On Thu, 08 Jan 2004 16:58:42 GMT, The Ghost In The Machine wrote:
In sci.logic, Dave Seaman
dseaman@no.such.host
wrote
on Wed, 7 Jan 2004 17:22:28 +0000 (UTC)
bthf8k$hk4$1@mozo.cc.purdue.edu>:
On Wed, 07 Jan 2004 16:59:22 GMT, The Ghost In The Machine wrote:

[3] The digit sequence .999... is meaningless as one must always
use a "trailing digit" approach, writing it as .999...9, where ...
can be either finite (which is allowed under the Standard
Representation) or infinite (which the Standard Representation
does not cover and dismisses). This means that operations
such as (.999...9)^2 = .999...99800...001 are possible.

Note that [3] would require a bit of work rewriting transfinite
theory, as omega (the ordinal analogue to the cardinal infinity)
is such that omega+1 = omega in the Standard Representation.

Omega+1 has the same cardinality as omega, but they are different
ordinals. If the (w+1)-strings of digits are ordered lexicographically,
then there is no order-preserving map of such strings to the reals.


Interesting. So expressions such as omega+1 and 2*omega actually do
make sense? If so, mea culpa.

Yes. In fact, w is identical to N, the set of natural numbers (including 0),
and w+1 = w U {w} = { 0, 1, 2, 3, ..., w }.

Addition and multiplication are noncommutative:

1+w = w < w+1, (w+1 has a last element but 1+w does not)
2*w = w < w*2 = w+w. (an w of pairs vs. a pair of w's)

However, if we do allow transfinites and infinitesimals into our
number system a la Garry Denke, I'm not sure how the rest of mathematics
is going to deal with it (the concept of limit in particular may have
to be redefined).

Hopefully it's a simple retrofit but I wonder.

There's nothing wrong with considering transfinite ordinal sequences of
digits in general, but as I said, there is no order-preserving map of
such objects to the reals for ordinals > w.

No, .999... is not equal to 1, therefore any operation
involving .999... may not substitute 1 therefor.

At least, one can fashion a semi-consistent math system
therewith, although it would take a lot of work and probably
would not be worth it.

(For starters, if .999... != 1, what is .999... - (.999... * 10 - 9)
equal to? Answer: 0.000...9. This sort of thing will get rather
weird very quickly.)

Of course, if one cubes .999... one can do something like the following:

.9 * .9 * .9 = .729

.99 * .99 *.99 = .729 + 3*.0729 + 3*.00729 + .000729
= .729 + .2187 + 0.02187 + .000729
= .970299

Note the 7 in the 10^-2 spot.

.999 * .999 * .999 = .970299 + 3*.0088209 + 3*.00008019 + .000000729
= .970299 + .0264627 + .00024057 + .000000729
= .997002999

and yet another 7, this time in the 10^-3 spot.

.9999 * .9999 * .9999 = .997002999 + 3*.0008982009 + 3*.0000008091 +
.000000000729
= .997002999 + .0026946027 + 3*.0000024273 +
.000000000729
= .999700029999

and there's that 7 again. Is that second term always going to be
.000...0002something ?

Of course it's probably simpler to abstract the problem and
equate .999...9 = 1 - 10^(-n), and then

(1 - 10^(-n))^3 = 1 - 3*10^(-n) + 3*10^(-2*n) - 10^(-3*n)

which proves that (.999...9)^3 = .999...99700...0299...9, in a
simpler fashion than your equations below, or for that matter
my algebra.

So far, no problem here. But one cannot conclude that

(.999...)^3 = .999...99700...0299...

unless one ascribes a definitive meaning to digit expansions
with infinite ellipses and defines their proper addition,
subtraction, etc. It is naive to conclude that the
arithmetic operation

10 * (.999...)^3 - 9 - (.999...)^3

= 9.999...97000...2999...990 - 9 - .999...99700...0299...999
= -.000...02699...7300...009.

makes any sense at all.

I have no idea how to write it in a way that anyone here would accept,
but perhaps there is a way to write the squares of 0.999... tending
toward 0, not toward 1, settling the matter. Simple arithmetic says...

0.9^2 = .81
0.9^3 = .729
0.9^4 = .6561

0.99^2 = .9801
.099^3 = .970299
0.99^4 = .96059601

0.999^2 = .998001
0.999^3 = .997002999
0.999^4 = .996005996001

0.9999^2 = .99980001
0.9999^3 = .999700029999
0.9999^4 = .9996000599960001

0.99999^2 = .9999800001
0.99999^3 = .999970000299999
0.99999^4 = .99996000059999600001

0.999999^2 = .999998000001
0.999999^3 = .999997000002999999
0.999999^4 = .999996000005999996000001

0.9999999^2 = .99999980000001
0.9999999^3 = .999999700000029999999
0.9999999^4 = .9999996000000599999960000001

0.99999999^2 = .9999999800000001
0.99999999^3 = .999999970000000299999999
0.99999999^4 = .99999998000000059999999600000001

0.999999999^2 = .999999998000000001
0.999999999^3 = .999999997000000002999999999
0.999999999^4 = .999999996000000005999999996000000001

0.9999999999^2 = .99999999980000000001
0.9999999999^3 = .999999999700000000029999999999
0.9999999999^4 = .9999999996000000000599999999960000000001

0.99999999999^2 = .9999999999800000000001
00.99999999999^3 = .999999999970000000000299999999999
0.99999999999^4 = .99999999996000000000059999999999600000000001

0.999999999999^2 = .999999999998000000000001
0.999999999999^3 = .999999999997000000000002999999999999
0.999999999999^4 = .999999999996000000000005999999999996000000000001

etc., etc., etc.

...the squares are heading toward 0.

Garry Denke, Geologist
Denoco Inc. of Texas

No, nothing at all like that. How about responding to what I *actually*
wrote, and even quoting the parts of it that you're "refuting".

""It now seems clear that the reason Garry is talking past everyone else,
and vice versa, is that he has a different meaning of "tends to 1" in
mind. This is illuminated by a subsequent post:

In the *abstract numerical value* view, the claim that .999999999999^4
is closer to 1 than .9^4 is (i.e. that |1-.999999999999^4| < |1-.9^4|)
is (I hope) indisputably true, so that clearly isn't the view that Garry
is meaning here.

ISTM he (and Charlie Volkstorf) hold a *concrete representational* view.
For Garry, the successively longer decimal fractions *look*
progressively less like the simple numeral "1".

A big problem with Garry's view is that the concrete decimal
representation of .999... is unending, yet he hold fast to the notion
that there is a "last" 9 digit (and hence a last 1 digit in (.999...)^2
and (.999...)^4). This reminds me of "Phil".

It turns out that if one takes the trouble to *precisely define*
numerical addition and multiplication on potentially infinite decimals,
the ONLY WAY to make it work consistently with the accepted arithmetic
for finite decimals (which can each be considered to have an infinite
sequence of "0"s to the right) results in the concrete decimal
representation of (.999...)^2 being exactly .999... There really is no
last one (or Last One). ISTM the reason Garry doesn't accept that fact
is that it's not immediately obvious, and he hasn't worked through the
detailed consequences of arithmetic for infinite decimals.

Note that another of these consequences is that in terms of concrete
decimals, 1.000... - .999... is exactly equal to 0.000... This is much
easier to show than examples involving multiplication, since this can be
done one digit at a time.""

--

Sequence .9^2, .9^3, .9^4, ...
Sequence .99^2, .99^3, .99^4, ...
Sequence .999^2, .999^3, .999^4, ...
Sequence .9999^2, .9999^3, .9999^4, ...
Sequence .99999^2, .99999^3, .99999^4, ...
Sequence .999999^2, .999999^3, .999999^4, ...
Sequence .9999999^2, .9999999^3, .9999999^4, ...
Sequence .99999999^2, .99999999^3, .99999999^4, ...
Sequence .999999999^2, .999999999^3, .999999999^4, ...
Sequence .9999999999^2, .9999999999^3, .9999999999^4, ...
Sequence .99999999999^2, .99999999999^3, .99999999999^4, ...
Sequence .999999999999^2, .999999999999^3, .999999999999^4, ...
Sequence .9999999999999^2, .9999999999999^3, .9999999999999^4, ...
etc., etc., etc.,

tend to 0.

Therefore: Sequence .999...^2, .999...^3, .999...^4, ... tends to 0

--

Garry Denke, Geologist
Denoco Inc. of Texas

In sci.logic, Dave Seaman

Omega+1 has the same cardinality as omega, but they are different
ordinals. If the (w+1)-strings of digits are ordered lexicographically,
then there is no order-preserving map of such strings to the reals.

Interesting. So expressions such as omega+1 and 2*omega actually do
make sense? If so, mea culpa.

Yes. In fact, w is identical to N, the set of natural numbers (including 0),
and w+1 = w U {w} = { 0, 1, 2, 3, ..., w }.

Addition and multiplication are noncommutative:

1+w = w < w+1, (w+1 has a last element but 1+w does not)
2*w = w < w*2 = w+w. (an w of pairs vs. a pair of w's)

Ugh. My brain's beginning to hurt now... :-)

So which one of these applies to such Garry-Denke-isque decimal
expansions as

(.999....)^2 = .999...99800...001 ?

It feels like 8 is in the w+2 or w+3 position but I want to make sure.
1 is presumably in the position w*2 + 2.

Presumably 1+w refers to things like

9,(9,9,9,9,...)

where one tacks things onto the left side of an infinite sequence.
If one computes .999... / 10 + .8 = .8999..., one gets a similar
sequence.

However, if we do allow transfinites and infinitesimals into our
number system a la Garry Denke, I'm not sure how the rest of mathematics
is going to deal with it (the concept of limit in particular may have
to be redefined).

Hopefully it's a simple retrofit but I wonder.

There's nothing wrong with considering transfinite ordinal sequences of
digits in general, but as I said, there is no order-preserving map of
such objects to the reals for ordinals > w.

That might be good enough. Of course that's probably related
to the impossibility of reliably defining things such as

.999... - .999...99800...001 = .000...?!?!?!...

:-)

In sci.logic, Dave Seaman <dseaman@no.such.host
wrote on Thu, 8 Jan 2004 17:43:55 +0000 (UTC)
[snip]

There's nothing wrong with considering transfinite ordinal sequences of
digits in general, but as I said, there is no order-preserving map of
such objects to the reals for ordinals > w.


That might be good enough. Of course that's probably related
to the impossibility of reliably defining things such as

.999... - .999...99800...001 = .000...?!?!?!...

:-)

.9^2 = .81
.99^2 = .9801
.999^2 = .998001
.9999^2 = .99980001
.99999^2 = .9999800001
.999999^2 = .999998000001
.9999999^2 = .99999980000001
.99999999^2 = .9999999800000001
.999999999^2 = .999999998000000001
.9999999999^2 = .99999999980000000001
.99999999999^2 = .9999999999800000000001
.999999999999^2 = .999999999998000000000001
.9999999999999^2 = .99999999999980000000000001
.99999999999999^2 = .9999999999999800000000000001
.999999999999999^2 = .999999999999998000000000000001
.9999999999999999^2 = .99999999999999980000000000000001
.99999999999999999^2 = .9999999999999999800000000000000001
.999999999999999999^2 = .999999999999999998000000000000000001
.9999999999999999999^2 = .99999999999999999980000000000000000001
.99999999999999999999^2 = .9999999999999999999800000000000000000001

Each position after the decimal point eventually becomes a "9" and stays
that way.

In sci.logic, Dave Seaman
dseaman@no.such.host
wrote

That might be good enough. Of course that's probably related
to the impossibility of reliably defining things such as

.999... - .999...99800...001 = .000...?!?!?!...

:-)

I wouldn't go so far as to say it's impossible. After all, nonstandard
analysis makes sense of infinitely large integers. But then, the
hyperintegers of NSA are not the same as the transfinite ordinals, and I
am not aware that anyone has tried to use digit strings indexed by the
hyperintegers to define values in the hyperreals.


NSA presumably routinely works with very large (1,024-bit)
integers, but they *are* integers; the challenge is mostly
in defining operations therewith in terms of arrays
of smaller integers, sort of like defining multidigit
arithmetic in terms of the "times table", except that
instead of a 10 × 10 matrix one can memorize by rote,
one has a 65536 × 65536 affair that is implemented by the
microprocessor's multiply instruction.

On Sat, 10 Jan 2004 20:57:46 GMT, The Ghost In The Machine wrote:
In sci.logic, Dave Seaman

Omega+1 has the same cardinality as omega, but they are different
ordinals. If the (w+1)-strings of digits are ordered lexicographically,
then there is no order-preserving map of such strings to the reals.


Interesting. So expressions such as omega+1 and 2*omega actually do
make sense? If so, mea culpa.

Yes. In fact, w is identical to N, the set of natural numbers (including 0),
and w+1 = w U {w} = { 0, 1, 2, 3, ..., w }.

Addition and multiplication are noncommutative:

1+w = w < w+1, (w+1 has a last element but 1+w does not)
2*w = w < w*2 = w+w. (an w of pairs vs. a pair of w's)

Ugh. My brain's beginning to hurt now... :-)

So which one of these applies to such Garry-Denke-isque decimal
expansions as

(.999....)^2 = .999...99800...001 ?

None of them, as far as I can see.

It feels like 8 is in the w+2 or w+3 position but I want to make sure.
1 is presumably in the position w*2 + 2.

That seems to be implied by the notation, but I'll resist any urge to try
to make sense of it.

Presumably 1+w refers to things like

9,(9,9,9,9,...)

where one tacks things onto the left side of an infinite sequence.
If one computes .999... / 10 + .8 = .8999..., one gets a similar
sequence.


However, if we do allow transfinites and infinitesimals into our
number system a la Garry Denke, I'm not sure how the rest of mathematics
is going to deal with it (the concept of limit in particular may have
to be redefined).

Hopefully it's a simple retrofit but I wonder.

There's nothing wrong with considering transfinite ordinal sequences of
digits in general, but as I said, there is no order-preserving map of
such objects to the reals for ordinals > w.


That might be good enough. Of course that's probably related
to the impossibility of reliably defining things such as

.999... - .999...99800...001 = .000...?!?!?!...

:-)

I wouldn't go so far as to say it's impossible. After all, nonstandard
analysis makes sense of infinitely large integers. But then, the
hyperintegers of NSA are not the same as the transfinite ordinals, and I
am not aware that anyone has tried to use digit strings indexed by the
hyperintegers to define values in the hyperreals.

panoptes@iquest.net (Daniel W. Johnson) wrote in message news:<1g79ml7.1fmle7h1f387w2N%panoptes@iquest.net>...
.9^2 = .81
.99^2 = .9801
.999^2 = .998001
.9999^2 = .99980001
.99999^2 = .9999800001
.999999^2 = .999998000001
.9999999^2 = .99999980000001
.99999999^2 = .9999999800000001
.999999999^2 = .999999998000000001
.9999999999^2 = .99999999980000000001
.99999999999^2 = .9999999999800000000001
.999999999999^2 = .999999999998000000000001
.9999999999999^2 = .99999999999980000000000001
.99999999999999^2 = .9999999999999800000000000001
.999999999999999^2 = .999999999999998000000000000001
.9999999999999999^2 = .99999999999999980000000000000001
.99999999999999999^2 = .9999999999999999800000000000000001
.999999999999999999^2 = .999999999999999998000000000000000001
.9999999999999999999^2 = .99999999999999999980000000000000000001
.99999999999999999999^2 = .9999999999999999999800000000000000000001

Each position after the decimal point eventually becomes a "9" and stays
that way.

Daniel W. Johnson has proven that 0.999... will never equal 1.
I knew someone here could do it.

Garry Denke, Geologist
Denoco Inc. of Texas

Daniel W. Johnson has proven that 0.999... will never equal 1.

On Sun, 11 Jan 2004 20:32:57 GMT, The Ghost In The Machine wrote:
In sci.logic, Dave Seaman
dseaman@no.such.host
wrote

That might be good enough. Of course that's probably related
to the impossibility of reliably defining things such as

.999... - .999...99800...001 = .000...?!?!?!...

:-)

I wouldn't go so far as to say it's impossible. After all, nonstandard
analysis makes sense of infinitely large integers. But then, the
hyperintegers of NSA are not the same as the transfinite ordinals, and I
am not aware that anyone has tried to use digit strings indexed by the
hyperintegers to define values in the hyperreals.


NSA presumably routinely works with very large (1,024-bit)
integers, but they *are* integers; the challenge is mostly
in defining operations therewith in terms of arrays
of smaller integers, sort of like defining multidigit
arithmetic in terms of the "times table", except that
instead of a 10 × 10 matrix one can memorize by rote,
one has a 65536 × 65536 affair that is implemented by the
microprocessor's multiply instruction.

1024 bits is not enough to hold a hyperinteger.

My "NSA" is nonstandard analysis. Is yours perhaps the National Security
Agency?

panoptes@iquest.net (Daniel W. Johnson) wrote in message news:<1g79ml7.1fmle7h1f387w2N%panoptes@iquest.net>...
.9^2 = .81
.99^2 = .9801
.999^2 = .998001
.9999^2 = .99980001
.99999^2 = .9999800001
.999999^2 = .999998000001
.9999999^2 = .99999980000001
.99999999^2 = .9999999800000001
.999999999^2 = .999999998000000001
.9999999999^2 = .99999999980000000001
.99999999999^2 = .9999999999800000000001
.999999999999^2 = .999999999998000000000001
.9999999999999^2 = .99999999999980000000000001
.99999999999999^2 = .9999999999999800000000000001
.999999999999999^2 = .999999999999998000000000000001
.9999999999999999^2 = .99999999999999980000000000000001
.99999999999999999^2 = .9999999999999999800000000000000001
.999999999999999999^2 = .999999999999999998000000000000000001
.9999999999999999999^2 = .99999999999999999980000000000000000001
.99999999999999999999^2 = .9999999999999999999800000000000000000001

Each position after the decimal point eventually becomes a "9" and stays
that way.

Daniel W. Johnson has proven that 0.999... will never equal 1.
I knew someone here could do it.