S = Lim S in this case, contradicts the fact that 1/3 has no representation in base 10.

However, it also leads to even more absurd definitions, that is, pi=3.14159...  when in fact it is known that pi has NO measure.


https://www.youtube.com/watch?v=5hulvl3GgGk

https://www.youtube.com/watch?v=w8s_8fNePEE


Your comments are unwelcome shit and will be ignored.

This is posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.

1/3 has a representation as g-adic.
Similarly, pi is a clearly defined value and also, as any real number, has a g adic represantation.

It has a representation in many other radix systems too you imbecile. Your point? I see. None.

On Sunday, 12 February 2017 08:30:17 UTC-6, shio...@googlemail.com  wrote:

Pi is a clearly defined value? What is you moron? Don't tell me circumference length divided by diameter.

Pi only has a number associated with it, if such a number describes its MEASURE you infinite FUCKING MORON!

Pi is clearly defined as value by many different expressions, for example as a certain integral.
Since that value is unique, it defines pi.

And my point is that the radix system in actual math IS the 10-adic system, and that 1/3 has an expression in it.
It is infinitely long, but that is no problem. It can be infinitely long, that's not against the law.

what is g-adic?...  anyway,
PI is the ratio of diameter to the area of the sphere,
which requires some thought, but it is about 32 hundredths;
the ratio of circmference to the diameter is 31 tenths; see?

> Pi only has a number associated with it, if such a number describes its MEASURE

<deletia excretia>

On Sunday, February 12, 2017 at 8:14:11 AM UTC-5, John Gabriel wrote:
> S = Lim S in this case, contradicts the fact that 1/3 has no representation in base 10.
>

JG also doesn't believe in the number pi, the square root of 2, negative numbers or even zero. JG is a full-on crank and a troll. He claims to be greatest mathematician ever, but cannot even prove that 2+2=4 in his goofy system! (See the thread, "The spamming troll John Gabriel in his own words.")


Dan

On Sunday, 12 February 2017 10:38:24 UTC-6, shio...@googlemail.com  wrote:
> On Sunday, 12 February 2017 16:29:36 UTC+1, John Gabriel  wrote:
> > On Sunday, 12 February 2017 08:30:17 UTC-6, shio...@googlemail.com  wrote:
> > > On Sunday, 12 February 2017 14:14:11 UTC+1, John Gabriel  wrote:
> > > > S = Lim S in this case, contradicts the fact that 1/3 has no representation in base 10.
> > > >
> > > > However, it also leads to even more absurd definitions, that is, pi=3.14159...  when in fact it is known that pi has NO measure.
> > > >
> > > >
> > > > https://www.youtube.com/watch?v=5hulvl3GgGk
> > > >
> > > > https://www.youtube.com/watch?v=w8s_8fNePEE
> > > >
> > > >
> > > > Your comments are unwelcome shit and will be ignored.
> > > >
> > > > This is posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.
> > >
> > > 1/3 has a representation as g-adic.
> > > Similarly, pi is a clearly defined value and also, as any real number, has a g adic represantation.
> >
> > Pi is a clearly defined value? What is you moron? Don't tell me circumference length divided by diameter.
> >
> > Pi only has a number associated with it, if such a number describes its MEASURE you infinite FUCKING MORON!
>
> Pi is clearly defined as value by many different expressions, for example as a certain integral.

No fool. It's not clearly defined as an integral. It's only definition is circumference / diameter.  Any integral you claim represents pi does not, because it is a "limit" that the integral returns, not the value of pi. In this case, no one knows what the fuck is the limit because pi has no measure.

> Since that value is unique, it defines pi.

Asshole. That it is unique means shit. Moooorrrrroooooon. 1/3 is also unique.

>
> And my point is that the radix system in actual math IS the 10-adic system, and that 1/3 has an expression in it.

By saying this, you are claiming that it is possible to represent 1/3 in base 10. A number theorem contradicts your claim. You are an idiot. Period.

> It is infinitely long, but that is no problem. It can be infinitely long, that's not against the law.

Bwaaa haaaaa haaaaa.

The only thing that is infinitely long is your STUPIDITY.

Chuckle. Fuck off you idiot. I piss and shit on you!

Last time I looked at pi it was 6*atan(sqrt(1/3)) and
somebody made a nice streaming algorithm from it.

Computing the Digits in π
Carl D. Offner - October 15, 2015
http://www.cs.umb.edu/~offner/files/pi.pdf

Am Sonntag, 12. Februar 2017 21:27:20 UTC+1 schrieb John Gabriel:

Street math:
“I knew this kid. He found the exact value of pi. He went nuts.”

JG went nuts without finding anything.

[3 divided into 1] scaled to [3 divided into 10]:

10 - 3 = 7, sum of counts of 1
7 - 3 = 4, sum of counts of 2
4 - 3 = 1, sum of counts of 3
1 - 3 < 0, remainder of 1

Result of 3 + 1/3

Then [3 divided into 10] scaled to [3 divided into 1]:

Result of 3/10 + (1/3 /10) = 3/10 + 1/30
.
.

 

[3 divided into 1] scaled to [3 divided into 10]:

10 - 3 = 7, sum of counts of 1
7 - 3 = 4, sum of counts of 2
4 - 3 = 1, sum of counts of 3
1 - 3 < 0, remainder of 1

Result of 3 + 1/3

Then [3 divided into 10] scaled to [3 divided into 1]:

Result of 3 + 1/3 divided by 10 = 3/10 + 1/30
.

""
> > Pi is clearly defined as value by many different expressions, for example as a certain integral.
>
> No fool. It's not clearly defined as an integral. It's only definition is circumference / diameter.  Any integral you claim represents pi does not, because it is a "limit" that the integral returns, not the value of pi. In this case, no one knows what the fuck is the limit because pi has no measure.""

No. Simply and flat out no.
The integral well defines a value (and so can a limit by the way).
There is no other value that integral could have than pi, so it defines it.

Also, as i said, defining something as the limit of a sequence bears no problem at all, since the only important thing in a definition is that it uniquely defines something.

For example, 1/3 can easily be defined as limit of the sequences 0,3, 0,33 and so on because that limit is unique and no other limit of any such sequence (the sequences which define g-adics) can have it as limit.

""
> > Since that value is unique, it defines pi.
>
> Asshole. That it is unique means shit. Moooorrrrroooooon. 1/3 is also unique.
> ""

That it is unique is the only important thing here, and yes, the same applies for 1/3 and its limit definition.

""
> >
> > And my point is that the radix system in actual math IS the 10-adic system, and that 1/3 has an expression in it.
>
> By saying this, you are claiming that it is possible to represent 1/3 in base 10. A number theorem contradicts your claim. You are an idiot. Period.
> ""

No i am not. I am saying that 1/3 has a representation as 10-adic, which may aswell be infinite.
You might want to rehash your knowledge of g-adics.
They are always infinite, even if infinitely many digits are 0.
Even 1 is defined by an infinitely long sequence as g-adic.

""
> The only thing that is infinitely long is your STUPIDITY.
>
> Chuckle. Fuck off you idiot. I piss and shit on you!
""

But how you a little man like you be able to do that?

Choke on ball-sack, you boring idiot.

On Sunday, 12 February 2017 16:01:36 UTC-6, shio...@googlemail.com  wrote:

> > No fool. It's not clearly defined as an integral. It's only definition is circumference / diameter.  Any integral you claim represents pi does not, because it is a "limit" that the integral returns, not the value of pi. In this case, no one knows what the fuck is the limit because pi has no measure.""
>
> No. Simply and flat out no.

Yes, you idiot. Yes.

> The integral well defines a value (and so can a limit by the way).

No and NO and NO again. An improper integral is a limit and limits require the existence of "irrational numbers" which DO NOT exist. Therefore, there is no
measure and hence no number.

> There is no other value that integral could have than pi, so it defines it.

Idiot. If a limit is not a rational number, then it is NOT a number. The only numbers are RATIONAL numbers. There are NO other numbers. Deal with it moron.

In mathematics, we talk about things being well defined. A number is well defined as the measure of a magnitude.


You are not a mathematician and have ZERO qualifications in mathematics. Plainly, you are an idiot.

>

<excrement>

it was said that a limit cannot be irrational,
such as the secondr00t of a half, or
the secondr00t(...9999.o)

> <deletia excretia>

""
> > The integral well defines a value (and so can a limit by the way).
>
> No and NO and NO again. An improper integral is a limit and limits require the existence of "irrational numbers" which DO NOT exist. Therefore, there is no
> measure and hence no number.""

What you call measure is completely irrelevant for the definition of a mathematical element.
And obviously irrational numbers exist.
To deny their existance means basically to deny the existance of infinite sequences and there is no basis for that.

>""
> > There is no other value that integral could have than pi, so it defines it.
>
> Idiot. If a limit is not a rational number, then it is NOT a number. The only numbers are RATIONAL numbers. There are NO other numbers. Deal with it moron.
>
> In mathematics, we talk about things being well defined. A number is well defined as the measure of a magnitude.
> ""

What you call number is irrelevant for mathematics.
Pi is a proper and well defined element of the ring of real numbers.

In mathematics, your definition of measure and number both have absolutely no place, because they are bad and lead to nothing.

If you want, yes, say that after your definition, pi is not a number.
It is not like anyone cares what you think is a number because pi as a ring element is absolutely well defined and useful.

""
> You are not a mathematician and have ZERO qualifications in mathematics. Plainly, you are an idiot.
> ""

I got a degree unlike you, which qualifies me to teach you.

Yes, if you keep on telling yourself that enough times, you will believe it. Chuckle.

Get new material. No one cares about how badly you don't understand anything.

yeah, get some new material.  like,
if the area of the sphere is the unit,
what is a)
the circumference, and b)
the diameter

well, I can't think if a sum to approach the r00t of a half,
although 2 is just 1/1 +1/2 +1/4 + some dots

On Sunday, February 12, 2017 at 8:24:00 PM UTC-8, shio...@googlemail.com wrote:
> ""
> > > The integral well defines a value (and so can a limit by the way).
> >
> > No and NO and NO again. An improper integral is a limit and limits require the existence of "irrational numbers" which DO NOT exist. Therefore, there is no
> > measure and hence no number.""
>
> What you call measure is completely irrelevant for the definition of a mathematical element.
> And obviously irrational numbers exist.
> To deny their existance means basically to deny the existance of infinite sequences and there is no basis for that.

I think John suffers from some mental condition. Nothing serious AFAICT but
just enough to make him produce those nonsensical posts here.

--
Jan

Don't need to tell it to myself, got it signed by my uni.

ya, most probably.

On Monday, 13 February 2017 16:47:43 UTC-6, Jan  wrote:
> On Sunday, February 12, 2017 at 8:24:00 PM UTC-8, shio...@googlemail.com wrote:
> > ""
> > > > The integral well defines a value (and so can a limit by the way).
> > >
> > > No and NO and NO again. An improper integral is a limit and limits require the existence of "irrational numbers" which DO NOT exist. Therefore, there is no
> > > measure and hence no number.""
> >
> > What you call measure is completely irrelevant for the definition of a mathematical element.
> > And obviously irrational numbers exist.
> > To deny their existance means basically to deny the existance of infinite sequences and there is no basis for that.
>
> I think John suffers from some mental condition. Nothing serious AFAICT but
> just enough to make him produce those nonsensical posts here.

The problem lies not with me but those who are too stupid to understand.

Euler blundered seriously by defining S = Lim S.

https://www.linkedin.com/pulse/eulers-worst-definition-lim-john-gabriel/

>
> --
> Jan

On Sunday, February 12, 2017 at 8:22:46 PM UTC-7, John Gabriel wrote:

> Idiot. If a limit is not a rational number, then it is NOT a number. The
> only numbers are RATIONAL numbers. There are NO other numbers. Deal with
> it moron.

We all have to deal with facts. Why we have to deal with the things you
say, however, is not apparent: you may claim that you know better than
the majority of the people who call themselves mathematicians, but except
for making loud and annoying claims, you really present no reason why we
should believe any of this stuff.

Now, it _is_ true that we don't need any irrational numbers to measure
distances in the real world, because we can't measure them accurately
enough to even need more than a tiny fraction of the rational numbers.

Maybe you have figured out a way to still do calculus while using a
number system that has holes in it like the rational number system does,
instead of one where stuff like Dedekind cuts are used to plug all the
wholes. Why it's worth taking the extra trouble to care about that kind
of stuff is more than merely unclear.

John Savard

On Sunday, February 12, 2017 at 8:22:46 PM UTC-7, John Gabriel wrote:

> You are not a mathematician and have ZERO qualifications in mathematics.
> Plainly, you are an idiot.

If it's plain, why is the whole world under the sway of mythmatics instead
of what you call mathematics?

John Savard

On Saturday, 30 September 2017 01:32:28 UTC-5, Quadibloc  wrote:
> On Sunday, February 12, 2017 at 8:22:46 PM UTC-7, John Gabriel wrote:
>
> > Idiot. If a limit is not a rational number, then it is NOT a number. The
> > only numbers are RATIONAL numbers. There are NO other numbers. Deal with
> > it moron.
>
> We all have to deal with facts. Why we have to deal with the things you
> say, however, is not apparent: you may claim that you know better than
> the majority of the people who call themselves mathematicians, but except
> for making loud and annoying claims, you really present no reason why we
> should believe any of this stuff.

Actually I do and there are mainstream academics who agree with me.

Dr. Ahmad Zainy Al-Yasry (PhD mathematics) stated on LinkedIn that my article

https://www.linkedin.com/pulse/how-we-got-numbers-john-gabriel-1

is the most accurate account of how we got numbers. He has never read anything with more clarity, accuracy and rigour.

Your statement is just an assertion.

>
> Now, it _is_ true that we don't need any irrational numbers to measure
> distances in the real world, because we can't measure them accurately
> enough to even need more than a tiny fraction of the rational numbers.

We can't measure incommensurable magnitudes. If we had irrational numbers, there would be no need to measure them because a number is the measure of a magnitude.

>
> Maybe you have figured out a way to still do calculus while using a
> number system that has holes in it like the rational number system does,

The expression "number system" is meaningless nonsense and those who use it clearly do not understand what is a number.  There are no holes because there is no such thing as a "real" number line - it is a myth.

> instead of one where stuff like Dedekind cuts are used to plug all the
> wholes.

By the ridiculous and illogical notion that every distance is related to some indistinguishable point? Absurd. ...

Most students have no influence or sway what their curricula entail.
The world is under the sway of the academic trash heap composed of ignorant, arrogant and incompetent academics who refuse to see the light.

In due time, the ideas of the New Calculus will replace the rot of mainstream calculus. As with everything, people are very slow to catch on to the truth. Already, there are thousands who have learned the New Calculus and have applied it. That group is growing daily.

Understanding and progress only comes through well-formed concepts, not anti-mathematical nonsense such as infinity, infinitesimals and limit theory tha...

Den söndag 12 februari 2017 kl. 14:14:11 UTC+1 skrev John Gabriel:
> S = Lim S in this case, contradicts the fact that 1/3 has no representation in base 10.
>
> However, it also leads to even more absurd definitions, that is, pi=3.14159...  when in fact it is known that pi has NO measure.
>
>
> https://www.youtube.com/watch?v=5hulvl3GgGk
>
> https://www.youtube.com/watch?v=w8s_8fNePEE
>
>
> Your comments are unwelcome shit and will be ignored.
>
> This is posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.

ANd so you continue your shit about focusing only on your own delusions, common, grow up.

On Sunday, February 12, 2017 at 8:14:11 AM UTC-5, John Gabriel wrote:
> S = Lim S in this case, contradicts the fact that 1/3 has no representation in base 10.
>

S = Lim S is nonsense to begin with -- one of your biggest blunders ever, Troll Boy. And that's saying a LOT!  

> However, it also leads to even more absurd definitions, that is, pi=3.14159...  when in fact it is known that pi has NO measure.
>

Like it or not, Troll Boy, pi is a real number. So is root 2, negative numbers, and 0. Your hero would have had you gassed and incinerated as a mental defective and for attempting to sabotage the education system with your kooky ideas.


Dan

Download my DC Proof 2.0 software at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

Your days are numbered moron. You will be ashamed of everything that you have accused me of - well, that's if you live long enough.  

Please, whatever you do, do not produce any offspring. There are enough stupid people in academia and we don't need more.

I don't know what is more stupid, to make "Euler S =
Lim S" a case, or to think Newton series work
without convergence.

Here is an example of a Newton rule, that doesn't work
when diverging series are invokved, we quickly get the
contradiction:

     0 = 1

I am using Newtons schema from Page 158/159: in his
THE METHOD OF FLUXIONS AND INFINITE SERIES. For
a picture of this schema see here:
https://gist.github.com/jburse/e60242e2e02e611e4373df55bfc37953#gistcomment-2211516

So I apply the schema:

    1 + 1 + 1 + 1 + 1 + 1 ...

      - 1 - 1 - 1 - 1 - 1 ...

    ==================================
    0 + 0 + 0 + 0 + 0 + 0 ... =   1

On Saturday, September 30, 2017 at 9:05:28 AM UTC-7, John Gabriel wrote:
> On Saturday, 30 September 2017 08:14:55 UTC-5, Zelos Malum  wrote:
> > Den söndag 12 februari 2017 kl. 14:14:11 UTC+1 skrev John Gabriel:
> > > S = Lim S in this case, contradicts the fact that 1/3 has no representation in base 10.
> > >
> > > However, it also leads to even more absurd definitions, that is, pi=3.14159...  when in fact it is known that pi has NO measure.
> > >
> > >
> > > https://www.youtube.com/watch?v=5hulvl3GgGk
> > >
> > > https://www.youtube.com/watch?v=w8s_8fNePEE
> > >
> > >
> > > Your comments are unwelcome shit and will be ignored.
> > >
> > > This is posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.
> >
> > ANd so you continue your shit about focusing only on your own delusions, common, grow up.
>

> Your days are numbered moron. You will be ashamed of everything that you have accused me of - we...

Am Samstag, 30. September 2017 23:14:36 UTC+2 schrieb mitchr...@gmail.com:
>
> .9 repeating and One share a sameness. They are quantities
> that are different by the infinitely small.
> .9 repeating is a transcendental One; the First quantity
> below one. The infinitely small difference means a shared
> sameness that is still not absolutely same.
>
> Mitchell Raemsch

If there is a quantity between 0.999... and 1 and, therefore, these are two different points on the number line then you should define the distance between these two points. If you don't, then your first quantity is simply undefined.

'infinitely small' is not a definition. There are no two distinct points on the number line 'infinite(simal)ly' far away from each other.

On Saturday, 30 September 2017 16:04:36 UTC-5, burs...@gmail.com  wrote:
> I don't know what is more stupid, to make "Euler S =
> Lim S" a case, or to think Newton series work
> without convergence.

Huh? You need to slow down with all that pot because your offspring will be affected. As it is I am worried about what kind of morons you will produce.

Who said Newton series works without convergence?

Jan, the rest of your comment is just too absurd.

In future, be "brief". No, I am not talking about your sex life man. I mean your comments just go on and on. If you want to hold a discussion, keep your comments short and focused. You are a scatterbrain. Hope this helps.

netzweltler explained on 9/30/2017 :
They do not differ
   by infinite small.
They differ only
   by none at all.

Well, if you define 0.999... to be equal to a brick, then a brick and 0.999... differ by none at all.

There is not a single support for this bullshit equality aside from S = Lim S and this is an ill-formed definition - the Eulerian Blunder.

Well you wrote here Newton didn't consider infinity,
and you say he can define partial sums without infinity.

Well this might be true, but you then go on and say
he used limits. But how do you get limits, without

https://groups.google.com/d/msg/sci.math/HIzzJSLsw60/vSOH7WnhAwAJ

knowing whether a series converges or not? For convergence
you need to make statement about infinitely many elements,

for example the Cauchy condition, is for infinitely many
pairs n,m, namely you need to know (or assume you know):

   forall n,m >= N(e) |an - am| =< e

The above looks like a pi-sentence, and is not verifiable
if we do not know much about {ak}. So you are in the waters of:

It is also familiar in the philosophy of science that most
hypotheses are neither verifiable nor refutable. Thus, Kant’s
antinomies of pure reason include such statements as that space
is infinite, matter is infinitely divisible, and the series of
efficient causes is infinite. These hypotheses all have the form

  forall x exists y P(x, y).

For example, infinite divisibility amounts to “for every
product of fission, there is a time by which attempts to cut
it succeed” and the infinity of space amounts to “for each
distance you travel, you can travel farther.”

https://www.an...

On Saturday, 30 September 2017 19:23:32 UTC-5, burs...@gmail.com  wrote:
> Well you wrote here Newton didn't consider infinity,
> and you say he can define partial sums without infinity.
>
> Well this might be true, but you then go on and say
> he used limits. But how do you get limits, without
>
> https://groups.google.com/d/msg/sci.math/HIzzJSLsw60/vSOH7WnhAwAJ

Very easy:  Limits don't care if the terms are all there or even there at all.

> For convergence you need to make statement about infinitely many elements,

Wrong. You arrive at the conclusion about convergence from only the general partial sum.

>
> for example the Cauchy condition, is for infinitely many
> pairs n,m, namely you need to know (or assume you know):

That is false. We use inference and the general term to arrive at the conclusion. Nothing about infinity anywhere.

>
>    forall n,m >= N(e) |an - am| =< e

The forall is true because of the conclusion. Nothing about infinity here also.

>
> The above looks like a pi-sentence, and is not verifiable
> if we do not know much about {ak}. So you are in the waters of:
>
> It is also familiar in the philosophy of science that most
> hypotheses are neither verifiable nor refutable. Thus, Kant’s
> antinomies of pure reason include such statements as that space
> is infinite, matter is infinitely divisible, and the series of
> efficient causes is infinite. These hypotheses all have the form
>
>   forall x exists y P(x, y).
>

> For example, infinite divisibility a...

So the product, its terms sn*tn might be observable,
but that the product is Cauchy is not observable directly.

That a series is Cauchy is neither effectively refutable
nor effectively verifiable. If you find e, n, m with:

    |an-am| > e

You still don't know whether there is N, where the series
behaves Cauchy. The full Cauchy condition is:

   forall e exists N forall n,m>=N |an-am|=<e

So it has the shape VEV.

Am Sonntag, 1. Oktober 2017 02:23:32 UTC+2 schrieb burs...@gmail.com:
> Well you wrote here Newton didn't consider infinity,
> and you say he can define partial sums without infinity.
>
> Well this might be true, but you then go on and say
> he used limits. But how do you get limits, without
>
> https://groups.google.com/d/msg/sci.math/HIzzJSLsw60/vSOH7WnhAwAJ
>
> knowing whether a series converges or not? For convergence
> you need to make statement about infinitely many elements,
>
> for example the Cauchy condition, is for infinitely many
> pairs n,m, namely you need to know (or assume you know):
>
>    forall n,m >= N(e) |an - am| =< e
>
> The above looks like a pi-sentence, and is not verifiable
> if we do not know much about {ak}. So you are in the waters of:
>
> It is also familiar in the philosophy of science that most
> hypotheses are neither verifiable nor refutable. Thus, Kant’s
> antinomies of pure reason include such statements as that space
> is infinite, matter is infinitely divisible, and the series of
> efficient causes is infinite. These hypotheses all have the form
>

>   forall x exists y...

This doesn't mean that math or Newton cannot identify
some series as Cauchy, but if we had a black box:

    +------+
    | a_n  |--->  You
    +------+

And you could take a ticket one by one with the
numbers of the sequence, you will never know whether

the series is Cauchy or not.
> > antinomies of pure reason inclu...

Math can even reason, that if these two boxes
are Cauchy, i.e. the series {an} and {bn}:

 
     +------+
     | a_n  |--->  You
     +------+

     +------+
     | b_n  |--->  You
     +------+

That then the following series will be Cauchy
as well, namely the product series {an*bn}:

     +----------+
     | a_n*b_n  |--->  You
     +----------+

You can construct the product series, this is
observable, you can just combine the two black boxes
and create a new black box. If each black box,

gives a rationl number, you can build a new
black box, for the product, but still being Cauchy
is not effectively refutable or effectively verifiable.

We only proved a formal implication:

  {an}, {bn} Cauchy ==> {an*bn} Cauchy

That {an*bn} is observable is rather trivial, and
that Newton could also do it is to expect. But if
you can construct {an*bn} doesn't mean you have solved
the limit problem for unknown series.
> > > you need to make ...

On Saturday, September 30, 2017 at 2:42:46 PM UTC-7, netzweltler wrote:
> Am Samstag, 30. September 2017 23:14:36 UTC+2 schrieb mitchr...@gmail.com:
> >
> > .9 repeating and One share a sameness. They are quantities
> > that are different by the infinitely small.
> > .9 repeating is a transcendental One; the First quantity
> > below one. The infinitely small difference means a shared
> > sameness that is still not absolutely same.
> >
> > Mitchell Raemsch
>
> If there is a quantity between 0.999... and 1

At some point there needs to be next quantities with
nothing in between.

and, therefore, these are two different points on the number line then you should define the distance between these two points. If you don't, then your first quantity is simply undefined.
>
> 'infinitely small' is not a definition.

It has a definition of being one divided by infinity
It can't be divided any further. It is The Infinitely divided One.
Their quantity difference is by the infinitely small.
This means there are no quantities in between them.

>There are no two distinct points on the number line 'infinite(simal)ly' far away from each other.

Mitchell Raemsch

Am Sonntag, 1. Oktober 2017 02:52:29 UTC+2 schrieb mitchr...@gmail.com:
> On Saturday, September 30, 2017 at 2:42:46 PM UTC-7, netzweltler wrote:
> > Am Samstag, 30. September 2017 23:14:36 UTC+2 schrieb mitchr...@gmail.com:
> > >
> > > .9 repeating and One share a sameness. They are quantities
> > > that are different by the infinitely small.
> > > .9 repeating is a transcendental One; the First quantity
> > > below one. The infinitely small difference means a shared
> > > sameness that is still not absolutely same.
> > >
> > > Mitchell Raemsch
> >
> > If there is a quantity between 0.999... and 1
>
> At some point there needs to be next quantities with
> nothing in between.

Here you say there is a quantity in between.

>
> and, therefore, these are two different points on the number line then you should define the distance between these two points. If you don't, then your first quantity is simply undefined.
> >
> > 'infinitely small' is not a definition.
>
> It has a definition of being one divided by infinity

> It can't be...

mitchr...@gmail.com has brought this to us :

> On Saturday, September 30, 2017 at 2:42:46 PM UTC-7, netzweltler wrote:
>> Am Samstag, 30. September 2017 23:14:36 UTC+2 schrieb mitchr...@gmail.com:
>>>
>>> .9 repeating and One share a sameness. They are quantities
>>> that are different by the infinitely small.
>>> .9 repeating is a transcendental One; the First quantity
>>> below one. The infinitely small difference means a shared
>>> sameness that is still not absolutely same.
>>>
>>> Mitchell Raemsch
>>
>> If there is a quantity between 0.999... and 1
>
> At some point there needs to be next quantities with
> nothing in between.

Not really, in fact in the reals if there are two different (nearly?)
adjacent real numbers then there is always a place between them for
another real number to occupy.

Even worse would be the case of two nearly adjacent rationals. If 0.999
repeating (a rational number) were to be taken as one of those numbers
and 1.000 repeating (another rational number) as the other then it is
easy to see that in the reals, many numbers (irrationals) would have to
exist between these two. The fact remains that these two numbers are
actually only two representations of the same exact number.

It seems counterintuitive when a number is viewed (or represented) as
an infinite unending '...

Am Sonntag, 1. Oktober 2017 13:56:01 UTC+2 schrieb FromTheRafters:
>
> It seems counterintuitive when a number is viewed (or represented) as

> an infinite unending 'process' of achieving better and better
> approximations, and that we can never actually reach the destination
> number. In my view, this sequence and/or infinite sum is a
> representation of the destination number "as if" we could have gotten
> there by that process.
If the process doesn't get us there then we don't get there. Where do you get your "as if" from?

> IOW "*After* infinitely many 'better'
> approximations" we reach the destination number *exactly* even if we
> cannot 'pinpoint' that number on the number line.
Please define "*After* infinitely many 'better' approximations". All we've got is infinitely many approximations - each approximation telling us that we get closer to 1 but don't reach 1. There is no *after* specified in this process.

> A 'limit' is not an
> approximation, it is the destination number (if there is one in that
> field) implied by the sequence or series in question.

On Saturday, 30 September 2017 19:32:08 UTC-5, burs...@gmail.com  wrote:
> So the product, its terms sn*tn might be observable,
> but that the product is Cauchy is not observable directly.
>
> That a series is Cauchy is neither effectively refutable
> nor effectively verifiable. If you find e, n, m with:
>
>     |an-am| > e
>
> You still don't know whether there is N, where the series
> behaves Cauchy. The full Cauchy condition is:
>
>    forall e exists N forall n,m>=N |an-am|=<e
>
> So it has the shape VEV.
>
> Am Sonntag, 1. Oktober 2017 02:23:32 UTC+2 schrieb burs...@gmail.com:
> > Well you wrote here Newton didn't consider infinity,
> > and you say he can define partial sums without infinity.
> >
> > Well this might be true, but you then go on and say
> > he used limits. But how do you get limits, without
> >
> > https://groups.google.com/d/msg/sci.math/HIzzJSLsw60/vSOH7WnhAwAJ
> >
> > knowing whether a series converges or not? For convergence
> > you need to make statement about infinitely many elements,
> >
> > for example the Cauchy condition, is for infinitely many
> > pairs n,m, namely you need to know (or assume you know):
> >
> >    forall n,m >= N(e) |an - am| =< e

Each time I think you can't say anything more stupid, you surprise me.
Listen stupid, the forall you have there is based on induction. There is no forall. A Cauchy sequence does not require "forall".  The "forall" is a result of inference.

> >
> > The above looks like a pi-sentence, and is not verifiable
> > if we do not know much about {ak}. So you are in the waters of:
> >
> > It is also familiar in the philosophy of science that most
> > hypotheses are neither verifiable nor refutable. Thus, Kant’s

> > antinomies of pure reason include such statements as that...

After serious thinking netzweltler wrote :

> Am Sonntag, 1. Oktober 2017 13:56:01 UTC+2 schrieb FromTheRafters:
>>
>> It seems counterintuitive when a number is viewed (or represented) as
>> an infinite unending 'process' of achieving better and better
>> approximations, and that we can never actually reach the destination
>> number. In my view, this sequence and/or infinite sum is a
>> representation of the destination number "as if" we could have gotten
>> there by that process.
> If the process doesn't get us there then we don't get there. Where do you get
> your "as if" from?

If you had sufficient time, then you would get there.

>> IOW "*After* infinitely many 'better'
>> approximations" we reach the destination number *exactly* even if we
>> cannot 'pinpoint' that number on the number line.
> Please define "*After* infinitely many 'better' approximations". All we've
> got is infinitely many approximations - each approximation telling us that we
> get closer to 1 but don't reach 1. There is no *after* specified in this
> process.

There is also no "time" mentioned, so why is there an assumption of a
process which takes time to complete? It is already completed (pi
exists as a n...

Am Sonntag, 1. Oktober 2017 15:20:16 UTC+2 schrieb FromTheRafters:
> After serious thinking netzweltler wrote :
> > Am Sonntag, 1. Oktober 2017 13:56:01 UTC+2 schrieb FromTheRafters:
> >>
> >> It seems counterintuitive when a number is viewed (or represented) as
> >> an infinite unending 'process' of achieving better and better
> >> approximations, and that we can never actually reach the destination
> >> number. In my view, this sequence and/or infinite sum is a
> >> representation of the destination number "as if" we could have gotten
> >> there by that process.
> > If the process doesn't get us there then we don't get there. Where do you get
> > your "as if" from?
>
> If you had sufficient time, then you would get there.

Show how time is involved in our process.

>
> >> IOW "*After* infinitely many 'better'
> >> approximations" we reach the destination number *exactly* even if we
> >> cannot 'pinpoint' that number on the number line.
> > Please define "*After* infinitely many 'better' approximations". All we've
> > got is infinitely many approximations - each approximation telling us that we
> > get closer to 1 but don't reach 1. There is no *after* specified in this
> > process.
>
> There is also no "time" mentioned, so why is there an assumption of a
> process which takes time to complete? It is already completed (pi

> exists as a number despite our inability to pinpoint it on the number
> line by using an infinite alternating sum or any of the other infinite
> processes) we just can't pinpoint it because we exist in a time
> constrained universe with processes ...

Not every forall is induction. Induction works
only for natural numbers.

But the epsilon is a forall over rational numbers
(or real number), what induction do you have in mind.

And why do you call Peano crapaxioms, when nevertheless
you say induction is needed here.

Ever incurred to you that Peano axioms capture
natural numbers and their induction?

The problem with bird brain John Gabriel is a kind
of simplicistic view of math. Things he can sometimes
do by himself are summed as

"You arrive at the conclusion about convergence from

only the general partial sum.".
This is like saying nothing.

Then the nonsense goes on:

"We use inference and the general term to arrive at the
conclusion. Nothing about infinity anywhere."

This is like saying nothing again.

It only focuses on the starting point, the series give
in a form {a_n}, which JG termed general partial sum,
and next JG termed it general term.

It completely neglects what the outcome is, what shape
it is, namely the convergence statement. And it also
completely neglects what "meas" there are to get from
A to B, i.e. form {a_n} for example to {a_n} Cauchy.

Now there came "induction", which is a nice guess,
but unfortunately it is probably only half an answer,
because "induction" is N, what do we need for Q?

The statement {a_n} Cauchy is:

   forall e exists N foral...

netzweltler formulated the question :

> Am Sonntag, 1. Oktober 2017 15:20:16 UTC+2 schrieb FromTheRafters:
>> After serious thinking netzweltler wrote :
>>> Am Sonntag, 1. Oktober 2017 13:56:01 UTC+2 schrieb FromTheRafters:
>>>>
>>>> It seems counterintuitive when a number is viewed (or represented) as
>>>> an infinite unending 'process' of achieving better and better
>>>> approximations, and that we can never actually reach the destination
>>>> number. In my view, this sequence and/or infinite sum is a
>>>> representation of the destination number "as if" we could have gotten
>>>> there by that process.
>>> If the process doesn't get us there then we don't get there. Where do you
>>> get  your "as if" from?
>>
>> If you had sufficient time, then you would get there.
> Show how time is involved in our process.

If you have to add a next number (like one quarter) to a previous
result of adding such previous numbers (like one plus one half) then
you have introduced time. Thee is a 'previous' calculation needed as
input to the next calculation. The idea that you 'never' get there (to
two) introduces time also. I'm with you, I don't think time has any
place in this.

>>>> IOW "*After* infinitely many 'better'
>>>> approximations" we reach the destination number *exactly* even if we

>>>> cannot 'pinpoint' that number on ...

On Sunday, October 1, 2017 at 12:22:43 AM UTC-7, netzweltler wrote:
> Am Sonntag, 1. Oktober 2017 02:52:29 UTC+2 schrieb mitchr...@gmail.com:
> > On Saturday, September 30, 2017 at 2:42:46 PM UTC-7, netzweltler wrote:
> > > Am Samstag, 30. September 2017 23:14:36 UTC+2 schrieb mitchr...@gmail.com:
> > > >
> > > > .9 repeating and One share a sameness. They are quantities
> > > > that are different by the infinitely small.
> > > > .9 repeating is a transcendental One; the First quantity
> > > > below one. The infinitely small difference means a shared
> > > > sameness that is still not absolutely same.
> > > >
> > > > Mitchell Raemsch
> > >
> > > If there is a quantity between 0.999... and 1
> >
> > At some point there needs to be next quantities with
> > nothing in between.
> Here you say there is a quantity in between.

It goes both ways. There are the in-betweens
until they reach to smallest or infinitely
small difference

Mitchell Raemsch

>
> >
> > and, therefore, these are two different points on the number line then you should define the distance between these two points. If you don't, then your first quantity is simply undefined.
> > >
> > > 'infinitely small' is not a definition.
> >
> > It has a definition of being one divided by infinity

> >...

Hello my little stupid.

Peano's Crapaxiom 5 is the induction axiom:

If a set S of numbers contains zero and also the successor of every number in S, then every number is in S.

Did you even ever bother to study Peano's crapaxioms? Or did you just memorise them by heart? Stupid boy you are.

On Sunday, October 1, 2017 at 9:32:32 PM UTC+2, John Gabriel wrote:

"It is a very bad idea and nothing less than stupid to define 1/3 = 0.333..."

Why?

"0.333..." is just short for

 lim  SUM_{k=1..n} (3/10^n)
n->oo

and the latter *is* (provable) 1/3.

Btw. Hence 1/3 = 0.333... is a THEOREM not a DEFINITION, idiot.

On Sunday, 1 October 2017 14:38:14 UTC-5, Me  wrote:
> On Sunday, October 1, 2017 at 9:32:32 PM UTC+2, John Gabriel wrote:
>
> "It is a very bad idea and nothing less than stupid to define 1/3 = 0.333..."
>
> Why?
>
> "0.333..." is just short for
>
>  lim  SUM_{k=1..n} (3/10^n)
> n->oo

Wrong. It is short for

SUM_{k=1...oo} (3/10^n)

That is what ensures you have the correct representation and the only justification for you using 0.333... for otherwise you could @.#*!...  
Ever asked yourself why you use the former? Yes idiot, because it signifies a mythical  "infinite" sum.

It is most definitely not the same as 1/3 because 1/3 is well defined. It is not shorter than 1/3 because any idiot can see that 0.333... has more literals. Chuckle.

>
> and the latter *is* (provable) 1/3.
>
> Btw. Hence 1/3 = 0.333... is a THEOREM not a DEFINITION, idiot.

No dipshit. It is a ***DEFINITION***.  There is ZERO to prove. The fact that the limit of the series 0.3+0.03+... is 1/3 proves NOTHING ...

On Sunday, October 1, 2017 at 9:48:24 PM UTC+2, John Gabriel wrote:
> On Sunday, 1 October 2017 14:38:14 UTC-5, Me  wrote:
> > On Sunday, October 1, 2017 at 9:32:32 PM UTC+2, John Gabriel wrote:
> >
> > "It is a very bad idea and nothing less than stupid to define 1/3 = 0.333..."
> >
> > Why?
> >
> > "0.333..." is just short for
> >
> >  lim  SUM_{k=1..n} (3/10^n)
> > n->oo
>
> Wrong. It is short for
>
> SUM_{k=1...oo} (3/10^n)

John, you really should seek some help.

I guess, you once KNEW that

SUM_{k=1...oo}

is just short for

 lim  SUM_{k=1..n} .
n->oo

Hint: We CAN'T actually "sum up" infinitely many terms.

On Sunday, 1 October 2017 14:48:24 UTC-5, John Gabriel  wrote:
> On Sunday, 1 October 2017 14:38:14 UTC-5, Me  wrote:
> > On Sunday, October 1, 2017 at 9:32:32 PM UTC+2, John Gabriel wrote:
> >
> > "It is a very bad idea and nothing less than stupid to define 1/3 = 0.333..."
> >
> > Why?
> >
> > "0.333..." is just short for
> >
> >  lim  SUM_{k=1..n} (3/10^n)
> > n->oo
>
> Wrong. It is short for
>
> SUM_{k=1...oo} (3/10^n)
>
> That is what ensures you have the correct representation and the only justification for you using 0.333... for otherwise you could @.#*!...  
> Ever asked yourself why you use the former? Yes idiot, because it signifies a mythical  "infinite" sum.
>
> It is most definitely not the same as 1/3 because 1/3 is well defined. It is not shorter than 1/3 because any idiot can see that 0.333... has more literals. Chuckle.
>
> >
> > and the latter *is* (provable) 1/3.
> >
> > Btw. Hence 1/3 = 0.333... is a THEOREM not a DEFINITION, idiot.
>

> No dipshi...

No idiot. It is DEFINED that way by Euler. It is short for your IGNORANCE and STUPIDITY.

>
> Hint: We CAN'T actually "sum up" infinitely many terms.

Of course you can't. So why use 0.333... as the representation and not asadaff... instead? Because you are a stupid imbecile who has never learned to think indep...

On Sunday, 12 February 2017 07:14:11 UTC-6, John Gabriel  wrote:
> S = Lim S in this case, contradicts the fact that 1/3 has no representation in base 10.
>
> However, it also leads to even more absurd definitions, that is, pi=3.14159...  when in fact it is known that pi has NO measure.
>
>
> https://www.youtube.com/watch?v=5hulvl3GgGk
>
> https://www.youtube.com/watch?v=w8s_8fNePEE
>
>
> Your comments are unwelcome shit and will be ignored.
>
> This is posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.

S = Lim S  

STOPS RIGHT HERE AND NOW. It is the Eulerian Blunder and has no place in rational thought, never mind mathematics.

On Sunday, 1 October 2017 15:01:36 UTC-5, Me  wrote:

> On Sunday, October 1, 2017 at 9:54:51 PM UTC+2, Me wrote:
> > On Sunday, October 1, 2017 at 9:48:24 PM UTC+2, John Gabriel wrote:
> > > On Sunday, 1 October 2017 14:38:14 UTC-5, Me  wrote:
> > > > On Sunday, October 1, 2017 at 9:32:32 PM UTC+2, John Gabriel wrote:
> > > >
> > > > "It is a very bad idea and nothing less than stupid to define 1/3 = 0.333..."
> > > >
> > > > Why?
> > > >
> > > > "0.333..." is just short for
> > > >
> > > >  lim  SUM_{k=1..n} (3/10^n)
> > > > n->oo
> > >
> > > Wrong. It is short for
> > >
> > > SUM_{k=1...oo} (3/10^n)
> >
> > John, you really should seek some help.
> >
> > I guess, you once KNEW that
> >
> > SUM_{k=1...oo}
> >
> > is just short for
> >
> >  lim  SUM_{k=1..n} .
> > n->oo
> >

> > Hint: We CAN'T actually "sum up" infinitely many terms.
>

> Additional comment:
>
> > > SUM_{k=1...oo} (3/10^n)
> > >
> > [...] It is most definitely not the same as 1/3
>
> Well, actually it is. There's a simple prove for...

On Sunday, October 1, 2017 at 9:59:14 PM UTC+2, John Gabriel wrote:

> why use "0.333..." as the representation and not "asadaff..."

Because it gives a hint which number I might mean in this case.

Same with

0.111...
0.222...
0.333...
:
0.999...
 
> You use 0.333... because it is supported by the idea of an infinite sum.

Guess so.

> By the same token [...] you would never write 3.14159... for pi if it
> did not have the support of the [...] infinite sum.

Guess so, yes. :-)

It's called /decimal representation/.

See: https://en.wikipedia.org/wiki/Decimal_representation

On Sunday, October 1, 2017 at 10:05:23 PM UTC+2, John Gabriel wrote:
> On Sunday, 1 October 2017 15:01:36 UTC-5, Me  wrote:
> > On Sunday, October 1, 2017 at 9:54:51 PM UTC+2, Me wrote:
> > > On Sunday, October 1, 2017 at 9:48:24 PM UTC+2, John Gabriel wrote:
> > > > On Sunday, 1 October 2017 14:38:14 UTC-5, Me  wrote:
> > > > > On Sunday, October 1, 2017 at 9:32:32 PM UTC+2, John Gabriel wrote:
> > > > >
> > > > > "It is a very bad idea and nothing less than stupid to define 1/3 = 0.333..."
> > > > >
> > > > > Why?
> > > > >
> > > > > "0.333..." is just short for
> > > > >
> > > > >  lim  SUM_{k=1..n} (3/10^n)
> > > > > n->oo
> > > >
> > > > Wrong. It is short for
> > > >
> > > > SUM_{k=1...oo} (3/10^n)
> > >
> > > John, you really should seek some help.
> > >
> > > I guess, you once KNEW that
> > >
> > > SUM_{k=1...oo}
> > >
> > > is just short for
> > >
> > >  lim  SUM_{k=1..n} .
> > > n->oo
> > >
> > > Hint: We CAN'T actually "sum up" infinitely many terms.
> >
> > Additional comment:
> >

...

On Sunday, 1 October 2017 15:05:23 UTC-5, John Gabriel  wrote:
> On Sunday, 1 October 2017 15:01:36 UTC-5, Me  wrote:
> > On Sunday, October 1, 2017 at 9:54:51 PM UTC+2, Me wrote:
> > > On Sunday, October 1, 2017 at 9:48:24 PM UTC+2, John Gabriel wrote:
> > > > On Sunday, 1 October 2017 14:38:14 UTC-5, Me  wrote:
> > > > > On Sunday, October 1, 2017 at 9:32:32 PM UTC+2, John Gabriel wrote:
> > > > >
> > > > > "It is a very bad idea and nothing less than stupid to define 1/3 = 0.333..."
> > > > >
> > > > > Why?
> > > > >
> > > > > "0.333..." is just short for
> > > > >
> > > > >  lim  SUM_{k=1..n} (3/10^n)
> > > > > n->oo
> > > >
> > > > Wrong. It is short for
> > > >
> > > > SUM_{k=1...oo} (3/10^n)
> > >
> > > John, you really should seek some help.
> > >
> > > I guess, you once KNEW that
> > >
> > > SUM_{k=1...oo}
> > >
> > > is just short for
> > >
> > >  lim  SUM_{k=1..n} .
> > > n->oo
> > >
> > > Hint: We CAN'T actually "sum up" infinitely many terms.
> >
> > Additional comment:
> >

> >...

On Sunday, 1 October 2017 15:06:56 UTC-5, Me  wrote:
> On Sunday, October 1, 2017 at 9:59:14 PM UTC+2, John Gabriel wrote:
>
> > why use "0.333..." as the representation and not "asadaff..."
>
> Because it gives a hint which number I might mean in this case.

A hint???!!! Chuckle. asadaff... is just as good if it is defined in the same way.  What hint do you get from this 3489833.2947293849879333311120000101000...  ??? Go on moron. Tell me what hint you get! Chuckle.

>
> Same with
>
> 0.111...
> 0.222...
> 0.333...
> :
> 0.999...

Nonsense. 1/9, 2/9, etc are very well defined and require nothing in the form of a hint. Only idiots like you need hints.

>  
> > You use 0.333... because it is supported by the idea of an infinite sum.
>
> Guess so.

Finally you admit that is a DEFINITION!  Good boy! Good boy! That is the first step. Chuckle.

>
> > By the same token [...] you would never write 3.14159... for pi if it
> > did not have the support of the [...] infinite sum.
>
> Guess so, yes. :-)
>
> It's called /decimal representation/.

You can call it decimal wankation. It is still the same shit.

On Sunday, October 1, 2017 at 10:05:23 PM UTC+2, John Gabriel wrote:

John, you are confused. Sorry.

> You define
>
>   oo
>  SUM (3/10^n) = 1/3   [A]
>  n=1
>
> as being equal to
>
>  lim  SUM_{k=1..n} (3/10^k) = 1/3  [B]
>  n->oo

Nope. You are talking nonsense, man. :-)

oo
SUM a_n
n=1

is defined as

 lim  SUM_{k=1..n} a_k
n->oo

(where a_n is "unspecified"; just a/any function from IN in IR).

So all we know (in this case) BY DEFINITION is

oo
SUM (3/10^n) =
n=1

lim  SUM_{k=1..n} (3/10^k) .
n->oo

We don't know (at this point) if

oo
SUM (3/10^n) = 1/3 .
n=1

Hence this has to be proved.

On Sunday, October 1, 2017 at 9:48:24 PM UTC+2, John Gabriel wrote:

> SUM_{k=1...oo} (3/10^n)

>
> It is most definitely not the same as 1/3

Well, actually it is. There's a simple prove for the theorem

 lim  SUM_{k=1..n} (3/10^k) = 1/3 .
n->oo

Hence

On Sunday, 1 October 2017 15:19:14 UTC-5, Me  wrote:
> On Sunday, October 1, 2017 at 10:05:23 PM UTC+2, John Gabriel wrote:
>
> John, you are confused. Sorry.
>
> > You define
> >
> >   oo
> >  SUM (3/10^n) = 1/3   [A]
> >  n=1
> >
> > as being equal to
> >
> >  lim  SUM_{k=1..n} (3/10^k) = 1/3  [B]
> >  n->oo
>
> Nope. You are talking nonsense, man. :-)

You are writing nonsense.

>
> oo
> SUM a_n
> n=1
>
> is defined as
>
>  lim  SUM_{k=1..n} a_k
> n->oo
>
> (where a_n is "unspecified"; just a/any function from IN in IR).

Makes no difference idiot. You are introducing irrelevant matters.

>
> So all we know (in this case) BY DEFINITION is
>
> oo
> SUM (3/10^n) =
> n=1
>
> lim  SUM_{k=1..n} (3/10^k) .
> n->oo

YES. It is defining the SERIES to be its LIMIT. Simply and plainly stupid.

>
> We don't know (at this point) if
>
> oo
> SUM (3/10^n) = 1/3 .
> n=1
>

Of course we know stupid!!  There is nothing to prove. The fact that 1/3 is a limit has ZERO to do with infinity. It is an inference concluded from the nth partial sum you moron!!!

> Hence this has to be proved.

Nonsense. Nothing to prove. S = Lim S  is a DEFINITION, not a theorem and requires NO proof.

On Sunday, October 1, 2017 at 4:02:07 PM UTC-4, John Gabriel wrote:
> On Sunday, 12 February 2017 07:14:11 UTC-6, John Gabriel  wrote:
> > S = Lim S in this case, contradicts the fact that 1/3 has no representation in base 10.
> >
> > However, it also leads to even more absurd definitions, that is, pi=3.14159...  when in fact it is known that pi has NO measure.
> >
> >
> >

> > Your comments are unwelcome shit and will be ignored.
> >
> > This is posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.
>
> S = Lim S  
>
> STOPS RIGHT HERE AND NOW. It is the Eulerian Blunder and has no place in rational thought, never mind mathematics.

You mean that was YOUR blunder, Troll Boy. As even you were recently forced to concede when confronted with the evidence, "Of course he [Euler] did not write 'Lim S'... He did not talk about S." (May 27, 2017)

With any sane person, that would be the end of it, but not with a psycho troll like you!


Dan

Download my DC Proof 2.0 software at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

Nope, you both are wrong: if at all, then the k
goes to the exponent of 10 and not n:


    lim  SUM_{k=1..n} (3/10^k)
    n->oo

Which is synonymous (by the usual convention what
oo should mean in this context, because k is integer) to:

    lim  SUM_{k=1..oo} (3/10^k)

So much to the dislexia of bird brain John Grabriel.
We might expect that he is neither able to read math,
nor able to write math,

and that he is completely confused.

Corr.:

Which is synonymous (by the usual convention what
oo should mean in this context, because k is integer) to:

    SUM_{k=1..oo} (3/10^k)

On Sunday, October 1, 2017 at 9:38:14 PM UTC+2, Me wrote:
> On Sunday, October 1, 2017 at 9:32:32 PM UTC+2, John Gabriel wrote:
>
> "It is a very bad idea and nothing less than stupid to define 1/3 = 0.333..."
>
> Why?
>
> "0.333..." is just short for
>
>  lim SUM_{k=1..n} (3/10^n)
> n->oo

Should read:

 lim SUM_{k=1..n} (3/10^k)

On Monday, October 2, 2017 at 12:26:38 AM UTC+2, burs...@gmail.com wrote:
> Nope, you both are wrong: if at all, then the k
> goes to the exponent of 10 and not n:
>
>
>     lim  SUM_{k=1..n} (3/10^k)
>     n->oo
>
> Which is synonymous (by the usual convention what
> oo should mean in this context, because k is integer) to:
>
>     lim  SUM_{k=1..oo} (3/10^k)

Huh?! Nope. No "lim".

 oo
SUM (3/10^n)
n=1

is short for

 lim  SUM_{k=1..n} (3/10^k)
n->oo

See: https://en.wikipedia.org/wiki/Series_(mathematics)

On Sunday, 1 October 2017 17:56:48 UTC-5, Me  wrote:
> On Monday, October 2, 2017 at 12:26:38 AM UTC+2, burs...@gmail.com wrote:
> > Nope, you both are wrong: if at all, then the k
> > goes to the exponent of 10 and not n:
> >
> >
> >     lim  SUM_{k=1..n} (3/10^k)
> >     n->oo
> >
> > Which is synonymous (by the usual convention what
> > oo should mean in this context, because k is integer) to:
> >
> >     lim  SUM_{k=1..oo} (3/10^k)
>
> Huh?! Nope. No "lim".

See? Even your fellow birdbrain is confused. It doesn't matter whether you write lim in front or not stupid. You are DEFINING

>
>  oo
> SUM (3/10^n)
> n=1
>
> is short for

As the LIMIT:

>
>  lim  SUM_{k=1..n} (3/10^k)
> n->oo
>

And these are two different objects. One is a series and the other is a limit.

S  =  Lim S

You are extremely dense!

We didn't write:

    SUM_{k=1..n} (3/10^k) = lim n->oo SUM_{k=1..n} (3/10^k)

I guess you need very quickly to see a doctor, bird
brain John Gabriel. You seem to have hallucinations
in the form of "S=Lim S" or something similar.

BTW: Here is a recommendation, since Latex is used
very often in math:

    Use _ for subscript:   foo_bar will put bar subscript to foo

    Use ^ for superscript: foo^bar will put bar superscript to foo

Now you can write the two dimensional sum in one dimensio¨n
as following text, lets begin with the partial sum:

    s(n) = sum_k=1^n 3/10^k

Guess what you can directly input the RHS into Wolfram Alpha:

    https://www.wolframalpha.com/input/?i=sum_k%3D1^n+3%2F10^k

And it will give you the closed form:

    s(n) = 1/3(1-10^(-n))

Next step, you can use underscore, subscript also for limes,
but there are more typographic tricks:

    Use -> for tends to: foo->bar will be read as foo arrow bar

    Use oo for infinity: oo will be read as infinity

So lets try the limes now, we have:

    a = lim_n->oo sum_k=1^n 3/10^k

      = lim_n->oo 1/3...

Now compare when you input SUM_{k=1..n} (3/10^k) into
Wolfram Alpha. It will give up, it does not recognize it:

    https://www.wolframalpha.com/input/?i=SUM_{k%3D1..n}+%283%2F10^k%29

Wolfram Alpha is close to how the sum is written:

        to
      Sigma summand
     var=from  

The flat linear representation (using underscore _ for
subscript and ^ for superscript) is (Sigma is
replaced by sum):

     sum_var=from^to summand

Interestingly the greek letter sigma works also
in Wolfram Alpha, here is a test:

    Σ_k=1^n 3/10^k

    https://www.wolframalpha.com/input/?i=%CE%A3_k%3D1^n+3%2F10^k

Wolfram Alpha even writes:

    Assuming "Σ" is a sum

P.S.: My Windows machine says:
Σ = U+03A3
http://www.fileformat.info/info/unicode/char/03a3/index.htm

Am Montag, 2. Oktober 2017 03:30:06 UTC+2 schrieb burs...@gmail.com:
> Now you can write the two dimensional sum in one dimension

On Saturday, September 30, 2017 at 7:01:04 AM UTC-6, John Gabriel wrote:

> Well, for starters, education of calculus is in a dire mess. Students
> never understand calculus. Even their professors never understand.
> They simply learn to use.

Well, even if I don't agree with much of what you've said, I can
certainly grant you this point...

since the ancient Greeks had so much trouble with what we call
"irrational numbers", why couldn't it be the case that today's math
students might have the same problem...

and thus benefit from a calculus that can work without them?

You are correct that 1/4 = 0.25 is the same as 1/4 = 25/100, too.

It may be wrong, in some sense, to say that 1/3 = 0.3333... but it is
correct that 1/3 is greater than 3333/10000 and less than 3334/10000.
Similar things can be said about pi - even if pi is not really a
number, sometimes, like when figuring out how much adhesive tape will
fit on a roll, we might want to do arithmetic with it.

However, I don't think I could get you to consider that perhaps you and
conventional mathematicians are simply using different words to talk
about the same thing.

John Savard

On Sunday, October 1, 2017 at 1:32:32 PM UTC-6, John Gabriel wrote:

> Peano's Crapaxiom 5 is the induction axiom:

> If a set S of numbers contains zero and also the successor of every
> number in S, then every number is in S.

Well, that's clearly wrong. 1/2 would not be in S, and we're all agreed
that 1/2 is a number.

However, I think this is really what Peano was saying:

If S contains zero,
and if it is also true that
if S contains x then it contains the successor of x

(where the successor of x is what we would normally call x+1, but Peano hadn't gotten around to defining addition yet)

then S will contain 0, 1, 2, 3, 4, 5... and indeed any non-negative integer.

For example, it contains 5 because

S contained zero
and because if it contains x, it contains x+1,
it contains 1 because it contains 0
it contains 2 because it contains 1
it contains 3 because it contains 2
it contains 4 because it contains 3
it contains 5 because it contains 4

Clearly, this will work for any non-negative integer, however large,
even if one doesn't bother to type that much.

John Savard

On Sunday, October 1, 2017 at 2:08:13 PM UTC-6, John Gabriel wrote:
 
> [A] is an infinite series.
> [B] is the limit of an infinite series.

> They are two different things which you are DEFINING to be the
> same.

[A] might not be real. It might not really be something that
makes sense.

[B], on the other hand, definitely does make sense, because
each of the finite series in the sequence that approaches
the limit certainly can be added up.

So what mathematicians are doing is saying, since [A] is
bad, but we either would like to have [A], or we would like
to refer to [B] by the name of [A] because it is shorter to
skip mentioning that we're "really" talking about a limit,
we will now say that whenever [A] is written, we will
understand that [B] is what is really meant whenever it is
necessary to concern oneself with that distinction.

This is not really such an awful thing for them to be doing.

John Savard

On Sunday, October 1, 2017 at 1:22:43 AM UTC-6, netzweltler wrote:
 
> Do you agree that 0.999... means infinitely many commands
> Add 0.9 + 0.09
> Add 0.99 + 0.009
> Add 0.999 + 0.0009
> …?
> Then following all of these infinitely many commands won’t get you to point
> 1. If you reached point 1 you have disobeyed those commands, because every
> single of those infinitely many commands tells you to get closer to 1 but
> NOT reach 1.

You would be correct if Zeno's paradoxes were correct. But they're not.
Achilles can and does overtake the tortoise every day.

0.9999... does *NOT* mean actually doing those infinitely many steps. There
is never time to do that many commands. Instead, it means the place that
doing them would take you, if you _could_ do them.

Yes, doing any _finite_ number of those commands would not get you to 1. You
would have to disobey them to get that far.

But you *can't* do an infinite number of commands. Period.

So that isn't the criterion you use to figu...

On Sunday, October 1, 2017 at 2:02:07 PM UTC-6, John Gabriel wrote:
 
> S = Lim S  

> STOPS RIGHT HERE AND NOW. It is the Eulerian Blunder and has no place in
> rational thought, never mind mathematics.

S = Lim S isn't a statement of fact. It's a definition. It means that (in
certain appropriate cases) when we write an S that doesn't really have a
legitimate meaning, but Lim S does have a value, we are simply using S as a
shorter way of writing Lim S.

So we use this not as a way of saying that the bad, invalid, S is now a good
thing, but as a way of *getting rid* of S (and stuff like infinitesimals) and
_only_ using real things which do have values - such as Lim S.

You have just misunderstood what Euler was trying to do. He wasn't trying to
introduce bad messy thinking into mathematics; instead, he found some there, and
he was using this method to get _rid_ of all the messy stuff quickly without
having to make mathematicians do a lot of extra work.

Euler was not your enemy. He wasn't saying the false thing that S is just as
good as Lim S. He knew that S was bad, but Lim S was good, and he was therefore
doing just what you are trying to do: fix mathematics so that it rests on the
good stuff and doesn't try to use the bad stuff.

John Savard

On Sunday, October 1, 2017 at 8:22:32 PM UTC-7, Quadibloc wrote:
> On Sunday, October 1, 2017 at 1:22:43 AM UTC-6, netzweltler wrote:
>  
> > Do you agree that 0.999... means infinitely many commands
> > Add 0.9 + 0.09
> > Add 0.99 + 0.009
> > Add 0.999 + 0.0009
> > …?
> > Then following all of these infinitely many commands won’t get you to point
> > 1. If you reached point 1 you have disobeyed those commands, because every
> > single of those infinitely many commands tells you to get closer to 1 but
> > NOT reach 1.
>
> You would be correct if Zeno's paradoxes were correct. But they're not.
> Achilles can and does overtake the tortoise every day.
>
> 0.9999... does *NOT* mean actually doing those infinitely many steps. There
> is never time to do that many commands. Instead, it means the place that
> doing them would take you, if you _could_ do them.
>
> Yes, doing any _finite_ number of those commands would not get you to 1. You
> would have to disobey them to get that far.

...

are you trying to insist that 0.(3) is equal to
lim{n \to infinity} s(n) = lim{n --> infinity} 0.3 + 0.03 + 0.003 + ... + 3*10^(-n) = 1/3 ?

Using the real numbers 0.(3) is not equal to 1/3.

But it seems you have chosen to prefer eating the cake and having it too,
meaning you have rejected the infinitesimals but using them too.
That's why you end up with a falsehood 0.(3) = 1/3 in real numbers.

You should be able to check a simple arithmetic
0.(3)3 + 0.(3)3 + 0.(3)4 = 0.(3)(3)+0.(3)(3)+0.(3)(3) = 1
and therefore 0.(3)(3) = 1/3

Den söndag 1 oktober 2017 kl. 15:17:49 UTC+2 skrev John Gabriel:
> On Saturday, 30 September 2017 19:32:08 UTC-5, burs...@gmail.com  wrote:
> > So the product, its terms sn*tn might be observable,
> > but that the product is Cauchy is not observable directly.
> >
> > That a series is Cauchy is neither effectively refutable
> > nor effectively verifiable. If you find e, n, m with:
> >
> >     |an-am| > e
> >
> > You still don't know whether there is N, where the series
> > behaves Cauchy. The full Cauchy condition is:
> >
> >    forall e exists N forall n,m>=N |an-am|=<e
> >
> > So it has the shape VEV.
> >
> > Am Sonntag, 1. Oktober 2017 02:23:32 UTC+2 schrieb burs...@gmail.com:
> > > Well you wrote here Newton didn't consider infinity,
> > > and you say he can define partial sums without infinity.
> > >
> > > Well this might be true, but you then go on and say
> > > he used limits. But how do you get limits, without
> > >
> > > https://groups.google.com/d/msg/sci.math/HIzzJSLsw60/vSOH7WnhAwAJ
> > >
> > > knowing whether a series converges or not? For convergence
> > > you need to make statement about infinitely many elements,
> > >
> > > for example the Cauchy condition, is for infinitely many
> > > pairs n,m, namely you need to know (or assume you know):
> > >
> > >    forall n,m >= N(e) |an - am| =< e
>
> Each time I think you can't say anything more stupid, you surprise me.
> Listen stupid, the forall you have there is based on induction. There is no forall. A Cauchy sequence does not require "forall".  The "forall" is a result of inference.
>
> > >
> > > The above looks like a pi-sentence, and is not verifiable
> > > if we do not know much about {ak}. So you are in the waters of:
> > >
> > > It is also familiar in the philosophy of science that most
> > > hypotheses are neither verifiable nor refutable. Thus, Kant’s
> > > antinomies of pure reason include such statements a...

Hey moron! Nice to see you :) I still see you are using horseshit notation that you cannot even define coherently!

!/3=0.(3) is trivial to prove for real numbers.

maanantai 2. lokakuuta 2017 8.45.49 UTC+3 Zelos Malum kirjoitti:
> Den måndag 2 oktober 2017 kl. 07:29:13 UTC+2 skrev 7777777:
> > sunnuntai 1. lokakuuta 2017 23.21.58 UTC+3 Me kirjoitti:
> > > On Sunday, October 1, 2017 at 9:48:24 PM UTC+2, John Gabriel wrote:
> > >
> > > > SUM_{k=1...oo} (3/10^n)
> > > >
> > > > It is most definitely not the same as 1/3
> > >
> > > Well, actually it is. There's a simple prove for the theorem
> > >
> > >  lim  SUM_{k=1..n} (3/10^k) = 1/3 .
> > > n->oo
> > >
> > > Hence
> > >
> > >  oo
> > > SUM (3/10^n) = 1/3 .
> > > n=1
> >
> > are you trying to insist that 0.(3) is equal to
> > lim{n \to infinity} s(n) = lim{n --> infinity} 0.3 + 0.03 + 0.003 + ... + 3*10^(-n) = 1/3 ?
> >
> > Using the real numbers 0.(3) is not equal to 1/3.
> >
> > But it seems you have chosen to prefer eating the cake and having it too,
> > meaning you have rejected the infinitesimals but using them too.

> > That's why you end up with a falsehood 0.(3) = 1/3 in re...

Am Sonntag, 1. Oktober 2017 17:45:39 UTC+2 schrieb FromTheRafters:
> netzweltler formulated the question :
> > Am Sonntag, 1. Oktober 2017 15:20:16 UTC+2 schrieb FromTheRafters:
> >> After serious thinking netzweltler wrote :
> >>> Am Sonntag, 1. Oktober 2017 13:56:01 UTC+2 schrieb FromTheRafters:
> >>>>
> >>>> It seems counterintuitive when a number is viewed (or represented) as
> >>>> an infinite unending 'process' of achieving better and better
> >>>> approximations, and that we can never actually reach the destination
> >>>> number. In my view, this sequence and/or infinite sum is a
> >>>> representation of the destination number "as if" we could have gotten
> >>>> there by that process.
> >>> If the process doesn't get us there then we don't get there. Where do you
> >>> get  your "as if" from?
> >>
> >> If you had sufficient time, then you would get there.
> > Show how time is involved in our process.
>
> If you have to add a next number (like one quarter) to a previous
> result of adding such previous numbers (like one plus one half) then
> you have introduced time. Thee is a 'previous' calculation needed as
> input to the next calculation. The idea that you 'never' get there (to
> two) introduces time also. I'm with you, I don't think time has any
> place in this.
>
> >>>> IOW "*After* infinitely many 'better'
> >>>> approximations" we reach the destination number *exactly* even if we
> >>>> cannot 'pinpoint' that number on the number line.
> >>> Please define "*After* infinitely many 'better' approximations". All we've
> >>> got is infinitely many approximations - each approximation telling us that
> >>> we  get closer to 1 but don't reach 1. There...

On Sunday, 1 October 2017 22:29:18 UTC-5, Quadibloc  wrote:
> On Sunday, October 1, 2017 at 2:02:07 PM UTC-6, John Gabriel wrote:
>  
> > S = Lim S  
>
> > STOPS RIGHT HERE AND NOW. It is the Eulerian Blunder and has no place in
> > rational thought, never mind mathematics.
>
> S = Lim S isn't a statement of fact. It's a definition. It means that (in
> certain appropriate cases) when we write an S that doesn't really have a
> legitimate meaning, but Lim S does have a value, we are simply using S as a
> shorter way of writing Lim S.

How ridiculous is that statement.  You think 0.333... is a shorter way of writing the well-defined number 1/3 ?  That's absurd. It's even more absurd when you consider that there is no way to write 1/3 in base 10.

>
> So we use this not as a way of saying that the bad, invalid, S is now a good
> thing, but as a way of *getting rid* of S (and stuff like infinitesimals) and
> _only_ using real things which do have values - such as Lim S.
>

> You hav...

On Monday, 2 October 2017 01:01:58 UTC-5, 7777777  wrote:
> maanantai 2. lokakuuta 2017 8.45.49 UTC+3 Zelos Malum kirjoitti:
> > Den måndag 2 oktober 2017 kl. 07:29:13 UTC+2 skrev 7777777:
> > > sunnuntai 1. lokakuuta 2017 23.21.58 UTC+3 Me kirjoitti:
> > > > On Sunday, October 1, 2017 at 9:48:24 PM UTC+2, John Gabriel wrote:
> > > >
> > > > > SUM_{k=1...oo} (3/10^n)
> > > > >
> > > > > It is most definitely not the same as 1/3
> > > >
> > > > Well, actually it is. There's a simple prove for the theorem
> > > >
> > > >  lim  SUM_{k=1..n} (3/10^k) = 1/3 .
> > > > n->oo
> > > >
> > > > Hence
> > > >
> > > >  oo
> > > > SUM (3/10^n) = 1/3 .
> > > > n=1
> > >
> > > are you trying to insist that 0.(3) is equal to
> > > lim{n \to infinity} s(n) = lim{n --> infinity} 0.3 + 0.03 + 0.003 + ... + 3*10^(-n) = 1/3 ?
> > >
> > > Using the real numbers 0.(3) is not equal to 1/3.
> > >
> > > But it seems you have chosen to prefer eating the cake and having it too,

> > > meaning y...

On Sunday, 1 October 2017 20:30:06 UTC-5, burs...@gmail.com  wrote:
> We didn't write:
>
>     SUM_{k=1..n} (3/10^k) = lim n->oo SUM_{k=1..n} (3/10^k)

You wrote and you believe that:

n->oo SUM_{k=1..n} (3/10^k) = 0.333... = lim n->oo SUM_{k=1..n} (3/10^k) = 1/3

In other words,   S = Lim S

S = n->oo SUM_{k=1..n} (3/10^k)
Lim S = lim n->oo SUM_{k=1..n} (3/10^k)

Am Montag, 2. Oktober 2017 05:22:32 UTC+2 schrieb Quadibloc:
> On Sunday, October 1, 2017 at 1:22:43 AM UTC-6, netzweltler wrote:
>  
> > Do you agree that 0.999... means infinitely many commands
> > Add 0.9 + 0.09
> > Add 0.99 + 0.009
> > Add 0.999 + 0.0009
> > …?
> > Then following all of these infinitely many commands won’t get you to point
> > 1. If you reached point 1 you have disobeyed those commands, because every
> > single of those infinitely many commands tells you to get closer to 1 but
> > NOT reach 1.
>
> You would be correct if Zeno's paradoxes were correct. But they're not.
> Achilles can and does overtake the tortoise every day.
>
> 0.9999... does *NOT* mean actually doing those infinitely many steps. There
> is never time to do that many commands. Instead, it means the place that
> doing them would take you, if you _could_ do them.
>
> Yes, doing any _finite_ number of those commands would not get you to 1. You
> would have to disobey them to get that far.

Even doing an _infinite_ number of those commands wouldn't get you to 1.

1. 0.99 + 0
2. 0.99 + 0
3. 0.99 + 0
...

Neither a finite num...

Den måndag 2 oktober 2017 kl. 08:01:58 UTC+2 skrev 7777777:
> maanantai 2. lokakuuta 2017 8.45.49 UTC+3 Zelos Malum kirjoitti:
> > Den måndag 2 oktober 2017 kl. 07:29:13 UTC+2 skrev 7777777:
> > > sunnuntai 1. lokakuuta 2017 23.21.58 UTC+3 Me kirjoitti:
> > > > On Sunday, October 1, 2017 at 9:48:24 PM UTC+2, John Gabriel wrote:
> > > >
> > > > > SUM_{k=1...oo} (3/10^n)
> > > > >
> > > > > It is most definitely not the same as 1/3
> > > >
> > > > Well, actually it is. There's a simple prove for the theorem
> > > >
> > > >  lim  SUM_{k=1..n} (3/10^k) = 1/3 .
> > > > n->oo
> > > >
> > > > Hence
> > > >
> > > >  oo
> > > > SUM (3/10^n) = 1/3 .
> > > > n=1
> > >
> > > are you trying to insist that 0.(3) is equal to
> > > lim{n \to infinity} s(n) = lim{n --> infinity} 0.3 + 0.03 + 0.003 + ... + 3*10^(-n) = 1/3 ?
> > >
> > > Using the real numbers 0.(3) is not equal to 1/3.
> > >
> > > But it seems you have chosen to prefer eating the cake and having it too,

> > > meaning you have rejec...

And they might be less dislexis, especially
concerning the ellipses and nonsense like Euler
did a blunder in the form of "S=Lim S".

Here is a nice other notation convention, lets
start with unterval description, its very common
in math to use for example:

   [3...] : An open interval, the set {3,4,5,...}
            the ellipses stands for a infinite continuation
            of the series

   [3...11] : A closed interval, the set {3,4,5,..,9,10,11}
              the ellipses stands for a finite continuation
              of the series.

We can use the same in decimal representation an will
get in very short notation that the partial sums are
different from the limit:

   0.333... = 1/3

   0.333...333 <> 1/3

Or in expanded math:

   lim n->oo sum_k=1^n 3/10^k = 1/3

   forall n sum_k=1^n 3/10^k <> 1/3

Or as bird brain John Gabriel puts it "You think

0.333... is a shorter way of writing the well-defined
number 1/3 ? That's absurd."

Its not absurd if you interpret the ... in 0.333...
correctly. It means two things:
- In...

Nope, I only believe:

  0.333... = 1/3

  0.333...333 <> 1/3

I never ever wrote somewhere:

  0.333...333 = 1/3

For an explanation see here:
https://groups.google.com/d/msg/sci.math/bgU-4JWvHbY/I3FJuzByBgAJ

> > > > mea...

> > > > meaning you have rejected the infinitesimals but using them too.

> > ...

On Monday, October 2, 2017 at 12:45:50 AM UTC-6, netzweltler wrote:

> Correct. Time is of no concern. So, let me modify the list:
>
> t = 0: write 0.9
> t = 0.9: append another 9
> t = 0.99: append another 9
> ...
>
> to
>
> 1. write 0.9
> 2. append another 9
> 3. append another 9
> ...
>
> Do you still agree that this is a _complete_ list of all the actions needed to write 0.999... (already present - in no time)? It is a list of additions as well. All the additions it takes to sum up to 0.999...
> Again the question:
> If your claim is, that we reach point 1, you need to show which step on this list of infinitely many steps accomplishes that.

The claim is not that 1 is reached during the doing of those actions, even at
the "end" of doing them - but an endless list of an infinite number of actions
*has* no end. That's why it's called "endless".

Instead, 1 is the limit of that endless list.

So 1 can stand outside the endless list, and yet still be the limit. Because the
limit is not part of the actions themselves.

John Savard

On Monday, October 2, 2017 at 12:46:00 AM UTC-6, John Gabriel wrote:
> You think 0.333... is a shorter way of writing the well-defined number 1/3 ?  
> That's absurd. It's even more absurd when you consider that there is no way to
> write 1/3 in base 10.

  .
0.3

or
  _
0.3

are other ways of writing 1/3, if the dot or the bar will line up properly over
the 3.

0.333... has to be understood to imply that it is 3's that are to be assumed
going on endlessly.

These are _other_ ways of writing 1/3. Whether they're shorter or not doesn't
matter.

In other cases, though, using the convention that "S" really means "Lim S" does
let things be written more compactly.

Thus, instead of saying "the limit as n goes to infinity of the sum from i = 1
to n of (expression)", we just say "the sum from i = 1 to infinity of
(expression)", even though adding an infinite number of terms can't really be
done.

John Savard

I guess nobody minds if you are claiming that the limit of the sequence { 0.9, 0.99, 0.999, ... } equals 1.

> >>>> line by using an infinite altern...

On Monday, 2 October 2017 07:30:49 UTC-5, Quadibloc  wrote:
> On Monday, October 2, 2017 at 12:46:00 AM UTC-6, John Gabriel wrote:
> > You think 0.333... is a shorter way of writing the well-defined number 1/3 ?  
> > That's absurd. It's even more absurd when you consider that there is no way to
> > write 1/3 in base 10.
>
>   .
> 0.3
>
> or
>   _
> 0.3
>
> are other ways of writing 1/3, if the dot or the bar will line up properly over the 3.

That is FALSE but I can see you have very little understanding of mathematics and your mind is made up. You didn't even read my response. Chuckle.

On 10/1/2017 3:22 AM, netzweltler wrote:

> Do you agree that 0.999... means infinitely many commands
> Add 0.9 + 0.09
> Add 0.99 + 0.009
> Add 0.999 + 0.0009

> ...?

0.999... does not mean infinitely many commands.

There is a set of results of certain finite sums, a set of
numbers. We can informally write that set as
    { 0.9, 0.99, 0.999, ... }
That is an infinite set, but we can give it a finite description.

    (Our finite description won't use '...'. The meaning of
    '...' depends upon it being obvious. If we are discussing
    what '...' means, it must not be obvious, so we ought to
    avoid using '...')

There is number which is the unique least upper bound of that set.
The least upper bound is a finite description of that number.

0.999... means "the least upper bound of the set
    { 0.9, 0.99, 0.999, ... }".
That number can be show to be 1, by reasoning in a finite manner
from these finite descriptions of what we mean.

If you give 0.999... some meaning other than what we mean,
and then it turns out there are problems of some sort with
your meaning, than that is your problem, not ours.

On Monday, 2 October 2017 10:59:21 UTC-5, Jim Burns  wrote:
> On 10/1/2017 3:22 AM, netzweltler wrote:
>
> > Do you agree that 0.999... means infinitely many commands
> > Add 0.9 + 0.09
> > Add 0.99 + 0.009
> > Add 0.999 + 0.0009
> > ...?
>
> 0.999... does not mean infinitely many commands.

In fact it means exactly infinitely many commands.
But of course if you define a series to be equal to its limit, then that's like defining an apple equal to an orange. That is your problem, not ours.

You'll have to re-examine what Euler passed on to you.

>
> There is a set of results of certain finite sums, a set of
> numbers. We can informally write that set as
>     { 0.9, 0.99, 0.999, ... }
> That is an infinite set, but we can give it a finite description.
>
>     (Our finite description won't use '...'. The meaning of
>     '...' depends upon it being obvious. If we are discussing
>     what '...' means, it must not be obvious, so we ought to
>     avoid using '...')
>

> There is number which i...

> >>>>> If you insist on introducing time to our process, try this:
> >>>>
> >>>> You misund...

Am Montag, 2. Oktober 2017 17:59:21 UTC+2 schrieb Jim Burns:
> On 10/1/2017 3:22 AM, netzweltler wrote:
>
> > Do you agree that 0.999... means infinitely many commands
> > Add 0.9 + 0.09
> > Add 0.99 + 0.009
> > Add 0.999 + 0.0009
> > ...?
>
> 0.999... does not mean infinitely many commands.

But that's exactly what it means. Infinitely many commands. Infinitely many additions. Infinitely many steps trying to reach a point on the number line.

> There is a set of results of certain finite sums, a set of
> numbers. We can informally write that set as
>     { 0.9, 0.99, 0.999, ... }
> That is an infinite set, but we can give it a finite description.
>
>     (Our finite description won't use '...'. The meaning of
>     '...' depends upon it being obvious. If we are discussing
>     what '...' means, it must not be obvious, so we ought to
>     avoid using '...')
>
> There is number which is the unique least upper bound of that set.

> The least upper bound is a finite description of that numb...

On 10/2/2017 1:58 PM, netzweltler wrote:
> Am Montag, 2. Oktober 2017 17:59:21 UTC+2 schrieb Jim Burns:
>> On 10/1/2017 3:22 AM, netzweltler wrote:

>>> Do you agree that 0.999... means infinitely many commands
>>> Add 0.9 + 0.09
>>> Add 0.99 + 0.009
>>> Add 0.999 + 0.0009
>>> ...?
>>
>> 0.999... does not mean infinitely many commands.
>
> But that's exactly what it means.

That's not the standard meaning.

You give it some other meaning, and then you find a problem
with the meaning you gave it. Supposing I wanted to sort out
what that other meaning was, and how to make sense of it, my
attention to your meaning would not affect the standard meaning.

I am not a math historian, but the impression I have
is that great care was taken in choosing the standard meaning
in order to avoid problems like the ones you are finding.

You have the ability to create and then wallow in whatever
problems you choose. No one is able to take that power away
from you. But you can't "choose" by an act of your will to
make your created problem...

Am Montag, 2. Oktober 2017 20:35:56 UTC+2 schrieb Jim Burns:
> On 10/2/2017 1:58 PM, netzweltler wrote:
> > Am Montag, 2. Oktober 2017 17:59:21 UTC+2 schrieb Jim Burns:
> >> On 10/1/2017 3:22 AM, netzweltler wrote:
>
> >>> Do you agree that 0.999... means infinitely many commands
> >>> Add 0.9 + 0.09
> >>> Add 0.99 + 0.009
> >>> Add 0.999 + 0.0009
> >>> ...?
> >>
> >> 0.999... does not mean infinitely many commands.
> >
> > But that's exactly what it means.
>
> That's not the standard meaning.

So, you disagree that

0.999... = 0.9 + 0.09 + 0.009 + ... ?

> You give it some other meaning, and then you find a problem
> with the meaning you gave it. Supposing I wanted to sort out
> what that other meaning was, and how to make sense of it, my
> attention to your meaning would not affect the standard meaning.
>
> I am not a math historian, but the impression I have
> is that great care was taken in choosing the standard meaning
> in order to avoid problems like the ones you are finding.
>
> You have the ability to create and then wallow in whatever
> problems you choose. No one is able to take that power away

> from you. But yo...

Well this is probably the greatest nonsense somebody
ever posted on sci.math. You know, you didn't say
rational number line.

So when it is the real number line, pi is of course
there. There is of course a point on the real number
line that is pi.

Am Montag, 2. Oktober 2017 19:54:44 UTC+2 schrieb netzweltler:
> Yes. pi is already there and we can exactly locate its position on the number line, but you cannot locate a point on the number line representing pi if this point would be the result of a stepwise process - neither a finite process nor an infinite.

Even the Greek knew that, for them there were
incomensurable rations and commensurable ratios,
the later had a number representation we

call today a rational number. The former is
for example pi, which is the ratio between
circumpherence and diameter.

The Greek already knew how to compare any
ratios, incommensurable and commensurable,
is this true? Lets make a check:

if nw >=< mx, then ny >=< mz.
https://mathcs.clarku.edu/~djoyce/elements/bookV/defV5.html

Can this be used to compare two incommensurable
ratios as well? Lets assume we have two
real numbers r1 and r2, which are irrational.

When is r1 < r2 or r1 = r2 or r1 > r2?

Well the decimal representation might help us,
lets consider the case r1 < r2, then some digits
are the but later a bigger diggit appears:

    r1 = d0.d1...dk dk+1 ...

    r2 = d0.d1...dk fk+1 ...

   dn+1 < fn+1.

Now do we have r1 != r2? We need to
show a violation of:

   if nw >=< mx, then ny >=< mz.

Lets represent r1 and r2 as r1:1 and
r2:1, now take n=10^(k+1) and
m = d0 d1 ... dk fk+1. We will have:

    nw < mx

Proof: nw=n*r1 ...

On Monday, October 2, 2017 at 10:09:44 PM UTC+2, burs...@gmail.com wrote:

> Well this is probably the greatest nonsense somebody
> ever posted on sci.math.

I don't thinks so. You know there are such luminaries like WM, AP or JG.

But actually "Netzweltler" already showed up in de.sci.mathematic posting nonsensical views concernig (infinitely many) "commands", "steps" and their "result" etc. etc.

Maybe he was a pupil of Mückenheim, poor guy.

On Monday, 2 October 2017 13:35:56 UTC-5, Jim Burns  wrote:
> On 10/2/2017 1:58 PM, netzweltler wrote:
> > Am Montag, 2. Oktober 2017 17:59:21 UTC+2 schrieb Jim Burns:
> >> On 10/1/2017 3:22 AM, netzweltler wrote:
>
> >>> Do you agree that 0.999... means infinitely many commands
> >>> Add 0.9 + 0.09
> >>> Add 0.99 + 0.009
> >>> Add 0.999 + 0.0009
> >>> ...?
> >>
> >> 0.999... does not mean infinitely many commands.
> >
> > But that's exactly what it means.
>
> That's not the standard meaning.

Correct Jim! The mainstream meaning is S = Lim S. Hence, by **definition**, not proof or theorem, 0.999... = 1.

>
> You give it some other meaning, and then you find a problem
> with the meaning you gave it. Supposing I wanted to sort out
> what that other meaning was, and how to make sense of it, my
> attention to your meaning would not affect the standard meaning.
>
> I am not a math historian, but the impression I have
> is that great care was taken in choosing the standard meaning
> in order to avoid problems like the ones you are finding.

Nope. Euler blundered badly!

>
> You have the ability to create and then wallow in whatever
> problems you choose. No one is able to take that power away

> from you. Bu...

On Monday, 2 October 2017 15:40:48 UTC-5, Me  wrote:
> On Monday, October 2, 2017 at 10:09:44 PM UTC+2, burs...@gmail.com wrote:
>
> > Well this is probably the greatest nonsense somebody
> > ever posted on sci.math.
>
> I don't thinks so. You know there are such luminaries like WM, AP or JG.

You may quote WM in the same sentence as me, but you are a scumbag to include AP who is a delusional crank like you. That is such a below the belt blow. Tsk, tsk. AP is clearly a nutjob. Heck, I have less respect for AP than I have for you.

>
> But actually "Netzweltler" already showed up in de.sci.mathematic posting nonsensical views concernig (infinitely many) "commands", "steps" and their "result" etc. etc.
>
> Maybe he was a pupil of Mückenheim, poor guy.

He'd be lucky!

Well these luminaries WM, AP and JG lack complete
credibility, you even don't need to read any of their
posts, you can always assume its just provocative nonsense.

But Netzweilers are of interest, since they might show
some very juvenile believes, maybe even not their own,
just some mish mash from dunno where,

maybe from the land of Oz?

JG definitely qualifies as the Scarecrow in
the land of Oz. Just some bird brain...

On Sunday, 12 February 2017 07:14:11 UTC-6, John Gabriel  wrote:
> S = Lim S in this case, contradicts the fact that 1/3 has no representation in base 10.
>
> However, it also leads to even more absurd definitions, that is, pi=3.14159...  when in fact it is known that pi has NO measure.
>
>
> https://www.youtube.com/watch?v=5hulvl3GgGk
>
> https://www.youtube.com/watch?v=w8s_8fNePEE
>
>
> Your comments are unwelcome shit and will be ignored.
>
> This is posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.

I just realised that when I first posted this comment, I had not included the link to my now world famous article:

https://www.linkedin.com/pulse/eulers-worst-definition-lim-john-gabriel

"In his first appearance, the Scarecrow reveals
that he lacks a brain and desires above all else
to have one. In reality, he is only two days
old and merely ignorant."

https://en.wikipedia.org/wiki/Scarecrow_%28Oz%29

Am Montag, 2. Oktober 2017 22:49:55 UTC+2 schrieb burs...@gmail.com:
> JG definitely qualifies as the Scarecrow in
> the land of Oz. Just some bird brain...
>
> Am Montag, 2. Oktober 2017 22:47:08 UTC+2 schrieb burs...@gmail.com:
> > Well these luminaries WM, AP and JG lack complete
> > credibility, you even don't need to read any of their
> > posts, you can always assume its just provocative nonsense.
> >
> > But Netzweilers are of interest, since they might show
> > some very juvenile believes, maybe even not their own,
> > just some mish mash from dunno where,
> >
> > maybe from the land of Oz?
> >
> > Am Montag, 2. Oktober 2017 22:40:48 UTC+2 schrieb Me:
> > > On Monday, October 2, 2017 at 10:09:44 PM UTC+2, burs...@gmail.com wrote:
> > >

> > > > Well t...

On Monday, October 2, 2017 at 10:47:08 PM UTC+2, burs...@gmail.com wrote:

> But Netzweltler's are of interest, since they might show

> some very juvenile believes, maybe even not their own,
> just some mish mash from dunno where, maybe from the land of Oz?

The problem with "Netzweltler" (as you will soon find out for yourself, I guess) is that he suffices the original definition of a crank:

"A crank is defined as a man who cannot be turned." (Nature, 8 Nov 1906, 25/2)

Hilbert was more positive about Netzweilers:

During 1900, in an address to the International Congress of Mathematicians in Paris, Hilbert suggested that answers to problems of mathematics are possible with human effort. He declared, "In mathematics there is no ignorabimus.",[2] and he worked with other formalists to establish foundations for mathematics during the early 20th century.

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

On Monday, October 2, 2017 at 10:46:30 PM UTC+2, John Gabriel wrote:

> you are a scumbag to include AP who is a delusional crank ...

Yes, I know that I'm a scumbag, and certainly AP is a delusional crank, but ... so are you. Sorry for that, John. Really. I guess, you defintely could do better.

> AP is clearly a nutjob.

Sure. Your case is different, I agree. (Just call you /nuts/ wouldn't do you justice.)

It's kind of sad and pathetic to see you and Jan Burse stroking each others' frail egos, especially when you are so ignorant from all the brainwashing you've received. Chuckle.

On Monday, October 2, 2017 at 11:01:38 PM UTC+2, burs...@gmail.com wrote:

> Hilbert was more positive about Netzweilers: [...]

"O brave new world. That has such people in't!" :-)

I mean it.

But I guess Hilbert didn't mean Netzweilers, rather
mathematics in general, and Gödel and others showed
that Hilberts program doesn't work...

On Monday, October 2, 2017 at 11:09:20 PM UTC+2, John Gabriel wrote:

> It's kind of sad and pathetic to see you and Jan Burse stroking each others'
> frail egos,

Just brothers in arms, I guess.

> especially when you are so ignorant from all the brainwashing you've received.

Some'd call it /education/, John.

(If performed properly it means that you (as a result of it) know what you are talking about (afterward), to a certain degree, John. Not a bad thing, actually.)

You mean you were immune to "brainwash" so far,
because you don't have any brains anyway?

bird brain John Gabriel, scarecrow from
the land of Oz.

Lets make a list, about bird brain John Gabriels
juvenile brain:
- Cannot read a text by Euler, which was especially
  tailored to the general public
- Thinks this text expresses S=Lim S, and writes
  dozen posts and articles about it
- Completely unable to see his mistake or even
  understand what Euler was doing
- This goes already on for years, so its he is
  not only an ignoramus, but also an ignorabimus

Am Montag, 2. Oktober 2017 23:15:35 UTC+2 schrieb burs...@gmail.com:
> You mean you were immune to "brainwash" so far,
> because you don't have any brains anyway?
>
> bird brain John Gabriel, scarecrow from
> the land of Oz.
>
> Am Montag, 2. Oktober 2017 23:09:20 UTC+2 schrieb John Gabriel:
> > On Monday, 2 October 2017 15:58:04 UTC-5, Me  wrote:
> > > On Monday, October 2, 2017 at 10:47:08 PM UTC+2, burs...@gmail.com wrote:
> > >
> > > > But Netzweltler's are of interest, since they might show
> > > > some very juvenile believes, maybe even not their own,

> > > > just...

netzweltler wrote on 10/2/2017 :

[...]

> Yes. pi is already there and we can exactly locate its position on the number
> line,

Okay, show me. Where *exactly* is pi on the real number line?

Use your stepwise method if you want to, take your time, and post the
answer when you finish the calculation.

Am Montag, 2. Oktober 2017 22:09:44 UTC+2 schrieb burs...@gmail.com:
> Well this is probably the greatest nonsense somebody
> ever posted on sci.math. You know, you didn't say
> rational number line.
>
> So when it is the real number line, pi is of course
> there. There is of course a point on the real number
> line that is pi.

It doesn't make sense to discuss the "number line" as long as the problem under discussion hasn't been fixed.

Am Montag, 2. Oktober 2017 22:40:48 UTC+2 schrieb Me:

> On Monday, October 2, 2017 at 10:09:44 PM UTC+2, burs...@gmail.com wrote:
>
> > Well this is probably the greatest nonsense somebody
> > ever posted on sci.math.
>
> I don't thinks so. You know there are such luminaries like WM, AP or JG.
>
> But actually "Netzweltler" already showed up in de.sci.mathematic posting nonsensical views concernig (infinitely many) "commands", "steps" and their "result" etc. etc.

So, what's wrong about (infinitely many) "commands", "steps" and their "result" etc. etc.?

Oh, I guess the problem lies not with *them*.

> > Maybe he was a pupil of Mückenheim, poor guy.

No comment?

netzweltler formulated the question :
Nothing really, it looks like a good way to get approximations as near
to a number as desired, but mathematics is not restricted to simple
arithmetic.

divide 9 by 9.
9 = 1 x 9 + 0. right?

9 = 0 x 9 + 9
90 = 9 x 9 + 9
900 = 99 x 9 + 9
9000 = 999 x 9 + 9
90000 = 9999 x 9 + 9

it means that

9 = 0 x 9 + 9
9.0 = 0.9 x 9 + 0.9
9.00 = 0.99 x 9 + 0.09
9.000 = 0.999 x 9 + 0.009
9.0000 = 0.9999 x 9 + 0.0009
9.00000 = 0.99999 x 9 + 0.00009
9.000000 = 0.999999 x 9 + 0.000009
9.0000000 = 0.9999999 x 9 + 0.0000009
9.00000000 = 0.99999999 x 9 + 0.00000009
9.000000000 = 0.999999999 x 9 + 0.000000009
9.0000000000 = 0.9999999999 x 9 + 0.0000000009
9.00000000000 = 0.99999999999 x 9 + 0.00000000009
9.000000000000 = 0.999999999999 x 9 + 0.000000000009

unless i've mistyped.

What do you want to fix? If pi is already there,
than a 0-step process is sufficient, the process says:

   hi I am at pi

Or if you want you can use a 1-step process, one
that starts with Euler number e:
 
   hi I am at e
   now I add pi-e to myself
   hi I am at pi

I guess you mean Q-series or something. Yes pi is
irrational, no element from Q. And a Q-series will
never hit pi on its way. Here is a proof:

Proof: Assume a Q-series would hit pi on its way.
Then there would be an index n, such that sn=pi,
the partial sum up to n summands would equal pi.

But each partial sum of a Q-series is from Q, and
pi is not from Q, so we would get a contradiction
saying pi is from Q, since it would be sn=pi.

So by proof by contradiction the
Q-series cannot hit pi.

Am Montag, 2. Oktober 2017 23:32:25 UTC+2 schrieb netzweltler:
> Am Montag, 2. Oktober 2017 22:09:44 UTC+2 schrieb burs...@gmail.com:
> > Well this is probably the greatest nonsense somebody

> > ever posted on sci.math. You know...

On Monday, 2 October 2017 17:15:35 UTC-4, Me  wrote:
> On Monday, October 2, 2017 at 11:09:20 PM UTC+2, John Gabriel wrote:
>
> > It's kind of sad and pathetic to see you and Jan Burse stroking each others'
> > frail egos,
>
> Just brothers in arms, I guess.

More like morons in arms.

>
> > especially when you are so ignorant from all the brainwashing you've received.
>
> Some'd call it /education/, John.

Bad education. Yes.

Because pi is an incommensurable ratio, a Greek notion,
it can never by a rational number. But I don't know
whether they really considered pi already as such.

There is this story with pythagoras and sqrt(2)
irrational. But I am also loss what concerns the
"lost book by Fibonacci", I must have lost it as well:

Were ratios of incommensurable magnitudes
interpreted as irrational numbers prior to Fibonacci?
https://hsm.stackexchange.com/questions/5582/were-ratios-of-incommensurable-magnitudes-interpreted-as-irrational-numbers-prio/5585

Am Dienstag, 3. Oktober 2017 00:33:48 UTC+2 schrieb burs...@gmail.com:
> What do you want to fix? If pi is already there,
> than a 0-step process is sufficient, the process says:
>
>    hi I am at pi
>
> Or if you want you can use a 1-step process, one
> that starts with Euler number e:
>  
>    hi I am at e
>    now I add pi-e to myself
>    hi I am at pi
>
> I guess you mean Q-series or something. Yes pi is
> irrational, no element from Q. And a Q-series will
> never hit pi on its way. Here is a proof:
>

> Proof: Assume a Q-series would hi...

True. There are the fallacies and ill-formed definitions such as S = Lim S.

Whoever thought that morons would still be equating a series to its limit in the 21st century...

If you say so.

That pi would be never hit by Q-steps, I guess I
dont need to explain this induction step:

   sn in Q and an in Q ==> sn+1 in Q

So if you add rational number summands to you get
rational number partial sums. But still who

suggested pi irrational first is an interesting
question. I guess it was not Leonardo de Pisa (Fibonnaci).

Here is another hint:

"There is a claim on the wikipedia article on irrational
numbers that Aryabhata wrote that pi was incommensurable
(5th century) but the question had to be asked as soon someone realized there was such numbers... (that was 5th century Before Christ)"
When was π first suggested to be irrational?
https://math.stackexchange.com/questions/177546/when-was-pi-first-suggested-to-be-irrational

(No use to ask bird brain John Gabriel, he
knows nothing, he still waits that somebody
bangs his anal warts for his claim S=Lim S)

Am Dienstag, 3. Oktober 2017 00:41:50 UTC+2 schrieb burs...@gmail.com:
> Because pi is an incommensurable ratio, a Greek notion,
> it can never by a rational number. But I don't know
> whether they really considered pi already as such.
>
> There is this story with pythagoras and sqrt(2)
> irrational. But I am also loss what concerns the
> "lost book by Fibonacci", I must have lost it as well:
>
> Were ratios of incommensurable magnitudes
> interpreted as irrational numbers prior to Fibonacci?
> https://hsm.stackexchange.com/questions/5582/were-ratios-of-incommensurable-magnitudes-interpreted-as-irrational-numbers-prio/5585
>
>
> Am Dienstag, 3. Oktober 2017 00:33:48 UTC+2 schrieb burs...@gmail.com:

> > What d...

Our friend Newton used for pi:

pi = 3 sqrt(3)/4 + 24 integ_0^(1/4) sqrt(x-x^2) dx

Can this be used to show pi irrational?
Doing the integral we find sin^(-1).

(*)
Issac Newton in 1665-1666
“I am ashamed to tell you to how many figures I
carried these computations, having not other
business at the time.”
http://www.math.tamu.edu/~dallen/masters/alg_numtheory/pi.pdf
> > interpreted as irrational numbers prior to Fibon...

For Ivan Niven the irrationality of pi,
is just Corollary 2.6. He is really a master:

"In this monograph, Ivan Niven provides a
masterful exposition of some central results
on irrational, transcendental, and normal
numbers. He gives a complete treatment by
elementary methods of the irrationality of
the exponential, logarithmic, and trigonometric
functions with rational arguments"

Irrational Numbers (Carus Mathematical Monographs,
Band 11) (Englisch) Taschenbuch – 18. August 2005
von Ivan Niven (Autor)
https://www.amazon.de/Irrational-Numbers-Carus-Mathematical-Monographs/dp/0883850389
> > > it can never by a rational number. ...

But I don't know whether the promise "elementary"
is true. For irrational numbers such as sqrt(2),

which are also algebraic, often number theoretic
approaches work, to show irrationality, we can

use the polynomial that makes it algebraic.
But pi is not algebraic, it is irrational,

but also transcendental, and we want to use it
to illustrate something about series,

how can be show pi irrational, without invoking
series? Is this possible? Would integrals be allowed?
> > > (No use to ask bird brain John Gabr...

I don't buy that. Who establishes the truth. I won't turn to
the Status Quo.

Mitchell Raemsch

I am trying to challenge the stupid "If the process
doesn't get us there then we don't get there."
If the Q-sequence is not fixed, like, 0.9, 0.99, 0.999, ...
if we can choose it, you can just do 0.25, 0.50, 0.75, 1.00

and you have reached 1. But for pi, you cannot even do
that, for Q-sequences. But since pi is transcendental, how
can I show this without invoking a Q-sequence definition
of pi itself. Besides that, the rest is uninteresting

every body knows:
- That sets need not have a maximum
- That sets can have an outside lub

Why pi? Well Netzweltler imported it from somewhere, and
somehow his post implies that he assumes pi irrational..
which means basically he posseses some processes that
"If the process doesn't get us there", he nevertheless

gets something, for example pi. Or what definition of
pi does he use? Pretty unclear for the moment.

Am Dienstag, 3. Oktober 2017 02:32:28 UTC+2 schrieb mitchr...@gmail.com:
> On Monday, October 2, 2017 at 1:58:04 PM UTC-7, Me wr...

Truth is never established. It simply is. Truth is never easy and always inconvenient.

> > > > (No use to ask bird...

Pathetic bullshit. Your comfort zone an obstacle
for you for doing real math. Bangging you head,

the S=Lim S is inconic for that, on:

- That sets need not have a maximum
- That sets can have an outside lub

Is not math. Thats just boring stuff. What happened
to your Archimedes studies, and your failed pi irrational

attempt. Nothing I guess. Only idiotic scarecrow
posts so far. Pathetic bullshit.

Am Dienstag, 3. Oktober 2017 02:52:16 UTC+2 schrieb John Gabriel:
> On Monday, 2 October 2017 20:32:28 UTC-4, mitchr...@gmail.com  wrote:
> > On Monday, October 2, 2017 at 1:58:04 PM UTC-7, Me wrote:
> > > On Monday, October 2, 2017 at 10:47:08 PM UTC+2, burs...@gmail.com wrote:
> > >
> > > > But Netzweltler's are of interest, since they might show
> > > > some very juvenile believes, maybe even not their own,
> > > > just some mish mash from dunno where, maybe from the land of Oz?
> > >

> > > The problem with "Netzweltler" (as you will soon find out for yourself, I guess) is that he...

Bird brain John Gabriel mongo math in two lines:
- Q: Is pi irrational. A: dunno.
- Q: What did Euler write. A: S=Lim S.

Am Dienstag, 3. Oktober 2017 02:57:01 UTC+2 schrieb burs...@gmail.com:
> Pathetic bullshit. Your comfort zone an obstacle
> for you for doing real math. Bangging you head,
>
> the S=Lim S is inconic for that, on:
> - That sets need not have a maximum
> - That sets can have an outside lub
>
> Is not math. Thats just boring stuff. What happened
> to your Archimedes studies, and your failed pi irrational
>
> attempt. Nothing I guess. Only idiotic scarecrow
> posts so far. Pathetic bullshit.
>
> Am Dienstag, 3. Oktober 2017 02:52:16 UTC+2 schrieb John Gabriel:
> > On Monday, 2 October 2017 20:32:28 UTC-4, mitchr...@gmail.com  wrote:
> > > On Monday, October 2, 2017 at 1:58:04 PM UTC-7, Me wrote:
> > > > On Monday, October 2, 2017 at 10:47:08 PM UTC+2, burs...@gmail.com wrote:
> > > >
> > > > > But Netzweltler's are of interest, since they might show

> > >...

On Monday, 2 October 2017 20:59:24 UTC-4, burs...@gmail.com  wrote:
> Bird brain John Gabriel mongo math in two lines:
> - Q: Is pi irrational. A: dunno.
> - Q: What did Euler write. A: S=Lim S.
>
> Am Dienstag, 3. Oktober 2017 02:57:01 UTC+2 schrieb burs...@gmail.com:
> > Pathetic bullshit. Your comfort zone an obstacle
> > for you for doing real math. Bangging you head,
> >
> > the S=Lim S is inconic for that, on:
> > - That sets need not have a maximum
> > - That sets can have an outside lub
> >
> > Is not math. Thats just boring stuff. What happened
> > to your Archimedes studies, and your failed pi irrational
> >
> > attempt. Nothing I guess. Only idiotic scarecrow
> > posts so far. Pathetic bullshit.
> >
> > Am Dienstag, 3. Oktober 2017 02:52:16 UTC+2 schrieb John Gabriel:
> > > On Monday, 2 October 2017 20:32:28 UTC-4, mitchr...@gmail.com  wrote:
> > > > On Monday, October 2, 2017 at 1:58:04 PM UTC-7, Me wrote:

> > > > > On Monday, October 2, 2017 at 10:47:08 PM...

Showing pi irrational works also with,
Eulers Basel problem:

   pi^2/6 = lim n->oo sum_k=1^n 1/k^2

If you can show the result transcendental.
Then pi cannot be algebraic, because pi^2

would be also algebraic, a contradiction to
pi^2 transcendental. But this is again a

Q-series which I would like to avoid.

Am Dienstag, 3. Oktober 2017 02:59:24 UTC+2 schrieb burs...@gmail.com:
> Bird brain John Gabriel mongo math in two lines:
> - Q: Is pi irrational. A: dunno.
> - Q: What did Euler write. A: S=Lim S.
>
> Am Dienstag, 3. Oktober 2017 02:57:01 UTC+2 schrieb burs...@gmail.com:
> > Pathetic bullshit. Your comfort zone an obstacle
> > for you for doing real math. Bangging you head,
> >
> > the S=Lim S is inconic for that, on:
> > - That sets need not have a maximum
> > - That sets can have an outside lub
> >
> > Is not math. Thats just boring stuff. What happened
> > to your Archimedes studies, and your failed pi irrational
> >

> > attempt. Nothing I guess. Only idiotic...

On 10/2/2017 2:47 PM, netzweltler wrote:
> Am Montag, 2. Oktober 2017 20:35:56 UTC+2
> schrieb Jim Burns:
>> On 10/2/2017 1:58 PM, netzweltler wrote:
>>> Am Montag, 2. Oktober 2017 17:59:21 UTC+2
>>> schrieb Jim Burns:
>>>> On 10/1/2017 3:22 AM, netzweltler wrote:

>>>>> Do you agree that 0.999... means infinitely many commands
>>>>> Add 0.9 + 0.09
>>>>> Add 0.99 + 0.009
>>>>> Add 0.999 + 0.0009
>>>>> ...?
>>>>
>>>> 0.999... does not mean infinitely many commands.
>>>
>>> But that's exactly what it means.
>>
>> That's not the standard meaning.
>
> So, you disagree that
> 0.999... = 0.9 + 0.09 + 0.009 + ... ?

Your '...' is not usable. If we say what we _really_ mean,
in a manner clear enough to reason about, then the '...'
disappears. Also, what  we are left with are finitely many
statements of finite length. You will not find infinitely
many commands in those finitely-many, finite-length
statements.

We sometimes write the set of natural numbers as
    { 0, 1, 2, 3, ... }
The '...' is informal. We do not use '...' in our reasoning,
we use a correct description of what the '...' stands for.

Do you see '...' anywhere in the following?

The set N contains 0, and for every element x in N, its
successor Sx is in N.

This is true of N but not true of any _proper_ subset of N.

_Therefore_ , if we can prove that B is a subset of N
which contains 0 and which, for element x of B, contains Sx,
then B is not a _proper_ subset of N.

B nonetheless is a subset of N, we just said so. The only subset
of N which B can be is N. Therefore, B = N.

This is finite reasoning about the infinitely many elements
in N. Note that there is no '...' in it.

I could continue and derive 0.999... = 1 from our definitions,
and nowhere in that deri...

Am Dienstag, 3. Oktober 2017 00:33:48 UTC+2 schrieb burs...@gmail.com:

> What do you want to fix? If pi is already there,
> than a 0-step process is sufficient, the process says:
>
>    hi I am at pi
>
> Or if you want you can use a 1-step process, one
> that starts with Euler number e:
>  
>    hi I am at e
>    now I add pi-e to myself
>    hi I am at pi
>
> I guess you mean Q-series or something. Yes pi is
> irrational, no element from Q. And a Q-series will
> never hit pi on its way. Here is a proof:
>
> Proof: Assume a Q-series would hit pi on its way.
> Then there would be an index n, such that sn=pi,
> the partial sum up to n summands would equal pi.
>
> But each partial sum of a Q-series is from Q, and
> pi is not from Q, so we would get a contradiction
> saying pi is from Q, since it would be sn=pi.
>
> So by proof by contradiction the
> Q-series cannot hit pi.
>
>
> Am Montag, 2. Oktober 2017 23:32:25 UTC+2 schrieb netzweltler:

> > Am Montag, 2. Oktober 201...

> and nowhere in that derivation will be '.....

netzweltler brought next idea :
>> and nowhere in that derivation will be '...'. There will not be
>> infinitely many commands in it either.
>>
>>>> You give it some other meaning, and t...

On Sunday, 12 February 2017 08:14:11 UTC-5, John Gabriel  wrote:
> S = Lim S in this case, contradicts the fact that 1/3 has no representation in base 10.
>
> However, it also leads to even more absurd definitions, that is, pi=3.14159...  when in fact it is known that pi has NO measure.
>
>
> https://www.youtube.com/watch?v=5hulvl3GgGk
>
> https://www.youtube.com/watch?v=w8s_8fNePEE
>
>
> Your comments are unwelcome shit and will be ignored.
>
> This is posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.

I have uploaded a new video which will probably be my last on this topic:

https://www.youtube.com/watch?v=NBOs-Xf_UIg

LinkedIn aarticle:

https://www.linkedin.com/pulse/eulers-worst-definition-lim-john-gabriel

If you think calculus needs this rot to work, think again!

https://www.youtube.com/watch?v=-8JAbBJ9a0w

Jim Burns used his keyboard to write :

>>> which contains 0 and which, for element x of B, contains Sx,
>>> then B is not a _proper_ subset of N.
>>>
>>> B nonetheless is a subset of N, we just said so. The only subset
>>> of N which B can be is N. Therefore, B = N.
>>>
>>> This is finite reasoning about the infinitely many elements
>>> in N. Note that there is no '...' in it.
>>>

>>> I could ...

On 10/3/2017 8:20 AM, FromTheRafters wrote:

> But 0.999 repeating is a rational number, no need for repeating
> decimals at all in the naturals. Repeating zeros is okay I guess,
> but why use them in the naturals. In the rationals and reals,
> repeating zeros are called 'terminating' decimal expansions and
> the trailing zeros are elided.

Infinite decimals represent real number "fractions".
The need for 0.999..., such as it is, is for _some real number_
to be assigned appropriately to every infinite decimal expansion.
It's the same reason we assign the infinite decimal 0.1234000...
to the terminating decimal 0.1234. We want 0.1234000... to be
_something_ . What else would it be?

> >> which contains 0 and which, for element x of B, contains Sx,
> >> then B is not a _proper_ subset of N.
> >>
> >> B nonetheless is a subset of N, we just said so. The only subset
> >> of N which B can be is N. Therefore, B = N.
> >>
> >> This is finite reasoning about the infinitely many elements
> >> in N. Note that there is no '...' in it.
> >>

> >> I could continue and derive 0.999... = 1 from our definitions,
> >> and nowhere in that derivation will be '...'. There will not be
> >> infinitely many commands in it either.
>

> > Sorry, no. The meaning of "..." is absolutely clear in this
> >  context and
>
> Is it clear to you? Reall...

On Tuesday, 3 October 2017 10:53:44 UTC-4, Jim Burns  wrote:
> On 10/3/2017 8:20 AM, FromTheRafters wrote:
>
> > But 0.999 repeating is a rational number, no need for repeating
> > decimals at all in the naturals. Repeating zeros is okay I guess,
> > but why use them in the naturals. In the rationals and reals,
> > repeating zeros are called 'terminating' decimal expansions and
> > the trailing zeros are elided.
>
> Infinite decimals represent real number "fractions".

Wrong. Strings of infinite decimals are a consequence of a long division algorithm gone bananas. Division by definition is a finite algorithm. It is completely unremarkable that repeated patterns are found for fractions not measurable in all bases. Since there is no such thing as a unique string which represents a given rational number, it follows clearly that such a decimal string is an ill-formed concept. You can't say that 0.333... is a unique string representation. After how many digits is it unique? 10 billio...

Jim Burns submitted this idea :

> On 10/3/2017 8:20 AM, FromTheRafters wrote:
>
>> But 0.999 repeating is a rational number, no need for repeating
>> decimals at all in the naturals. Repeating zeros is okay I guess,
>> but why use them in the naturals. In the rationals and reals,
>> repeating zeros are called 'terminating' decimal expansions and
>> the trailing zeros are elided.
>
> Infinite decimals represent real number "fractions".
> The need for 0.999..., such as it is, is for _some real number_
> to be assigned appropriately to every infinite decimal expansion.

I agree, if you meant infinite 'repeating' decimal expansions, but
netzweltler wrote about little n in N (n an element of the naturals)
and was using decimal expansions which aren't needed in whole numbers.
Sure, in the rationals and the reals decimal expansions certainly make
sense.

> It's the same reason we assign the infinite decimal 0.1234000...

> to the terminating decimal 0.1234. We want 0.1234000... to be...

On 10/3/2017 12:15 PM, FromTheRafters wrote:
> Jim Burns submitted this idea :

>> It's the same reason we assign the infinite decimal 0.1234000...

>> to the terminating decimal 0.1234. We want 0.1234000... to be
>> _something_ . What else would it be?
>

> I think you missed my point.

You're very likely right about that.

I think I would have to study this thread more closely in order
to appreciate this better from netzweltler's point of view.
He seems determined to use things in a non-standard way.
I just want to point out that it's non-standard and (I hope)
explain the standard way.

Jim Burns was thinking very hard :
It is good to do so just in case somebody comes here to learn
something. There is a lot of bad information being regurgitated here.
It is as if they want to mislead those who don't already know how to
tell the difference between intuitive sounding bullshit and mathematics
which can at times seem to be very counterintuitive.

Also, I...

> >>>> You give it some other meaning, and then you find a problem
> >>>> with the me...

> >> which contains 0 and which, for element x ...

Nope, doesn't make any sense at all. Just plain
crazy. Why would you write a formula for a partial
sum as a number:

If ... means n-th place, then this here:

    0.999...

Would mean:

    1-10^(-n)

So it would not be a number, but an expression with
a varying place holder n. Making this exxpression
here also dependent on n:

    0.999... = 1

But fact is that the following expression:

    0.999...

Has the meaning:

    lim n->oo (1-10^(-n))

Which has the value:

    1

Am Dienstag, 3. Oktober 2017 21:15:40 UTC+2 schrieb netzweltler:
> The meaning of "..." is absolutely clear in this context and we both know that there is a nth decimal place for each n ∈ N in 0.999...

netzweltler laid this down on his screen :
>>>>>> with the meaning you gave it. Supposing I wanted to sort out
>>>>>> what that other mea...

On Tuesday, 3 October 2017 15:40:54 UTC-4, burs...@gmail.com  wrote:
> Nope, doesn't make any sense at all. Just plain
> crazy. Why would you write a formula for a partial
> sum as a number:
>
> If ... means n-th place, then this here:
>
>     0.999...
>
> Would mean:
>
>     1-10^(-n)
>
> So it would not be a number, but an expression with
> a varying place holder n. Making this exxpression
> here also dependent on n:
>
>     0.999... = 1
>
> But fact is that the following expression:
>
>     0.999...
>
> Has the meaning:
>
>     lim n->oo (1-10^(-n))

No.  0.999... = \sum_{k=1}^{\infty}  9/(10^k)

>
 lim_{n \to \infty} \sum_{k=1}^{n}  9/(10^k) has the value:

On Tuesday, 3 October 2017 13:33:55 UTC-4, FromTheRafters  wrote:
> Jim Burns was thinking very hard :
> > On 10/3/2017 12:15 PM, FromTheRafters wrote:
> >> Jim Burns submitted this idea :
> >
> >>> It's the same reason we assign the infinite decimal 0.1234000...
> >>> to the terminating decimal 0.1234. We want 0.1234000... to be
> >>> _something_ . What else would it be?
> >>
> >> I think you missed my point.
> >
> > You're very likely right about that.
> >
> > I think I would have to study this thread more closely in order
> > to appreciate this better from netzweltler's point of view.
> > He seems determined to use things in a non-standard way.
> > I just want to point out that it's non-standard and (I hope)
> > explain the standard way.
>
> It is good to do so just in case somebody comes here to learn
> something. There is a lot of bad information being regurgitated here.

True. Most of it is in your comments and those of Jim Burns.

> It is as if they want to mislead tho...

Since  \sum_{k=1}^n 9/(10^k) = 1 - 10^(-n),
and \sum_{k=1}^oo a(k) = lim n->oo \sum_{k=1}^n a(k)

The two are synonymous:

   lim n->oo (1-10^(-n)) = \sum_{k=1}^{\infty}  9/(10^k)

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

BTW: The following authors here on sci.math already
explained this two you like a dozen times:
- Dan
- Me
- Markus Klyver
- Zelos Malum
- Etc..

You, bird brain John Gabriel, probably qualify
for the most stupid human being on the planet.

Should we call the Guiness book of records?

On Tuesday, 3 October 2017 17:21:40 UTC-4, burs...@gmail.com  wrote:
> Since  \sum_{k=1}^n 9/(10^k) = 1 - 10^(-n),
> and \sum_{k=1}^oo a(k) = lim n->oo \sum_{k=1}^n a(k)
>
> The two are synonymous:
>
>    lim n->oo (1-10^(-n)) = \sum_{k=1}^{\infty}  9/(10^k)

No birdbrain. They are not synonymous.

\sum_{k=1}^oo a(k)  is a SERIES you moron. It is the justification for 0.abc...

lim n->oo \sum_{k=1}^n a(k) is a LIMIT you moron. It is the UPPER BOUND of all the partial sums of the series.

MOOOOOROOOOOOON!!!!!

<crapola>

Why are they the same? Simply because oo is not
an element from the natural numbers. So

   \sum_{k=1}^oo 9/(10^k)

cannot mean something where the last summand is
9/(10^oo) because a value 9/(10^oo) doesn't exist

in Q, since there is no k=oo. So all you have is
the partial sums that go on and on,

    \sum_{k=1}^n 9/(10^k)
    0.9
    0.99
    0.999
    ...

And you cannot identify a final value. To extract
a value from the series you need the limes.

On Tuesday, 3 October 2017 17:35:54 UTC-4, burs...@gmail.com  wrote:
> Why are they the same?

Are you talking to yourself again you delusional dimwit?

<crapola>

Nope, thats not how this notation is used.
Series are indicated by {sn}. You don't
need the infinity symbols, since the indexes
of the partial sums range over the natural

numbers, no infinity involved for the series
itself, in the sense that we need to express
something tens to infinity or similar. You find
the notation already explained here, if you use

sum_i=1^oo, you turn a series into a limit,
see for yourself:

an expression like

    a1 + a2 + a3 + ⋯

or, using the summation sign,

    sum_i=1^oo ai

https://en.wikipedia.org/wiki/Series_%28mathematics%29

I have uploaded the picture
https://gist.github.com/jburse/e60242e2e02e611e4373df55bfc37953#gistcomment-2219906

There is also a certain logic behind using
{an} or (an) for series, and

a1 + a2 + a3 + ⋯ or sum_i=1^oo ai for the
limit, since {an} or (an) wants to

indicate a multiplicity of values, but
a1 + a2 + a3 + ⋯ or sum_i=1^oo ai

is just one value.

Am Dienstag, 3. Oktober 2017 23:43:44 UTC+2 schrieb burs...@gmail.com:
> Nope, thats not how this notation is used.
> Series are indicated by {sn}. You don't
> need the infinity symbols, since the indexes
> of the partial sums range over the natural
>
> numbers, no infinity involved for the series
> itself, in the sense that we need to express
> something tens to infinity or similar. You find
> the notation already explained here, if you use
>
> sum_i=1^oo, you turn a series into a limit,
> see for yourself:
>
> an expression like
>
>     a1 + a2 + a3 + ⋯
>
> or, using the summation sign,
>
>     sum_i=1^oo ai
>
> https://en.wikipedia.org/wiki/Series_%28mathematics%29
>
> I have uploaded the picture

> https://gist.github.com/jburse/e60242e2e02...

Corr.:

There is also a certain logic behind using

{an} or (an) for *sequence*, and

Am Dienstag, 3. Oktober 2017 23:51:43 UTC+2 schrieb burs...@gmail.com:
> There is also a certain logic behind using
> {an} or (an) for series, and
>
> a1 + a2 + a3 + ⋯ or sum_i=1^oo ai for the
> limit, since {an} or (an) wants to
>
> indicate a multiplicity of values, but
> a1 + a2 + a3 + ⋯ or sum_i=1^oo ai
>
> is just one value.
>
> Am Dienstag, 3. Oktober 2017 23:43:44 UTC+2 schrieb burs...@gmail.com:
> > Nope, thats not how this notation is used.
> > Series are indicated by {sn}. You don't
> > need the infinity symbols, since the indexes
> > of the partial sums range over the natural
> >
> > numbers, no infinity involved for the series
> > itself, in the sense that we need to express
> > something tens to infinity or similar. You find
> > the notation already explained here, if you use
> >
> > sum_i=1^oo, you turn a series into a limit,
> > see for yourself:
> >
> > an expression like
> >

> >    ...

On Tuesday, 3 October 2017 17:43:44 UTC-4, burs...@gmail.com  wrote:
> Nope, thats ...

Shhhh moron. Shhhh.

So you won the stupidity award with:
lim n->oo (1-10^(-n)) <> \sum_{k=1}^{\infty} 9/(10^k)

Your parrot tourett gets in your way,
formal multiplication, sums, you get everything

wrong, thats quite a bad track record for
the greatest mathematician of all time...

> >>>> in a manner clear enough to reason about, then th...

On Monday, October 2, 2017 at 2:50:45 PM UTC-6, John Gabriel wrote:

> I just realised that when I first posted this comment, I had not included the
> link to my now world famous article:

> https://www.linkedin.com/pulse/eulers-worst-definition-lim-john-gabriel

And in that article, Euler did indeed use oversimpllified language, by talking
about continuing the sum of the series to infinity, so that the error becomes
zero.

Instead of the proper rigorous language, where the error can be made smaller
than any finite value by continuing the sum of the series to a large enough, but
finite number of terms.

But, although you correctly caught a slip of Euler's tongue, or, rather, pen,
that still has exactly zero relevance to the validity of the proper rigorous
language.

John Savard

keskiviikko 4. lokakuuta 2017 0.21.40 UTC+3 burs...@gmail.com kirjoitti:
> Since  \sum_{k=1}^n 9/(10^k) = 1 - 10^(-n),
> and \sum_{k=1}^oo a(k) = lim n->oo \sum_{k=1}^n a(k)
>
> The two are synonymous:
>
>    lim n->oo (1-10^(-n)) = \sum_{k=1}^{\infty}  9/(10^k)

are you trying to insist that  0.(9) is equal to
lim{n \to infinity} s(n) = lim{n --> infinity} 0.9 + 0.09 + 0.009 + ... + 9*10^(-n) = 1 ?

Using the real numbers 0.(9) is not equal to 1

and you have not demonstrated how do you get S(∞) = 1. You have only shown
S(∞) = 0.(9)

and because you are not willing to tell what you are doing, I have to
explain everything myself:

S(∞) = 1 is possible only if you have included infinitesimals into the real numbers, but since you all have rejected them you are not supposed to use them too, right? Would you agree that it is reasonable to demand
that one cannot eat a cake and have it too? Or perhaps doing so brings
some benefits to all of you so you are not going be so strict when
you give some li...

On Monday, October 2, 2017 at 5:52:16 PM UTC-7, John Gabriel wrote:
> On Monday, 2 October 2017 20:32:28 UTC-4, mitchr...@gmail.com  wrote:
> > On Monday, October 2, 2017 at 1:58:04 PM UTC-7, Me wrote:
> > > On Monday, October 2, 2017 at 10:47:08 PM UTC+2, burs...@gmail.com wrote:
> > >
> > > > But Netzweltler's are of interest, since they might show
> > > > some very juvenile believes, maybe even not their own,
> > > > just some mish mash from dunno where, maybe from the land of Oz?
> > >
> > > The problem with "Netzweltler" (as you will soon find out for yourself, I guess) is that he suffices the original definition of a crank:
> > >
> > > "A crank is defined as a man who cannot be turned." (Nature, 8 Nov 1906, 25/2)
> >
> > I don't buy that. Who establishes the truth. I won't turn to
> > the Status Quo.
> >
> > Mitchell Raemsch
>
> Truth is never established. It simply is. Truth is never easy and always inconvenient.

Who has it then?
What about God's absolute truth?
And who is willing to be corrected by it?

Mitchell Raemsch

corr.

keskiviikko 4. lokakuuta 2017 8.31.03 UTC+3 7777777 kirjoitti:
> multiply it by 3 and arrive at
> 0.(9)9 + 0.(9)9 + 3*0.(3)4 = 0.(9)(9)+0.(9)(9)+0.(9)(9) = 1

0.(9)9 + 0.(9)9 + 3*0.(3)4 = 0.(9)(9)+0.(9)(9)+0.(9)(9) = 3

Am Dienstag, 3. Oktober 2017 21:40:54 UTC+2 schrieb burs...@gmail.com:
> Nope, doesn't make any sense at all. Just plain
> crazy. Why would you write a formula for a partial
> sum as a number:

Partial Sum? 0.999... means an infinite sum.

> If ... means n-th place,

No, it doesn't.

> >>>> in a manner clear enough to reason about, then the '...'
> >>>> disappears. Also, what  we are left with are finitely many

> >>>> statements of finite length. You will not find infi...

On Tuesday, 3 October 2017 22:13:36 UTC-4, Quadibloc  wrote:
> On Monday, October 2, 2017 at 2:50:45 PM UTC-6, John Gabriel wrote:
>
> > I just realised that when I first posted this comment, I had not included the
> > link to my now world famous article:
>
> > https://www.linkedin.com/pulse/eulers-worst-definition-lim-john-gabriel
>
> And in that article, Euler did indeed use oversimpllified language, by talking about continuing the sum of the series to infinity, so that the error becomes zero.

If assertions were pennies, you would be very rich! Chuckle.

>
> Instead of the proper rigorous language, where the error can be made smaller
> than any finite value by continuing the sum of the series to a large enough, but finite number of terms.

Nonsense. Euler was very articulate and he wrote in his home language.

>
> But, although you correctly caught a slip of Euler's tongue, or, rather, pen,
> that still has exactly zero relevance to the validity of the proper rigorous
> language.

FALSE.

>
> John Savard

On Wednesday, 4 October 2017 01:39:51 UTC-4, mitchr...@gmail.com  wrote:
> On Monday, October 2, 2017 at 5:52:16 PM UTC-7, John Gabriel wrote:
> > On Monday, 2 October 2017 20:32:28 UTC-4, mitchr...@gmail.com  wrote:
> > > On Monday, October 2, 2017 at 1:58:04 PM UTC-7, Me wrote:
> > > > On Monday, October 2, 2017 at 10:47:08 PM UTC+2, burs...@gmail.com wrote:
> > > >
> > > > > But Netzweltler's are of interest, since they might show
> > > > > some very juvenile believes, maybe even not their own,
> > > > > just some mish mash from dunno where, maybe from the land of Oz?
> > > >
> > > > The problem with "Netzweltler" (as you will soon find out for yourself, I guess) is that he suffices the original definition of a crank:
> > > >
> > > > "A crank is defined as a man who cannot be turned." (Nature, 8 Nov 1906, 25/2)
> > >
> > > I don't buy that. Who establishes the truth. I won't turn to
> > > the Status Quo.
> > >
> > > Mitchell Raemsch
> >

> > Truth is never established. It simply...

>In fact it means exactly infinitely many commands.
>But of course if you define a series to be equal to its limit, then that's like defining an apple equal to an orange. That is your problem, not ours.

It doesn't because it is not operations upon operations, it is just a representation of one element in real numbers.

It does. And 0.875 is representing 3 operations, e.g.
0.8 + 0.07 + 0.005 or
0.5 + 0.25 + 0.125.

It can be an element AND represent some number of operations.

Your skull is too thick. Dry up and die.

netzweltler was thinking very hard :
I agree. The thing is that a finite number of steps (or commands) can,
at best, give a good enough approximation of some numbers. By adding
'ad infinitum' to the 'end' of these, and assuming it can be completed
in that (NaN) number of steps, you can arrive at the exact answer. This
is a case where it is not about the trip, but about the (eventual)
destination being defi...

> destination being defined exactly.
>
> As you already know, an arrow traveling from zero toward a target at
> two which can be described as going halfway there then halfway the
> remaining distance, then halfway a...

> > remaining distance, then halfway again, may seem like an unending
> > process with only better and better approximations being attainable.
> > But, if I can get t...

Still struggling with "S=Lim S", your own
blunder, bird brain John Gabriel birdbrains?
> > two which can ...

On 10/4/2017 4:19 AM, netzweltler wrote:
> Am Mittwoch, 4. Oktober 2017 01:29:50 UTC+2
> schrieb Jim Burns:

>> <Burns<netzweltler>>

>>
>> > Do you agree that 0.999... means infinitely many
>> > commands
>> > Add 0.9 + 0.09
>> > Add 0.99 + 0.009
>> > Add 0.999 + 0.0009
>> > ...?
>>
>> 0.999... does not mean infinitely many commands.
>>

>> </Burns<netzweltler>>
>
> To me it looks like that we don't even agree, that there
> are infinitely many 9s following.

Maybe we agree, maybe we don't. We might be using the same
words and meaning different things by them.

I say there are infinitely many nines following the '.'
What I mean by "infinitely many" here is that there is a
map, one-to-one but not onto, from those after-dot decimal
places to those after-dot decimal places. And '9' is in every
place.

I could say more, and I should, in order to say what 0.999...
means, but that is what "infinitely many 9s following" means.

What do you mean? That there is a '...' at the end?
How would someone reason from that to any conclusion?

> If there are infinitely many 9s following then there is a
> bijection between N and the decimal places in 0.999...

True enough for our purposes here. "Infinitely many" is broader
than we really want here. There are uncountable ordinals,
for example. If "9s following" is a map from an uncountable
ordinal to {0,1,2,3,4,5,6,7,8,9}, then there would not be
such a bijection, N to "decimal places".

But there is a bijection between N and decimal places.
Even better, there is an order-preserving bijection.
I think we mean the same thing here.

> If there are infinitely many 9s following then we are
> dealing with an infinite stepwise process as described
> above and I can't see why this shouldn't mean
> "infinitely many commands".

Even though you don't see why, it doesn't mean that.
I'm trying to explain, but if you still don't get it, that
will not mean that 0.999... means infinitely many commands.

Your argument is basically "This thing that you guys mean
does not make sense". You're right (I think) that "this thing"
does not make sense, but you're wrong that "this thing" is
what we mean.

----
When I write
    0.999...
I mean that there is a map (very uninteresting) fro...

Den tisdag 3 oktober 2017 kl. 22:04:52 UTC+2 skrev John Gabriel:

> On Tuesday, 3 October 2017 15:40:54 UTC-4, burs...@gmail.com  wrote:
> > Nope, doesn't make any sense at all. Just plain
> > crazy. Why would you write a formula for a partial
> > sum as a number:
> >

> > If ... means n-th place, then this here:

> >
> >     0.999...
> >
> > Would mean:
> >
> >     1-10^(-n)
> >
> > So it would not be a number, but an expression with
> > a varying place holder n. Making this exxpression
> > here also dependent on n:
> >
> >     0.999... = 1
> >
> > But fact is that the following expression:
> >
> >     0.999...
> >
> > Has the meaning:
> >
> >     lim n->oo (1-10^(-n))
>

> No.  0.999... = \sum_{k=1}^{\infty}  9/(10^k)
>
> >
>  lim_{n \to \infty} \sum_{k=1}^{n}  9/(10^k) has the value:

> >
> >     1
> >
> > Am Dienstag, 3. Oktober 2017 21:15:40 UTC+2 schrieb netzweltler:
> > > The meaning of "..." is absolutely clear in this context and we both know that there is a nth decimal place for each n ∈ N in 0.999...

\sum_{k=1}^{\infty}  9/(10^k) is defined as lim_{n \to \infty} \sum_{k=1}^{n}  9/(10^k). Hence, 0.999... = 1.

On Wednesday, 4 October 2017 15:06:33 UTC-4, Markus Klyver  wrote:
> Den tisdag 3 oktober 2017 kl. 22:04:52 UTC+2 skrev John Gabriel:
> > On Tuesday, 3 October 2017 15:40:54 UTC-4, burs...@gmail.com  wrote:
> > > Nope, doesn't make any sense at all. Just plain
> > > crazy. Why would you write a formula for a partial
> > > sum as a number:
> > >
> > > If ... means n-th place, then this here:
> > >
> > >     0.999...
> > >
> > > Would mean:
> > >
> > >     1-10^(-n)
> > >
> > > So it would not be a number, but an expression with
> > > a varying place holder n. Making this exxpression
> > > here also dependent on n:
> > >
> > >     0.999... = 1
> > >
> > > But fact is that the following expression:
> > >
> > >     0.999...
> > >
> > > Has the meaning:
> > >
> > >     lim n->oo (1-10^(-n))
> >
> > No.  0.999... = \sum_{k=1}^{\infty}  9/(10^k)
> >
> > >
> >  lim_{n \to \infty} \sum_{k=1}^{n}  9/(10^k) has the value:
> > >
> > >     1
> > >

> > > Am Dienstag, 3. Oktober 2017 21:15:40 UTC+...

There is no end. Nothing follows after infinitely many 9s. "infinitely many 9s following" replaces '...'.
> I mean that there is a map (very uninteresting) from N to digits
>     0 |-> 9
>     1 |-> 9
>     2 |-> 9
>        ...
> and this map represents the infinite decim...

Am Mittwoch, 4. Oktober 2017 13:43:34 UTC+2 schrieb FromTheRafters:
> netzweltler was thinking very hard :
> > Am Mittwoch, 4. Oktober 2017 11:44:35 UTC+2 schrieb Zelos Malum:
> >>> In fact it means exactly infinitely many commands.
> >>> But of course if you define a series to be equal to its limit, then that's
> >>> like defining an apple equal to an orange. That is your problem, not ours.
> >>
> >> It doesn't because it is not operations upon operations, it is just a
> >> representation of one element in real numbers.
> >
> > It does. And 0.875 is representing 3 operations, e.g.
> > 0.8 + 0.07 + 0.005 or
> > 0.5 + 0.25 + 0.125.
> >
> > It can be an element AND represent some number of operations.
>
> I agree. The thing is that a finite number of steps (or commands) can,
> at best, give a good enough approximation of some numbers.

The 3-step operations above give the number exactly. You can get this number after any finite number of operations. Whereas you can't get the ...

keskiviikko 4. lokakuuta 2017 17.31.48 UTC+3 burs...@gmail.com kirjoitti:
> Still struggling with "S=Lim S", your own
> blunder, bird brain John Gabriel birdbrains?

why don't you answer to my criticism? How do you end up with S(∞) = 1?
You have only shown how S(∞) = 0.(9)
Do you just simply put 0.(9) = 1, but why? It is just because you say so?
In doing so, can you just put everything that you want to be equal to anything you want? Just because that's what you want.

On 10/4/2017 3:58 PM, netzweltler wrote:
> Am Mittwoch, 4. Oktober 2017 18:27:18 UTC+2
> schrieb Jim Burns:
>> On 10/4/2017 4:19 AM, netzweltler wrote:

>>> To me it looks like that we don't even agree, that there
>>> are infinitely many 9s following.
>>
>> Maybe we agree, maybe we don't. We might be using the same
>> words and meaning different things by them.
>>
>> I say there are infinitely many nines following the '.'
>> What I mean by "infinitely many" here is that there is a
>> map, one-to-one but not onto, from those after-dot decimal
>> places to those after-dot decimal places. And '9' is in every
>> place.
>>
>> I could say more, and I should, in order to say what 0.999...
>> means, but that is what "infinitely many 9s following" means.
>>
>> What do you mean? That there is a '...' at the end?
>
> There is no end.

I mean a '...' at the end of the description.
When one writes
    0.9, 0.99, 0.999, ...
one puts '...' at the end of _that_ but what does it mean?

What I mean by "infinitely many 9s following" is broken down
into concepts that we already share in order to explain what
I mean -- which might not be the same as what you mean, even
though we use the same words, "infinitely many 9s following".

You raised this question. Do we agree? This is a question
which we can answer.

>  Nothing follows after infinitely many 9s.
>  "infinitely many 9s following" replaces '...'.

How do you say "inf...

There is no S(oo), there is no index n=oo.
Only lim n->oo S(n), which happens is can

be written as sum_i=1^oo a(i). For a definition
see this picture here:

Guiness book of records in stupidity,
bird brain John Gabriel entry
https://gist.github.com/jburse/e60242e2e02e611e4373df55bfc37953#gistcomment-2219906

Its really a definition, there is nothing to
guess, its a mathematical notational convention.

On 10/4/2017 3:58 PM, netzweltler wrote:

> While it is true that there is no line on the list
>    0 |-> write 0.9
>    1 |-> append another 9
>    2 |-> append another 9
>        ...
> that produces other than a finite string, it is true that
>  the complete list produces an infinite string.
> "We are dealing with infinitely many finite stepwise processes"
>  is true for each line of the list.
> "we are dealing with an infinite stepwise process" is true
>  for the complete list. We wouldn't get the resulting
>  0.999... otherwise.

This is backwards. The finite initial segments
    0.9, 0.99, 0.999, ...
_do not_ result in the infinite decimal
    0.999...

We have the infinite decimal _first_ , and then we trim it
to finite length, operate on the finite decimals in the
familiar way, and define the unique least upper bound of
all of the finite sums as the value of the infinite decimal.

_We don't do what you're describing_

Am Donnerstag, 5. Oktober 2017 15:02:00 UTC+2 schrieb Jim Burns:
> On 10/4/2017 3:58 PM, netzweltler wrote:
> > Am Mittwoch, 4. Oktober 2017 18:27:18 UTC+2
> > schrieb Jim Burns:
> >> On 10/4/2017 4:19 AM, netzweltler wrote:
>
> >>> To me it looks like that we don't even agree, that there
> >>> are infinitely many 9s following.
> >>
> >> Maybe we agree, maybe we don't. We might be using the same
> >> words and meaning different things by them.
> >>
> >> I say there are infinitely many nines following the '.'
> >> What I mean by "infinitely many" here is that there is a
> >> map, one-to-one but not onto, from those after-dot decimal
> >> places to those after-dot decimal places. And '9' is in every
> >> place.
> >>
> >> I could say more, and I should, in order to say what 0.999...
> >> means, but that is what "infinitely many 9s following" means.
> >>
> >> What do you mean? That there is a '...' at the end?
> >
> > There is no end.
>
> I mean a '...' at the end of the description.
> When one writes
>     0.9, 0.99, 0.999, ...
> one puts '...' at the end of _that_ but what does it mean?

(1-(1/10)^n)n∈N

> What I mean by "infinitely many 9s following" is broken down
> into concepts that we already share in order to explain what
> I mean -- which might not be the same as what you mean, even
> though we use the same words, "infinitely many 9s following".
>
> You raised this question. Do we agree? This is a question
> which we can answer.
>
> >  Nothing follows after infinitely many 9s.
> >  "infinitely many 9s following" replaces '...'.
>

> How do you say "infinitely many 9s following" without merely
> trading one thing tha...

Doesn't make any sense, since it is a sequence
and not a number. But obviously the decimal

notation 0.999... refers to a number. So its
simply lim n->oo (1-(1/10)^n), from the

mathematical notational convention, that the
... in the above context includes the limit.

...

On 10/5/2017 10:00 AM, netzweltler wrote:
> Am Donnerstag, 5. Oktober 2017 15:22:35 UTC+2
> schrieb Jim Burns:

[...]

>> _We don't do what you're describing_
>

> Nevertheless,

"Nevertheless"?
Do you agree that what you're describing is not what we're doing?

> the process
> 0 |-> write 0.9
> 1 |-> append another 9 (to the 0.9 already written)
> 2 |-> append another 9 (to the 0.99 already written)
>     ...
> results in 0.999...
>
> Whereas the process you specified earlier
> 0 |-> 0.9
> 1 |-> 0.99
> 2 |-> 0.999
>     ...
> is nothing else but an infinite list of terminating decimals.

Right. Nothing else but an infinite list of terminating decimals,
which presents no problem, right?

And we (meaning _we_ whether or not you include yourself)
assign the value of the least upper bound of that list
to the non-terminating decimal 0.999...

I'm guessing you don't have a problem with the LUB either,
because you talk about other things instead.
_But this is what we do_

On 10/5/2017 9:59 AM, burs...@gmail.com wrote:

> Doesn't make any sense, since it is a sequence
> and not a number. But obviously the decimal
> notation 0.999... refers to a number. So its
> simply lim n->oo (1-(1/10)^n), from the
>
> mathematical notational convention, that the
> ... in the above context includes the limit.

I am not familiar with a convention that includes the
limit as part of the sequence. That sounds like a bad idea
to me, not least because we can't assume that a particular
sequence has a limit.

In this discussion, no matter what you may be familiar with,
it's important to keep clear which we're talking about at
any point: the sequence or the limit of the sequence.

Anyway, what netzwelter wrote out is what I asked for,
that sequence defined without '...' My hope is that he
does not take your advice.

On Thursday, 5 October 2017 12:15:22 UTC-4, Jim Burns  wrote:
> On 10/5/2017 9:59 AM, burs...@gmail.com wrote:
>
> > Doesn't make any sense, since it is a sequence
> > and not a number. But obviously the decimal
> > notation 0.999... refers to a number. So its
> > simply lim n->oo (1-(1/10)^n), from the
> >
> > mathematical notational convention, that the
> > ... in the above context includes the limit.
>
> I am not familiar with a convention that includes the
> limit as part of the sequence.

Wow. And yet you will say on the other hand that every converging decimal sequence has a limit. For finite sequences the limit is the sum. For sequences whose limits cannot be expressed in a given radix system, you treat the series as the limit. Finally for sequences whose limits are not rational numbers, you conjure up "irrational numbers". You are very confused.

> That sounds like a bad idea to me, not least because we can't assume that a particular sequence has a limit.

But you do! ...

> I'm gues...

1) Do you understand how decimals expansions are defined as limits?
2) Do you understand limits?

> But you do! Every converging decim...

> > But you do! Every converging decimal sequence has a limit. What's more absurd is that you use the sequence as the representation of the limit. Chuckle.
> >
> > >
> > > In this discussi...

On 10/5/2017 3:12 PM, netzweltler wrote:
> Am Donnerstag, 5. Oktober 2017 17:59:25 UTC+2
> schrieb Jim Burns:
>> On 10/5/2017 10:00 AM, netzweltler wrote:
>>> Am Donnerstag, 5. Oktober 2017 15:22:35 UTC+2
>>> schrieb Jim Burns:

>> [...]
>>>> _We don't do what you're describing_
>>>
>>> Nevertheless,
>>
>> "Nevertheless"?
>> Do you agree that what you're describing
>> is not what we're doing?

*NETZWELTLER*
DO YOU AGREE THAT WHAT YOU'RE DOING
IS NOT WHAT WE'RE DOING?

I think you do agree.
This a pretty fundamental requirement:
When you criticize what someone is doing,
criticize _what they are doing_ and not something else.

>>> the process
>>> 0 |-> write 0.9
>>> 1 |-> append another 9 (to the 0.9 already written)
>>> 2 |-> append another 9 (to the 0.99 already written)
>>>      ...
>>> results in 0.999...
>>>
>>> Whereas the process you specified earlier
>>> 0 |-> 0.9
>>> 1 |-> 0.99
>>> 2 |-> 0.999
>>>      ...
>>> is nothing else but an infinite list of terminating decimals.
>>
>> Right. Nothing else but an infinite list of terminating decimals,
>> which presents no problem, right?
>>
>> And we (meaning _we_ whether or not you include yourself)
>> assign the value of the least upper bound of that list
>> to the non-terminating decimal 0.999...
>>

>> I'm guessing you don't have a problem with the LUB either,
>> because you talk about other things instead.
>> _But this is what we do_
>

> We obviously agree that the process you specified earlier

> 0 |-> 0.9
> 1 |-> 0.99
> 2 |-> 0.999
>     ...
> is nothing else but an infinite list of terminating decimals.

It think it is also obvious that ...

Jim Burns has brought this to us :
> It think it is also obvious that you have no problem with

> an infinite list of terminating decimals.
>

>> What you don't want to see is, that the process

>> 0 |-> write 0.9
>> 1 |-> append another 9 (to the 0.9 already written)
>> 2 |-> append another 9 (to the 0.99 already written)
>>     ...
>> results in 0.999...
>>

>> Maybe you cannot see that I am not writing a new number
>> in a new line at each step - as in your process. I am
>> appending the 9s in the same line. So I am not creating
>> an infinit...

Den torsdag 5 oktober 2017 kl. 05:56:03 UTC+2 skrev 7777777:

We've been through this, real numbers are cauchy sequences modulo null sequences. That is why and for the sequence for 1 <1> and the sequence for S/0.999... we have <1-10^-n>, their differens is <10^-n> which is a null sequence, ergo they are equal.

Am Freitag, 6. Oktober 2017 02:54:40 UTC+2 schrieb Jim Burns:
> On 10/5/2017 3:12 PM, netzweltler wrote:
> > Am Donnerstag, 5. Oktober 2017 17:59:25 UTC+2
> > schrieb Jim Burns:
> >> On 10/5/2017 10:00 AM, netzweltler wrote:
> >>> Am Donnerstag, 5. Oktober 2017 15:22:35 UTC+2
> >>> schrieb Jim Burns:
>
> >> [...]
> >>>> _We don't do what you're describing_
> >>>
> >>> Nevertheless,
> >>
> >> "Nevertheless"?
> >> Do you agree that what you're describing
> >> is not what we're doing?
>
> *NETZWELTLER*
> DO YOU AGREE THAT WHAT YOU'RE DOING
> IS NOT WHAT WE'RE DOING?

Let's say I agree. Doesn't mean that it is obvious to me what *you* are doing.
All I've seen so far is, that you define 0.999... to be the LUB or limit (I guess you use these expressions interchangeably in this context) of the sequence
(1-1/10^n)n∈N. That's it.

Do we agree that a LUB or a limit is a point on the number line - or to simplify it in case of a positive number - a point on a geometric line from 0 to +infinity? I haven't seen any proof for that yet other than the claim that it is. For some reason coincident with point 1.

> I think you do agree.
> This a pretty fundamental requirement:
> When you criticize what someone is doing,
> criticize _what they are doing_ and not something else.
>
> >>> the process
> >>> 0 |-> write 0.9
> >>> 1 |-> append another 9 (to the 0.9 already written)
> >>> 2 |-> append another 9 (to the 0.99 already written)
> >>>      ...
> >>> results in 0.999...
> >>>
> >>> Whereas the process you specified earlier
> >>> 0 |-> 0.9
> >>> 1 |-> 0.99
> >>> 2 |-> 0.999
> >>>      ...
> >>> is nothing else but an infinite list of terminating decimals.
> >>
> >> Right. Nothing else but an infinite list of terminating decimals,
> >> which presents no problem, right?
> >>
> >> And we (meaning _we_ whether or not you include yourself)
> >> assign the value of the least upper bound of that list
> >> to the non-terminating decimal 0.999...
> >>

> >> I'm guessing you don't have a problem with th...

> > > > In this discussion, no matter what you may be familiar with,
> > > > it's important to keep clear which we're talking about at
> > > > any point: the sequence or the limit of the sequence.
> > >

> > > Really?!!!!  But you call 0.333... the limit!! And yet, 0.333... cannot exist without the sequence. It IS the sequence, N...

Den torsdag 5 oktober 2017 kl. 16:00:25 UTC+2 skrev netzweltler:
> Am Donnerstag, 5. Oktober 2017 15:22:35 UTC+2 schrieb Jim Burns:
> > On 10/4/2017 3:58 PM, netzweltler wrote:
> >
> > > While it is true that there is no line on the list
> > >    0 |-> write 0.9
> > >    1 |-> append another 9
> > >    2 |-> append another 9
> > >        ...
> > > that produces other than a finite string, it is true that
> > >  the complete list produces an infinite string.
> > > "We are dealing with infinitely many finite stepwise processes"
> > >  is true for each line of the list.
> > > "we are dealing with an infinite stepwise process" is true
> > >  for the complete list. We wouldn't get the resulting
> > >  0.999... otherwise.
> >
> > This is backwards. The finite initial segments
> >     0.9, 0.99, 0.999, ...
> > _do not_ result in the infinite decimal
> >     0.999...
> >
> > We have the infinite decimal _first_ , and then we trim it
> > to finite length, operate on the finite decimals in the

...

As conincidental as the following:

  lim n->oo (1-1/10^n) = 1

> > >> I'm guessing you don't have a problem with the LUB either,
> > >> because you talk about other things instead.
> > >> _But this is what we do_
> > >

> > > We obviously agree that the process you specified earlier
> > > 0 |-> 0.9
> > > 1 |-> ...

...

> > > familiar ...

> > > > 1...

> > > > > Really?!!!!  But you call 0.333... the limit!! And yet, 0.333... cannot exist without the sequence. It IS the sequence, NOT the limit. The limit is a well-formed rational number 1/3.

> > > > >
> > > > > >
> > > > > > Anyway, what netzwelter wrote out is what I asked for,
> > > > > > that sequence defined without '...' My hope is that he
> > > > > > does not take your advice.
> > > > > >
> > > > > > > Am Donnerstag, 5. Oktober 2017 15:50:38 UTC+2
> > > > > > > schrieb netzweltler:
> > > > > > >>> I mean a '...' at the end of the description.
> > > > > > >>> When one writes
> > > > > > >>>      0.9, 0.99, 0.999, ...
> > > > > > >>> one puts '...' at the end of _that_ but what does it mean?
> > > > > > >>
> > > > > > >> (1-(1/10)^n)n∈N
> > > >

> > > > 0.333. is not a sequence, Gabriel. 0.333... is a limit.
> > >
> > > Wrong. 0.333... IS a sequence. 1/3 is a limit.  S = Lim S does not make 0.333... a sequence.
> > >
> > > > And every decimal expansion is a limit,
> > >
> > > Wrong here also. Only converging decimal expansions are limits, that is, they end in zeroes or nines - according to the theory you claim to know and are so ignorant of...
> > >
> > > > because that's how they are defined. Finite decimal expansions are also infinite decimal expansions. 0.25 = 0.2500000.... for example, so every real number can be represented by an infinite decimal expansion.
> > >
> > > FALSE. You have not answered any of the questions I put to you. If you try, then you might see that you are mistaken.
> >
> > N...

> > > No, we don't.
> >
> > Yes, you do. You are Eulerites and your strict observance to S = Lim S is proof.
> >
> > > We define real nu...

> > > > We define real numbers as equivalence classes of rational Cauchy sequences.
> > > ...

On 10/6/2017 6:03 AM, netzweltler wrote:
> Am Freitag, 6. Oktober 2017 02:54:40 UTC+2
> schrieb Jim Burns:
>> On 10/5/2017 3:12 PM, netzweltler wrote:
>>> Am Donnerstag, 5. Oktober 2017 17:59:25 UTC+2
>>> schrieb Jim Burns:
>>>> On 10/5/2017 10:00 AM, netzweltler wrote:
>>>>> Am Donnerstag, 5. Oktober 2017 15:22:35 UTC+2
>>>>> schrieb Jim Burns:

>>>> [...]
>>>>>> _We don't do what you're describing_
>>>>>
>>>>> Nevertheless,
>>>>
>>>> "Nevertheless"?
>>>> Do you agree that what you're describing
>>>> is not what we're doing?
>>
>> *NETZWELTLER*
>> DO YOU AGREE THAT WHAT YOU'RE DOING
>> IS NOT WHAT WE'RE DOING?
>
> Let's say I agree. Doesn't mean that it is obvious to me
> what *you* are doing.

Great. Let's say you agree. Will you stop saying that
    "0.999... means infinitely many commands"?

> All I've seen so far is, that you define 0.999... to be
> the LUB or limit (I guess you use these expressions
> interchangeably in this context) of the sequence
> (1-1/10^n)n∈N. That's it.

LUB and limit are not interchangeable, but they are closely
related. It happens that, in this case, that
    LUB{ 1 - 1/10^k | k e N }
is the same number as
    lim_n->oo  1 - 1/10^n

In this context, I prefer LUB because the argument is very
short:
    That set of real numbers is bounded and non-empty.
    For every bounded and non-empty set of real numbers,
    there is a least upper bound. (This is what we mean by
    "real numbers".) That set has a least upper bound, which
    is 1. This is why we say 0.999... = 1.

Also, conceptually, LUB "happens all at once". There is no
progression from point to point. Limits also "happen all at
once" but some people don't get that. I like LUB in this context
because there should be no way to mistake it as happening
progressively in some infinite way.

----
(i)
Let us define the set P of decimal places as a set
    (which we can think of as  { -1, -2, -3, ... } )
such that
    -1 e P
    (Ax)( x e P  ->  x-1 e P )
    (all B sub P)
       ( -1 e B ) &
          (Ax)( x e B  ->  x-1 e B )
       ->  ( B = P )

   (What I _did not_ do in this definition is enumerate explicitly
    what the elements of P are. What we have instead is a way to
    prove in some cases whether some set we are interested in is P.
    This does not require infinite many statements to do.)

(ii)
Let us define the set D of decimal digits
    D = {0,1,2,3,4,5,6,7,8,9}

(iii)
Let us define an _infinite decimal fraction_ f as a map from
places P to digits D.

We can represent that map as a subset f of the Cartesian
product PxD such that
    (all x e P)(exists unique y e D)( (x,y) e f )

    (In order...

torstai 5. lokakuuta 2017 16.12.34 UTC+3 burs...@gmail.com kirjoitti:
> There is no S(oo), there is no index n=oo.


fail.

There is S(∞), there is index n = ∞
look at:
http://amsi.org.au/ESA_Senior_Years/SeniorTopic1/1d/1d_2content_5.html


> Only lim n->oo S(n), which happens is can
>
> be written as sum_i=1^oo a(i)

you contradict yourself.
sum_i=1^oo a(i) is S(∞)

Am Samstag, 7. Oktober 2017 01:42:15 UTC+2 schrieb Jim Burns:
> On 10/6/2017 6:03 AM, netzweltler wrote:
> > Am Freitag, 6. Oktober 2017 02:54:40 UTC+2
> > schrieb Jim Burns:
> >> On 10/5/2017 3:12 PM, netzweltler wrote:
> >>> Am Donnerstag, 5. Oktober 2017 17:59:25 UTC+2
> >>> schrieb Jim Burns:
> >>>> On 10/5/2017 10:00 AM, netzweltler wrote:
> >>>>> Am Donnerstag, 5. Oktober 2017 15:22:35 UTC+2
> >>>>> schrieb Jim Burns:
>
> >>>> [...]
> >>>>>> _We don't do what you're describing_
> >>>>>
> >>>>> Nevertheless,
> >>>>
> >>>> "Nevertheless"?
> >>>> Do you agree that what you're describing
> >>>> is not what we're doing?
> >>
> >> *NETZWELTLER*
> >> DO YOU AGREE THAT WHAT YOU'RE DOING
> >> IS NOT WHAT WE'RE DOING?
> >
> > Let's say I agree. Doesn't mean that it is obvious to me
> > what *you* are doing.
>
> Great. Let's say you agree. Will you stop saying that
>     "0.999... means infinitely many commands"?

No. Because it is not obvious to me why the equation
0.999... = 0.9 + 0.09 + 0.009 + ...
should be wrong.
>     (In order to construct the particular map which
>     represents 0.999... we would just say y = 9 for all x.)
>
> (iv)
> Define D^P as the set of all functions P --> D.
> We're using D^P to represent all the decimal fractions of
> real numbers -- that is, [0,1].
> ...

> > (v)
> > Define, for each map to digits f:...

On Saturday, 7 October 2017 08:18:44 UTC-4, FromTheRafters  wrote:
> netzweltler wrote :
> > Am Samstag, 7. Oktober 2017 01:42:15 UTC+2 schrieb Jim Burns:
> >> On 10/6/2017 6:03 AM, netzweltler wrote:
> >>> Am Freitag, 6. Oktober 2017 02:54:40 UTC+2
> >>> schrieb Jim Burns:
> >>>> On 10/5/2017 3:12 PM, netzweltler wrote:
> >>>>> Am Donnerstag, 5. Oktober 2017 17:59:25 UTC+2
> >>>>> schrieb Jim Burns:
> >>>>>> On 10/5/2017 10:00 AM, netzweltler wrote:
> >>>>>>> Am Donnerstag, 5. Oktober 2017 15:22:35 UTC+2
> >>>>>>> schrieb Jim Burns:
> >>
> >>>>>> [...]
> >>>>>>>> _We don't do what you're describing_
> >>>>>>>
> >>>>>>> Nevertheless,
> >>>>>>
> >>>>>> "Nevertheless"?
> >>>>>> Do you agree that what you're describing
> >>>>>> is not what we're doing?
> >>>>
> >>>> *NETZWELTLER*
> >>>> DO YOU AGREE THAT WHAT YOU'RE DOING
> >>>> IS NOT WHAT WE'RE DOING?
> >>>
> >>> Let's say I agree. Doesn't mean that it is obvious to me
> >>> what *you* are doing.
> >>

> >> Great. Let's say you agree. Will you stop s...

Not always "no last term", here the ellipsis
means "no last term" AND "limit". It is

defined as such in math when the infinite
operation is plus (aka series)(*):

   a1 + a2 + a3 + ...

When the infinite operation is comma (aka sequence)
then no limit is involved:

   s1, s2, s3, ...

Since you have this choice, using ellipsis with
or without limit, i.e. by using a plus or using
a comma, there is no harm by this convention.

Because you don't understand this simple mechanism,
of noting a series or a sequence, this is why we
call you bird brain John Gabriel, and why everybody

thinks you are dumb as bread. Even Cantor could
do it, in his paper here:

    Ueber die Ausdehnung eines Satzes aus
    der Theorie der trigonometrischen Reihen
    Von G. Cantor in Halle a. S. [Math. Annalen 5, 123–132 (1872).]
    http://www.maths.tcd.ie/pub/HistMath/People/Cantor/Ausdehnung/Ausdehnung.pdf

(*)
See here:
https://gist.github.com/jburse/e60242e2e02e611e4373df55bfc37953#gistcomment-2219906
https://en.wikipedia.org/wiki/Series_%28mathematics%29

Am Samstag, 7. Oktober 2017 14:57:50 UTC+2 schrieb John Gabriel:
> On Saturday, 7 October 2017 08:18:44 UTC-4, FromTheRafters  wrote:
> > netzweltler wrote :
> > > Am Samstag, 7. Oktober 2017 01:42:15 UTC+2 schrieb Jim Burns:
> > >> On 10/6/2017 6:03 AM, netzweltler wrote:
> > >>> Am Freitag, 6. Oktober 2017 02:54:40 UTC+2
> > >>> schrieb Jim Burns:
> > >>>> On 10/5/2017 3:12 PM, netzweltler wrote:
> > >>>>> Am Donnerstag, 5. Oktober 2017 17:59:25 UTC+2
> > >>>>> schrieb Jim Burns:
> > >>>>>> On 10/5/2017 10:00 AM, netzweltler wrote:

> > >>>>>>> A...

Harry Stoteles explained on 10/7/2017 :

> Not always "no last term", here the ellipsis
> means "no last term" AND "limit". It is
>
> defined as such in math when the infinite
> operation is plus (aka series)(*):
>
>    a1 + a2 + a3 + ...
>
> When the infinite operation is comma (aka sequence)
> then no limit is involved:
>
>    s1, s2, s3, ...
>
> Since you have this choice, using ellipsis with
> or without limit, i.e. by using a plus or using
> a comma, there is no harm by this convention.
>
> Because you don't understand this simple mechanism,
> of noting a series or a sequence, this is why we
> call you bird brain John Gabriel, and why everybody
>
> thinks you are dumb as bread. Even Cantor could
> do it, in his paper here:
>
>     Ueber die Ausdehnung eines Satzes aus
>     der Theorie der trigonometrischen Reihen
>     Von G. Cantor in Halle a. S. [Math. Annalen 5, 123–132 (1872).]
>    
> http://www.maths.tcd.ie/pub/HistMath/People/Cantor/Ausdehnung/Ausdehnung.pdf
>
> (*)

...

> > See here:
> > https://gist.github.com/jburse/e60242e2e02e611e4373df55bfc37953#gistcomment-2219906
> > https://en.wikipedia.org/wiki/Series_%28mathematics%29
> >
>

> Lists of the largest primes have an ellipsis in the middle of the
> displayed numerals. It means neither no last term nor limit. It means
> that some (rather large) number of numerals have been omitted.
>
> It this context however, I always tak...

> > > Def...

On Saturday, 7 October 2017 10:29:16 UTC-4, Jim Burns  wrote:
> On 10/7/2017 4:06 AM, netzweltler wrote:
> > Am Samstag, 7. Oktober 2017 01:42:15 UTC+2
> > schrieb Jim Burns:
> >> On 10/6/2017 6:03 AM, netzweltler wrote:
> >>> Am Freitag, 6. Oktober 2017 02:54:40 UTC+2
> >>> schrieb Jim Burns:
> >>>> On 10/5/2017 3:12 PM, netzweltler wrote:
> >>>>> Am Donnerstag, 5. Oktober 2017 17:59:25 UTC+2
> >>>>> schrieb Jim Burns:
> >>>>>> On 10/5/2017 10:00 AM, netzweltler wrote:
> >>>>>>> Am Donnerstag, 5. Oktober 2017 15:22:35 UTC+2
> >>>>>>> schrieb Jim Burns:
>
> >>>>>> [...]
> >>>>>>>> _We don't do what you're describing_
> >>>>>>>
> >>>>>>> Nevertheless,
> >>>>>>
> >>>>>> "Nevertheless"?
> >>>>>> Do you agree that what you're describing
> >>>>>> is not what we're doing?
> >>>>
> >>>> *NETZWELTLER*
> >>>> DO YOU AGREE THAT WHAT YOU'RE DOING
> >>>> IS NOT WHAT WE'RE DOING?
> >>>
> >>> Let's say I agree. Doesn't mean that it is obvious to me
> >>> what *you* are doing.
> >>

> >> Great. Let's say you agree. Wil...

Well we were not discussing elllipsis in the middle,
it was about ellipsis at the end of a list.

Of course a ellipsis has a further meaning when
it is not at the end but somewhere in the middle.

Do you doubt that?

No he shouldn't, since decimal representations
are a special case, where we have:

   lim n->oo sn = lub {sn}

This is because decimal representations are
montonic. For a decimal representation we have:

   s1 < s2 < s3 < s4 < ...

For a geometric series as Euler was discussing,
we don't have montonicity, since the r might
be negative, for example using r= -1/2 gives:

   1 - 1/2 + 1/4 - 1/8 + 1/16 -+ ... = 2/3

So basicall a sequence:

   1
   1/2
   3/4
   5/8
   11/16
   ...

But decimal representations, have always a
montonic sequence, this decimal representation:

   d0.d1 d2 d3 d4 ...

Leads to this montonic sequence:

   d0
   d0.d1
   d0.d1 d2
   d0.d1 d2 d3
   d0.d1 d2 d3 d4
   ...

So for decimal representations we can focus on
the lub when trying to find the limit.

Am Samstag, 7. Oktober 2017 16:49:43 UTC+2 schrieb genm...@gmail.com:
> > By definition,
> >    0.999... = LUB{ 0.9, 0.9 + 0.09, 0.9 + 0.09 + 0.009, ... }
>

> You should rather quote S = Lim S:
>
>      1 = Lim_{n \to \infty}  1 - 10^(-n)

burs...@gmail.com formulated the question :

> Well we were not discussing elllipsis in the middle,
> it was about ellipsis at the end of a list.

I was.

> Of course a ellipsis has a further meaning when
> it is not at the end but somewhere in the middle.

Of course.

> Do you doubt that?

No, and I didn't say that I did. In fact I also stated the context was
different in the part of my post which you neglected to include.

Why did you only take the first part and state the obvious?

On 10/7/2017 10:49 AM, genm...@gmail.com wrote:
> On Saturday, 7 October 2017 10:29:16 UTC-4,
> Jim Burns  wrote:

>> Am I wrong about that paraphrase? Will you show me the
>> infinitely many additions which I do? Which _I_ do,
>> not which _you_ do.
>

> I'll show you:  0 .  9  9  9  .  .  .
>
> Without the infinite steps, what is your justification
> for using that representation, especially when 1 is
> perfectly valid?
>
> <too much nonsense>

"too much nonsense" is your way of saying that you don't
want your question answered. But snipping the answer doesn't
make it go away.

We represent the decimal 0.999... as a map from negative decimal
places to decimal digits, all 9s.

 From that map, we can define a set of the infinitely many
finite sums 0.9, 0.99, 0.999, ... , all of which are perfectly
well-defined using rational addition and multiplication.

The _value_ of 0.999... is _defined_ as the least upper bound
of that set { 0.9, 0.99, 0.999, ... }.

In other news, that value is 1. No inf...

you're gonna discuss this infinite bull shit all hell long

don't make this list my hell : - )

can't you just learn it and pass by it like all other babies?

one third goo goo

thirty three ninety ninth dah dah

1/3 = 3/9
= 33/99
= 333/999
= 3333/9999
= 33333/99999
= 3...3 / 9...9

Thats just:

  (10^k-1)/3 / (10^k-1)

Has nothing to do with:

  0.333... = 1/3

Because:

  0.3       = 3/10          <> 3/9
  0.33      = 33/100        <> 33/99
  0.333     = 333/1000      <> 333/999
  0.3333    = 3333/10000    <> 3333/9999
  ...

FromTheRafters already wrote that the ellipsis
in the middle is something else. I already asked
him whether he has some doubts.

So you have some doubts?

What do you mean by S(∞) if it's not meant to be interpreted as a limit?

Den lördag 7 oktober 2017 kl. 10:06:29 UTC+2 skrev netzweltler:
> Am Samstag, 7. Oktober 2017 01:42:15 UTC+2 schrieb Jim Burns:
> > On 10/6/2017 6:03 AM, netzweltler wrote:
> > > Am Freitag, 6. Oktober 2017 02:54:40 UTC+2
> > > schrieb Jim Burns:
> > >> On 10/5/2017 3:12 PM, netzweltler wrote:
> > >>> Am Donnerstag, 5. Oktober 2017 17:59:25 UTC+2
> > >>> schrieb Jim Burns:
> > >>>> On 10/5/2017 10:00 AM, netzweltler wrote:
> > >>>>> Am Donnerstag, 5. Oktober 2017 15:22:35 UTC+2
> > >>>>> schrieb Jim Burns:
> >
> > >>>> [...]
> > >>>>>> _We don't do what you're describing_
> > >>>>>
> > >>>>> Nevertheless,
> > >>>>
> > >>>> "Nevertheless"?
> > >>>> Do you agree that what you're describing
> > >>>> is not what we're doing?
> > >>
> > >> *NETZWELTLER*
> > >> DO YOU AGREE THAT WHAT YOU'RE DOING
> > >> IS NOT WHAT WE'RE DOING?
> > >
> > > Let's say I agree. Doesn't mean that it is obvious to me
> > > what *you* are doing.
> >

> > pla...

burs...@gmail.com explained :

> Thats just:
>
>   (10^k-1)/3 / (10^k-1)
>
> Has nothing to do with:
>
>   0.333... = 1/3
>
> Because:
>
>   0.3       = 3/10          <> 3/9
>   0.33      = 33/100        <> 33/99
>   0.333     = 333/1000      <> 333/999
>   0.3333    = 3333/10000    <> 3333/9999
>   ...
>
> FromTheRafters already wrote that the ellipsis
> in the middle is something else. I already asked
> him whether he has some doubts.
>
> So you have some doubts?

Maybe I misunderstood what you meant by doubts. I am agreeing that
context is king is this respect.

http://www.solving-math-problems.com/ellipsis.html

On Saturday, 7 October 2017 11:35:20 UTC-4, Jim Burns  wrote:
> On 10/7/2017 10:49 AM, genm...@gmail.com wrote:
> > On Saturday, 7 October 2017 10:29:16 UTC-4,
> > Jim Burns  wrote:
>
> >> Am I wrong about that paraphrase? Will you show me the
> >> infinitely many additions which I do? Which _I_ do,
> >> not which _you_ do.
> >
> > I'll show you:  0 .  9  9  9  .  .  .
> >
> > Without the infinite steps, what is your justification
> > for using that representation, especially when 1 is
> > perfectly valid?
> >
> > <too much nonsense>
>
> "too much nonsense" is your way of saying that you don't
> want your question answered.

False. It means you are writing a lot of irrelevant junk.

> But snipping the answer doesn't make it go away.
>
> We represent the decimal 0.999... as a map from negative decimal
> places to decimal digits, all 9s.

What?!!!! What bullshit is this? Never heard of it before.

>
>  From that map, we can define a set of the infinitely many

> finite sums 0.9, 0.99, 0.999, ... , all o...

Because he is a troll and a moron. Your response to it isn't helping.

Den lördag 7 oktober 2017 kl. 22:14:39 UTC+2 skrev John Gabriel:
> On Saturday, 7 October 2017 11:35:20 UTC-4, Jim Burns  wrote:
> > On 10/7/2017 10:49 AM, genm...@gmail.com wrote:
> > > On Saturday, 7 October 2017 10:29:16 UTC-4,
> > > Jim Burns  wrote:
> >
> > >> Am I wrong about that paraphrase? Will you show me the
> > >> infinitely many additions which I do? Which _I_ do,
> > >> not which _you_ do.
> > >
> > > I'll show you:  0 .  9  9  9  .  .  .
> > >
> > > Without the infinite steps, what is your justification
> > > for using that representation, especially when 1 is
> > > perfectly valid?
> > >
> > > <too much nonsense>
> >
> > "too much nonsense" is your way of saying that you don't
> > want your question answered.
>
> False. It means you are writing a lot of irrelevant junk.
>
> > But snipping the answer doesn't make it go away.
> >
> > We represent the decimal 0.999... as a map from negative decimal
> > places to decimal digits, all 9s.
>

> What?!!!! What bullshit is this? Never h...

On 10/7/2017 4:14 PM, John Gabriel wrote:
> On Saturday, 7 October 2017 11:35:20 UTC-4,
> Jim Burns  wrote:

>> But snipping the answer doesn't make it go away.
>>
>> We represent the decimal 0.999... as a map from
>> negative decimal places to decimal digits, all 9s.
>
> What?!!!! What bullshit is this? Never heard of it before.

Well, at least you admit it.

There isn't any point of the rest of your, um, "output", then.

Go read a book or something.

Hardly. It's NOT done that way at all. So there isn't any point discussing it because it's not relevant. "Negative decimal places"? That's just nonsense.

>
> Go read a book or something.

I'd tell you the same but I have read most books and found them to be severely wanting.

On Saturday, 7 October 2017 17:38:13 UTC-4, Markus Klyver  wrote:
> Den lördag 7 oktober 2017 kl. 22:14:39 UTC+2 skrev John Gabriel:
> > On Saturday, 7 October 2017 11:35:20 UTC-4, Jim Burns  wrote:
> > > On 10/7/2017 10:49 AM, genm...@gmail.com wrote:
> > > > On Saturday, 7 October 2017 10:29:16 UTC-4,
> > > > Jim Burns  wrote:
> > >
> > > >> Am I wrong about that paraphrase? Will you show me the
> > > >> infinitely many additions which I do? Which _I_ do,
> > > >> not which _you_ do.
> > > >
> > > > I'll show you:  0 .  9  9  9  .  .  .
> > > >
> > > > Without the infinite steps, what is your justification
> > > > for using that representation, especially when 1 is
> > > > perfectly valid?
> > > >
> > > > <too much nonsense>
> > >
> > > "too much nonsense" is your way of saying that you don't
> > > want your question answered.
> >
> > False. It means you are writing a lot of irrelevant junk.
> >
> > > But snipping the answer doesn't make it go away.
> > >

> > > We represent the decimal 0.9...

> > > > We represent the...

Am Samstag, 7. Oktober 2017 14:18:44 UTC+2 schrieb FromTheRafters:
> netzweltler wrote :
> > Am Samstag, 7. Oktober 2017 01:42:15 UTC+2 schrieb Jim Burns:
> >> On 10/6/2017 6:03 AM, netzweltler wrote:
> >>> Am Freitag, 6. Oktober 2017 02:54:40 UTC+2
> >>> schrieb Jim Burns:
> >>>> On 10/5/2017 3:12 PM, netzweltler wrote:
> >>>>> Am Donnerstag, 5. Oktober 2017 17:59:25 UTC+2
> >>>>> schrieb Jim Burns:
> >>>>>> On 10/5/2017 10:00 AM, netzweltler wrote:
> >>>>>>> Am Donnerstag, 5. Oktober 2017 15:22:35 UTC+2
> >>>>>>> schrieb Jim Burns:
> >>
> >>>>>> [...]
> >>>>>>>> _We don't do what you're describing_
> >>>>>>>
> >>>>>>> Nevertheless,
> >>>>>>
> >>>>>> "Nevertheless"?
> >>>>>> Do you agree that what you're describing
> >>>>>> is not what we're doing?
> >>>>
> >>>> *NETZWELTLER*
> >>>> DO YOU AGREE THAT WHAT YOU'RE DOING
> >>>> IS NOT WHAT WE'RE DOING?
> >>>
> >>> Let's say I agree. Doesn't mean that it is obvious to me
> >>> what *you* are doing.
> >>

> >> Great. Let's sa...

> > > places P to digits D.
> > >

> > > W...

The digits after the point have negative base exponents,
its perfectly legit to call them like that:

Digits     1  2  3 .  4  5  6
Places     2  1  0   -1 -2 -3

123.456 = 1*10^2 + 2*10^1 + 3*10^0 + 4*10^(-1) + 5*10^(-1) + 6*10^(-2)

But I guess this is already heavy stuff for bird brain
John Gabriel, he doesnt have a very flexible mind.

Corr.:

123.456 = 1*10^2 + 2*10^1 + 3*10^0 + 4*10^(-1) + 5*10^(-2) + 6*10^(-3)

Am Sonntag, 8. Oktober 2017 12:38:26 UTC+2 schrieb burs...@gmail.com:
> Digits     1  2  3 .  4  5  6
> Places     2  1  0   -1 -2 -3
>

> 123.456 = 1*10^2 + 2*10^1 + 3*10^0 + 4*10^(-1) + 5*10^(-) + 6*10^(-2)

It happens that netzweltler formulated :

>>>> Great. Let's say you agree. Will you stop saying that
>>>>     "0.999... means infinitely many commands"?
>>>
>>> No. Because it is not obvious to me why the equation
>>> 0.999... = 0.9 + 0.09 + 0.009 + ...
>>> should be wrong.
>>

>> It's not wrong. The first one is a representation of the number one.
>> The second is a representation of the number one. Two things equal to
>> the same thing are equal to each other.
>>
>> [...]
>
> I'd have to look it up: Did you say that 0.999... IS the result of infinitely
> many addition operations or IS NOT the result of infinitely many addi...

> > > > > We represent the decimal 0.999... as a map from negative d...

How does any of this have anything to do with 0.999... you baboon?

> > many addition operations or IS NOT the result of infinitely many addition
> > operations?
>
> If you had an oracle with enough time which could do the arithmetic and
> hand you an answer, then yes. Without ...

Know nothing and understand nothing, exemplified
by bird brain John Gabriel himself. Here is a hint
birdy brain:

  0 . 9  9  9  9  ...
     -1 -2 -3 -4

On Sunday, 8 October 2017 08:20:56 UTC-4, burs...@gmail.com  wrote:
> Know nothing and understand nothing, exemplified
> by bird brain John Gabriel himself. Here is a hint
> birdy brain:
>
>   0 . 9  9  9  9  ...
>      -1 -2 -3 -4

What is the relevance of this O stupid one?  It is true for any representation. What is its ***RELEVANCE*** ????

On Sunday, 8 October 2017 08:24:12 UTC-4, John Gabriel  wrote:
> On Sunday, 8 October 2017 08:20:56 UTC-4, burs...@gmail.com  wrote:
> > Know nothing and understand nothing, exemplified
> > by bird brain John Gabriel himself. Here is a hint
> > birdy brain:
> >
> >   0 . 9  9  9  9  ...
> >      -1 -2 -3 -4
>
> What is the relevance of this O stupid one?  It is true for any representation. What is its ***RELEVANCE*** ????

The ignoramus Jim Burns makes a comment and you simply assume it has relevance?  What a moron!

>
> >
> > Am Sonntag, 8. Oktober 2017 14:18:21 UTC+2 schrieb John Gabriel:
> > > On Sunday, 8 October 2017 06:38:26 UTC-4, burs...@gmail.com  wrote:
> > > > The digits after the point have negative base exponents,
> > > > its perfectly legit to call them like that:
> > > >
> > > > Digits     1  2  3 .  4  5  6
> > > > Places     2  1  0   -1 -2 -3
> > > >
> > > > 123.456 = 1*10^2 + 2*10^1 + 3*10^0 + 4*10^(-1) + 5*10^(-1) + 6*10^(-2)
> > > >

> > > > But I guess this is already hea...

You wrote:
"Hardly. It's NOT done that way at all. So there isn't any point discussing it because it's not relevant. "Negative decimal places"? That's just nonsense."

Of course we have for S(n), note the negative exponent:

   0.999...999  = 1 - 10^(-n)
     \---n---/

And then for S, we have:

   0.999... = lim n->oo (1 - 10^(-n)) = 1

BTW nobody says that S(n) generates S. Normally the verb
is only used for something that has multiple outcomes,
for example 2^j*5^k generates natural numbers, whos inverse
have a finite decimal repesentation.

The operator that is applied on the function form
S(n) is the limit operator lim, so what we say S is
the limit of S(n) when n tends to infinity.

Here have a banana, do you
recognize how stupid you are?

Banana Song (I'm A Banana)
https://www.youtube.com/watch?v=LH5ay10RTGY

Am Sonntag, 8. Oktober 2017 14:24:12 UTC+2 schrieb John Gabriel:
> On Sunday, 8 October 2017 08:20:56 UTC-4, burs...@gmail.com  wrote:

> > Know nothing and understand nothing, exemplified ...

> > > by bird brain John Gabriel himself. Here is a hint
> > > birdy brain:
> > >
> > >   0 . 9  9  9  9  ...
> > >      -1 -2 -3 -4
> >

> > What is the relevance of this O stupi...

On Sunday, 8 October 2017 08:32:26 UTC-4, burs...@gmail.com  wrote:

Have some time to kill so I will make a fool of you and ridicule you because you are a baboon.

> You wrote:
> "Hardly. It's NOT done that way at all. So there isn't any point discussing it because it's not relevant. "Negative decimal places"? That's just nonsense."
>
> Of course we have for S(n), note the negative exponent:
>
>    0.999...999  = 1 - 10^(-n)
>      \---n---/
>

We use S(n) to generate the SEQUENCE you stupid kaffir!

S(n)=1 - 10^(-n) GENERATES ALL  the terms of the sequence Jim Burns uses:

0.999... = {0.9; 0.99; 0.999; ...}

Try it moron!!! 0.9 = S(1),  0.99 = S(2), etc.

 
> And then for S, we have:
>
>    0.999... = lim n->oo (1 - 10^(-n)) = 1

That is not for S you monkey!! Lim S = lim n->oo (1 - 10^(-n)) = 1

So that S = 0.999... and then monkey Euler does a trick:

0.999... = Lim S = 1.

Get it stupid? Bwaaa haaaa haaaaa. I piss an shit on you moron.

>
> BTW nobody says that S(n) generates S...

You need something with a parameter n, the
limit operator needs a sequence as argument, so
thats why you need something with a parametr n.

> We use S(n) to generate the SEQUENCE you stupid kaffir!
> S(n)=1 - 10^(-n) GENERATES ALL  the terms of the sequence Jim Burns uses:
> 0.999... = {0.9; 0.99; 0.999; ...}
> Try it moron!!! 0.9 = S(1),  0.99 = S(2), etc.

Yes what you write above is something of the few
true insights by you already for the last 7 years.
Now the limit operator must know on what
generated sequeces it operators,

it is not the limit operators takes something and then
generates something. It is the other way around,
the limit takes something which is generated, it
takes a multitude and outputs a singletude:

   lim n->oo S(n) = S

Or in your exampe of 0.999... = 1:

   lim n->oo (1-10^(-n)) = 1.

You see very simple what the limit operator does:

    Input: A generate sequence, S(n), 1-10^(-n)
    Operation: Limit (lim n-oo)
    Output: A value, S, 1

So there is no "S=Lim ...