Deary, deary me...
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crank wrote: I had difficulties with some bad arguments I tried to counter
crank wrote:with some wrong assumptions on my part, and other issues.
crank wrote: Chaitin's claim, if I understand it, is that reals are infinite precision,
crank wrote:but that isn't a mathematical term, so whatever, since uncomputable numbers will have an infinite, non-repeating decimal representation, they in effect contain infinite information,
scott1328 wrote:Here is a diagram of how the mapping works Nicko.
scott1328 wrote:crank wrote: I had difficulties with some bad arguments I tried to counter
What bad arguments?crank wrote:with some wrong assumptions on my part, and other issues.
this is truecrank wrote: Chaitin's claim, if I understand it, is that reals are infinite precision,
That wasn't his claim.crank wrote:but that isn't a mathematical term, so whatever, since uncomputable numbers will have an infinite, non-repeating decimal representation, they in effect contain infinite information,
uncomputable numbers have no decimal representation whatsoever, if they did, they would be computable. Uncomputables contain no information whatsoever.
But when you work on a computer, the last thing on earth you're ever going to see is a real number, because a real number has an infinite number of digits of precision,
crank wrote:scott1328 wrote:crank wrote: I had difficulties with some bad arguments I tried to counter
What bad arguments?crank wrote:with some wrong assumptions on my part, and other issues.
this is truecrank wrote: Chaitin's claim, if I understand it, is that reals are infinite precision,
That wasn't his claim.crank wrote:but that isn't a mathematical term, so whatever, since uncomputable numbers will have an infinite, non-repeating decimal representation, they in effect contain infinite information,
uncomputable numbers have no decimal representation whatsoever, if they did, they would be computable. Uncomputables contain no information whatsoever.
You got one right. Uncomputables have decimal representation, you can't right one out, but you can begin to generate one, in principal, the easiest would be a random number generator [not the computer algorithmic ones, true random number generators].
Chaitin:But when you work on a computer, the last thing on earth you're ever going to see is a real number, because a real number has an infinite number of digits of precision,
scott1328 wrote:
Chaitin was clearly speaking in the context of computing real numbers. He was not making the claim you have put into his mouth.
But to say you can begin writing an uncomputable number out is false because all you have done with your random number generator is generate a number that equally identifies countably infinite rational number, countably infinite computable numbers, and uncountably infinite non-computable numbers.
In what sense of the word have you identified any number whatsoever with your random number generator?
So in fact there are more uncomputable reals than computable reals. From Cantor's theory of infinite sets, we see that the set of uncomputable reals is just as big as the set of all reals, while the set of computable reals is only as big as the set of whole numbers. The set of uncomputable reals is much bigger than the set of computable reals.
#{uncomputable reals} = #{all reals} = ℵ1
#{computable reals} = #{computer programs} = #{whole numbers} = ℵ0
ℵ1 > ℵ0
scott1328 wrote:I have listened to the video and read the paper you quoted. And in a video and paper aimed at popularizing deeply technical ideas, I would expect such language and metaphors to make his ideas accessible. I do not believe he is claiming what you think he is.
And if he is claiming that reals have infinite precision, then he is playing fast and loose with the mathematics.
Chaitin does make a particularly egregious error in the paper you citedSo in fact there are more uncomputable reals than computable reals. From Cantor's theory of infinite sets, we see that the set of uncomputable reals is just as big as the set of all reals, while the set of computable reals is only as big as the set of whole numbers. The set of uncomputable reals is much bigger than the set of computable reals.
#{uncomputable reals} = #{all reals} = ℵ1
#{computable reals} = #{computer programs} = #{whole numbers} = ℵ0
ℵ1 > ℵ0
In the text I've quoted Chaitin asserted that the cardinality of the reals, typically referred to as C, is equivalent to Aleph-1. This assertion, called the Continuum Hypothesis is formally undecidable in standard formulations of mathematical set theory, that he would assert it as true in his popular paper, leads me to doubt the seriousness of this entire paper.
\aleph_1 is the cardinality of the set of all countable ordinal numbers, called ω1 or (sometimes) Ω. This ω1 is itself an ordinal number larger than all countable ones, so it is an uncountable set. Therefore \aleph_1 is distinct from \aleph_0. The definition of \aleph_1 implies (in ZF, Zermelo–Fraenkel set theory without the axiom of choice) that no cardinal number is between \aleph_0 and \aleph_1. If the axiom of choice (AC) is used, it can be further proved that the class of cardinal numbers is totally ordered, and thus \aleph_1 is the second-smallest infinite cardinal number.
Evolving wrote:I wouldn't say it was "trivial". It took Cantor to think of it!
Obviously, once you've seen it, it's natural to think, "I could have thought of that!"
In fact: see my sig.
Newmark wrote:But it's factually wrong. There is a one-to-one correspondence between the integers and the rationals, as they are both countable sets of the same cardinality (aleph-0)*. Infinities may be a bit counter-intuitive, as a set may contain as many elements as a proper subset of said set, but anyone who can't be bothered to look this up has no business writing a math book.
For anyone who's interested, wiki on countable sets is a good place to start...
[EDIT] *Or rather, it's the other way around, but the cardinality of the sets should given you a hint even if you don't understand the proof.
LjSpike wrote:Newmark wrote:But it's factually wrong. There is a one-to-one correspondence between the integers and the rationals, as they are both countable sets of the same cardinality (aleph-0)*. Infinities may be a bit counter-intuitive, as a set may contain as many elements as a proper subset of said set, but anyone who can't be bothered to look this up has no business writing a math book.
For anyone who's interested, wiki on countable sets is a good place to start...
[EDIT] *Or rather, it's the other way around, but the cardinality of the sets should given you a hint even if you don't understand the proof.
I'm way out of my league, and technically we do have infinite numbers of each one, but that is a bit of a cheats way out of it isn't it, just to write down that numbers go on infinitely...
See this is a secondary schoolers approach to your problem:
There is an infinite number of integers, as fractions normally have the denominator and numerator represented by an integer
So we could take the area between 0 and 1.
We have 2 integers here, 0 and 1. However, we have an infinite number of real rational numbers between these integers, as we have 1 over every integer that exists (so 1 over x infinite times over) but we also have 1 being able to be replaced by anything up to x and above 0 and it still being between 0 and 1 or 0 and 1. So we have an infinite number of integers, but we have an infinity*infinity number of rationals between each increment in integers.
So the relationship is 1:infinity*infinity
This is working of simpler sequences solutions / proofs logic. (I've done so many of these sorts of questions and variants on a GCSE level its slightly dull at times now... "When will both trains be back at the station?")
Any sequence is infinite, unless it has a specified reason not to be, so to prove it on an easier level look at a single setup in the first sequence (integers) and look at the equivalent numbers of steps up in the second sequence, and you give yourself the ratio (or relationship) in a 1:x format. Its a fairly universal solution I've found, as both trains will be in the station when the multiple of x is an integer.
To be fair, LJSpike admitted it was a GCSE perspective (one you typically obtain by studying maths between 14-16 years old in the UK). And maths is taught pretty shit here IMO.scott1328 wrote:This is merely incoherent wibble.
liftA2 (%) nats nats
LjSpike wrote:
Yup. Anyway I could shorten my description now, which I shall do. After using a lot of arrows on paper to try and find the most efficient description.
Integers go on infinitely, we could just say that rationals go on infinitely too, but then again, were not giving a 'concrete figure' to either of them. So if we take an increment of 1 in the integers chain, we can then express all of the rationals in this chunk as
Y/X (Y over X). X is any number from 1 to infinity. Y is any number from X * the lower end of our 'chunk' of the integer chain, to infinity. So we can say X and Y both have a range of infinite size.
So to find every combination of them we find all of the possible values of X, which we know to be infinity. We then find all the possible values of Y for all the possible values of X, as we know Y can have a range of infinite size, but can also have a range of say 1, we can do X * Y, which is Infinity * (Infinity/2) as halfway between 1, and infinity is a half of infinity, which is still infinity (as infinity is infinitely large) so we could say that its infinity * infinity or infinity^2
https://en.wikipedia.org/wiki/Countable_set
https://en.wikipedia.org/wiki/Countable_set
So for a range of 1 in integers, we have a range of infinity^2 (infinity squared) of rationals.
If we were to choose a range of 2 in integers, we can simply extrapolate the pattern found, to be 2*infinity^2 in rationals. So for a range of infinite integers, we multiply infinity^2 by infinity, giving us infinity*infinity*infinity or infinity^3
So for the total amount of integers to rationals we can say the ratio is infinity:infinity^3
Simplifying this ratio down would divide it by infinity, as that is a common factor of both of them. So we then get 1 : infinity^2 (as anything divided by itself = 1, and dividing a^x by a results in a^(x-1))
I would personally be happy to leave it at that, but as we (theoretically) can't have a number bigger than infinity, we can handle this now at the end (just like rounding in a trig question, or such, you leave it till the end to gain the best accuracy).
So we can say, infinity^2, is still infinity as nothing can be larger than infinity, giving us the answer of 1:infinity.
scott1328 wrote:LjSpike wrote:
Yup. Anyway I could shorten my description now, which I shall do. After using a lot of arrows on paper to try and find the most efficient description.
Integers go on infinitely, we could just say that rationals go on infinitely too, but then again, were not giving a 'concrete figure' to either of them. So if we take an increment of 1 in the integers chain, we can then express all of the rationals in this chunk as
Y/X (Y over X). X is any number from 1 to infinity. Y is any number from X * the lower end of our 'chunk' of the integer chain, to infinity. So we can say X and Y both have a range of infinite size.
So to find every combination of them we find all of the possible values of X, which we know to be infinity. We then find all the possible values of Y for all the possible values of X, as we know Y can have a range of infinite size, but can also have a range of say 1, we can do X * Y, which is Infinity * (Infinity/2) as halfway between 1, and infinity is a half of infinity, which is still infinity (as infinity is infinitely large) so we could say that its infinity * infinity or infinity^2
https://en.wikipedia.org/wiki/Countable_set
https://en.wikipedia.org/wiki/Countable_set
So for a range of 1 in integers, we have a range of infinity^2 (infinity squared) of rationals.
If we were to choose a range of 2 in integers, we can simply extrapolate the pattern found, to be 2*infinity^2 in rationals. So for a range of infinite integers, we multiply infinity^2 by infinity, giving us infinity*infinity*infinity or infinity^3
So for the total amount of integers to rationals we can say the ratio is infinity:infinity^3
Simplifying this ratio down would divide it by infinity, as that is a common factor of both of them. So we then get 1 : infinity^2 (as anything divided by itself = 1, and dividing a^x by a results in a^(x-1))
I would personally be happy to leave it at that, but as we (theoretically) can't have a number bigger than infinity, we can handle this now at the end (just like rounding in a trig question, or such, you leave it till the end to gain the best accuracy).
So we can say, infinity^2, is still infinity as nothing can be larger than infinity, giving us the answer of 1:infinity.
Read this Wiki article, it explains how the rationals can be put into a one-to-one correspondence with the natural numbers.
http://en.wikipedia.org/wiki/Countable_set
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