Real numbers and the Stern-Brocot tree

TonesInDeepFreeze

the journey corresponding to RRL converges to the point corresponding to the golden ratio. — keystone

That is egregious.

First you said reals are paths. And that is plausible.

Then you said reals are not paths but they're nodes. That is absurd since there are no nodes for irrationals.

Then you said reals are nodes that are in "row infinity". That is absurd since there is no "row infinity".

Then you said reals are programs. That is plausible, and I mentioned that there might be something like that in the mathematical literature.

Then you said you'd clean up your act and reply with something more clear.

Then you come back only to circle back again to saying reals are (or "correspond to", whatever that means) nodes in "row infinity".

Like I said, bait and switch. But with you, it's bait and switch and switchbait and switch again back to one of the earlier switches. In another thread I learned a term "Motte and Bailey fallacy", for when someone can't defend their position so they switch to an easier position to defend but not acknowledging that they've switched. But you actually switch back to the earlier impossible to defend position!

Bait: reals are paths
Switch: reals are nodes in "row infinity" as those nodes are converged on
SwitchBait: reals are programs
Switch: reals are nodes in "row infinity" as those nodes are converged on

So you SKIPPED all the explanation I gave you why the notion of "row infinity" is nonsense. And you SKIPPED extensive explanation why it is question begging to say that such nodes are the limits (per convergence) of sequences: For x to be a limit of a sequence, x has to exist to even be a limit. It is nonsense to say that a sequence converges to a point Phi when you haven't proven that there IS such a point. I've told you about that over and over, but you blithely and dishonestly SKIP it.

Then the piece de resistance:

I'll use 'actualized' for what I mean. In my view, an object is actualized if it is present in the memory of a 'computer'. I am actualized because I am present in the 'computer of the universe'. I'm thinking of a purple cow so that purple cow is actualized because it is present in the 'computer of my mind'. This statement is actualized as I type because it is present in the memory of my laptop. When a memory of an object is flushed, it becomes 'potentialized'. — keystone

That is not progress when the subject is a mathematical account of reals as programs.

TonesInDeepFreeze

Let's contrast with mathematics:

We have axioms and definitional axioms.

We provide an algorithm by which to determine whether a given string is or is not an axiom. We provide an algorithm to determine whether a given sequence of strings is or is not a proof from the axioms.

We provide an algorithm to determine whether a given formula is or is not a definitional axiom.

By asserting 'there exists an x such that blah blah about x' we at least mean that we have a proof of that formalized as a theorem.

We define 'is a natural number'.

We prove that there exist natural numbers.

We prove that there exists {n | n is a natural number}.

We prove that there exists an equivalence relation on the natural numbers to form the set of integers as a set of certain equivalence classes of ordered pairs of natural numbers.

We prove that there exists an equivalence relation on the integers to form the set of rationals as a set of certain equivalence classes on ordered pairs of integers.

We define 'is a Cauchy sequence of rationals'.

We prove there exists an equivalence relation on the Cauchy sequences of rationals to form the set of reals as a set of certain equivalence classes on the Cauchy sequences of rationals.

We define 'less than', addition and multiplication on the the reals.

We define 'is a complete ordered field'.

We prove that <R 'less than' addition multiplication> is a complete ordered field.

We prove that all complete ordered fields are isomorphic with one another.

We prove the theorems and provide definitions used in real analysis (including calculus and mathematics for the sciences).

Mic drop.

TonesInDeepFreeze

But, sincerely, I do like your new version of the tree, accounting for the negative rational numbers. Also, in context of standard math, I like the elegance of Phi generated by such a simple instruction (I'm taking your word for it that that sequence of rationals does converge to Phi, and I'm assuming we can prove that).

keystone

Do you consider this a valid function?
— keystone

I haven't the foggiest. — TonesInDeepFreeze

def endless_loop()
	while True:
	    print("Looping indefinitely...")
	Return 1

I do plan to respond to all of your comments (it may take some time for me to collect my thoughts) but please allow me to dwell on this point some more as your response will help me compose a response to your earlier comments. In case you 'haven't the foggiest' simply because you're not familiar with Python, what this function is designed to do is loop indefinitely (printing the string 'Looping indefinitely...' with each iteration) and at the end return (i.e. output) 1.

This function follows all of the required Python syntax and so it can be interpreted and run. It is a valid program.

What I want is your view on whether it makes any sense to say what this function returns. I want to distinguish between the design of the code and the execution. The code can never be executed to completion so it can never actually return 1. However, as described above the code is designed to return 1.

I would phrase this by saying that the output of the function is potentially 1, but it is never actually 1.

What do you think?

TonesInDeepFreeze

I'm not looking forward to a response in which I would read you compounding your confusions due to ignorance of the basic mathematics and dodges to fancifulness instead while SKIPPING the decisive refutations I provide you.

/

"I haven't the foggiest" - TonesInDeepFreeze

I revised the above response, since actually I did know what you were driving at. My revised response:

(1) Print "Hello". Go to (2).
(2) Go to (1).
(3) Print "Goodbye". Halt.

That's an algorithm. The execution of it successively prints "Hello", and it never prints "Goodbye", and it never halts.

Algorithms may have an instruction that is never executed [...] — TonesInDeepFreeze

/

What I want is your view on whether it makes any sense to say what this function returns. — keystone

I'm not interested in evaluating computer code. I do know what you have in mind. It's more readable (does not require that anyone know the particulars of a particular computer language) the way I've described it.

The algorithm never halts. It prints "Hello" over and over again. It never prints "Goodbye".

as described above the code is designed to return 1. — keystone

"Designed to" is not defined. Without a definition, a reasonable sense would be psychological: what is the intention of the programmer, which is mental. That's not mathematics.

So, not to yet again detour through such subjectivity, we merely observe an objective and mathematical fact: There is an instruction in the algorithm that is never executed.

I would phrase this by saying that the output of the function is potentially 1, but it is never actually 1. — keystone

Definitions:

T is a potential output per an algorithm G if and only there is an instruction in G to print T but execution of G does not print T.

T is an actual output per an algorithm G if and only execution of G prints T.

We don't need to say 'actual output' as we can just say 'output'.

But now we're YET AGAIN back to the same mistake you keep making:

For Phi to be a potential output, Phi has to first exist. Your confused fanciful musings claim Phi is a potential output, but you don't first prove existence of Phi to have Phi as a potential output.

But you revert to an analogy with '1' as a potential output.

In context of your finitist proposal, the number 1 (or "Goodbye") already exists to be a potential outupt, but Phi (or an infinite string of "Peach Desk Happiness Sink Rain Courage [...]") does not already exist. Your analogy is clearly inapt.

As in other threads, you just keep rephrasing your most basic misconception. You shift from one imaginary picture to another in which, somehow, your own mentation allows impossible things. The imaginary pictures are not mathematics, no matter that you're able to shift from one to another after another.

The basic misconception in this thread is the same as in the other threads, such as with Thompson's lamp. In Thompson's lamp, it's an impossible situation. There is no final state in the infinite sequence of states, and the sequence does not converge to a limit state. Here, there is no final row in the tree, but there is a sequence that converges to the limit Phi, but we must FIRST have the existence of Phi for it to be the limit. And with "Hello Goodbye", successive finite strings of "Hello"s are printed, but never an infinite string of them, and "Goodbye", though it exists, is never printed. "Goodbye" is NOT like a limit there.

TonesInDeepFreeze

↪keystone

I asked you a question:

Do you agree that in principle any algorithm can be stated as a Turing machine?

Please answer that.

The set theoretic formulation of Turing machines is the lingua franca of computability. We don't need to evaluate particular computer code in particular computer languages that discussants and readers of the thread might not know. Moreso, for simple algorithms, we can convey the ideas in plain English such as:

(1) Print "Hello". Go to (2).
(2) Go to (1).
(3) Print "Goodbye". Halt. — TonesInDeepFreeze

But if you do need a real formalization, then I suggest Turning machines, which is common ground in computability.

TonesInDeepFreeze

And I would like your answer whether or not you understand this:

"[...] the limit of infinite SB paths are nodes corresponding to real numbers."
— keystone

In your proposal, you don't prove that there IS such limit. [Saying it is "potential" but not "actual" is just more hand waving by you; or, in the sense we can mathematically define 'potential' and 'actual', they still don't obviate that you have not proved that there is such a limit.]

In contrast with your mental pictures, in mathematics with Cauchy sequences, there we DO prove that there is an object that is the limit.

You can't just hand wave to a say that there is a limit; it's not enough to say that the terms of the sequence get closer to each the next - you have to PROVE that there is an object such that the terms get arbitrarily close to it.

Consider sqrt(2). In the rational numbers, there is no limit to the sequence 1, 1.4, 1.41, 1.414, 1.4142 ... But we prove that in the reals there is a limit, viz. the least upper bound of the range of the sequence. And that least upper bound is the square root of 2.

That is, we prove that, in the reals, every bounded set has a least upper bound. {x | x^2 < 2} is bounded. So it has a least upper bound. Then we prove that the least upper bound of {x | x^2 < 2} squared is 2, thus the least upper bound of {x | x^2 < 2} = sqrt(2) . And the sequence of rationals that are approximations converges to the limit, which is sqrt(2).

You can't just say, "The terms of the sequence get closer, so PRESTO POOF OF MAGIC, there's this mystical, fictional "mirage" thing that I say is the limit!" You have to PROVE that there is such a limit.

sime

↪keystone

You are asking basic questions that concern the topic of "Denotational Semantics", which use partially ordered sets (more specifically, Scott Domains) to denote partial states of evaluation with respect to the computation of a term such as a number. Terms of any type are represented as having a totally undefined value prior to evaluation, a partially defined value during the course of evaluation, and in the case of finite terms that can be fully evaluated, a totally defined value after evaluation known as a "normal form".

In denotational semantics, the Type corresponding to 'Computable real numbers' refers to the set of fix-point equations that if iteratively applied on a given rational number, generates a sequence of prefixes that are Cauchy convergent. To obtain an extensional value for a term of 'computable real number type' requires iteratively evaluating the term and then terminating the iterative evaluation abruptly after an arbitrary number of finite iterations, to produce a finite prefix representing a rational number that is very misleadingly said , to "approximate" the real number concerned (it is misleading since we are comparing apples, namely fix-point equations that are defined intensionally in terms of equations and that refer to types, to oranges that are observable states of computation that refer to terms.

A question remains as to who gets to decide when to terminate the iterative evaluation : the interpreter/compiler, or the user of the program? In programming languages with strict semantics, their respective interpreters and compilers always evaluate the term of every type to the fullest extent possible, meaning that real numbers cannot exist as types in such languages, since their terms have no "normal form" and would cause programs to loop endlessly if evaluated. In such languages, real number constants tend to be denoted by rational numbers with a priori fixed values decided at compile time.

By contrast, in a language with lazy semantics such as Haskell, terms can be used and passed around in partially evaluated form. This means that real numbers can exist in the sense of partially-evaluated "infinite lists" consisting of an evaluated prefix and an unevaluated tail. These lazy languages allow runtime conditions to decide what rational value is used in place of a term of real-number type, which is allowed to vary during the course of computation and which corresponds more closely to the notion of "potential infinity".

TonesInDeepFreeze

I'll take your word for it that R RL[...] yield successively more accurate approximations of Phi.

Algorithm [named 'Chasing mirage-Phi']

(1) X = R. Go to (2).
(2) X = XR. Go to (3).
(3) X = XL. Go to (2).
(4) X = mirage-Phi. [whatever that is!]

In keystoneland is mirage-Phi an infinite sequence RRLRLRL... ? No, because in keystoneland, there are no infinite sequences.

In keystoneland, is mirage-Phi a "potential" output? It would be, if it were something that exists in keystoneland.

Lesson: Of course an algorithm may have an instruction that is never executed. But if the instruction mentions an object as an output, then that object has to exist, whether as an output or a "potential" output.

Please, no more or your handwaving! You don't have a coherent proposal. If you are sincere, then state axioms, rules of inference, definitions, and prove the theorems that justify a logical construction.

keystone

@TonesInDeepFreeze

You raise some good criticisms. I have reformulated my proposal to address them.

The object that I want to talk about moving forward is not the actually infinite S-B tree (which you're right, I don't believe in), but instead the blue line as illustrated below.

Don't think of this figure as depicting an infinite collection of blue lines, but instead as a single line, and each row describes a finite set of cuts which can be performed on the line. Although I'm not talking about the S-B tree, the cuts are closely linked with the S-B tree, which I'll call the S-B algorithm moving forward to avoid any confusion on my philosophy).

This line is not the real number line (which is composed of uncountably infinite points). On the very top row we have segment [] as described by the interval (-inf,+inf). There are no points on this line. I recognize that intervals across continua are typically associated with the infinite set of points between them so I'm not using the term 'interval' as it is typically used. To avoid any confusion, moving forward, in this context I will use pseudo-interval instead of interval.

In row 0, the line is cut into two segments:
(1) segment L corresponding to the pseudo-interval (-inf,0)
(2) segment R corresponding to the pseudo-interval (0,+inf)

With each cut, a point is introduced on the line corresponding to a rational number. This point is not an actual object, but instead only demarks the space between adjacent line segments.

With each successive row, the line is cut more and more times.

If asked, what number is a solution to 'phi^2-phi-1=0?' I would say that I cannot provide a number. But what I can provide is the pseudo-interval within which it lies at any row. For example,

Row 0: R
Row 1: RR
Row 2: RRL
Row 3: RRLR
Row 4: RRLRL
Row 5: RRLRLR
Etc.

We can proceed arbitrarily far down the tree, continually shrinking the pseudo-interval within which the solution lies. However, since there is no bottom of the tree, perfect precision cannot be achieved (i.e. the pseudo-interval can never be reduced to a single number, and the line cannot be reduced to a point). The solution to this problem is therefore not a number, but instead a vanishingly small pseudo-interval.

One might say that the sequence of pseudo-intervals describing phi is (R,RR,RRL,RRLR,RRLR,RRLRL,RRLRLR,…) but this sequence is infinite and so I do not believe it exists. What exists is the potentially infinite algorithm to generate successive cuts to narrow the pseudo-interval. In this case, the algorithm can be succinctly captured by the string: RRL. Similarly, the algorithms corresponding to 1 are RLR and RRL. Infinite S-B strings are algorithms. Reals are algorithms.

The claim that there are only countably many algorithms/programs does not imply that this view has gaps in the line. The line is there in full from the start. It simply means that it is impossible for any computer to hone in on some sections of the line.

It is my suspicion that this view is so closely tied with Cauchy sequences that maybe few new proofs are needed. Perhaps only a proof to shown that Cauchy sequences are actually talking about this. All that's really changed is the philosophy - but even that hasn't changed as much as one might think. For example, when explaining integrals, we use ever shrinking rectangles - this is not unlike the ever shrinking pseudo-intervals used here.

P.S. I've moved on from talking about the output of an unending program, but I do agree that any program you can write can in principle be stated as a Turing machine.

keystone

By contrast, in a language with lazy semantics such as Haskell, terms can be used and passed around in partially evaluated form. This means that real numbers can exist in the sense of partially-evaluated "infinite lists" consisting of an evaluated prefix and an unevaluated tail. These lazy languages allow runtime conditions to decide what rational value is used in place of a term of real-number type, which is allowed to vary during the course of computation and which corresponds more closely to the notion of "potential infinity". — sime

Actually, the algorithm used to perform arithmetic on the Stern-Brocot tree was originally written in Haskell (I rewrote it in Python) and it does just this!

https://www.sciencedirect.com/science/article/pii/S1570866706000311#:~:text=The%20Stern%E2%80%93Brocot%20tree%20is,rope%20between%20zero%20and%20infinity'.

TonesInDeepFreeze

The solution to this problem is therefore not a number, but instead a vanishingly small pseudo-interval. — keystone

You STILL don't get it. You just keep putting new clothes on an old pig. Every time, you reformulate but you retain the essential fallacy.

There is no single "vanishingly small pseudo-interval". There are only successively smaller pseudo-intervals on successive rows.

This is still essentially the same subject, as you still (after I asked twice) SKIPPED answering my question*, though I can see for myself that the correct answer is 'no'.

* https://thephilosophyforum.com/discussion/comment/806060

I really should not continue to reply when you so obnoxiously continue to apply the same fallacy clothed differently each time though I have explained it over and over and over and even asked you whether you understand, yet you don't reply even to that question itself.

But I'll look under the hood yet again:

As I understand, there is just one line, and then each row is a different set of pseudo-intervals for that line.

I don't see a problem with that. Except that I don't see that it improves the more simple approach:

There is an algorithm that successively outputs the S-B rows. And there are algorithms to output successive S-B finite paths, each path ending in a node. We take the comp-reals to be those algorithms, calling them real-ithms.

Then, define '<', '+' and '*' on real-ithms, and you'd be on our way to something.

In other words, for arbitrary real-ithms G and R:

G < R <-> [fill in definiens]

G+R = [fill in definiens]

G*R = [fill in definiens]

But then this:

The solution to this problem is therefore not a number, but instead a vanishingly small pseudo-interval. — keystone

No, throw that out. There is no such thing. There are only successive rows with smaller and smaller pseudo-intervals.

Your "vanishingly small pseudo-interval" is just a variation on the "mirage last row" fallacy.

The claim that there are only countably many algorithms/programs does not imply that this view has gaps in the line. — keystone

"Gaps". What I actually say: There are only countably many comp-reals, so they don't provide a complete ordered field, thus there is no isomorphism with the continuum. But the standard calculus depends on the completeness of the reals. So if you claim your proposal can do mathematics for the sciences, then you have to show how it does that. And by 'show' I don't ostensive examples, handwaving or picture stories. I mean axioms and theorems - formulas. And formulas then used to solve problems of science.

The line is there in full from the start. — keystone

You are SUCH a self-contradicting liar. You said:

There are no points on this line. — keystone

[bold ORIGINAL]

You seem to have no compunctions about insulting intelligence.

All that's really changed is the philosophy — keystone

Changed from standard analysis? No, your proposal is radically different from standard analysis, from the start: Standard analysis uses infinite sets; you disclaim infinite sets. Standard analysis has a continuum; you don't. Standard analysis has uncountably many reals; with you, it's not clear how many comp-reals there are, since there are denumerably many real-ithms but you disclaim that there are denumerable sets (but maybe you could argue that there is not a set of all real-ithms but instead a program that itself generates real-ithms).

I've moved on from talking about the output of an unending program — keystone

Right, you dressed it up differently again, now as "a vanishingly small pseudo-interval". Same pig, new dress.

TonesInDeepFreeze

In sum:

Your 'pseudo-intervals' are okay, up to, but not including, a 'vanishingly small pseudo-interval'.

But it's unnecessarily complicated when instead all we need is:

There is an algorithm that successively outputs the S-B rows. And there are algorithms to output successive S-B finite paths, each path ending in a node. We take the comp-reals to be those algorithms, calling them real-ithms.

Then, define '<', '+' and '*' on real-ithms, and you'd be on our way to something.

In other words, for arbitrary real-ithms G and R:

G < R <-> [fill in definiens]

G+R = [fill in definiens]

G*R = [fill in definiens]

/

Your move to this yet next iteration of you proposal (same pig, new dress) only serves to DISTRACT from the fact that AGAIN you want to eat your cake and have it too. You want only finite objects, but you also want real numbers, and you wanted them to be at the "mirage last row", or the last output of after an infinite loop, but now courtesy of a "vanishingly small pseudo-interval" while those DON'T EXIST with just your finite-only approach.

jgill

Here's an explanation of this process converging to the Golden Ratio. Leaving out the details, we begin with the

cf = $1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots }}}$ , corresponding to the pattern of movement down the SB chart. This cf can be generated by iterating the function

$f(x)=\frac{1}{1+x},\text{ }{{f}^{(2)}}(x)=f(f(x))=\frac{1}{1+f(x)}=\frac{1}{1+\frac{1}{1+x}},\text{ etc}\text{.}$

Setting $f(x)=x\text{ }\Rightarrow \text{ }x=\alpha ,\beta$ , where the first of these fixed points

is an attractor: $f(x)-\alpha =K(x-\alpha ),\text{ }\left| K \right|\lt1$ , $K=K(x)$ , for most values of x.

And ${{f}^{(n)}}(x)-\alpha ={{K}^{n}}(x-\alpha )\to 0$ as $n\to \infty$

Now, the value of the cf is $1+\alpha$ = Golden Ratio

This curiosity has a relation to arguments about endless cause/effect chains and first causes. Viewing the cf as going back in time, there is no end to the process, but stopping at any value of n , the x we have chosen at random, say, is a first cause.

TonesInDeepFreeze

↪jgill

What does 'the GR' stand for there?

jgill

↪TonesInDeepFreeze

Fixed. :cool:

I've left out details to give an outline.

TonesInDeepFreeze

↪jgill

I shoulda gleaned it meant the golden ratio. I was distracted by fact that coincidentally I used 'G' and 'R' for something different.

keystone

I really should not continue to reply when you so obnoxiously continue to apply the same fallacy, clothed differently, each time though I have explained it over and over and over and even asked you whether you understand, yet you don't reply even to that question itself. — TonesInDeepFreeze

I appreciate you continuing this discussion. I'm getting plenty of value out of this dialogue. I don't think your infinite loop programming example of me not listening was a fair representation as I believe I am learning (perhaps not fast enough).

You STILL don't get it. You just keep putting new clothes on an old pig. Every time, you reformulate but you retain the essential fallacy.

There is no single "vanishingly small pseudo-interval". There are only successively smaller pseudo-intervals on successive rows. — TonesInDeepFreeze

You're right. I shouldn't have used 'vanishingly small pseudo-interval'. That makes it sound like I'm talking about an object at the bottom of the tree. The only object is the line. There is no bottom of the tree.

Let me rephrase a paragraph from my earlier post here:
If asked, what number is a solution to 'phi^2-phi-1=0?' I would say that I cannot provide a number. But what I can provide is an algorithm (which inputs integer, ROW, and outputs pseudo-interval, PI) such that for any positive rational number, EPS, there is an input ROW which outputs a PI, whose potential cuts (rational number, m) would all yield |m^2-m-1| < EPS. This algorithm can be described as RRL and I call it the golden ratio.

Except that I don't see that it improves the more simple approach — TonesInDeepFreeze

The idea behind this proposal is that the fundamental object is the line, not the point. This difference is significant.

But it is the standard calculus depends on the completeness of the reals. — TonesInDeepFreeze

But doesn't that mean that standard calculus depends on there being no gaps on the line? That's what this suggests:
https://en.wikipedia.org/wiki/Completeness_of_the_real_numbers

There are no points on this line.
— keystone
[bold ORIGINAL]

You seem to have no compunctions about insulting intelligence. — TonesInDeepFreeze

Infinite sets are so deeply embedded in your thinking that you're not even willing to imagine the possibility that points are not fundamental. Flip your thinking upside down just for a moment. In my proposal I start with a line and points get added as we progress down the rows. I'm starting with an unmarked 'stick', not a perfect precision 'ruler'. If I'm starting with a 'perfect precision ruler' on the top row then please tell me the coordinate of even a single specific point on that line [].

Changed from standard analysis? No, your proposal is radically different from standard analysis, from the start: Standard analysis uses infinite sets, you disclaim infinite sets. Standard analysis has a continuum; you don't. Standard analysis has uncountably many reals. With you, it's not clear how many comp-reals there are, since there are denumerably many real-ithms but you disclaim that there are denumerable sets (but maybe you could argue that there is not a set of all real-ithms but instead a program that itself generates real-ithms). — TonesInDeepFreeze

Are my proposed algorithms that different from Cauchy sequences?
Is my proposed line that different from the real number line?
Is my proposed line not continuous?
Standard analysis achieves length by having uncountably many points. Is length not also achieved by having pseudo-intervals?
Of course I'm proposing something different, but we must remember that calculus came before set theory. I'm only proposing a different foundational underpinning. If you don't think that's philosophy then sure.

Then, define '<', '+' and '*' on real-ithms, and you'd be on our way to something.

In other words, for arbitrary real-ithms G and R:
G < R <-> [fill in definiens]
G+R = [fill in definiens]
G*R = [fill in definiens] — TonesInDeepFreeze

This is sound advice. I need to research this. I appreciate you advising me on next steps.

You want only finite objects, but you also want real numbers — TonesInDeepFreeze

I don't want all real numbers, only the computable ones when they're needed. And to calculate the area between y=1, y=0, x=0 and x=1, I don't need to there to exist infinite points. As long as I know that there are no gaps across the lines then I can calculate the area just fine. Would anyone claim to explicitly make use of infinite points in their calculation?

TonesInDeepFreeze

I'm getting plenty of value out of this dialogue. — keystone

I've supplied some pretty good posts.

I don't think your infinite loop programming example of me not listening was a fair representation — keystone

The infinitude of it was a joke to express that it feels interminable. More than fair in that way.

The idea behind this proposal is that the fundamental object is the line, not the point. — keystone

The simple version is programs. Bringing in a concept of an initial object that is determined solely by -inf and +inf and then pseudo-intervals is extraneous. Everything you need is captured by:

"There is an algorithm that successively outputs the S-B rows. And there are algorithms to output successive S-B finite paths, each path ending in a node. We take the comp-reals to be those algorithms, calling them real-ithms."

This algorithm can be described as R RL[...] and I call it phi. — keystone

Exactly. The algorithm is comp-Phi. That's what I said. And the algorithm only needs to print successive rows - not worrying about cuts and pseudo-intervals. My guess is that you wanted to gussy it all up so that it seduces us to think that in a prettier getup there's some kind of vanishing point horizon mirage-limit. But at least now (hopefully, sincerely and stably*) you've backed off from claiming that.

* Let's set the stopwatch to see how long before you relapse and try to smuggle it in dressed up differently again.

But doesn't that mean that standard calculus depends on there being no gaps on the line? — keystone

What? I just said, "gaps" is not defined by you. The mathematics meanwhile is that there is a continuum. A continuum requires that the ordering is complete, meaning that for every bounded set, there is a least upper bound.

Infinite sets are so deeply embedded in your thinking that you're not even willing to imagine the possibility that points are not fundamental. — keystone

You don't get to incorrectly say what I'm willing to imagine. I don't even have notion of "points are fundamental", let alone that I won't imagine that it's not the case.

I'm happy to see any mathematics or formal logic - no matter how different from set theory and standard logic - as long as there are axioms, rules of inference, and definitions such that we can rigorously verify whether something is an axiom, an application of an inference rule or a definition. Or any algorithm, hopefully that can be conveyed as a Turing machine.

Please don't make pronouncements about what I am willing to imagine.

In my proposal — keystone

I understand your proposal. I am just saying that you don't need all those gooey toppings. Plain vanilla does the job just as well.

Are my proposed algorithms that different from Cauchy sequences? — keystone

Indeed they are! I EXPLAINED this. I don't understand what you don't understand in my explanation.

(1) An algorithm is finite. A Cauchy sequence is denumerable. And an equivalence class of Cauchy sequences has the uncountable cardinality of the set of equivalence classes of Cauchy sequences.*

* I think that sentence is right.

(2) There are only denumerably many algorithms, but uncountably many equivalence classes of Cauchy sequences.

(3) Cauchy sequences have a limit. But if we somehow defined the limit of an algorithm, then that would be infinitistic (unless some actual rigorous workaround could be formulated).

Why don't you already know this?

Is my proposed line that different from the real number line? — keystone

Your use of 'line' is only a figure of speech. It's not a line. It has nothing on it; it's a placeholder only - as YOU said. It's not a line in the sense of geometry or analytic geometry.

Already, we have confusion because you use a word in an utterly personal way and it gets conflated with the actual mathematical sense. So you should call it 'the k-line' so that it doesn't get mixed up again with 'line' in the mathematical sense.

And at no output does the cutting remotely resemble the continuum. First, at every output, there are only finitely many cuts and thus only finitely many rationals described. Second, there are no irrationals described. That is VERY different from the continuum that has both rational reals and irrational reals and altogether not just finitely many, but uncountably many, and proving a continuum.

And even if we took the UNION of the rows (which you can't do, because that is infinitistic) we still would have only the denumerable set of rationals described and none of the uncountably many irrationals.

Is my proposed line not continuous? — keystone

What? Are you TROLLIING me? Your questions are so ignorant and stupid that I can't help but suspect that you are.

Your "line", the k-line, has NOTHING on it, as YOU said. So 'continuous' is not even applicable.

And there is no infinite set of cuts on the k-line that comes after all the rows. You just now admitted that.

Standard analysis achieves length by having uncountably many points. — keystone

Wrong. You don't know ANYTHING about this. You don't even know high school mathematics.

Length is the absolute value of a difference. Even without irrationals, we have length with just rationals. Uncountability is not required to define length. Sheesh!

Is length not also achieved by having pseudo-intervals? — keystone

I don't know. You can do the arithmetic to see whether differences restricted to only those between rationals all on a row work out as desired.

But even if it worked out, calculus needs more than just lengths. It seems you don't know what calculus is. Do you?

we must remember that calculus came before set theory — keystone

So what? It used infinitisitic methods. Set theory provided axioms to make those methods rigorously derived from axioms.

I'm only proposing a different foundational underpinning. If you don't think that's philosophy then sure. — keystone

I respect and encourage philosophical frameworks for various notions of finitism. But your own mathematical proposal inspired by your particular finitism is incoherent. You are have a massive mental block that doesn't allow you to understand the basic illogic in your thinking. You keep wanting to have both only finite objects but also objects that exist only as provided as an end of an infinite process, while refusing in different forms to recognize that there is no such end hence no such objects.

/

I haven't tried to formalize your latest idea, but the rough sketch I come up with is this:

There is a primitive object, called 'the k-line'.

There are two more primitive objects, called '-inf' and '+inf. They are ordered so that -inf is less than +inf.

There are two more primitive objects: R and L.

The k-line but also associated with -inf and +inf is the base row.

There is an algorithm, call it the 'k-S-B algorithm', that generates rows, starting with the base row, then to the next row that is row 0, ad infinitum. The k-S-B algorithm recursively exhausts all "turn decisions" of R and L.

The k-S-B algorithm also associates each row with a set of fractions and an ordering on them, and the fractions are grouped in "cuts" which provide "pseudo-intervals".

So, a row is the k-line, with associated fractions, along with associated cuts and associated pseudo-intervals.

A real-ithm is an algorithm that executes non-terminatingly and each successive output is a finite sequence of Rs and Ls depending on a sequence of "turn decisions".

A k-real is a real-ithm.

NOTE:

We don't need the k-line. It is extraneous to capturing the information we want. We can just say a row is the set of cuts.

We don't need cuts. They are extraneous to capturing the information we want. We can just mention the fractions and their ordering.

I think the reason you want all that is to give the illusion that it amounts to a kind of pseudo-"continuum". But it doesn't. Essentially it's a big red herring. Toss out the red herring and simplify as I showed you, which is basically what you proposed yesterday.

NOTE:

There is no final row.

No real-ithm outputs a denumerable sequence.

Only computable reals are described.

A continuum cannot be described (as I explained a few posts ago and again here).

TonesInDeepFreeze

↪keystone

I've responded to all your main points and nearly all your secondary points. And I answered the exact questions you asked me.

Meanwhile, I've asked you three times now whether you understand this post:

https://thephilosophyforum.com/discussion/comment/806060

But you still say not a word about it.

What's up with that?

keystone

@TonesInDeepFreeze

To continue to clean up some of the language:

A point in the context of my view is a k-point
A line in the context of my view is a k-line
An interval in the context of my view is a k-interval
An algorithm which takes as input any natural number (corresponding to the row) and always outputs a k-interval (in finite time) is a k-algorithm
A function which takes as input rational number(s) and outputs rational number(s) is a k-function
A string of L's and R's (and underlines to represent repeats) is a k-string
A k-line which has no gaps is k-continuous.
..and so on.

Meanwhile, I've asked you three times now whether you understand this post:
https://thephilosophyforum.com/discussion/comment/806060
But you still say not a word about it. — TonesInDeepFreeze

Sorry, I thought I was answering this question indirectly but let me be more clear. The successive outputs of a k-algorithm do not converge to any object. Ever. The S-B algorithm does not terminate (or to someone who believes in actual infinity - there is no bottom of the S-B tree).

Bringing in a concept of an initial object that is determined solely by -inf and +inf and then pseudo-intervals is extraneous. — TonesInDeepFreeze

The problem is that we've only been talking about numbers so far. The k-line becomes important when exploring higher dimensions. Consider the following k-functions:

y=3x^2+2
y=0
x=1
x=0

These k-functions can be illustrated as k-lines as depicted in the k-graph below. Please do not concern yourself with the shape of these k-lines. All properties of these k-lines are invariant under any continuous deformation.

In this figure there are only four k-points: (0,0), (0,2), (1,5), and (1,0). Notice how these k-points emerge when the k-lines cut each other.

The question is: What is the area between these k-lines?

One can estimate the area by introducing more k-functions and using rectangles.

So the area is approximately (5)(0.5)+(2.75)(0.5) = 3.875.
I can continue adding more k-functions and summing the areas of smaller and smaller rectangles but this process will never be exhausted. I cannot provide an exact answer of the area between the four k-lines.

However, because the above k-lines are k-continuous, I can using integration and provide you with the k-algorithm which can output a k-interval of arbitrarily narrow width. This k-algorithm can be described by the k-string RRRLR. Therefore the k-area between the k-lines is the k-string RRRLR.

Of course, k-algorithms are just one type of algorithm. If people were to adopt my view I wouldn't expect them to transition to using the S-B algorithm. I could have used decimals to describe algorithms and the decimal which would describe the equivalent algorithm is 2.9.

The reason why I say that my proposition is philosophical and not mathematical is because if you take away all of the k's you basically have standard calculus.

Are my proposed algorithms that different from Cauchy sequences?
— keystone
Indeed they are! I EXPLAINED this. — TonesInDeepFreeze

I think it's a matter of perspective by what one means by 'that different'. What I believe though is that if this approach ever gets formalized it's going to use a lot of similar language as Cauchy sequences.

Your use of 'line' is only a figure of speech. It's not a line. It has nothing on it — TonesInDeepFreeze

k-lines are associated with k-functions that describe their infinite potential.

So what? It used infinitisitic methods. Set theory provided axioms to make those methods rigorously derived from axioms. — TonesInDeepFreeze

Maybe one day you will see set theory as the mathematics of the bottom of the S-B tree…the bottom which you (rightfully) claim doesn't exist. Perhaps it is you who wants to eat your cake and have it too.

There is an algorithm, call it the 'k-S-B algorithm', that generates rows, starting with the base row, then to the next row that is row 0, ad infinitum. The k-S-B algorithm recursively exhausts all "turn decisions" of R and L. — TonesInDeepFreeze

No, I realized after contemplating about our debate on outputting something after an infinite loop that it wasn't necessary. k-algorithms only take natural numbers (corresponding to rows) as input and always output a k-interval in finite time.

TonesInDeepFreeze

Meanwhile, I've asked you three times now whether you understand this post:
https://thephilosophyforum.com/discussion/comment/806060
But you still say not a word about it.
— TonesInDeepFreeze

Sorry, I thought I was answering this question indirectly but let me be more clear. The successive outputs of a k-algorithm do not converge to any object. Ever. The S-B algorithm does not terminate (or to someone who believes in actual infinity - there is no bottom of the S-B tree). — keystone

It's good that you've conceded that there is no convergence to an object, that the algorithm does not terminate, and there is no last row. And it seems that the post helped you to that. But I don't know actually know what role the post had in that. I was asking whether you understand the post, which includes the various aspects of its explanations. Knowing your answer would let me know how much communication is taking place here.

TonesInDeepFreeze

AMAZING obtuseness right here:

Maybe one day you will see set theory as the mathematics of the bottom of the S-B tree…the bottom which you (rightfully) claim doesn't exist. Perhaps it is you who wants to eat your cake and have it too. — keystone

And this answers the question: No you did NOT understand the post mentioned in my post above.

I went out of my way to distinguish between standard mathematics and keystone musings.

In standard mathematics, there ARE objects to serve as limits. I gave an exact example of that. In standard mathematics, with infinite sets, there IS a limit to the sequence of successive finite approximations of Phi. But in keystone musings, without infinite sets, there is NOT. I get to say, "Phi is the limit", because set theory proves there IS such a limit. You do not, because your framework PRECLUDES that infinitistic limit.

Your admonishment about this is a product of you getting completely backwards.

But at least it's good to have an answer: No, you did not understand that post.

TonesInDeepFreeze

To continue to clean up some of the language: — keystone

I'm not inclined to indulge you with a formulation that is more complicated than it needs to be. I offered you a more simple outline. You can follow up on it if you like.

Your use of 'line' is only a figure of speech. It's not a line. It has nothing on it
— TonesInDeepFreeze

k-lines are associated with k-functions that describe their infinite potential. — keystone

Wow, you just turn on a dime away from what you say previously. I've explicated enough. Go back and read my posts and think about them rather than driving right over them.

TonesInDeepFreeze

The k-line becomes important when exploring higher dimensions. — keystone

You haven't even figured out the first "dimension". But carry on, though I will very likely not be subscribing.

TonesInDeepFreeze

I think it's a matter of perspective by what one means by 'that different'. — keystone

What? I listed the CRUCIAL, ESSENTIAL ways in which they are different. Rather than recognize that, you cop out with "it's a matter of perspective what one means by 'that different'."

TonesInDeepFreeze

I believe though is that if this approach ever gets formalized it's going to use a lot of similar language as Cauchy sequences. — keystone

So what? Lots of things use similar language, but say RADICALLY different things.

TonesInDeepFreeze

↪keystone

You are still confused. You still SKIP the MAIN points I post. You SKIP over the explanations about how you're mathematically wrong (not wrong to eschew standard mathematics, but wrong about the implications of your OWN framework). Often enough, your notions are incoherent. You make ridiculous jejune arguments (see posts above). And you lie about me. You're a sinkhole.

keystone

I was asking whether you understand the post, which includes the various aspects of its explanations. Knowing your answer would let me know how much communication is taking place here. — TonesInDeepFreeze

I suppose you would like me to paraphrase so you can judge my comprehension. Fair enough. Earlier on in the discussion, I incorrectly claimed that the S-B paths converged to a limit. But all that there is in the tree are nodes and paths. Each node corresponds to a rational number. You used sqrt(2) but let me use phi for simplicity since it is RRL. The nodes along the phi path are 1, 2, 3/2, 5/3, … But none of the nodes along the path are the limit, the limit in fact is an irrational number. In this sense, the rational numbers are incomplete because the limit of some sequences of rational numbers are not rational numbers themselves. So then you asked, what object on the tree does this limit correspond to? Such an object must be a part of the system for it to be complete. Reals do not have this problem. The limit of every sequence of real numbers is also a real number.

One way to rephrase this is to say that in the rationals there are some bounded sets for which there is no least upper bound. Essentially what this means is that if I were to have a set of rational numbers as described by the red coloring, there are many upper bounds (in blue) but there is no least upper bound (member of blue) because the number that we need is sqrt(2), which is not a rational number. Again, reals do not have this problem. "Every non-empty subset of X with an upper bound has a least upper bound in X."

I think this may be an important point to you because you are stressing the importance of completeness to calculus. If in my view the number system is incomplete, how is it possible to have the continuum needed to do much of calculus?

I get to say, "Phi is the limit", because set theory proves there IS such a limit. You do not, because your framework PRECLUDES that infinitistic limit. — TonesInDeepFreeze

I understand this. I wasn't challenging this. What I'm suggesting is that by starting with uncountably infinite objects (corresponding to real numbers) you are effectively starting with the 'bottom of the tree'. And that agreeing to the former and not the latter is wanting your cake and having it too. I do not challenge the fact that one can prove a lot of things using a system built on infinite sets.

Of course, I'm sure you do not see set theory as starting from the bottom of the tree. And that's why I mused that maybe one day…

keystone

So what? Lots of things use similar language, but say RADICALLY different things. — TonesInDeepFreeze

Okay, fine. It's significantly different. I suppose I'm coming from the applied side of mathematics and I don't see how my view changes anything in my day to day computations. But yes, if a mathematician were to go from a system based on actual infinities to one based on potential infinites there will be significant differences in the formalization. I should not understate this.

Okay, let me retrace my steps responding more thoroughly to some of your comments from your earlier post.

A continuum requires that the ordering is complete, meaning that for every bounded set, there is a least upper bound. — TonesInDeepFreeze

A continuum defined by numbers requires that the ordering is complete. I'm not proposing this. I'm attacking this from the other direction - numbers defined by a continuum. The ordering of numbers in this system does not need to be complete.

You don't get to incorrectly say what I'm willing to imagine. I don't even have notion of "points are fundamental", let alone that I won't imagine that it's not the case….
Please don't make pronouncements about what I am willing to imagine. — TonesInDeepFreeze

Okay, I take back my comment. But let me ask this. Would you agree to either of the following?
1) A continuum is defined completely by numbers.
2) A line is made up entirely of points.
I'm sure you will not agree to the exact phrasing, but perhaps you could say it in your words.

Cauchy sequences have a limit. But if we somehow defined the limit of an algorithm, then that would be infinitistic — TonesInDeepFreeze

I don't think there's a need to define the limit of an algorithm.

our use of 'line' is only a figure of speech. It's not a line. It has nothing on it; it's a placeholder only - as YOU said. It's not a line in the sense of geometry or analytic geometry. — TonesInDeepFreeze

In my response yesterday I wanted to show you that the k-line is not just a placeholder. By moving up a dimension, it becomes clear that even the uncut line holds a lot of information. For example, if you draw a plot including the k-lines y=1 and y=2, the y=2 line will be above the k-line y=1 even if they do not have any points on them.

And at no output does the cutting remotely resemble the continuum. First, at every output, there are only finitely many cuts and thus only finitely many rationals described. Second, there are no irrationals described. That is VERY different from the continuum that has both rational reals and irrational reals and altogether not just finitely many, but uncountably many, and proving a continuum. — TonesInDeepFreeze

You keep coming back to this point, which comes from a very number-centric view. I agree that a number system based on rational numbers cannot be continuous. I am not proposing that. I'm proposing an algorithm (e.g. program-based, equation-based) system. The restrictions on number systems do not apply to algorithm systems.

Your "line", the k-line, has NOTHING on it, as YOU said. So 'continuous' is not even applicable. And there is no infinite set of cuts on the k-line that comes after all the rows. You just now admitted that. — TonesInDeepFreeze

Over and over you repeat the same point, as if I'm not understanding you. I understand what you're saying. I just thought my last response most concisely addressed this point, but I'm responding here because I actually value your input and you think I skip over points. In my view the equation (line) is fundamental. In the standard view the number (point) is fundamental. I say start with an equation (line) and then create numbers (points). The standard view is to start with infinite numbers (points) and then create the equation (line). You are wrong to say that the k-line has nothing on it. It has an equation (algorithm). And I wanted to go up a dimension to demonstrate this.

Length is the absolute value of a difference. Even without irrationals, we have length with just rationals. Uncountability is not required to define length. Sheesh! — TonesInDeepFreeze

True. But in higher dimensions what about arc length.

But even if it worked out, calculus needs more than just lengths. It seems you don't know what calculus is. Do you? — TonesInDeepFreeze

In my post yesterday, I tried to demonstrate to you that my thinking extends beyond length to include, for example, area.

You keep wanting to have both only finite objects but also objects that exist only as provided as an end of an infinite process, while refusing in different forms to recognize that there is no such end hence no such objects. — TonesInDeepFreeze

At one point I was arguing this, but now everything that I'm working with is finite. The k-line fits entirely on the page (and so does the higher dimensional plot), there is no infinite tree, I'm not waiting infinite time for the k-algorithm to spit something out. What objects are you referring to?

We don't need the k-line. It is extraneous to capturing the information we want. We can just say a row is the set of cuts.

We don't need cuts. They are extraneous to capturing the information we want. We can just mention the fractions and their ordering.

I think the reason you want all that is to give the illusion that it amounts to a kind of pseudo-"continuum". But it doesn't. Essentially it's a big red herring. Toss out the red herring and simplify as I showed you, which is basically what you proposed yesterday. — TonesInDeepFreeze

It is impossible to formalize a number system that completely describes a continuum without actual infinity. Such a system would be dead on arrival. I truly appreciate you drafting out some ideas, but I don't think that's the right way to go. I don't think that my algorithmic/interval/line approach is a red herring. Rather, I see it as the only viable path a finitist can go to make sense of calculus.

So when you say that you’re unwilling to go to higher dimensions because I haven’t worked out the first dimension what I’m hearing is that you’re unwilling to deal with functions because I haven’t worked out the number system. And what I’m trying to show you is that you have it backwards. The algorithms (e.g. functions) come first and the numbers follow. I’m building an algorithm system, not a number system. If you’d consider my last message you might understand this.

So in sum, I am responding to your main points, I’m just not spelling it out.

Real numbers and the Stern-Brocot tree

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