Left to right in arithmetic, right to left in math?

So you are suggesting that (.999...[infinite]...)^2
= .999...[infinite]...998000...[infinite]...001?

No, I'm cubing the subject "0.999..." to show the kids
that the subject "0.999... = 1" is a false statement.

The cube of .999... is not equal to 1,
therefore .999... is not equal to 1.

9^3 = 729
.9^3 = .729

99^3 = 970299
.99^3 = .970299

999^3 = 997002999
.999^3 = .997002999

9999^3 = 999700029999
.9999^3 = .999700029999

The subject "0.999... = 1" is a false statement.
Please discontinue lying to kids in textbooks.

Thank you.

Garry Denke, Geologist
Denoco Inc. of Texas

What is the square of .9999.... then?

Pick a vowel, because its not equal to 1. Neither is its cube.

The cube of .999... is not equal to 1,
therefore .999... is not equal to 1.

9^3 = 729
.9^3 = .729

99^3 = 970299
.99^3 = .970299

999^3 = 997002999
.999^3 = .997002999

* Tim Wouters
Correction:

0.999999.....99 < 0.999999.....995 < 0.999999.....999


But isn't 0.999999.....99 = 0.999999.....999=0.99...?
And thus 0.999999.....995 = 0.999999.....99 = 0.99...

No, because:

0.999...999 = nonsense <> 0.9999 = 1

there are no poxitions after infinitely many nines, and particularly,
there is no last one. Thus your "easy" is easily wrong.

there are no poxitions after infinitely many nines, and particularly,
there is no last one. Thus your "easy" is easily wrong.

Squares of 0.999... tend away from Last One

9^4 = 6561
.9^4 = .6561

99^4 = 96059601
.99^4 = .96059601

[snip rest for brevity]

Challenge: prove Last One wrong.

0.999...999 = nonsense <> 0.9999 = 1

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?

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.

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.

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.

Another problem are its squares, like 0.999...^4, which do not
tend towards 1
^^^^^^^^^^^^^^
as falsely stated in this thread, but do
tend away from 1
^^^^^^^^^^^^^^^^

.9^4 = .6561
.99^4 = .96059601
[snip]
.99999999999^4 = .99999999996000000000059999999999600000000001
.999999999999^4 = .999999999996000000000005999999999996000000000001

there are no poxitions after infinitely many nines, and particularly,
there is no last one. Thus your "easy" is easily wrong.

Squares of 0.999... tend away from Last One

.9^4 = .6561
[snip]
.999999999999^4 = .999999999996000000000005999999999996000000000001

[snip rest for brevity]

Challenge: prove Last One wrong.

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.

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

David C. Ullrich

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

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.

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.