Alan Mackenzie <acm@muc.de> writes:

> Skep Dick <skepdick22@gmail.com> wrote:

>> That's one way to defend your religion from critique.
>
> I'd hardly characterise the content of your last few posts as
> critique.

It's odd. Mr/Ms Dick comes across as a teenager[1]. In the old days,
young people would come a troll like this on Usenet every now and then,
but I'd be very surprised if Skep really is a teenager. If I'm wrong
he/she is the only one to have posted on Usenet since the early 90s.

[1] I'll admit I remember playing the "nothing means anything" game when
I was 13, but that was before the Internet so I could only annoy the
other lads in my gang. And, anyway, they played it as well (we were,
unsurprisingly, what later became know as nerds).

--
Ben.

On Thursday, 18 August 2022 at 22:07:07 UTC+2, Alan Mackenzie wrote:
> As I said in my last post, such is beyond your understanding. Look at it
> this way: The foundatations of mathematics are based on set theory. If
> sets were not defined, the whole theory would collapse like a house of
> cards. This is seen not to have happened.
You don't seem to understand a word of English.

The epression "THE foundations of Mathematics" is nonsense.

Set theory is ONE possible foundation of Mathematics.
And even then the use of the word "foundation" is a bit of a lie.

Set theory doesn't have a built-in deductive system. So it's founded upon first order logic.

You know which system has a built-in deductive system? Type theory.

> > Stating facts is "making moves"?!?!?
> You lack the understanding to know what a mathematical fact is.
You keep lying about my understanding. It's getting pretty tiresome. I haven't called you a "fucking idiot" in a while.

Mathematics doesn't deal with facts! Mathematics deals with proofs!
Implications of axioms and the like.

> You haven't produced any such evidence.
Your inability to recognize the evidence is not the same thing as me failing to produce it.

> > Sets are NOT defined in Mathematics.
> You're wrong, here. Mathematicians know, accept, and use the definition
> of sets.
You are confusing set theory with Mathematics. Let me say this in a language that will understand...

ALL set theory is Mathematics, not ALL Mathematics is set theory.

> > Ooooh. Is that how it works?
> I'm afraid it is. Only cranks and mathematicians care about it. The
> former get it wrong.

> > Everybody who understands the fact that sets are not formally defined
> > is a crank.
> Not quite. Many people pass through that state of misunderstanding.
> Some get stuck on it.
From where I am looking you seem like the kind who's stuck on it..

>They tend to become cranks. A defining
> characteristic of a crank is to disparage what is beyond his
> understanding.
You mean like you are disparading me? Despite the fact that my understanding is significantly more comprehensive than yours?

>This is what you are doing.
Wel. This is awkward, but... that is what YOU are doing.

I guess one of us doesn't understand decidability?

> I'd hardly characterise the content of your last few posts as critique.
I'd hardly characterise anything you posess as "understanding", but hey...

I mean you even have an axiom to RESTRICT comprehension in set theory.
It sure seems like that might get in your way of your COMPREHENSION.

On Thursday, 18 August 2022 at 22:18:54 UTC+2, Ben Bacarisse wrote:
> Alan Mackenzie <a...@muc.de> writes:
> > Skep Dick <skepd...@gmail.com> wrote:
>
> >> That's one way to defend your religion from critique.
> >
> > I'd hardly characterise the content of your last few posts as
> > critique.
> It's odd. Mr/Ms Dick comes across as a teenager[1]. In the old days,
> young people would come a troll like this on Usenet every now and then,
> but I'd be very surprised if Skep really is a teenager. If I'm wrong
> he/she is the only one to have posted on Usenet since the early 90s.
>
> [1] I'll admit I remember playing the "nothing means anything" game when
> I was 13, but that was before the Internet so I could only annoy the
> other lads in my gang. And, anyway, they played it as well (we were,
> unsurprisingly, what later became know as nerds).
You think I am playing the "nothing means anything game? Hahahahahaha! idiot!

I am literally pointing out arbitrary choices in your systems! The implication of arbitrary choices being that had you made a different set of choices - your system, and your symbols would take on a different meaning!
I am pointing out that axioms are an attempt to impose constraints on meaning (of which there is infinite supply).

Pretty much ALL operators in Mathematics are polymorphic - they have multiple meanings!
The expression "x = x" can take on an infinite number of meanings in ANY infinitary logic!

Idiot++

I am playing the "Symbols mean what they are INTERPRETED to mean" game. It's a game about constructing self-interpreting systems.

http://compilers.cs.ucla.edu/popl16/popl16-full.pdf

Skep Dick <skepdick22@gmail.com> wrote:
> On Thursday, 18 August 2022 at 22:18:54 UTC+2, Ben Bacarisse wrote:
>> Alan Mackenzie <a...@muc.de> writes:
>> > Skep Dick <skepd...@gmail.com> wrote:

>> >> That's one way to defend your religion from critique.

>> > I'd hardly characterise the content of your last few posts as
>> > critique.
>> It's odd. Mr/Ms Dick comes across as a teenager[1]. In the old days,
>> young people would come a troll like this on Usenet every now and then,
>> but I'd be very surprised if Skep really is a teenager. If I'm wrong
>> he/she is the only one to have posted on Usenet since the early 90s.

>> [1] I'll admit I remember playing the "nothing means anything" game when
>> I was 13, but that was before the Internet so I could only annoy the
>> other lads in my gang. And, anyway, they played it as well (we were,
>> unsurprisingly, what later became know as nerds).
> You think I am playing the "nothing means anything game? Hahahahahaha!
> idiot!

> I am literally pointing out arbitrary choices in your systems! The
> implication of arbitrary choices being that had you made a different
> set of choices - your system, and your symbols would take on a
> different meaning!

No. I already refuted this by pointing out that, were there
arbitrariness, different cultures would end up with different
mathematics. Group theory and graph theory are the same in Germany as in
the USA as in China, or South Korea, or at the South Pole. The natural
languages used vary, of course, and the mathematical symbols used also
vary somewhat. But the theorems are the same.

> I am pointing out that axioms are an attempt to impose constraints on
> meaning (of which there is infinite supply).

They're more than an attempt, they're a success. A good set of axioms
specifies exactly what it is intended to, without redundancy.

[ .... ]

> Idiot++

Not helpful.

> I am playing the "Symbols mean what they are INTERPRETED to mean" game.

It's also an unproductive game. Symbols with standard meanings allow for
thought and discussion about meaning. Even (especially) in a computer
programming language where if there weren't such standard meanings, the
writing of software would be grossly impeded.

> It's a game about constructing self-interpreting systems.

Ah, OK. I wish you all the best in this, but it's outside of where I
want to go.

> http://compilers.cs.ucla.edu/popl16/popl16-full.pdf

--
Alan Mackenzie (Nuremberg, Germany).

On Thursday, 18 August 2022 at 23:09:31 UTC+2, Alan Mackenzie wrote:
> No. I already refuted this by pointing out that, were there
> arbitrariness, different cultures would end up with different
> mathematics.
And I pointed out to you (over and over) that this is PRECISELY what happened.

There were cultures with 0-indexed number systems, and cultures with 1-indexed number systems.
Cultures with negative numbers and cultures without them.

When different cultures work IN ISOLATION - they arrive at different mathematics!

>Group theory and graph theory are the same in Germany as in
> the USA as in China, or South Korea, or at the South Pole.
No shit! Because You live in 2022 and nobody works IN ISOLATION anymore!

>The natural languages used vary, of course, and the mathematical symbols used also
> vary somewhat. But the theorems are the same.
No shit sherlock! That is because when you are practicing REVERSE Mathematics (e.g Computer Science)
you start with the theorems ONLY. https://en.wikipedia.org/wiki/Reverse_mathematics

You WANT (x=x) to be a theorem.
You DON'T want not(x=x) to be a theorem!
You WANT an identity axiom.

Not everybody WANTS an identity axiom!

Schrodinger didn't!

https://en.wikipedia.org/wiki/Schr%C3%B6dinger_logic

> They're more than an attempt, they're a success. A good set of axioms
> specifies exactly what it is intended to, without redundancy.
Oh really? So why don't you axiomatize your success criteria?

So that I can understand exactly what it is that you intended with your axioms.

Is your formal language incomplete... or something?

> Ah, OK. I wish you all the best in this, but it's outside of where I
> want to go.
😂😂😂😂😂😂😂😂😂😂😂😂

Coudl you axiomatize where you want to go? Or is it a theorem that you can't state in a formal language?

On 18/08/2022 20:06, Ben Bacarisse wrote:
> If RL(LR) = 1/3, why is R(L) not 0?

Because, viewed as a Hackenbush string, the player of R wins with
or without the first move, so R(L) is > 0. It's a surreal number, but not
a real number.

> R(L) would indeed be, as you say,
> less than any normal positive fraction, but not special in any way.
> The simple Hackenbush numbers have a "rounding up" bit in the conversion
> rule,

Rounding up /or/ down, depending who has to cut [down if R, up if
L, as playing in a number always makes your position worse].

> so RL(LR) is not exactly 1/3

Yes it is. If you play

RL(LR) + RL(LR) + RL(LR) + L,

then the first player "has to" choose how far into one of the "(LR)"s to
cut, then the second player can cut another one further along [and so
closer to 1/3]. If you complete the analysis, we deduce that the game
is a second-player win, ie is zero [exactly].

> but 1/3 + 1/𝜔 and R(L) = 1/𝜔.

Talking about "1/𝜔" is a bit deeper! Left as an exercise.

--
Andy Walker, Nottingham.
Andy's music pages: www.cuboid.me.uk/andy/Music
Composer of the day: www.cuboid.me.uk/andy/Music/Composers/Grieg

On Thursday, August 18, 2022 at 12:33:58 PM UTC-7, Skep Dick wrote:
> On Thursday, 18 August 2022 at 21:28:40 UTC+2, dklei...@gmail.com wrote:
> > On Thursday, August 18, 2022 at 12:11:26 PM UTC-7, Skep Dick wrote:
> > > On Thursday, 18 August 2022 at 20:29:11 UTC+2, dklei...@gmail.com wrote:
> > > > It means just what it says. X is a countably
> > > > infinite sequence of numbers such each
> > > > member is larger than the previous one.
> > > >
> > > > Infinity is a weird beast.
> > > I don't understand what you mean by "countably infinite sequence". Are you confusing the discreteness of the symbols in the sequence (their topology) with the ability to finish counting the elements in the sequence?
> > Your background with digital computers is
> > coloring your understand of mathematics.
> > Mathematics has no concept of process.
> Buuuuulshit!
>
> https://en.wikipedia.org/wiki/%CE%A0-calculus

The first sentence acknowledges that this is not
mathematics. It is computer science. This is a
very cute little "language" I need to look into but
it is not mathematics.
>
> >A reference to X in lim (X -> Y) implies some
> > single unchanging entity X not any kind of
> > variable. What mathematics does with process
> > is work with the completed process as a
> > single whole.
> Please explain to me what you mean by "completed". when dealing with an infinite stream of symbols?
>
The usual meaning - all the symbols even if you
have to run past several infinities to get there.

On Friday, 19 August 2022 at 00:36:38 UTC+2, dklei...@gmail.com wrote:
> The first sentence acknowledges that this is not
> mathematics. It is computer science.
There is no difference. It's all formalism and symbol juggling.

Mathematical proofs are computer programs.
The notation, language and narratives will be different - the implications will be the same.

> The usual meaning - all the symbols even if you
> have to run past several infinities to get there.
So when you arrive there. Past ALL of the infinities.

Why can't you tell me the last digit of pi?

On Thursday, August 18, 2022 at 3:45:34 PM UTC-7, Skep Dick wrote:
> On Friday, 19 August 2022 at 00:36:38 UTC+2, dklei...@gmail.com wrote:
> > The first sentence acknowledges that this is not
> > mathematics. It is computer science.
> There is no difference. It's all formalism and symbol juggling.
>
Once upon a time people really believed that but only
a few years of work with real digital devices have
convinced people they are not the same. This, it
seems, is a question for the philosophers.

To me the test is whether or not there are any variable.
A variable being a symbol that had more than one value
during the course of its use. Mathematics has no
variables (in this sense).
>
> Mathematical proofs are computer programs.
> The notation, language and narratives will be different - the implications will be the same.
> > The usual meaning - all the symbols even if you
> > have to run past several infinities to get there.
> So when you arrive there. Past ALL of the infinities.
>
> Why can't you tell me the last digit of pi?

Meaningless. One does not (cannot?) calculate the
digits of pi by applying some process to the first
K digits to get the K+1 digit. One uses some other
iterative process that continues forever and therefore
has no last step.

On Thursday, 18 August 2022 at 18:51:38 UTC+2, Andy Walker wrote:
> On 18/08/2022 12:29, Malcolm McLean wrote:
> > Base zero [...]
>
> If you only have one symbol to use, then maths gets a little
> boring. Balanced binary is more interesting. Basic idea: start from
> O [ == zero, or the origin]. If some point X on the number line is to
> the right of where you are, go right, indicated by R; if to the left,
> go left, indicated by L. For suitable choices of how far you go, every
> point on the line can be reached by a sequence such as RRRLLRLLRRR.
> "Suitable" has certain desirable properties:

This game is not interesting at all if we admit the axiom of choice.

For a suitable choice of how far right, or left I have to go I give you this algorithm:

if X > 0 then go X to the right. Because 0 + X = X
If X < 0 then go X to the left. Because 0 - X = - X
If X = 0 then do nothing.

First mover always wins in 1 step.

On 8/18/22 8:17 AM, Skep Dick wrote:
> On Thursday, 18 August 2022 at 14:00:15 UTC+2, richar...@gmail.com wrote:
>> (And refusing to define what you means, means your statement is
>> meaningless).
> So your refusal to define "define", "means" and "meaningless" mustmean that the above sentence is meaningless, right?
>
>> Yep, YOU are the discontinuity, and show you don't understand the math
>> you are spouting.
> **I** am not the discontinuity, you fucking retard.
>
> Any ARBITRARILY CHOSEN LOCATION (colloquially refered to as the number 0) on the continuum is the discontinuity!
>
>

Nope, You don't follow the rules, you can't claim to have found a hole
in the rules.

The Real Numbers do not include the "value" of "infinity", so any "math"
done with a "value" that is "infinite" doesn't show anything about the
Real Numbers, as you are not working inside the Real Number System.

If you are working in some extension of the Real Numbers that includes
Infinity, you need to indicate which one (so we know what infinity you
are using) and then you need to obey the restrictions of THAT system.

You have, so you have shown nothing.

On 8/18/22 8:47 AM, Skep Dick wrote:
> On Thursday, 18 August 2022 at 02:54:17 UTC+2, richar...@gmail.com wrote:
>> There IS a decidable < operation on elements of R.
> What do you mean by "decidable"!?! You must have somen notion of "decidability" far removed from the usual notion pertaining to Turing Machine-decidable.
>
>> The fact that "closer to infinity", since infinity isn't a normal value
>> of the field is defined differently than "closer to x" where x IS a
>> member of the field isn't a problem.
>>
>> That is just operator overloading.
> I understand generics/overloading just fine, you sophist.
>
> You overloaded the notion of "decidability" in your previous comment to mean something other than Turing Decidable.
>
>> x is closer to infinity than y if x > y, which holds for all x and y
>> that are actually members of the Real Number system.
> But x > y doesn't tell you ANYTHING about "closeness to infinity" ?!?!?!?
> x > y is about DISTANCE FROM ZERO!

Nope.

-1 > -2 but -1 is CLOSER to 0 the -2
2 > 1 and 1 is CLOSER to 0 than 2.

So x > y doesn't tell us anything about which number is closer to 0.

>
> x > y is just a proposition.

Nope, depends on the context.

It could be a proposition.
It could be a restriction,
It could be a fact.

>
> If x is further away from 0 than y then (x>y) is a theorem!
> If x is NOT further away from 0 than y then not(x>y) is a theorem!
>

On Friday, 19 August 2022 at 01:06:53 UTC+2, dklei...@gmail.com wrote:
> Once upon a time people really believed that but only
> a few years of work with real digital devices have
> convinced people they are not the same. This, it
> seems, is a question for the philosophers.
It's not a philosophical question. It's a pragmatic one.

https://en.wikipedia.org/wiki/Curry%E2%80%93Howard_correspondence

> To me the test is whether or not there are any variable.
> A variable being a symbol that had more than one value
> during the course of its use. Mathematics has no
> variables (in this sense).
Of course it does. They are called free variables.

Every expression in Mathematics has a free variable - its truth-value!

> > Why can't you tell me the last digit of pi?
> Meaningless. One does not (cannot?) calculate the
> digits of pi by applying some process to the first
> K digits to get the K+1 digit.
Of course we can! We have algorithms to generate ALL the digits of pi.

Its just that for every extra digit - you need extra time.

>One uses some other
> iterative process that continues forever and therefore
> has no last step.
Yeah! So what? These are your exact words "What mathematics does with process is work with the completed process as a single whole."

If the process for generating the digits of pi is "completed" then tell me the last digit!

On Friday, 19 August 2022 at 01:21:14 UTC+2, richar...@gmail.com wrote:
> Nope.
>
> -1 > -2 but -1 is CLOSER to 0 the -2
> 2 > 1 and 1 is CLOSER to 0 than 2.
Dickhead. Those aren't real numbers. Those are natural numbers.

You can't have decidable < on the Real numbers because that implies solving the halting problem.

Take values which are infinitely-close to zero. e.g compare 0 < 0.000...(there may be a non-0 in here keep looking)...
You can't traverse infinitely many digits in finite time.

This applies to any x < y ! because you can just re-write it to 0 < y - x

> So x > y doesn't tell us anything about which number is closer to 0.
Maybe! If the numbers are computable.

> Nope, depends on the context.
No it doesn't. The context is R.

> It could be a proposition.
> It could be a restriction,
> It could be a fact.
But in the context of comparing two real numbers - it's a proposition.

It only becomes a "fact" (or a negation of a fact) if the proof terminates.

On 8/18/22 9:13 AM, Skep Dick wrote:
> On Thursday, 18 August 2022 at 14:01:23 UTC+2, richar...@gmail.com wrote:
>> Nothing.
>> lim, as an operator, needs a function to apply to.
> No shit, sherlock! That's how all higher order functions work.
>
> To avoid confusing yourself any further apply it to the identity function: f(x) = x

or could otherwise be written lim(x -> ∞) x

>
> lim(x -> ∞) f(x)

That results in the results that the Limit does not exist.

Not all limits exist.

A limit exists only if the value it approach is a finite value, i.e a
actual Real Number.

>
>> What is "1 + " mean?
> Do you mean "+ 1" or "1 +"? Those are different things!
>
> "1 +" means "1 applied to +".

Which means what?

> "+ 1" means "+ applied to 1"
>
> "lim(x -> ∞) f(x)" is not the same thing as "f(x)lim(x -> ∞)
>

No, just like 1 / 2 is not 2 / 1

On 8/18/22 8:20 AM, Skep Dick wrote:
> On Thursday, 18 August 2022 at 14:11:25 UTC+2, richar...@gmail.com wrote:
>> And if you think Humans are computers you think too little of yourself.
> The USE of the English word "I" exemplifies self-reference. Recursion!
>
> I compute. But the "compute" is superfluous.
>
>> Hard to make that point when Mathematics vastly predate Computers.
> Mathematics predates the humans (who compute) ?!?!? Idiot.
>
>> What Computer did Pythagoras use to make his proofs?
> You think people before Pythagoras didn't compute?!?

I didn't say that, and you didn't answer the question

>
> You fucking idiot!

Your just proving your admission of being a Dick.

>
>> I don't think you actually understand computer science either.
> Oh, I think I understand recursion/self-reference just fine.
>

In other words, we need to ignore what you say because you don't
sactually mean what you say.

On 8/18/22 8:48 AM, Skep Dick wrote:
> On Thursday, 18 August 2022 at 14:02:14 UTC+2, richar...@gmail.com wrote:
>> No, I think that is likely, as you prove you are dumber that others.
> Dude! I am infinitely dumb.
>
> And you are still dumber.

Nope, but your too dumb to understand that. Good thing you admit it though.

On Friday, 19 August 2022 at 01:33:46 UTC+2, richar...@gmail.com wrote:
> That results in the results that the Limit does not exist.
That's such a peculiar claim! It means you must have decidable existence for limits?!?!

Please show me your algorithm.

> A limit exists only if the value it approach is a finite value, i.e a
> actual Real Number.
So if it approaches pi? Does that limit exist or ?

How would you tell if it approaches Pi or a number very close to pi?

> Which means what?
It means SyntaxError. "+" is a binary function.

> No, just like 1 / 2 is not 2 / 1
Obviously. Some operators commute and some don't.

Even "strict" logical operators like AND don't have to commute if you are using Linear Logic.

On Friday, 19 August 2022 at 01:36:45 UTC+2, richar...@gmail.com wrote:
> On 8/18/22 8:48 AM, Skep Dick wrote:
> > On Thursday, 18 August 2022 at 14:02:14 UTC+2, richar...@gmail.com wrote:
> >> No, I think that is likely, as you prove you are dumber that others.
> > Dude! I am infinitely dumb.
> >
> > And you are still dumber.
>
> Nope, but your too dumb to understand that. Good thing you admit it though.
I mean if you want to beat your chest and declare victory, go right ahead.

I've admitted over and over that you won. Loser.

The fact that you conceptualise every interaction as some turn-based adverserial game is pretty idiotic to begin with.

When you figure out what concurrency means you'll figure out that are the kind of interactive games where the players don't take turns - they work in parallel. And play cooperatively.

Is just that it's super-hard to define what it means "to win" in such games.

On Friday, 19 August 2022 at 01:35:32 UTC+2, richar...@gmail.com wrote:
> On 8/18/22 8:20 AM, Skep Dick wrote:
> > On Thursday, 18 August 2022 at 14:11:25 UTC+2, richar...@gmail.com wrote:
> >> And if you think Humans are computers you think too little of yourself.
> > The USE of the English word "I" exemplifies self-reference. Recursion!
> >
> > I compute. But the "compute" is superfluous.
> >
> >> Hard to make that point when Mathematics vastly predate Computers.
> > Mathematics predates the humans (who compute) ?!?!? Idiot.
> >
> >> What Computer did Pythagoras use to make his proofs?
> > You think people before Pythagoras didn't compute?!?
>
> I didn't say that, and you didn't answer the question
Didn't require answering.

> > You fucking idiot!
> Your just proving your admission of being a Dick.
Axioms don't need proof.

> In other words, we need to ignore what you say because you don't
> sactually mean what you say.
You are just proving to be a bigger Dick than I am. Every single person means PRECISELY what they way.

What else could they mean?

The fact that you can't understand their meaning goes to show you don't understand what meaning is.

On 8/18/22 7:33 PM, Skep Dick wrote:
> On Friday, 19 August 2022 at 01:21:14 UTC+2, richar...@gmail.com wrote:
>> Nope.
>>
>> -1 > -2 but -1 is CLOSER to 0 the -2
>> 2 > 1 and 1 is CLOSER to 0 than 2.
> Dickhead. Those aren't real numbers. Those are natural numbers.

-1 and -2 are NOT Natural Numbers

And All the Natural Numbers, and the Integers are part of the Real
Numbers, so they are Real Numbers too.

>
> You can't have decidable < on the Real numbers because that implies solving the halting problem.

Nope,

>
> Take values which are infinitely-close to zero. e.g compare 0 < 0.000...(there may be a non-0 in here keep looking)...
> You can't traverse infinitely many digits in finite time.

No real number is infinitely close to zero. Those are the infinitesimals
which are NOT part of the Reals.

All Real Numbers are Finite.

>
> This applies to any x < y ! because you can just re-write it to 0 < y - x

And that is a Finite number. Remember, you keep saying we are talking
about R, which is a well defined set.

>
>> So x > y doesn't tell us anything about which number is closer to 0.
> Maybe! If the numbers are computable.

Which number in my example wasn't computable.

-1, -2, 1, 2?

>
>> Nope, depends on the context.
> No it doesn't. The context is R.
>
>> It could be a proposition.
>> It could be a restriction,
>> It could be a fact.
> But in the context of comparing two real numbers - it's a proposition.
>
> It only becomes a "fact" (or a negation of a fact) if the proof terminates.
>

No, it can be a restriction.

In the limit definition we talk about something needing to be true for
all x > X, so that statement is a restriction on what x's we need to
look at.

It says the statement that followed doesn't need to hold for ALL x in R,
just all those x in R that are greater than X.

On 8/18/22 7:44 PM, Skep Dick wrote:
> On Friday, 19 August 2022 at 01:33:46 UTC+2, richar...@gmail.com wrote:
>> That results in the results that the Limit does not exist.
> That's such a peculiar claim! It means you must have decidable existence for limits?!?!
>
> Please show me your algorithm.
>
>
>> A limit exists only if the value it approach is a finite value, i.e a
>> actual Real Number.
> So if it approaches pi? Does that limit exist or ?

Pi is a Finite Number.

Finite Number doesn't mean it has a finite representation

>
> How would you tell if it approaches Pi or a number very close to pi?

Is the limit - Pi zero, or a non-zero number.

>
>> Which means what?
> It means SyntaxError. "+" is a binary function.

Right, just like a lim operation without an expresion to take the limit of.

If you mean the limit of x, say the limit of x.

>
>> No, just like 1 / 2 is not 2 / 1
> Obviously. Some operators commute and some don't.
>
> Even "strict" logical operators like AND don't have to commute if you are using Linear Logic.

On Friday, 19 August 2022 at 01:59:18 UTC+2, richar...@gmail.com wrote:
> -1 and -2 are NOT Natural Numbers
Idiot.!1 is a natural number. 2 is a natural number.

Every time you see "-X" just remember it's just a shorthand for (0 - (X))

> And All the Natural Numbers, and the Integers are part of the Real
> Numbers, so they are Real Numbers too.
Which has what to do with comparing infinite-precision Real numbers?

> > You can't traverse infinitely many digits in finite time.
> No real number is infinitely close to zero. Those are the infinitesimals
> which are NOT part of the Reals.
Yep! You are a fucking idiot!

The reals are smooth ( https://en.wikipedia.org/wiki/Smoothness )
If if there is no Real number infinitely close to any other Real number then there are gaps in the continuum.

Ooops?

> All Real Numbers are Finite.
Soo the Diagonalization argument doesn't mean anything to you?

> > This applies to any x < y ! because you can just re-write it to 0 < y - x
> And that is a Finite number. Remember, you keep saying we are talking
> about R, which is a well defined set.
OK well. You are as dumb as Olcott.
> >
> >> So x > y doesn't tell us anything about which number is closer to 0.
> > Maybe! If the numbers are computable.
> Which number in my example wasn't computable.
Your example is not R. Your example is a tiny subset of R.

> > It only becomes a "fact" (or a negation of a fact) if the proof terminates.
> >
> No, it can be a restriction.
It can be. It isn't.

> In the limit definition we talk about something needing to be true for
> all x > X, so that statement is a restriction on what x's we need to
> look at.
The > in the limit statement is as undecidable as the > in x > y.

Because > is undecidable for the Reals.

> It says the statement that followed doesn't need to hold for ALL x in R,
> just all those x in R that are greater than X.
And "greater than X" is undecidable in the Reals. Because for any pair x,y it's possible that (x > y) OR (x == y) OR (x < y)

It's your algorithm's job to CHOOSE the correct theorem.

Andy Walker <anw@cuboid.co.uk> writes:

> On 18/08/2022 20:06, Ben Bacarisse wrote:
>> If RL(LR) = 1/3, why is R(L) not 0?
>
> Because, viewed as a Hackenbush string, the player of R wins with
> or without the first move, so R(L) is > 0. It's a surreal number, but not
> a real number.

Right (I happen to know this) but I'm not sure you gave enough
information for anyone who does /not/ know the rules to work it out.

The only information you gave about the number represented by a string
was based on summing the digit values. And this included an example of
what /looks/ like an infinite sum:

RL(LR) = 1 - 1/2 - 1/4 + 1/8 - 1/16 + 1/32 ... = 1/3.

And yet,

R(L) = 1 - 1/2 - 1/4 - 1/8 ... =/= 0.

Again, I know why, by how could anyone else work out that from the
rules?

Calling this "balanced binary" also somewhat loaded the dice. It
strongly suggested that summing digit values (even in limit cases) is
what gives a digit string its denoted value.

--
Ben.

On Friday, 19 August 2022 at 02:07:13 UTC+2, richar...@gmail.com wrote:
> >> A limit exists only if the value it approach is a finite value, i.e a
> >> actual Real Number.
> > So if it approaches pi? Does that limit exist or ?
> Pi is a Finite Number.
No it isn't. The are finite computable representations of PI.

Dumb luck!

> Finite Number doesn't mean it has a finite representation
All the worse for you! What do you mean by "finite number" if NOT. "finite representation" ?

Are you meaning the numbersystem with base pi in which 1 means pi?

> > How would you tell if it approaches Pi or a number very close to pi?
> Is the limit - Pi zero, or a non-zero number.
You don't even know if the limit exists.

> > It means SyntaxError. "+" is a binary function.
> Right, just like a lim operation without an expresion to take the limit of.
So you don't understand how functors; or currying works? Gotcha!

> If you mean the limit of x, say the limit of x.
I do mean the limit of x ! Where x is a free variable.

EXACTLY like I mean "g = + 1 x"

g 5 = 6

# let plus x y = x + y;;
val plus : int -> int -> int = <fun>
# let g = plus 1;;
val g : int -> int = <fun>
# g 5;;
- : int = 6