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Dragon Star
2006-Jan-06, 12:12 AM
1,000th post, CLOSE IT NOW!!!!

Taks
2006-Jan-06, 12:13 AM
You still have an infinite series, which is what confuses people. Some also seem to confuse infinite series with infinite numbers, though, so it's a good thing you pointed out the distinction between the two.agreed. i realized the confusion after your last post.

Do I get to have the last word? :Dnever. :)

taks

ToSeek
2006-Jan-06, 01:30 AM
1,000th post, CLOSE IT NOW!!!!

Oops, too late! (Ironically, if you hadn't posted this, we would have been at exactly 1,000 now. ;) )

Dragon Star
2006-Jan-06, 01:42 AM
Oops, too late! (Ironically, if you hadn't posted this, we would have been at exactly 1,000 now. ;) )

What? We are at 1002, not 1001, nice try to put this all on me!:naughty: ;)

pst...Close it anyways!

umop ap!sdn
2006-Jan-06, 02:09 AM
Infinity possibly being the amount of time until my next move in our chess game at Scotson's Shack...
I'm gonna have to write down my next few moves just so as I can remember my strategy. :D

But .999~ is not an infinite number, is it? I mean, it's right there between 0 and 1.
I meant adding or subtracting from the quantity of nines.

See if you can close it with the 1000th post :)
That was suggested with a similar thread (albeit on a different subject) at another board... they ended up closing it a little late. :D

Monique
2006-Jan-06, 02:23 AM
Thread is joke, yes?? Proof is well known!!!

HenrikOlsen
2006-Jan-06, 03:06 AM
What? We are at 1002, not 1001, nice try to put this all on me!:naughty: ;)

pst...Close it anyways!
Your post was #1001, putting it over.
Oh well, just close it at 2000 then

Thomas(believer)
2006-Jan-06, 04:43 AM
again, it is proved. do the math yourself. long division, 9 into 1...

that you would "choose" not to believe simply means you choose not communicate in the accepted mathematical language the rest of us are using.

except that pi is not exact. at least, we cannot describe it exactly in numerical terms. the complete stream of digits of pi are unknown as it is a transcendental number.

taks
These are my points too!

Can you proove that 0.0(~0's)1 <> 0.00~ ?
Is it not mathematical convention (or axiom maybe?) to say, that something times an infinitisimal small number is zero?

umop ap!sdn
2006-Jan-06, 04:47 AM
Can you proove that 0.0(~0's)1 <> 0.00~ ?
There cannot be any such number as .000~1, because the ~ means the previous digits repeat to infinity. Since infinity has no end, the 1 is never tacked on and we are left with zero.

Otherwise, it would be easy to say that .999~ + .000~1 = 1. Since .000~1 is meaningless, it just doesn't hold up. :)

Taks
2006-Jan-06, 07:28 AM
Can you proove that 0.0(~0's)1 <> 0.00~ ?it is proved within the statement itself. 0.000~ is 0, not approaching zero, not almost zero, but zero. put a 1 on the end and it is greater than zero by definition.

Is it not mathematical convention (or axiom maybe?) to say, that something times an infinitisimal small number is zero?
Thanks for your repliesinfinitesimal is essentially zero, but not quite. different concept.

taks

snarkophilus
2006-Jan-06, 07:57 AM
it is proved within the statement itself. 0.000~ is 0, not approaching zero, not almost zero, but zero. put a 1 on the end and it is greater than zero by definition.

You can't put a one on the end. Like the world sitting on a turtle sitting on a turtle ~, it's turtles all the way down.

It's a countable sequence, right? So what's the position of the 1? It must be a finite number, if there is one. So what real number can you add to 0 to get 0.0~1? There isn't one, so 0.0~1 isn't a real number (because the reals are closed under addition).

tofu
2006-Jan-06, 04:16 PM
http://seaofcrisis.com/ext/babb/irony.jpg

I grabbed a screenshot way back when. I guess the irony of this is lost on some people.

JohnW
2006-Jan-06, 04:19 PM
Thread is joke, yes?? Proof is well known!!!
Sadly, Monique, it's not a joke. Depressing, isn't it?

Moose
2006-Jan-06, 04:19 PM
Re Tofu's screenshot: Oh yeah. Now that's a moment of Zen. :)

Lance
2006-Jan-06, 04:29 PM
Question:

Accepting that .999~ = 1

Does:

.8999~ = .999~ ?

If not, why not?

Moose
2006-Jan-06, 04:31 PM
Because .9999~ - .8999~ = .1

SeanF
2006-Jan-06, 04:35 PM
Question:

Accepting that .999~ = 1

Does:

.8999~ = .999~ ?

If not, why not?
What moose said.

However, it is true that .8999~ = .9

Taks
2006-Jan-06, 04:35 PM
You can't put a one on the end. Like the world sitting on a turtle sitting on a turtle ~, it's turtles all the way down.uh, not sure if you noticed, but I didn't put a 1 on the end... the believer did. if you meant him, you should have quoted him.

taks

Lance
2006-Jan-06, 04:35 PM
Okay, so:

.8999~ = .9 ?

That makes sense.

Dumb original question.

Double post.

Yeah, I get it. Thanks.

Thomas(believer)
2006-Jan-06, 05:50 PM
You can't put a one on the end.

But I did put it there. (and sqeeuzed an infinite number of zero's in between) :razz:

HenrikOlsen
2006-Jan-06, 06:05 PM
But I did put it there. (and sqeeuzed an infinite number of zero's in between) :razz:
You didn't, what you did was to write a malformed expression, which was just as meaningless as asking what "3+7-" is.

peteshimmon
2006-Jan-06, 11:10 PM
You can specify a one,one,root two triangle
as irrational but get ever near with integer
values hundreds, thousands, millions of figures
long. Pointless thinking about it what?

Taks
2006-Jan-07, 12:33 AM
But I did put it there. (and sqeeuzed an infinite number of zero's in between) :razz:snarkophilus' point is that your notation does not make mathematical sense. i.e. there can be no end to an infinite series (it is infinite) therefore there can be no "1 on the end"! if there were a "1 on the end" then that would imply an end, which would imply a less than infinite series.

taks

Taks
2006-Jan-07, 12:35 AM
btw, i forgot to mention this a bit ago but...

woohoo for me being the sneaky 1000th poster! :)

taks

montebianco
2006-Jan-07, 02:08 AM
Here's one. (http://www.bautforum.com/showpost.php?p=371396&postcount=280)

How barbaric!

(Or you can always cheat, and use a base greater than 10. ;))

:D

Dragon Star
2006-Jan-07, 03:36 AM
Your post was #1001, putting it over.
Oh well, just close it at 2000 then

I don't get what your saying, Taks jumped ahead of ToSeek, so it would have been at 1,001 for Taks.

Don't blame this on me...:hand:

Chuck
2006-Jan-07, 04:25 AM
I hate it too, but it must continue until everyone is in agreement.

Dragon Star
2006-Jan-07, 04:29 AM
I hate it too, but it must continue until everyone is in agreement.

:cry:

"The thread that crashed the Internet in 2006":rolleyes:

Taks
2006-Jan-07, 08:38 AM
is this where we start posting evil, maniacal laughter? :)

taks

Thomas(believer)
2006-Jan-07, 08:54 AM
I hope it never closes, or when the number of replies reaches ~ .

Thomas(believer)
2006-Jan-07, 07:25 PM
I'm sorry to bother you again. But I got myself into trouble.
Again, I have no problems with the fact that 0.999~ equals 1.

But my own joke got myself into trouble.

Suppose I want to create a real number as follows:

0.99(a lot of 9's)8

I don't think there is something mathematical wrong with that.
As some people pointed out the number of 9's cannot reach infinity because there is that 8 (or any other sequence, except a sequence of 9's in the end)
Therefor this sequence can get as close to 1 as I like but never can equal 1.
For this sequence to become an infinite sequence of 9's the 8 would have to turn into a 9 somewhere. But now I have the following problem. If I follow this reasoning then the last 9 in the sequence of 9's should turn into a 0 somewhere.

I hope someone understands what I am trying to say here and can explain to me where my reasoning goes wrong.

Thanks.

montebianco
2006-Jan-07, 07:31 PM
I'm sorry to bother you again. But I got myself into trouble.
Again, I have no problems with the fact that 0.999~ equals 1.

But my own joke got myself into trouble.

Suppose I want to create a real number as follows:

0.99(a lot of 9's)8

I don't think there is something mathematical wrong with that.
As some people pointed out the number of 9's cannot reach infinity because there is that 8 (or any other sequence, except a sequence of 9's in the end)
Therefor this sequence can get as close to 1 as I like but never can equal 1.
For this sequence to become an infinite sequence of 9's the 8 would have to turn into a 9 somewhere. But now I have the following problem. If I follow this reasoning then the last 9 in the sequence of 9's should turn into a 0 somewhere.

I hope someone understands what I am trying to say here and can explain to me where my reasoning goes wrong.

Thanks.

If what you mean is that the sequence

1) 0.998
2) 0.9998
3) 0.99998
4) 0.999998

and so on, approaches one, that is correct. About the 8 turning into a 9, note that from 1) to 2), the 8 turned into a 9 and a new 8 was added at the end. The same thing happens as 2) changes to 3), as 3) changes to 4), and so on...

Moose
2006-Jan-07, 07:36 PM
Suppose I want to create a real number as follows:

0.99(a lot of 9's)8

Sorry, no, your reasoning isn't right.

You can have lots and lots of nines, then an eight. You can have as many nines as you want, and trail it with any digit you want. You simply cannot have an infinite number of nines with a terminating digit. Infinite means infinite. There is no terminating digit. Pretty much everybody who'd choked on the 0.999~ = 1 thing had trouble understanding the scope of infinity.

The moment you bring in a trailing digit, it's not an infinite string. It's finite, and it behaves just like any other finite number.

That said, sticking a (lots and lots) in the middle of the finite number makes it finite but indeterminate, and therefore essentially useless.

Is 0.999(lots)8 greater than, equal, or less than 0.999(many)8?

I (really) hope this helps clarify things for you.

Skipjack
2006-Jan-07, 07:42 PM
Well, I voted for No.
It is infinitely close to 1, but then still not 1.
For the same reason the largest number you can think of, is just as far away from infinity as is the number 1.
CU
Skipjack

cjl
2006-Jan-07, 07:51 PM
That is incorrect. You cannot have a number that is infinitely close to one that is not one. There is either a finite separation, or there is no finite separation. In the case of .9~, there is no finite separation, and therefore it is one. This proof has probably been shown before in this thread (I didn't feel like digging through all of it), but this can be shown with a simple method:
let's define x = .999~
therefore 10x = 9.999~
subtract x to get 9x = 9
therefore x = 1

Thomas(believer)
2006-Jan-07, 07:55 PM
A more mathematical writing for my sequence is:

9x10^-1 + 9x10^-2 + .... + 9x10^-n + 8x10^-(n+1)

I don't see something wrong with this mathematical representation and I have the feeling it converges to 1 as n goes to infinity.
I think my problem at the moment is that we represent numbers with digits and that you always have to make a discrete step to go to the next number, intuitivaly this violates continuity.

replaced converts to converges. Excuse me, for my bad english

Skipjack
2006-Jan-07, 08:02 PM
Ok, thats a good point and I understand that.
hmmm...

Moose
2006-Jan-07, 08:46 PM
And in any case, an empirical demonstration that's easy enough to teach to middle school kids:

1/9 = .111~
2/9 = .222~
3/9 = .333~
4/9 = .444~
5/9 = .555~
6/9 = .666~
7/9 = .777~
8/9 = .888~
9/9 = .999~

And what does 9/9 also equal? There's no sleight of hand up there.

montebianco
2006-Jan-07, 08:49 PM
A more mathematical writing for my sequence is:

9x10^-1 + 9x10^-2 + .... + 9x10^-n + 8x10^-(n+1)

I don't see something wrong with this mathematical representation

I don't either.

and I have the feeling it converges to 1 as n goes to infinity.

And your feeling is correct :D To put it differently, 1-<this number> is 2x10^-(n+1), and this goes to zero as n goes to infinity

I think my problem at the moment is that we represent numbers with digits and that you always have to make a discrete step to go to the next number, intuitivaly this violates continuity.

I am not quite sure what the problem is. Perhaps it is a matter of distinguishing between numbers and the symbols that represent those numbers. Real numbers are defined as limits of sequences of rational numbers. There are many different sequences of rational numbers that converge to the same real number. For example:

0.9
0.99
0.999
0.9999
0.99999
...

1.1
1.01
1.001
1.0001
1.00001
...

0.8
0.98
0.998
0.9998
0.99998
...

are three different sequences of rational numbers that all converge to the same real number (that is, one, despite what a large number of people here seem to think). When we write something like 0.999~, that is a symbol representing the first sequence (or really, the limit of the first sequence). There is no conventional symbol used to represent the other two. I suppose we could write 1.000~1 and 0.999~8 to represent the latter two sequences; nothing particularly wrong with that, it's just not standard usage...

Moose
2006-Jan-07, 08:56 PM
I suppose we could write 1.000~1 and 0.999~8 to represent the latter two sequences; nothing particularly wrong with that, it's just not standard usage...

And it's ambiguous to boot.

Take 1.000~1 and 1.000~2. Which is greater? You don't know.

To represent this, you'd be better off using a sum:

1 + (1 x 10^-H) > 1 + (2 x 10^-(H+1) )

And in this way, you know what value they represent and can actually do arithmetic and meaningful comparisons on them.

montebianco
2006-Jan-07, 09:06 PM
And it's ambiguous to boot.

Take 1.000~1 and 1.000~2. Which is greater? You don't know.

Well, with the definition I had in mind, they'd be equal, since both are equal to one.

To represent this, you'd be better off using a sum:

1 + (1 x 10^-H) > 1 + (2 x 10^-(H+1) )

And in this way, you know what value they represent and can actually do arithmetic and meaningful comparisons on them.

Well, this is certainly what I would do to represent a particular number for a specific value of H (or just right the correct number of zeroes). I'm just thinking there could be a reasonable way to assing meaning to symbols like 1.000~1, specifically, the sequence in my earlier post. Of course, the limit is still equal to one, but now someone could start a thread with the poll, "Is 1.000~1 exactly equal to one" and about 40% of responders could get it wrong...

N

Moose
2006-Jan-07, 09:10 PM
Well, with the definition I had in mind, they'd be equal, since both are equal to one.

What? No they're not. That's an ambiguously long yet finite series. And they're both definitely greater than one.

The reason 0.999~ = 1 is because 0.999~ is an infinite series. That's a property 1.000[lots]1 doesn't share.

Edit to add: Proof: 1.000[lots]1 > 1.000[lots]05 > 1 where [lots] determines an arbitrarily long, yet identical string of zeroes.

[Edit again:] And so long as you have a terminating digit, I can split the difference.

montebianco
2006-Jan-07, 09:17 PM
What? No they're not. That's an ambiguously long yet finite number. And they're both definitely greater than one.

If you use the definition I proposed, they are both equal to one. If you use some other definition, then maybe they are not.

The reason 0.999~ = 1 is because 0.999~ is an infinite series. That's a property 1.000[lots]1 doesn't share.

If you identify a symbol such as 1.0~1 with the infinite sequence:

1.1
1.01
1.001
1.0001
...

then it has a limit of one. Again, if you use some definition different than the one I proposed, then you may come to a different answer...

Lance
2006-Jan-07, 09:21 PM
And in any case, an empirical demonstration that's easy enough to teach to middle school kids:

1/9 = .111~
2/9 = .222~
3/9 = .333~
4/9 = .444~
5/9 = .555~
6/9 = .666~
7/9 = .777~
8/9 = .888~
9/9 = .999~

And what does 9/9 also equal? There's no sleight of hand up there.
Not to further the argument, but this proof never did work for me because it's a circular argument.

In order to believe that .999~ = 1, you have to also believe that 1/9=.111~. But as I was taught way back on grade school (60s), .333~ is the closest decimal representation of 1/3 but it can never be exactly equal.

Just my 2 cents. Carry on.

(Not meant to infer that I doubt the other proofs. It's just this one that is circular.)

Moose
2006-Jan-07, 09:24 PM
Montebianco: Hmm. Yeah, okay, I could see that, but you could never get away from the limit notation and have it function. 1.000~1 cannot logically exist, as it necessarily remains a finite value. But the limit of the series you describe, sure, that's one.

Moose
2006-Jan-07, 09:28 PM
In order to believe that .999~ = 1, you have to also believe that 1/9=.111~. But as I was taught way back on grade school (60s), .333~ is the closest decimal representation of 1/3 but it can never be exactly equal.

And I learned in grade school that 1/3 = 0.333~ exactly. *shrug* Like grade school teachers have never been wrong before.

It's not a circular argument at all. If there is an infinite string of threes after the decimal, it's exactly equal to 1/3. If there's a finite string of threes, then it's an approximation.

0.333~ denotes an infinite series. Not a finite one.

Thomas(believer)
2006-Jan-07, 09:29 PM
1 = 1+{sum(0x10-h)h>1} < 1+{sum(0x10^-h)h>1}+1*10^-(h+1)

or

1 <1+10^(-h+1)

but 10^-h is exactly 0 as h goes to infinity????

and then 1 = 1.0000~1

Why am I doing this to myself?

montebianco
2006-Jan-07, 09:30 PM
Not to further the argument, but this proof never did work for me because it's a circular argument.

In order to believe that .999~ = 1, you have to also believe that 1/9=.111~. But as I was taught way back on grade school (60s), .333~ is the closest decimal representation of 1/3 but it can never be exactly equal.

Just my 2 cents. Carry on.

(Not meant to infer that I doubt the other proofs. It's just this one that is circular.)

Certainly I agree that most of the "proofs" offered are not proofs at all, although since many of the no-people are running on intuition only, intuition may work better as a persuasive tool than rigorous argument.

I think some of the confusion here is for sequences vs. limits of those sequences. The two sequences:

0.9
0.99
0.999
0.9999
0.99999
...

and

1.0
1.00
1.000
1.0000
1.00000

are obviously different sequences, but they each tend to the same limit (I'm not going to babble on about Cauchy sequences, probably nobody wan't to hear that...). The standard definition of real numbers identifies all sequences which approach the same limit, even if the sequences themselves are different...

Lance
2006-Jan-07, 09:32 PM
If there is an infinite string of threes after the decimal, it's exactly equal to 1/3.

That's kind of my point...

You have to believe that before .999~ can equal 1.

It is a circular argument, not a proof.

montebianco
2006-Jan-07, 09:36 PM
1 = 1+{sum(0x10-h)h>1} < 1+{sum(0x10^-h)h>1}+1*10^-(h+1)

or

1 <1+10^(-h+1)

All looks good to me for any real value of h, although I think you have a missing "^" above.

but 10^-h is exactly 0 as h goes to infinity????

Well, the mathematically precise way of saying it is the limit of 10^-h is exactly 0 as h goes to (positive) infinity.

and then 1 = 1.0000~1

It depends on how you define 1.0000~1, which isn't exactly standard mathematical notation. If you define it as I did above, the limit of the sequence

1.1
1.01
1.001
1.0001
1.00001
...

then, yes, the limit of this sequence is equal to one.

Why am I doing this to myself?

montebianco
2006-Jan-07, 09:41 PM
That's kind of my point...

You have to believe that before .999~ can equal 1.

It is a circular argument, not a proof.

Lance, I agree with your point here. (Again, though, the argument might be persuasive even if it is not rigorous :D)

It seems to me this proceeds in two steps. The first would be, what is the definition of the symbol 0.999~? Moose appears to prefer an infinite sum formulation, 0+9/10+9/100+9/1000+9/10000+..., whereas I use a sequence formulation (the limit of 0, 0.9, 0.99, 0.999, 0.9999,...). The second step is, what is the limit of the sequence/sum (whichever formulation you prefer)? If someone answers the second question as "something different than one" then they are using a different definition of a limit than I am, although none of the people who argued thusly seemed inclined to state precisely what that definition was...

N

Roy Batty
2006-Jan-07, 09:41 PM
You know, if only a 3rd option on the original poll had said 'As near as damn it'
I think this thread might not as gone as long... apologies if this has said before but I could'nt be bothered to re-read it all again :doh: :)

Moose
2006-Jan-07, 09:41 PM
That's kind of my point...

You have to believe that before .999~ can equal 1.

It is a circular argument, not a proof.

Ah. I see what you're getting at. Still, 1/3 = .333~ or 1/9 = .111~ is an easier intuitive stretch for most people, which is useful considering the problem is that people are getting tripped up by their intuition to cope with this problem.

Some (most) of the formal proofs are somewhat beyond my ability to do more than scratch my head over. 1/3 = .333~ was the first one I understood, and the one that instantly persuaded me that 1 really must be equal to 0.999~, despite my intuitive mistake. (I did come to understand some of the other proofs and demonstrations, but that took a bit longer.)

Thomas(believer)
2006-Jan-07, 10:07 PM
1/13 = 0,076923076923(~076923) sequence ends always on 3

1/13 = 0,0769230769(~230769) sequence ends always on 9

Well, what can I say? Digits are ugly things to represent continuum.

montebianco
2006-Jan-07, 11:20 PM
You know, if only a 3rd option on the original poll had said 'As near as damn it'
I think this thread might not as gone as long... apologies if this has said before but I could'nt be bothered to re-read it all again :doh: :)

Perhaps it would not have gone on so long, but I don't need a third option - either two numbers are equal, or they are not equal.

montebianco
2006-Jan-07, 11:31 PM
Some (most) of the formal proofs are somewhat beyond my ability to do more than scratch my head over. 1/3 = .333~ was the first one I understood, and the one that instantly persuaded me that 1 really must be equal to 0.999~, despite my intuitive mistake. (I did come to understand some of the other proofs and demonstrations, but that took a bit longer.)

I think an issue is that many in this thread treat numbers, which are creations of a system of axioms, as if they real objects that can be examined empirically. 0.999~ equals one because we define the symbol 0.999~ to be the limit of 0, 0.9, 0.99, 0.999, 0.9999, ... (or you seem to prefer the infinite sum method, which is equivalent), and from the definition of a limit, that is equal to one. Now, our system of mathematics is the way it is because it has been found to be useful. But when you get down to it, the two numbers are equal because we define them that way.

Certainly, it would be possible to create a different mathematical system, and some in this thread have pointed out systems of mathematics in which there do exist infinitesmal numbers. I'm sure all of the no-people said "no" because they have understood the standard definition of the real numbers, have carefully pondered its implications, and decided that some other mathematical system is better, and not because they've never studied basic analysis :D

Thomas(believer)
2006-Jan-08, 07:55 AM
I have possible proof for inequality. I understand this is very controversial and accept the consequences. :D

Imagine this sequence 1.00~ . To get a higher number, I have to change one of the zero's onto a 1. The problem is that I really need to do this at some moment else the number won't change. But by doing this, I need to break infinity.
The next higher number is always something like: 1.( n 0's)1(00~). Where n is countable and not infinite. That's why I believe our number system is not good for describing a continuum.
Now take the sequence 0.999~. I can follow the same logic here.
The next higher number will look something like: 1.(n+1 0's)999~
For 1.000 and 0.999~ to be equal 1.(n 0's)1(00~)-1.(n+1 0's)999~ needs to equal zero. Or 0.(n+1 0's)1(000~) needs to equal zero.
But n is countable and not infinite. Therefor the difference is not equal to zero. So 1.000<>0.999~ QED???

It has been 17 years ago, I learned about Cauchy series and I have lost the details of this theory.
I think it defines that digit's which are placed on a position in inifinty behind the comma (I think that's dot in american system) are insignificant and are treated the same. i.e. they get blurred to make continuum possible.

Please correct me if I'm wrong.

Candy
2006-Jan-08, 08:08 AM
You will be crushed.

Thomas(believer)
2006-Jan-08, 08:12 AM
You will be crushed.
Thanks for the warning.

Thomas(believer)
2006-Jan-08, 12:42 PM
I have possible proof for inequality. I understand this is very controversial and accept the consequences. :D

Imagine this sequence 1.00~ . To get a higher number, I have to change one of the zero's onto a 1. The problem is that I really need to do this at some moment else the number won't change. But by doing this, I need to break infinity.
The next higher number is always something like: 1.( n 0's)1(00~). Where n is countable and not infinite. That's why I believe our number system is not good for describing a continuum.
Now take the sequence 0.999~. I can follow the same logic here.
The next higher number will look something like: 1.(n+1 0's)999~
For 1.000 and 0.999~ to be equal 1.(n 0's)1(00~)-1.(n+1 0's)999~ needs to equal zero. Or 0.(n+1 0's)1(000~) needs to equal zero.
But n is countable and not infinite. Therefor the difference is not equal to zero. So 1.000<>0.999~ QED???

It has been 17 years ago, I learned about Cauchy series and I have lost the details of this theory.
I think it defines that digit's which are placed on a position in inifinty behind the comma (I think that's dot in american system) are insignificant and are treated the same. i.e. they get blurred to make continuum possible.

Please correct me if I'm wrong.
:wall: I have made mistake. Ah, was middle in the night euraka feeling.
The difference between the two numbers above is off course 0.(~0)1
But if this number has no meaning does that mean it equals zero or is it blur?

montebianco
2006-Jan-08, 01:22 PM
I have possible proof for inequality. I understand this is very controversial and accept the consequences. :D

It's not clear to me why it is controversial. If people are familiar with the real number system, then they know the answer. If they are using their own systems of mathematics, well, that's up to them...

Imagine this sequence 1.00~ . To get a higher number, I have to change one of the zero's onto a 1. The problem is that I really need to do this at some moment else the number won't change. But by doing this, I need to break infinity.
The next higher number is always something like: 1.( n 0's)1(00~).

Well, that's a higher number. It's not the next higher number (the next higher number doesn't exist, you can always find one between the two).

Where n is countable and not infinite.

I would say that n is finite. A countable set is one which has a one-to-one correspondence with the set of integers.

That's why I believe our number system is not good for describing a continuum.

If you like, but I'm not sure how else to do it...

Now take the sequence 0.999~. I can follow the same logic here.
The next higher number will look something like: 1.(n+1 0's)999~
For 1.000 and 0.999~ to be equal 1.(n 0's)1(00~)-1.(n+1 0's)999~ needs to equal zero. Or 0.(n+1 0's)1(000~) needs to equal zero.
But n is countable and not infinite. Therefor the difference is not equal to zero. So 1.000<>0.999~ QED???

The difference between the two numbers is not zero for finite n, certainly. 0.999 is not equal to 1.001. The original poll question does not ask if 0.<some finite number of 9s> is equal to one; the answer to that would be "no." It asks if 0.999~ is equal to one. This number is defined by the limit. The limit of the difference above, as n goes to positive infinity, is zero.

It has been 17 years ago, I learned about Cauchy series and I have lost the details of this theory.

Loosely speaking, a Cauchy sequence is any sequence such that the elements get closer and closer together. Specifically, for any number d>0, however small, there must be some point in the sequence such that all later numbers differ from each other by no more than d.

I think it defines that digit's which are placed on a position in inifinty behind the comma (I think that's dot in american system) are insignificant and are treated the same. i.e. they get blurred to make continuum possible.

Symbols such as 0.999~1 don't have any meaning in conventional terminology. I took my stab at defining such a symbol earlier. By my definition, 0.999~1 would be the limit of the sequence:

0.9991
0.99991
0.999991
0.9999991
...

This is a Cauchy sequence, and its limit is equal to one. So if you use my definition, then the 1 is indeed insignificant and doesn't change the value of the number. But in conventional terminology, 0.999~1 has no meaning.

N

Edit - changed "significant" in the last sentence to "insignificant"

worzel
2006-Jan-08, 01:43 PM
montebianco's comment sums it up very well. You simply can't argue that within the real number system 1>0.999...

The most convincing demonstration of this for me is that between any two non-equal real numbers there is a third real number between them that is not equal to either. So what real number is between 0.999... and 1?

You could add the two together and divide by two to find this number.

0.999... + 1 = 1.999...
1.999... / 2 = 0.999...

So for 0.999... to be less than 1, 0.999... must be less than 0.999... which is obviously false, so 0.999... < 1 also must be false, in the real number system.

tofu
2006-Jan-08, 01:59 PM
worzel and montebianco. Everything that you just said is wrong. I'm sorry. Don't blame me. It's not my fault. It just is.

If you have a pie, and you want to divide it equally among your three friends, you will give each of them 1/3rd of the pie.

With me so far? Please stop me if you don't follow any of this.

1/3 is 0.333...

One of your friends get's 0.333...
Two of your friends together get twice that much, 2/3rds or 0.666...
Three of your friends together get 3 times 1/3. 3/3rds. 9.999... one. 100%. All of it.

This poll is kind of silly. It's like asking, "do you believe in gravity?" or "do you think that atoms exist?" it doesn't reallly matter what you think. If you can't understand gravity or atoms, that's one thing. That's sad. But that just means that YOU don't understand. That doesn't mean they don't exist.

worzel
2006-Jan-08, 02:01 PM
worzel and montebianco. Everything that you just said is wrong. I'm sorry.
...
Three of your friends together get 3 times 1/3. 3/3rds. 9.999... one. 100%. All of it.
Then why do you agree with us?

montebianco
2006-Jan-08, 02:04 PM
worzel and montebianco. Everything that you just said is wrong. I'm sorry. Don't blame me. It's not my fault. It just is.

If you have a pie, and you want to divide it equally among your three friends, you will give each of them 1/3rd of the pie.

With me so far? Please stop me if you don't follow any of this.

Given that worzel and I both have said that 0.999~ is equal to one, I don't think we're the ones having trouble following things...

tofu
2006-Jan-08, 02:07 PM
lol. sorry. worzel said 1>0.999...

tell you what, I prommise to stay away from the forum before breakfast ok.

montebianco
2006-Jan-08, 02:10 PM
lol. sorry. worzel said 1>0.999...

tell you what, I prommise to stay away from the forum before breakfast ok.

OK :D For the record, the full statement was

You simply can't argue that within the real number system 1>0.999...

Relmuis
2006-Jan-08, 03:05 PM
I suppose that, though somebody might wonder whether 0.9999999... is exactly equal to one, nobody will wonder whether 0.9999999.../0.9999999... is exactly equal to one. Or whether 3.1415653.../3.1415653... is exactly equal to one. In general, I assume that everyone here will agree that any decimal number divided by itself will yield unity.

Suppose that 0.99999999... is not exactly equal to 1. It must then be equal to some other number, which we may call q. If q is not equal to 1 it must be either larger than 1 or smaller than 1. But we will assume q to be a positive number (i.e. larger than 0).

If q is positive and larger than 1, q squared must be larger than q, and therefore larger than 1. (q2>q>1)
If q is positive and smaller than 1, q squared must be smaller than q, and therefore smaller than 1. (q2<q<1)
Therefore, if q is positive and unequal to 1, q squared must be unequal to 1.

Therefore if q squared is equal to 1, q must be equal to 1.

We now merely have to show that q squared is equal to 1.

To do this, we consider that q = 9/10 + 9/100+ 9/1000 + 9/10000 + ...
Squaring this sequence, we get:

q2 = 81/100 + 162/1000 + 243/10000 + 324/100000 + ...

Or, more illuminatingly, q2 = 81/10 * (0.1 + 0.02 + 0.003 + 0.0004 + ...) = 81/10 * 0.123456790123456790123456790...
This is, of course, equal to 0.123456790123456790123456790.../(10/81).
But 10/81 is itself equal to 0.12345679023456790123456790..., as one can see by doing large division.

Therefore, q2 = 0.123456790123456790.../0.123456790123456790... = 1.

Therefore q = 1, or 0.9999999... = 1.

Lance
2006-Jan-08, 05:55 PM
It looks like Microsoft may be paying attention to this thread.

Open the calculator accessory and enter:

1 / 9 =

0.11111111111111111111111111111111

Now, without clearing or hitting anything else, enter:

* 9 =

And get the unexpected (for a calculator):

1

However, if you begin your calculation:

0.11111111111111111111111111111111 * 9 =

You get the expected:

0.999999999999999999999999999999

Moose
2006-Jan-08, 06:30 PM
It looks like Microsoft may be paying attention to this thread.

Nah, it's not unexpected at all.

See, when you do 1 / 9, you get that long string of ones, along with another three or so in memory that aren't actually displayed. So that when you then multiply it by nine, the trailing nine (the one that isn't displayed) gets rounded off. And the carry flips all the nines until the decimal.

But if you type it in yourself, those three hidden ones aren't there, and the calculator has no need to round off the trailing nine as it can display the exact finite product you entered in its entirety.

How a computer rounds off numbers is something programmers have to keep firmly in mind, especially when you're coding things like matrix math, especially larger matrices.

01101001
2006-Jan-08, 06:48 PM
See, when you do 1 / 9, you get that long string of ones, along with another three or so in memory that aren't actually displayed.

It might be doing a bit more than that.

enter 1 / 9 =
displays 0.11111111111111111111111111111111

enter - 0.11111111111111111111111111111111 =
displays 1.1111111111111111111111111111111e-31

enter - 1.1111111111111111111111111111111e-31 =
displays 1.1111111111111111111111111111111e-62

enter - 1.1111111111111111111111111111111e-62 =
displays 1.1111111111111111111111111111111e-93

enter - 1.1111111111111111111111111111111e-93 =
displays - 1.1111111111110898886057792700708e-124

mickal555
2006-Jan-08, 06:58 PM
It's funny...

My maths teacher reakons they are not equil... :(

peteshimmon
2006-Jan-08, 07:17 PM
Got it! They are not the same. There is an
infinitely small difference! Now quickly...

Lance
2006-Jan-08, 07:29 PM
Got it! They are not the same. There is an
infinitely small difference! Now quickly...
Infinitely small?

Converging on zero?

peteshimmon
2006-Jan-08, 08:34 PM
But never getting there! Quickly..lock.. last
word!

umop ap!sdn
2006-Jan-08, 09:23 PM
There is an infinitely small difference!
And the way to write out the infinitely small difference is .000~. It just so happens that .000~ = 0. :D

Thomas(believer)
2006-Jan-08, 11:12 PM
It must have something to do with ordening. 99->100
.99->1.00 etc. It doesn't matter how many 9's you have. If you make things equal it disrupts the way we order the numbers. That's not good.
Even in infinity the pattern stays the same. Like a fractal maybe.
But to be honest, I am very tired and need to go to bed.
Oh, and although I don't have any problems with the mathematics I have been teached and which I understand. I just have the feeling something ain't right and I like to find out what that is. It's very likely I'm wrong.

worzel
2006-Jan-09, 01:33 AM
11 < 22, does that mean 1/1 < 2/2?

2006-Jan-09, 02:00 AM
It must have something to do with ordening. 99->100
.99->1.00 etc. It doesn't matter how many 9's you have. If you make things equal it disrupts the way we order the numbers. That's not good.
Even in infinity the pattern stays the same. Like a fractal maybe.
But to be honest, I am very tired and need to go to bed.
Oh, and although I don't have any problems with the mathematics I have been teached and which I understand. I just have the feeling something ain't right and I like to find out what that is. It's very likely I'm wrong.

Actually it's to do with ordering that 0.999... = 1. As it can be shown that 1 is the lowest number such that x > 0.9...9 for any decimal of the form 0.9...9 it means that 0.999.. = 1. i.e. if we think of 0.999.. as a decimal series then 1 is the value the sequence of it's partial sums converge to.

montebianco
2006-Jan-09, 03:18 AM
It's funny...

My maths teacher reakons they are not equil... :(

Hi Mickal,

Assuming that the problem has been properly explained to your maths teacher, then I am sad to hear this. Perhaps I can have someone at U. Brisbane have a word with him/her...

N

montebianco
2006-Jan-09, 03:20 AM
It must have something to do with ordening. 99->100
.99->1.00 etc. It doesn't matter how many 9's you have. If you make things equal it disrupts the way we order the numbers. That's not good.
Even in infinity the pattern stays the same. Like a fractal maybe.
But to be honest, I am very tired and need to go to bed.
Oh, and although I don't have any problems with the mathematics I have been teached and which I understand. I just have the feeling something ain't right and I like to find out what that is. It's very likely I'm wrong.

Thomas,

I'm not really sure if I understand what is bothering you. Is it that some numbers have two decimal representations and others have only one? I agree that's not a desirable feature, but I don't really know how to do better...

N

Monique
2006-Jan-09, 04:14 AM
Thomas,

I'm not really sure if I understand what is bothering you. Is it that some numbers have two decimal representations and others have only one? I agree that's not a desirable feature, but I don't really know how to do better...

N
Some numbers are irrational, cannot be represented. Such is life.

montebianco
2006-Jan-09, 04:17 AM
Some numbers are irrational, cannot be represented. Such is life.

Well, not with a finite decimal. But then some rational numbers can't be represented with a finite decimal either, and those are the rational numbers that have only one representation :D

Monique, what type of mathematics do you work on?

Nick

HenrikOlsen
2006-Jan-09, 05:07 AM
I think part of the problem may be that some people haven't learned all there is to learn about irrational numbers. :)

ToSeek
2006-Jan-09, 05:12 AM
Well, not with a finite decimal. But then some rational numbers can't be represented with a finite decimal either, and those are the rational numbers that have only one representation :D

You can always represent a rational number with a finite decimal - you just have to choose the right number base. (0.33333... base 10 = 0.1 base 3, for example.)

2006-Jan-09, 09:37 AM
I love oranges, I'm sure you do too. So lets go back to grade one and check our question with oranges.

Say you have 9 oranges and you give one to a friend. Now what
fraction of the total number of oranges have you given away? Right, 1/9.
How do we represent that in decimals? Right, 0.11111 (repeating).
Now, how many oranges does your friend have? She has:
(total number of oranges) times (her fraction of the total number) =
= (9) times (0.11111 (repeating)) = .99999 (repeating) = 1 orange.

So 0.9(repeating) DOES equal 1.

Thomas(believer)
2006-Jan-09, 10:31 AM
Thinking about infinity drives the mind crazy sometimes.
When I try to visualise this endless string of 9's, I get hang up on it. It's dynamic, it never ends. To me it seems a whole universe not just one thing.

This reminds me of a poem:

moving slowly through
endless lowlands,
rows of unthinkably
thin poplars
standing as high plumes
on the horizon;
and sunken within
wonderful space,
farm houses
scattered throughout the land,
clusters of trees, villages,
cropped towers,
churches and elms
in one great association.
the air hangs low
and the sun is slowly
muffled in a gray
mottled fog,
and in all the many provinces
the voice of the water
with its eternal calamities
is feared and heard.

Hendrik Marsman

Matthew
2006-Jan-09, 10:38 AM
Ah, but the poem does end.

montebianco
2006-Jan-09, 02:52 PM
You can always represent a rational number with a finite decimal - you just have to choose the right number base. (0.33333... base 10 = 0.1 base 3, for example.)

Well, then it wouldn't be a decimal, would it :D

Actually, that's a good point. If the not-equal people believe 0.999~ is less than one, but that there is no such number just smaller than 1/3, then their beliefs about the existence of numbers have to change when they change bases...

montebianco
2006-Jan-09, 02:56 PM
I think part of the problem may be that some people haven't learned all there is to learn about irrational numbers. :)

Well, I think everyone may fall into that category :D

It's about time someone brought up the existence of a function which is continuous on the irrational numbers and discontinuous on the rational numbers...

Relmuis
2006-Jan-09, 03:05 PM
In post 1068 I presented a proof that 0.99999999... = 1.

There is another, seemingly shorter proof, but it is invalid, unless one assumes that 1.00000000 = 1. If it isn't self-evident that 0.99999999... = 1, then it can't be self-evident either that 1.00000000... = 1.

So I will prove this first. Let 1.00000000... = p.
This means that p is the sum of a sequence: p = 1 + 0/10 + 0/100 + 0/1000 + 0/10000 + ...
Now one might say that all but the first of these terms is equal to zero, and adding zero to a running total doesn't change anything, so the sequence should sum to 1, the only non-zero term. But there are infinitely many of these zero terms, and infinity times zero might not come to zero at all.

Therefore we square p, by squaring the entire sequence.

This yields p2= 1 + (0+0)/10 + (0+0+0)/100 + (0+0+0+0+0)/1000 + (0+0+0+0+0)/10000 +...

Any finite number of zero's will undoubtedly add up to zero, therefore:

p2= 1 + 0/10 + 0/100 + 0/1000 + 0/10000 + ... = p

Assuming that p is not equal to zero we may divide both sides of this equation by p, yielding p = 1.

So 1.00000000... = 1.

Now we are almost there. Let 0.99999999... = q.
This means that q is the sum of a sequence: q = 0 + 9/10 + 9/100 + 9/1000 + 9/10000 +...
Multiplying both sides of this equation with 10, we find:
10*q = 0 + 9 + 9/10 + 9/100 + 9/1000 + 9/10000 +... = 9.00000000... + q
Subtracting q from both sides of this equation, we find 9q = 9.00000000....
Dividing both sides of this equation by 9, we find q = 1.00000000... = 1.

So 0.99999999... = 1.

worzel
2006-Jan-09, 03:53 PM
Actually, that's a good point. If the not-equal people believe 0.999~ is less than one, but that there is no such number just smaller than 1/3, then their beliefs about the existence of numbers have to change when they change bases...
I made that point way back at the beginning of this discussion (either this thread or the one on FWIS) and got told I was confusing the issue :sad:

Lance
2006-Jan-09, 04:03 PM
I love oranges, I'm sure you do too. So lets go back to grade one and check our question with oranges.

Say you have 9 oranges and you give one to a friend. Now what
fraction of the total number of oranges have you given away? Right, 1/9.
How do we represent that in decimals? Right, 0.11111 (repeating).
Now, how many oranges does your friend have? She has:
(total number of oranges) times (her fraction of the total number) =
= (9) times (0.11111 (repeating)) = .99999 (repeating) = 1 orange.

So 0.9(repeating) DOES equal 1.
This is that same circular argument again.

.999~ only equals 1 in this example if you already believe "0.11111 (repeating)" = 1/9.

If you were taught in school that .333~ is "the best you can do in decimal to represent 1/3 because decimal cannot exactly equal 1/3", then what follows proves nothing.

hhEb09'1
2006-Jan-09, 04:19 PM
If you were taught in school that .333~ is "the best you can do in decimal to represent 1/3 because decimal cannot exactly equal 1/3", then what follows proves nothing.One cannot believe everything one is taught. :)

Because that is false. What they should have said is that is "the best you can do in decimal to represent 1/3 because a finite decimal cannot exactly equal 1/3"

Lance
2006-Jan-09, 04:23 PM
One cannot believe everything one is taught. :)
I understand that.

My point is that some of the proofs offered require previous belief in the concept and so are no proof at all, just circular arguments.

I am not arguing that .999~ != 1

2006-Jan-09, 04:26 PM
This is that same circular argument again.

.999~ only equals 1 in this example if you already believe "0.11111 (repeating)" = 1/9.

If you were taught in school that .333~ is "the best you can do in decimal to represent 1/3 because decimal cannot exactly equal 1/3", then what follows proves nothing.

If you we're taught that at school I'd be bitterly disappointed about the eductaion level of my teachers in that they were either unware of the defintion of decimals or had no knowledge of even the most basic analysis.

Lance
2006-Jan-09, 04:36 PM
If you we're taught that at school I'd be bitterly disappointed about the eductaion level of my teachers in that they were either unware of the defintion of decimals or had no knowledge of even the most basic analysis.
I find that just a little inflamitory. If the subject weren't commonly taught that way, this discussion would not exist.

If this is "the most basic analysis", why is this thread still going on?

Disinfo Agent
2006-Jan-09, 04:41 PM
If the WTC buildings weren't demolished with bombs, why is this thread (http://www.bautforum.com/showthread.php?t=35175) still going on?

I advised the doubters here to read this thread and the other threads where this matter was discussed before making more objections. Have you done that, Lance?

Lance
2006-Jan-09, 04:45 PM
I advised the doubters here to read this thread and the other threads where this matter was discussed before making more objections. Have you done that, Lance?
And I suggest you reread my posts.

I am not trying to make a case for .999~ <> 1 and have said it several times.

My only point is that one form of the proof is a circular argument.

Please understand what you are replying to before assuming I am making a case that I am not.

Thomas(believer)
2006-Jan-09, 04:46 PM
I found this on the net: http://users.forthnet.gr/ath/kimon/Continuum.htm

I'm sure it has something to do with this thread. Maybe someone with more mathematical background than me, can give some enlightment.

Disinfo Agent
2006-Jan-09, 04:47 PM
I am not trying to make a case for .999~ <> 1 and have said it several times.

My only point is that one form of the proof is a circular argument.You are mistaken about that. Read the previous discussions; your objection has been addressed already.

worzel
2006-Jan-09, 04:53 PM
Aren't all mathematical arguments essentially circular? To a certain extent you can chose which statements are to be axioms and do away with your original axioms that are implied by the new ones. Deciding where to start (with axioms that seem self evident) and how to progress to less obvious results is just a matter of taste :)

Carnifex
2006-Jan-09, 04:54 PM
A number 0.99999... can be viewed as a sum of geometrical progression 0.9, 0.09, 0.009, 0.0009... (b1 = 0.9, q = 0.1) The sum of infinite geometrical progression is proved to be b1 / (1 - q) when q is less than 1. When q is greater than or equals 1, the sum diverges and therefore does not exist.

b1 / (1 - q) = 0.9 / (1 - 0.1) = 0.9 / 0.9 = 1

One may argue that this sum is never reached and therefore we should think that 0.999999... is different from by an infinitesimal quantity. This leaves us with a dilemma - theoretically they are not the same, practically they are the same, because the difference is infinitesimal.

I voted for equality, because in the real life "infinitesimal" can be viewed as zero without many complications. For example, one can safely assume, that if f(x) = sin x / x, then f(0) = 1, even though in reality it is infinitisemaly close to 1 when x is infinitesimaly close to 0.

EDIT. By the way, I just noticed that 0.99999... actually is not infinitesimaly smaller than 1, because I've calculated the sum of infinite geometrical progression, and 0.9999... is a sum of infinite geometrical progression.

worzel
2006-Jan-09, 04:57 PM
I voted for equality, because in the real life "infinitesimal" can be viewed as zero without many complications. For example, one can safely assume, that if f(x) = sin x / x, then f(0) = 1, even though in reality it is infinitisemaly close to 1 when x is infinitesimaly close to 0.

Yep, we've all pretty much reached our limit with this thread :)

Lance
2006-Jan-09, 05:01 PM
Clearly there is something I have said that you are not understanding but I can't offer further explaination since I don't know where the disconnection is.

I agree that .999~ = 1

I am not trying to argue otherwise.

My only point here is that one form of proof relies on a circular argument, and I have demonstrated it. If you read back a couple of pages you will see where others have agreed with this.

Let me try again:

To state that .999~ = 1 by relying on the proof that:

1/9 = .111~ AND 1/9 * 9 = 9/9 = 1

Requires you to already believe that:

1/9 = .111~

This is a circular argument and proves nothing.

Disinfo Agent
2006-Jan-09, 05:02 PM
Aren't all mathematical arguments essentially circular?Tautological.

Disinfo Agent
2006-Jan-09, 05:03 PM
Clearly there is something I have said that you are not understanding but I can't offer further explaination since I don't know where the disconnection is.

I agree that .999~ = 1

I am not trying to argue otherwise.

My only point here is that one form of proof relies on a circular argument, and I have demonstrated it. If you read back a couple of pages you will see where others have agreed with this.

Let me try again:

To state that .999~ = 1 by relying on the proof that:

1/9 = .111~ AND 1/9 * 9 = 9/9 = 1

Requires you to already believe that:

1/9 = .111~

This is a circular argument and proves nothing.I knew that was what you were talking about, and no, the argument is not circular.

Lance
2006-Jan-09, 05:05 PM
I knew that was what you were talking about, and no, the argument is not circular.
I have just demonstrated that it is.

Prove your claim to the contrary.

worzel
2006-Jan-09, 05:08 PM
I see Lance's point though. Circular or not, that demonstration only works if you believe that 1/9=0.111~ (something many inequality believers don't accept).

hhEb09'1
2006-Jan-09, 05:09 PM
If the subject weren't commonly taught that way, this discussion would not exist.I would tend to disagree with that.

First, I don't think it is commonly taught that way, but, secondly, even if it weren't, this discussion could still exist because some people insist on misunderstanding their teachers. :)

Lance
2006-Jan-09, 05:13 PM
I see Lance's point though. Circular or not, that demonstration only works if you believe that 1/9=0.111~ (something many inequality believers don't accept).
Thank you.

I will say again that I am not a disbeliever. I was, at first, but the proofs here convinced me early on that .999~ does in fact equal 1. My only desire is to see proofs that are really proofs. Proofs that fall apart when scrutinized by non-believers do no one any good.

JMV
2006-Jan-09, 05:14 PM
Why isn't long division enough to convince people that 1/9=0.111~?

Lance
2006-Jan-09, 05:15 PM
Why isn't long division enough to convince people that 1/9=0.111~?
Because it keeps going, and never quite gets there. Get out your #2 and a few legal pads and you can see for yourself.

That is the biggest single reason why the non-believers don't believe.

hhEb09'1
2006-Jan-09, 05:22 PM
Because it keeps going, and never quite gets there. Get out your #2 and a few legal pads and you can see for yourself.

That is the biggest single reason why the non-believers don't believe.Because they need more than one legal pad to convince themselves? :)

Disinfo Agent
2006-Jan-09, 05:26 PM
I have just demonstrated that it is.

Prove your claim to the contrary.An argument is circular when it assumes what one wishes to prove. But what you wish to prove, in this case, is that 0.999~=1, not that 1/9=0.111~.
So again, Lance, you have not demonstrated that the argument is circular.

As to Worzel's point, the funny thing is that most people only think of doubting that 1/9=0.111~ after they realise that it leads them to conclude that 0.999~=1.

Moose
2006-Jan-09, 05:26 PM
Because they need more than one legal pad to convince themselves? :)

Insanity (http://www.quotationspage.com/quote/26032.html): doing the same thing over and over again and expecting different results.

Albert Einstein, (attributed)
US (German-born) physicist (1879 - 1955)

worzel
2006-Jan-09, 05:27 PM
The trouble is, I would imagine that a formal proof working from the axioms of the real number system would probably be completely unintelligable to an EB. They would ask for something more intuitive, which when provided will necessarily be incomplete and open to being called circular.

2006-Jan-09, 05:30 PM
funny somebody would think 1/9 doesnt have infinite decimals. :)

Carnifex
2006-Jan-09, 05:43 PM
Worzel - I believe "geometrical progression" proof is quite simple and unbeatable. However, it seems quite weird that people don't get why 1/9 = 0.111111...... Neither do I understand Lance's statement "never quite gets here". If the results are uniformic and predictable, you can skip as many steps as you want, even an infinite number of them, and get answer from that. Most theorems in calculus are proved this way - by doing some operation (usually division of an interval [a, b]) several times, proving, that the result is predictable, than skiping an infinitely large number of steps and making conclusions out of this result.

An example:
0 x 9 < 1 < 1 x 9
0.1 x 9 < 1 < 0.2 x 9
0.11 x 9 < 1 < 0.12 x 9
0.111 x 9 < 1 < 0.112 x 9
0.1111 x 9 < 1 < 0.1112 x 9
(at this point a person with some brain working notices a pattern and writes: )
Process is repeated infinitely large number of steps N.
After Nth step we get:
0.111..N_ones..111 x 9 < 1 < 0.111..N_ones..112 x 9
From this we deduce, that after an infinte number of steps we will get to
0.11111.....infinitely_many_ones x 9 = 1

Now, of course, you can say that it's all based on believing that if a x b = c, that c / b = a....... But hey, ain't this at least one thing that is absolutely obvious???

Lance
2006-Jan-09, 05:48 PM
An argument is circular when it assumes what one wishes to prove.
Yes, on that we agree.

But what you wish to prove, in this case, is that 0.999~=1, not that 1/9=0.111~.
Okay, this is where you are not understand it. You disagree with my use of the term "circular argument" and not that the proof itself is faulty. Fine, I can live what that. I thought you were disputing the invalidity of the proof itself.

So again, Lance, you have not demonstrated that the argument is circular.
As above, semantics. I have still disproved the validity of the proof, which was my only point in the first place. You may call it what ever you wish.

Moose
2006-Jan-09, 05:57 PM
If the results are uniformic and predictable, you can skip as many steps as you want, even an infinite number of them, and get answer from that.

Proof by induction. One of the few methods that doesn't make my head asplode.

Now, of course, you can say that it's all based on believing that if a x b = c, that c / b = a....... But hey, ain't this at least one thing that is absolutely obvious???

No, even these things had to be proved at some point. Still, it's usually best if you refrain from proving mathematics from the very first axiom since the dawn of time: 1.

2006-Jan-09, 06:08 PM
If you use sequences (such as is needed to offer a comphrehensive proof that 0.99.. =1) you can use mathematical induction as a sequence (in this context) is a function from N to R and induction is axiomatic in N.

The easiet proof that 0.99.. I know is proof by contradiction, I might post it later.

Disinfo Agent
2006-Jan-09, 06:14 PM
An argument is circular when it assumes what one wishes to prove.
Yes, on that we agree.Great.

But what you wish to prove, in this case, is that 0.999~=1, not that 1/9=0.111~.
Okay, this is where you are not understand it. You disagree with my use of the term "circular argument" and not that the proof itself is faulty.No, it's you who doesn't understand. We agree on the definition of circular argument, but the argument you're complaining about is not circular . Got it, now?

For the meaning of 'circular', see the definition with which Lance has just agreed, quoted above.

Fine, I can live what that. I thought you were disputing the invalidity of the proof itself.I don't know about you, but I am.

So again, Lance, you have not demonstrated that the argument is circular.As above, semantics. I have still disproved the validity of the proof [...]No, you haven't. Call that semantics, if you like, but you have not disproved the validity of the proof.

hhEb09'1
2006-Jan-09, 06:16 PM
As above, semantics. I have still disproved the validity of the proof, which was my only point in the first place. You may call it what ever you wish.I don't think it's a disproof so much as a call for more detail, right?
Still, it's usually best if you refrain from proving mathematics from the very first axiom since the dawn of time: 1.It's a fine line. :)

The easiet proof that 0.99.. I know is proof by contradiction, I might post it later.What could be easier than x = .999..., so 10x = 9.999..., and 10x - x = 9.? :)

OK, I left out a couple details.

2006-Jan-09, 06:16 PM
I learned in school to use the word "recurring" to describe numbers such as 0.999~ and 1/9.
If you ask ANY mathematician they will tell you 1/9 equals 0.1 (recurring). No doubt about that. If you question that then you are questioning the basics of mathematics which is something like questioning the existence of logic.

Mathematics is right. Therefore,
1/9 = 0.1 (recurring). Therefore,
0.9 (recurring) = 1.

Thomas(believer)
2006-Jan-09, 06:25 PM
Moreover, that answer would contradict a theorem in the reals which says: let x, y be reals such that x < y; then x < (x+y)/2 < y. Taking x = 1, y = 0.999…, we have, 0.999…9 less than y alright but also less than x, contradicting the theorem. One may argue that 1 = 0.999… to evade this contradiction. However, present construction of the reals has no proof of this statement, mainly, because 0.999… is ill-defined and does not belong to the domain of the additive and multiplicative operations. Moreover, in the natural ordering of the decimals, we have 0.999… < 1. Therefore, since some nonzero reals, terminating or nonterminating, do not have multiplicative inverse in this domain, it follows that this so-called complete ordered field is, after all, neither complete, nor ordered, nor a field. In fact, it is not even well-defined – it does not exist.

I got the quote from this site: http://home.iprimus.com.au/pidro/infinite.html

It says that there is no proof.

Lance
2006-Jan-09, 06:28 PM
No, you haven't. Call that semantics, if you like, but you have not disproved the validity of the proof.
Of course I did. QED

Now I will ask again, provide proof to the contrary if you disagree.

If you cannot provide such proof then this discussion is settled.

I am not interested in hearing any more "no you didn't" arguments.

Disinfo Agent
2006-Jan-09, 06:30 PM
I got the quote from this site: http://home.iprimus.com.au/pidro/infinite.html

It says that there is no proof.

Monique
2006-Jan-09, 06:30 PM
Well, not with a finite decimal. But then some rational numbers can't be represented with a finite decimal either, and those are the rational numbers that have only one representation :D

Monique, what type of mathematics do you work on?

Nick
I believe someone else ask also. My area is real and complex analysis.

I got the quote from this site: http://home.iprimus.com.au/pidro/infinite.html

It says that there is no proof.

Many proofs exist. Many demonstrate in thread. Suggest you examine proofs presented, show flaws if you believe website.

Disinfo Agent
2006-Jan-09, 06:33 PM
Of course I did. QEDYour "proof" is flawed, because it does not use the definition of 'circular argument' to which you agreed above.

If you cannot provide such proof then this discussion is settled.Unfortunately for you, I can, and I have.

I am not interested in hearing any more "no you didn't" arguments.Then stop making them.

hhEb09'1
2006-Jan-09, 06:34 PM
I got the quote from this site: http://home.iprimus.com.au/pidro/infinite.html

It says that there is no proof.You have to be careful. That site says "Among the defective concepts in mathematics is the complex number i", and concludes that Andrew Wiles proof of Fermat's Last Theorem is defective because he brought "in the defect of complex analysis".

Did Marilyn Vos Savant have anything to do with that?

2006-Jan-09, 06:34 PM
I don't think it's a disproof so much as a call for more detail, right?It's a fine line. :)
What could be easier than x = .999..., so 10x = 9.999..., and 10x - x = 9.? :)

OK, I left out a couple details.

Nothinng really, all we need to do is just add a few dfeitnions and prove a few of the steps (which ins't that difficult).

Taks
2006-Jan-09, 06:35 PM
Of course I did. QEDQED means proof by demonstration, which i have not seen.

I am not interested in hearing any more "no you didn't" arguments.but you haven't shown that 1/9 != 0.111~. there is no "believing" that 1/9 = 0.111~, you can actually work it out by hand. once you divide 9 into 1 first time and get 9, you have immediately proven that the infinite series is equal to 1/9 because the rules you've used to get the first 1, still apply to the second step of the divide so you'll get the second 1 in the series by definition and the process repeats ad infinitum. that's QED. there is then no circular argument.

taks

Thomas(believer)
2006-Jan-09, 06:36 PM

Does someone has a link for me?

montebianco
2006-Jan-09, 06:36 PM
An argument is circular when it assumes what one wishes to prove. But what you wish to prove, in this case, is that 0.999~=1, not that 1/9=0.111~.
So again, Lance, you have not demonstrated that the argument is circular.

Did anyone prove that 0.111~ == 1/9? The proof of that is from the limiting definition, which we can simply apply directly to 0.999~ to get the answer. The proof by multiplication, incidentally, rests on the assumption that the rules of arithmetic can be applied to an infinite series term by term. In this case, they can, because the series is convergent at a geometric rate. A lot of the standard "proofs" of nonsense facts, such as 1 == 2, are based on application of the rules of arithmetic to a divergent series. So I would say the proof by multiplication requires not only 0.111~ == 1/9, but also that (0.111~)*9 == 0.999~. The proof of the first fact alone is essentially equivalent to a proof that 0.999~ == 1.

I'm with Lance on this one, most of the "proofs" offered here are really heuristics. I don't see how to construct a truly rigorous proof without going straight to the definition of the real numbers, which is by equivalence classes of Cauchy sequences of rational numbers...

worzel
2006-Jan-09, 06:37 PM
I got the quote from this site: http://home.iprimus.com.au/pidro/infinite.html

It says that there is no proof.
The crux of that argument is that 0.999... must come before 1 in the natural ordering of the decimals. If by "natural" he means ordered according to magnitude then he is simply (and incorrectly) assuming that 0.999... is less than 1.

I don't know what he means when he says that 0.999... doesn't belong to the domain of certain operations, sure it does, it's just another representation of 1 after all.

Monique
2006-Jan-09, 06:37 PM
You have to be careful. That site says "Among the defective concepts in mathematics is the complex number i", and concludes that Andrew Wiles proof of Fermat's Last Theorem is defective because he brought "in the defect of complex analysis".

Did Marilyn Vos Savant have anything to do with that?
Who??

Taks
2006-Jan-09, 06:40 PM
the problem is when you think of integers and reals in the same fashion. the difference is countability, i think. i.e. integers are countable, 1, 2, 3, ... and order is obvious. reals are not countable (and therefore constitute a larger form of "infinity", btw) and order becomes a little more ambiguous.

taks

Taks
2006-Jan-09, 06:44 PM
Who??marily vos savant was the woman listed with the world's highest IQ. however, a couple notes. 1. guiness no longer recognizes "world's highest IQ" as it turns out various tests show different results. 2. marily scored a 224 on a ratio test as a child, which has a standard deviation of 24. more accurate adult tests have a standard deviation of 15 or 16 which puts her actual adult IQ in the 180-190 range. 3. marilyn apparently scored 46 out of 48 on ronald hoeflin's analogy test (entrance for the MEGA society, a minimum of 176 IQ which is 44 out of 48 i think), which confirms the 180+ range... however, others have achieved perfect scores as i recall.

taks

Moose
2006-Jan-09, 06:45 PM
Who??

Marilyn Vos Savant (http://en.wikipedia.org/wiki/Marilyn_vos_Savant) is a columnist who claims to have the highest IQ in the world.

She does a question and answer column.

hhEb09'1
2006-Jan-09, 06:48 PM
I'm with Lance on this one, most of the "proofs" offered here are really heuristics. I don't see how to construct a truly rigorous proof without going straight to the definition of the real numbers, which is by equivalence classes of Cauchy sequences of rational numbers...How much do we get to start with? Are we going to have to prove convergence theorems? :)

Who??Marilyn Vos Savant (http://en.wikipedia.org/wiki/Marilyn_vos_Savant) wrote the book The World's Most Famous Math Problem, in which she claimed that Andrew Wiles's proof of Fermat's Last Theorem was flawed because (http://en.wikipedia.org/wiki/Marilyn_vos_Savant#Fermat.27s_last_theorem) "if we reject a hyperbolic method of squaring the circle, we should also reject a hyperbolic proof of Fermat's last theorem".

Taks
2006-Jan-09, 06:48 PM
By hand, I can work out that: *snip*yes. exactly my point.

i think we must also keep in mind the few things i brought up earlier about the axiomatic construction of arithmetic. i.e. you have to start out with certain known rules. rules that we all believe to be true before we start any discussion on proofs. the concepts of division, infinite series and convergence/divergence are all built on those axiomatic beliefs.

if one chooses not to accept those beliefs, then there is no point in debating any of us that do, as said individual is attempting to compare apples to oranges. speaking in a different language, in fact.

taks

montebianco
2006-Jan-09, 06:50 PM
By long hand division, I get:

1/9 == 0.<n digits of 1> + 1/(9*10^n)

So to prove that 1/9 == 0.111~, you need to prove (or assume) that the limit of 1/(9*10^n) is zero as n goes to positive infinity. Which is remarkably similar to the thing one must prove to show that 0.999~ == 1 directly...

Moose
2006-Jan-09, 06:50 PM
How much do we get to start with? Are we going to have to prove convergence theorems? :)

More importantly, are we going to have to prove arithmatic? :shifty:

Disinfo Agent
2006-Jan-09, 06:51 PM
Did anyone prove that 0.111~ == 1/9? The proof of that is from the limiting definition, which we can simply apply directly to 0.999~ to get the answer.Or you can use the division algorithm, as Taks has done.
But that's beside the point. The structure of the proof is the following:

Hypothesis: 1/9=0,111~ and (0.111~)*9=0.999~.

Thesis: 0.999~=1.

As long as the hypothesis is true, the thesis is also true, and the proof of this implication does not rely on the condition that 0.999~=1. There is no circularity here.

So I would say the proof by multiplication requires not only 0.111~ == 1/9, but also that (0.111~)*9 == 0.999~. The proof of the first fact alone is essentially equivalent to a proof that 0.999~ == 1.The expressions 0.111~ and 0.999~ can be defined through the same general formula, yes. However, you don't have to use that formula to prove that 0.999~=1.

I'm with Lance on this one, most of the "proofs" offered here are really heuristics. I don't see how to construct a truly rigorous proof without going straight to the definition of the real numbers, which is by equivalence classes of Cauchy sequences of rational numbers...How many truly rigorous proofs did you see in high school? This is high school math. Rigorous enough is all you need.

montebianco
2006-Jan-09, 06:53 PM
yes. exactly my point.

I deleted my post, because the formulae were wrong. I posted a new one, which now appears below the post in which it is quoted :D

i think we must also keep in mind the few things i brought up earlier about the axiomatic construction of arithmetic. i.e. you have to start out with certain known rules. rules that we all believe to be true before we start any discussion on proofs. the concepts of division, infinite series and convergence/divergence are all built on those axiomatic beliefs.

I agree with that, although the usual axiomatic construction would be that the real numbers are defined by limits of Cauchy sequences, with the consequence that long division applied to infinite decimal representations works just as well as it does with finite decimal representations. Some of the proofs above appear to take an assumption that long division works on infinite decimals, and derive the equality of various real numbers by consequence.

Lance
2006-Jan-09, 06:54 PM
but you haven't shown that 1/9 != 0.111~.
And it was never my intention to do so. That is not my point.

Taks
2006-Jan-09, 06:55 PM
By long hand division, I get:

1/9 == 0.<n digits of 1> + 1/(9*10^n)uh, actually, you get:

1/9 == 0.<n digits of 1> + 1/(9*10^n) + ...

it is infinitely long, not n digits +1.

taks

Taks
2006-Jan-09, 06:56 PM
And it was never my intention to do so. That is not my point.you said QED yet have not demonstrated anything?

taks

Disinfo Agent
2006-Jan-09, 06:57 PM

Thomas(believer)
2006-Jan-09, 07:01 PM
Thanks

montebianco
2006-Jan-09, 07:03 PM
uh, actually, you get:

1/9 == 0.<n digits of 1> + 1/(9*10^n) + ...

it is infinitely long, not n digits +1.

taks

I get exactly what I wrote:

1/9 == 0.<n digits of 1> + 1/(9*10^n)

Exact. Not approximate.

Lance
2006-Jan-09, 07:03 PM
you said QED yet have not demonstrated anything?

taks
Yes, I did, a few pages back.

It is not my claim that .999~ <> 1.

I agree that .999==1.

My only point is that one of the proofs used in this thread is flawed, and I have demonstrated that.

hhEb09'1
2006-Jan-09, 07:03 PM
I get exactly what I wrote:I was going to agree with montebianco but he beat me to it

hhEb09'1
2006-Jan-09, 07:06 PM
My only point is that one of the proofs used in this thread is flawed, and I have demonstrated that.I've gone back through the last pages, and I can't find it. Which post?

PS: four! five! this thread is growing too fast :)

montebianco
2006-Jan-09, 07:09 PM
Or you can use the division algorithm, as Taks has done.

Comments on that elsewhere. If you take it as axiomatic that the long division algorithm works on infinite decimals, then you can do that.

The structure of the proof is the following:

Hypothesis: 1/9=0,111~ and (0.111~)*9=0.999~.

Thesis: 0.999~=1.

As long as the hypothesis is true, the thesis is also true, and the proof of this implication does not rely on the condition that 0.999~=1. There is no circularity here.

True enough. If you assume the two conditions in the hypothesis, one of which is pretty much along the same lines as the thesis, then you get the thesis as an implication.

How many truly rigorous proofs did you see in high school? This is high school math. Rigorous enough is all you need.

I agree that the proof is not rigorous.

Disinfo Agent
2006-Jan-09, 07:14 PM
True enough. If you assume the two conditions in the hypothesis, one of which is pretty much along the same lines as the thesis, then you get the thesis as an implication.Not quite. The real problem with the thesis -- what shocks so many people -- is that it equates two different decimal representations (0.999~ and 1.000~) with the same number, 1. This does not happen in the case of 0.111~.

Lance
2006-Jan-09, 07:16 PM
I've gone back through the last pages, and I can't find it. Which post?

PS: four! five! this thread is growing too fast :)
My original comment on the subject was here (http://www.bautforum.com/showpost.php?p=647133&postcount=1044):

And in any case, an empirical demonstration that's easy enough to teach to middle school kids:

1/9 = .111~
2/9 = .222~
3/9 = .333~
4/9 = .444~
5/9 = .555~
6/9 = .666~
7/9 = .777~
8/9 = .888~
9/9 = .999~

And what does 9/9 also equal? There's no sleight of hand up there.
Not to further the argument, but this proof never did work for me because it's a circular argument.

In order to believe that .999~ = 1, you have to also believe that 1/9=.111~. But as I was taught way back on grade school (60s), .333~ is the closest decimal representation of 1/3 but it can never be exactly equal.

Just my 2 cents. Carry on.

(Not meant to infer that I doubt the other proofs. It's just this one that is circular.)

And I later added here (http://www.bautforum.com/showpost.php?p=648859&postcount=1105):

I agree that .999~ = 1

I am not trying to argue otherwise.

My only point here is that one form of proof relies on a circular argument, and I have demonstrated it. If you read back a couple of pages you will see where others have agreed with this.

Let me try again:

To state that .999~ = 1 by relying on the proof that:

1/9 = .111~ AND 1/9 * 9 = 9/9 = 1

Requires you to already believe that:

1/9 = .111~

This is a circular argument and proves nothing.

If there is any fault here, it is with my ability to express the concept. Not with the concept itself. For that, I apologize.

Lance
2006-Jan-09, 07:19 PM
Not quite. The real problem with the thesis -- what shocks so many people -- is that it equates two different decimal representations (0.999~ and 1.000~) with the same number, 1. This does not happen in the case of 0.111~.
The argument works perfectly well when someone already agrees with you. It fails, however, when they don't. The point of a proof, I would think, is to prove something. If it leaves questions unanswered for those you are trying to convince, what have you gained?

montebianco
2006-Jan-09, 07:24 PM
Not quite. The real problem with the thesis -- what shocks so many people -- is that it equates two different decimal representations (0.999~ and 1.000~) with the same number, 1. This does not happen in the case of 0.111~.

I'm in full agreement that this is a persuasive tool that may convince a lot of people. I believe Lance's point (with which I am in agreement) is that it replaces one limiting argument to be proven with another.

Disinfo Agent
2006-Jan-09, 07:27 PM
The argument works perfectly well when someone already agrees with you. It fails, however, when they don't.Have you done a poll on that?

hhEb09'1
2006-Jan-09, 07:33 PM
I'm in full agreement that this is a persuasive tool that may convince a lot of people. I believe Lance's point (with which I am in agreement) is that it replaces one limiting argument to be proven with another.but how far do you want to take that? Lance's observation is not a disproof of the validity, as is clear from his quotes.

If you read back through this thread (ha ha), you'll find that I had repeatedly asked for a response from anyone who disbelieved that 1/3 = .333...

Anyone else, having seen no response, could have assumed that that sort of thing would then be a reasonible place to start.

PS: I never did, but that's just me.

2006-Jan-09, 07:33 PM
I'm in full agreement that this is a persuasive tool that may convince a lot of people. I believe Lance's point (with which I am in agreement) is that it replaces one limiting argument to be proven with another.

Of course this is always going to be true of any proof which is not a proof from first principles. Basically we just have to breakdown a proof such thta the assumptions are not contested.

Monique
2006-Jan-09, 07:41 PM
When finished, PM me please. If consensus determine proposition is false, I must look for new field for study... :)

hhEb09'1
2006-Jan-09, 07:44 PM
Anyone else, having seen no response, could have assumed that that sort of thing would then be a reasonible place to start.

Basically we just have to breakdown a proof such thta the assumptions are not contested.Exactly.

What Lance has done, is contested one of the assumptions. He wants to see more detail there. Nothing wrong with that.

Thomas(believer)
2006-Jan-09, 07:50 PM
10^-n > 10^-(n+1)

As n goes to infinity

0 > 0 ??

This is not true.

Conclusion 10^-n <>0 as n goes to infinity.

What's wrong with my logic?

Further, for 0.1111~ == 1/9, the infinite string must be unique.
I really don't know how to prove the string is unique. The example above seems to indicate it is not unique.
If it is not unique then I guess that 0.111~ = 1/9.

2006-Jan-09, 07:51 PM
Was it 1/9 = 0.111..?

That's fairly easy 0.111.. is defined as being equal to the value that the sum of the terms (1/10)^k (k=1,2,3..) converges to. A geometric series is a sequence whose terms are of the form r^k (k=1,2,3..) and the sum converges to r/(1-r) which in the case of the above series is 1/9.

Disinfo Agent
2006-Jan-09, 07:55 PM
10^-n > 10^-(n+1)

As n goes to infinity

0 > 0 ??

This is not true.

Conclusion 10^-n <>0 as n goes to infinity.

What's wrong with my logic?When you take both sides of an inequality to the limit, you must change > to >=, because the difference between the two sides may go to zero (they "meet" at infinity).

10^-n > 10^-(n+1) for all n does not imply 0>0,
but it does imply 0>=0, which is true.

hhEb09'1
2006-Jan-09, 07:56 PM
Conclusion 10^-n <>0 as n goes to infinity.

What's wrong with my logic?Nothing. 10^-n can never equal 0, no matter what n is.

However, the limit of 10^-n is not of the form 10^-n, it can equal zero.

2006-Jan-09, 08:01 PM
The limit in this case is simply the largest number x such that x<10^-n for (n=1,2,3..), that just coems staright form the definition of a limit.

Thomas(believer)
2006-Jan-09, 08:04 PM
Maybe I'm wrong, but this looks like the rules have to be changed for convenience. That's not fair;)

Moose
2006-Jan-09, 08:05 PM
Oh, Lance, since this argument is over my post, I should point out that what I offered wasn't a proof, but a demonstration. I called it that because I fully acknowledge it wasn't rigorous. It wasn't intended to be.

That said, while your point is somewhat valid, the fact that your hypothetical someone doesn't believe that 1/9 = .111~ doesn't make it any less mathematically true. The breakdown isn't in the proof, but in the person evaluating the proof.

As Grapes pointed out, the problem you raise is that you want the "proof" to start further back. That's fine, but the question becomes where do you draw the line between what is known and what needs to be proven? At some point, you have to accept that mathematics is right, or you roll right back to the state of math in stone age.

2006-Jan-09, 08:11 PM
As Grapes pointed out, the problem you raise is that you want the "proof" to start further back. That's fine, but the question becomes where do you draw the line between what is known and what needs to be proven? At some point, you have to accept that mathematics is right, or you roll right back to the state of math in stone age.

The most comphrehensive proof possible is one form firts principles i.e. you start with the axioms, that's as far as you can go 'back'. Axioms by defintion are true in the theory which they are axioms in.

I'd dispute whether mathematics is 'right' or not (for a start Godel shown that any sufficently complex mathematical system is incomplete and cannot prove it's own consistency), as it comes down to the old debate about whether platonism is correct.

Disinfo Agent
2006-Jan-09, 08:15 PM
Maybe I'm wrong, but this looks like the rules have to be changed for convenience. That's not fair;)No, it's just the old adage that what works for finite numbers and sequences doesn't always carry over to infinite numbers and series. You can check it, yourself.

0.1^n is definitely > 0.1^(n+1) for every integer n

However, lim (0.1^n)=0 and lim (0.1^(n+1))=0.

So, clearly you can't say that lim (0.1^n)>lim (0.1^(n+1)). The inequality breaks down at the limit.

But the inequality still holds as long as you write >= instead of just > (we can prove this). This is a general result for convergent sequences.

hhEb09'1
2006-Jan-09, 08:19 PM
This is a general result for convergent sequences.My daughter asked for my help with a Pinching Theorem problem just a few days ago :)

Disinfo Agent
2006-Jan-09, 08:21 PM
Ah, thanks! I couldn't remember the name of the damn thing.

Pinching Theorem (http://mathworld.wolfram.com/PinchingTheorem.html)

montebianco
2006-Jan-09, 08:25 PM
Was it 1/9 = 0.111..?

That's fairly easy 0.111.. is defined as being equal to the value that the sum of the terms (1/10)^k (k=1,2,3..) converges to. A geometric series is a sequence whose terms are of the form r^k (k=1,2,3..) and the sum converges to r/(1-r) which in the case of the above series is 1/9.

Yes, that's correct. We can use the same method to establish that 0.999~ == 1 :D

montebianco
2006-Jan-09, 08:26 PM
When finished, PM me please. If consensus determine proposition is false, I must look for new field for study... :)

Most of the discussion is now among people who agree that the proposition is true, but disagree about what consistutes a valid proof of the proposition :D

montebianco
2006-Jan-09, 08:31 PM
Maybe I'm wrong, but this looks like the rules have to be changed for convenience. That's not fair;)

Others have responded, but what rule has been changed? The statement you made is true for any positive value of n. What rule tells you that you can apply a limit to both sides, and expect that the inequality still holds?

A much simpler example of how you cannot simply perform any operation you like on both sides of an inequality, and expect the inequality still to hold:

1 < 2

(now multiply both sides by -1)

-1 < -2

It doesn't work that way; you can't multiply both sides of an inequality by a negative number, and expect the inequality still to hold. Similarly, you can't take limits of a sequence of inequalities, and expect the inequality to hold in the limit.

Monique
2006-Jan-09, 08:32 PM
Is great relief. I do not wish to start new career!!

Thomas(believer)
2006-Jan-09, 08:34 PM
For every e>0 there exists a n>0 such that:

10^-n - 10^-(n+1)<e

That's I think the definition of a limit. And so per definition they are equal in the limit. It's somehow remarkable that e>0.

Disinfo Agent
2006-Jan-09, 08:35 PM
Yes, that's correct. We can use the same method to establish that 0.999~ == 1 :D

If you take it as axiomatic that the long division algorithm works on infinite decimals, then you can do that [prove that 1/9=0.111~].Which is what we do in high school. ;)

hhEb09'1
2006-Jan-09, 08:43 PM
Which is what we do in high school. ;)I'm not so sure that we even have to "take it as axiomatic that the long division algorithm works on infinite decimals." Finite decimals is enough. :)

Still, after that, it's pretty clear to ten year-olds that 1 divided by 3 results in a finite string of 3s that could go out forever (a conclusion reached long before the second page of a legal pad, I'd bet :) ). And what else would .333... equal?

Thomas(believer)
2006-Jan-09, 08:52 PM
And what else would .333... equal?
Everything in in the neighbourhood of 1/3 which is not a real number?
That is, if there exist things in the neighbourhood of 1/3 which are not real numbers. I'm sure someone can link me to some interesting site for having said this.:razz:

hhEb09'1
2006-Jan-09, 08:57 PM
That is, if there exist things in the neighbourhood of 1/3 which are not real numbers. What set is your neighbourhood in?

Thomas(believer)
2006-Jan-09, 09:03 PM
What set is your neighbourhood in?
A set which represents a continuous line. R has to be a subset from this set.
Does such a set exist?

hhEb09'1
2006-Jan-09, 09:07 PM
A set which represents a continuous line. R has to be a subset from this set.
Does such a set exist?R does represent a continuous line. It can be the entire set.

Thomas(believer)
2006-Jan-09, 09:10 PM
R does represent a continuous line. It can be the entire set.

I understand that. But are there other sets which represent a continuous line and which include R.

Disinfo Agent
2006-Jan-09, 09:12 PM
Why complicate things? :p

Thomas(believer)
2006-Jan-09, 09:23 PM
Because it is fun? :D

Disinfo Agent
2006-Jan-09, 09:23 PM
Ew, you're sounding like a mathematician. :D

Thomas(believer)
2006-Jan-09, 09:29 PM
Not really. But I love it!
I believe that all people who still reply to this thread, much somehow love it. ;)

HenrikOlsen
2006-Jan-09, 10:22 PM
Well, I think everyone may fall into that category :D

It's about time someone brought up the existence of a function which is continuous on the irrational numbers and discontinuous on the rational numbers...
f(x)=x

HenrikOlsen
2006-Jan-09, 10:30 PM
Aren't all mathematical arguments essentially circular? To a certain extent you can chose which statements are to be axioms and do away with your original axioms that are implied by the new ones. Deciding where to start (with axioms that seem self evident) and how to progress to less obvious results is just a matter of taste :)
Not really, an argument is only circular when you try to prove something by using the thing to be proven as part of the argument.

Mathematics do not prove axioms, they are by definition the assumptions used to define a mathematical system. Everything else within that system is build either on the axioms(and definitions) or on postulates already proven.

montebianco
2006-Jan-09, 10:59 PM
f(x)=x

That function is continuous everywhere...

Edit - I think I understand what went wrong here. We are looking for a function f(x) defined on the real numbers (an interval is good enough), which is continuous where x is irrational, and not continuous where x is rational.

2006-Jan-09, 11:12 PM
That function is continuous everywhere...

Edit - I think I understand what went wrong here. We are looking for a function f(x) defined on the real numbers (an interval is good enough), which is continuous where x is irrational, and not continuous where x is rational.

There isn't such a function, any function that is discontinious on an everywhere dense set of the reals is discontinious everywhere.

montebianco
2006-Jan-09, 11:13 PM
Still, after that, it's pretty clear to ten year-olds that 1 divided by 3 results in a finite string of 3s that could go out forever (a conclusion reached long before the second page of a legal pad, I'd bet :) ). And what else would .333... equal?

The equality of the number represented by the string 0.333~ and the number 1/3 follows from the definition of a real number as a limit of Cauchy sequences, the same one that establishes the equality of the number represented by the string 0.999~ and 1. In the discussion of the circularity or lack thereof of the argument offered, there was a certain amount of talk about how far "back" the proof would go. But I would dearly love to know in what axiomatic system the equality of 0.333~ and 1/3 is obvious, but the equality of 0.999~ and 1 is not. Perhaps it is pretty clear to ten year-olds, but I suppose it is pretty clear to a lot of them that 0.999~ is not equal to one. If the set of axioms used here is whatever is clear to ten-year olds, I think I'll stick with mine...

montebianco
2006-Jan-09, 11:14 PM
There isn't such a function, any function that is discontinious on an everywhere dense set of the reals is discontinious everywhere.

There most definitely is such a function.

2006-Jan-09, 11:20 PM
There most definitely is such a function.

Hmm, actually yes I may of been a bit rash (I thought that as the function is discontinious at some poitn in evry neihbourhood of an irrational numebr that would be sufficent to prove that it is discontious at every irrational, but it's not).

worzel
2006-Jan-09, 11:47 PM
Aren't all mathematical arguments essentially circular? To a certain extent you can chose which statements are to be axioms and do away with your original axioms that are implied by the new ones. Deciding where to start (with axioms that seem self evident) and how to progress to less obvious results is just a matter of taste Not really, an argument is only circular when you try to prove something by using the thing to be proven as part of the argument.Yeah, I was being flippant. But the point was simply that all theorems of a formal system are tautologies (which are loosely speaking circular arguments outside of maths). You could only conceivably take issue with them by taking issue with the axioms of the formal system. As the inequalitiy believers have yet to state a position on any axiom of the real number system they are free to run around in circles disagreeing with whatever proof or demonstration they want.

Mathematics do not prove axioms, they are by definition the assumptions used to define a mathematical system. Everything else within that system is build either on the axioms(and definitions) or on postulates already proven.Yes, but my point was that to a certain extent you can choose your axioms. Some theorems of formulation A might be the axioms of an equivalent formulation B, and those axioms of B might might prove theorems in B which are actually axioms in A (in fact, they'd have to if they weren't axioms of B and the two formulations were equivalent).

EG The simplist complete form of boolean logic that I've seen has only three axioms, one logical connective (implication), and one rule of inference (plus propositional variables and the constant FALSE). I think it went something like:

Axiom Schema 1: P=>(Q=>P)
Axiom Schema 2: (P=>(Q=>R))=>((P=>Q)=>(P=>R))
Axiom Schema 3: P=>((P=>FALSE)=>FALSE)
Rule of inference: P, P=>Q |- Q (modus ponens)

Not so sure that the 3rd one I gave is adequate to make it complete, but dropping the third gives you intuitionistic logic (no proof by contradiction).

Not exactly the most self evident set of axioms or the easiest set of inference rules to use, but possibly useful if you want to write an automated theorem prover :)

Compare that to this definition of Boolean Algebra (http://en.wikipedia.org/wiki/Boolean_algebra#Formal_definition).

montebianco
2006-Jan-10, 12:00 AM
Yeah, I was being flippant. But the point was simply that all theorems of a formal system are tautologies (which are loosely speaking circular arguments outside of maths). You could only conceivably take issue with them by taking issue with the axioms of the formal system. As the inequalitiy believers have yet to state a position on any axiom of the real number system they are free to run around in circles disagreeing with whatever proof or demonstration they want.

Just to be clear, the recent flurry of activity seemed to be mostly (entirely) among people solidly in the 0.999~ == 1 camp. Any disagreement appeared to me to be about the soundness of a proof, not on the validity of the proposition itself. :D

Lance
2006-Jan-10, 12:02 AM
Just to be clear, the recent flurry of activity seemed to be mostly (entirely) among people solidly in the 0.999~ == 1 camp. Any disagreement appeared to me to be about the soundness of a proof, not on the validity of the proposition itself. :D
Agreed.

worzel
2006-Jan-10, 12:07 AM
Also agreed :D

Just pointing out that we could go round in circles for ever if we never commit ourselves to some assumptions to start us off.

montebianco
2006-Jan-10, 12:18 AM
Also agreed :D

Just pointing out that we could go round in circles for ever if we never commit ourselves to some assumptions to start us off.

Just so it's clear, my assumptions are the basic axioms about addition and multiplication on the rational numbers. The real numbers are then defined as equivalence classes of Cauchy sequences of rational numbers, and the equality of 0.999~ and 1 follows straight away.

Of course, this is probably just so much gobbly-gook to a lot of the not-equal people...

N

Taks
2006-Jan-10, 01:18 AM
Just pointing out that we could go round in circles for ever if we never commit ourselves to some assumptions to start us off.that's what the point, and definition, of the axioms is for. i.e. we all have to be on an equal footing, a common language, in order for any debate to even make sense, irrespective of any truths or falsehoods.

taks

worzel
2006-Jan-10, 01:29 AM
indeed

Dark Jaguar
2006-Jan-10, 02:07 AM

What is .6666~?

Wouldn't 1 - .9999~ = 0.0~1? That was an odd way to phrase it, but it would be an infinite number of zeroes with a 1 at the end as I see it. The fact that a physical writer could never reach that 1 is irrelevent, only because the fact that a physical writer could never finish 0.999~ doesn't seem to get in the way.

I really don't get this. What is meant by "it tends to one"? Sure it TENDS to it, but isn't that different than actually BEING it?

Musashi
2006-Jan-10, 02:11 AM
There is no way to write an infinte number of zeros and then put a one at the "end".

Taks
2006-Jan-10, 02:12 AM
the 0.000~1 question has been answered recently... within the last 5 or 6 pages i think. i mean, other than musashi's pithy response. :)

taks

worzel
2006-Jan-10, 02:13 AM

What is .6666~?
2/3

Wouldn't 1 - .9999~ = 0.0~1?
There's no such number as 0.0~1. What could it mean to have infinite 0s and then a 1? That's very different to just having infinte 0s, or infinite 9s.

I really don't get this. What is meant by "it tends to one"? Sure it TENDS to it, but isn't that different than actually BEING it?
It means that how ever close to 1 you chose your number such that it's less than 1, 0.999~ will awlays be bigger than it. I.e. 0.999~ is infinitely close to one, there are no numbers between 0.999~ and 1. Try and find one if you don't believe it.

Taks
2006-Jan-10, 02:15 AM
I really don't get this. What is meant by "it tends to one"? Sure it TENDS to it, but isn't that different than actually BEING it?you are sort of correct, it does not "tend to 1." however, as n approaches infinity, the series sum(9/10^n) approaches 1 (or "tends" to 1) in the limit. but the shorthand notation 0.999~ isn't really a limit, i.e. all the digits are there because n is already at infinity, sort of. if that makes any sense. :)

as worzel pointed out, there is no number between 0.999~ and 1.

taks

montebianco
2006-Jan-10, 02:17 AM

What is .6666~?

Well, if you ask me, 2/3. If you ask the people who answered "no" to the poll, maybe something else...

Wouldn't 1 - .9999~ = 0.0~1? That was an odd way to phrase it, but it would be an infinite number of zeroes with a 1 at the end as I see it. The fact that a physical writer could never reach that 1 is irrelevent, only because the fact that a physical writer could never finish 0.999~ doesn't seem to get in the way.

The symbol 0.0~1 is not usually defined as meaningful. I gave this one a stab in an earlier post, defining it as the limit of the sequence:

0.1
0.01
0.001
0.0001
0.00001
0.000001
....

Since the limit of the sequence is zero, I defined 0.0~1 as zero. But do keep in mind that this is a completely non-standard definition.

I really don't get this. What is meant by "it tends to one"? Sure it TENDS to it, but isn't that different than actually BEING it?

I think the question we have to contemplate is, what is the meaning of an infinite sequence? The standard definition of the symbol 0.999~ is the limit of the sequence:

0.999
0.9999
0.99999
0.999999
0.9999999
...

The limit of this sequence is equal to one. This is the way real numbers are defined, as sequences of rational numbers, and two sequences which tend to the same limit are identified with the same real number.

That's standard math. Is this the only possible way to do things? Certainly, you could define a number system in which two sequences of rational numbers define different numbers, even if they tend to the same limit. But if you do that, a lot of basic rules of arithmetic (many of which I am sure most posters here would hold as self-evident) won't work any more.

So the number 0.999~ is defined as one, because that's the limit of the sequence of digits. You can use a different definition (i.e., the difference between tending to one and being one, as in your post), but in that case, a lot of pretty basic math doesn't work anymore. I suspect it's going into more detail than many want to go into here, but it is possible to show that the real number system defined the standard way (e.g., 0.999~ is defined as one) is the only number system with a fairly short list of basic properties we would expect it to have...

Hope that helps, follow up if you like...

Nick

worzel
2006-Jan-10, 02:20 AM
Just a tip for those who have difficulty with limits. Try reading the first few chapters of an elemtary calculus text book (or the calculus chapter of a maths book). Typically they start with approximations to the gradient of a curve through simple geometry and then consider what happens in the limit as you make your triangle ever smaller. This graphical approach makes it very intuitive to follow IMO.

montebianco
2006-Jan-10, 02:20 AM
but the shorthand notation 0.999~ isn't really a limit, i.e. all the digits are there because n is already at infinity, sort of. if that makes any sense. :)

I would say it is a limit, because it is defined that way, and from the definition of a limit, the limit of:

0.9
0.99
0.999
0.9999
0.99999
...

is equal to one. (Use the infinite sum formulation if you prefer.) This is the definition of 0.999~

Dark Jaguar
2006-Jan-10, 02:45 AM
I think I am getting a handle on it now. It seemed like some silly mathematical joke, or a weakness in mathematical expression, at first. Now it makes sense. Oh and, I suppose I should have picked something else instead of .666~...

Your idea of .00~1 equaling zero makes sense too. It certainly makes a lot more sense than what I keep hearing about how "you just can't dot the end of infinity". Sure you can't, but you can't write out infinity either. The other resolution, it "tending to" zero, makes more sense.

Here's a question. Why does one need to "define" 1 in such a manner? It seems self evident what "one-ness" is to me.

montebianco
2006-Jan-10, 03:28 AM
Here's a question. Why does one need to "define" 1 in such a manner? It seems self evident what "one-ness" is to me.

The various procedures in the thread are not needed to define one, since one is rational. The history of mathematical thought on this sort of thing begins with the rational numbers, which can be defined as p/q where p and q are integers and q is not zero. You can solve a lot of problems just with the rational numbers, and rules for the behavior of rational numbers (how they add, how they multiply, things like x*(y+z) == x*y + x*z) had been worked out. But people noticed certain gaps in the number system, for example, there is no number equal to the square root of two. This means that the length of the sides of a square can be rational, and the length of the diagonal of a square can be rational, but you can't have both for the same square. The ratio of the circumference of a circle to its diameter (pi) is also not rational. So this procedure of defining real numbers as sequences of rational numbers was designed to give us the irrational numbers; but you get the rational numbers back as a by-product. Some sequences of rational numbers approach a rational number as a limit:

0.9
0.99
0.999
0.9999
...

approaches one, whereas other sequences:

1
1+1/(1)
1+1/(1)+1/(1*2)
1+1/(1)+1/(1*2)+1/(1*2*3)
1+1/(1)+1/(1*2)+1/(1*2*3)+1/(1*2*3*4)
1+1/(1)+1/(1*2)+1/(1*2*3)+1/(1*2*3*4)+1/(1*2*3*4*5)
...

do not - this sequence approaches the number e, which is irrational. So there really isn't any need to define one as a limit of sequences of other numbers, it just happens as a by-product of the procedure we use to define numbers like the square root of two, pi, e, and other irrational numbers.

Strictly speaking, once the real numbers are defined, the rules of arithmetic such as addition and multiplication, which we already know for the rational numbers, need to be proven again for the real numbers, before we can claim that the real number '1' defined as limits of sequences such as 0.999~ is really the same as the rational number '1' :D A rigorous course in analysis will do exactly that...

N

Lance
2006-Jan-10, 03:45 AM
Infinite: (http://www.merriam-webster.com/cgi-bin/dictionary-tb?book=Dictionary&va=infinite)
3 : subject to no limitation or external determination
So if .999~ is infinitely close to 1, it is not externally determinable to be different than one. Therefore they must be equal.

worzel
2006-Jan-10, 08:14 AM
Your idea of .00~1 equaling zero makes sense too. It certainly makes a lot more sense than what I keep hearing about how "you just can't dot the end of infinity". Sure you can't, but you can't write out infinity either. The other resolution, it "tending to" zero, makes more sense.
I agree. I think montebianco's attempt to define 0.0~1 as the limit to a sequence is very instructive. Presumably everyone agrees that 1-0.9~=0.0~1, and 0.0~1=0 might be easier to see than 0.9~=1. In fact, I think that's how our maths teacher taught us limits at high school.

2006-Jan-10, 02:23 PM
The thing is 0.00~1, doesn't make alot of sense, as the '~' implies that there are an infinite number of '0's between the deciaml point and the '1', yet as I said earlier we can regard a sequence (a sequence of digts in this case) in this instance as being a function from N to R and between any two naturals there are only ever a finite number of natural numbers. So in this case the '1' is not and cannot be part of the decimal.

The only way we could possibly attach menaing to it is as the limit of the sequence 0.1, 0.01, 0.001 as has been done above in which case it is just 0 by definition.

hhEb09'1
2006-Jan-10, 04:56 PM
I understand that. But are there other sets which represent a continuous line and which include R.But that's why I asked which set you were using--I thought we were mostly working in the reals.
The only way we could possibly attach menaing to it is as the limit of the sequence 0.1, 0.01, 0.001 as has been done above in which case it is just 0 by definition.I'm not sure of that. I'm pretty sure someone has made some sort of sense out of things like 1.000...1, which is probably different from (not equal to) 1.000...2 :)

HenrikOlsen
2006-Jan-10, 05:03 PM
But not while using the same mathematical system as the rest of us.
If we allow that 1.00~1 is the limit of 1+1*10-n as n→∞, then 1.00~2 is the limit of 1+2*10-n as n→∞, which is also 1.

2006-Jan-10, 05:07 PM
I'm pretty sure someone has made some sort of sense out of things like 1.000...1, which is probably different from (not equal to) 1.000...2 :)

Your tlakign about non-standrad analysis and sets (like the surreal numbers) which are not the real numbers, but even in this context soemthing like 1.000...1 doesn't make a great deal of sense without alot of clarification.

Taks
2006-Jan-10, 07:15 PM
I would say it is a limit, because it is defined that way, and from the definition of a limit, the limit of:

0.9
0.99
0.999
0.9999
0.99999
...

is equal to one. (Use the infinite sum formulation if you prefer.) This is the definition of 0.999~no, it's really not a limit. the function of a limit has an ever increasing length of digits, but always finite. even the list you show above contains numbers with finite lengths. as n approaches infinity, there are always a finite number of digits as well, ever increasing, but always finite. putting the bar over the last digit is sort of like saying you're done with the limit function... all the digits are there.
taks

Carnifex
2006-Jan-10, 08:15 PM
It is a limit of a function. To be more exact, it is a limit of a sequence Xn when n approaches infinity. This number, the limit, has one and only one meaning, which we can (of course, it is a simplification, but kids in the school are taught that way and it does make some sence) interpret as "the infinityth member of the sequence Xn".

By defining sequence as 0.9, 0.99, 0.999 we are able to find it's limit, which is a number 0.9999~ (which we can find by simply adding digits infinitely), or which is 1 (which we can find from the definition of the limit)

By the way, the definition of the limit of a sequence, which is accepted in the whole world of mathematics is:

"A number a is called the limit of the sequence Xn, when for every no matter how small positive e we can find a number N so that for every n > N the following is true: |Xn - a| < e"

I'm quite sure that it is quite obvious that according to the definition the sequence 0.9, 0.99, 0.999 has a limit, which is 1.

I even might add: Resistance is futile :D

agingjb
2006-Jan-10, 08:32 PM
Good summary Carnifex. But when the posters on a site such as this vote 40% for a proposition that any elementary text on real analysis would show to be flawed, then I agree "resistance is futile".

Vaelroth
2006-Jan-10, 08:33 PM
I would say no, you can argue limits all you want but in a numerical sense they are not the same digits. .999~ != 1 just the same as 2/3 != .666~. Why? The numbers are representative of each other, but they are not the same.

The number pi cannot be completely calculated because it goes on to infinity, it can however be represented by the ratio that creates the number pi, much the same as what is going on here. I can't disagree with the calculus logic of limits and how .999~ = 1 based on mathematical theorems, but I will disagree that because they are not the same numerals that they are not the same number in recognition. None of us would look at .999~ and call it 1, just the same as none of us would look at .666~ and call it 2/3. We would have to prove each situation before we could refer to the infinite numerical value to the fractional ratio.

Taks
2006-Jan-10, 08:47 PM
I will disagree that because they are not the same numerals that they are not the same number in recognition.but they are the same number in recognition.

None of us would look at .999~ and call it 1, just the same as none of us would look at .666~ and call it 2/3.yes we would, and we do.

We would have to prove each situation before we could refer to the infinite numerical value to the fractional ratio.but that's already been done.

btw, monte and carnifex, i think i do actually agree with what you're saying. rather, the representation is the limit, or the result of the limiting function. no further explanation needed.

taks

Carnifex
2006-Jan-10, 08:52 PM
Basically what I see here is denying that we can determine the result of an infinite operation from several steps albeit we can, because we're smart apes and we can see patterns.

For example, why one can state that 0.(142857) is equal 1/7? Because infinite dividing of 1 by 7 shows that. Yes, we are not physically able to carry out this operation infinite times, but we can predict its outcome after an infinite number of stages, because there is a repeating pattern. Therefore, 0.(142857) is equal 1/7, 0.(3) is equal 1/3, 0.(9) is equal 1 and so one.

Vaelroth
2006-Jan-10, 08:59 PM
Taks: Yes its already been done and I have seen the proof and with the proof the two numbers are equal. However, without the proof they are not necessarily equal. That was my point. So in reality they are both equal and inequal.

I don't really know how to phrase this any better, but thats my best shot.

agingjb
2006-Jan-10, 09:12 PM
I wonder, in what sense is IV (in Roman numerals) not equal to 4? I would have said that they were equal in any mathematical sense.

Monique
2006-Jan-10, 09:59 PM
Taks: Yes its already been done and I have seen the proof and with the proof the two numbers are equal. However, without the proof they are not necessarily equal. That was my point. So in reality they are both equal and inequal.

I don't really know how to phrase this any better, but thats my best shot.
Bold statement does not make mathematical sense. Quantity define as Infinite series (0.99~) and quantity define as 1.0 are equal. How represented not important.

Example for representation: 1.25 = 5/4

Different representation same value. Proof for infinite series converge to finite value exist, equality true. Representation for infinite series independent to mathematical properties.

worzel
2006-Jan-10, 10:33 PM
Or in short: while they are syntactically different, they are semantically the same :)

Vaelroth
2006-Jan-10, 10:37 PM
Monique, 5/4 isn't an infinite series though. :think:

Worzel, I think your post sums up what I'm trying to say (and maybe its what other people are trying to tell me).

hhEb09'1
2006-Jan-10, 10:47 PM
I would say no, you can argue limits all you want but in a numerical sense they are not the same digits. .999~ != 1 just the same as 2/3 != .666~. Why? The numbers are representative of each other, but they are not the same.

The number pi cannot be completely calculated because it goes on to infinity, it can however be represented by the ratio that creates the number pi, much the same as what is going on here. I can't disagree with the calculus logic of limits and how .999~ = 1 based on mathematical theorems, but I will disagree that because they are not the same numerals that they are not the same number in recognition. None of us would look at .999~ and call it 1, just the same as none of us would look at .666~ and call it 2/3. We would have to prove each situation before we could refer to the infinite numerical value to the fractional ratio.2/4 and 1/2 are not the same numerals either but I have no problem saying that they are equal, in the sense that we are talking about.

Would you say that they are not the same either?

Monique
2006-Jan-10, 10:51 PM
Monique, 5/4 isn't an infinite series though. :think:

Worzel, I think your post sums up what I'm trying to say (and maybe its what other people are trying to tell me).
Point is same, quantity independent of representation. Some infinite series provably converge to finite value. In such case, infinite series is alternate representation for finite quantity.

worzel
2006-Jan-10, 11:10 PM
Monique, 5/4 isn't an infinite series though. :think:
Hold on, isn't that the sum of the infinite series:

5/8 + 5/16 + 5/32 + 5/64 + ... + 5/2n+2 + ...

and therefore the limit of the infinite sequence:

5/8, 15/16, 35/32, 75/64, ..., 5(2n-1)/2n+2, ...

Taks
2006-Jan-10, 11:10 PM
Monique, 5/4 isn't an infinite series though. :think: yes it is. 1.250~

and as monique pointed out, saying that something is equal and inequal (sp?) at the same time does not make sense.

taks

Taks
2006-Jan-10, 11:14 PM
i like worzel's infinite representation better. :)

taks

2006-Jan-10, 11:29 PM
All real numbers (in what is probably the easiest construction) are equivalence classes of Cauchy sequences of rational numbers, in each equivalence class that represents a real number there is more than one Cauchy sequence (it's easy to show infact there are an infinite number of Cauchy sequences in each equivalence class).

The decimal digits of a real number are infact the terms in a certain class of series whose sequence of partial sums belong to the equivalence class of Cauchy sequences representing that number. In some cases there are two such decimal series whose sequence of partial sums belong to that equivalence class so there are two different decimal representations of that number.

Whilst 0.999... and 1.000... (I've added the trailing zeroes as they do represent terms in the related decimal series) do not represent the same decimal series the two sequences that they represent both belong to the same equivalence class and so they both represent the same real number.

worzel
2006-Jan-10, 11:34 PM
It's been said a few times on this thread already, but it seems the problem for many is that because most numbers have a unique decimal representation (if we ignore leading and trailing 0s) people have difficulty accepting that not all have numbers have only one decimal representation, even though they're happy with many ratios of integers representing the same number.

01101001
2006-Jan-10, 11:38 PM
yes it is. 1.250~

Or for instance in another base, 1.(02)3

I don't understand the aversion some seem to have for there being different names for a number. They seem OK with the various language names for number one, un, uno, eine, yi, ichi...

It is frequently represented, named, symbolized, "1" but also such as:

01
1.0
1.00
1.(0)
1.0~
1.0...

and infinitely many more, as:

0+1
9/9
10^0
1!

All are just more names of one.

0.999~ is simply yet another name of one.

montebianco
2006-Jan-10, 11:39 PM
It's been said a few times on this thread already, but it seems the problem for many is that because most numbers have a unique decimal representation (if we ignore leading and trailing 0s) people have difficulty accepting that not all have numbers have only one decimal representation, even though they're happy with many ratios of integers representing the same number.

Yep, and as has been pointed out, if they change bases, they would then seem to believe they are killing off a whole bunch of numbers and creating another whole bunch of new numbers :D

For 1.25, if you want to stay in decimal, just write 1.24999~

montebianco
2006-Jan-10, 11:40 PM
Or for instance in another base, 1.(02)3

I don't understand the aversion some seem to have for there being different names for a number. They seem OK with the various language names for number one, un, uno, eine, yi, ichi...

It is frequently represented, named, symbolized, "1" but also such as:

01
1.0
1.00
1.(0)
1.0~
1.0...

and infinitely many more, as:

0+1
9/9
10^0
1!

All are just more names of one.

0.999~ is simply yet another name of one.

Don't forget exp(2*pi*i) :D

Lance
2006-Jan-10, 11:41 PM
For 1.25, if you want to stay in decimal, just write 1.24999~
Oh, what fun...

Does anyone have a kid in school working on decimals now?

2.0
-.5
___
1.4999~

And see what the teacher has to say.

2006-Jan-10, 11:43 PM
Perhaps the problem is that most real numbers only have one decimal representation, so it seems strange that a few should have two.

Carnifex
2006-Jan-10, 11:48 PM
Um... Because my teacher introduced me this problem and gave me several simple and obvious proofs that 0.(9) is 1 (or 0.(1) in binary, or 0.(2) in ternary etc), I actually assume there wouldn't be any problems with the answer you provided. However, the shorter form is used a lot times more often. Actually, I can't recall any serious scientist that would have used 0.(9) as a number, because it may be represented as 1 with a similar success...

EDIT (UPDATE):
Bad jcsd - actually, the funniest part of it is that every number, which is represented in a finite number of digits, has an alternative representation with a series of nines, and all the numbers which are represented by an infite number of digits do not have an alternative representation. And that counts for every positional system (only you change "nines" with another digit, which is last in that system).

Mathematics has a tendency to go against common sense. That's why it's wonderful.

snarkophilus
2006-Jan-10, 11:54 PM
0+1
9/9
10^0
1!

All are just more names of one.

I'm going to argue this, even though I agree that 0.9~ = 1.

0+1, 1!, etc. are expressions, not numbers. When you evaluate those expressions, you find they all equal 1, but each expression itself is a different creature, consisting of both operators and numbers.

This is an important distinction because, supposing you have defined addition but not division, you might be able to say that 0+1 = 1, but you couldn't say that 9/9 = 1. When you're dealing with the real numbers, where these things have been nicely defined, it doesn't make much of a difference, but if you were to take that notation to some other system, you might get in trouble.

It's also important from the point of view of your computer, for a variety of reasons. You might be able to enter 1.0 exactly as a floating point number, but when you calculate 9/9, you might get 0.9999(to machine precision). It's a subtle distinction, but if it happens a lot it can wreak havoc on sensitive calculations. There are also parser considerations (both natural language parsers and compilers).

</picking at nits>

2006-Jan-10, 11:58 PM
Bad jcsd - actually, the funniest part of it is that every number, which is represented in a finite number of digits, has an alternative representation with a series of nines, and all the numbers which are represented by an infite number of digits do not have an alternative representation. And that counts for every positional system (only you change "nines" with another digit, which is last in that system).

Yep, but most real numbers still only have a single decimal representation (most rela numbers are irrational and therefore only have one representation).

worzel
2006-Jan-11, 12:01 AM
EDIT (UPDATE):
Bad jcsd - actually, the funniest part of it is that every number, which is represented in a finite number of digits, has an alternative representation with a series of nines,
Oops, you're right of course (although you were replying to Bad jcsd so hopefully you missed my post).

and all the numbers which are represented by an infite number of digits do not have an alternative representation.Except those that are represented with a trailing infinity of 9s somewhere after the decimal point, but I guess you meant those that require an infinite number of digits.