6 ms·

Actually it's possible to represent 1/3 perfectly accurately, but what about all the numbers that it's theoretically impossible to compute? (Almost all real num

Actually it's possible to represent 1/3 perfectly accurately, but what about all the numbers that it's theoretically impossible to compute? (Almost all real numbers have this property)

mcny 9y ago
> Almost all real numbers have this property > Almost all I love this comment because it brings back memories of school. In a layman's terms, I think almost all in this case means all but a finite amount can we say almost all real numbers are irrational? 1 1/1 2/1 3/1 4/1 5/1 ... 2 1/2 2/2 3/2 4/2 5/2 ... 3 ... so clearly we can count all the rational numbers but how many irrational numbers are there? are there (many) more irrational numbers than there are rational numbers?
- saltcured 9y ago
  Well, if I remember correctly, a product of an irrational and a rational is still irrational, right? If so, here's a simple argument: If you have N rational numbers and I have K irrationals to start with, I can produce roughly N*K additional irrationals. Since we know at least two irrationals (pi and e) we can make at least twice as many irrationals as rationals.
  - jjgreen 9y ago
    Unfortunately this argument fails -- there are twice as many whole numbers as there are odd numbers right? but they are in 1:1 correspondence 0 <-> 1 1 <-> 3 2 <-> 5 : so the odd and whole numbers have the same cardinality: there are the same number of each.
- Ma8ee 9y ago
  The rational numbers are clearly countable. The irrational numbers are uncountable, which means that for every rational number there are infinitely many irrational numbers.
  - btilly 9y ago
    But for every integer there are also an infinite number of rationals. And since we can put the rationals into 1-1 correspondence with integers, that means that for every rational there are an infinite number of rationals. Infinity is funny like that. The conclusion that there are somehow more irrationals than rationals depends on subtle philosophical points that have no possible proof or disproof and usually get glossed over. Accepting that philosophy also leads to the conclusion that not only do numbers which can in no way ever be represented exist, but there are more of them than numbers which we can explicitly name. Now I ask you, in what sense do they REALLY exist?
    - umanwizard 9y ago
      > The conclusion that there are somehow more irrationals than rationals depends on subtle philosophical points that have no possible proof or disproof and usually get glossed over Can you elaborate? The proof that there are more irrationals than rationals is very straightforward, from my perspective.
      - btilly 9y ago
        The easiest way to understand it is to look at the question from a philosophical framework where it makes no sense to claim that there are "more" irrationals than rationals. And then untangle why it came to a different answer. In Constructivism, all statements have 3 possible values, not 2. They are true, false, and not proven. All possible objects must have a construction. So instead of talking about a vague "Cauchy sequence", we might instead have a computer program that given n will return a rational within 1/n of the answer. The first thing to notice is that all possible things that could exist is contained within a countable set of all possible constructions. There can't be "more" irrationals than rationals. But what about diagonalization? That proof still works. You still can't enumerate the reals. But why not? The answer is because determining whether a given program represents a real is a decision problem that cannot in general be solved by any algorithm. You are running into the same category of challenges that lie behind Gödel's theorem and The Halting Problem. It is not that there are "more" irrationals than rationals. It is that there are specific programs which you can't decide whether they represent reals. From a constructivist's eyes, the traditional proof that there are more irrationals falls apart because you're reasoning about unprovable statements about arbitrary sequences whose logic could have been constructed with the same sort of logic that you're applying to them. Is it any wonder that you wind up concluding the "existence" of things that clearly don't actually exist? Now stepping back from BOTH philosophies, the differences between them lie in different attitudes about existence and truth. Attitudes that underly the axioms which we use, and cannot possibly be proven one way or another. (Gödel actually proved that. Any contradiction in Constructivism is immediately a contradiction in classical mathematics. But conversely there is a purely mechanical transformation of any proof in classical mathematics which resulted in a contradiction, into a constructive proof that also results in a contradiction.)
- jjgreen 9y ago
  There are, see Cantor's diagonal argument.
- monktastic1 9y ago
  The rationals have measure zero, so almost all reals are irrational. The Cantor set is an example of an uncountable set which also has measure zero. http://austinrochford.com/posts/2013-12-31-almost-no-rationals.html http://austinrochford.com/posts/2013-12-31-almost-no-rationa...
- roywiggins 9y ago
  You can count computable irrationals by numbering the algorithms that calculate (successive approximations to) them and lining them up in order. This is what Turing used his "Turing Machines" to do. So all of the irrationals that we use or could ever use in calculations are countable. The uncountability of the irrationals comes entirely from the uncomputable ones, which we will probably never see. The reals are deeply weird.
- ska 9y ago
  It's not a finite amount, but an amount with measure zero.
- umanwizard 9y ago
  Yes, almost all real numbers are irrational, but I was talking about non-computable numbers, not irrational numbers.
yarg 9y ago
If they can't be computed, what are you planning on using them for? The biggest problem is going to be with the useful transcendentals.
craftyguy 9y ago
>Actually it's possible to represent 1/3 perfectly accurately I'm curious, how?
- ars 9y ago
  > I'm curious, how? As 1/3 - exactly as it's on your screen. All rational numbers can be represented exactly.
  - craftyguy 9y ago
    Hmm, yes, it can be represented in ASCII. But you still have to approximate when storing it in a way that is actually useful for computation using a finite number of bits.
    - umanwizard 9y ago
      struct rational { int numerator; int denominator; };
    - dragonwriter 9y ago
      Many real programming languages support arbitrary precision decimal and/or rational numbers. Sure, there are times when space or speed concerns favor inexact representation over correctness, but that's an optimization that ought to be properly evaluated; the fact that lots of languages are designed in a way which makes it the default everyone reaches for contributes to lots of errors.
    - LASR 9y ago
      No. of course not. A lot of scientific calculators do this. They have a fractional arithmetic mode that keeps numbers you input as fractions as fractions in the output. Scientific calculators even have irrational modes where they represent irrational numbers such as sqrt(2) as sqrt(2) and can do 100% precise operations on it. Like sqrt(2) * sqrt(2) will give you a result of 2 https://i.imgur.com/l83fwr2.png https://i.imgur.com/l83fwr2.png
- mamcx 9y ago
deleted 9y ago
[deleted]
dllthomas 9y ago
At the risk of explaining the joke, obviously if I can't compute a number I won't need to represent it...

Actually it's possible to represent 1/3 perfectly accurately, but what about all the numbers that it's theoretically impossible to compute? (Almost all real numbers have this property)

mcny 9y ago

> Almost all real numbers have this property > Almost all I love this comment because it brings back memories of school. In a layman's terms, I think almost all in this case means all but a finite amount can we say almost all real numbers are irrational? 1 1/1 2/1 3/1 4/1 5/1 ... 2 1/2 2/2 3/2 4/2 5/2 ... 3 ... so clearly we can count all the rational numbers but how many irrational numbers are there? are there (many) more irrational numbers than there are rational numbers?

saltcured 9y ago
Well, if I remember correctly, a product of an irrational and a rational is still irrational, right? If so, here's a simple argument: If you have N rational numbers and I have K irrationals to start with, I can produce roughly N*K additional irrationals. Since we know at least two irrationals (pi and e) we can make at least twice as many irrationals as rationals.
- jjgreen 9y ago
  Unfortunately this argument fails -- there are twice as many whole numbers as there are odd numbers right? but they are in 1:1 correspondence 0 <-> 1 1 <-> 3 2 <-> 5 : so the odd and whole numbers have the same cardinality: there are the same number of each.
Ma8ee 9y ago
The rational numbers are clearly countable. The irrational numbers are uncountable, which means that for every rational number there are infinitely many irrational numbers.
- btilly 9y ago
  But for every integer there are also an infinite number of rationals. And since we can put the rationals into 1-1 correspondence with integers, that means that for every rational there are an infinite number of rationals. Infinity is funny like that. The conclusion that there are somehow more irrationals than rationals depends on subtle philosophical points that have no possible proof or disproof and usually get glossed over. Accepting that philosophy also leads to the conclusion that not only do numbers which can in no way ever be represented exist, but there are more of them than numbers which we can explicitly name. Now I ask you, in what sense do they REALLY exist?
  - umanwizard 9y ago
    > The conclusion that there are somehow more irrationals than rationals depends on subtle philosophical points that have no possible proof or disproof and usually get glossed over Can you elaborate? The proof that there are more irrationals than rationals is very straightforward, from my perspective.
    - btilly 9y ago
      The easiest way to understand it is to look at the question from a philosophical framework where it makes no sense to claim that there are "more" irrationals than rationals. And then untangle why it came to a different answer. In Constructivism, all statements have 3 possible values, not 2. They are true, false, and not proven. All possible objects must have a construction. So instead of talking about a vague "Cauchy sequence", we might instead have a computer program that given n will return a rational within 1/n of the answer. The first thing to notice is that all possible things that could exist is contained within a countable set of all possible constructions. There can't be "more" irrationals than rationals. But what about diagonalization? That proof still works. You still can't enumerate the reals. But why not? The answer is because determining whether a given program represents a real is a decision problem that cannot in general be solved by any algorithm. You are running into the same category of challenges that lie behind Gödel's theorem and The Halting Problem. It is not that there are "more" irrationals than rationals. It is that there are specific programs which you can't decide whether they represent reals. From a constructivist's eyes, the traditional proof that there are more irrationals falls apart because you're reasoning about unprovable statements about arbitrary sequences whose logic could have been constructed with the same sort of logic that you're applying to them. Is it any wonder that you wind up concluding the "existence" of things that clearly don't actually exist? Now stepping back from BOTH philosophies, the differences between them lie in different attitudes about existence and truth. Attitudes that underly the axioms which we use, and cannot possibly be proven one way or another. (Gödel actually proved that. Any contradiction in Constructivism is immediately a contradiction in classical mathematics. But conversely there is a purely mechanical transformation of any proof in classical mathematics which resulted in a contradiction, into a constructive proof that also results in a contradiction.)
jjgreen 9y ago
There are, see Cantor's diagonal argument.
monktastic1 9y ago
The rationals have measure zero, so almost all reals are irrational. The Cantor set is an example of an uncountable set which also has measure zero. http://austinrochford.com/posts/2013-12-31-almost-no-rationals.html http://austinrochford.com/posts/2013-12-31-almost-no-rationa...
roywiggins 9y ago
You can count computable irrationals by numbering the algorithms that calculate (successive approximations to) them and lining them up in order. This is what Turing used his "Turing Machines" to do. So all of the irrationals that we use or could ever use in calculations are countable. The uncountability of the irrationals comes entirely from the uncomputable ones, which we will probably never see. The reals are deeply weird.
ska 9y ago
It's not a finite amount, but an amount with measure zero.
umanwizard 9y ago
Yes, almost all real numbers are irrational, but I was talking about non-computable numbers, not irrational numbers.

yarg 9y ago

If they can't be computed, what are you planning on using them for? The biggest problem is going to be with the useful transcendentals.

craftyguy 9y ago

>Actually it's possible to represent 1/3 perfectly accurately I'm curious, how?

ars 9y ago
> I'm curious, how? As 1/3 - exactly as it's on your screen. All rational numbers can be represented exactly.
- craftyguy 9y ago
  Hmm, yes, it can be represented in ASCII. But you still have to approximate when storing it in a way that is actually useful for computation using a finite number of bits.
  - umanwizard 9y ago
    struct rational { int numerator; int denominator; };
  - dragonwriter 9y ago
    Many real programming languages support arbitrary precision decimal and/or rational numbers. Sure, there are times when space or speed concerns favor inexact representation over correctness, but that's an optimization that ought to be properly evaluated; the fact that lots of languages are designed in a way which makes it the default everyone reaches for contributes to lots of errors.
  - LASR 9y ago
    No. of course not. A lot of scientific calculators do this. They have a fractional arithmetic mode that keeps numbers you input as fractions as fractions in the output. Scientific calculators even have irrational modes where they represent irrational numbers such as sqrt(2) as sqrt(2) and can do 100% precise operations on it. Like sqrt(2) * sqrt(2) will give you a result of 2 https://i.imgur.com/l83fwr2.png https://i.imgur.com/l83fwr2.png
mamcx 9y ago

deleted 9y ago

[deleted]

dllthomas 9y ago

At the risk of explaining the joke, obviously if I can't compute a number I won't need to represent it...