1. 0.999 ... is 1 for the same reason 0.333... is 1/3, it's just a different expression for the same number.

2. Originally Posted by laserlight
Prove it. (The qualifier "any number of nines" sounds strange to me, given the context: we're talking about the 9s repeating without end.)
I can't proof it, but I can proof the reverse, that 0.9999..... = 1. I haven't read the entire thread, but I couldn't find the actual proof in it glancing over it.

10*0.99999.... = 9.99999....
Subtract 0.99999.... from 9.99999... As there are an infinite number of 9's in both, each 9 in 0.999... has a matching 9 in 9.9999... it "removes". Hence:
9.9999... - 0.9999... = 9

So let's say a = 0.99999...
I just showed that 10*a - a = 9.
9*a = 9
a = 1

Not my own proof, but I've always particularly liked this proof...

3. The problem here is there is in quantifying infinity. I think what you're trying to say is that as the number of nines approaches infinity, the value approaches 1. (?)

Been a while since I've done any real math, but I'm much more comfortable with that than saying 0.999... = 1.

edit: btw, I like the identity proof.

4. Originally Posted by Perspective
The problem here is there is in quantifying infinity. I think what you're trying to say is that as the number of nines approaches infinity, the value approaches 1. (?)
You're right saying that: as the number of nines approaches infinity, the value approaches one. However, if there is an INFINITE amount of 9's, as indicated by 0.999..., then the proof I presented holds.

There's the difference between the number of 9's approaching infinity, or there actually being an infinite number. They're not the same thing.

5. Originally Posted by Mario F.
Isn't it self-evident that an irrational number composed entirely 9s is smaller than its closest higher integer? Maybe you math folks have some way to formalize this proof. I do wonder though if it is required when we have already been expressing such things in other ways. Like with an interval, for instance ( [0, 1[ ).
But if it is smaller you should be able to find a number betwean 0.999.... and 1.

6. Originally Posted by Perspective
If there's one thing I've learned about math, it's that Sang-drax is always right.

What's the average of infinity and 12? You can't plug concepts like that into normal math equations. The idea of 0.999... is different from a specific number.

And certainly 0.999... is not equal to 1. Just because you can't find a number in between them doesn't make them the same. The inequality 0.999... < 1 still holds for any number of nines.
.999... is a number, nothing more, nothing less. It is well defined as .999.... But applying the concept of infinity and generalizing that you can expect something true for infinity to be true for "any number" [arbitrarily large] is a fallacy.

7. Originally Posted by Mario F.
Isn't it self-evident that an irrational number composed entirely 9s is smaller than its closest higher integer? Maybe you math folks have some way to formalize this proof. I do wonder though if it is required when we have already been expressing such things in other ways. Like with an interval, for instance ( [0, 1[ ).
That is true for any finite number of nines. You can prove it with nested interval theorem. EDIT: Actually, I'm not sure I can . However, you can infinitely nest intervals with 1 and .999... always in them. Nested interval theorem states there is exactly one number in every infinitely nested interval.

8. Originally Posted by Shakti
But if it is smaller you should be able to find a number betwean 0.999.... and 1.
Of course I can. It's 0.999...

Real numbers are uncountable. The "you cannot find a number" is simply an exploitation of our inability to represent these numbers to their infinite precision. It does not mean however such a number can't exist. Many theorems have proven the infinite set of rational numbers. The same way we use oo to represent infinity, for the purposes of this debate we can simply stipulate that 0.(9) is not a number, but the representation of a number that has an infinite number of 9s.

As such, 0.(9) stands between 0.(9) and 1. Or, if you want a more complete answer, oo stands between 0.(9) and 1.

9. Originally Posted by Perspective
What's the average of infinity and 12? You can't plug concepts like that into normal math equations. The idea of 0.999... is different from a specific number.
Where is this "infinity" thing coming from? The fact that the number of digits is infinite is irrelevant.

How do you express the value 1/10 in decimal? It's 0.1. How do you express it in binary? It requires an unending number of digits. Thus, the infinitude of the representation is an artifact of the chosen base, not anything to do with the number itself.

And certainly 0.999... is not equal to 1. Just because you can't find a number in between them doesn't make them the same. The inequality 0.999... < 1 still holds for any number of nines.
It's a tough concept and there are many ways of looking at it, but 0.999... is indisputably equal to 1. Period.

The real numbers are dense. This means that between any two unequal real numbers there are an infinite number of real numbers. Thus if 0.999... and 1 are not equal, there are an infinite number (uncountably infinite, actually) of values between the two. Name just one of them.

10. Originally Posted by brewbuck
The real numbers are dense. This means that between any two unequal real numbers there are an infinite number of real numbers. Thus if 0.999... and 1 are not equal, there are an infinite number (uncountably infinite, actually) of values between the two. Name just one of them.
I think I have on the previous post; if you want one such number, then I'll give you 0.(9). If you want more, then I'll give you oo.

11. Originally Posted by Mario F.
I think I have on the previous post; if you want one such number, then I'll give you 0.(9). If you want more then I'll give you oo.
You're saying 0.(9) is a number which is unequal to itself?

EDIT: Okay, let's start the long process of going through each of the thousands of explanation of why 0.999... == 1.

Let x = 0.999...
Thus 10 * x = 9.999...
Thus 10 * x - x = 9.999... - 0.999...
All the nines on the right of the decimal subtract away:
10 * x - x = 9
9 * x = 9
x = 1

The only way the proof fails is if 0.999... != 0.999...

12. It follows on the same concept that oo does not equal oo, brewbuck. It's another of infinity paradoxes; also somewhat explored by Hippasus when he proved that the square root of 2 is both odd and even.

Conceptually you know very well that for every precision, no mater how large, that you can think for 0.(9), you need always to add another decimal place. You'll do this indefinitely. However, for 0.(9) to become 1 you need to add 0.(0)1. You are required an infinitesimal to reach 1. You are thus required another infinity.

But I also believe your axiom to be wrong:

Let x = 0.999...
Thus 10 * x = 9.999...
Thus 10 * x - x = 9.999... - 0.999...
And here you will make your first mistake:
All the nines on the right of the decimal subtract away:
10 * x - x = 9
They don't really. Not for the "last" decimal place, represented by an infinitesimal. If we replace the above for a finite quantity it's easy to see why:

Let X = 0.999
Thus 10 * x = 9.99.
Thus 10 * x - x = 9.99 - 0.999
Oops!

Next you make your second mistake, I believe:
10 * x - x = 9
But oo-oo != 0. It's undefined. It's both 0 and 1, or 12 and 13. Or any other number you can imagine. One of the properties of infinity is that it does not equal itself. Mathematically your expression becomes undefined at this point. The thought is that we know there is a number for your expression, but we can never reach it. As such it's undefined.

10 * x - x = undefined

I suppose I can be corrected somewhere since my math skills are tremendously limited. But more than my gross attempt at correcting you, I think it stands that 0.(9) is the representation of a number that never reaches 1. On the right side of an interval, this number would be represented as "1[" and this is a crucial hint to the nature of this number and the fact it does not equal 1.

Therefore

1/3 + 1/3 + 1/3 = 1
Iff 1/3 = 0.333... then 0.333... + 0.333... + 0.333... = 0.999...
QED.

There are real numbers that result in a repeated sequence in any base. 0.2 doesn't stop being 0.2 when you express it in binary. But I think 1 is special because it's a unit. I think that in many bases (above binary perhaps, unless you entertain fractional bits) 1 has alternate expressions.

But again... pants on head retarded here.

14. Ah, I see your point now. Trouble is it has been long established that irrational numbers cannot be represented as fractions. So, 0.333... cannot be really represented as 1/3. Why? Well, exactly because of the example you gave. Many other examples exist where the irrationality of irrational numbers is evident every time you try to represent them as fractional numbers, like the aforementioned Hippasus proof.

We may do so on a daily-basis for convenience sake because the margins of error we impose become acceptable. But when it comes down to discussing the nature of 0.333..., or 0.999... for that matter, bringing in fractional numbers will not do because it will introduce an error margin that simply isn't acceptable.

15. 1/3 is not irrational on the basis that I just expressed it in rational form. (e.g. sqrt 2 is irrational. ) But I see you're point: 0.333... is an approximation of 1/3, just as 0.999... would be of 1. As an approximation, it couldn't be exactly equal, but the difference is infinitesimally small, so by all means of proof it would be equal.

EDIT: By that I mean 0.999... or 0.333... carried out to some finite place would be an approximation but the difference between the actual numbers and 1 and 1/3 respectively are so small.