Is Arithmetic Circular?

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wtf
Posts: 1178
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Re: Is Arithmetic Circular?

Post by wtf »

Skepdick wrote: Sun Mar 27, 2022 10:06 am Are you not agreeing with, and therefore repeating/amplifying Peano's definitions?

Are you not repeating, and therefore saying the same thing as Peano?
Actually not. If you would re-read the thread, you would see that @Eodnhoj7 said that "S(0),S(S(0)),S(S(S(0))) is the circularity of S". I asked him what was circular about that.

Then @Impenitent said, "I see circularity at equidistant points around the center," and I asked him what he meant by that.

Neither poster has yet favored me with a response.

I did use, in passing, the common and universally accepted shorthands for S0, SS0, SSS0, ... (omitting the parens) of 1, 2, 3, etc. That's about it. Perhaps you should take up your concerns with the math department at the local university. My usage of these symbols is quite standard. I don't understand your concerns.

In this context the use of the symbols 1, 2, 3, ... is totally incidental. S0, SS0, SSS0, ... is sufficient, and if that is your concern, so be it. Nevermind 1, 2, 3, ...

Skepdick wrote: Sun Mar 27, 2022 10:06 am
Why do you constantly keep tripping up over notation/syntax?
Well, if I wrote S0 = 1 and you said, "That's a tuna sandwichification! Why are you doing that?" I'd point out that 1 = S0 is a definition, not a tuna sandwichifacation. And you'd be quite disingenuous to claim I'm "tripping up over notation/syntax."

In fact, you have used the wrong word. From my reading of the Wiki page you linked, stratification is not the same as what's commonly known as a definition. But, I'm not too familiar with stratification. Perhaps this is another word for definition, or another aspect of definitions. Either way, I just don't understand the concern you're raising.

Do you object to my usage of the symbols 1, 2, 3, .. for S0, SS0, SSS0, ...? Or do you object to Peano's having done that? Or all the mathematicians and logicians in the world doing that? What exactly is the concern here?

Again, I made no assertions. I merely used the common convention, and asked these two posters to explain what they think is circular about the sequence S0, SS0, SSS0, ...

If you object to my use of the symbols 1, 2, 3, ... then I won't use them. I'm not wedded to them. S0, SS0, SSS0, ... are perfectly fine for me in this context.
Skepdick wrote: Sun Mar 27, 2022 10:06 am
Peano's definitions ARE stratifications by definition of what stratification is. None of the above is controversial.
Ok. If so, then I consider myself educated on the topic. If definitions are stratifications, or stratifications are definitions (I'm still unclear on this), then "today I learned," and I thank you for the information.

But what concerns you about my use of the common symbols 1, 2, 3, ... for the positive integers? Surely you know that pretty much everyone does this from age 3 or 4 onward.
Skepdick wrote: Sun Mar 27, 2022 10:06 am What's controversial is the following missing definitions:

define(0, ???)
define(S, x, ???)

Which is why I gave you the generator s() which is not stratified - it has no interpretation in Peano's framework.
You truly lost me here. Are you arguing that the Peano axioms are wrong? Or that they're ok but that I should not have used the common shorthands 1, 2, 3, ...? What exactly is your concern? You are being totally unclear. You are unhappy about something I said, but all I did was use the symbols 1, 2, 3, ... as the names of the positive integers as given by the Peano axioms. And for my purpose, I could have used S0, SS0, SSS0, ... just as well. Would you still be unhappy with that? Or is it just the symbols 1, 2, 3, ... that you don't like? Roman numerals would do as well, we're only talking about particular representations of numbers.

Is it possible for you to be crystal clear about your objection(s), so that I can address them?
Impenitent
Posts: 4333
Joined: Wed Feb 10, 2010 2:04 pm

Re: Is Arithmetic Circular?

Post by Impenitent »

wtf wrote: Fri Mar 25, 2022 10:57 pm
Impenitent wrote: Fri Mar 25, 2022 9:09 pm I see circularity at equidistant points around the center
What is the center of the sequence 0, 1, 2, 3, 4, 5, ...? Or if that's not what you mean, what do you mean?
you asked "Where do you see circularity?"

-Imp
wtf
Posts: 1178
Joined: Tue Sep 08, 2015 11:36 pm

Re: Is Arithmetic Circular?

Post by wtf »

Impenitent wrote: Mon Mar 28, 2022 3:49 pm
wtf wrote: Fri Mar 25, 2022 10:57 pm
Impenitent wrote: Fri Mar 25, 2022 9:09 pm I see circularity at equidistant points around the center
What is the center of the sequence 0, 1, 2, 3, 4, 5, ...? Or if that's not what you mean, what do you mean?
you asked "Where do you see circularity?"

-Imp
Yes. That's what I asked. In the Peano sequence 0, S0, SS0, SSS0, SSSS0, ..., where do you see circularity?

Or if you are talking about something other than the Peano sequence, what are you talking about? Just wondering.

ps ... Or ... do you mean you were just talking about circularity in general, and you meant that you see circularity in circles? I guess one could read it that way.
Skepdick
Posts: 14366
Joined: Fri Jun 14, 2019 11:16 am

Re: Is Arithmetic Circular?

Post by Skepdick »

wtf wrote: Tue Mar 29, 2022 2:54 am Yes. That's what I asked. In the Peano sequence 0, S0, SS0, SSS0, SSSS0, ..., where do you see circularity?
There is no circularity in the syntax. The circularity is in the semantics of the generator-function which produces the Peano sequence!

Of course, we probably can't even come to an agreement on what it means "to define" something. In particular the problem is with the meaning of "..." in your prefered notation/syntax. By using "..." you are quietly invoking induction without being explicit about the fact that induction is defined in terms of recursion and implemented in terms of addition!

Here is the circular/recursive definition of ℕ in Haskell:

Code: Select all

data ℕ =
   Zero |
   S ℕ
   deriving Show
In context of the above definition "S0" and "SS0" are, in fact meaningless because 0 is undefined in ℕ.
But whether we call it (S Zero), or (S0) - potato/potatoh. It's just bickering over notation/symbols.

Right until we actually attempt to make 0, 1, 2 and 3 meaningful. By defining them...

Code: Select all

fromPeano Zero = 0
fromPeano (S Zero) = 1
fromPeano (S (S Zero)) = 2
fromPeano (S ( S ( S Zero) ) ) = 3

Code: Select all

*Main> fromPeano Zero
0
*Main> fromPeano (S Zero)
1
*Main> fromPeano (S (S Zero))
2
*Main> fromPeano (S (S (S Zero)))
3
Again, nothing controversial here, except for the fact that the shorthands 0,1,2 and 3 are defined, but 4,5,6, ... are NOT defined.
Being undefined the symbols 4,5,6 etc. are all meaningless!

Code: Select all

*Main> fromPeano (S (S (S (S Zero))))
*** Exception: main.hs:(8,1)-(11,34): Non-exhaustive patterns in function fromPeano
In order for you to "define" the shorthands of ALL members of ℕ you actually have to define toPeano as a total function.
Do you have any notations/conventions for defining total functions?

Here's my recursive attempt in Haskell...

Code: Select all

fromPeano Zero = 0
fromPeano (S x) = 1 + fromPeano x

*Main> fromPeano (S (S (S (S (S (S (S (S (S (S Zero))))))))))
10
But how does that even work ?!?!? I am using the symbol "+" in my definition of fromPeano (S x) = 1 + fromPeano x.
Where is the term "+" defined? Surely it's a form of cheating to use addition while we haven't even defined the numbers, let alone arithmetic itself ?!?!

And, naturally, when I start Haskell without its implicit Prelude (knowledge of arithmetic) the compiler complains precisely about that!

Code: Select all

➜  ~ ghci -XNoImplicitPrelude main.hs
GHCi, version 8.10.7: https://www.haskell.org/ghc/  :? for help
[1 of 1] Compiling Main             ( main.hs, interpreted )

main.hs:8:21: error:
    • Variable not in scope: (+) :: t0 -> t -> t
    • Perhaps you want to add ‘+’ to the import list in the import of
      ‘Prelude’ (main.hs:1:1-20).
  |
8 | fromPeano (S x) = 1 + fromPeano x
  |                     ^
So that's (at least) three circularities.
The Natural numbers are recursively defined.
The Numerical shorthands for the Natural numbers are recursively defined.
The the total function which relates the Peano sequences to the corresponding numerical shorthands are recursively defined in terms of addition.

So, please, please, please! Help me define fromPeano x as a total function without invoking arithmetic. All of that circular stuff is hiding right behind your "..."
Last edited by Skepdick on Tue Mar 29, 2022 2:08 pm, edited 10 times in total.
Impenitent
Posts: 4333
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Re: Is Arithmetic Circular?

Post by Impenitent »

wtf wrote: Tue Mar 29, 2022 2:54 am
Impenitent wrote: Mon Mar 28, 2022 3:49 pm
wtf wrote: Fri Mar 25, 2022 10:57 pm

What is the center of the sequence 0, 1, 2, 3, 4, 5, ...? Or if that's not what you mean, what do you mean?
you asked "Where do you see circularity?"

-Imp
Yes. That's what I asked. In the Peano sequence 0, S0, SS0, SSS0, SSSS0, ..., where do you see circularity?

Or if you are talking about something other than the Peano sequence, what are you talking about? Just wondering.

ps ... Or ... do you mean you were just talking about circularity in general, and you meant that you see circularity in circles? I guess one could read it that way.
I was making a general reference

-Imp
Skepdick
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Joined: Fri Jun 14, 2019 11:16 am

Re: Is Arithmetic Circular?

Post by Skepdick »

wtf wrote: Mon Mar 28, 2022 4:17 am In this context the use of the symbols 1, 2, 3, ... is totally incidental. S0, SS0, SSS0, ... is sufficient, and if that is your concern, so be it. Nevermind 1, 2, 3, ...
To make it really easy for you. I have no problem with S0, SS0, SSS0; or 1, 2, 3; or ٠ ١ ٢ ٣ ٤ ٥ ٦ ٧ ٨ ٩; or egyptian numerals. I general - I have no problem with any alphabet, because in all of those sequences you seem to be far more interested in the total order of the sequence and the inequality relations between elements, and not the elements themselves. e.g you'd be pretty surprised if I defined 0 > S0 > SS0 > SSS0,...

All of the above aside it's your "..." that's definitionally vague - the sequence generator is undefined.
If "S0, SS0, SSS0, ..." is a sequence, what function would generate that sequence?

The one below does, but it's defined in terms of co-recursion/subtraction.

Code: Select all

data ℕ =
   Zero |
   S ℕ
   deriving Show

toPeano 0 = Zero
toPeano x = S (toPeano (x-1))

peanoSequence :: [ℕ]
peanoSequence = map toPeano [0..]
If you don't actually care about generating the sequence, then why not just use the null-function? Why introduce 0 when S is sufficient?

S, SS, SSS, ...

Code: Select all

In [1]: def S(x): pass 
In [2]: S(S) 
In [3]: S(S(S))
In [4]: S(S(S(S)))
And still - the question is whetrher the inequality relation is defined or undefined for elements of the sequence.

Are "The Natural Numbers" still "The Natural Numbers" (as we known them) if we define S < SSS < SS< SSSS, or is the total order implicit in ℕ ?
wtf
Posts: 1178
Joined: Tue Sep 08, 2015 11:36 pm

Re: Is Arithmetic Circular?

Post by wtf »

Impenitent wrote: Tue Mar 29, 2022 1:50 pm
I was making a general reference

Ok, thanks.
Skepdick wrote: Tue Mar 29, 2022 10:54 am
There is no circularity in the syntax. The circularity is in the semantics of the generator-function which produces the Peano sequence!
I found this post of yours incomprehensible in its entirety. I'm sure the fault is all mine. I was better able to respond to the points you made in your subsequent response, so I'll just focus on that.
Skepdick wrote: Tue Mar 29, 2022 2:19 pm
To make it really easy for you. I have no problem with S0, SS0, SSS0; or 1, 2, 3; or ٠ ١ ٢ ٣ ٤ ٥ ٦ ٧ ٨ ٩; or egyptian numerals. I general - I have no problem with any alphabet, because in all of those sequences you seem to be far more interested in the total order of the sequence and the inequality relations between elements, and not the elements themselves. e.g you'd be pretty surprised if I defined 0 > S0 > SS0 > SSS0,...
I wouldn't be surprised at all. The natural numbers in their usual order are called ω, that's the Greek lower-case omega. If you reverse the order, you get an ordering of the natural numbers called ω*. The '*' says to take ω and just reverse its order. As you can see it's familiar enough to mathematicians to have a name. Ain't no thang. In fact it can be represented as the non-positive integers in their usual order 0 > -1 > -2 > -3 > ... so this is a very familiar order relation.

Of course those would no longer be the usual Peano natural numbers. They'd be an alternative way of ordering the set-theoretic natural numbers.

I do appreciate that you wrote "0 > S0 > SS0 > SSS0,..." freely using "..." for its obvious shorthand meaning.
Skepdick wrote: Tue Mar 29, 2022 2:19 pm
All of the above aside it's your "..." that's definitionally vague -
Oh I see. You only object to MY usage of "..." Got it.
Skepdick wrote: Tue Mar 29, 2022 2:19 pm the sequence generator is undefined.
The sequence is defined perfectly well by the Peano axioms. As I asked before, are you objecting to the work of Giuseppe Peano? Don't shoot me, I'm only the Peano player!
Skepdick wrote: Tue Mar 29, 2022 2:19 pm If "S0, SS0, SSS0, ..." is a sequence, what function would generate that sequence?
I don't see why this is relevant. Suppose I generate a random bit, 1, or 0, say by flipping a fair coin. I do it once for each natural number, to generate a bitstring like 0010101010111000101...

What function "generates" that bitstring? It's true that there is a function that maps each positive integer n to a bit, 1 or 0. But there is no computable function that does so, since by assumption we have generated these bits randomly.

A nice computer science-y example is the Chaitin's constant Omega. It's a definable real number; but there is no computable function that determines the bits, because such a computable function would in effect solve the Halting problem, which can't be done.

So you are simply wrong that given a sequence there must be a computable generating function. Some sequences (almost all of them in fact, in the usual technical sense) are entirely random and not generated by any computable process. You probably know that, so I wonder why you would act as if you don't.
Skepdick wrote: Tue Mar 29, 2022 2:19 pm The one below does, but it's defined in terms of co-recursion/subtraction.
Are you saying that Peano's axioms are flawed? I wonder why you don't take this up with the Wiki editors or perhaps the estate of Giuseppe Peano. Why are you going on about this with me? All I did was ask @Eodnhoj7 why he thought S0, SS0, etc., is circular. That's all I said. Then you jumped in with these irrelevant code blocks. I have no idea where you're coming from or what point you're trying to make.
Skepdick wrote: Tue Mar 29, 2022 2:19 pm
Are "The Natural Numbers" still "The Natural Numbers" (as we known them) if we define S < SSS < SS< SSSS, or is the total order implicit in ℕ ?
ℕ is a set. It has no inherent order. Any order is as good as another.

ω is the pair (ℕ, <) where < is the usual order 0, 1, 2, 3, ... Forgive the dots, I'm sure you know what I mean.

Now the order 0, 1, 3, 2, 4, ... is a different order. But if you think about it, it's order-isomorphic to the usual order 0, 1, 2, 3, 4, ... That is, if we simply call 2 by the name '3', and 3 by the name '2', it's exactly the same order. So it's a different order, but the same order type.

To give an example of a different order type, what if I ordered the natural numbers like this:

0, 1, 2, 4, 5, 6, 7, ..., 3

Then everything is as it usually is, except that 3 is the largest number. Since the natural numbers in their usual order have no largest element, this funny order is a different order type. In fact it's the order type ω + 1. That is, it's the successor to ω. It's what you get when you count all the natural numbers and then keep on going.

This is all part of the subject known as the ordinal numbers, a fascinating object of study.

But if you use the notation 0, S0, SS0, etc., the implication is that S is the Peano successor function, which uniquely gives (generates if you like) the sequence 0, 1, 2, 3, 4, etc.

By the way at some point you asked why we need 0, why can't we just say S, SS, SSS. The reason is that your induction or recursive definition requires a base case! I suspect that you know this too. Your sequence S, SSS, SS, SSSS, ... isn't well-defined since the symbols have been given no definition.
Skepdick
Posts: 14366
Joined: Fri Jun 14, 2019 11:16 am

Re: Is Arithmetic Circular?

Post by Skepdick »

wtf wrote: Wed Mar 30, 2022 2:57 am I wouldn't be surprised at all. The natural numbers in their usual order are called ω, that's the Greek lower-case omega. If you reverse the order, you get an ordering of the natural numbers called ω*. The '*' says to take ω and just reverse its order.
Now that's a super-peculiar idea. I have no problem reasoning about and reversing objects of finite size, but I have no idea what it means to "reverse" something with a start and no end.

Because after all "first n elements of X" means "last n elements of reversed X". Is just that "last element" is a rather strange notion in an inductive sequence, where there is always another element. And another. And another.

Code: Select all

In [1]: order = [0,1,2,3,4,5,6]

In [2]: list(reversed(order))[-1::-2] == order[0::2]
Out[2]: True
So if I am to use Python's notation here I expect the first two elements in ω to correspond to the the last two element in ω*

Code: Select all

ω[0::2] == ω*[-1::-2]
Mapping an finite sequence to its reverse is quite trivial.

Code: Select all

In [3]: list(zip(order, list(reversed(order))))
Out[3]: [(0, 6), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 0)]
None of this bothers me with finite sequences, but I am super curious what the first two pairs of the following function are...

Code: Select all

In [3]: zip(ω, ω*)
Could you state them explicitly? I am having trouble with the 2nd element of each pair.

(0, ???), (S0, ???)
wtf wrote: Wed Mar 30, 2022 2:57 am In fact it can be represented as the non-positive integers in their usual order 0 > -1 > -2 > -3 > ... so this is a very familiar order relation.

Of course those would no longer be the usual Peano natural numbers. They'd be an alternative way of ordering the set-theoretic natural numbers.
I notice that in addition to the Peano numbers you've also brought the set-theoretic natural numbers in to the discussion - this doesn't really help anything because they are recursively/inductively defined also. You know, that pesky "...".

0 = {} , 1 = {0}, 2 = {0,1}

Or in general:
n = {0, 1, …, n − 1}
n + 1 = n ∪ {n}

So the set-theoretic definition of "the natural numbers" is appealing to arithmetic in order to define the natural numbers! Is arithmetic a priori numbers, or isn't it? And I won't even get stuck into the (lacking) definition of ∪.

I mean, if set theoreticians are happy to use the expressions "n", "n-1", "n+1" and "..." in defining "the numbers" then surely I can write down: ..., (n + 1) + 1 < n + 1 < n < n - 1 < (n - 1) - 1, ....

What does that make it? ℤ*?

And, of course - I just have to ask the question: What are the first and last elements of zip(ℤ,ℤ*)?
wtf wrote: Wed Mar 30, 2022 2:57 am I do appreciate that you wrote "0 > S0 > SS0 > SSS0,..." freely using "..." for its obvious shorthand meaning.
Skepdick wrote: Tue Mar 29, 2022 2:19 pm
All of the above aside it's your "..." that's definitionally vague -
Oh I see. You only object to MY usage of "..." Got it.
Difference being that I am opnenly admitting my usage of "..." is circular. It's recursive/inductive.

In ordered theories with a mimimum element I use "..." to indicate recursion: n, n+1, n+2, ...
In ordered theories with a maximum element I use "..." to indicate corecursion: n, n-1, n-2, ...
And in ordered theories without a minimum/maximum elements I am using it corecursively and recursively respectively: ..., -3, -2, -1, 0, 1, 2, 3, ...

Something which you seem unwilling to admit.
wtf wrote: Wed Mar 30, 2022 2:57 am The sequence is defined perfectly well by the Peano axioms. As I asked before, are you objecting to the work of Giuseppe Peano? Don't shoot me, I'm only the Peano player!
So what's the axiomatic definition of "..."?
wtf wrote: Wed Mar 30, 2022 2:57 am I don't see why this is relevant. Suppose I generate a random bit, 1, or 0, say by flipping a fair coin. I do it once for each natural number, to generate a bitstring like 0010101010111000101...

What function "generates" that bitstring?
This function does.

Code: Select all

def generator():
	return "0010101010111000101..."
For any sequence, number, set; or any entity whatsoever in the Mathematical universe that you claim "exists" I can trivially assume the unit type.
wtf wrote: Wed Mar 30, 2022 2:57 am It's true that there is a function that maps each positive integer n to a bit, 1 or 0. But there is no computable function that does so, since by assumption we have generated these bits randomly.
I don't understand what you mean by "computable" here. Perhaps you mean "deterministic"? Reproducible? A pure function?
A function that returns a static mapping every time?

Your nomenclature is confusing from the get go. There can be no such thing as a randomness in pure mathematics.
A mathematical function must return the same result for the same input - which literally defeas the purpose of random number generators.

However, outside of the domain of "purity" there is a function which is computable in the sense that it runs on a computer which generates an infinite-length random bitstring.
In a lazy-evaluated language such as Haskell you can define such function, you can bind that function to a variable and you can access its n-th element of such a function which is as close to Turing's notion of "computable function" as you can get.

The caveat being you can only get the n-th element of a random bitstring just once.

If you want to do it twice then you have to record the random string and replay it. Bit by bit.

wtf wrote: Wed Mar 30, 2022 2:57 am A nice computer science-y example is the Chaitin's constant Omega. It's a definable real number; but there is no computable function that determines the bits, because such a computable function would in effect solve the Halting problem, which can't be done.
Definable real number. Lol. We've had that discussion.
wtf wrote: Wed Mar 30, 2022 2:57 am So you are simply wrong that given a sequence there must be a computable generating function.
How could I possibly be wrong? Given any sequence where did the sequence originate from?
Even if that something is a Mathematician the sequence has a generator.

Sequences don't appear from thin air.
wtf wrote: Wed Mar 30, 2022 2:57 am Some sequences (almost all of them in fact, in the usual technical sense) are entirely random and not generated by any computable process.
Now THAT is an extraordinary claim and I am going to demand extraordinary evidence.

1. What is the origin of the sequences you are speaking about?
2. Even if we ignore the origin and assume that all of those sequences "just exist"... how do you know they are "random"?

In particular I would love to see your random(X) predicate which returns a Boolean for any given sequence X.

Because from my limited knowledge, I am pretty damn sure randomness is not provable. In the usual, technical sense of "provable".
wtf wrote: Wed Mar 30, 2022 2:57 am You probably know that, so I wonder why you would act as if you don't.
Probably because I don't know that.

There is literally no way for me know that any particular sequence is random. Not without understanding how the sequence was produced.
Least I (eventually) find a computable function that happens to generate that particular sequence.

But, of course, there is one particular function that can generate ANY particular sequence it has already seen!

Code: Select all

def f(x):
   y = x
   return y
You have invinite memory, right? Variable assignment (data copying) is free, right?
wtf wrote: Wed Mar 30, 2022 2:57 am Are you saying that Peano's axioms are flawed? I wonder why you don't take this up with the Wiki editors or perhaps the estate of Giuseppe Peano. Why are you going on about this with me? All I did was ask @Eodnhoj7 why he thought S0, SS0, etc., is circular. That's all I said. Then you jumped in with these irrelevant code blocks. I have no idea where you're coming from or what point you're trying to make.
I am not saying that at all. I am saying ALL of Mathematics is circular/recursive.

I explained my point as precisely as possible.

"0, S0, SS0" is not circular.
"0, S, SS0, ..." is circular.
"n, n+1, n+2 , ..." is circular.

Because infinite induction is the same thing as non-halting recursion. Computer scientists call them generators.

Obviously - because that is how production rules work!
wtf wrote: Wed Mar 30, 2022 2:57 am ω is the pair (ℕ, <) where < is the usual order 0, 1, 2, 3, ... Forgive the dots, I'm sure you know what I mean.
Indeed, I know what you mean. Too bad you can't define it without falling into circularity.
wtf wrote: Wed Mar 30, 2022 2:57 am Now the order 0, 1, 3, 2, 4, ... is a different order. But if you think about it, it's order-isomorphic to the usual order 0, 1, 2, 3, 4, ... That is, if we simply call 2 by the name '3', and 3 by the name '2', it's exactly the same order. So it's a different order, but the same order type.
Well. Why can't we say the exact same thing about ALL numbers then?

Any order is order-isomorphic to any random shuffling of itself. Is just permutations/combinatorics!

Code: Select all

In [1]: from random import shuffle
In [2]: order = [1,2,3,4,5,6]
In [3]: shuffle(order); order
Out[3]: [5, 1, 3, 6, 4, 2]
In [4]: shuffle(order); order
Out[4]: [3, 5, 4, 6, 1, 2]
In [5]: shuffle(order); order
Out[5]: [2, 4, 1, 3, 5, 6]
Heck, why even stop there?

Any and all permutations of representations of ℕ, ℤ, ℝ, ℂ will be order-isomorphic if you remember the new names of everything.
wtf wrote: Wed Mar 30, 2022 2:57 am To give an example of a different order type, what if I ordered the natural numbers like this:

0, 1, 2, 4, 5, 6, 7, ..., 3

Then everything is as it usually is, except that 3 is the largest number. Since the natural numbers in their usual order have no largest element, this funny order is a different order type. In fact it's the order type ω + 1. That is, it's the successor to ω. It's what you get when you count all the natural numbers and then keep on going.
An infinity with a largest AND a smallest element makes no sense to me. And your "..." really is not helping.

If you grammar allows you to say 7, ..., 3 then surely you also say 0,...,1? Then everything is as it usually is, except 0 is the smallest element and 1 is the largest element.

And since 1 is the successor to 0, ω+1 is the successor to ω then how about this ordering: ω, ..., ω + 1 ?

That pesky "..."
wtf wrote: Wed Mar 30, 2022 2:57 am By the way at some point you asked why we need 0, why can't we just say S, SS, SSS. The reason is that your induction or recursive definition requires a base case! I suspect that you know this too. Your sequence S, SSS, SS, SSSS, ... isn't well-defined since the symbols have been given no definition.
That is a super super-strange criticism.

0 is the base case in the sequence 0, S0, SS0, SSS0, ....
What is the definition of the symbol 0?
What is the definition of the symbol S?

If we are going to define S, why do we have to define 0 also? Just make S recursively call itself.

S, SS, SSS, SSSS, ....

Unit types all the way down. Or up. Whatever.

Code: Select all

In [1]: def S(n=None):
   ...:     while True:
   ...:         yield S(n)
   ...:

In [2]: S()
Out[2]: <generator object S at 0x7fef8933e6d0>

In [3]: S(S())
Out[3]: <generator object S at 0x7fef892e7ac0>

In [4]: S(S(S()))
Out[4]: <generator object S at 0x7fef892e7b30>
Skepdick
Posts: 14366
Joined: Fri Jun 14, 2019 11:16 am

Re: Is Arithmetic Circular?

Post by Skepdick »

wtf wrote: Wed Mar 30, 2022 2:57 am I wouldn't be surprised at all. The natural numbers in their usual order are called ω,
...
Now the order 0, 1, 3, 2, 4, ... is a different order. But if you think about it, it's order-isomorphic to the usual order 0, 1, 2, 3, 4, ...
wtf wrote: Wed Mar 30, 2022 2:57 am ℕ is a set. It has no inherent order. Any order is as good as another.
All in all - I just can't reconcile the above ideas into anything coherent.

If ℕ is a set which has no order whatsoever and 0,1,2,3,... are just arbitrary placeholders then any number is exactly the same as every other number, therefore it's absolutely fair to say 0 = 1 = 2 = 3 = ... ; or 0 = S0 = SS0 = SSS0; or a = b = c = d =... in ℕ but not in ω.

e.g if x and y are unique natural numbers then x = y in ℕ ; and x != y in ω
Eodnhoj7
Posts: 8595
Joined: Mon Mar 13, 2017 3:18 am

Re: Is Arithmetic Circular?

Post by Eodnhoj7 »

wtf wrote: Fri Mar 25, 2022 7:47 pm
Eodnhoj7 wrote: Thu Mar 24, 2022 11:44 pm S(0),S(S(0)),S(S(S(0))) is the circularity of S
Starting at 0, this gives you the sequence 0, 1, 2, 3, 4, 5, 6, ...

I see no circularity. Where do you see circularity?
The repetition of 1 within 1, 2, 3, 4 as well as the repetition of S in S(0),S(S(0)),S(S(S(0))); repetition is circularity.

That leaves us with 0 at first glance being non-circular, however this is not the case. The zero of zero, as the void of void (considering 0 and void are the same), is a self-negating statement that results in 1 thus is circular.
Last edited by Eodnhoj7 on Wed Mar 30, 2022 11:52 pm, edited 1 time in total.
Eodnhoj7
Posts: 8595
Joined: Mon Mar 13, 2017 3:18 am

Re: Is Arithmetic Circular?

Post by Eodnhoj7 »

wtf wrote: Fri Mar 25, 2022 10:57 pm
Impenitent wrote: Fri Mar 25, 2022 9:09 pm I see circularity at equidistant points around the center
What is the center of the sequence 0, 1, 2, 3, 4, 5, ...? Or if that's not what you mean, what do you mean?
The sequence 0, 1, 2, 3, 4, 5, ... is incomplete. In reality it is -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ... thus the center is zero; the center is empty.
Eodnhoj7
Posts: 8595
Joined: Mon Mar 13, 2017 3:18 am

Re: Is Arithmetic Circular?

Post by Eodnhoj7 »

Skepdick wrote: Wed Mar 30, 2022 2:39 pm
wtf wrote: Wed Mar 30, 2022 2:57 am I wouldn't be surprised at all. The natural numbers in their usual order are called ω,
...
Now the order 0, 1, 3, 2, 4, ... is a different order. But if you think about it, it's order-isomorphic to the usual order 0, 1, 2, 3, 4, ...
wtf wrote: Wed Mar 30, 2022 2:57 am ℕ is a set. It has no inherent order. Any order is as good as another.
All in all - I just can't reconcile the above ideas into anything coherent.

If ℕ is a set which has no order whatsoever and 0,1,2,3,... are just arbitrary placeholders then any number is exactly the same as every other number, therefore it's absolutely fair to say 0 = 1 = 2 = 3 = ... ; or 0 = S0 = SS0 = SSS0; or a = b = c = d =... in ℕ but not in ω.

e.g if x and y are unique natural numbers then x = y in ℕ ; and x != y in ω
Agreed.
wtf
Posts: 1178
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Re: Is Arithmetic Circular?

Post by wtf »

Eodnhoj7 wrote: Wed Mar 30, 2022 11:50 pm
The sequence 0, 1, 2, 3, 4, 5, ... is incomplete. In reality it is -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ... thus the center is zero; the center is empty.
Since the sequence is infinite in both directions, how do you know the center isn't, say, 6? After all there's a bijection between the set of integers less than 6, and the set of integers greater than 6.
wtf
Posts: 1178
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Re: Is Arithmetic Circular?

Post by wtf »

Skepdick wrote: Wed Mar 30, 2022 10:40 am
Now that's a super-peculiar idea.
I may not respond any more after this post. I understand that you're expressing an objection to the Peano axioms or to mathematics in general, but I am not the lord high defender of the realm. I'm just trying to explain a few things. I supplied a number of links you might find of interest.

If you think math is circular, you're probably not the only one. And if you don't think the Peano axioms are ultimately sensible, there are finitists and ultra-finitists who agree with you. I could never talk you out of those opinions, nor do I even want to.

Skepdick wrote: Wed Mar 30, 2022 10:40 am I have no problem reasoning about and reversing objects of finite size, but I have no idea what it means to "reverse" something with a start and no end.
Reversing a binary relation is easy. A binary relation is a set of ordered pairs (a,b) indicating that a and b are in a given relation. The reverse relation is the set of corresponding pairs (a,b). If a < b, then in the reverse order, b < a. That is, if (a,b) is an ordered pair in the relation, then (b,a) is an ordered pair in the inverse relation. This is extremely basic.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
Because after all "first n elements of X" means "last n elements of reversed X". Is just that "last element" is a rather strange notion in an inductive sequence, where there is always another element. And another. And another.
ω* is not an inductive relation, obviously.
Skepdick wrote: Wed Mar 30, 2022 10:40 am I am super curious what the first two pairs of the following function are...

Code: Select all

In [3]: zip(ω, ω*)
Since ω* is not a Python list, this expression is undefined and would throw a runtime error. Surely you can see that ω* is not defined inductively, and is not a Python list.

Skepdick wrote: Wed Mar 30, 2022 10:40 am
(0, ???), (S0, ???)
You've already gotten a Python error, since the second argument to zip() is not a legal Python list. Nor are we even talking about programming languages. You want to force a mathematical discussion into the context of programming languages, but it's a poor fit. This effort is the source of the confusions you're experiencing, and the frustration I'm having trying to respond.

In any event the < relation is a binary relation, and a binary relation is a set of ordered pairs, and you can reverse any binary relation by reversing all the ordered pairs. And of course it's perfectly clear that if you reverse an inductive relation, the reversed relation is not going to be inductive.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
I notice that in addition to the Peano numbers you've also brought the set-theoretic natural numbers in to the discussion - this doesn't really help anything because they are recursively/inductively defined also. You know, that pesky "...".
The dots are only an informal shorthand. They're not a part of the formal Peano axioms.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
Or in general:
n = {0, 1, …, n − 1}
n + 1 = n ∪ {n}

So the set-theoretic definition of "the natural numbers" is appealing to arithmetic in order to define the natural numbers!
I see no arithmetic here, only the set-theoretic operations of going from n to {n}, and forming a union.
Skepdick wrote: Wed Mar 30, 2022 10:40 am Is arithmetic a priori numbers, or isn't it? And I won't even get stuck into the (lacking) definition of ∪.
It's perfectly well defined by the axiom of union.
Skepdick wrote: Wed Mar 30, 2022 10:40 am I mean, if set theoreticians are happy to use the expressions "n", "n-1", "n+1" and "..." in defining "the numbers" then surely I can write down: ..., (n + 1) + 1 < n + 1 < n < n - 1 < (n - 1) - 1, ....

What does that make it? ℤ*?
What? Not sure what you're getting at. Given the usual order on Z you can certainly define Z*, the reverse order. What's the problem.
Skepdick wrote: Wed Mar 30, 2022 10:40 am And, of course - I just have to ask the question: What are the first and last elements of zip(ℤ,ℤ*)?
In this case neither Z nor Z* are Python lists, so you'd get two runtime errors here. Z does not have a first element so it is not a Python list. Likewise for Z*.
Skepdick wrote: Wed Mar 30, 2022 10:40 am All of the above aside it's your "..." that's definitionally vague -
And yours isn't. As I've noted.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
Difference being that I am opnenly admitting my usage of "..." is circular. It's recursive/inductive.
What? Recursive/inductive constructions are not circular. You've made this error a couple of times. The whole point of recursion/induction is the base case to avoid circularity and lack of well-foundedness.

Skepdick wrote: Wed Mar 30, 2022 10:40 am In ordered theories with a mimimum element I use "..." to indicate recursion: n, n+1, n+2, ...
In ordered theories with a maximum element I use "..." to indicate corecursion: n, n-1, n-2, ...
And in ordered theories without a minimum/maximum elements I am using it corecursively and recursively respectively: ..., -3, -2, -1, 0, 1, 2, 3, ...

Something which you seem unwilling to admit.
I don't recall your asking me to admit it. Doesn't seem connected to the rest of the conversation, you lost me.

Skepdick wrote: Wed Mar 30, 2022 10:40 am
This function does.

Code: Select all

def generator():
	return "0010101010111000101..."
Really? Where are you going to store those infinitely many bits? Are you going to keep running down to Best Buy for more memory? Crucial has better prices.

Skepdick wrote: Wed Mar 30, 2022 10:40 am
For any sequence, number, set; or any entity whatsoever in the Mathematical universe that you claim "exists" I can trivially assume the unit type.
Actually not. If a sequence is algorithmically compressible -- that is, if it can be generated by a program or Turing machine -- then you can store it in a computer and generate as many bits of it as you like, subject to time/space constraints. However very few bitstrings are algorithmically compressible. After all there are only countably many Turing machines and uncountably many bitstrings.

Give this a read.

https://en.wikipedia.org/wiki/Algorithm ... ion_theory

Skepdick wrote: Wed Mar 30, 2022 10:40 am I don't understand what you mean by "computable" here.
Really? Computable real numbers were defined by Turing in 1936. A real number is computable if its digits can be cranked out by a Turing machine; that is, if there is a TM that, when the positive integer n is input, halts with the n-th digits of the decimal (or binary if you like) digit of the number.

Since there are only countably many TMs, and uncountably many real numbers, "almost all," in the sense of all but countably many real numbers, are not computable.

Skepdick wrote: Wed Mar 30, 2022 10:40 am Perhaps you mean "deterministic"? Reproducible? A pure function?
A function that returns a static mapping every time?
When I said computable, I meant computable in the sense of Turing 1936. What other definition is there? It's the only one.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
Your nomenclature is confusing from the get go.
Turing 1936. As the poet said, "That is all ye know, and all ye need to know."
Skepdick wrote: Wed Mar 30, 2022 10:40 am
There can be no such thing as a randomness in pure mathematics.
Oh my gosh. Ever hear of random variables? But in this context, algorithmic information theory is the operative context. A bitstring is random if it's not computable. That's from Kolmogorov. Please give this a read.

The idea is that if I have 10101010101010... I can say, "Start with 1 and alternate 1's and 0's." That's a short string that generates an infinite sequence. So the sequence is compressible, and not random. But a random sequence can't be compressed. That's literally the definition.

https://en.wikipedia.org/wiki/Kolmogorov_complexity
Skepdick wrote: Wed Mar 30, 2022 10:40 am A mathematical function must return the same result for the same input - which literally defeas the purpose of random number generators.
The bits of Chaitin's constant are determined (in the sense of being definable) but noncomputable. It's in the Wiki page on Chaitin's constant.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
However, outside of the domain of "purity" there is a function which is computable in the sense that it runs on a computer which generates an infinite-length random bitstring.
In a lazy-evaluated language such as Haskell you can define such function, you can bind that function to a variable and you can access its n-th element of such a function which is as close to Turing's notion of "computable function" as you can get.
Here's what gets me. You toss out Haskell and Python and generators etc. but you seem to have no familiarity with binary relations or algorithmic randomness or computability. From my point of view there's a mismatch. But I hope some of my links are helpful.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
The caveat being you can only get the n-th element of a random bitstring just once.
Well how many times do you need to get it? It's an arbitrary bitstring.

Skepdick wrote: Wed Mar 30, 2022 10:40 am
If you want to do it twice then you have to record the random string and replay it. Bit by bit.
Yes ok. I have a lot of pencils, I'll write down the bitstring then play it back. I think you're kind of hung up on the wrong thing here. There's some function that inputs a natural number n and outputs a bit, one or zero. That's a function, because it's deterministic. Give me n and I'll give you back the n-th bit. But there is no PROGRAM that can generate the bits. It's a function, but it's not a computable function. In other words God (or an oracle, if you prefer) can tell you the n-th bit; but there is no program that can tell you the n-th bit. Is that a helpful analogy?

Skepdick wrote: Wed Mar 30, 2022 10:40 am
Definable real number. Lol. We've had that discussion.
I don't recall that, but it's a perfectly sensible notion. You should read the Wiki page on Chaitin's constant. It turns out that there are numbers that are first-order definable in logic, yet not computable. Chaitin's constant is one such.

The distinction between definability and computability is quite subtle, and I'm not an expert. But there are numbers that are definable yet not computable.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
How could I possibly be wrong?
LOL. Thanks for the chuckle.

Skepdick wrote: Wed Mar 30, 2022 10:40 am
Given any sequence where did the sequence originate from?
Coin flip, cosmic ray, who cares? There are uncountably many bitstrings, and only countably many Turing machines or Python programs. So all but countably many bitstrings are uncomputable; that is, Kolmogorov-random.

But you know there's an important distinction here. You want every bitstring to be generated or to "come from" somewhere. In math, we are not so much concerned about that. We have a set, N, the set of natural numbers. We have the set of functions N -> {0,1}, the set of all functions that input a natural number and output a bit. The entire set of all possible such functions has mathematical existence in set theory. There are uncountably many such functions. We are not required to give an account of where these functions "come from." There's nobody in the basement hammering them out. They exist because the set N exists, and the set {0,1} exists, and the set of all functions from N to {0,1} exists.

Here "exists" means, "exists according to the rules of set theory." That's of course a very nebulous form of existence. But if you don't like it or don't believe in it, I hope you can at least accept that mathematicians believe in it and, for the sake of discussion, accept it on its own terms.

Skepdick wrote: Wed Mar 30, 2022 10:40 am
Even if that something is a Mathematician the sequence has a generator.
No, because a generator is a thing in programming language theory and must be computable. But most sequences are not computable hence can not have a generator.

Skepdick wrote: Wed Mar 30, 2022 10:40 am Sequences don't appear from thin air.
The set of bitstrings exists as an abstract set. It's the set of functions from the natural numbers to the set {0.1}. It's an uncountable set.

Skepdick wrote: Wed Mar 30, 2022 10:40 am
Now THAT is an extraordinary claim and I am going to demand extraordinary evidence.
I've already given the proof several times. There are uncountably many bitstrings but only countably many Turing machines, hence "almost all" -- defined in this context as "all but countably many" -- are noncomputable. They can't have generators, they can't be cranked out by any algorithm.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
1. What is the origin of the sequences you are speaking about?
Just pick any noncomputable sequence. But for concreteness, use the very special case of Chaitin's constant, which is definable but not computable.

But again, you raise the question of the "origin" of these sequences. Their origin is set-theoretic. We have a set, which is commonly denoted 2^N, where ^ denotes exponentiation, and which represents the set of all functions from N to the set {0,1}. There are uncountably many such functions, via the standard diagonal argument. I agree with you that mathematicians posit existence with breathless grandiosity. Where do they "come from?" It's not a bad question. They come from the axioms of set theory.
Skepdick wrote: Wed Mar 30, 2022 10:40 am 2. Even if we ignore the origin and assume that all of those sequences "just exist"... how do you know they are "random"?
Because there are uncountably many sequences of 0's and 1's, and only countably many Turing machines, Python programs, Haskell programs, or algorithms. That's the proof.
Skepdick wrote: Wed Mar 30, 2022 10:40 am In particular I would love to see your random(X) predicate which returns a Boolean for any given sequence X.
After all this, not a bad question. Actually not sure. Given a bitstring, how do I know whether it corresponds to some algorithm. Not sure at the moment. But why does it matter? I've already shown several times that these sequences exist in great abundance.
Skepdick wrote: Wed Mar 30, 2022 10:40 am Because from my limited knowledge, I am pretty damn sure randomness is not provable. In the usual, technical sense of "provable".
That would be news to Kolmogorov and Chaitin. And Turing. And everyone else who's taken the trouble to understand their work. Or even glance at the Wiki pages describing their work. I don't want to belabor this point already, I've explained it several times.

Skepdick wrote: Wed Mar 30, 2022 10:40 am
Probably because I don't know that.
Ok. Well I hope I've been able to shed some light. Turing, Kolmogorov, Chaitin. Uncountably many bitstrings, only countably many Turing machines. Hence "almost all" bitstrings have no generator, can't have their digits cranked out by a TM.
Skepdick wrote: Wed Mar 30, 2022 10:40 am There is literally no way for me know that any particular sequence is random.
This is true. We'd have to see infinitely many of its bits, and we can't do that. This is an epistemological problem, but not an ontological one. Even given an infinite sequence I'm not sure how to determine whether it's computable or not.

There are analogous problems in math. For example we don't know whether pi + e (where e = 2.71etc) is rational or not. An amazing thing not to be able to know, but it's an open problem.

So I agree with you that given an infinite bitstring, we can't look at it and know whether it's computable or not. But that does not matter! Why do you think it does?

Skepdick wrote: Wed Mar 30, 2022 10:40 am
ot without understanding how the sequence was produced.
Well if it's produced by an algorithm, then it's not random!

Skepdick wrote: Wed Mar 30, 2022 10:40 am Least I (eventually) find a computable function that happens to generate that particular sequence.
I agree, but it doesn't matter. A sequence is random or not independently of our ability to know that fact. Just like pi + e is rational or not even though it's an impossibly difficult problem to solve.

Skepdick wrote: Wed Mar 30, 2022 10:40 am But, of course, there is one particular function that can generate ANY particular sequence it has already seen!
What?
Skepdick wrote: Wed Mar 30, 2022 10:40 am

Code: Select all

def f(x):
   y = x
   return y
You have invinite memory, right? Variable assignment (data copying) is free, right?
Of course, made several trips to Best Buy. In any event this seems to be the identity function. I am not following your point.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
I am not saying that at all. I am saying ALL of Mathematics is circular/recursive.
But circular and recursive are not the same thing at all. What do you mean?
Skepdick wrote: Wed Mar 30, 2022 10:40 am
I explained my point as precisely as possible.

"0, S0, SS0" is not circular.
"0, S, SS0, ..." is circular.
"n, n+1, n+2 , ..." is circular.
Well there are philosophers who consider the Peano axioms to be circular, in the sense that if I ask for the Peano representation of 11 trillion, one would have to go SSSSS...SSSS0, but how did I know how many S's to count? So there's perhaps a valid point to be made, or at least some finitist or ultra-finitist philosophers would so argue.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
Because infinite induction is the same thing as non-halting recursion. Computer scientists call them generators.
Generators are by definition algorithms, and there aren't enough algorithms to generate all the possible bitstrings. Do you take this point?
Skepdick wrote: Wed Mar 30, 2022 10:40 am
Obviously - because that is how production rules work!
Productions (for example in formal language or automata theory) are algorithms. There are only countably many of them. There are uncountably many binary sequences or bitstrings.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
Indeed, I know what you mean. Too bad you can't define it without falling into circularity.
The dots are never part of the formal descriptions, either in set theory or the Peano axioms.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
wtf wrote: Wed Mar 30, 2022 2:57 am Now the order 0, 1, 3, 2, 4, ... is a different order. But if you think about it, it's order-isomorphic to the usual order 0, 1, 2, 3, 4, ... That is, if we simply call 2 by the name '3', and 3 by the name '2', it's exactly the same order. So it's a different order, but the same order type.
Well. Why can't we say the exact same thing about ALL numbers then?
There is only one order type equivalent to ω, even though there are many distinct representations.
wtf wrote: Wed Mar 30, 2022 2:57 am Any order is order-isomorphic to any random shuffling of itself. Is just permutations/combinatorics!
No! I already gave you the counterexample. 0, 1, 2, 3, ... is not the same order type as 1, 2, 3, 4, ..., 0. There is no order-isomorphism between those two ordered sets, even though there is a bijection aka permutation.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
Any and all permutations of representations of ℕ, ℤ, ℝ, ℂ will be order-isomorphic if you remember the new names of everything.
No, I just showed the counterexample. Besides, there is no order relation on the complex numbers that respects the algebraic operations.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
An infinity with a largest AND a smallest element makes no sense to me. And your "..." really is not helping.
You just study some order theory and the ordinal numbers. But here is an algorithm for the order type 0, 1, 2, 4, 5, 6, ..., 3, which we'll call <' for "funny less than."

* If n and m are not 3, then n <' m if n < m.

* If m = 3 and n isn't three, then n <' m.

[See below for a Python example that makes this more clear].

You could program this and it would compute <' perfectly. Everything's funny-less-than 3 (except for 3) and other than that, everything's the way it usually is. This simply defines an alternate order relation on the natural numbers. And in particular, it's NOT the same order type as the usual order, even though it's in bijection with the usual order.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
If you grammar allows you to say 7, ..., 3 then surely you also say 0,...,1? Then everything is as it usually is, except 0 is the smallest element and 1 is the largest element.
You lost me here. And perhaps I lost you.

0, 1, 2, 4, 5, 6, ..., 7 is just a shorthand for the funny-order that says that everything is the same as it usually is, but everything other than 7 is funny-less-than 7. You could easily code that up in Python, right? No problem at all.

Code: Select all

def funny_less_than(n, m) :
    if n == 7 :
       return False
    if m == 7 and n != 7 then :
        return True
    return n < m
Is that more clear? Now this imposes an alternate order on the natural numbers with the property that there's a smallest element, namely 0, which is smaller than everything except itself; and a largest element, namely 7, which is larger than everything except itself. So we have permuted the natural numbers and not only changed the order, but also changed the order type.

Conclusion: All non-identity permutations change the order. But SOME of them change the order type, and others don't. For example 0, 1, 3, 2, 4, 5, 6, ... changes the order, but the order type is the same. There's an order-preserving bijection between this and the usual order.
Skepdick wrote: Wed Mar 30, 2022 10:40 am
And since 1 is the successor to 0, ω+1 is the successor to ω then how about this ordering: ω, ..., ω + 1 ?

That pesky "..."
Well that's not how ordinals work. ω + 1 is the immediate successor to ω, that's how it's defined. There's nothing between them.

And, in case you are concerned that I am using the symbol +, which implies that arithmetic is prior, this too can be fixed. Actually ω+1 is the set-theoretic union of ω. Without going into detail, you can take my word for it that the set theory is logically prior to the arithmetic without any logical circularity.
Last edited by wtf on Thu Mar 31, 2022 7:01 am, edited 13 times in total.
wtf
Posts: 1178
Joined: Tue Sep 08, 2015 11:36 pm

Re: Is Arithmetic Circular?

Post by wtf »

Skepdick wrote: Wed Mar 30, 2022 2:39 pm
If ℕ is a set which has no order whatsoever and 0,1,2,3,... are just arbitrary placeholders then any number is exactly the same as every other number, therefore it's absolutely fair to say 0 = 1 = 2 = 3 = ... ; or 0 = S0 = SS0 = SSS0; or a = b = c = d =... in ℕ but not in ω.

e.g if x and y are unique natural numbers then x = y in ℕ ; and x != y in ω
No, that's not right at all.

I'm sure you know that a set, without any additional structure, has no inherent order. The set {banana, orange, apple} is the same as the set {orange, apple, banana}. A set is characterized only by its elements.

Now apple, orange, and banana are not mere "placeholders." They are elements of the set. Just because the order doesn't matter in the set, doesn't mean the elements can't be distinguished.

If you're at the grocery store, it doesn't matter if they put the apple, orange and banana in the grocery bag in that order, or in some other order. Either way, the set of groceries is exactly the same because it always has exactly the same contents.

But the elements themselves, the apple, orange, and banana, are still distinct elements different from one another. Surely you see that, right?

Likewise, if we merely called an apple a banana and a banana an apple, it also wouldn't change the contents of the bag. And there would still be one of each fruit, they'd just have different names. And they'd still all be distinct from one another no matter what name you call them and no matter what order you place them in the grocery bag.
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