0.999
master1729
how about that ?
Marshall
> [0 , 1[ = [0 , 0.999...]
>
> how about that ?
snore
Frederick Williams
>
> [0 , 1[ = [0 , 0.999...]
>
> how about that ?
If you can sum a geometric series, you can work out what 0.999... is.
--
http://indology.info/papers/gombrich/uk-higher-education.pdf
master1729
if you can read a book , you can work out the difference between open and closed set.
David C. Ullrich
wrote:
>[0 , 1[ = [0 , 0.999...]
>
>how about that ?
Black equals white. How about _that_?
spudnik
where this is the explicit ambiguity between 3.14159999... and
3.14160000...;
you have to use "the 'real' one."
netzweltler
(d is the distance between 1 and the position of the mark):
time 0: I am putting a mark at position 0 (d = 1)
time 0.9: I am putting a mark at position 0.9 (d = 0.1)
time 0.99: I am putting a mark at position 0.99 (d = 0.01)
time 0.999: I am putting a mark at position 0.999 (d = 0.001)
…
time 1: I am putting a mark at position 1 (d = 0)
The only time where d = 0 is time 1.
How many marks did I put in the interval [0,1]?
Jesse F. Hughes
> I am putting marks on the number line like this
> (d is the distance between 1 and the position of the mark):
>
> time 0: I am putting a mark at position 0 (d = 1)
> time 0.9: I am putting a mark at position 0.9 (d = 0.1)
> time 0.99: I am putting a mark at position 0.99 (d = 0.01)
> time 0.999: I am putting a mark at position 0.999 (d = 0.001)
> ...
>
> time 1: I am putting a mark at position 1 (d = 0)
>
> The only time where d = 0 is time 1.
> How many marks did I put in the interval [0,1]?
You placed countably many marks. The order type of the times you marked
is omega + 1.
--
Jesse F. Hughes
"That's cool for us in Alabama. 'Cause you know, it's either this or
the monster truck rally." -- An Alabaman expresses appreciation for
local repertory theater on NPR
Bart Goddard
be05-a80...@w17g2000yqh.googlegroups.com:
> On 29 Dez., 12:28, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> If the marks are countable, shouldn't I be able to assign a natural
> number to every mark, even to the mark at position 1?
Yes. And it's easy to do so. Use the map that assigns 1 to the
mark at 1, and assigns n to the mark you indicate with d=1/10^{n-2}.
--
Cheerfully resisting change since 1959.
netzweltler
> netzweltler <reinhard_fisc...@arcor.de> wrote in news:2c533e12-f47a-4444-
> be05-a80cf2ce1...@w17g2000yqh.googlegroups.com:
>
> - Zitierten Text anzeigen -
Does it make some kind of sense to say, from time 0 to time 1 I placed
countably infinitely many marks, and the last mark I placed at time 1?
--
netzweltler
Math1723
> [0 , 1[ = [0 , 0.999...]
>
> how about that ?
Incorrect, since 0.999... = 1.
Bart Goddard
801e-2ee...@v17g2000yqv.googlegroups.com:
> mber to every mark, even to the mark at position 1?
>>
>> Yes. �And it's easy to do so. �Use the map that assigns 1 to the
>> mark at 1, and assigns n to the mark you indicate with d=1/10^{n-2}.
>>
>> --
>> Cheerfully resisting change since 1959.- Zitierten Text ausblenden -
>>
>> - Zitierten Text anzeigen -
>
> Does it make some kind of sense to say, from time 0 to time 1 I placed
> countably infinitely many marks, and the last mark I placed at time 1?
Yes, it makes perfect sense. In order to deal with infinity,
you made up a model to help you think about it. In that model,
someone makes all those marks. You keep asking questions with
obvious and easy answers. If _your_ introduction of time into
the model confuses you, then you may need a different model.
netzweltler
> netzweltler <reinhard_fisc...@arcor.de> wrote in news:31c9b520-26d9-48c1-
> 801e-2ee987d3a...@v17g2000yqv.googlegroups.com:
I've made up another model that confuses me even more. Your answers
might clarify it.
I am putting marks on the number line like this
(d is the distance between adjacent marks):
time 0: I am putting a mark at position 0, and one at position 1 (d =
1)
time 0.5: I am putting a mark at position 0.5 (d = 0.5)
time 0.75: I am putting a mark at position 0.25, and one at position
0.75 (d = 0.25)
time 0.875: I am putting marks at position 0.125, 0.375, 0.625, and
0.875 (d = 0.125)
...
What is the distance d at time 1?
--
netzweltler
Jesse F. Hughes
> I've made up another model that confuses me even more. Your answers
> might clarify it.
>
> I am putting marks on the number line like this
> (d is the distance between adjacent marks):
>
> time 0: I am putting a mark at position 0, and one at position 1 (d =
> 1)
> time 0.5: I am putting a mark at position 0.5 (d = 0.5)
> time 0.75: I am putting a mark at position 0.25, and one at position
> 0.75 (d = 0.25)
> time 0.875: I am putting marks at position 0.125, 0.375, 0.625, and
> 0.875 (d = 0.125)
> ...
>
> What is the distance d at time 1?
Let's clarify things a bit. Let d_t be the distance between adjacent
marks at time t. Thus, d_0 = 1 and d_0.875 = 0.125. Then we can say
lim_{t -> 1} d_t = 0.
Nonetheless, d_1 is not defined by anything you said above, since
there's no clear concept of adjacency for the marks that have been
placed prior to t = 1. For instance, consider the mark at 0. There is
also a mark at 1/2^n for every n, and none of these marks are adjacent
to 0, so it makes no sense to ask how far the mark adjacent to 0 is.
--
"Being who I am, I know that's a solution that will run in polynomial
time, but for the rest of you, it will take a while to figure that out
and know why [...But] it's the same principle that makes n! such a
rapidly growing number." James S. Harris solves Traveling Salesman
spudnik
master1729
indeed.
master1729
> I am putting marks on the number line like this
> (d is the distance between 1 and the position of the
> mark):
>
> time 0: I am putting a mark at position 0 (d = 1)
> time 0.9: I am putting a mark at position 0.9 (d =
> 0.1)
> time 0.99: I am putting a mark at position 0.99 (d =
> 0.01)
> time 0.999: I am putting a mark at position 0.999 (d
> = 0.001)
> ...
>
> time 1: I am putting a mark at position 1 (d = 0)
>
> The only time where d = 0 is time 1.
> How many marks did I put in the interval [0,1]?
in standard set theory there are aleph_0 digits after the '.'.
0.999... -> card(...) = w , card (w) = aleph_0.
for supporting evidence of that claim consider cantors diagonal :
0.1326541...
0.3841371...
0.3451314...
..
the ... clearly must have card aleph_0 , otherwise the reals cannot have cardinality 2^aleph_0.
i aware of criticism that the cardinality of '...' is chosen arbitrary.
your post reminds me of thompsons lamp and zeno's paradox.
i suppose you have a specific intention with your question.
regards
tommy1729
Jesse F. Hughes
> notation, [0,1[, means that 1 is not included?
More commonly written [0,1), at least in my day.
--
Jesse F. Hughes
"Being wrong is easy, knowing when you're right can be hard, but
actually being right and knowing it, is the hardest thing of all."
-- James S. Harris
Jesse F. Hughes
> i aware of criticism that the cardinality of '...' is chosen
> arbitrary.
I'm not aware of that criticism. Who complains about that?
--
Jesse F. Hughes
Did you lay down in heaven? Did you wake up in hell?
I bet you never guessed that it would be so hard to tell.
-- The Flatlanders, /Judgment Day/
Bart Goddard
news:82cb11af-a0ed-4128...@m37g2000vbn.googlegroups.com:
> I've made up another model that confuses me even more. Your answers
> might clarify it.
>
> I am putting marks on the number line like this
> (d is the distance between adjacent marks):
>
> time 0: I am putting a mark at position 0, and one at position 1 (d 1)
> time 0.5: I am putting a mark at position 0.5 (d = 0.5)
> time 0.75: I am putting a mark at position 0.25, and one at position
> 0.75 (d = 0.25)
> time 0.875: I am putting marks at position 0.125, 0.375, 0.625, and
> 0.875 (d = 0.125)
> ...
>
> What is the distance d at time 1?
You will remain confused as long as you insist on thinking
that infinite sets must behave like finite sets. The limit
of continuous functions is not continuous, as you hopefully
learned in calculus. There's no theorem that says "the limit
of things with propery P has to have property P."
Your finite sets of marks have "adjacency". The infinite set
of marks does not. So your d is not well-defined.
Suppose you have a cube of granite which measures 10 light-years
on a side. Once every billion years, an immortal raven, which
is fed exclusively on bland custard, comes to sharpen its beak
on the cube of granite. After the block of granite has been
completely worn away by the raven's abrasions, we'll still have
barely started on our way to the end of time.
Infinity is whopping, freaking big. Take whatever you can
imagine and raise it to the whatever-you-can-imagine-factoral
power, and you still don't have it right. In fact you're
99.99999....% short of imagining infinity. So these finite
games you're playing with marks on sticks will never do the trick for
you.
master1729
is that so ?
for fractions it is so.
but what if 0.999... = 1 - infinitesimal ?
most teachers ' prove ' that 0.99... = 1 by the following :
0.99... x 1 = 0.99... x ( 10 - 1 ) / 9
=> 0.99... = (9.99... - 0.99...) / 9
=> 0.99... = (9) / 9
=> 0.99... = 1
however this is clearly false :
1) 0.999... x 10 = 9.99...0
2) 9.99...0 - 0.999... = 0.000...9
3) 0.000...9 / 9 = 0.000...1
4) 0.000...1 is at least infinitesimally different from 0.
this doesnt mean i accept 0.000...1 as a valid number , but it shows the classic proof of 0.999... = 1 has holes.
two possible interpretations of 0.999... are
1) 1 - h , where h is a kind of infinitesimal
e.g. nilpotent h , dx , multiplicative inverse of an ordinal , ...
2) [0 , 1[ = [0 , 0.999...]
of course you will disagree whatever i say ...
tommy1729
master1729
>
> > notation, [0,1[, means that 1 is not included?
>
> More commonly written [0,1), at least in my day.
>
both notations are used in europe.
> --
> Jesse F. Hughes
> "Being wrong is easy, knowing when you're right can
> be hard, but
> actually being right and knowing it, is the hardest
> thing of all."
> --
> -- James
> -- James S. Harris
Mike
MikeB
Jesse F. Hughes
> In my opinion, the notation ]0,1[ is much better that (0,1) for
> evident reasons. I also guess that it is the oldest.
Somehow, I doubt that it is the older notation, but that's just my
guess.
> It is widely used over the world. Nevertheless, the two notations were
> defined as standart (see Wikipedia).
--
Conservative, n:
A statesman who is enamored of existing evils, as distinguished
from the Liberal who wishes to replace them with others.
-- Ambrose Bierce
netzweltler
> netzweltler <reinhard_fisc...@arcor.de> wrote innews:82cb11af-a0ed-4128...@m37g2000vbn.googlegroups.com:
>
> > I've made up another model that confuses me even more. Your answers
> > might clarify it.
>
> > I am putting marks on the number line like this
> > (d is the distance between adjacent marks):
>
> > time 0: I am putting a mark at position 0, and one at position 1 (d 1)
> > time 0.5: I am putting a mark at position 0.5 (d = 0.5)
> > time 0.75: I am putting a mark at position 0.25, and one at position
> > 0.75 (d = 0.25)
> > time 0.875: I am putting marks at position 0.125, 0.375, 0.625, and
> > 0.875 (d = 0.125)
> > ...
>
> > What is the distance d at time 1?
>
> You will remain confused as long as you insist on thinking
> that infinite sets must behave like finite sets. The limit
> of continuous functions is not continuous, as you hopefully
> learned in calculus. There's no theorem that says "the limit
> of things with propery P has to have property P."
>
> Your finite sets of marks have "adjacency". The infinite set
> of marks does not. So your d is not well-defined.
My assumption is I can apply the same rules as in my first model:
> I am putting marks on the number line like this
> (d is the distance between 1 and the position of the mark):
> time 0: I am putting a mark at position 0 (d = 1)
> time 0.9: I am putting a mark at position 0.9 (d = 0.1)
> time 0.99: I am putting a mark at position 0.99 (d = 0.01)
> time 0.999: I am putting a mark at position 0.999 (d = 0.001)
> ...
> time 1: I am putting a mark at position 1 (d = 0)
There I was able to define d(1) = 0. And I was able to put a last mark
at position 1.
What else am I doing in my second model?
Even in my second model I am placing a last mark at position 1 at time
1. So, the distance between the last mark I am placing at time 1 and
the mark at position 1, which I placed there at time 0, is 0 (and they
are not adjacent, if they occupy the same point, of course). Why is
d(1) well-defined in my first model and bad-defined in the second?
--
netzweltler
Math1723
> > On Dec 23, 10:48 am, master1729 <tommy1...@gmail.com>
> > wrote:
> > > [0 , 1[ = [0 , 0.999...]
>
> > > how about that ?
>
> > Incorrect, since 0.999... = 1.
>
> is that so ?
>
> for fractions it is so.
>
> but what if 0.999... = 1 - infinitesimal ?
There is no number called "infinitesimal". There is a class of
numbers called infinitesimal, but they do not exist amongst the real
numbers.
If you are using a different number system than the default reals,
then you need to explicitly do so.
But even when doing so, what you wrote is incorrect.
> most teachers ' prove ' that 0.99... = 1 by the following :
>
> 0.99... x 1 = 0.99... x ( 10 - 1 ) / 9
>
> => 0.99... = (9.99... - 0.99...) / 9
>
> => 0.99... = (9) / 9
>
> => 0.99... = 1
I don't know about "most", but it is certainly A way to prove the
point.
> however this is clearly false :
>
> 1) 0.999... x 10 = 9.99...0
>
<snip>
There is no such number as "9.99..0". This is most easily
demonstrated by my merely asking which decimal place the 0 belongs.
In the real number system, a decimal expansion displays a digit in the
10^-nth location for every positive finite n. Obviously, there is no
positive finite n which qualifies in your description.
> of course you will disagree whatever i say ...
Certainly not. I won't disagree if, for example, you say something
correct.
Bart Goddard
887d-e6f...@o14g2000yqe.googlegroups.com:
>> Your finite sets of marks have "adjacency". ĸThe infinite set
>> of marks does not. ĸSo your d is not well-defined.
>
> My assumption is I can apply the same rules as in my first model:
>
It's a bad assumption. It comes from not thinking about
infinite sets correctly.
netzweltler
> your post reminds me of thompsons lamp and zeno's paradox.
>
Yes. I can also choose 1/2 + 1/4 + 1/8 + 1/16 + ...
time 0: I am putting a mark at position 0 (d = 1)
time 0.5: I am putting a mark at position 0.5 (d = 0.5)
time 0.75: I am putting a mark at position 0.75 (d = 0.25)
time 0.875: I am putting a mark at position 0.875 (d = 0.125)
...
time 1: I am putting a mark at position 1 (d = 0)
The arrow has reached the target at time 1. It is moving according to
the function d(t) = 1 - t.
--
netzweltler
master1729
>
> > There is no such number as "9.99..0". This is
> most
> > easily
> > demonstrated by my merely asking which decimal
> place
> > the 0 belongs.
> > In the real number system, a decimal expansion
> > displays a digit in the
> > 10^-nth location for every positive finite n.
> > Obviously, there is no
> > positive finite n which qualifies in your
> > description.
>
>
> n = oo = w = aleph_0
>
> and i already said that i dont necc accept 0.999...0
> as a number , but ofcourse you snipped that ( what a
> coincidence ).
>
> tommy1729
0.999...0 is not archimedean if anyone wonders about the objections to it.
master1729
??? you probably meant )0,1( ?
Math1723
> > anonym wrote :
>
> > > There is no such number as "9.99..0". This is
> > most
> > > easily
> > > demonstrated by my merely asking which decimal
> > place
> > > the 0 belongs.
> > > In the real number system, a decimal expansion
> > > displays a digit in the
> > > 10^-nth location for every positive finite n.
> > > Obviously, there is no
> > > positive finite n which qualifies in your
> > > description.
>
> > ofcourse there is no positive finite n.
>
> > n = oo = w = aleph_0
oo, w and aleph_0 are not natural numbers. A real number has decimal
places only in positions of finite natural numbers.
> > and i already said that i dont necc accept 0.999...0
> > as a number , but ofcourse you snipped that ( what a
> > coincidence ).
Not a coincidence. It was snipped since nothing else made any further
sense.
And if you "dont necc accept 0.999...0 as a number", why did you
mention it?
> 0.999...0 is not archimedean if anyone wonders about the objections to it.
I don't know what that means. The Archimedean Property is something
that may (or may not) hold for a given group, field or algebraic
structure. I don't know what it means to say a given number has that
property. Please define your terms. Thanks.
spudnik
I'd never seen the other, before.
spudnik
subtracting from the "last place" of the real number,
1.0000..., period.
netzweltler
> Suppose you have a cube of granite which measures 10 light-years
> on a side. Once every billion years, an immortal raven, which
> is fed exclusively on bland custard, comes to sharpen its beak
> on the cube of granite. After the block of granite has been
> completely worn away by the raven's abrasions, we'll still have
> barely started on our way to the end of time.
>
> Infinity is whopping, freaking big. Take whatever you can
> imagine and raise it to the whatever-you-can-imagine-factoral
> power, and you still don't have it right. In fact you're
> 99.99999....% short of imagining infinity.
Suppose the universe exists infinitely many years, and the elements in
the universe can arrange in infinitely many combinations during that
time, one of these combinations will be planet earth, just by
coincidence. This is not short of imaging infinity.
Just for the case it is:
Then imagine, if there are infinitley many elements in an infinitely
big universe, you can find a second planet earth in this universe,
exactly like the first one, just one atom arranged differently.
--
netzweltler
netzweltler
> netzweltler <reinhard_fisc...@arcor.de> writes:
> > I've made up another model that confuses me even more. Your answers
> > might clarify it.
>
> > I am putting marks on the number line like this
> > (d is the distance between adjacent marks):
>
> > time 0: I am putting a mark at position 0, and one at position 1 (d =
> > 1)
> > time 0.5: I am putting a mark at position 0.5 (d = 0.5)
> > time 0.75: I am putting a mark at position 0.25, and one at position
> > 0.75 (d = 0.25)
> > time 0.875: I am putting marks at position 0.125, 0.375, 0.625, and
> > 0.875 (d = 0.125)
> > ...
>
> > What is the distance d at time 1?
>
> Let's clarify things a bit. Let d_t be the distance between adjacent
> marks at time t. Thus, d_0 = 1 and d_0.875 = 0.125. Then we can say
>
> lim_{t -> 1} d_t = 0.
Why not using the function d(t) = 1 - t to define what d is? This
function is good for every t e { 0, 0.5, 0.75, 0.875 ... }, why not
for d(t) = d(1)?
> Nonetheless, d_1 is not defined by anything you said above, since
> there's no clear concept of adjacency for the marks that have been
> placed prior to t = 1. For instance, consider the mark at 0. There is
> also a mark at 1/2^n for every n, and none of these marks are adjacent
> to 0, so it makes no sense to ask how far the mark adjacent to 0 is.
If d(1) is undefined, what choice do we have? It can either be 0 as
the function d(t) = 1 - t tells us, or greater than 0 (I am not taking
into account infinitesimals, since they are not defined in the real
number system). What we definitely know is, that we´ve got infinitely
many marks in [0,1] at time 1. How can infinitely many d(1)'s sum up
to 1, if d(1) is greater than 0 (or not infinitesimal, what I am not
taking into account here)?
And again, how can we easily accept, that d(1) is 0 in the first model
(and prove that 0.999... = 1 this way) and at the same time ignore,
that d(1) is 0 in the second?
--
netzweltler
bacle
Maybe if you can show that d_t is continuous
on {0.5, 0.75,...} , it would follow that
lim_t->1 d_t =d_1 =1-1 =0 . You may also be able to
argue that d_t is defined on a dense neighborhood
( in 1-eps.)
of 1 in {0,0.5,..,..} , so that ( I think we actually
need actual uniform continuity) it has a unique
continuous extension , given by the above description
David C. Ullrich
<je...@phiwumbda.org> wrote:
>spudnik <Spac...@hotmail.com> writes:
>
>> notation, [0,1[, means that 1 is not included?
>
>More commonly written [0,1), at least in my day.
Yes, although [0,1[ is better, because it avoids the
ambiguity with (a,b).
David C. Ullrich
wrote:
No, _that_ notation is something you just made up.
Bart Goddard
8a8e-bb1...@c2g2000yqc.googlegroups.com:
>> Infinity is whopping, freaking big. Take whatever you can
>> imagine and raise it to the whatever-you-can-imagine-factoral
>> power, and you still don't have it right. In fact you're
>> 99.99999....% short of imagining infinity.
>
> Suppose the universe exists infinitely many years, and the elements in
> the universe can arrange in infinitely many combinations during that
> time, one of these combinations will be planet earth, just by
> coincidence. This is not short of imaging infinity.
Yes it is. That combination will happen in a finite amount of
time. There will still be 99.999...% of the time remaining.
> Just for the case it is:
>
> Then imagine, if there are infinitley many elements in an infinitely
> big universe, you can find a second planet earth in this universe,
> exactly like the first one, just one atom arranged differently.
And there will still be 99.999...% of the time remaining.
There is a huge difference between "arbitrarily large" and
"infinity." Indeed, this is the difference which the
".999... doesn't equal 1" people trip on. An arbitrarily
long string of 9's can never equal 1. But an infinitely
long string of 9's does equal 1. (Because it's whopping,
freaking longer.)
Bart Goddard
87c2-65f...@j29g2000yqm.googlegroups.com:
>> > What is the distance d at time 1?
>>
>> Let's clarify things a bit. ĸLet d_t be the distance between adjacent
>> marks at time t. ĸThus, d_0 = 1 and d_0.875 = 0.125. ĸThen we can
> say
>>
>> ĸlim_{t -> 1} d_t = 0.
>
> Why not using the function d(t) = 1 - t to define what d is?
But it pre-supposes the answer. You're searching for
a certain value which you call d_0. You then define it
to be 0 and assert that you've "discovered" the value...?
That's no way to do science.
netzweltler
> netzweltler <reinhard_fisc...@arcor.de> wrote in news:912b1414-e4a1-45df-
> 8a8e-bb1d3195e...@c2g2000yqc.googlegroups.com:
I fully agree. 0.9 + 0.09 + 0.009 + ... never reaches 1. Since "never"
is of no concern if we are taking into account infinity we can omit
the "never" and it results:
0.9 + 0.09 + 0.009 + ... reaches 1.
Or we can say 0.9 + 0.09 + 0.009 + ... at no stage of computation
reaches 1. But infinity means after all stages.
--
netzweltler
Jussi Piitulainen
The best is the [a .. b) family. No ambiguity with any other notation,
parentheses open in the right direction only, and even [a, b] is left
free for some other use.
In proper typesetting, use thin spaces around .. .
Jesse F. Hughes
> On 29 Dez., 18:10, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> netzweltler <reinhard_fisc...@arcor.de> writes:
>> > I've made up another model that confuses me even more. Your answers
>> > might clarify it.
>>
>> > I am putting marks on the number line like this
>> > (d is the distance between adjacent marks):
>>
>> > time 0: I am putting a mark at position 0, and one at position 1 (d =
>> > 1)
>> > time 0.5: I am putting a mark at position 0.5 (d = 0.5)
>> > time 0.75: I am putting a mark at position 0.25, and one at position
>> > 0.75 (d = 0.25)
>> > time 0.875: I am putting marks at position 0.125, 0.375, 0.625, and
>> > 0.875 (d = 0.125)
>> > ...
>>
>> > What is the distance d at time 1?
>>
>> Let's clarify things a bit. Let d_t be the distance between adjacent
>> marks at time t. Thus, d_0 = 1 and d_0.875 = 0.125. Then we can say
>>
>> lim_{t -> 1} d_t = 0.
>
> Why not using the function d(t) = 1 - t to define what d is? This
> function is good for every t e { 0, 0.5, 0.75, 0.875 ... }, why not
> for d(t) = d(1)?
You've already defined what d(t) is. It's the distance between adjacent
marks at time t. Since there are no adjacent marks at time t=1, it
follows that d(1) is undefined.
There's really no other option here, if you have already stipulated that
d(t) is the distance between adjacent marks at time t.
>> Nonetheless, d_1 is not defined by anything you said above, since
>> there's no clear concept of adjacency for the marks that have been
>> placed prior to t = 1. For instance, consider the mark at 0. There is
>> also a mark at 1/2^n for every n, and none of these marks are adjacent
>> to 0, so it makes no sense to ask how far the mark adjacent to 0 is.
>
> If d(1) is undefined, what choice do we have? It can either be 0 as
> the function d(t) = 1 - t tells us, or greater than 0 (I am not taking
> into account infinitesimals, since they are not defined in the real
> number system). What we definitely know is, that we've got infinitely
> many marks in [0,1] at time 1. How can infinitely many d(1)'s sum up
> to 1, if d(1) is greater than 0 (or not infinitesimal, what I am not
> taking into account here)?
No. d(1) is simply undefined. You can't declare that it is *any* value
at all (zero or otherwise) without contradicting the definition of d
you've already stipulated: that it is the distance between adjacent
marks.
> And again, how can we easily accept, that d(1) is 0 in the first model
> (and prove that 0.999... = 1 this way) and at the same time ignore,
> that d(1) is 0 in the second?
If d is defined the same way in the first model (as the distance between
adjacent marks), then d(1) is also undefined in the first model (since
no mark is adjacent to 1). If, on the other hand, d is the distance
between the new mark and 1 in the first model, then d(1) = 0 in that
model.
See, what we accept about d depends on its definition. Funny, huh?
--
Jesse F. Hughes
"If anything is true in general about Usenet, it's that people can go
on and on about just about anything." -- James Harris speaks the
truth.
netzweltler
> netzweltler <reinhard_fisc...@arcor.de> wrote in news:dccb2772-0035-46ae-
> 87c2-65fb4e3c4...@j29g2000yqm.googlegroups.com:
>
> >> > What is the distance d at time 1?
>
> >> marks at time t. ÿThus, d_0 = 1 and d_0.875 = 0.125. ÿThen we can
> > say
>
> >> ÿlim_{t -> 1} d_t = 0.
>
> > Why not using the function d(t) = 1 - t to define what d is?
>
> But it pre-supposes the answer. You're searching for
> a certain value which you call d_0. You then define it
> to be 0 and assert that you've "discovered" the value...?
> That's no way to do science.
We define d(1) when we postulate 0.999... = 1.
On 30 Dez., 11:49, bacle <h...@here.com> wrote:
> Maybe if you can show that d_t is continuous
>
> on {0.5, 0.75,...} , it would follow that
>
> lim_t->1 d_t =d_1 =1-1 =0 .
Nobody doubts, that Zeno's arrow reaches the target according to the
function d(t) = 1 - t (even if Zeno dared to quantize the movement of
the arrow). When I am marking the position of the arrow at different
times, the marks are following the arrow up to the target at position
1.
t = 0: I am putting a mark at position 0 (d = 1)
t = 0.5: I am putting a mark at position 0.5 (d = 0.5)
t = 0.75: I am putting a mark at position 0.75 (d = 0.25)
t = 0.875: I am putting a mark at position 0.875 (d = 0.125)
...
t = 1: I am putting a mark at position 1 (d = 0)
I can still calculate the variuos positions using the function d(t) =
1 - t. But you are right. This is a continuous movement.
But nobody minds if I do the same here:
t = 0: I am putting a mark at position 0 (d = 1)
t = 0.9: I am putting a mark at position 0.9 (d = 0.1)
t = 0.99: I am putting a mark at position 0.99 (d = 0.01)
t = 0.999: I am putting a mark at position 0.999 (d = 0.001)
...
t = 1: I am putting a mark at position 1 (d = 0)
Even here I can calculate the variuos positions using the function
d(t) = 1 - t, including d(1) = 0.
So, I am postulating d(1) = 0, for Zeno, for 0.999 = 1, and for this
model:
t = 0: I am putting a mark at position 0, and one at position 1 (d =
1)
t = 0.5: I am putting a mark at position 0.5 (d = 0.5)
t = 0.75: I am putting a mark at position 0.25, and one at position
0.75 (d = 0.25)
t = 0.875: I am putting marks at position 0.125, 0.375, 0.625, and
0.875 (d = 0.125)
...
--
netzweltler
Math1723
"almost nothing" is not a real number. "Nothing" on the other hand
is. And 1 minus nothing = 1.
Jim Burns
>> On Dec 23, 10:48 am, master1729<tommy1...@gmail.com>
>> wrote:
>>> [0 , 1[ = [0 , 0.999...]
>>>
>>> how about that ?
>>
>> Incorrect, since 0.999... = 1.
>
> is that so ?
>
> for fractions it is so.
>
> but what if 0.999... = 1 - infinitesimal ?
Then you were wrong earlier to leave out the fairly
important fact that you mean something else by 0.999...
and you compounded your error by not saying (by /still/
not saying) what it is you do mean.
Let's leave that for a minute, though. You want some
sort of non-standard real numbers, apparently. I would
like to know which properties of the real numbers you
are willing to give up, in order to get the properties
you want.
The real numbers are a complete ordered field, that is
to say, they follow seventeen (by my count) specific axioms.
I am told that every complete ordered field is isomorphic,
so that, not only are the real numbers a complete ordered
field, but every complete ordered field is the real numbers,
apart from features that we agree are irrelevant.
The real numbers are designed to be optimally convenient
to those who wish to calculate. To that end, we know
without checking that adding, subtracting, multiplying,
or dividing (except by 0) two real numbers /has an answer/.
We know that we can compare any two real numbers, and so on.
We take these properties for granted, but they didn't appear
automatically. They were designed into our number system.
Two properties in particular you seem to dislike:
(1) If we carry out a sequence of calculations that agree
with each other better and better without bound (there is
no distance larger than zero that they stay apart), then
the calculations approach an answer in the real numbers.
(Pythagorus famously proved that we can't always do this
in the rational numbers.)
(2) If you and I carry out two completely different
sequences of calculations that agree with each other
better and better without bound (understood as before),
then we are calculating the same real number.
If your number system has those properties, then it is
complete, and if it also treats addition, subtraction,
multiplication, division and comparison in the usual
fashion, then your number system is the real numbers,
not just something like the real numbers.
When you object to 0.999... being called the same number
as 1, you object to completeness. Have you really thought
about what calculations would be like without completeness?
Suppose that I am performing a series of calculations
of greater and greater precision: 0.9, 1.01, 0.999,
1.0001, 0.99999, and so on. Which number am I calculating,
0.999... or 1? Or perhaps there is no number instead of
too many?
Consider all the different methods we have that we today
call "calculating pi". What does it even mean (without
completeness!) to say that these very different methods
calculate "the same number"? What does the infinite sum
4/1 - 4/3 + 4/5 - 4/7 + 4/9 - ...
have in common with the infinite product
2(2/1)(2/3)(4/3)(4/5)(6/5)(6/7)...
other than the closeness of the partial results?
In this new system you want, which properties of
the real numbers, all of which we have come to
depend upon, are you willing to do without?
Jim Burns
netzweltler
>
> - Zitierten Text anzeigen -
Ok. I agree. According to the definition "distance between adjacent
marks" d(1) is undefined. But why do the marks loose the adjacency?
Are they as dense as the points in [0,1]? Did I mark every point in
[0,1]?
--
netzweltler
Jesse F. Hughes
>
> Ok. I agree. According to the definition "distance between adjacent
> marks" d(1) is undefined. But why do the marks loose the adjacency?
Because, by your own description, there is no mark adjacent to 0 at time
t=1. For any mark x (other than the mark at 0), there is some other
mark between 0 and x.
You have read your own description of the thought experiment, right?
> Are they as dense as the points in [0,1]? Did I mark every point in
> [0,1]?
No. The points you marked are all rationals.
You *did* read what you wrote, right?
Look, consider the set
{ x in [0,1] | x is not equal to 1/2 }.
Clearly, this set is not the same set as [0,1]. Just as clearly, the
set is dense.
--
"You gotta use two hands...to make one note. Does that make sense? Why
would you want to do that? Why would you want to use two hands to
make one note?" -- Jazz Accordionist Sam Franco explains the inferiority
of electric guitars on This American Life.
Bart Goddard
news:16241728-a4b7-4b38...@fm22g2000vbb.googlegroups.com:
> Ok. I agree. According to the definition "distance between adjacent
> marks" d(1) is undefined. But why do the marks loose the adjacency?
Because adjacency is a property of the finite set of points.
And infinity is so much bigger than any finite sent of points
that it's unreasonable to think a property like "adjacency"
would survive for soooo long.
> Are they as dense as the points in [0,1]? Did I mark every point in
> [0,1]?
No and no.
Bart Goddard
a281-062...@32g2000yqz.googlegroups.com:
>> > Why not using the function d(t) = 1 - t to define what d is?
>>
>> But it pre-supposes the answer. ÿYou're searching for
>> a certain value which you call d_0. ÿYou then define it
>> to be 0 and assert that you've "discovered" the value...?
>> That's no way to do science.
>
> We define d(1) when we postulate 0.999... = 1.
>
No, you defined d(1) to be the distance between
adjacent points. You're hoping to show that 0.999...
and 1 are "adjacent". No one is going to let you
assume they're adjacent in order to prove they're
adjacent. Most of us think that 0.999... and 1
represent the _same_ point on the real number line, and
so we don't think a point can be adjacent to itself.
We'd all agree that the distance between 0.999... and
1 is 0 (because most of think that the distance between
a point and itself is 0.) But your function d(t) is
not defined at t=1.
bacle
> netzweltler <reinhard...@arcor.de> wrote in
> news:30b45367-5224-4b94-
> a281-062...@32g2000yqz.googlegroups.com:
>
>
> 1 is 0 (because most of think that the distance
> between
> a point and itself is 0.)
You can also argue that d( 1, 0.99999...) is indefinitely small, so that by the Archimedean
property, it must be zero. This may be the same as
saying that 1, and 0.99... are indefinitely close,
which cannot happen for different real numbers, which
may end up being what you first said.
BURT
> On 29 Dez., 20:48, Bart Goddard <goddar...@netscape.net> wrote:
>
>
>
>
>
> > netzweltler <reinhard_fisc...@arcor.de> wrote innews:82cb11af-a0ed-4128...@m37g2000vbn.googlegroups.com:
>
> > > I've made up another model that confuses me even more. Your answers
> > > might clarify it.
>
> > > I am putting marks on the number line like this
> > > (d is the distance between adjacent marks):
>
> > > time 0.5: I am putting a mark at position 0.5 (d = 0.5)
> > > time 0.75: I am putting a mark at position 0.25, and one at position
> > > 0.75 (d = 0.25)
> > > time 0.875: I am putting marks at position 0.125, 0.375, 0.625, and
> > > 0.875 (d = 0.125)
> > > ...
>
> > > What is the distance d at time 1?
>
> > that infinite sets must behave like finite sets. The limit
> > of continuous functions is not continuous, as you hopefully
> > learned in calculus. There's no theorem that says "the limit
> > of things with propery P has to have property P."
>
> > Your finite sets of marks have "adjacency". The infinite set
> > of marks does not. So your d is not well-defined.
>
> My assumption is I can apply the same rules as in my first model:
>
> > time 0: I am putting a mark at position 0 (d = 1)
> > time 0.99: I am putting a mark at position 0.99 (d = 0.01)
> > time 0.999: I am putting a mark at position 0.999 (d = 0.001)
> > ...
> > time 1: I am putting a mark at position 1 (d = 0)
>
> There I was able to define d(1) = 0. And I was able to put a last mark
> at position 1.
>
> What else am I doing in my second model?
>
> Even in my second model I am placing a last mark at position 1 at time
> 1. So, the distance between the last mark I am placing at time 1 and
> the mark at position 1, which I placed there at time 0, is 0 (and they
> are not adjacent, if they occupy the same point, of course). Why is
> d(1) well-defined in my first model and bad-defined in the second?
>
> --
> netzweltler- Hide quoted text -
>
> - Show quoted text -
.9 repeating is a quantity infinitely close to One or being the same.
The addition of the smallest quantity to it gives one.
Mitch Raemsch
Virgil
BURT <macro...@yahoo.com> wrote:
> .9 repeating is a quantity infinitely close to One or being the same.
> The addition of the smallest quantity to it gives one.
And how big is half of your allegedly smallest quantity?
BURT
are added three times.
3/3 yields the result of One which is infinitely close to it and next
absolute quantity after.
One stands on its own and 3/3 really is lacking any meaning. Math must
be as simmple as One.
Nobody would do it that way.
The real meaning is in the infinitely close but not the same.
Mitch Raemsch
netzweltler
> > news:30b45367-5224-4b94-
> > a281-062308bed...@32g2000yqz.googlegroups.com:
>
> > We'd all agree that the distance between 0.999... and
> > 1 is 0 (because most of think that the distance
> > between
> > a point and itself is 0.)
>
> You can also argue that d( 1, 0.99999...) is indefinitely small, so that by the Archimedean
> property, it must be zero. This may be the same as
> saying that 1, and 0.99... are indefinitely close,
> which cannot happen for different real numbers, which
> may end up being what you first said.
What is the difference between saying d(1, 0.999...) is indefinitely
small, and saying it is infinitesimally small? This sounds like, d(1,
0.999...) is an infinitesimal, and since infinitesimals are not
defined in the real number system, d(1, 0.999...) is undefined, or
indefinitely small.
--
netzweltler
adamk
> > Bart Goddard wrote, in part:
> >
> > > netzweltler <reinhard_fisc...@arcor.de> wrote in
> > > news:30b45367-5224-4b94-
> > > a281-062308bed...@32g2000yqz.googlegroups.com:
> >
> > > We'd all agree that the distance between 0.999...
> and
> > > 1 is 0 (because most of think that the distance
> > > between
> > > a point and itself is 0.)
> >
> > You can also argue that d( 1, 0.99999...) is
> indefinitely small, so that by the Archimedean
> > property, it must be zero. This may be the same as
> > saying that 1, and 0.99... are indefinitely close,
> > which cannot happen for different real numbers,
> which
> > may end up being what you first said.
>
> What is the difference between saying d(1, 0.999...)
> is indefinitely
> small, and saying it is infinitesimally small?
If you are working with standard real numbers, there
is no such thing as infinitesimally small. As Bacle
said, a standard real number cannot be indefinitely
small without being 0.
I thing what he had in mind was that:
d(1, 0.999...)<1/n for all integers n
From the Archimedean principle, it follows that
d(1, 0.999...)=0 , and by properties of metric
spaces, d(x,y)=0 iff x=y
This
> sounds like, d(1,
> 0.999...) is an infinitesimal, and since
> infinitesimals are not
> defined in the real number system, d(1, 0.999...) is
> undefined, or
> indefinitely small.
Yes, and this is precisely the point: a standard
real number cannot be indefinitely small without
being zero , by the Archimedean property.
In addition, the distance between two different
numbers in a metric space cannot be indefinitely small
without being zero.
>
> --
> netzweltler
Virgil
BURT <macro...@yahoo.com> wrote:
If they are not the same, then the absolute difference must be strictly positive.
And half of that difference will still be positive.
And half of that half will still be positive.
And so on ad infinitum.
So if there are any numbers at all between 0.999... and 1.000, there are infinitely many.
BURT
> In article <157540bf-17aa-43f0-885b-d7d6931b6...@h17g2000pre.googlegroups.com>,
>
> BURT <macromi...@yahoo.com> wrote:
> > .9 repeating is a quantity infinitely close to One or being the same.
> > The addition of the smallest quantity to it gives one.
>
> And how big is half of your allegedly smallest quantity?
It's smallest and indivisble.
Mitch Raemsch
Virgil
BURT <macro...@yahoo.com> wrote:
In a field containing the integers, nothing is indivisible.
Meaning both that zero is not divisible and that everything else is.
Tim Little
> Ok. I agree. According to the definition "distance between adjacent
> marks" d(1) is undefined. But why do the marks loose the adjacency?
> Are they as dense as the points in [0,1]?
In the sense that between every two points in [0,1] there is a marked
point (in fact infinitely many of them), yes.
In the sense that the set of marked points has measure zero, no.
Which sense do you mean?
> Did I mark every point in [0,1]?
No, not even all the rationals in [0,1]. Just the dyadic rationals.
- Tim
netzweltler
> On 2010-12-30, netzweltler <reinhard_fisc...@arcor.de> wrote:
>
> > Ok. I agree. According to the definition "distance between adjacent
> > marks" d(1) is undefined. But why do the marks loose the adjacency?
> > Are they as dense as the points in [0,1]?
>
> In the sense that between every two points in [0,1] there is a marked
> point (in fact infinitely many of them), yes.
>
> In the sense that the set of marked points has measure zero, no.
>
> Which sense do you mean?
According to Adam
On 31 Dez., 00:03, adamk <ad...@adamk.net> wrote:
> In addition, the distance between two different
> numbers in a metric space cannot be indefinitely small
> without being zero.
So I conclude, if there is an indefinitely small distance between 1
and 0.999... (let's say, from a geometrical point of view) which is 0
in the reals, the minimum distance of two points in [0,1] is
indefinitely small (from a geometrical point of view) and 0 in the
reals. Did I get this right?
> > Did I mark every point in [0,1]?
>
> No, not even all the rationals in [0,1]. Just the dyadic rationals.
So between the marks there are infinitely many other (non-dyadic)
rationals and infinitely many other reals?
>
> - Tim
--
netzweltler
at1with0
BURT
> In article <09e7fbfe-5c2f-438a-8aa9-62ec5e585...@f20g2000prn.googlegroups.com>,
>
> BURT <macromi...@yahoo.com> wrote:
> > On Dec 30, 12:12 pm, Virgil <VIR...@VIRGIL.NET> wrote:
> > > In article
> > > <157540bf-17aa-43f0-885b-d7d6931b6...@h17g2000pre.googlegroups.com>,
>
> > > BURT <macromi...@yahoo.com> wrote:
> > > > .9 repeating is a quantity infinitely close to One or being the same.
> > > > The addition of the smallest quantity to it gives one.
>
> > > And how big is half of your allegedly smallest quantity?
>
> > It's smallest and indivisble.
>
> In a field containing the integers, nothing is indivisible.
>
> Meaning both that zero is not divisible and that everything else is.
It's not an integer. Zero is not a quantity.
Mitch Raemsch
Tim Little
> So I conclude, if there is an indefinitely small distance between 1
> and 0.999... (let's say, from a geometrical point of view) which is
> 0 in the reals,
A false premise since 1 and 0.999... are just different names for the
same point. There is no "indefinitely small" distance between them,
because the distance is very definitely zero.
> Did I get this right?
Perhaps yes, in the sense of your implication being a vacuous truth.
http://en.wikipedia.org/wiki/Vacuous_truth
However, if you did not intend it to be vacuous (an implication with a
false premise) then it wasn't right.
> So between the marks there are infinitely many other (non-dyadic)
> rationals and infinitely many other reals?
Yes, in the sense that if you pick any two marks, between them there
are infinitely many non-dyadic rationals and irrationals. However, do
be aware that there are also infinitely many other marks between them.
In your infinite set of marks, no two marks are "adjacent".
- Tim
netzweltler
> On 2010-12-31, netzweltler <reinhard_fisc...@arcor.de> wrote:
>
> > So I conclude, if there is an indefinitely small distance between 1
> > and 0.999... (let's say, from a geometrical point of view) which is
> > 0 in the reals,
>
> A false premise since 1 and 0.999... are just different names for the
> same point. There is no "indefinitely small" distance between them,
> because the distance is very definitely zero.
>
> > Did I get this right?
>
>
> However, if you did not intend it to be vacuous (an implication with a
> false premise) then it wasn't right.
I better stay with d(1, 0.999...) is definitely zero.
> > So between the marks there are infinitely many other (non-dyadic)
> > rationals and infinitely many other reals?
>
> Yes, in the sense that if you pick any two marks, between them there
> are infinitely many non-dyadic rationals and irrationals. However, do
> be aware that there are also infinitely many other marks between them.
> In your infinite set of marks, no two marks are "adjacent".
>
> - Tim
Another question:
t = 0: I am putting a mark at position 0, and one at position 1 (d =
1)
t = 0.5: I am putting a mark at position 0.5 (d = 0.5)
t = 0.75: I am putting a mark at position 0.25, and one at position
0.75 (d = 0.25)
t = 0.875: I am putting marks at position 0.125, 0.375, 0.625, and
0.875 (d = 0.125)
...
The number of marks I am placing is increasing each step (for t >
0.5). How many marks am I placing at t = 1?
--
netzweltler
Math1723
> It's not an integer. Zero is not a quantity.
It most certainly is. Zero is both an integer and a quantity.
Math1723
>
> Another question:
>
> t = 0: I am putting a mark at position 0, and one at position 1 (d =
> 1)
> t = 0.5: I am putting a mark at position 0.5 (d = 0.5)
> t = 0.75: I am putting a mark at position 0.25, and one at position
> 0.75 (d = 0.25)
> t = 0.875: I am putting marks at position 0.125, 0.375, 0.625, and
> 0.875 (d = 0.125)
> ...
>
> The number of marks I am placing is increasing each step (for t >
> 0.5). How many marks am I placing at t = 1?
Countably infinitely many marks have been made by the time you get to
t=1. But since all the integers have been used up by t=1, your
description leaves ambiguous what to do at time t=1.
Jesse F. Hughes
> Another question:
>
> t = 0: I am putting a mark at position 0, and one at position 1 (d =
> 1)
> t = 0.5: I am putting a mark at position 0.5 (d = 0.5)
> t = 0.75: I am putting a mark at position 0.25, and one at position
> 0.75 (d = 0.25)
> t = 0.875: I am putting marks at position 0.125, 0.375, 0.625, and
> 0.875 (d = 0.125)
> ...
>
> The number of marks I am placing is increasing each step (for t >
> 0.5). How many marks am I placing at t = 1?
You haven't specified whether you are placing any marks at t=1 at all.
Your rule above indicates how you place marks at t = 1 - 1/2^n, and 1 is
not equal to 1 - 1/2^n.
So, if you place marks only when the above rule says that you should,
then you place no marks at t = 1.
--
Jesse F. Hughes
"Surround sound is going to be increasingly important in future
offices."
-- Microsoft marketing manager displays his keen insight
netzweltler
If I want to place another countably infinitely many marks at t = 1
right over the marks which are already there. I am starting with the
first mark at position 0. At which position do I have to place the
second mark, if I want to place the marks from the left to the right?
Where to place the 200th mark, the 2000th mark? What is the index of
the mark, I am placing at position 0.125 then?
--
netzweltler
Math1723
>
> If I want to place another countably infinitely many marks at t = 1
> right over the marks which are already there. I am starting with the
> first mark at position 0. At which position do I have to place the
> second mark, if I want to place the marks from the left to the right?
> Where to place the 200th mark, the 2000th mark? What is the index of
> the mark, I am placing at position 0.125 then?
Not sure I am following. You mean at time t=1, place a second mark on
0 and 1, at time t=1.5 place a second mark at 0.5, and for every time
t=1+x, place a second mark on the same positions you did at time x?
If so, then at time t=2, you are back to where you were originally,
except that all the dyadic rationals now have two marks instead of
one.
I suppose you could continue this process indefinitely. But to what
end?
Jesse F. Hughes
There is no such position.
Really, you seem to be asking similarly stupid questions over and over.
There is no mark adjacent to 0 at time t=1.
> Where to place the 200th mark, the 2000th mark? What is the index of
> the mark, I am placing at position 0.125 then?
You simply cannot do what you want to do. Simple as that.
--
Jesse F. Hughes
"But you have to support spyware if you're going to have free
file-sharing applications. Fair's fair."
-- NYU student Keith Caron in Wired
Bart Goddard
news:b5de6035-85b3-468f...@39g2000yqa.googlegroups.com:
> Another question:
>
> t = 0: I am putting a mark at position 0, and one at position 1 (d 1)
> t = 0.5: I am putting a mark at position 0.5 (d = 0.5)
> t = 0.75: I am putting a mark at position 0.25, and one at position
> 0.75 (d = 0.25)
> t = 0.875: I am putting marks at position 0.125, 0.375, 0.625, and
> 0.875 (d = 0.125)
> ...
>
> The number of marks I am placing is increasing each step (for t >
> 0.5). How many marks am I placing at t = 1?
No, it's the same question. Zeno's paradox is still beating you
up. Achilles has to go through countably infinitely many
steps before he can pass the tortoise. Your Achilles can
never pass the tortoise, even though he can run 10 times
faster.
Tim Little
> 0.5). How many marks am I placing at t = 1?
You haven't defined what happens at t = 1, only what happens at
t = 1 - 2^-n. Since you've left it open, I'm going to say "two": one
at position 1/3 and one at position 2/3.
- Tim
Tim Little
> place the marks from the left to the right?
You can't, because there is no second mark from left to right. You've
already been informed of that.
- Tim
netzweltler
On 1 Jan., 02:56, Tim Little <t...@little-possums.net> wrote:
> On 2010-12-31, netzweltler <reinhard_fisc...@arcor.de> wrote:
>
> > At which position do I have to place the second mark, if I want to
> > place the marks from the left to the right?
>
> You can't, because there is no second mark from left to right.
What if I am searching for the mark next to position 0
time 0: I am searching at position 1
time 0.5: I am searching at position 0.5
time 0.75: I am searching at position 0.25
time 0.875: I am searching at position 0.125
…
Shouldn't I have found the mark I am looking for at time 1 and be able
to assign my second mark to it?
--
netzweltler
Jesse F. Hughes
> ...
>
> Shouldn't I have found the mark I am looking for at time 1 and be able
> to assign my second mark to it?
You're being purposely obtuse.
What do you expect us to say? There is no such mark, but you think
you've found it at time 1? Yeah, okay, if it makes you feel better: you
will find the mark that doesn't exist at time 1.
That's not true, of course, but so what?
--
Jesse F. Hughes
"I send papers to math journals and I damn well get a reply. Sure,
they're polite rejections but they had better reply to me."
-- James S. Harris, on influence.
Marshall
> On 31 Dez. 2010, 17:17, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> > >> So, if you place marks only when the above rule says that you should,
> > >> then you place no marks at t = 1.
>
> > > right over the marks which are already there. I am starting with the
> > > first mark at position 0. At which position do I have to place the
> > > second mark, if I want to place the marks from the left to the right?
>
> > There is no such position.
>
> > Really, you seem to be asking similarly stupid questions over and over.
> > There is no mark adjacent to 0 at time t=1.
>
> > > place the marks from the left to the right?
>
> > You can't, because there is no second mark from left to right.
>
> What if I am searching for the mark next to position 0
Apparently you think if you rephrase the question the answer
will change. It won't.
What part of "There is no mark adjacent to 0 at time t=1"
did you not understand?
What part of "There is no such position" did you not
understand?
What part of "You can't, because there is no second mark
from left to right" did you not understand?
> time 0: I am searching at position 1
> time 0.5: I am searching at position 0.5
> time 0.75: I am searching at position 0.25
> time 0.875: I am searching at position 0.125
> …
>
> Shouldn't I have found the mark I am looking for at time 1 and be able
> to assign my second mark to it?
Since the thing you describe doesn't exist, you will not find
the mark you are looking for, ever.
Furthermore, your thought experiments are needlessly
complex, obscuring the underlying issue. Instead of
messing around with time and rational numbers, just
ask yourself what is the largest natural number. See
if you can figure out the answer to that question; it
is the same as the answer to your more confusing
question.
Marshall
netzweltler
What Math1723 said is so true:
On 31 Dez. 2010, 15:55, Math1723 <anonym1...@aol.com> wrote:
> Countably infinitely many marks have been made by the time you get to
> t=1. But since all the integers have been used up by t=1, your
> description leaves ambiguous what to do at time t=1.
So all integers have been used up.
What I don't understand I can only explain this way:
At t = 1 I have finished the task of setting infinitely many marks. No
additional marks will be added at t = 1 and after. Every mark is in
position, the set is complete. All marks I have set have been set
before t = 1. And for each step at t < 1 is valid, that I added a new
mark, which is closer to position 0 as the closest mark I added the
step before, i.e. at each step I was setting a new "second mark". What
I am searching for is these marks and each of them does exist and is
present on the number line in [0,1] at t = 1. That's all I want to
find. I am not looking for something non-existent. I am not looking
for something I haven't placed there before.
--
netzweltler
Jesse F. Hughes
> At t = 1 I have finished the task of setting infinitely many marks. No
> additional marks will be added at t = 1 and after. Every mark is in
> position, the set is complete. All marks I have set have been set
> before t = 1. And for each step at t < 1 is valid, that I added a new
> mark, which is closer to position 0 as the closest mark I added the
> step before, i.e. at each step I was setting a new "second mark". What
> I am searching for is these marks and each of them does exist and is
> present on the number line in [0,1] at t = 1. That's all I want to
> find. I am not looking for something non-existent. I am not looking
> for something I haven't placed there before.
You are changing your story.
Earlier, you said: What if I am searching for the mark next to position
0
There is no such mark.
Are you merely confused or are you a troll?
--
Jesse F. Hughes
"To be honest, I don't have enough interest in math to spend the time
it would take to clean up the mess that I believe has been created in
the past 100 or so years." -- Curt Welch lets the world down.
netzweltler
> netzweltler <reinhard_fisc...@arcor.de> writes:
> > At t = 1 I have finished the task of setting infinitely many marks. No
> > additional marks will be added at t = 1 and after. Every mark is in
> > position, the set is complete. All marks I have set have been set
> > before t = 1. And for each step at t < 1 is valid, that I added a new
> > mark, which is closer to position 0 as the closest mark I added the
> > step before, i.e. at each step I was setting a new "second mark". What
> > I am searching for is these marks and each of them does exist and is
> > present on the number line in [0,1] at t = 1. That's all I want to
> > find. I am not looking for something non-existent. I am not looking
> > for something I haven't placed there before.
>
> You are changing your story.
No.
> Earlier, you said: What if I am searching for the mark next to position
> 0
At each step for t < 1 there is a mark next to position 0. By t = 1 I
finished setting marks. I just don't understand why there is no mark
next to position 0 now. When we stop switching Thomson's lamp by t =
1, the state of the lamp (on/off) is not defined. That´s because we
cannot tell, if we stop switching after an odd or an even number of
steps. But I have defined, that for "each" step is valid, I am setting
a new "second mark". And this is valid up to the point when I finished
setting marks. So the state at t = 1 should be well-defined. And this
means: There is a "second mark".
> There is no such mark.
>
> Are you merely confused or are you a troll?
>
> --
> Jesse F. Hughes
> "To be honest, I don't have enough interest in math to spend the time
> it would take to clean up the mess that I believe has been created in
> the past 100 or so years." -- Curt Welch lets the world down.
--
netzweltler
Jesse F. Hughes
> On 1 Jan., 22:43, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> netzweltler <reinhard_fisc...@arcor.de> writes:
>> > At t = 1 I have finished the task of setting infinitely many marks. No
>> > additional marks will be added at t = 1 and after. Every mark is in
>> > position, the set is complete. All marks I have set have been set
>> > before t = 1. And for each step at t < 1 is valid, that I added a new
>> > mark, which is closer to position 0 as the closest mark I added the
>> > step before, i.e. at each step I was setting a new "second mark". What
>> > I am searching for is these marks and each of them does exist and is
>> > present on the number line in [0,1] at t = 1. That's all I want to
>> > find. I am not looking for something non-existent. I am not looking
>> > for something I haven't placed there before.
>>
>> You are changing your story.
>
> No.
>
>> Earlier, you said: What if I am searching for the mark next to position
>> 0
>
> At each step for t < 1 there is a mark next to position 0. By t = 1 I
> finished setting marks. I just don't understand why there is no mark
> next to position 0 now.
Then you are purposely obtuse.
There are marks at each 1/2^n for every n in N. Hence, for every mark
placed, there is another mark closer to 0. Thus, there is no mark next
to 0 at time 1.
There's really nothing more to it than that.
> When we stop switching Thomson's lamp by t =
> 1, the state of the lamp (on/off) is not defined. That's because we
> cannot tell, if we stop switching after an odd or an even number of
> steps. But I have defined, that for "each" step is valid, I am setting
> a new "second mark". And this is valid up to the point when I finished
> setting marks. So the state at t = 1 should be well-defined. And this
> means: There is a "second mark".
Quite simply: bullshit. It does not follow that there is a mark next to
0 at t=1, and this is plain to see.
I ask again:
>> There is no such mark.
>>
>> Are you merely confused or are you a troll?
--
Jesse F. Hughes
"To all Leaders of the World, buy or rent the movie 'The Day
After'[...] I assure you will have a new perspective on WMDs."
-- practical advice from online petitions
Bart Goddard
b4ec-2df...@o4g2000yqd.googlegroups.com:
> I just don't understand why there is no mark
> next to position 0 now.
Yes, infinity is tricky. It's way different from finity.
Suppose there is a motel with infinitely many rooms numbered
from 1 to infinity. Also suppose the motel is full, with a
tenent in every room. If someone shows up wanting a room,
the clerk, not wanting to term them away, gets on the intercom
and has everyone move down one room. Now room 1 is empty
and everyone who had a room still has a room and the newcomer
also gets a room.
Even when infinity is full, there's still room.
Likewise, when you think you've found the "point next to 0",
a simple argument shows that you had to be wrong, since
half of that value is even closer to 0.
netzweltler
> netzweltler <reinhard_fisc...@arcor.de> writes:
> > On 1 Jan., 22:43, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> netzweltler <reinhard_fisc...@arcor.de> writes:
> >> > At t = 1 I have finished the task of setting infinitely many marks. No
> >> > additional marks will be added at t = 1 and after. Every mark is in
> >> > position, the set is complete. All marks I have set have been set
> >> > before t = 1. And for each step at t < 1 is valid, that I added a new
> >> > mark, which is closer to position 0 as the closest mark I added the
> >> > step before, i.e. at each step I was setting a new "second mark". What
> >> > I am searching for is these marks and each of them does exist and is
> >> > present on the number line in [0,1] at t = 1. That's all I want to
> >> > find. I am not looking for something non-existent. I am not looking
> >> > for something I haven't placed there before.
>
> >> You are changing your story.
>
> > No.
>
> >> Earlier, you said: What if I am searching for the mark next to position
> >> 0
>
> > At each step for t < 1 there is a mark next to position 0. By t = 1 I
> > finished setting marks. I just don't understand why there is no mark
> > next to position 0 now.
>
> Then you are purposely obtuse.
>
> There are marks at each 1/2^n for every n in N. Hence, for every mark
> placed, there is another mark closer to 0.
Yes, but I have placed all these closer marks at t = 1. I did forget
not even one of them to place.
And yes, I am purposely obtuse and I am merely confused and I am a
troll - from your point of view.
--
netzweltler
Jesse F. Hughes
Er, right. So, this makes it perfectly apparent that at the end of your
task, there is no mark adjacent to 0.
> And yes, I am purposely obtuse and I am merely confused and I am a
> troll - from your point of view.
It's a trifecta!
--
Jesse F. Hughes
"Those two boys sound kind of like you."
-- Quincy P. Hughes to his father while listening to Car Talk for
the first time.
adamk
I don't know if this will help clarify (not that I fully get infinity myself):
Consider the sequence S={1/2^n } , which converges
to 0. 0 is a limit point (the only one, actually),
of S. By definition, for all eps.>0 , there is a
point of S in B(0,eps.).
Marshall
Again, you have made this needlessly complex. Simplify it and
see if it's any easier to understand.
Let's eliminate the concept of time, and the whole business about
1/2^n. Make it simple.
Suppose at step n, you have all the number from 0 to n inclusive.
Thus:
At step 0, the largest number in your collection is 0.
At step 5, the largest number in your collection is 5.
etc.
Got it?
At every step, there is a largest number in the collection.
After you have done all the steps, you have the collection
of all natural numbers. What is the largest natural number
in this collection? Or more simply, what is the largest
natural number?
This "puzzle" is simpler than the thing you're asking about,
but keeps its essence. Solve this one first.
Marshall
Tim Little
> Shouldn't I have found the mark I am looking for at time 1 and be
> able to assign my second mark to it?
You aren't doing anything at time 1, because you've only said you're
doing things at times of the form 1 - 2^-n. At every time where you
have said you're doing something, there is a later time where you find
a closer mark.
- Tim
netzweltler
>
> - Zitierten Text anzeigen -
There is no trick in your puzzle. You will never come to an end. The
trick is to show the state, when the set of natural numbers is
complete. As with Thomson's lamp. We want to know the state of the
lamp, when the lamp has been switched infinitely often, which is
undefined in this case. And as with 0.999... . We want to know the
state, when all the 9's are complete, which is 1 in this case.
On 2 Jan., 03:23, Tim Little <t...@little-possums.net> wrote:
> On 2011-01-01, netzweltler <reinhard_fisc...@arcor.de> wrote:
>
> > Shouldn't I have found the mark I am looking for at time 1 and be
> > able to assign my second mark to it?
>
> You aren't doing anything at time 1, because you've only said you're
> doing things at times of the form 1 - 2^-n. At every time where you
> have said you're doing something, there is a later time where you find
> a closer mark.
>
> - Tim
Yes, but at t = 1 I can state I have found every mark. Every mark I
have placed in [0,1] in my task before. I will find no additional mark
at t = 1, because there is no mark left to find, i.e. no closer mark
to find.
--
netzweltler
Tim Little
> i.e. no closer mark to find.
Closer mark than what? Pick a mark, *any* mark. There are infinitely
many closer ones. That's what makes it exactly like the natural
numbers. All your useless frills like supertasks, time ordering, and
powers of 1/2 are irrelevant. A closer mark corresponds to a larger
natural number.
There is no closest mark to zero for *exactly* the same reason as
there is no largest natural number. You appear to understand the
latter, but appear to have some bizarre mental block against the
former.
- Tim
netzweltler
On 31 Dez. 2010, 15:55, Math1723 <anonym1...@aol.com> wrote:
> Countably infinitely many marks have been made by the time you get to
> t=1. But since all the integers have been used up by t=1, your
> description leaves ambiguous what to do at time t=1.
So, "all" natural numbers have been used up by t = 1.
Do you agree with that?
--
netzweltler
netzweltler
> On 2 Jan., 02:10, Marshall <marshall.spi...@gmail.com> wrote:
> > Again, you have made this needlessly complex. Simplify it and
> > see if it's any easier to understand.
>
> > Let's eliminate the concept of time, and the whole business about
> > 1/2^n. Make it simple.
>
> > Suppose at step n, you have all the number from 0 to n inclusive.
> > Thus:
>
> > At step 0, the largest number in your collection is 0.
>
> > At step 5, the largest number in your collection is 5.
>
> > etc.
>
> > Got it?
>
> > At every step, there is a largest number in the collection.
>
> > After you have done all the steps, you have the collection
> > of all natural numbers. What is the largest natural number
> > in this collection? Or more simply, what is the largest
> > natural number?
>
> > This "puzzle" is simpler than the thing you're asking about,
> > but keeps its essence. Solve this one first.
>
> > Marshall- Zitierten Text ausblenden -
>
> > - Zitierten Text anzeigen -
>
> There is no trick in your puzzle. You will never come to an end.
I am wrong. Your puzzle is kind of tricky. What I am missing is, you
didn't show, how you come to an end. And that's what I show by
compressing infinity (as there are infinitely many points compressed
in [0,1]). There is no need to define a largest integer in my model,
because I have used up all these infinitely many integers by t = 1.
--
netzweltler
Jussi Piitulainen
> didn't show, how you come to an end. And that's what I show by
> compressing infinity (as there are infinitely many points compressed
> in [0,1]). There is no need to define a largest integer in my model,
> because I have used up all these infinitely many integers by t = 1.
Mark +-n at time t = (2/3)sum((1/3)^k, k=0,...,n) for each n >= 0.
As long as t < 1, there are infinitely many unmarked integers, but at
t = 1, all integers have been marked. Is there a largest integer?
netzweltler
No. As long as you define a largest integer you are dealing with
finite sets. Trust me, I can see the contradiction, that follows from
what I am saying. A mark next to 0 at t = 1 implies, that I have set a
"last" mark in a "last" step. We cannot define a last step, if we are
dealing with countable infinite sets. Nevertheless, I have stopped
placing marks by time 1. I have stopped adding marks to the set of
marks by time 1, because it is complete already.
There is nothing like, time 1 has come, and I am still placing marks
at t < 1. I am not travelling in time.
--
netzweltler
Jussi Piitulainen
> On 2 Jan., 13:38, Jussi Piitulainen wrote:
> > netzweltler writes:
> > > didn't show, how you come to an end. And that's what I show by
> > > compressing infinity (as there are infinitely many points compressed
> > > in [0,1]). There is no need to define a largest integer in my model,
> > > because I have used up all these infinitely many integers by t = 1.
> >
> > Mark +-n at time t = (2/3)sum((1/3)^k, k=0,...,n) for each n >= 0.
> >
> > As long as t < 1, there are infinitely many unmarked integers, but at
> > t = 1, all integers have been marked. Is there a largest integer?
>
> No. As long as you define a largest integer you are dealing with
> finite sets. Trust me, I can see the contradiction, that follows from
> what I am saying. A mark next to 0 at t = 1 implies, that I have set a
> "last" mark in a "last" step. We cannot define a last step, if we are
> dealing with countable infinite sets. Nevertheless, I have stopped
> placing marks by time 1. I have stopped adding marks to the set of
> marks by time 1, because it is complete already.
Equally, if you mark the integers +n and -n, for n >= 0, at time t =
(2/3)sum((1/3)^k, k=0,...,n), you will have stopped adding marks by
time 1, because then all integers have a mark.
So at t = 1 you have placed all marks without ever placing the last
mark.
> There is nothing like, time 1 has come, and I am still placing marks
> at t < 1. I am not travelling in time.
I never implied that.
Brian Chandler
Right, the end. The natural numbers go on and on without end;
according to my tiny dictionary, 'unendlich' is one German word for
"infinite", which is only a Latin dressing-up of the English for
"without end". So Jesse cannot have "come to an end", since there
isn't one. Your proper question to Jesse would be "Since these things
never end, how can you have done all of them??"
There are two (sensible) answers to this question: (a) is to say
"Since they never end, we cannot do all of them". This essentially is
a refusal consider the entity (or collection if you like) "All of the
natural numbers".
Answer (a) is a bit limiting, so many people choose to answer (b)
"Well, we cannot consider actually a "process" of going through all
natural numbers until we have completed them, but we can still
_consider_ them as a collection or a totality." This is in practice
Jesse's line. After all, even though we could never "go through" all
of them, we can make many statements which we can prove true of any
one them, and thus of the whole collection. For example, we can say
(trivially) that we know there is no number which is a prime number,
which is greater than 2 and which is even. After a very long and hard
proof, we can say there are no three of them, a, b, and c, such that
a^3 + b^3 = c^3. We can phrase such statements using quantifiers like
"any", but often it is simpler if we can also use statements using the
quantifier "all". None of this suggests that the numbers ever get to
an "end".
You are confusing yourself with stuff about reciprocals, too. One of
the algebraic properties of the positive rationals, is that they form
a group under multiplication. So any natural number (considered as a
positive rational) has a reciprocal which is a positive rational.
"Positive" (in English; have to be careful in French!) means "greater
than (and thus different from) zero". So one thing you know is that
the reciprocals of the naturals do not include zero. But just as there
is no largest natural (or largest positive rational), there is no
smallest positive rational, either. If for example you place marks on
number line for the reciprocals 1, 1/2, 1/3, 1/4, ... you know that
there is no smallest one, no mark at zero, and no mark closest to zero
either.
And that's what I show by
> compressing infinity (as there are infinitely many points compressed
> in [0,1]). There is no need to define a largest integer in my model,
> because I have used up all these infinitely many integers by t = 1.
Beware of talking about "compressing" infinity. An unendlich string of
things goes on with no end, so "compressing" it really makes no
difference, since there is no end to get closer.
HTH
Brian Chandler
Marshall
> On 2 Jan., 10:56, netzweltler <reinhard_fisc...@arcor.de> wrote:
> > On 2 Jan., 02:10, Marshall <marshall.spi...@gmail.com> wrote:
> > >
> > > Again, you have made this needlessly complex. Simplify it and
> > > see if it's any easier to understand.
>
> > > Let's eliminate the concept of time, and the whole business about
> > > 1/2^n. Make it simple.
>
> > > Suppose at step n, you have all the number from 0 to n inclusive.
> > > Thus:
>
> > > At step 0, the largest number in your collection is 0.
>
> > > At step 5, the largest number in your collection is 5.
>
> > > etc.
>
> > > Got it?
>
> > > At every step, there is a largest number in the collection.
>
> > > After you have done all the steps, you have the collection
> > > of all natural numbers. What is the largest natural number
> > > in this collection? Or more simply, what is the largest
> > > natural number?
>
> > > This "puzzle" is simpler than the thing you're asking about,
> > > but keeps its essence. Solve this one first.
>
>
> I am wrong. Your puzzle is kind of tricky. What I am missing is, you
> didn't show, how you come to an end.
This is math. There is no "how." There is only "what."
Are you a programmer by any chance? (I am.) Programming,
especially imperative programming, biases one against being
able to work in this declarative manner.*
"They cannot remember their ignorance, who have to use
their knowledge so often." --Thoreau
> And that's what I show by
> compressing infinity (as there are infinitely many points compressed
> in [0,1]). There is no need to define a largest integer in my model,
> because I have used up all these infinitely many integers by t = 1.
In your version, the point closest to zero is the exact analog of
the largest natural number. Your search for a point closest to
zero is EXACTLY THE SAME AS a search for a largest natural.
The one search succeeds if and only if the other search does.
You can prove this by putting your marks (those marks <= .5)
in one-to-one correspondence with the naturals.
Marshall
* The exception being declarative programming, of course.
Marshall
> netzweltler wrote:
>
> > I am wrong. Your puzzle is kind of tricky. What I am missing is, you
> > didn't show, how you come to an end.
>
> Right, the end. The natural numbers go on and on without end;
> according to my tiny dictionary, 'unendlich' is one German word for
> "infinite", which is only a Latin dressing-up of the English for
> "without end". So Jesse cannot have "come to an end", since there
> isn't one. Your proper question to Jesse would be "Since these things
> never end, how can you have done all of them??"
>
> There are two (sensible) answers to this question: (a) is to say
> "Since they never end, we cannot do all of them". This essentially is
> a refusal consider the entity (or collection if you like) "All of the
> natural numbers".
>
> Answer (a) is a bit limiting, so many people choose to answer (b)
> "Well, we cannot consider actually a "process" of going through all
> natural numbers until we have completed them, but we can still
> _consider_ them as a collection or a totality."
Cue the ever-insipid Don Stockbauer.
Marshall
Brian Chandler
> Cue the ever-insipid Don Stockbauer.
(1) Yes
(2) No
(How am I doing?)
Brian Chandler
Marshall
Nailed it.
Marshall