1

# 0.99999.....=1

Tryagain

2
Thu 7 Nov, 2013 04:49 pm
David, it is good to see you have a sense of humor lurking within your empirical intellect. May the climate of your new domain be advantageous for your health.

Now guys; I have been called many things, such as, ‘Hey hillbilly, haul your fat ass off my lawn – and your truck too.’
Butt I have never been called a matherma wotsits.

Quote: “But infinitely many 9's after the decimal WILL get you to 1.”

I demure!

Do we agree that when you subtract a number from itself, the result is zero?

For instance, 4 – 4 = 0.

Good, I can see we iz going to getalong just fine.

So what is the result when you subtract 0.999... from 1?

For shorter subtractions, you get:

1.0 1.00 1.000 1.0000 1.00000
0.9 0.99 0.999 0.9999 0.99999
__ ___ ____ ______ _______
0.1 0.01 0.001 0.0001 0.00001

Then what about 1.000... – 0.999...?
You'll get an infinite string of zeroes. But you ask, "what about that '1' at the end?"

Ah, but 0.999... is an infinite decimal; there is no "end", and thus there is no "1 at the end". The zeroes go on forever. And 0.000... = 0.
Then 1 – 0.999... = 0.000... = 0, and 1 = 0.999....

But if two numbers are different, then you can fit another number between them, such as their average. Am I write or am I right!

But what number could you possibly fit between 0.999... and 1.000...?

1
Thu 7 Nov, 2013 07:42 pm
@Tryagain,
The fundamental question is whether 1/3 = .3333....
Kolyo

1
Thu 7 Nov, 2013 08:07 pm
@Tryagain,
[LOL at what precedes this except. ]

Tryagain wrote:

But if two numbers are different, then you can fit another number between them, such as their average. Am I write or am I right!

But what number could you possibly fit between 0.999... and 1.000...?

Yup. That's where the "0.999 < 1" crowd gets into trouble. If they are different then their average should lie between them. If they are real numbers, so is it. As a real number it should have some kind of decimal representation. But what would that be, to fall between 0.999... and 1.000... ?
Kolyo

1
Thu 7 Nov, 2013 08:37 pm
Pardon the digression...

The fundamental question is whether 1/3 = .3333....

If you make .999... the kind of hyperreal that would put it between 1 and any real number smaller than 1, then you should be consistent and make .333... the kind of hyperreal that would put it between 1/3 and any real number less than 1/3.

In other words, you should define .333... as (.3, ,33, .333, ...). It would then be less than 1/3 by an infinitesimal amount.

In hyperreals, you can actually add and subtract infinitesimals just like normal numbers. (1/30, 1/300, 1/3000, ...) is the hyperreal form of an infinitesimal, as is every sequence of reals that approaches zero in the limit.

There is one more thing about hyperreals that ... anyone who cares about them ... should know -- without which you can't really understand how they work. You know that one hyperreal is less than another if all except a finite number of the terms in the sequence of reals that represents the first hyperreal are smaller than the corresponding terms in the sequence of reals that represents the second hyperreal.

So for example (0.9, 0.99, 0.999, ...) < (1, 1, 1, ...),
because 0.9 < 1, 0.99 < 1, 0.999 < 1, etc.

But for any real number x < 1,
(x, x, x, ...) < (0.9, 0.99, 0.999, ...),
because eventually the terms in the second sequence overtake the terms in the first sequence for good as they rise above x.
So if x were, say, 0.999997, the first five terms in the first sequence would be greater than the corresponding terms in the second. But except for those five, all the terms in the second sequence would be larger than the corresponding terms in the first.

0 Replies

Kolyo

1
Thu 7 Nov, 2013 08:38 pm
@Kolyo,
Kolyo wrote:

[LOL at what precedes this except. ]

"except" should be "excerpt"
0 Replies

Kolyo

1
Thu 7 Nov, 2013 09:27 pm
@Kolyo,
Kolyo wrote:

That's where the "0.999 < 1" crowd gets into trouble.

Also, I meant the "0.999... < 1" crowd.
OmSigDAVID

1
Fri 8 Nov, 2013 12:41 am
OmSigDAVID wrote:
accountant.

Ah. You're discussing arithmetic. We're discussing math.
Not all math is arithmetic; all arithmetic is math.

David
0 Replies

OmSigDAVID

1
Fri 8 Nov, 2013 12:50 am
@Kolyo,
Kolyo wrote:

Kolyo wrote:

That's where the "0.999 < 1" crowd gets into trouble.

Also, I meant the "0.999... < 1" crowd.
is that ALMOST good enuf, but not quite IS good enuf
and shud not be challenged by Orthodox Tea Party types like ME.

Therefore, if an impoverished worker who curries your favor
leave work an hour early, then that is EXACTLY what he is supposed to DO
and it is perfect execution of his contract because O.99 = 1.

BULLoney! How many bridges r u selling ??

David
markr

1
Fri 8 Nov, 2013 02:15 am
@OmSigDAVID,
Why do you insist on using finite series in your arguments instead of the infinite series which is the topic of discussion? Nobody claimed that any finite number of nines to the right of the decimal equals one.
0 Replies

OmSigDAVID

1
Fri 8 Nov, 2013 02:57 am
Practical applications of mathematical evil purported in O.9999 = 1:

Defendant is arrested for using pliers to remove the first finger of victim
(without his consent) on a given Tuesday, but only up to its first knuckle.
It is stipulated by all counsel that as of Monday, the victim was left un-disturbed
and that he then had 1O fingers intact. The victim 's testimony is taken.
He inculpates the defendant. On cross-exam, he admits that he lost
only 1 of his fingers to defendant's violence, and even that finger
remains intact, except beyond its first knuckle.

Defense counsel qualifies an expert witness in evil liberal math,
who attests that O.9999 = 1.
Defense counsel argues in exculpation that .9 = 1 and indeed,
the victim still had MORE than
.9 of all of his fingers remaining intact. Since .9 + more of victim 's fingers
remain intact: ALL of his fingers were left alone, un-troubled and intact.
Hence, there was no crime.

An evil, liberal judge will dismiss the case, taking judicial notice
of the expert 's mathematical testimony and judicially declaring
that no harm was done and that the victim remains fully intact because .9 = 1.

Defendant is liberated; he leaves jail
and joining the ranks of recidivism, he finds new victims n he does it again,
repeating the cycle in criminal litigation.

David
markr

2
Fri 8 Nov, 2013 11:20 am
@OmSigDAVID,
Yet another finite example.
0 Replies

1
Mon 18 Nov, 2013 07:52 am
@OmSigDAVID,
OmSigDAVID wrote:
who attests that O.9999 = 1.

Except nobody here is attesting that. There are an infinity of numbers that fit between .9999 and 1.

0 Replies

OmSigDAVID

1
Mon 18 Nov, 2013 10:32 pm
The fundamental question is whether 1/3 = .3333....
The fundamental question is whether the decimal format
will enable u to express one-third, or not.

1
Wed 20 Nov, 2013 11:16 am
@OmSigDAVID,

I can express 1/3 in decimal notation just fine. Stick that in your vinculum and smoke it.

1
Wed 20 Nov, 2013 11:16 am

http://en.wikipedia.org/wiki/0.999%E2%80%A6

Quote:
The equality 0.999... = 1 has long been accepted by mathematicians and is part of general mathematical education. Nonetheless, some students find it sufficiently counterintuitive that they question or reject it, commonly enough that the difficulty of convincing them of the validity of this identity has been the subject of numerous studies in mathematics education.
OmSigDAVID

1
Wed 20 Nov, 2013 01:36 pm
That 's a lie; a liberal lie.
Its not MY problem. If that format is dysfunctional,
then I will not persist in beating a square peg into a round hole,
as u r obsessively-compulsively doing.

I can do this: 1/3.

I can express 1/3 in decimal notation just fine.
That 's another liberal lie; (DD contradicts himself)
"don 't look behind that curtain", DD screams. "Pay no attention to that charlatan back there."

David

1
Wed 20 Nov, 2013 01:41 pm
@OmSigDAVID,
OmSigDAVID wrote:
That 's another liberal lie

I see you've been reading 1984 again....

Oldthinker unbellyfeels newmath.
0 Replies

OmSigDAVID

1
Wed 20 Nov, 2013 02:06 pm
If that number is subtracted from 1,
then according to u, the result will be O, right, DD???

http://en.wikipedia.org/wiki/0.999%E2%80%A6

Quote:
The equality 0.999... = 1 has long been accepted by mathematicians and is part of general mathematical education. Nonetheless, some students find it sufficiently counterintuitive that they question or reject it, commonly enough that the difficulty of convincing them of the validity of this identity has been the subject of numerous studies in mathematics education.

Its not just "counterintuitive"; its false. 1 does not equal .9

David
raprap

1
Wed 20 Nov, 2013 08:36 pm
@OmSigDAVID,
Question David---If I'm standing ten feet from a wall and with every step I halve my distance to the wall, how many steps will be required to get to the wall?

Rap
tomr

1
Wed 20 Nov, 2013 11:20 pm
@raprap,
Question raprap -- If I cut a length of string up into pieces of roughly equal length shouldn't those pieces add up to the original length of the string?

.999... = lim n->∞ (1- (1/10^n))
.999... = lim n->∞ 1 - lim n->∞ (1/10^n)
.999... = 1 - lim n->∞ (1/10^n)

Now lets look at the term lim n->∞ (1/10^n). The limit L = 0 as n->∞. But if we cut up a string of length 1 into roughly equal lengths based on the expression (1/10^n) for finite values of n we always can sum the pieces up and get the total string length back. So if n= 1, I make 9 cuts in the string and each string length is 1/10 of the total and there are 10 pieces at that length. If n= 2 then there are 100 pieces with a length 1/100. This can be done for any value of n. Then we can sum the individual pieces to get the original length of the string. When you apply that logic as n->∞ a problem occurs. You start with a string of length 1, but the length of each cut by the limit laws must be zero [lim n->∞ (1/10^n) = 0], so when you sum up an infinite number of zero length pieces you get nothing for your total string length. So there is a contradiction in equating the quantity lim n->∞ (1/10^n) with zero I think.

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