Proposal: Definition of Infinity

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wij

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Aug 14, 2022, 7:35:27 PMAug 14
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The vague, no-logic concept of infinity seems dominated people's mind.
What is infinity? What does "lim(x→∞) f(x)" mean?

If infinity is merely a 'concept', not a number, what does x approach to?
If x is not getting "closer" to ∞? What does 'approach' mean?
Therefore, ∞-(x+1) < ∞-x must be valid inequality to mean x+1 is closer than x to infinity ∞.

But valid what? Most people agree ∀n∈ℕ, n<∞.

Is x+1 not closer than x to infinity?
So, infinity ∞ must have arithmetic meaning. Here is one:
The multiplicative inverse of ∞ is 1/∞, the additive inverse is -∞

All in all, that is the definition of infinity (the symbol '∞') proposed.
All is that simple, the usage treating ∞ as if it is a unique number is
safe-guaranteed, what left is interpretation. Though I think I figured this
part (merely means a procedure never terminate), there may be lots more
instances to test its interpretation in various scenario.

Richard Damon

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Aug 14, 2022, 8:34:39 PMAug 14
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If we are talking the real number system, as implied by the limit
operator, then the definition of what lim(x->inf) f(x) means

is there a number L, such that for ANY error e > 0, no matter how small,
can we find an X such that for all x > X that |f(x)-L| < e

If L exists, then it is the value of lim(x->inf) f(x)

Generally, we will find some bounding formula of some X(e) where we can
prove that | F(x) - L | < e for all x > X(e),

Jeff Barnett

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Aug 14, 2022, 8:48:38 PMAug 14
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On 8/14/2022 5:35 PM, wij wrote:
> The vague, no-logic concept of infinity seems dominated people's mind.
> What is infinity? What does "lim(x→∞) f(x)" mean?

It has a well-defined meaning in standard analysis. There is no mystery.

> If infinity is merely a 'concept', not a number, what does x approach to?
> If x is not getting "closer" to ∞? What does 'approach' mean?
> Therefore, ∞-(x+1) < ∞-x must be valid inequality to mean x+1 is closer than x to infinity ∞.

First off, 'number' is merely a concept also. In standard analysis, you
don't get closer to infinity, you merely have larger and larger numbers.

> But valid what? Most people agree ∀n∈ℕ, n<∞.

Agreed. That's the nice thing about having a standard definition: not
only is what you said true; it's true by /definition/!

> Is x+1 not closer than x to infinity?

No it isn't.

> So, infinity ∞ must have arithmetic meaning. Here is one:

Infinity does have arithmetic meaning but not necessarily as a number.
There are ways to adjoin it to the numbers but then you must change the
axioms defining real fields.

> The multiplicative inverse of ∞ is 1/∞, the additive inverse is -∞

Both claims are incorrect for standard analysis. These statements if
allowed, would lead to inconsistencies. And that means you would have a
broken system.

> All in all, that is the definition of infinity (the symbol '∞') proposed.
> All is that simple, the usage treating ∞ as if it is a unique number is
> safe-guaranteed, what left is interpretation. Though I think I figured this
> part (merely means a procedure never terminate), there may be lots more
> instances to test its interpretation in various scenario.

Look up "limit" in a standard (calculus or analytic geometry) text book
and the mysteries that so plague you will disappear.

There are interesting systems that do some of what you want or believe
that have been developed. But, note well, they were developed by folks
who know well what the pitfalls were and they went into the work with
enough background to not make an utter mess of it.

What I'm suggesting is that you (1) learn standard analysis, (2) study
alternative formulations, and (3) then and only then do you try to
invent a system to your own taste. You might even enjoy the learning
experience.
--
Jeff Barnett

Keith Thompson

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Aug 14, 2022, 10:37:13 PMAug 14
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wij <wyni...@gmail.com> writes:
> The vague, no-logic concept of infinity seems dominated people's mind.
> What is infinity? What does "lim(x→∞) f(x)" mean?
>
> If infinity is merely a 'concept', not a number, what does x approach to?
> If x is not getting "closer" to ∞? What does 'approach' mean?
> Therefore, ∞-(x+1) < ∞-x must be valid inequality to mean x+1 is closer than x to infinity ∞.
>
> But valid what? Most people agree ∀n∈ℕ, n<∞.

Typically the "<" relationship is defined over the real numbers. Since
∞ is not a real number, n<∞ is no more valid than n<♫.

Of course you can define < over other sets. Exactly what set did you
have in mind as the domain of the "<" relationship in your statement?

> Is x+1 not closer than x to infinity?

If it's "closer", can you define how much closer? Is ∞-(x+1) different
from ∞-x?

> So, infinity ∞ must have arithmetic meaning. Here is one:
> The multiplicative inverse of ∞ is 1/∞, the additive inverse is -∞
>
> All in all, that is the definition of infinity (the symbol '∞') proposed.
> All is that simple, the usage treating ∞ as if it is a unique number is
> safe-guaranteed, what left is interpretation. Though I think I figured this
> part (merely means a procedure never terminate), there may be lots more
> instances to test its interpretation in various scenario.

Is that supposed to be a *definition* of ∞? Just that it's
multiplicative inverse is 1/∞ and its additive inverse is -∞?

OK, but if I were to define ∞, I'd probably try to come up with a
definition that doesn't apply equally well to 8. (Just in case my point
wasn't clear, the multiplicative inverse of 8 is 1/8 and its additive
inverse is -8.)

--
Keith Thompson (The_Other_Keith) Keith.S.T...@gmail.com
Working, but not speaking, for Philips
void Void(void) { Void(); } /* The recursive call of the void */

wij

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Aug 15, 2022, 5:17:38 AMAug 15
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On Monday, 15 August 2022 at 10:37:13 UTC+8, Keith Thompson wrote:
> wij <wyni...@gmail.com> writes:
> > The vague, no-logic concept of infinity seems dominated people's mind.
> > What is infinity? What does "lim(x→∞) f(x)" mean?
> >
> > If infinity is merely a 'concept', not a number, what does x approach to?
> > If x is not getting "closer" to ∞? What does 'approach' mean?
> > Therefore, ∞-(x+1) < ∞-x must be valid inequality to mean x+1 is closer than x to infinity ∞.
> >
> > But valid what? Most people agree ∀n∈ℕ, n<∞.
> Typically the "<" relationship is defined over the real numbers. Since
> ∞ is not a real number, n<∞ is no more valid than n<♫.
>
> Of course you can define < over other sets. Exactly what set did you
> have in mind as the domain of the "<" relationship in your statement?
> > Is x+1 not closer than x to infinity?
> If it's "closer", can you define how much closer? Is ∞-(x+1) different
> from ∞-x?

I cannot really figure out what you mean.
It seems the definition is not properly presented caused your problems, sorry:

'∞' ::=
1. ∀n∈ℕ, n<∞
2. The multiplicative inverse of ∞ is 1/∞, the additive inverse is -∞

Thus, ∞ denotes a unique number. x+1 is 1 closer than x to ∞ (note that it is
illegal for limit theory to say this way).

> > So, infinity ∞ must have arithmetic meaning. Here is one:
> > The multiplicative inverse of ∞ is 1/∞, the additive inverse is -∞
> >
> > All in all, that is the definition of infinity (the symbol '∞') proposed.
> > All is that simple, the usage treating ∞ as if it is a unique number is
> > safe-guaranteed, what left is interpretation. Though I think I figured this
> > part (merely means a procedure never terminate), there may be lots more
> > instances to test its interpretation in various scenario.
> Is that supposed to be a *definition* of ∞? Just that it's
> multiplicative inverse is 1/∞ and its additive inverse is -∞?
>
> OK, but if I were to define ∞, I'd probably try to come up with a
> definition that doesn't apply equally well to 8. (Just in case my point
> wasn't clear, the multiplicative inverse of 8 is 1/8 and its additive
> inverse is -8.)
>
> --
> Keith Thompson (The_Other_Keith) Keith.S.T...@gmail.com
> Working, but not speaking, for Philips
> void Void(void) { Void(); } /* The recursive call of the void */

I might just talk to myself.
Definition is an arbitrary thing. Like 'programming theory', We ask what kind of
problem it solves, efficiency, benefit and deficiency. With number theory, logic
is a must. I think the notion conveyed by limit is accepted is because of these
qualities except the last one mentioned.

wij

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Aug 15, 2022, 5:38:43 AMAug 15
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The issue has been discussed many times. This proposal is primarily about the
definition of infinity.

Pythagorean's real number is Q, they could use the infinite-approaching argument
very validly deducing that all numbers are ratio number. Anyone can use Q to
approach any number and deduce that all real numbers are rational (sure modern
people won't do this).

Snippet from https://groups.google.com/g/comp.theory/c/DaybI0JY4Vc
...
To add more material came up to me (not well ordered):

----------------------------
There are quite a number of proofs of "repeating decimals are irrational".
The basic is the correct equation of 1/3 and its decimal form from long
division (kids understand this 'infinity' with no problem) should be:

1/3= 0.333... + nonzero_remainder.

----------------------------
To translate the 0.999... problem to limit:

Let A= lim(n->∞) 1-1/2^n = 0.999...
B= lim(n->∞) 1-1/10^n = 0.999...

Assume A=B
<=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
<=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n
<=> lim(n->∞) 1 = lim(n->∞) 1/5^n
<=> 1=0

[Note] I just demonstrate an instance. The limit theory can evolve as it does
(e.g. one-sided limit... There are many slightly different versions of
interpretation of limit as it evolves). Readers might find different
authors use different rules.
Limit is a technic to find its 'limit', it cannot form a logically
consistent theory for real number, e.g. the result of limit in general
must be verified, e.g. numerically, one cannot absolutely trust the
result of limit arithmetic. And at final, lim(x->c) f(c)= L does not
'deduce' f(c)=L (In text book, probably just reads "lim(x->c) f(c)= L, SO
WRITTEN as f(c)=L"). Limit theory only says the limit of 0.999... is 1,
the theory does not say 0.999...=1. There is no equality concept in the
ε-δ theory.
If one resorts to Dedekind-cut-like theories (I did not really read it),
from the knowledge that all the combinations of discrete symbols cannot
represent all the real numbers, I can conclude what those theories
claim are false, let alone I suspect there should be circular arguments
there, because many terms there must be well defined as a fundamental
theory, are undefined (prove me wrong).

The limit example above demonstrated "0.999..." cannot denote a specific number,
which also means "repeating decimal" cannot specify a unique number (A!=B).
Using limit is invalid for me (for this question) but the result is correct,
see the provided reference (I found a typo there).

-----------------------
Simple arithmetic (this should also be a valid way 2.718... is calculated):
(0.999....)^n approaches 1/e
(1.000...1)^n approaches e (or defined as e)
A possible rebuttal might be that the (1-1/n) in lim(n->∞) (1-1/n)^n is an invalid
number (approximated like 0.999...), or it is a 'concept' etc...
But if it is not a number, the whole equation is broken.

-----------------------
A[0]=0
A[n]=(A[n-1]+1)/2

The density property says (implicitly) n can enumerate infinitely (otherwise, it
won't be a rule) and A[∞] never be 1. A[n] infinitely approaches 1 in form
like 0.999.... This is like in the case of the interval [0,1), infinite numbers
of 0.999...s are located near the open end of [0,1).
Can we infinitely refine the scale of a ruler and the last scale never touches
the scale of 1? I think, yes, something like the √2 story, otherwise all numbers
can be 'proved' rational.

wij

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Aug 15, 2022, 5:47:33 AMAug 15
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You are addressing mostly about unrelated stuff.
If your are talking the inconsistency and incorrectness hidden in your standard
analysis, I agree. In my theory and view, many kind of standard-analysis are
actually trying to solve problems caused by the limit theory. No need and
unnecessary to build lies on lies.

The proposed definition of infinity is super simple and safe-guaranteed, and, it
SOLVED many infinity related paradoxes: Classes of liar's paradoxes, Zeno's
paradoxes, Supertask paradox, myth of infinite series,... and can build a
"one-point slope theory", Your choice. What the standard analysis solves?
Inconsistencies from limit, exam/thesis/paper/degree/title/money, such things I guess.

People can use the proposed definition of infinity as an 'informal' option to test.
It is super simple, safe-guaranteed, no need to say more.

Andy Walker

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Aug 15, 2022, 7:22:25 AMAug 15
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On 15/08/2022 01:48, Jeff Barnett wrote:
> On 8/14/2022 5:35 PM, wij wrote:
[...]
> What I'm suggesting is that you (1) learn standard analysis, (2)
> study alternative formulations, and (3) then and only then do you try
> to invent a system to your own taste. You might even enjoy the
> learning experience.

Seconded. But, sadly, experience is that Wij and the others
here who seem to think that they have new insights into mathematics
will take no notice, and instead convince themselves that everyone
else is simply hidebound, a "learned-by-rote", and incapable of new
thoughts.

Meanwhile, I commend [yet again] the study of the surreal
numbers; much easier than the hyperreals, and more useful. The
links with games can be explained to anyone who is not totally
innumerate, and infinity/infinitesimals arise in a very natural
and practical way. For some hints and further references, see

https://www-cs-faculty.stanford.edu/~knuth/sn.html

or, of course, google. Also, any book by [or co-authored by] the
late, great John Conway is worth reading.

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

Richard Damon

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Aug 15, 2022, 8:02:58 AMAug 15
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Actually, the real Pythagorean's eventually realized (c 5th century BC)
that the length of the hypotenuse of a right triangle with the two legs
having length 1 was not a rational number, and this caused them problem.

Yes, it took them a while, but that is the irrationality of Man when he
sticks to wrong ideas.


> Snippet from https://groups.google.com/g/comp.theory/c/DaybI0JY4Vc
> ...
> To add more material came up to me (not well ordered):
>
> ----------------------------
> There are quite a number of proofs of "repeating decimals are irrational".
> The basic is the correct equation of 1/3 and its decimal form from long
> division (kids understand this 'infinity' with no problem) should be:
>
> 1/3= 0.333... + nonzero_remainder.
>
> ----------------------------
> To translate the 0.999... problem to limit:
>
> Let A= lim(n->∞) 1-1/2^n = 0.999...
> B= lim(n->∞) 1-1/10^n = 0.999...
>
> Assume A=B
> <=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
> <=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n

And this step is invalid. You either multiplied by a "non-number" or
divided by zero depending on the steps you did to make that transition.

This is the problem of assuming that "infinity" is a number.

Ben Bacarisse

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Aug 15, 2022, 8:32:00 AMAug 15
to
wij <wyni...@gmail.com> writes:

> There are quite a number of proofs of "repeating decimals are irrational".
> The basic is the correct equation of 1/3 and its decimal form from long
> division (kids understand this 'infinity' with no problem) should be:
>
> 1/3= 0.333... + nonzero_remainder.
>
> ----------------------------
> To translate the 0.999... problem to limit:
>
> Let A= lim(n->∞) 1-1/2^n = 0.999...
> B= lim(n->∞) 1-1/10^n = 0.999...
>
> Assume A=B

(Technically, it's the two lines above that are the assumptions. Once
you have written Let A = <stuff> = 0.999... and B = <other stuff> =
0.999... the line A=B is not an assumption anymore since it follows from
the earlier two.)

> <=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
> <=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n
> <=> lim(n->∞) 1 = lim(n->∞) 1/5^n

Your algebra is wrong here. What rule of algebra do you think let's you
take this step? It's not multiplying by 2, and you can't multiply by
2^n as n is a bound variable.

> <=> 1=0

This was pointed out before (by at least two posters). Why do you keep
posting "proofs" with basic mistakes in them?

> -----------------------
> Simple arithmetic (this should also be a valid way 2.718... is calculated):
> (0.999....)^n approaches 1/e

(0.999...)^n = 1 for all n >= 0.

> (1.000...1)^n approaches e (or defined as e)

1.000...1 is a non-standard definition. I know of two meanings one might
give it, and in neither case does (1.000...1)^n approaches e. Anyway,
you should not use ambiguous notations without a some explanation of
what the notation means.

> -----------------------
> A[0]=0
> A[n]=(A[n-1]+1)/2
>
> The density property says (implicitly) n can enumerate infinitely
> (otherwise, it won't be a rule) and A[∞] never be 1.

Inductive definitions like this define A[k] for all k in N. They don't
define A[z] for any z not in N. You can extend the definition to
include something called oo (so that, for example, A[oo] = 42 if you
like), but a more natural extension would be to define A[oo] = 1.
Either way, it's up to anyone making the extension to defend it.

> A[n] infinitely
> approaches 1 in form like 0.999....

> This is like in the case of the interval [0,1), infinite numbers
> of 0.999...s are located near the open end of [0,1).

Yes. And open intervals in R have a least upper bound in R. For
well-behaved sequences like the partial sums 0.999... the limit is the
same as the least upper bound.

--
Ben.

wij

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Aug 15, 2022, 10:52:32 AMAug 15
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Why do you point to where you seem not addressing.
Andy probably just missed a point, I provided a reference.
Ben made an error and (assume he saw my reply to Andy) made an error again.

Let A= lim(n->∞) 1-1/2^n = 0.999...
B= lim(n->∞) 1-1/10^n = 0.999...

Assume A=B
<=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
<=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n
<=> (lim(n->∞) 2^n)*(lim(n->∞) 1/2^n) = (lim(n->∞) 2^n)*(lim(n->∞) 1/10^n)
<=> lim(n->∞) 2^n/2^n = lim(n->∞) 2^n/10^n
<=> lim(n->∞) 1 = lim(n->∞) 1/5^n
<=> 1=0

I wonder how much does you guys really understand you are talking?

Ben Bacarisse

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Aug 15, 2022, 11:45:59 AMAug 15
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I made no mistake.

> Let A= lim(n->∞) 1-1/2^n = 0.999...
> B= lim(n->∞) 1-1/10^n = 0.999...
>
> Assume A=B
> <=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
> <=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n
> <=> (lim(n->∞) 2^n)*(lim(n->∞) 1/2^n) = (lim(n->∞) 2^n)*(lim(n->∞) 1/10^n)

No. This step is invalid. You can't multiply a limit by anything but a
real number. lim(n->∞) 2^n is not a real number.

> <=> lim(n->∞) 2^n/2^n = lim(n->∞) 2^n/10^n

The product law for limits is only valid when both limits are real.

> <=> lim(n->∞) 1 = lim(n->∞) 1/5^n
> <=> 1=0
>
> I wonder how much does you guys really understand you are talking?

Obviously if you don't understand the basics of real analysis, you will
doubt anyone who points them out.

--
Ben.

Skep Dick

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Aug 15, 2022, 11:54:20 AMAug 15
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On Monday, 15 August 2022 at 02:48:38 UTC+2, Jeff Barnett wrote:
> What I'm suggesting is that you (1) learn standard analysis
Why are you directing him to standard analysis when the system DOESN'T do anything of what he's asking/saying ?!? Are you just trying to waste his time?

Why aren't you suggesting that he learns NONstandard analysis instead, which does PRECISELY what he's trying to do?

https://en.wikipedia.org/wiki/Nonstandard_analysis


Skep Dick

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Aug 15, 2022, 12:00:40 PMAug 15
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You really need to stop listening to all the fossils in the group. They can't help you - they don't know any better.

You are right. Don't let them guilt you into believing otherwise - they don't understand how to compute/program with Real numbers!

They don't understand parametricity. https://en.wikipedia.org/wiki/Parametricity and why it's directly relevant to the syntax of "lim(x->∞)" if you are passing "∞" as a parameter to a function then "∞" is BOUND TO A VARIABLE.

If you are binding objects to free variables - you are treating those objects as numbers/values. If "∞" is not a number then "lim(x->∞)" is a syntax error.


Skep Dick

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Aug 15, 2022, 12:02:31 PMAug 15
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On Monday, 15 August 2022 at 17:45:59 UTC+2, Ben Bacarisse wrote:
> Obviously if you don't understand the basics of real analysis, you will
> doubt anyone who points them out.

Obviously, if you don't understand the basics of syntax, semantics, bound and unbound variables
you will doubt anyone who points out that "lim(x -> ∞)" is a syntax error IF "∞ is not a number"

Skep Dick

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Aug 15, 2022, 12:14:19 PMAug 15
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You are 100% correct when using the Hyperreal numbers! That is *R not R.

I will repeat it until the cows come home. Despite having said it three times already.

You can't parametrize functions by outside the domain of thefunction!
So you can parametrize lim(x -> y) by y = ∞ then ∞ is in the domain of lim(x -> y).
If ∞ is in the domain of lim(x -> y) then ∞ is a number!


wij

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Aug 15, 2022, 12:20:56 PMAug 15
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Ben Bacarisse

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Aug 15, 2022, 12:25:25 PMAug 15
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You don't get to say what the syntax means. lim(x->a) and lim(x->oo)
are well-established form used to mean two quite different kinds of
limits, despite the similarity in the syntax.

--
Ben.

Skep Dick

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Aug 15, 2022, 12:27:34 PMAug 15
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Yes, I do get to say it. And if you don't like me saying it - I will repeat it louder.

lim(x -> a) means EXACTLY

let a = ∞
lim(x -> a)


Skep Dick

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Aug 15, 2022, 12:29:56 PMAug 15
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On Monday, 15 August 2022 at 18:25:25 UTC+2, Ben Bacarisse wrote:
Lets try that again... without the errors.

Yes, I do get to say it. And if you don't like me saying it - I will repeat it louder.

lim(x -> ∞) means EXACTLY THE SAME THING AS

Ben Bacarisse

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Aug 15, 2022, 12:34:41 PMAug 15
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Yes of course I do. But what is the point of citing a document that
does not support your erroneous algebra? In fact that document says
nothing at all about /any/ of the limits in your so-called proof (though
you'd have to know a bit of the subject to see that).

--
Ben.

Ben Bacarisse

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Aug 15, 2022, 12:39:33 PMAug 15
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Wij is not working in *R. He cites standard rules about limits in R
(rules he or she does not properly understand) to support this bogus
proof. Cranks don't want to be correct in some "other" system (though,
as it happens, 0.999... = 1 in *R as well as in R), they want
conventional wisdom to be wrong.

--
Ben.

wij

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Aug 15, 2022, 12:40:15 PMAug 15
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Thank you. I kind of lost, wondering what the world is. Luckily, computers give
me confidence I am not crazy (and I used to play electronics. I think I understand
how real thing works).

I feel ℝ is not closed and incomplete. But I am a programmer, just learn what I
feel need to learn (for time/learning efficiency reason).

You have mentioned Hyperreal several times. After seeing what my idea is,
should I really learn it? What would I get?

Skep Dick

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Aug 15, 2022, 12:43:06 PMAug 15
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On Monday, 15 August 2022 at 18:39:33 UTC+2, Ben Bacarisse wrote:
> Skep Dick <skepd...@gmail.com> writes:
>
> > On Monday, 15 August 2022 at 16:52:32 UTC+2, wyni...@gmail.com wrote:
>
> >> Ben made an error and (assume he saw my reply to Andy) made an error again.
> >> Let A= lim(n->∞) 1-1/2^n = 0.999...
> >> B= lim(n->∞) 1-1/10^n = 0.999...
> >>
> >> Assume A=B
> >> <=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
> >> <=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n
> >> <=> (lim(n->∞) 2^n)*(lim(n->∞) 1/2^n) = (lim(n->∞) 2^n)*(lim(n->∞) 1/10^n)
> >> <=> lim(n->∞) 2^n/2^n = lim(n->∞) 2^n/10^n
> >> <=> lim(n->∞) 1 = lim(n->∞) 1/5^n
> >> <=> 1=0
> >> I wonder how much does you guys really understand you are talking?
> >
> > You are 100% correct when using the Hyperreal numbers! That is *R not
> > R.
> Wij is not working in *R. He cites standard rules about limits in R
> (rules he or she does not properly understand) to support this bogus
> proof.
He has openly told you what theorems he is interested in!
None of the theorems he wants hold in R.
Most of the theorems he wants hold in *R

Why are you dragging him down instead of pulling him up?!?

>Cranks don't want to be correct in some "other" system (though,
> as it happens, 0.999... = 1 in *R as well as in R), they want
> conventional wisdom to be wrong.
Bullshit. 0.999... is meaningless in *R. It's a syntax error.



Skep Dick

unread,
Aug 15, 2022, 12:48:17 PMAug 15
to
On Monday, 15 August 2022 at 18:39:33 UTC+2, Ben Bacarisse wrote:
> Cranks don't want to be correct in some "other" system (though,
> as it happens, 0.999... = 1 in *R as well as in R), they want
> conventional wisdom to be wrong.
You have no idea what "they" actually want! Because you are a tone-deaf zealot.

What "cranks" actually want is for you to stop assaulting their intuitions with nonsense like 0.999... = 1
Because "cranks" actually have a killer intuition about Mathematics developed empirically, not through the usual academic indoctrination.

What "cranks" want is NOT the theorem 0.999... = 1.
What "cranks" want is the theorem 0.999... = 1 - ε

wij

unread,
Aug 15, 2022, 12:50:43 PMAug 15
to
It is 4 times. I think I am qualified to call you IDIOT and limit responding to you.

Skep Dick

unread,
Aug 15, 2022, 1:26:18 PMAug 15
to
On Monday, 15 August 2022 at 18:40:15 UTC+2, wyni...@gmail.com wrote:
> Thank you. I kind of lost, wondering what the world is.
The Mathematics world is full of academics who have never done a minute of engineering in their lives.

They mostly don't unverstand the value of closures/closed sets, mean while that is the essence of control theory/engineering.

> Luckily, computers give me confidence I am not crazy (and I used to play electronics. I think I understand
> how real thing works).
Wonderful! Then you already have the analogue (continuous) intuition of what R/*R is like.
You are a step ahead of most programmers who only have good intuition for the natural numbers.

> I feel ℝ is not closed and incomplete.
Asolutely! This is a well-known fact. The Real numbers are closed under addition, subtraction and multiplication, but not division because it's undefined for x/0.

And a fundamental fact of ALL Mathematics. No number system is closed under equality!
x == x is a Boolean, not a number!

What it seems to me is that you desperately want to be able to do computation with Real numbers (despite the limits of those pesky discrete computers!). And you are not alone.

>But I am a programmer, just learn what I feel need to learn (for time/learning efficiency reason).
Perfect! That's an indispensable intuition for engineers. We do Just-In-Time learning.

It works. Most of the time. Except when you desperately need a number system different to the status quo, and all the idiot-Mathematicians are trying to indoctrinate you instead of help you solve your pragmatic problems.

> You have mentioned Hyperreal several times. After seeing what my idea is,
> should I really learn it? What would I get?
You would get most of the theorems you are looking for and most of the answers you seek in your original post.

You will get to treat infinity as just-another-number.
You will get to understand the meaning of lim(x→∞) f(x) in terms of infinitesimals and infinites (they are complementary).

Like I said, the fndamental theorem of *R is 1/ε = ω/1.
In English: 1 divided by an infinitesimal quantity is an infinite quantity. Infinity multiplied by a really small quantity is 1.

This is a really really nice setting for an engineer, because you can reason about quotients, proportions, logarithmic functions, and all the usual stuff we want out of information theory/signal processing!

Mr Flibble

unread,
Aug 15, 2022, 1:39:35 PMAug 15
to
Infinitesimals don't exist: 1 / infinity = 0

/Flibble

Skep Dick

unread,
Aug 15, 2022, 1:52:13 PMAug 15
to
On Monday, 15 August 2022 at 19:39:35 UTC+2, Mr Flibble wrote:
> Infinitesimals don't exist: 1 / infinity = 0
Numbers don't exist!

0 is undefined.
The successor function is undefined.

Now fuck off.

Mr Flibble

unread,
Aug 15, 2022, 2:18:59 PMAug 15
to
Spicy.

/Flibble

Keith Thompson

unread,
Aug 15, 2022, 2:25:36 PMAug 15
to
wij <wyni...@gmail.com> writes:
> On Monday, 15 August 2022 at 10:37:13 UTC+8, Keith Thompson wrote:
>> wij <wyni...@gmail.com> writes:
>> > The vague, no-logic concept of infinity seems dominated people's mind.
>> > What is infinity? What does "lim(x→∞) f(x)" mean?
>> >
>> > If infinity is merely a 'concept', not a number, what does x approach to?
>> > If x is not getting "closer" to ∞? What does 'approach' mean?
>> > Therefore, ∞-(x+1) < ∞-x must be valid inequality to mean x+1 is closer than x to infinity ∞.
>> >
>> > But valid what? Most people agree ∀n∈ℕ, n<∞.
>> Typically the "<" relationship is defined over the real numbers. Since
>> ∞ is not a real number, n<∞ is no more valid than n<♫.
>>
>> Of course you can define < over other sets. Exactly what set did you
>> have in mind as the domain of the "<" relationship in your statement?
>> > Is x+1 not closer than x to infinity?
>> If it's "closer", can you define how much closer? Is ∞-(x+1) different
>> from ∞-x?
>
> I cannot really figure out what you mean.
> It seems the definition is not properly presented caused your problems, sorry:
>
> '∞' ::=
> 1. ∀n∈ℕ, n<∞
> 2. The multiplicative inverse of ∞ is 1/∞, the additive inverse is -∞

ℕ denotes the set of natural numbers, which is either the set of
non-negative integers or the set of positive integers (the difference
doesn't matter here).

Your "definition" implies that ∞ is not a natural number, since every
natural number is less than ∞.

> Thus, ∞ denotes a unique number. x+1 is 1 closer than x to ∞ (note that it is
> illegal for limit theory to say this way).

I suggest that your definition isn't a complete definition. It doesn't
imply that ∞ is unique. You could have two distinct infinite valuess
say aleph0 and aleph1, that both satisfy your definition.

∀n∈ℕ, n<aleph0
∀n∈ℕ, n<aleph1
aleph0 ≠ aleph1

And I still don't see how your definition implies that x+1 is "closer"
than x to ∞.

7 is closer than 6 to 10. 10-7 is 3; 10-6 is 4. 3 and 4 are two
distinct values, and comparing them shows us that 7 is closer than 6 to
10, and how much closer.

Do you claim that that same reasoning leads to the conclusion that x+1
is closer than x to ∞? Do you claim that ∞-(x+1) is different from ∞-x?
If so, is that claim consistent with your claim that ∞ is a unique
value?

I get the impression that you're trying to use common sense rules that
apply to the integers, and apply them to ∞. That doesn't work. For
example, common sense tells us that x+1 > x (and we can prove it given
the right axioms). But that's not true if x=∞ *and*, as you assert, ∞
is a unique value.

You can certainly define systems in which ∞ is a distinct value, and
with some effort you can define the results of various operations on ∞
and finite numbers and make them work consistently. I don't see that
you've actually done so.

[...]

--
Keith Thompson (The_Other_Keith) Keith.S.T...@gmail.com
Working, but not speaking, for Philips
void Void(void) { Void(); } /* The recursive call of the void */

Skep Dick

unread,
Aug 15, 2022, 2:38:14 PMAug 15
to
On Monday, 15 August 2022 at 20:25:36 UTC+2, Keith Thompson wrote:
> You can certainly define systems in which ∞ is a distinct value, and
> with some effort you can define the results of various operations on ∞
> and finite numbers and make them work consistently. I don't see that
> you've actually done so.
Why does every single system need re-defining from first principles?
Why do we constantly have to re-invent the wheel?

Why is the concept of "importing libraries/dependencies" so under-utilised in Mathematics?!?

Keith Thompson

unread,
Aug 15, 2022, 2:53:59 PMAug 15
to
If there are multiple possible libraries you could import, you still
have to specify which one you're using.

Skep Dick

unread,
Aug 15, 2022, 3:09:10 PMAug 15
to
On Monday, 15 August 2022 at 20:53:59 UTC+2, Keith Thompson wrote:
> If there are multiple possible libraries you could import, you still
> have to specify which one you're using.
Well yeah! But your search algorithm is going to take significantly longer if you aren't even bothering to eliminate the systems which **definitely** can't satisfy the necessary theorems!

dklei...@gmail.com

unread,
Aug 15, 2022, 3:11:31 PMAug 15
to
In most cases mathematicians study specific subject matter and there is
no importing involved. For example: I studied Perturbations of Operators
on Banach Spaces. Nothing extraneous was needed.

Ben Bacarisse

unread,
Aug 15, 2022, 3:16:22 PMAug 15
to
> You have mentioned Hyperreal several times. After seeing what my idea is,
> should I really learn it? What would I get?

Learning is always a good idea, but the hyperreals won't give you what
you want as far as the limits you presented go. And 0.999... = 1 in *R
as well. In fact, the convergence of such well-behaved series is almost
the "poster boy" case for using *R since *R's infinitesimals formalise
Euler's convergence criterion.

--
Ben.

Skep Dick

unread,
Aug 15, 2022, 3:17:57 PMAug 15
to
On Monday, 15 August 2022 at 21:11:31 UTC+2, dklei...@gmail.com wrote:
> In most cases mathematicians study specific subject matter and there is
> no importing involved. For example: I studied Perturbations of Operators
> on Banach Spaces. Nothing extraneous was needed.

When you say "study specific field" do yo mean that you spent time examining other people's no constructions, or did you construct a new field from first principles?

And I am not going to beat around the bush here... By "construct" I mean "invent".

Did you invent anything; or did you marvel at other people's inventions?

Ben Bacarisse

unread,
Aug 15, 2022, 3:21:08 PMAug 15
to
> It is 4 times. I think I am qualified to call you IDIOT and limit
> responding to you.

You can call me what you like. And do please limit your responses
(ideally to zero), but that will not change the fact that the document
you cite does not support your derivation. If you want to know why the
multiplication rule you've seen does no apply, just ask. Someone you
have not insulted might explain it.

--
Ben.

wij

unread,
Aug 15, 2022, 4:14:14 PMAug 15
to
No, 2<3.3, 3.3 is not a natural number.

> since every natural number is less than ∞.

Definition: ∀n∈ℕ, n<∞

I still don't quite understand you.

> > Thus, ∞ denotes a unique number. x+1 is 1 closer than x to ∞ (note that it is
> > illegal for limit theory to say this way).
>
> I suggest that your definition isn't a complete definition. It doesn't
> imply that ∞ is unique.

Yes, it is implied (see below)

> You could have two distinct infinite valuess
> say aleph0 and aleph1, that both satisfy your definition.
>
> ∀n∈ℕ, n<aleph0
> ∀n∈ℕ, n<aleph1
> aleph0 ≠ aleph1
>

There is only one symbol '∞' denoting infinity. ∞^2, ∞^∞ are also infinite
numbers. sin(∞), x^2+∞=0,... are valid numbers (usage is safe-guaranteed, and
there is practical meaning).
I am not considering aleph0/aleph1. In my understanding, the length of a point
is zero, Any_Infinity*0=0, ℝ cannot be stuffed by points (or numbers).

> And I still don't see how your definition implies that x+1 is "closer"
> than x to ∞.
>
> 7 is closer than 6 to 10. 10-7 is 3; 10-6 is 4. 3 and 4 are two
> distinct values, and comparing them shows us that 7 is closer than 6 to
> 10, and how much closer.
>
> Do you claim that that same reasoning leads to the conclusion that x+1
> is closer than x to ∞? Do you claim that ∞-(x+1) is different from ∞-x?
> If so, is that claim consistent with your claim that ∞ is a unique
> value?

closer::= if x-n < y-n, then, x is closer than y to n. (x,y,n>=0)

∞-(x+1) < ∞-x
<=> -(x+1) < -x
<=> x+1 > x
<=> 1>0
<=> true

> I get the impression that you're trying to use common sense rules that
> apply to the integers, and apply them to ∞. That doesn't work. For
> example, common sense tells us that x+1 > x (and we can prove it given
> the right axioms). But that's not true if x=∞ *and*, as you assert, ∞
> is a unique value.
>
> You can certainly define systems in which ∞ is a distinct value, and
> with some effort you can define the results of various operations on ∞
> and finite numbers and make them work consistently. I don't see that
> you've actually done so.
>

'∞' is a symbol like i,e,π,√2. Other arithmetic properties are the same as
usual numbers, implied and sufficient. Defining them will cause ambiguous problems.
Did these answer you?

wij

unread,
Aug 15, 2022, 4:50:43 PMAug 15
to
On Tuesday, 16 August 2022 at 01:26:18 UTC+8, Skep Dick wrote:
> On Monday, 15 August 2022 at 18:40:15 UTC+2, wyni...@gmail.com wrote:
> > Thank you. I kind of lost, wondering what the world is.
> The Mathematics world is full of academics who have never done a minute of engineering in their lives.
>
> They mostly don't unverstand the value of closures/closed sets, mean while that is the essence of control theory/engineering.

Yes, many people I know had never get their hand dirty in their life and become
construction/mechanics engineers or doctors. But that alone is not really that
bad, their environment caused that. Their 'abstract' idea may also be very
intelligent and work fine.

> > Luckily, computers give me confidence I am not crazy (and I used to play electronics. I think I understand
> > how real thing works).
> Wonderful! Then you already have the analogue (continuous) intuition of what R/*R is like.
> You are a step ahead of most programmers who only have good intuition for the natural numbers.

1/3= 0.333... + nonzero_remainder is intuition. Kids understand this naturally without question.

> > I feel ℝ is not closed and incomplete.
> Asolutely! This is a well-known fact. The Real numbers are closed under addition, subtraction and multiplication, but not division because it's undefined for x/0.
>
> And a fundamental fact of ALL Mathematics. No number system is closed under equality!
> x == x is a Boolean, not a number!
>
> What it seems to me is that you desperately want to be able to do computation with Real numbers (despite the limits of those pesky discrete computers!). And you are not alone.

Actually, I might be able to not speaking for 10 years. All are in my mind.
Actually, what I want to achieve is a program that can do reasoning
(calculation, proving and reasoning are the same thing, harder to explain)
My hunch is that computer may be able to do math. problems, proving theorems.
I know my math. is not good comparing to MANY (everybody is good at something).
And there are already many symbolic computation systems, but I guess they
should be stuck by inconsistency issues.

> >But I am a programmer, just learn what I feel need to learn (for time/learning efficiency reason).
> Perfect! That's an indispensable intuition for engineers. We do Just-In-Time learning.

Yes, we know what we work for.

> It works. Most of the time. Except when you desperately need a number system different to the status quo, and all the idiot-Mathematicians are trying to indoctrinate you instead of help you solve your pragmatic problems.
> > You have mentioned Hyperreal several times. After seeing what my idea is,
> > should I really learn it? What would I get?
> You would get most of the theorems you are looking for and most of the answers you seek in your original post.
>
> You will get to treat infinity as just-another-number.
> You will get to understand the meaning of lim(x→∞) f(x) in terms of infinitesimals and infinites (they are complementary).
>
> Like I said, the fndamental theorem of *R is 1/ε = ω/1.
> In English: 1 divided by an infinitesimal quantity is an infinite quantity. Infinity multiplied by a really small quantity is 1.
>
> This is a really really nice setting for an engineer, because you can reason about quotients, proportions, logarithmic functions, and all the usual stuff we want out of information theory/signal processing!

Terrific. I got motive.

We may have many idea to exchange. You can email me or my C++ library project
at https://sourceforge.net/projects/cscall/

wij

unread,
Aug 15, 2022, 4:54:53 PMAug 15
to
Forgot to say thanks. Originally, I did not want to talk much. But you have handled
many loud noises (speech was like Laozi https://en.wikipedia.org/wiki/Laozi)
So, I decided to take the opportunity. You did helped a lot.

Keith Thompson

unread,
Aug 15, 2022, 5:36:34 PMAug 15
to
I'm not sure what it is you don't understand. When I said that ∞ is not
a natural number, I wasn't trying to point out a flaw.

What your definition does not provide is uniqueness. How do we know
that there isn't more than one value X such that ∀n∈ℕ, n<∞? There are
certainly models in which there is more than one such value. I don't
believe you've specified the model you're using precisely enough.

>> > Thus, ∞ denotes a unique number. x+1 is 1 closer than x to ∞ (note that it is
>> > illegal for limit theory to say this way).
>>
>> I suggest that your definition isn't a complete definition. It doesn't
>> imply that ∞ is unique.
>
> Yes, it is implied (see below)

I don't believe it is (see above). If you want ∞ to be unique, that
should be part of its definition, or an axiom, or a fact derivable from
axioms. Giving a partial definition and then saying "Oh, by the way,
it's unique" is not sufficiently precise.
If ∞-(x+1) < ∞-x, then you have multiple infinite values, even if you
only use the term "∞" to refer to one of them.

More concretely, you seem to be saying that ∞-1 and ∞-2 are distinct
values. They're clearly both infinite, right? But neither of them is
equal to ∞?

Certainly there are systems in which that's all true -- but I don't know
what system you're working with.

If you're talking about hyperreals, you can save a lot of time and
effort by saying so. Likewise if you're talking about some other well
defined system in which ∞ is treated as a unique number. There are a
number of such systems.

Jeff Barnett

unread,
Aug 15, 2022, 6:12:15 PMAug 15
to
On 8/15/2022 3:47 AM, win wrote:

> The proposed definition of infinity is super simple and safe-guaranteed, and, it
> SOLVED many infinity related paradoxes: Classes of liar's paradoxes, Zeno's
> paradoxes, Supertask paradox, myth of infinite series,... and can build a
> "one-point slope theory", Your choice. What the standard analysis solves?
> Inconsistencies from limit, exam/thesis/paper/degree/title/money, such things I guess.
>
> People can use the proposed definition of infinity as an 'informal' option to test.
> It is super simple, safe-guaranteed, no need to say more.

What raucous bull shit! You haven't proposed a definition of infinity
nor solved any paradox. Do as I suggested: go read a High School math
book and then express an opinion. But I must admit that you are very
humorous. Thanks for the entertainment. The only improvement is when you
and Peter O. start debating. The sound is like castrated wolves howling
at the full moon. When you are on your own like here, it brings up that
old Buddhist question: what is the sound of one hand clapping? Or is
that howling? Enjoy your self.
--
Jeff Barnett

Richard Damon

unread,
Aug 15, 2022, 6:50:41 PMAug 15
to
On 8/15/22 12:02 PM, Skep Dick wrote:
> On Monday, 15 August 2022 at 17:45:59 UTC+2, Ben Bacarisse wrote:
>> Obviously if you don't understand the basics of real analysis, you will
>> doubt anyone who points them out.
>
> Obviously, if you don't understand the basics of syntax, semantics, bound and unbound variables
> you will doubt anyone who points out that "lim(x -> ∞)" is a syntax error IF "∞ is not a number"
>

Nope, because the law of limits has a special definition for limits to
infinity.


Just because you don't understand the definitions doesn't mean they
don't exist.

Richard Damon

unread,
Aug 15, 2022, 6:53:59 PMAug 15
to
On 8/15/22 12:27 PM, Skep Dick wrote:
> On Monday, 15 August 2022 at 18:25:25 UTC+2, Ben Bacarisse wrote:
>> Skep Dick <skepd...@gmail.com> writes:
>>
>>> On Monday, 15 August 2022 at 17:45:59 UTC+2, Ben Bacarisse wrote:
>>>> Obviously if you don't understand the basics of real analysis, you will
>>>> doubt anyone who points them out.
>>>
>>> Obviously, if you don't understand the basics of syntax, semantics,
>>> bound and unbound variables you will doubt anyone who points out that
>>> "lim(x -> ∞)" is a syntax error IF "∞ is not a number"
>> You don't get to say what the syntax means. lim(x->a) and lim(x->oo)
>> are well-established form used to mean two quite different kinds of
>> limits, despite the similarity in the syntax.
> Yes, I do get to say it. And if you don't like me saying it - I will repeat it louder.
>
> lim(x -> a) means EXACTLY
>
> let a = ∞
> lim(x -> a)
>
>

Except there is a SPECIAL rule for limits to the non-finite-number infinity.

Thus you can't use the the rule for limits to a real number.

Definition aren't just syntax, but can refer to semantics

lim(x -> a) has two different definitions.

One for a being a member of the Real Numbers (or a Complex Number), and
another for a being an infinity.

Richard Damon

unread,
Aug 15, 2022, 7:15:55 PMAug 15
to
On 8/15/22 10:52 AM, wij wrote:
> On Monday, 15 August 2022 at 20:02:58 UTC+8, richar...@gmail.com wrote:
>> On 8/15/22 5:38 AM, wij wrote:
>>> On Monday, 15 August 2022 at 08:34:39 UTC+8, richar...@gmail.com wrote:
>>>> On 8/14/22 7:35 PM, wij wrote:
>>>>> The vague, no-logic concept of infinity seems dominated people's mind.
>>>>> What is infinity? What does "lim(x→∞) f(x)" mean?
>>>>>
>>>>> If infinity is merely a 'concept', not a number, what does x approach to?
>>>>> If x is not getting "closer" to ∞? What does 'approach' mean?
>>>>> Therefore, ∞-(x+1) < ∞-x must be valid inequality to mean x+1 is closer than x to infinity ∞.
>>>>>
>>>>> But valid what? Most people agree ∀n∈ℕ, n<∞.
>>>>>
>>>>> Is x+1 not closer than x to infinity?
>>>>> So, infinity ∞ must have arithmetic meaning. Here is one:
>>>>> The multiplicative inverse of ∞ is 1/∞, the additive inverse is -∞
>>>>>
>>> Let A= lim(n->∞) 1-1/2^n = 0.999...
>>> B= lim(n->∞) 1-1/10^n = 0.999...
>>>
>>> Assume A=B
>>> <=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
>>> <=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n
>> And this step is invalid. You either multiplied by a "non-number" or
>> divided by zero depending on the steps you did to make that transition.
>>
>> This is the problem of assuming that "infinity" is a number.
>>> <=> lim(n->∞) 1 = lim(n->∞) 1/5^n
>>> <=> 1=0
>>>
> Ben made an error and (assume he saw my reply to Andy) made an error again.
>
> Let A= lim(n->∞) 1-1/2^n = 0.999...
> B= lim(n->∞) 1-1/10^n = 0.999...
>
> Assume A=B
> <=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
> <=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n
> <=> (lim(n->∞) 2^n)*(lim(n->∞) 1/2^n) = (lim(n->∞) 2^n)*(lim(n->∞) 1/10^n)

As Ben points out lim(n->inf)*2^n) is Not a Number, so the limit doesn't
exist so you can't multiply by it.

If you don't follow the rules of the Math you are using, you get
unreliable results.

> <=> lim(n->∞) 2^n/2^n = lim(n->∞) 2^n/10^n
> <=> lim(n->∞) 1 = lim(n->∞) 1/5^n
> <=> 1=0
>
> I wonder how much does you guys really understand you are talking?


I wonder if YOU know what you are talking about.

dklei...@gmail.com

unread,
Aug 15, 2022, 8:36:01 PMAug 15
to
There were a couple of hundred relevant papers. I read them and tried a
slightly different approach. I uncovered a dozen or more new relationships
and theorems. The most noteworthy is still used from time to time (called
the Kleinecke-Shirokov Theorem). I am a Platonist in the sense I believe
mathematical entities exist and are discovered. Not invented.

If you want to wonder at mathematics admire the monster - the largest
sporatic simple group. Found not invented.

Ben Bacarisse

unread,
Aug 15, 2022, 9:20:20 PMAug 15
to
Skep Dick <skepd...@gmail.com> writes:

> On Monday, 15 August 2022 at 18:39:33 UTC+2, Ben Bacarisse wrote:
>> Skep Dick <skepd...@gmail.com> writes:
>>
>> > On Monday, 15 August 2022 at 16:52:32 UTC+2, wyni...@gmail.com wrote:
>>
>> >> Ben made an error and (assume he saw my reply to Andy) made an error again.
>> >> Let A= lim(n->∞) 1-1/2^n = 0.999...
>> >> B= lim(n->∞) 1-1/10^n = 0.999...
>> >>
>> >> Assume A=B
>> >> <=> lim(n->∞) 1-1/2^n = lim(n->∞) 1-1/10^n
>> >> <=> lim(n->∞) 1/2^n = lim(n->∞) 1/10^n
>> >> <=> (lim(n->∞) 2^n)*(lim(n->∞) 1/2^n) = (lim(n->∞) 2^n)*(lim(n->∞) 1/10^n)
>> >> <=> lim(n->∞) 2^n/2^n = lim(n->∞) 2^n/10^n
>> >> <=> lim(n->∞) 1 = lim(n->∞) 1/5^n
>> >> <=> 1=0
>> >> I wonder how much does you guys really understand you are talking?
>> >
>> > You are 100% correct when using the Hyperreal numbers! That is *R not
>> > R.
>> Wij is not working in *R. He cites standard rules about limits in R
>> (rules he or she does not properly understand) to support this bogus
>> proof.
> He has openly told you what theorems he is interested in!
> None of the theorems he wants hold in R.
> Most of the theorems he wants hold in *R

Not so.

> Why are you dragging him down instead of pulling him up?!?
>
>>Cranks don't want to be correct in some "other" system (though,
>> as it happens, 0.999... = 1 in *R as well as in R), they want
>> conventional wisdom to be wrong.
>
> Bullshit. 0.999... is meaningless in *R. It's a syntax error.

No. 0.990... means the same in R and in *R. It's an infinite sum. The
sum is 1 in both R and in *R. Just saying stuff is not how to do
mathematics.

--
Ben.

Ben Bacarisse

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Aug 15, 2022, 9:46:58 PMAug 15
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Skep Dick <skepd...@gmail.com> writes:

> On Monday, 15 August 2022 at 18:25:25 UTC+2, Ben Bacarisse wrote:
>> Skep Dick <skepd...@gmail.com> writes:
>>
>> > On Monday, 15 August 2022 at 17:45:59 UTC+2, Ben Bacarisse wrote:
>> >> Obviously if you don't understand the basics of real analysis, you will
>> >> doubt anyone who points them out.
>> >
>> > Obviously, if you don't understand the basics of syntax, semantics,
>> > bound and unbound variables you will doubt anyone who points out that
>> > "lim(x -> ∞)" is a syntax error IF "∞ is not a number"
>> You don't get to say what the syntax means. lim(x->a) and lim(x->oo)
>> are well-established form used to mean two quite different kinds of
>> limits, despite the similarity in the syntax.
>>
> Lets try that again... without the errors.
>
> Yes, I do get to say it. And if you don't like me saying it - I will
> repeat it louder.

Why would you do that? Anyway, don't worry; I do like you saying it.
Say it again and again, please.

> lim(x -> ∞) means EXACTLY THE SAME THING AS
>
> let a = ∞
> lim(x -> a)

The two kinds of limit are defined differently for rather obvious
reasons.

--
Ben.

Skep Dick

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Aug 16, 2022, 2:54:40 AMAug 16
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For rather obvious reaons the two different notations are semantically identical.

Either "∞" is a bound symbol; or "∞" is an ubound symbol in the expression "lim(x -> ∞)".

This is undergraduate computer science stuff. Compiler theory.

https://en.wikipedia.org/wiki/Name_binding



Skep Dick

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Aug 16, 2022, 4:42:45 AMAug 16
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On Tuesday, 16 August 2022 at 03:46:58 UTC+2, Ben Bacarisse wrote:
> Why would you do that? Anyway, don't worry; I do like you saying it.
> Say it again and again, please.
At your service...

The expression "lim(x -> 0)" means exactly the same thing as "lim(x-> ∞)" when 0 is bound to ∞
The expression "lim(x -> 1)" means exactly the same thing as "lim(x-> ∞)" when 1 is bound to ∞
The expression "lim(x -> 2)" means exactly the same thing as "lim(x-> ∞)" when 2 is bound to ∞
... <------------- Once we ignore the discontinuity in the continuum.
The expression "lim(x -> ∞)" means exactly the same thing as "lim(x-> ∞)" when ∞ is bound to ∞


Fred. Zwarts

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Aug 16, 2022, 4:56:55 AMAug 16
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Op 15.aug..2022 om 01:35 schreef wij:
> The vague, no-logic concept of infinity seems dominated people's mind.
> What is infinity? What does "lim(x→∞) f(x)" mean?
>
> If infinity is merely a 'concept', not a number, what does x approach to?
> If x is not getting "closer" to ∞? What does 'approach' mean?
> Therefore, ∞-(x+1) < ∞-x must be valid inequality to mean x+1 is closer than x to infinity ∞.
>
> But valid what? Most people agree ∀n∈ℕ, n<∞.
>
> Is x+1 not closer than x to infinity?
> So, infinity ∞ must have arithmetic meaning. Here is one:
> The multiplicative inverse of ∞ is 1/∞, the additive inverse is -∞
>
> All in all, that is the definition of infinity (the symbol '∞') proposed.
> All is that simple, the usage treating ∞ as if it is a unique number is
> safe-guaranteed, what left is interpretation. Though I think I figured this
> part (merely means a procedure never terminate), there may be lots more
> instances to test its interpretation in various scenario.

Nice idea. Let's forget what was told us by our teachers and follow our
intuition. ∞ can be used in equations just as any other number.
1/∞ = 0 and when 1/y=x it follows that 1/x=y, so 1/0=∞.
Similarly, 1/-∞ = 0, so 1/0 = -∞ , therefore -∞ = 1/0 = ∞ .
exp(∞) = ∞ and exp(-∞) = 0, so, because ∞ = -∞, 0 = ∞ .
exp(0) = 1 and exp(-∞) = 0, so 1 = 0 = ∞ = -∞ .
1+0=1 and 1+1=2, so 2 = 1 = 0 = ∞ = -∞.
We can continue and show that all numbers are equal.
This leads to what our intuition already expected: a better society,
where numbers are no longer discriminated, but all numbers are equal.
Wouldn't this mathematical model solve a lot of problems in our world?
It is suspected that capitalistic mathematical teachers try to hide this
truth from us to let us believe that people with more money are richer
and can pay more than people with less money, but there is no real
difference.

Skep Dick

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Aug 16, 2022, 6:31:39 AMAug 16
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On Tuesday, 16 August 2022 at 01:15:55 UTC+2, richar...@gmail.com wrote:
> If you don't follow the rules of the Math you are using, you get
> unreliable results.
If you can't make "The Rules of the the Math" explicit, and enforceable by a compiler - it's nobody's fault that the rules can be interpreted as anyone chooses.


Richard Damon

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Aug 16, 2022, 7:22:14 AMAug 16
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Nope, they have different meanings so are semantically different.

>
> Either "∞" is a bound symbol; or "∞" is an ubound symbol in the expression "lim(x -> ∞)".
>
> This is undergraduate computer science stuff. Compiler theory.
>
> https://en.wikipedia.org/wiki/Name_binding
>

Oh, "∞" is a "bound" symbol, it just isn't bound to a "value" but a concept.

You don't seem to understand the difference.

The limit operator is overloaded on the "type" of its operator.

If the limit term has a finite value, it means one thing, if it is an
infinite value it means another.

The two meanings have a lot of similarity, but are subtlety different,
because the definition for finite values isn't applicable for infinite
values.

Richard Damon

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Aug 16, 2022, 7:23:41 AMAug 16
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Nope, because you CAN'T just ignore that discontinuity, because the
universe radically changes when you do.


Richard Damon

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Aug 16, 2022, 7:29:22 AMAug 16
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Lets see your complier enforce a speed limit.

The rules are well established, and well known.

And it IS possible to express at least most of them (not sure if all) as
a set of rules that you can put into an appropriate rule system.

Perhaps one issue is that some are semantic, so can't always be detected
at compile time depending on your type system.

For instance, we don't know if 1/x is a valid operation unless we know
that x doesn't have the value 0.

And, it IS your fault for not following them, as YOU are the one
claiming to be working inside them when you make a statement about how
the system work.

You need to actually know a system before you can make actual statements
about it.

Skep Dick

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Aug 16, 2022, 7:54:49 AMAug 16
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On Tuesday, 16 August 2022 at 13:23:41 UTC+2, richar...@gmail.com wrote:
> Nope, because you CAN'T just ignore that discontinuity, because the
> universe radically changes when you do.
Noooo! Waaaaay! Are you serious!?!

Here is the continuum for you: |------------------------------->∞

Could you please draw me a line (a discontinuity) which splits the continuum into "finite" and "infinite" parts ?

Skep Dick

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Aug 16, 2022, 8:08:20 AMAug 16
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On Tuesday, 16 August 2022 at 13:22:14 UTC+2, richar...@gmail.com wrote:
> Oh, "∞" is a "bound" symbol, it just isn't bound to a "value" but a concept.
Shut the fuck up, sophist.

1 is symbol bound to a concept.
2 is symbol bound to a concept.
3 is symbol bound to a concept.
∞ is symbol bound to a concept.

> You don't seem to understand the difference.
I think it's time you stopped lying about my understanding.

Either symbols is a pointer to a valid object, or it isn't.

And you know what happens when you de-reference null-pointers...

> The limit operator is overloaded on the "type" of its operator.
You are running for the hills of polymorphism and type theory. It's not going to help you.
Yes the lim() operator is overloaded. So lets pretend we are talking about limits on Real numbers.

lim( x:ℝ -> a) with a bound to ∞ means the EXACT SAME THING as lim( x:ℝ -> ∞ )

Here is your number line ℝ: 0 |-----------------------------------------> ∞

Could you please draw me a line where a finite number in ℝ begins approaching ∞?

> If the limit term has a finite value, it means one thing, if it is an
> infinite value it means another.
Bullshit. I don't care about the value of the limit.

I care about the meaning of lim( x:ℝ -> ∞ )
if ALL numbers in ℝ are finite, and NO finite number is closer to infinity than any other then what the fuck does it mean for a real number to approach infinity?

> The two meanings have a lot of similarity, but are subtlety different,
> because the definition for finite values isn't applicable for infinite
> values.

Here is your number line ℝ: 0 |-----------------------------------------> ∞

Could you please put a mark on it (adiscontinuity!) where the "finite values" stop and the "infinite values" start?







Skep Dick

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Aug 16, 2022, 8:28:52 AMAug 16
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On Tuesday, 16 August 2022 at 13:29:22 UTC+2, richar...@gmail.com wrote:
> Lets see your complier enforce a speed limit.
Ooooh. You think I am talking about a concrete compiler?
I am talking about the abstract compiler in your head.

> The rules are well established, and well known.
Bullshit!

Stop talking about "The Rules" and show them to me already.
Stop talking about your God and show him to me.

> And it IS possible to express at least most of them (not sure if all) as
> a set of rules that you can put into an appropriate rule system.
You haven't expressed ANY of them yet.

> Perhaps one issue is that some are semantic, so can't always be detected
> at compile time depending on your type system.
Nonsense. Encode your semantics into your type system then!

What is the domain of the lim(x -> y) type?

> For instance, we don't know if 1/x is a valid operation unless we know
> that x doesn't have the value 0.
Whether X is allowed to be bound to 0 in 1/x is an arbitrary choice.

You can forbid it. And mandate that 1/0 is undefined.
You can allow it. And mandate that 1/0 is defined.

Both arguments (arbitrary choices) are an appeal to authority.

> And, it IS your fault for not following them, as YOU are the one
> claiming to be working inside them when you make a statement about how
> the system work.
> You need to actually know a system before you can make actual statements
> about it.

Bullshit. I have repeated it over and over. That I am NOT working INSIDE your system.
I have told you that I am working OUTSIDE of your system.

I am not making POSITIVE statements about how your system works.
I am making NEGATIVE statements about your system DOESN'T work.



Alan Mackenzie

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Aug 16, 2022, 8:35:22 AMAug 16