From: cbrown on

mueckenh(a)rz.fh-augsburg.de wrote:
> cbrown(a)cbrownsystems.com schrieb:
>
> > > >
> > > > > I can calculate 1 + 1/2.
> > > >
> > > > Good for you!
> > >
> > > Bad for you, if you can't.
>
> Can't you?
>
> > > I know that 2/3 < 1 is a convenient way to express 2 < 3.
> >
> > For heaven's sake. "2/3 < 1" is not just "a convenient way" of
> > expressing "2 < 3".
>
> It is *also*, among others, a convenient way of expressing "2 < 3
>

Let G be the additive group of rational numbers. Let H be the additive
group of dyadic rationals (i.e., rationals of the form x/2^y). Then H <
G. Do you think it then makes sense to say "1 < G/H"? Are you aware
that it is more usual to say "G/H is isomporphic to the additive group
of rationals with odd denominator?

> >
> > E(n)/P(n) -> 2 is a feature only of finite n. |E(n)| < |P(n)| is a
> > feature only of infinite n. /Many/ things that can be said in the
> > finite case are not true in the infinite case;
>
> But 1 + 1/2 + 1/4 + ... < 1 is *not* among those.
>

You say this because you "feel" it, not because you have a proof of it.
What exactly do you mean by "... < 1"?

> > for example E(n) is
> > finite for natural n, and not finite for infinite n.
>
> For example {1,2,3,...n} <--> {2,4,6,...2n} is a bijection for finite
> sets. It is obviously false for infinite sets.

Huh? There /is/ a bijection between the set of all naturals and the set
of even naturals. It is the function f(x) = 2*x.

> Never thought about that topic? Presumably not, because you had to learn
> that stuff from authorities. Here you see how harmful belief in
> authorities can be.
>

What "authorities" do you claim I am quoting when I state "the function
f(x) = 2*x is a bijection between the set of all naturals, and the set
of all even naturals"?

I am simply applying specific definitions of the terms "function",
"bijection", "the set of all naturals" and "the set of all even
naturals", using the usual logic.

> It is when you try to argue "lim n-> oo E(n)/P(n) = 2 contradicts
> > E(n)/P(n) < 1 in the infinite case" that your confusing notation
> > "looks" like a contradiction (to you; to me it looks like nonsense).
>
> Then you really do accept 1 + 1/2 + 1/4 + ... < 1 ?

Of course not. How do you infer that from my statements?

Cheers - Chas

From: Franziska Neugebauer on
,----[ <458b675c$0$97264$892e7fe2(a)authen.yellow.readfreenews.net> ]
| mueckenh(a)rz.fh-augsburg.de wrote:
|
| > Franziska Neugebauer schrieb:
|
| ,----[<4589b948$0$97268$892e7fe2(a)authen.yellow.readfreenews.net> ]
| |
>> ,----[ <4588194e$0$97248$892e7fe2(a)authen.yellow.readfreenews.net> ]
| | >> >> | >> >> > If an unbounded diagonal exists, then obviously an
| | >> >> | >> >> > unbounded line must exist. [(*)]
| | [...]
| | >> >> That does not prove (*) since the whole diagonal is not
| | >> >> bounded
| | >> >> by any such n.
| | >> >
| | >> > If you doubt my proof,
| | >>
| | >> Which proof? You simply stated:
| | >>
| | >> (1) forAll elem in diag Exists line (m is in line)
| | >> (2) forAll elem in diag (elem is finite natural number)
| | >> (3) forAll natural n (forAll elem in diag <= n
| | >> is in line n)
| | >>
| | >> So please show us step-by-step how (*), which states
| | >>
| | >> (*) notExists natural n (forAll m in diag (m <= n))
| | >> -> Exists natural n' (notExists natural n(forAll m
| | >> in line n' (m
| | >> <=n)),
| | >>
| | >> follows from (1), (2) and (3).
| | >
| | > See my recent reply to William Hughes.
| |
| | Please copy and paste. I am not able to guess what you have in mind.
| `----
|
| [...]
| > We talked about the potentially infinite set N.
| >> Which means that for any n we can find a single line L(n) that
| >> works
| >> for this n. It does not mean that that we can find a single
| >> line L_D, that works for all n.
| >
| > That is impossible, beause there is no "all n".
|
| If you can't proof (*) you should write so.
|
| > If "all n" were somewhere, then we had no problem to put them in one
| > single line.
|
| We are not disputing "all n are somewhere" but the validity of (*).
| Until now (*) is not valid.
|
| > Due to the finiteness of every n, every position of the
| > line was finite. And due to the fact that a number with only finite
| > indexes is a natural number, the whole number was a natural number.
|
| I don't know hat "whole number" you are writing about but it seems to
| me that you compulsively try to apply your well-known unproved AND
| invalid convictions ("N cannot be infinite because A n e N (n is
| finite)").
`----

mueckenh(a)rz.fh-augsburg.de wrote:

> Franziska Neugebauer schrieb:
>
>> > Recently we had the following claim (by Virgil)
>> >
>> > The number of equations
>> >
>> > 1+1+1+...+1 = n
>> >
>> > with finitely many 1 is infinite. Considering the fact that there
>> > are exactly as many 1 as different numbers, it is impossible that
>> > the number of numbers can be infinite while the sme number of 1 is
>> > finite.
>>
>> The number of sums (left to the "=") and the number of n's (right to
>> the "=") is not bounded. What precisely is your problem?
>
> The infinitely many sumands in
>
> 1
> 1+1
> 1+1+1
> ...

1. Every line contains finitely many summands.
2. Every n e N is finite. Do you agree?

> yield an infinitude of sums but not an infinitude of summands.

3. Every single sum contains only finitely many summands.
4. You may revisit my posting of November 21:

,----[ <4562fc49$0$97261$892e7fe2(a)authen.yellow.readfreenews.net> ]
| Condider the following scenario:
|
| state 1: Given a parking space with *two* cars having their
| wheels mounted. (each car has 4 wheels)
|
| state 2: Unmount all the wheels of both cars and put them on one
| stack of wheels. (stack of wheels has 8 wheels)
|
| Proposition 1: "There are cars in state 1 which have as many wheels
| as the stack of wheels in state 2 has."
|
| Proposition 2: "There is *a* car in state 1 which has as many
| wheels as the stack of wheels in state 2 has."
|
| Now my interpretation: Claim 1 states that there are at least two (due
| to the plural form "cars" and due to the "are" instead of "is") cars
| each of which have the property of having as many wheels as the stack
| of wheels has (8). That is wrong in my view.
|
| Claim 2 states that there is (at least) one car which has as many
| wheels as the stack of wheels has (8). Which is wrong, too.
|
| Question 1: Do you agree with my interpretation?
`----

> How can finitely many summands in this linear order yield an
> infinitude of sums?

That is your problem?

F. N.
--
xyz
From: Franziska Neugebauer on
context:
> > mueckenh(a)rz.fh-augsburg.de wrote:
> >
> > > Franziska Neugebauer schrieb:
> >
> > ,----[<4589b948$0$97268$892e7fe2(a)authen.yellow.readfreenews.net> ]
> > |
>> ,----[ <4588194e$0$97248$892e7fe2(a)authen.yellow.readfreenews.net> ]
> > | >> >> | >> >> > If an unbounded diagonal exists, then obviously an
> > | >> >> | >> >> > unbounded line must exist. [(*)]
> > | [...]
> > | >> >> That does not prove (*) since the whole diagonal is not
> > | >> >> bounded
> > | >> >> by any such n.
> > | >> >
> > | >> > If you doubt my proof,
> > | >>
> > | >> Which proof? You simply stated:
> > | >>
> > | >> (1) forAll elem in diag Exists line (m is in line)
> > | >> (2) forAll elem in diag (elem is finite natural number)
> > | >> (3) forAll natural n (forAll elem in diag <= n
> > | >> is in line n)
> > | >>
> > | >> So please show us step-by-step how (*), which states
> > | >>
> > | >> (*) notExists natural n (forAll m in diag (m <= n))
> > | >> -> Exists natural n' (notExists natural n(forAll m
> > | >> in line n' (m
> > | >> <=n)),
> > | >>
> > | >> follows from (1), (2) and (3).
> > | >
> > | > See my recent reply to William Hughes.
> > |
> > | Please copy and paste. I am not able to guess what you have in
> > | mind.
> > `----
> >
> > [...]
> > > We talked about the potentially infinite set N.
> > >> Which means that for any n we can find a single line L(n) that
> > >> works
> > >> for this n. It does not mean that that we can find a single
> > >> line L_D, that works for all n.
> > >
> > > That is impossible, beause there is no "all n".
> >
> > If you can't proof (*) you should write so.
> >
> > > If "all n" were somewhere, then we had no problem to put them in
> > > one
> > > single line.
> >
> > We are not disputing "all n are somewhere" but the validity of (*).
> > Until now (*) is not valid.
> >
> > > Due to the finiteness of every n, every position of the
> > > line was finite. And due to the fact that a number with only
finite
> > > indexes is a natural number, the whole number was a natural
number.
> >
> > I don't know hat "whole number" you are writing about but it seems
to me
> > that you compulsively try to apply your well-known unproved AND
invalid
> > convictions ("N cannot be infinite because A n e N (n is finite)").
> >
> > > Recently we had the following claim (by Virgil)
> > >
> > > The number of equations
> > >
> > > 1+1+1+...+1 = n
> > >
> > > with finitely many 1 is infinite. Considering the fact that there
are
> > > exactly as many 1 as different numbers, it is impossible that the
> > > number of numbers can be infinite while the sme number of 1 is
finite.
> >
> > The number of sums (left to the "=") and the number of n's (right to
the
> > "=") is not bounded. What precisely is your problem?

mueckenh(a)rz.fh-augsburg.de wrote:
> Franziska Neugebauer schrieb:
>> > In short: If any natural number n does exist, then there exists a
>> > line in which this and any smaller number does exist.
>> > This holds for every number n.
>>
>> Every natural n.
>>
>> > You know that the logical symbol for every is forall.
>>
>> I don't see how this could support your "famous" quantifier reversal.
>> Hence that does not prove the existence of a _single_ line (*).
>
> Show me a number which is not in a single line.

I do not claim that there is a _natural_ number which is not in a single
line. Hence I do not have to show what you have asked for.

> Or show me two lines, each of which lacks a number the other one does
> not lack.

I do not claim that there are two lines each of which mutually lacks a
number the other one does not lack. Hence I do not have to show what
you have asked for.

I still wait to read your proof of (*) which is derived from (1), (2)
and (3).

F. N.
--
xyz
From: Franziska Neugebauer on
mueckenh(a)rz.fh-augsburg.de wrote:

> no irrational number can be approximated better than to about 10^-100.

Linguistic Question: Is it meaningfull to speak of an "approximation" if
one denies the existence of the thing which is approximated (the
irrational number)?

F. N.
--
xyz
From: mueckenh on

stephen(a)nomail.com schrieb:

> Newberry <newberry(a)ureach.com> wrote:
>

> > What is supposed to follow from all this?
>
> That what is true "in the limit" is not necessarily true for the infinite
> case.

For instance the diagonal number of Cantor's proof differs from any
list entry of a finite list. But it certainly does not differ from any
list entry of an infinite list. (Compare the case 1.000...1 = 0.999...)


For instance the bijection {1,2,3,...,n} <--> {2,4,6,...,2n} holds for
every finite sequence, but it does not hold for an infinite sequence
(obviously, because there are more natural numbers than even atural
numbers). It is a pity that mathematicians have not known your theorem
earlier. It would have spared us much confusion.

Regards, WM