.999999999999 repeating = 1

[Amaroq]

I think I understand and accept it now. I'm not 100% sure, but the explanation that .999... is a concept rather than a concrete makes sense to me. I so vehemently rejected it because I thought it had metaphysical implications, which I now know that it does not. This wouldn't be the first time this person has tried to use high level science or math to attack Oism.

[Rockefeller]

First off, 0.99 repeating is not the same as 0.(9).

...

The only reason you're arguing is because you don't know that definition.

You are right, I was not familiar with the notation 0.(9). But you are wrong in assuming that that is the reason I am "arguing". Eiuol's post as well as his question had nothing to do with 0.(9) and everything to do with 0.999 repeating (which is a valid notation, as I'll explain below in this post).

But a number is not a concept, it is an instance of a concept.

Wrong. Numbers are concepts formed from groups of existents. A given number refers to a group of existents having "something" in common. This "something" is the criterion (i.e. range of a characteristic) by which existents are being grouped. What is retained in forming concept of a number is the relationship between group members, and what is omitted is their all other measurements.

0.99 repeating is nothing (unless you care to mention how many times it's repeating ...)

0.99-repeating (or more popularly 0.999...) is a proper and legitimate notation. It has a specific meaning, which is that 9 is to be repeated "large" number of times. Exactly how large is "large" depends on the context in which the notation is being used. That is why I said in my previous post that this notation combines concept of method and concept of numbers. Let me repeat: 0.99-repeating is not a decimal notation unless one specifies how many times 9 is written.

That's absurd. Those are all concepts. Numbers are not even concepts, math is not a language, and 0.(9) is most definitely not the English to the Hindi of 1.

There nothing even remotely similar about the two examples.

Not absurd because numbers are concepts as I just explained. Further, the analogy was about relationship between concepts and symbols. A given concept (number) may be described in one language (notation) but not in another. Got that?

Jesus Christ, with the "analogous". The definition of decimals and fractions is that they are different representations of the same thing: real numbers.

Jesus Christ indeed! Decimals and fractions are representations of not-quite the "same thing" because some numbers can be expressed as decimals, others can't.

On a separate note, how do you feel about this statement, by the creator of C++:

"Proof by analogy is fraud. --Bjarne Stroustrup"?

I feel happy because, on a not-so-separate note, my analogies were not proofs but means of providing clarity. But more importantly I feel happy because I enjoy all (well... most) of your thought-provoking posts.

Edit: Grammar

Edited October 18, 2009 by Rockefeller

[Jake_Ellison]

0.99-repeating (or more popularly 0.999...) is a proper and legitimate notation. It has a specific meaning, which is that 9 is to be repeated "large" number of times.

I'm not aware of the definition of such a notation. What are its uses, how does it relate to the rest of math, is it used in Statistics?

Jesus Christ indeed! Decimals and fractions are representations of not-quite the "same thing" because some numbers can be expressed as decimals, others can't.

Sure they are exactly the same thing: rational real numbers. (I did correct that, in the edit, from real to rational real..even though the old version is noy factually incorrect -- real numbers which are fractions of natural numbers can also be represented as decimals, with or without that () sign meaning infinite repetition. Irrational real numbers cannot be represented by either, only estimated with the decimal representation, with any given precision. That's why we use letters for them.) Your concept is not an irrational number, it can be represented with fractions, with the help of a variable (which is implied in your decimal representation too) just fine.

Not absurd because numbers are concepts as I just explained. Further, the analogy was about relationship between concepts and symbols. A given concept (number) may be described in one language (notation) but not in another. Got that?

That certainly isn't true with your concept (see how I acknowledge I misspoke?) of 0.999... (repeated a large amount of times, make that n) is just 10^n-1 / 10^n, with n being a large number.

That is a fraction, which by the way uses terms that are well defined in math, so I am mystified as to why anyone would resort to saying 0.999....repeated large amount of times. It certainly doesn't help teaching it to children (since it isn't being taught to children), or using it with computers (since the other one can be used just fine, and takes far less space--for large n's, which is what we are talking about).

[Rockefeller]

I'm not aware of the definition of such a notation. What are its uses, how does it relate to the rest of math, is it used in Statistics?

I think you are right - my definition (which is to determine the extent of repetition contextually) is rejected by Mathematicians rather overwhelmingly.. However, I am not convinced about the rationale behind including recurring decimals (defined by indefinite repetition) in the decimal number system. Because notice that in order to perform arithmetic operations (addition, multiplication) with decimal numbers, one necessarily needs finite number of digits. To put it philosophically, doesn't the popular usage imply using a concept of method (infinity) to refer to measurements which actually exist in reality (rational real numbers)?

Sure they are exactly the same thing: rational real numbers.

Right, I see your point now.

It certainly doesn't help teaching it to children (since it isn't being taught to children), or using it with computers (since the other one can be used just fine, and takes far less space--for large n's, which is what we are talking about).

OK that makes sense. So the floating-point representation of 0.999-repeating would be 1.0E-1 and not 9.999<some_more_digits>E-1.

[Steve D'Ippolito]

1.0E-1 is 0.1. the floating point representation of 0.999999..... is 1.0E0=1.

[Rockefeller]

1.0E-1 is 0.1. the floating point representation of 0.999999..... is 1.0E0=1.

Sorry, I typoed.

[th3ranger]

I understand this an academic discussion, but I have doubts over whether this relates to reality. How many decimal places do you need before the data becomes meaningless? Would an engineer when measuring something differentiate between .99999999... and 1? Are there very many things that matter in measurement down to the millionth of any normal measuring unit? I have difficulty imagining it mattering a whole lot unless you were talking about light-years and/or the speed of light. Let me tell ya, .9999999999c is tons faster than .9c, at least if you want to get across the galaxy within your lifetime or not...

[Jake_Ellison]

How many decimal places do you need before the data becomes meaningless?

There isn't such a number, in general. In some cases, the number is very, very high.

Would an engineer when measuring something differentiate between .99999999... and 1?

If you mean between 0.(9) and 1, then no, because 0.(9)=1. Otherwise, yes, he might.

Are there very many things that matter in measurement down to the millionth of any normal measuring unit?

Yep.

[brian0918]

Are there very many things that matter in measurement down to the millionth of any normal measuring unit?

The relativity correction for GPS is down to the trillionth (4 parts in 10^12, or ~38,000 nanoseconds per day).

Edited October 19, 2009 by brian0918

[Amaroq]

Alas, I've gotta change my position again. xD It wasn't long before I again rejected this, but up until now, I never felt like posting about it again.

I believe that my intuitions about why 1 cannot equal .99999... were correct, I just didn't know explicitly enough why.

I will now attempt to explain.

I'm going to focus on the proof that uses thirds and show why it is fallacious.

The proof goes as follows:

1/3 = .33333...

1/3 * 3 = 3/3 = 1

.33333... * 3 = .99999...

Therefore 1 = .99999...

Here's why it's wrong.

There's a reason why 1/3 creates an endlessly repeating set of 3's. We're dealing with the number 1 in base-10. The integer 1 in base-10 cannot be evenly divided into thirds. You can divide it pretty closely into three parts if you leave a 1 as a remainder. But then you've got to resolve the remainder. Every time you attempt this, you add another 3 to the end.

I'm not a big math guy, so I don't know if there's a mathematical concept to describe this phenomenon. There certainly isn't one that I know of, so I'm going to make one up. Let's call it, oh, say... Base-Fraction Irresolution.

While I'm at it, I'm going to define a law to describe how to correctly work with it. Let's call it Amaroq's Law of Base-Fraction Irresolution.

Any number that is divided into fractions that does not divide evenly must retain a base-fraction irresolution equal to the fraction if it is represented as an infinitely repeating number. Mathematical operations performed on the infinitely repeating number must also be applied to its base-fraction irresolution. If at any point the numerator of the base-fraction irresolution can be evenly divided by its denominator, the irresolution has come to a resolution and the corresponding number must represent the resolved fraction.

All this does basically is keeps track of the fraction that you're representing. It makes sure that you're correctly applying your math to the fraction itself, and prevents you from twisting the number that you've represented the fraction by.

Amaroq's Law of Base-Fraction Irresolution invalidates the above "proof".

Let's look at it again, this time I'm going to point out what you're doing wrong.

1/3 = .33333...

The resulting .33333... has a Base-Fraction Irresolution of 1/3. Let's represent this like so.

IBR of .33333... = 1/3

Next, you multiply both sides by 3.

3/3 = .99999...

Here's where the fallacy lies. You multiplied the 3's in .33333..., but ignored the phenomenon that caused them to repeat:

The Base-Fraction Irresolution.

IBR of .33333... = 1/3

1/3 * 3 = 3/3.

IBR of .99999... = 3/3.

The IBR has been resolved, therefore it is impossible to arrive at .99999... by this method. When you multiply the number .33333... (which has an IBR of 1/3 since it was arrived at by fractioning) by 3, the only possible answer you can arrive at is 1.

If you continue to use this as a proof, you are either substituting an explicitly defined set of repeating 3's that was not arrived at through a fraction, or you are ignoring the IBR. Therefore, you're working with a .33333... that holds no IBR. If you multiply .33333... which has no IBR, it can no longer be said to equal 1. It can only equal .99999...

99999... cannot equal 1. The majority of the mathematical world is wrong, and I am right.

(I wonder if I can get an award for this.)

Edited December 10, 2009 by Amaroq

[brian0918]

Equal signs invalidate any notion of "arriving at" some way of displaying of a number being different in any way from another way of displaying a number.. You simply have to accept the fact that "..." represents something specific that you are misrepresenting in your own mind. 0.999... = 1 because the difference between them is infinitely small, which in mathematics means zero.

Edited December 10, 2009 by brian0918

[Amaroq]

I assert that there's an intangible difference between .33333... explicitly defined and .33333... arrived at as a representation of a fraction.

What I've defined here is a way to tell the difference between the two.

The former is arrived at by simply expressing the number. Or perhaps a sequence such as:

.3 + .03 + .003 ... (Infinitely tacking on more 3's at the end.)

The latter is arrived at by infinitely trying to resolve an uneven fraction. No matter how infinitely long those .333...'s get, there's always technically going to be a remainder.

I believe that the difference should be taken into account when math is performed on the number.

[Lucio]

I guess the concept of "infinite" is what is not-natural, maybe your're at odds with that.

0.9... = 1 because "9..." means "infinite". It is NOT a "large" amount ot 9's. its "infinte" amounts of nines. "Infinite" is the non-natural concept here (for me at least)

--

Question to the forum:

There is no such thing as "infinite" in nature?

There is no such thing as "infinite" in objetive observable reality?

--

You may see 0.9... as a different "notation" for 1, to cope with a "representation" problem of base 10.

It is not an "identity law"contradiction, just diferent notations,

3*1/3 = 1 = 0.9...

For example, if you use base 3 (i'm guessing here, please verify)

Base 10 / Base 3

1 1

2 2

3 10

4 11

5 12

6 20

7 21

8 22

9 100

10 101

So 1/3 = 0.3...

becomes (in base 3)

1 / 10 = 0.1

(no infinite digits in base 3)

So the infinite repeating 9 is a result from the base (10) choosen for representation of the fraction.

[Amaroq]

Yes, you've pretty much got what I mean Lucio. 1/3 ends up as repeating 3's because you can't evenly represent thirds in base 10. I think that you can't multiply such a representation and get repeating 9's, because at that point, you've come back around to a whole number again, and a whole number can't be represented as an infinitely repeating decimal. (Unless it's infinitely repeating 0's.)

EDIT:

I guess I can accept that .999... is just a notation for 1. But I can't understand how it can be so. I think that when someone multiplies the decimal representation of 1/3 by 3, they're ignoring the fact that, technically, there's a remainder, and this allows them to arrive at .999...

Edited December 10, 2009 by Amaroq

[brian0918]

I assert that there's an intangible difference between .33333... explicitly defined and .33333... arrived at as a representation of a fraction.

What I've defined here is a way to tell the difference between the two.

The former is arrived at by simply expressing the number. Or perhaps a sequence such as:

.3 + .03 + .003 ... (Infinitely tacking on more 3's at the end.)

The latter is arrived at by infinitely trying to resolve an uneven fraction. No matter how infinitely long those .333...'s get, there's always technically going to be a remainder.

I believe that the difference should be taken into account when math is performed on the number.

What is the basis for your belief? You assert a difference between them, yet equate them.

I could do the same with anything: "3/1 is arrived at by dividing 3 by 1. However 3.0 is arrive at by adding 0.0 to 3. They are equal, but intangibly different. This difference should be taken into account, because while 3/1 = 3.0, 1/1+1/1+1/1 doesn't equal 1.0+1.0+1.0, at least not for me it doesn't."

Edited December 10, 2009 by brian0918

[Amaroq]

The difference between my "intangible difference" and yours is that mine actually shows a real difference that will result in two different outcomes if these differences are taken into account when math is performed on the number. I'd like to see any single mathematical operation performed on your "differences" that results in different outcomes.

The basis for my belief is that, technically, .333... arrived at by a fraction still has a remainder. It's repeating because it hasn't resolved its remainder yet. Therefore, I think it's fallacious to perform math on it without taking this remainder into account as well.

EDIT:

Maybe it was wrong for me to say it's an intangible difference. It's certainly a difference that you can't detect just by taking the resulting numbers at face value.

Edited December 10, 2009 by Amaroq

[brian0918]

The difference between my "intangible difference" and yours is that mine actually shows a real difference

Where? Show me the equation that, when equal things are added to each side, results in an equation that is not true.

Edited December 10, 2009 by brian0918

[Lucio]

I guess I can accept that .999... is just a notation for 1. But I can't understand how it can be so. I think that when someone multiplies the decimal representation of 1/3 by 3, they're ignoring the fact that, technically, there's a remainder, and this allows them to arrive at .999...

The problem is there is no remainder. 1/3 has no remainder.

If you like to use base 10 for representation (and you don't want to use fractions) you must come with an "idea" to represent 1/3.

That idea is 0.3... and since "..." means "infinte", there is no remainder, so 0.3... * 3 = 1

Is the representation of dividing an object in 3 equal parts in nature, and the rejoining the parts.

Question to the forum:

one divided infinte equals zero ?

HA! HA! ... I answer my self by google, and at the same time found a mathematical explanation of socialism...

http://mathforum.org/library/drmath/view/62486.html

"In words, if 1 chocolate bar is divided among an infinite number of people, no one gets anything!"

[Amaroq]

brian0918,

By the rules of math, you apparently can't represent an infinitely repeating number with a different number tacked onto the end. So you'll have to allow me to use the concept I defined instead.

(Whoops, IBR should probably be BFI, to stand for Base-Fraction Irresolution. What was I thinking?)

Let's try this.

1/3 = .33333...

.33333... BFI 1/3 = .33333...

Multiply everything by 3.

The .33333...'s that has a BFI now has a BFI of 3/3. It has resolved its uneven fractioning into a whole number.

1 = .99999...

Oh wait, you wanted me to add something. Technically, multiplication is adding multiple times.

So let's try it this way.

1/3 = .33333...

.33333... BFI 1/3 = .33333...

Let's add .33333... to both sides. We're adding the non-fractioned one, so there's no change in the BFI.

.66666... BFI 1/3 = .66666...

I can already see where I'd need to make my "Law" more explicit. I think .33333... BFI 1/3 should be written (.33333... BFI 1/3). Then the distributive property would apply.

But anyway, let's divide them by two.

.33333... BFI 1/6 = .33333...

Uh oh, something's wrong here... The .33333... on the left is no longer a perfect representation of the fraction 1/3. When I added a non-fractioned .33333... to both sides, everything looked fine except that BFI. Adding a number by itself and then dividing it by two should get you to where you started, shouldn't it? But here it doesn't, because on the left side, I didn't really add the same number to itself. I added two different numbers together. A .33333... with a BFI, and one without. Now everything's screwy.

I hope I've at least illustrated that there is a difference though. I'll have to come back to this when I've figured it out.

Lucio, I think I understand what you're saying. And I think I understand something I'm taking for granted without explaining clearly.

Here's what I've been taking for granted and expecting everyone else to see.

1 ÷ 3 = 1/3

If you do long division, 1 divided by 3, the result is that you infinitely add 3's after the decimal point. At any point, if you stop, you're left with a remainder. That remainder makes you keep going. It's the cause of the infinite 3's. What I see in the fraction 1/3 is the same thing I see in the division. Those 3's are repeating because you're infinitely making the division into three parts more and more precise. You're approaching a division into three precisely equal parts, but you're never going to get there.

The concept I named is supposed to take this into account. The fact that there'll always be that little remainder at the end, is what I'm trying to show here.

1 ÷ 3 = .33333...

10

9

10

9

10...

(Shoot, the board destroyed my indentation. Pretend you're looking at long division here in which I showed my work.)

This long division example shows that you're always going to have a 1 leftover, which means bringing down the next 0 in line and dividing into the resulting "10".

The reason why 3 ÷ 3 can never come out to be .99999... should be obvious. You would have to do invalid division to arrive at anything other than 1. The invalid division argument is an argument that I've already used here in this thread that was promptly ignored/evaded. Though one person did at least explain that "That's not rigorous enough." I think that that's an admission that there's a contradiction in math, and that when there is a contradiction in math, the more abstract concepts are given precedence over the ones that are closer to validation to sense perception by reduction. If that's true, there are floating abstractions at work here...

3 ÷ 3 = 3/3 ≠ .99999...

[Jake_Ellison]

1/3 = .33333...

1/3 * 3 = 3/3 = 1

.33333... * 3 = .99999...

Therefore 1 = .99999...

You're right, that's not proof (the third line is meaningless), and I'll explain why at the end.

There's a reason why 1/3 creates an endlessly repeating set of 3's. We're dealing with the number 1 in base-10. The integer 1 in base-10 cannot be evenly divided into thirds. You can divide it pretty closely into three parts if you leave a 1 as a remainder. But then you've got to resolve the remainder. Every time you attempt this, you add another 3 to the end.

I'm not a big math guy, so I don't know if there's a mathematical concept to describe this phenomenon. There certainly isn't one that I know of, so I'm going to make one up. Let's call it, oh, say... Base-Fraction Irresolution.

There's an English concept to describe it: your method of dividing 1 (in base ten or whatever other base) by 3 is wrong. 1/3 does not equal 0.333, or 0.3333 or any other 0.3 number short enough to store on a hard drive.

1 cannot be divided by 3 for a precise result, in base 10. An inexact result can be very helpful, when it comes to calculating stuff with computers, but it cannot be used as proof against something that is true by definition. (namely that 0.999... = 1.)

guess I can accept that .999... is just a notation for 1. But I can't understand how it can be so. I think that when someone multiplies the decimal representation of 1/3 by 3, they're ignoring the fact that, technically, there's a remainder, and this allows them to arrive at .999...

The rules of multiplication were defined long before the creation of this 0.999... notation. So, the moment when you create the notation 0.333..., there is no use for it, other than to say 1/3=0.333...., by definition. If you want to multiply it, you have either to define "*"(multiplication) for these type of numbers, or don't multiply it. There is no need to define separate rules of multiplication, and even if there were, they would prove nothing. The reason why there isn't a need is because 0.333... equals 1/3 by definition, and you can just multiply 1/3. Or, if you prefer to be inexact but not deal with fractions, you can multiply 0.33333333333333333 or even longer, by 3, which is also defined.

Either way, neither 0.333333333333333 times 3, nor 0.333.... times 3 are useful in proving stuff about what 0.333.... is. It is what we said it is, 1/3. And 0.999.... is what we said it is, 1. ( it is definitely not 3 * 0.333..., that's not a defined method)

[Amaroq]

Ah, that's actually very interesting Jake. I've been arguing this whole time against a proof that was already false for completely different reasons, it appears?

From what I observed, everyone was taking this:

.33333... * 3

to be valid math. You're the first person to ever tell me that it's not. I assumed that the infinite sequence of 3's was automatically multiplied, every single one of them.

The concept I was trying to create and define was meant to do basically what you just said. To be exact and just perform the math on the fraction instead of being inexact and doing it on the infinite sequence of numbers.

Edited December 10, 2009 by Amaroq

[Robert J. Kolker]

I think I understand and accept it now. I'm not 100% sure, but the explanation that .999... is a concept rather than a concrete makes sense to me. I so vehemently rejected it because I thought it had metaphysical implications, which I now know that it does not. This wouldn't be the first time this person has tried to use high level science or math to attack Oism.

0.(9) meaning the 9s repeat indefinitely is a short hand notion of the following infinite series

SUM (n = 1, inf) [9/10^n]. (A)

This in turn is the limit of a sequence of finite partial sums as follows.

Let S\sub n = SUM (i = 1,...,n) [9/10^1]. ( B )

The infinite series A above is lim (n -> inf) [s\sub n]

The notion of limit is well defined. Let {a\sub n} be an infinite sequence of real numbers. This is a function from the set of integers {n} into the reals. Thus for each integer n, a\sub n is a real number determined by a rule. For example the sequence {1/n} is given by the rule n -> 1/n for n >= 1.

Now what about limits? We say lim (n -> inf) [a\sub n] = L if and only if

for any small positive real number e, there exists an integer N\sub a such that for all n > e |a\sub n - L\ < e.

So the notion of a limit as the variable "approaches infinity" has a perfectly definite meaning.

Now apply this definition of limit to the sequence of partial sums S\sub n and you have it.

Please refer to http://en.wikipedia.org/wiki/1_%3D_0.999...

for a clear elementary treatment of the matter.

Bob Kolker

[brian0918]

By the rules of math, you apparently can't represent an infinitely repeating number with a different number tacked onto the end.

On the end of what? A repeating sequence has no end.

I'm still waiting for the equation, alone, without any commentary, that shows what you're describing.

While you're mulling that, I'll ask you what is different between 1 and 1.000...

[Lucio]

Here's what I've been taking for granted and expecting everyone else to see.

1 ÷ 3 = 1/3

If you do long division, 1 divided by 3, the result is that you infinitely add 3's after the decimal point. At any point, if you stop, you're left with a remainder. That remainder makes you keep going. It's the cause of the infinite 3's. What I see in the fraction 1/3 is the same thing I see in the division. Those 3's are repeating because you're infinitely making the division into three parts more and more precise. You're approaching a division into three precisely equal parts, but you're never going to get there.

The concept I named is supposed to take this into account. The fact that there'll always be that little remainder at the end, is what I'm trying to show here.

1 ÷ 3 = .33333...

10

9

10

9

10...

(Shoot, the board destroyed my indentation. Pretend you're looking at long division here in which I showed my work.)

This long division example shows that you're always going to have a 1 leftover, which means bringing down the next 0 in line and dividing into the resulting "10".

The reason why 3 ÷ 3 can never come out to be .99999... should be obvious. You would have to do invalid division to arrive at anything other than 1. The invalid division argument is an argument that I've already used here in this thread that was promptly ignored/evaded. Though one person did at least explain that "That's not rigorous enough." I think that that's an admission that there's a contradiction in math, and that when there is a contradiction in math, the more abstract concepts are given precedence over the ones that are closer to validation to sense perception by reduction. If that's true, there are floating abstractions at work here...

3 ÷ 3 = 3/3 ≠ .99999...

"You're approaching a division into three precisely equal parts, but you're never going to get there."

You're confusing "representation" with division. You cannot "represent" the result in base 10 with a finite number of digits right of the decimal point. But you CAN divide 1 by 3, WITHOUT REMAINDER.

Using Base 3: 1 / 10 = 0.1

Simple, with no remainder.

"If" you select base 10, and follow the algorithm, it has no end. You recognize that and write "0.333..." as the result.

"..." means INFINITE, and is different from "a large amount"

INFINITE ≠ LARGE AMOUNT

1/3 = 0.333... WITH NO REMAINDER, because "..." means INFINITE.

-----

Dividing by 3 Using base 10 means dividing by 10 and the arrange in 3 groups.

It's like this:

--Ok, I need to divide this brick in 3 parts.

--Hey!, I have a brillant idea, since I have 10 fingers, I'll divide it in 10 parts and after that I'll form 3 groups.

-- Lets see... 3 groups of 3... and one part remainig

--Hey!, I have a brillant idea, since I have 10 fingers, I'll divide the remaining part in 10 and after that I'll distribute them in the 3 groups.

-- Lets see... 3 groups of 3... and one remainig

--Hey!, I have a brillant idea, since I have 10 fingers, I'll divide the remaining part in 10 and after that I'll distribute them in the 3 groups...

...

-----

3 ÷ 3 = 3/3 ≠ .99999...

If you accept 0.999... as a different notation for 1,

then 3 ÷ 3 = 3/3 = .999... = 1

The reason why 3 ÷ 3 can never come out to be .99999... should be obvious

Sorry, but if "0.999..." is a different notation for 1 then 3 ÷ 3 = 0.999...

as true as 3 ÷ 3 = 1

I agree that is stupid to have a complicated confusing notation as 0.999... to mean 1 (in base 10)

Saying 0.9(INFINTE) to mean 1 is like taking a magic trick out of a hat,

but stupid complicated confusing magic trick out of a hat notation ≠ false

http://en.wikipedia.org/wiki/0.999...

[Robert J. Kolker]

On the end of what? A repeating sequence has no end.

I'm still waiting for the equation, alone, without any commentary, that shows what you're describing.

While you're mulling that, I'll ask you what is different between 1 and 1.000...

As real numbers there is no difference. However 1 taken as an integer has a different mathematical context than 1 taken as a real number. You can add a teeny tiny real number to the real number 1. You can only add integers to the integer 1.

When talking about a mathematical object it is useful to specify the set or domain to which the object belongs. With the set or domain comes all the postulates and rules for dealing with members of the set or domain.

Bob Kolker