1. Originally Posted by whiteflags I don't think you want to make a linguistic argument out of this, either.
No, in such a long topics everything that becomes a "linguistic argument" can be fairly dropped and we can focus on the more important part.
So I would stick on "0.9999.... = 1" issue. 2. Originally Posted by C_ntua As far as I can tell both are wrong assumptions. So lets start again more thoroughly

1.
Code:
`0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )`
We first solve the "lim" part as
Code:
`0.9999... =  9*10^-1 + 9*10^-2 + 9*10^-3 + ...`
which is correct.

2.
Code:
`1 - lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )`
again solve the lim part
Code:
`1 - 9*10^-1 + 9*10^-2 + 9*10^-3 + ...`
then using the above
Code:
`1 - 0.9999..`
all of this are simple and clear. But I fail to see how the above actually equals to zero.
Okay let me start again from this point. Apparently you agree the following:
Code:
```0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )
1 - 0.9999... = 1 - lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )
1 - 0.9999... = lim(x -> inf) ( 1 - 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )```
Am I right in that you agree on that?
Now the first two term are "1 - 9*10^-1". Combining them, by actually carrying out the subtraction, we get "1*10^-1" (to say: 1 - 0.9 = 0.1).
Now we do it for the second term: "1*10^-1 - 9*10*-2" which is 0.1 - 0.09 = 0.01.
Etcetera.
We repeat that over and over again until we're left with:
Code:
`1 - 0.9999... = lim(x -> inf) ( 1*10^-(x-1) - 9*10^-x )`
And then once more:

Code:
`1 - 0.9999... = lim(x -> inf) ( 1*10^-x )`
Now do you agree that the limit of 10^-x as x goes to infinity is 0? So, that:
Code:
`0 = lim(x -> inf) 10^-x`
Just as the limit of 1/x for x goes to infinity is 0, which is probably the first limit anyone learns.

So finally:
Code:
```We know:
1 - 0.9999... = lim(x -> inf) ( 1*10^-x )
0 = lim(x -> inf) 10^-x

Hence:
1 - 0.9999... = 0
1 = 0.9999...```

I'm not even interested in proving my right here, honestly, or proving you wrong. I would honestly try to find out where the disagreement exist and find out where it is, for either or maybe both of us to learn something. 3. Although I've found this thread entertaining and educational, I'm not really trying to get involved. I'm just going to throw out what I find wrong with the proof and see what happens.

Now, I'm good with some mathematics, but not great with calculus (where I know what I know about limits from), so I may have missed something you are doing with the algebra or the limits.

Code:
```0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )
1 - 0.9999... = 1 - lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )
1 - 0.9999... = lim(x -> inf) ( 1 - 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )```
Now, all of these involved the limit of many (`x' minus one assuming starting at zero) constant functions (which would be a constant) plus the limit of an infinitesimal (having the limit of zero).

I'm fine with those bits even if the equation is justified by the implication of the "..." bits; I'm just stating where I'm coming from.

[Code]1 - 0.9999... = lim(x -> inf) ( 1*10^-x )[/Quote]

Yea, this doesn't work for me. Because you've lost the constants the "equation" doesn't hold because the limit of the function `1*10^-x' as x approaches infinity is simply zero and not some unknown quantity as implied by the "..." bits.

Now, I agree that the limit of the function, for lack of a better word, `1-0.9999...' is zero. That said, the limit of a function doesn't imply anything about the results of the function at the limit (*). The result of a function applied to a given value doesn't imply anything about the limit. You seem to be saying that they are the same with the above "equation". If you were using `lim' on both sides, I'd be right with you.

Anyway, since this doesn't work for me, what follows from the argument also doesn't work for me.

Soma

(*) I feel that sounded weird, but the results had done been seen and may be quoted before I get done with the edit.

With that in mind, I'm getting at the fact that a given application of a function may or may not be defined at the limit, may be a value relative to the value associated with the limit, or may be any other value.
[/Edit] 4. Well I don't see why we couldn't compute 0.999... with limits but it's probably better stated as a sum anyway:

0.999... = Σ(x = 0, x->infinity) 9*10^-x
1 - Σ(x = 0, x->infinity) 9*10^-x = 0
1 = 0.999...

Interestingly if you did the subtraction for any finite length of 0.999... you would eventually have to write that last 1, but for forever places, things are different. No last digit. 5. EVOEx there is not disagreement yet, it is just what you presented wasn't fully clear.

So basically the "unclear point" was
Code:
`lim(x->inf) (1 - (9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x)) = lim(x->inf)(1*10^-x) = 0`
but now you clear it, sorry if this wasn't obvious for me.

The key point here is:
Code:
`lim(x->inf) (1 - (9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x)) = lim(x -> inf) ( 1*10^-(x-1) - 9*10^-x )`
since it is the point you get rid of the "..."

Thinking out loud:
Code:
```1*10^0 - 9*10^-1 = 1*10^-1
...
1*10-(n-2) - 9*10^-(n-1) = 1*10^-(n-1)
1*10-(n-1) - 9*10^-n = 1*10^-n```
But to get what you are you need to add up all these equations.
The blue will be canceled out so you will end up with
Code:
`1 -  (9*10^-1 + ... + 9*10^-n) = 1*10^-n`
which is what you want.

So the only thing left actually is
Code:
`0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )`
since this is where the 0.999... come in place.

To prove this is true I would say you can claim
Code:
`0.9999... = 9*10^-1 + 9*10^-2 + 9*10^-3 + ...`
Code:
`0.9999... = 9*10^-1 + 9*10^-2 + 9*10^-3 + ... +  lim(x -> inf)(9*10^-x)`
but then you add a "lim(x -> inf)" to everything and get what you want.
EDIT: Sorry, scratch that and I think that is where I got confused. I don't think you can add a limit to everything and get what you want. I believe the limit has to be defined to do this which is not really clear here.
Code:
`lim(x -> inf)(9*10^-x) + lim(x -> inf)(9*10^-(x-1)) + .... + lim(x -> inf)(9*10^-(x-2)) + ....`
so everything is zero. In the end you have
Code:
`0.999... = Σ(i=0...inf)(lim(x -> inf)(9*10^-(x-i)))`
where it is actually:
Code:
`0.999... =Σ(i=0...inf)(9*10^-i)`
you see the difference? The thing is that if you use a limit you cannot claim you have
Code:
`9*10^-x + 9*10^-(x-1) + ... + 9*10^-1`
think of the "x-i" part. You have "i->inf" and "x->inf". You assume that "x is getting smaller and smaller that it gets to 1. Right? But if x is near infinite it will never get to 1 no matter how many times you subtract from it. So concluding:
Code:
`0.9999... = lim(x -> inf) ( 9*10^-1 + 9*10^-2 + 9*10^-3 + ... + 9*10^-x )`
is NOT true.

Well, that is not a "passionate" NOT, please feel free to prove it, I am just thinking out loud here, maybe I am talking crazy and after 5min re-edit...

EDIT2: To clarify I would say that
Code:
`0.9999... = 9*10^-1 + 9*10^-2 + 9*10^-3 + ... +  9*10^-i where i->inf`
is the proper way to right it, but it is not a limit. The same error as you would do if you had
Code:
```lim(1/x) = 0 where x->inf
is equal to
1/x = 0 where x->inf```
where this is not true. The limit means "it approaches" a number as is already stated, not that it is equal, so using despite the math in the end putting the "limit" to "0.999..." is not right. Since you can say that the "limit of 0.999... is 1" but that doesn't mean that the equal 1. I believe there is the confusion? 6. Code:
`0.999... = Σ(x = 0, x->infinity) 9*10^-x`
Yep.

Code:
`1 - Σ(x = 0, x->infinity) 9*10^-x = 0`
Nope. (And again, using that as proof doesn't work if I don't accept that bit.) You are still only using the definition of the limit to prove the point. (And, I'm not arguing against the limit or the value of the limit in this case.) It is only the that, intuitively, the limit of a thing is not the thing itself, so to speak. (See the earlier post if I've confused you.)

Even if you combine them (Using both limit and sum with a bit of algebra which makes the most intuitive sense to me. Just not the one C_ntua posted which is wrong.), you still don't get the equation. The notion that "you never get to one" is fine. On the other hand, the notion that "you never get to move away from one" is also fine. To me that doesn't imply the equation.

Note also, that I have no problem with this being the case in formalism (I'm a computer scientist, I'm used to taking things on faith.) or the implied limit. (That's simply what it is so, yea.) I may have missed that part of this debate if that's the case. I didn't read every single post. I simply assumed we were talking about the values intuitively as an infinitesimal and an integer. In which case, ignore me and I'll go back to doing other things.
[/Edit]

Soma 7. The forum ate the first attempt to edit and the first attempt at posting.

I believe there is the confusion?
Probably. Possibly from the standpoint or formalism versus intuition, the notion or limits, and the nature of infinitesimals?

Meh. Anyway...

Soma 8. First: Σ(x = 0, x->infinity) 9*10^-x = 9.999...

Second: .999... = Σ(x = 0, x->infinity) .9*10^-x = .9/(1-.1) = 1 9. Originally Posted by phantomotap Probably. Possibly from the standpoint or formalism versus intuition, the notion or limits, and the nature of infinitesimals?
Indeed. The issue isn't on the formal validity of the so-called proof. There is really no much point arguing over that. The problem on the conclusion that is being drawn from that limit. This is valid in calculus, is often also accepted that the result of a limit can transit into algebra, but for this particular case in which the identity of a sequence is trying to be established, this simply cannot be used as proof. It's circular reasoning. It's of no use as a means to form proof. For that matter, I'd take a fraction-based proof any day before I can accept the result of a limit that is then transported into algebra.

As for infinitesimals... it's a mess. Real numbers cannot have them because the language wasn't properly constructed (or wasn't intended?) to accept them. It lacks resolution. Yet, Cauchy sequences which lie deeply ingrained in the concept of limits and sequences, depend on the concept of an infinitesimal. There's inconsistencies in the language and the axioms. It's no wonder these debates can then emerge and pretty much go on forever.

My real beef isn't so much about the idea that 0.999... is 1. I can live with that; Have been saying so for 4 pages in this thread. The world has live with that! My real beef is with the fact this notion actually developed a group of partisans who pretend to prove what cannot be proven in R. No proof can exist in R that establishes the identity of a rational number with a non terminating sequence, no more than the impossibility of establishing the identity of an irrational number. Still wikipedia articles are written on the matter in condescending tones, as if any of the so-called proofs had any mathematical validity whatsoever. Pretty amazing.

I'd rather prefer some humility. Something like "0.999... is 1 in R. It cannot be proven. But this is why it must be [insert formulaic explanation the likes of which we have seen on this thread]" 10. The issue isn't on the formal validity of the so-called proof.
I was talking about the formalism of the implied limit used in the proof.

*shrug*

Same difference I guess.

Soma 11. Originally Posted by User Name: First: Σ(x = 0, x->infinity) 9*10^-x = 9.999...
Second: .999... = Σ(x = 0, x->infinity) .9*10^-x = .9/(1-.1) = 1
Again, sticking only on the mathematical part, I wouldn't agree on the second. Though of course all my posts are completely pointless, would have been easier if I actually opened a math book rather than thinking and "arguing" but I will skip this part....

So if the second is believed to be true you can elaborate a bit...

I also can live with "0.999.... = 1", but nevertheless if it can be proven and how it can be proven are interesting parts for me. 12. Originally Posted by C_ntua I also can live with "0.999.... = 1", but nevertheless if it can be proven and how it can be proven are interesting parts for me.
All limit theory says, in this case, is that as the number of 9's increases then the value on the left hand side of the expression continually gets closer to 1. Or, to put it another way, we can get as close as we like to 1 by adding more 9's.

Conventionally mathematicians state that that the values will be equal in the limiting case where the number of 9's is truly infinite.

This thread appears to be taking the rather novel stance that conclusions derived using the theory of limits, and associated conventions, are invalid. According to that stance, the result is unprovable in the context of this thread. 13. Originally Posted by grumpy All limit theory says, in this case, is that as the number of 9's increases then the value on the left hand side of the expression continually gets closer to 1. Or, to put it another way, we can get as close as we like to 1 by adding more 9's.

Conventionally mathematicians state that that the values will be equal in the limiting case where the number of 9's is truly infinite.
Yes, but you are adding every time a smaller amount. So in the end you are adding something very close to 0. Which in the case of a truly infinite amount of 9s that will imply that you are actually adding 0 thus you will never get to 1. In other words the amount you are adding can get closer and closer to 0 and in the limiting case it will equal to 0. 14. Something like "0.999... is 1 in R. It cannot be proven. But this is why it must be [insert formulaic explanation the likes of which we have seen on this thread]"
When it is not specified otherwise, R is implied. In *R, .999... may not equal 1, but who considers *R when it is not specified? No one. 15. Originally Posted by C_ntua Yes, but you are adding every time a smaller amount. So in the end you are adding something very close to 0. Which in the case of a truly infinite amount of 9s that will imply that you are actually adding 0 thus you will never get to 1. In other words the amount you are adding can get closer and closer to 0 and in the limiting case it will equal to 0.
It is true that lim (x->infinity) of 10^-x is zero. And so is 9 * {lim (x->infinity) of 10^-x} ..... 9 times zero is still zero, whether that zero is a limiting case or not.

Or, each time a digit 9 is added, it is adding less than was added by the preceding digit 9, because it is being divided by a larger power of 10. Popular pages Recent additions 