## Question from a Texan Textbook

Deary, deary me...

Discuss the language of the universe.

### Re: Question from a Texan Textbook

crank wrote:
scott1328 wrote:
I am arguing with your benighted terminology. you keep saying integers are not real numbers, but they are, you keep saying other very odd things. Like 2.0000... is not the same as 2, that discrete objects cannot be counted with real numbers, they can. For some reason you seem to think that the "non-existance" of reals implies a discrete universe, it doesn't.

Or at least that is what you appear to be saying to me.

If you're doing some kind of mathematics that is restricted to integers, then to me, the 2 used there is not real, it's integer. Yes, integers are a subset of reals, but when you're doing something where reals don't apply, then you can't use 2.000..., you use 2. There's some fundamental difference when using an integer verses using a real, even if they're the same number. It's the difference between continuous and discrete, how an integer can only change by whole multiples of 1, and reals can change by infinitesimals. I apologize that I don't know the technical language, but I can't believe these distinctions aren't made in some way.

I think, in my far from qualified way, that infinite precision can't be real, from the informational aspects, which you don't agree exist, but there's a lot of mathematicians and physicists who do. Also, because of how there's a lot of physics that imply a quantized universe, I don't even know if that could technically be equivalent to 'discrete universe', though it has some real basic similarities if it isn't equivalent.

I think the problem here is that you are thinking of irrational numbers in terms of their expression/expansion rather than in terms of their position on a continuous line. All real numbers partition a line. Take for example sqrt(2) this quantity partitions the line into two subsets: set A which contains all the negative numbers and all the numbers whose square is less than 2, and set B which contains all those numbers whose squares are greater than or equal to two. If you think about it this way, you realize that it is not about calculating the decimal positions, but rather of deciding to which partition a quantity belongs.

In the same way you can see that the integer 2 partitions a line in exactly the same way the irrational sqrt(2) partitions a line. In set A are all those quantities that are less than two, in set B are all those quantities that are greater than or equal to 2.

This method called a Dedekind cut is in fact used in the definition and construction real numbers as an extension of the the rational numbers.

Given a set Q of all rational numbers, b is a real number if there is a partitioning of set Q into set A that contains all elements of Q that are less than b, and set B that contains all the elements of Q that are greater than or equal to b. If b is an element of set B then b is itself rational. If b is not an element of B then b is irrational.

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### Re: Question from a Texan Textbook

scott1328 wrote:
crank wrote:
scott1328 wrote:
I am arguing with your benighted terminology. you keep saying integers are not real numbers, but they are, you keep saying other very odd things. Like 2.0000... is not the same as 2, that discrete objects cannot be counted with real numbers, they can. For some reason you seem to think that the "non-existance" of reals implies a discrete universe, it doesn't.

Or at least that is what you appear to be saying to me.

If you're doing some kind of mathematics that is restricted to integers, then to me, the 2 used there is not real, it's integer. Yes, integers are a subset of reals, but when you're doing something where reals don't apply, then you can't use 2.000..., you use 2. There's some fundamental difference when using an integer verses using a real, even if they're the same number. It's the difference between continuous and discrete, how an integer can only change by whole multiples of 1, and reals can change by infinitesimals. I apologize that I don't know the technical language, but I can't believe these distinctions aren't made in some way.

I think, in my far from qualified way, that infinite precision can't be real, from the informational aspects, which you don't agree exist, but there's a lot of mathematicians and physicists who do. Also, because of how there's a lot of physics that imply a quantized universe, I don't even know if that could technically be equivalent to 'discrete universe', though it has some real basic similarities if it isn't equivalent.

I think the problem here is that you are thinking of irrational numbers in terms of their expression/expansion rather than in terms of their position on a continuous line. All real numbers partition a line. Take for example sqrt(2) this quantity partitions the line into two subsets: set A which contains all the negative numbers and all the numbers whose square is less than 2, and set B which contains all those numbers whose squares are greater than or equal to two. If you think about it this way, you realize that it is not about calculating the decimal positions, but rather of deciding to which partition a quantity belongs.

In the same way you can see that the integer 2 partitions a line in exactly the same way the irrational sqrt(2) partitions a line. In set A are all those quantities that are less than two, in set B are all those quantities that are greater than or equal to 2.

This method called a Dedekind cut is in fact used in the definition and construction real numbers as an extension of the the rational numbers.

Given a set Q of all rational numbers, b is a real number if there is a partitioning of set Q into set A that contains all elements of Q that are less than b, and set B that contains all the elements of Q that are greater than or equal to b. If b is an element of set B then b is itself rational. If b is not an element of B then b is irrational.

Let integer 2 partition the line, now let real 2.0... partition, but the '...' is not all zeros, but an omission of unknown length. Can you tell me if that is the same partition without using the fact that integer two is 2.00... with all zeros to infinity, or knowing the real out to infinity?

Or, another way, if all reals are not infinite precision, what makes a number an integer? How can you test for integerness if you don't know the the number to infinite precision? Like you said .9999... is an integer because it's equal to 1. But it isn't if you don't have all the '9's out to infinity. If you don't know all the digits past the decimal, out to infinity, you don't know the number.
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### Re: Question from a Texan Textbook

crank wrote:
scott1328 wrote:
crank wrote:
scott1328 wrote:
I am arguing with your benighted terminology. you keep saying integers are not real numbers, but they are, you keep saying other very odd things. Like 2.0000... is not the same as 2, that discrete objects cannot be counted with real numbers, they can. For some reason you seem to think that the "non-existance" of reals implies a discrete universe, it doesn't.

Or at least that is what you appear to be saying to me.

If you're doing some kind of mathematics that is restricted to integers, then to me, the 2 used there is not real, it's integer. Yes, integers are a subset of reals, but when you're doing something where reals don't apply, then you can't use 2.000..., you use 2. There's some fundamental difference when using an integer verses using a real, even if they're the same number. It's the difference between continuous and discrete, how an integer can only change by whole multiples of 1, and reals can change by infinitesimals. I apologize that I don't know the technical language, but I can't believe these distinctions aren't made in some way.

I think, in my far from qualified way, that infinite precision can't be real, from the informational aspects, which you don't agree exist, but there's a lot of mathematicians and physicists who do. Also, because of how there's a lot of physics that imply a quantized universe, I don't even know if that could technically be equivalent to 'discrete universe', though it has some real basic similarities if it isn't equivalent.

I think the problem here is that you are thinking of irrational numbers in terms of their expression/expansion rather than in terms of their position on a continuous line. All real numbers partition a line. Take for example sqrt(2) this quantity partitions the line into two subsets: set A which contains all the negative numbers and all the numbers whose square is less than 2, and set B which contains all those numbers whose squares are greater than or equal to two. If you think about it this way, you realize that it is not about calculating the decimal positions, but rather of deciding to which partition a quantity belongs.

In the same way you can see that the integer 2 partitions a line in exactly the same way the irrational sqrt(2) partitions a line. In set A are all those quantities that are less than two, in set B are all those quantities that are greater than or equal to 2.

This method called a Dedekind cut is in fact used in the definition and construction real numbers as an extension of the the rational numbers.

Given a set Q of all rational numbers, b is a real number if there is a partitioning of set Q into set A that contains all elements of Q that are less than b, and set B that contains all the elements of Q that are greater than or equal to b. If b is an element of set B then b is itself rational. If b is not an element of B then b is irrational.

Let integer 2 partition the line, now let real 2.0... partition, but the '...' is not all zeros, but an omission of unknown length. Can you tell me if that is the same partition without using the fact that integer two is 2.00... with all zeros to infinity, or knowing the real out to infinity?

Or, another way, if all reals are not infinite precision, what makes a number an integer? How can you test for integerness if you don't know the the number to infinite precision? Like you said .9999... is an integer because it's equal to 1. But it isn't if you don't have all the '9's out to infinity. If you don't know all the digits past the decimal, out to infinity, you don't know the number.

Your comment reads pretty much like word salad. And once again you are thinking again in decimal expansion. stop this.

One need not know all the digits out to infinity to know a number this is plainly FALSE.

The golden ratio Phi is defined to be the positive solution to the equation p2=p+1
The base of the natural logarithm is defined to be that unique number such that the function ex is its own derivative.
The value of pi is defined to be the sum ratio of the circumference of a circle to its diameter.
There are many others.

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### Re: Question from a Texan Textbook

The decimal expansion is the only way I know how to think, and it is perfectly adequate.
The rest of it, you are looking at the problem wrong. You say we know these numbers without knowing their decimal expansion. OK, now tell me, how do you know if a number you think is one of these is one? E.g., how do you prove some x = pi? Or, much easier, how do you prove some x = 2 the integer?
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### Re: Question from a Texan Textbook

x = y if and only if x-y = 0.
x = y if and if x/y = 1 and y<>0
x = y if and only if x ~< y and x~> y

And any number of other methods of proof. Depends on context.

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### Re: Question from a Texan Textbook

If you won't believe me, will you believe Cali? From his earlier post:

Actually, I can write down an expression that constitutes the definition of a real number. Viz:

where for all possible values of i, ai is an element of the set {0,1,2,3,4,5,6,7,8,9}.

If ai=0 for all i>N, where N is some suitable finite integer, then you have a rational number. But not all rational numbers fall into this category. 1/7 being a classic example. The decimal expansion of 1/7 continues indefinitely:

(1/7) = 0.124857124857124857124857124857124857124857124857124857124857...

But in this and other cases, the coefficients ai are cyclically periodic for all i>N, where N is some suitable finite integer.

The red text shows that the summation goes out to infinity, but for, e.g., an integer, they will be zero. This is infinite precision.
Last edited by crank on Sep 11, 2015 5:14 am, edited 1 time in total.
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### Re: Question from a Texan Textbook

scott1328 wrote:x = y if and only if x-y = 0.
x = y if and if x/y = 1 and y<>0
x = y if and only if x ~< y and x~> y

And any number of other methods of proof. Depends on context.

Surely, you see what's going on? 'x' is an arbitrary number, it is a string of decimals digits, infinitely long, you see zeros, but is it all zeros?

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### Re: Question from a Texan Textbook

scott1328 wrote:x = y if and only if x-y = 0.
x = y if and if x/y = 1 and y<>0
x = y if and only if x ~< y and x~> y

And any number of other methods of proof. Depends on context.

This doesn't say anything. I had to edit my previous post, I forgot the example.

how can you tell if .9999...=1? The ellipses isn't all '9's, it's some list of digits, it might be all '9's, it might not. How can you prove that this number, .9999... = 1??
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### Re: Question from a Texan Textbook

You are not asking questions that are meaningful. I have given you what any high school would tell you about real numbers. it is not my fault if you reject the definitions that actual mathematicians use and instead create useless strawman caricatures and then find out that those caricatures don't make any sense to you.

Why don't you write up a nice mathematical paper detailing exactly how you think 400 year of mathematicians got it all wrong and how you alone have discerned TRUTH. Let's see that Nobel prize in math.

By the way Cali's definition is only one of several different formulations. I happen to think it is a clumsy and useless formulation. Nevertheless each of these formulations can be used to derive the others.

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### Re: Question from a Texan Textbook

if the unexpressed digits .99999..... Are not all nines then the value is less than 1 and not equal to 1. It's not hard.

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### Re: Question from a Texan Textbook

scott1328 wrote:if the unexpressed digits .99999..... Are not all nines then the value is less than 1 and not equal to 1. It's not hard.

My god, you really don't understand this most basic pont. You have to know all the infinite digits to the right of the decimal to know it's all '9', in other words, it's infinite precision. To say 'if it's not all nines, it not equal to one. It's not hard' and not understand you have to know all infinity of them to determine this.

or see wiki:

A real number may be either rational or irrational; either algebraic or transcendental; and either positive, negative, or zero. Real numbers are used to measure continuous quantities. They may be expressed by decimal representations that have an infinite sequence of digits to the right of the decimal point; these are often represented in the same form as 324.823122147… The ellipsis (three dots) indicates that there would still be more digits to come.

This is so obvious, I can only think you have some block where you don't understand the question.
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### Re: Question from a Texan Textbook

16 year olds understand this. You erect barriers to your own understanding. Wallow in your own ignorance.

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### Re: Question from a Texan Textbook

Wallow in your own blindness. To not grasp that a real is an infinite precision number while posting "if the unexpressed digits .99999..... Are not all nines then the value is less than 1 and not equal to 1. It's not hard." ? That is seriously blind. And then accuse me of erecting barriers to understanding? You don't believe wiki, you don't believe Cali. And you accuse me of wallowing in ignorance? Whoever got stupid enough to up-vote you, too bad they're too much of a coward to add their take on this. I wish someone with brains could step in and help settle this. I don't know how to help someone who won't see what is plainly in front of them. I'll ask the forum.
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### Re: Question from a Texan Textbook

And how is that working out for you?

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### Re: Question from a Texan Textbook

“When you're born into this world, you're given a ticket to the freak show. If you're born in America you get a front row seat.”
-George Carlin, who died 2008. Ha, now we have human centipedes running the place

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### Re: Question from a Texan Textbook

crank wrote:Wallow in your own blindness. To not grasp that a real is an infinite precision number while posting "if the unexpressed digits .99999..... Are not all nines then the value is less than 1 and not equal to 1. It's not hard." ? That is seriously blind. And then accuse me of erecting barriers to understanding? You don't believe wiki, you don't believe Cali. And you accuse me of wallowing in ignorance? Whoever got stupid enough to up-vote you, too bad they're too much of a coward to add their take on this. I wish someone with brains could step in and help settle this. I don't know how to help someone who won't see what is plainly in front of them. I'll ask the forum.

The impression I get is that you seem to think that integers are not "real" numbers due to the impossibility of infinite precision in the "real" world.

Would this be accurate?
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### Re: Question from a Texan Textbook

Anyway back to the OP, christ on a bike - diagonalisation trivially shows that rational numbers are indeed countable, and how to count them. It's the sort of thing you might show a primary school child to teach them that maths isn't just about increasingly difficult sums, but about solving prolems by approaching them in different ways.

How can such an error possibly make it onto an exam paper? Who would think that, and how the fuck did they find themselves writing maths exams using only their ignorant and wrong intuitions?
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### Re: Question from a Texan Textbook

No, I have only said I think Chaitin's claim is interesting. I then got various arguments about aspects of reals vs integers and infinite precision. I had difficulties with some bad arguments I tried to counter, with some wrong assumptions on my part, and other issues. It's been mostly resolved. Chaitin's claim, if I understand it, is that reals are infinite precision, 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, and that isn't something that could exist in the real world. Part of what Chaitin discusses is how uncomputables are most of the reals, in that computables are countable, and uncomputables are not. So if you could pick a real at random, your chance of picking a computable is zero.

There was also a somewhat side issue of whether the integer x is the same as the real x. I argued it didn't make sense for them to be the same, but in maths, for the most part, they are but with conditions attached. So OK. I still contend that they are different, if not in maths, then in a philosophical and/or conceptual way. I would bet that our brains have a different, separate area that is in use when counting discrete objects than when applying numbers to continuous variables. That's just what I think, I have no evidence and haven't looked.

The issue is more or less resolved in a threadI started to see if I could get broader input.
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### Re: Question from a Texan Textbook

ED209 wrote:Anyway back to the OP, christ on a bike - diagonalisation trivially shows that rational numbers are indeed countable, and how to count them. It's the sort of thing you might show a primary school child to teach them that maths isn't just about increasingly difficult sums, but about solving prolems by approaching them in different ways.

How can such an error possibly make it onto an exam paper? Who would think that, and how the fuck did they find themselves writing maths exams using only their ignorant and wrong intuitions?

This is Texas, where textbook publishers fear to go and not go. The irrational is the norm. The creationists are still fighting, the head of the group that decides textbooks is a creationist, I think, it's an appointment by the gov, and it's a hotly debated issue, don't know if it's changed recently.

Aronra has a lot of videos that discuss this, he's been trying to bring sanity to the issue for a long time, both him and his wife.
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### Re: Question from a Texan Textbook

ED209 wrote:Anyway back to the OP, christ on a bike - diagonalisation trivially shows that rational numbers are indeed countable, and how to count them. It's the sort of thing you might show a primary school child to teach them that maths isn't just about increasingly difficult sums, but about solving prolems by approaching them in different ways.

How can such an error possibly make it onto an exam paper? Who would think that, and how the fuck did they find themselves writing maths exams using only their ignorant and wrong intuitions?

Perhaps you should be asking what group in the US has a stake in the notion that "actual infinities are impossible".
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