therefore R is countable. Corollary, irrationals are countable?
I hope I understand this right
Fraggle Rocker
Staff member
I don't understand the notation but I agree with the words.Absane said:Are you asking if the set {.1, .01, .11, .001, .101, .011, .111, .0001, .1001, ...} Forms the interval (0,1) . . . R?
And also, if that is the case, since we found a way to line up the <b>all</b> the numbers in (0,1), we can expand on that for all R and therefore R is countable. Corollary, irrationals are countable?
I hope I understand this right
Just as you can line up all the integers, regardless of the number of digits after the leading zeros, this way you can line up all the fractions, regardless of the number of digits before the trailing zeros. Am I correct that all fractions include all irrational numbers? If so, then it appears that you can list all irrational numbers and the list is exactly the same size (in both length and width, as it turns out) as the list of integers.
Absane, you're thinking of Cantor's diagonal argument? I think it relies on the countably infinite number of digits you can have in a real number in [0, 1). So as far as you count the numbers to try and include a new one, you can count the digits and exclude it again.
Fraggle, you can say that R is bigger than N because it's easy to count N using R but impossible the other way around. And you can prove that finding power sets gives you even bigger sets. Although mathematicians who constantly talk about the size of their sets may be compensating for something...
Fraggle, you can say that R is bigger than N because it's easy to count N using R but impossible the other way around. And you can prove that finding power sets gives you even bigger sets. Although mathematicians who constantly talk about the size of their sets may be compensating for something...
Zephyr, yes that is what I am talking about. I am just trying to explain it 20 different ways. I provided a link in one of my posts that explains the diagonal argument. What it really has turned into now is the fact that we can never "line up" the irrationals so they may be counted.
Fraggle Rocker
Staff member
This list contains all numbers if the integrity of our numbering system is taken for granted. Any number can be expressed in binary notation. (I could haved done it in decimal but I assume that we're all scientists here and binary was easier.) I think the problem you're driving at is that some finite numbers have an infinite number of digits. I can describe to you with complete precision how to navigate down my list to the number 1/3, which is .01010101... in binary and therefore has a sequence number of 10101010... And if you take my instructions you will indeed find it. Except of course for the fact that finding it will take an infinite amount of time.Pete said:
The fact that my way of sorting fractions gives a one-to-one mapping between integers and fractions troubles me. If I can count integers, then why doesn't this automatically imply that I can also count anything they can map to?
I guess it's because an integer with an infinite number of digits has an infinite value, whereas all fractions have finite values. Does it violate the definition of countability to say that you have to count to infinity to count 1/3, much less pi? Or is assigning fractions this kind of countability just another way of saying that the infinity of fractions is of a larger order than the infinity of integers?
Fraggle.. the thing with counting, in the natural sense, is that if I asked you "what was the n<sup>th</sup> number you counted" you could tell me. Make sense? It should also make sense that if I pointed at something that was counted at some point, you should be able to tell me when you counted it. So if I ask when did you count the square root of 2, when did you count it? 1? 10? 1929? Maybe you say infinity.. but infinity is not a subset of the natural numbers.
I am still working on that damned reply.. I am feeling pretty ill right now because I ate almost nothing all day and consumed too many sugary drinks
I am still working on that damned reply.. I am feeling pretty ill right now because I ate almost nothing all day and consumed too many sugary drinks
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