Sorry, I didn't get your point. Do you mean you can't add a 9 at the end
of 0.99... because then you get 1?

You can't add a 9 at the end of 0.99.. because it has no end.

But you did add a 5 at the end of 0.99.. Why's that allowed? There ain't
no difference.

(To be sure: , but I remember seeing this "argument" before.)

...the squares are heading toward 0.

Garry Denke, Geologist
Denoco Inc. of Texas

Tim Wouters <tim@wina.be> wrote in message news:<Pine.LNX.4.56.0401081826520.31511@sponamia.kotnet.org>...
Sorry, I didn't get your point. Do you mean you can't add a 9 at the end
of 0.99... because then you get 1?

You can't add a 9 at the end of 0.99.. because it has no end.

But you did add a 5 at the end of 0.99.. Why's that allowed? There ain't
no difference.

If you read carefully, he added a '5' at the end of an arbitrary, but
FINITE, sequence of '9'. The sequence 0.999... has no end. karl m

David C. Ullrich <ullrich@math.okstate.edu> wrote
On 8 Jan 2004 03:35:49 -0800, chvol@aol.com (Charlie-Boo) wrote:

In this case there _is_ a _completely_ _standard_ definition of what
these symbols mean already. An author is not required to restate
every standard definition he uses. But the definition of what these
symbols mean _has_ been given in this thread, several times by
several people.

A number is not a character string - this is not quibbling, this is
the whole point. One more time: .999... is not a character string,
it is a number. It is _true_ that a real number can have more than
one representation as a character string. So what? We're not talking
about what the notation _should_ mean, we're talking about what it _does_
mean. .999... is not a symbol at all, it is a number. ".999..." is a symbol
for this number. It _is_ a symbol for the _sum_ of a certain
infinite series. That means that it _is_ a symbol for number.

With all due respect, you completely miss my point. I don't know how
plainly I can state it. The problem or question isn't what the
"standard" meaning of .999… or ".999…" is or should be. The problem
is that in the "proof" given that .999…=1, the .999… is used in two
different ways: as a number and as an infinite series with a limit of
1.

Those are the two cases that I listed. I am not asking which is
correct or standard. I am simply saying that the "proof" to which I
responded is inconsistent as to which it is, with adverse
consequences.

If .999… and 1.000… are numbers in the "proof", then two different
numbers are equal and there is no longer a unique representation for a
number.

(And I said that I'm talking about the literal
representation, not expressions in general. I'm not denying that
1+1=2. My gawd.)

If .999… and 1.000… are infinite series in the "proof", then the
references to subtracting one from the other, and looking for a number
inbetween them, don't make any sense.

Those are clearly references to
the 9's as being digits of a real number.

Enough time has been spent on this high school math exercise.

Google
Groups has much more interesting topics than this, which is where I
will now turn my attention (if anyone is looking for me.)

Charlie Volkstorf

************************

David C. Ullrich

David C. Ullrich <ullrich@math.okstate.edu> wrote:
And a related question: If I wanted to understand Baroque
music, I should go to school and study astrophysics, right?

Rather, that's what you should study if you want to understand the Music
of the Spheres.

With all due respect, you completely miss my point. I don't know how
plainly I can state it. The problem or question isn't what the
"standard" meaning of .999… or ".999…" is or should be. The problem
is that in the "proof" given that .999…=1, the .999… is used in two
different ways: as a number and as an infinite series with a limit of
1.

If .999... and 1.000... are numbers in the "proof", then two different
numbers are equal and there is no longer a unique representation for a
number.

Rather, that's what you should study if you want to understand the Music
of the Spheres.

Just started a blog on this:
Music of the Spheres weblog: http://esolutionswork.com/Designer/

So what you're saying is the sequence .9^2, .9^3, .9^4 ...

.9^2 = .81
.9^3 = .729
.9^4 = .6561
etc

the sequence .99^2, .99^3, .99^4 ...

.99^2 = .9801
.99^3 = .970299
.99^4 = .96059601
etc

the sequence .999^2, .999^3, .999^4 ...

.999^2 = .998001
.999^3 = .997002999
.999^4 = .996005996001
etc

tend to 1 making the sequence .999...^2, .999...^3, .999...^4 tend to 1.

Is that what you are saying?

Garry Denke, Geologist
Denoco Inc. of Texas

chvol@aol.com (Charlie-Boo) writes:

With all due respect, you completely miss my point. I don't know how
plainly I can state it. The problem or question isn't what the
"standard" meaning of .999... or ".999..." is or should be. The problem
is that in the "proof" given that .999...=1, the .999... is used in two
different ways: as a number and as an infinite series with a limit of
1.

Then your point is wrong, wrong, wrong.

.999.... is a number. Namely, it is the sum of the series which has
been repeatedly mentioned here. It is *not* the series itself, but
the sum of the series. That sum is a real number, so there is no
contradiction at all in claiming

(1) .999... is a real number;
(2) .999... is the sum of the infinite series;
(3) .999... is not an infinite series.

My comment was addressing the "proof" that was given early in this
thread - nothing else. It has nothing to do with what other
mathematicians say or do or use for syntax and semantics or anything
else. It is completely self-contained in the "proof" on which I was
commenting.

Personally, I think part of the problem is the plethora of people who
live in the world where the published paper is the altar at which they
worship and its author is their messiah. All they know how to do is
to quote from their scripture.

There are two ways that he treats .999...

# 1. As a literal number. (To anyone who harps about my use of the
word literal: Get a life.) By literal I mean how you write down a
number - not a formula or a function or any operators. Just a number.
.999... means the decimal number that has 9 for the first digit after
the decimal point, and 9 for the second digit, and 9 for the third
digit, etc. etc. etc. Ok? Like 12.345 but only longer. That is one
interpretation and use of .999...

Under this interpretation, we can treat .999... as a number with which
we can calculate directly using the arithmetic operators. We can
divide it by 3. 3 goes into 9 3 times. 3 times 3 is 9. 9 from 9 is
0. Bring down the next 9. The quotient is .333...

# 2. As a symbol for an infinite series. In this case, the 9's don't
mean anything. It could just as well be XQZ. It just means 9/10 +
9/100 + 9/1000 + . . . . And please, drop the stupid obvious
statements like "that's the same thing as .999..." or "well that's equal
to 1" and "1 is a number". We know that. That's 1st grade
mathematics. That's not the point. The point is how the author is
using .999... in the "proof". He is simply manipulating .999... as an
infinite series the same as if it were XQZ. It's being treated as a
symbol that represents that series. He is not taking a number
consisting of a decimal point followed by all 9's and performing
arithmetic on them as in case # 1 above.

He writes:

Other person: 0.999.... does NOT equal 1.

Faulty proof: It certainly does! Just try to subtract 0.999... from
1: 1 - 0.999... = 0 Reason: There is no real number between
0.999... and 1, and, therefore, "they" must be one and the same
number!

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".

In sci.logic, Dave Seaman wrote:
[re: ordinal numbers]
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.

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...?!?!?!...

The Ghost In The Machine <ewill@sirius.athghost7038suus.net> wrote:
In sci.logic, David W Cantrell
DWCantrell@sigmaxi.org
wrote
on 08 Jan 2004 05:54:31 GMT
20040108005431.441$aW@newsreader.com>:
The Ghost In The Machine <ewill@sirius.athghost7038suus.net> wrote:
[snip]
The standard question for equations such as x^2 = x is:

Give the set of values that x can be set to, such that the
equation holds.

This set is {0, 1}; both 0 and 1 satisfy the equation.
It's a little harder to show that nothing else will,
though one can try algebra and get lucky in this case:

x^2 = x
x^2 - x = 0
x(x - 1) = 0

In an infinite field such as the reals, it's now obvious that
0 and 1 are the only solutions. (In the ring of integers
mod 6, though, things get interesting, as the solution set
is {0,1,3,4}.}

FWIW, the original equation also has another solution if our number
system is the extended reals. Then the solution set is {0, 1, +oo}.

Good point. :-)

Is that the best you can say? Many mathematicians call the point +oo
"ideal", which makes it sound far better than merely "good"!

It might also be noted that x = +oo is not a solution of
either x^2 - x = 0 or x(x - 1) = 0. (But that shouldn't be surprising.
After all, the extended reals don't form a field.)

David