User talk:ConMan/Proof that 0.999... does not equal 1

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This is a place to question the proofs as to why 0.999... does not equal 1. Please be civil and discuss the proof, not the prover. If you have submitted a proof and it is questioned, you may either respond here or give further explanation in your proof.

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[edit] Please give proof of this assertion

Please give proof of this assertion:

"In fact it is not defined at all because our rules of arithmetic for multiplication apply to rational numbers only."

Firstly, 0.999... is a rational number, all repeating decimals (which includes terminating decimals, since it's just repeating 0's) are rational numbers. The key thing is that Σ9 / 10n converges, so basic arithmetic is well-defined on it. Secondly, multiplication applies to many things beyond rational numbers. The square root of 2 times the square root of 2 equals 2, for example - do you dispute that? --Tango (talk) 23:53, 13 December 2007 (UTC)

0.999... is not a rational number because it cannot be expressed as a rational number. Numbers that are expressed in any radix system with or without repeating patterns in a non-finite representation are not rational numbers. What is the square root of 2?

Please give proof that 0.999... is a rational number. Do not tell me it converges to a limit because pi also converges to a limit and it is not a rational number.

Also, convergence has 'nothing' to do with arithmetic. The concept of convergence came along centuries after the birth of arithmetic.

98.195.24.26 (talk) 22:57, 14 December 2007 (UTC)

0.999... can be expressed as a fraction - it's 1/1. If you are going to say otherwise, you need to give proof. You need to prove that there are no two integers that, when you divide one by the other, you get 0.999... (which you can't do, since it's not true). What I was saying about convergence doesn't relate to just rational numbers, it relates to decimal expressions in general. A decimal expression is a way of denoting a power series - it's the coefficients of all the powers of ten. You can do basic arithmetic on a power series by doing it on individual terms if and only if it converges (that would probably be covered in a first year undergraduate analysis course, although maybe not with full rigour). All decimal expressions are convergent power series, therefore you can do arithmetic with them. And what you do mean "What is the square root of 2?"? It's the number that, when squared, gives 2. It's about 1.4, and is an irrational number (I imagine there is a proof of that in a Wikipedia article somewhere - if you can't find one, I can reproduce the proof, it's very simple). --Tango (talk) 01:24, 15 December 2007 (UTC)
If you are going to say that 0.999... can be expressed as a fraction, you will need to prove it. You can't say 1/1 because then you are 'assuming' they are equal. If you can perform 'basic arithmetic' on individual terms of power series, then you are performing finite arithmetic on their partial sums, not the limit of the partial sums. You say all decimal expressions are convergent
power series - this is untrue. You cannot tell me that the square root of two is 'that number' which when multiplied gives 2. It's about 1.4 you say. You cannot find a limit w say, such that w*w = 2 but you perform approximated arithmetic on its 'about value' as you call it. So you say that square root of 2 is an irrational number and you can prove it. Hmmm, interesting. I believe
I have seen this 'proof' of your's. I can prove that it is rational according
to your previous statement that the sqrt(2) can be expressed as a power series.
Proof: All the partial sums of this power series are rational. By simple induction, if I add each term, the result is another rational number in the form a/b. Start with a/b + c/d where these are the first two terms. The result is: (ad+bc)/bd. Consider the next term e/f. Adding this term we have (adf+bcf+ebd)/bdf - once again a rational number. The rest I leave to you as
an exercise in mathematical induction. Third year high school mathematics I think, but it's been such a long time, I am not certain. Now if the above reasoning is true, what does this say about pi? Is it possible that pi
is also rational because of the misinterpretation of the original definition of a rational number? So Tango, seems like you made a lot of statements
and you have provided no proof. I want to see proof only. I cannot accept phrases like "about 1.4" or "0.999... is equal to 1/1 by definition" - it is not equal to 1 by any definition. Also, the fact that you have passed real analysis courses does not mean you understand these details. I have passed these courses too but do not agree with everything that is taught therein. I would say that over 70% of what is taught is incorrect and useless. 98.195.24.26 (talk) 14:15, 15 December 2007 (UTC)
There are numerous proofs that 0.999...=1 in the article, I have no need to provide any more. The definition I gave *is* the definition of root 2, and there are sequences of rational numbers (for example, truncated decimal expansions) which tend to it (although, I don't see how that is relevant). Decimal expansions are convergent power series, have you taken a course in elementary real analysis? If not, then I suggest you go and find a decent text book on the subject and inform yourself. As for performing basic arithmetic on power series by performing it on the individual terms - it's a widely known mathematical theorem. I can't remember the proof off the top of my head, but it will be in any good analysis text (I think it's a corollary of the Calculus of Limits Theorem). Your proof is flawed - induction works for arbitrarily large finite numbers, not for infinity. If it quite possible for the limit of a sequence not to have a property that the individual terms of the sequence have. Mathematics is not a matter of opinion - if you disagree with your lecturers, you are simply wrong. Mathematical proofs are absolute, you can't choose whether or not to agree with them. --Tango (talk) 15:26, 15 December 2007 (UTC)
I have not seen one valid proof in the 0.999...=1 article. You say that all decimal expansions converge - does 1 + 1/2 + 1/3 + 1/4 + .... converge? It is a decimal expansion. You gave no definition of root 2 anywhere - only what you think it is. Try answering the questions Tango. State a proof that you think is valid and I will show you that it is invalid. How wrong you are about mathematics - I can see you are a product of the flawed system of education. You state my proof is flawed, yet you do not provide any valid reason, only muddled nonsense. No proof is absolute - there is no such thing. It exists only in your imagination. 98.195.24.26 (talk) 16:55, 15 December 2007 (UTC)
The harmonic series (which is what the series you mention is called) is not a decimal expansion. Please let me now when you've found out what a decimal expansion is, and we can continue this discussion. Until that point, I am wasting my time. --Tango (talk) 18:09, 15 December 2007 (UTC)
Oh really? 1 + 0.5 + 0.333... + 0.25 + 0.2 + .... The harmonic series *is* a decimal expansion because it is a sum of decimal expansions. It is I who am wasting my time, not you. 98.195.24.26 (talk) 22:52, 15 December 2007 (UTC)
It's a sum of numbers which can be expressed as decimal expansions (ie. real numbers), it's not a decimal expansion. A decimal expansion is a series of powers of 10, for example 0.999...=9*1/10 + 9*(1/10)^2 + 9*(1/10)^3 + ..., that's what decimal notation means. Since 1/10 is less than 1 and the coefficients are constant (in this example - in general, they are bounded above [by 9], which I think it enough to guarantee convergence, but I can't immediately see how to prove it), it converges by the ratio test. --Tango (talk) 23:17, 15 December 2007 (UTC)
I've worked out how to prove the general case - just use the comparison test with 0.999.... --Tango (talk) 23:28, 15 December 2007 (UTC)
Of course *it is* a decimal expansion: 1 + 0.5 + 0.333... + is expressed as a series in powers of 10 because its sum is expressed in powers of 10 - this is what a decimal expansion is. And this decimal expansion does not converge 'contrary' to your claim that all decimal expansions converge. Now how about getting back to the subject? Show me any proof that 0.999... = 1 and I will show you that it is false. You sure know how to beat about the bush just like your fellow wikipedians. You say to use the comparison test with 0.999... - this test only shows that 0.999... converges to 1. It does not prove equality. A number is 'not equal' to the 'limit of a Cauchy sequence'. Cauchy did not exist when the foundations of arithmetic were laid. 98.195.24.26 (talk) 14:38, 16 December 2007 (UTC)
You're using "decimal expansion" wrong. A decimal expansion is not a type a number, it a way of expressing real numbers. Each of the numbers in your sum can be expressed as a decimal expansion (ie. they are real numbers), but the sum itself is not one (and cannot even be written as one, since, as you say, it doesn't converge). Convergence of 0.999... doesn't prove it equals one, I agree, it just proves you can do arithmetic with it. That arithmetic (x=0.999... => 10x=9.999... => 10x-x=9 => 9x=9 => x=1) proves it equals one. --Tango (talk) 19:14, 16 December 2007 (UTC)

Time to jump in here - taking the text straight from the introduction of decimal representation:

A decimal representation of a non-negative real number r is an expression of the form

r=\sum_{i=0}^\infty \frac{a_i}{10^i}

where a0 is a nonnegative integer, and a_1, a_2, \dots are integers satisfying 0\leq a_i\leq 9.

Hence, the harmonic series is not a decimal expansion, because the individual terms cannot be written in the form an / 10n, where an is an integer between 0 and 9. 0.999... does follow these restrictions, and it's fairly easy to prove that the sum will converge for such an expression. As to User:98.195.24.26, "Show me any proof that 0.999... = 1 and I will show you that it is false." - 0.999... has at least six proofs, of varying levels of sophistication and rigidity, covering several different fields of mathematics. Several more proofs have been presented on the Talk and Arguments pages, although you'll have to go back through the archive pages to find them all. One of the proofs I offered explicitly demonstrated that, under a few assumptions that allow you to actually take the infinite series to be an arithmetically manipulatable number, then you could prove that for the infinite series to have any meaningful value it had to be equal to the limit of the partial sums. If you are going to go through every single proof offered there, and demonstrate it to be critically false - i.e. not just invalid, but such that it cannot be amended to be correct, please go ahead. That's what this page is for. Confusing Manifestation(Say hi!) 23:20, 16 December 2007 (UTC)

Okay, let's get back to the main topic. Tango suggested the 10X proof is correct. Let's see why this proof is false.
     x = 0.999...
   10x = 10 (since 9.999... = 10)

=> 9x = 10 - 0.999... => x = (9.000...)/9 => x > 1

This cannot be true because the limit of 0.999... equals 1. So you see, this 'proof' must be false, because it can be used to show that x > 1. So what's the next 'proof' you want me to disprove?98.195.24.26 (talk) 00:38, 17 December 2007 (UTC)

9.000... equals 9, so 9.000.../9=1, it's not greater than 1. You haven't disproved anything. --Tango (talk) 01:57, 17 December 2007 (UTC)
I have to agree. You seem to be arguing that 10 - 0.999... = 9.000... > 9, and hence implicitly assuming that 0.999... < 1, which is as false as a proof that shows that 0.999... = 1 by assuming that 0.999... = 1. On the other hand, notice how your assumption that 0.999... < 1 leads to the solution that 0.999... > 1, both disproving your assumption and leading towards the conclusion that it must = 1.
And also, 0.999... doesn't have a limit - a sequence (such as (0.9, 0.99, 0.999, ...)) has a limit, 0.999... is a single number, which may be defined as the limit of that sequence - which happens to be 1.
If you're keen on proving the inequality, see if you can give a decent answer to this question: If 0.999... is a real number not equal to 1, then their difference 1 - 0.999... must be a non-zero real number, call it ε. What is the decimal representation of ε? And why is it that ε seems to behave like an infintesimal, even though the real numbers are supposed to be Archimedean? Confusing Manifestation(Say hi!) 02:48, 17 December 2007 (UTC)
What rubbish you have written. My above argument is completely clear. You are interpreting it incorrectly. I have shown you clearly in my argument what happens in such a 'proof' even when one assumes a 'fact' you consider to be true, i.e. 0.999... = 1 - it leads to all sorts of contradictions and errors. ConMan - you talk about an infinitesimal - can you exhibit even one infinitesimal number? The whole concept of infinitesimal is absolute rubbish. Infinitesimals are not well-defined, never have been well-defined and evidently never will be. My argument shows you what happens when idiots in high places think they understand certain concepts - they end up leading themselves into further darkness. This is what happens to fools and of course Wikipedia is a consortium of fools and idiots who are by no means competent, let alone experts. Radix systems were designed to represent numbers in a unique way. It was later discovered that not all numbers can be represented in a given radix system, in fact most numbers cannot be respresented except by approximation. Let's move on to your next so-called 'proof'.98.195.24.26 (talk) 03:35, 17 December 2007 (UTC)
Please, note the warning up the very top about being civil. Now, of course infintesimals are poorly defined and so forth, but my point was that if you allow 1 - 0.999... = \epsilon \ne 0, then it behaves just like the kind of infintesimal described in the article Archimedean property - given a positive number y, you can prove that no matter how many epsilons you add up you'll never reach y. As such, epsilon is an infintesimal, and there are no infintesimals in the real numbers, so the assumption that 1 - 0.999... doesn't equal zero is false.
Back to the previous proof. Here is what you've given, line by line, with my commentary.
     x = 0.999...                 // Definition, fine
   10x = 10 (since 9.999... = 10) // By assumption that 0.999... = 1, fine
=>  9x = 10 - 0.999...            // Subtracting the two previous lines, fine
=>   x = (9.000...)/9             // So 10 - 0.999... = 9.000..., fine
=>   x > 1                        // HOLD YOUR HORSES!

What is the difference between 9.000... and 9? Are you saying that 9.000... > 9? Then what's 9.000... - 9? By saying that 10 - 0.999... > 9, you are saying that 0.999... < 1, and hence you are not following your initial assumption! Confusing Manifestation(Say hi!) 05:07, 17 December 2007 (UTC)

Yes, I am saying that 9.000... > 9. This follows from the previous steps in complete agreement with previous assumptions. I will not discuss the infinitesimal because it is nonsense. There are no infinitesimals in the real numbers or any other numbers. Non-standard numbers are also nonsense. I have illustrated clearly what happens when you treat a 'false' assumption as true, you end up with a contradiction. The false assumption here is that 0.999...=1.
It is not possible to find an epsilon = 1-0.999... just as it is impossible for you to completely represent an irrational number. You don't seem to have a problem with using pi or other irrational numbers as approximations but you have the audacity (or rather foolishness) to make wild claims like the ones in your main article. By definition, 1 > 0.999... There is no duplicate representation. It is your attempt to have this equality that gives rise to duplicate representation. Quite frankly it does not matter that we can't find an epsilon = 1 - 0.999... because our arithemetic deals only with *finitely represented* quantities or approximations in all other calculations where we cannot completely represent given numbers. So which proof do you want me to illustrate as false next - your 'Archimedean-type' proof? 98.195.24.26 (talk) 19:04, 17 December 2007 (UTC)
How does 9.000...>9 follow from the previous steps? Please give a step-by-step proof. We know there are no real non-zero infinitesimals, but if you are correct 1-0.999... would be one. Give me an integer that, when multiplied by 1-0.999..., is greater than 1. If you can't do that, then 1-0.999... is, by definition, infinitesimal, and therefore equal to 0 (since that's the only infinitesimal in the real numbers). --Tango (talk) 21:47, 17 December 2007 (UTC)
Before you touch the Archimedean proof, I want to clarify something you've said - "It is not possible to find an epsilon = 1-0.999...". So you're saying that there is no non-zero number that equals 1-0.999...? Because if 1 is a real number, and 0.999... is a real number, then there must be a number which is the difference between them, because the reals are closed under subtraction.
And by saying there are no infinitesimals in the reals (excepting zero, of course), you are agreeing with both of us. Look, I'll break it down into a syllogism:
(1) There are no non-zero infinitesimals in the real numbers
(2) 1-0.999... is an infinitesimal

Therefore,
(3) 1-0.999... = 0.

(1) has been agreed by all parties as true, (2) is easily proven (and has been before), so (3) follows. And that's my "Archimedean-type" proof in a nutshell. Confusing Manifestation(Say hi!) 22:04, 17 December 2007 (UTC)

I do not agree with any of your statements. (1) There is no such thing as an *infinitesimal*. (2) 1-0.999... is not an infinitesimal, it is an indeterminate real value greater than 0. If this shocks you, why does it not also shock you that pi is also indeterminate. You can only approximate pi. (3) Therefore 1-0.999... > 0.

Tango: There are infinitely many numbers between 0.999... and 1 - check the archives and you will find a clear example. You are incorrectly assuming that the limit of a series determines a number. If you insist on defining a number by a series (in fact this is circular), you must be consistent: what is the limit of any irrational number defined as a Cauchy sequence? You do not know, therefore you cannot use this definition. Furthermore, the apparatus for comparing numbers in decimal form must be consistent. You start wih the most significant digit in the representation and continue the comparison until you find two digits that are not equal. Having 0.999... = 1, contradicts this method. You must also remember that it is impossible to represent all numbers in any radix system. Most numbers can only be approximated. And if this is so, why would anyone foolishly insist on having 0.999... = 1? 98.195.24.26 (talk) 01:58, 18 December 2007 (UTC)

Fine. There are no infinitesimals, I have no problem with that. But there are also no indeterminates in the reals, unless you're keen on breaking just about every other property the reals have. As long as we disagree on that, then all we're going to do is continue in this pattern of "you call me an idiot, I present another proof that you disbelieve". You are Logomath1 and I claim my 5 pounds, goodbye. Confusing Manifestation(Say hi!) 02:31, 18 December 2007 (UTC)
If there are infinitely many numbers between 0.999... and 1, give me an example. I'm only asking for one. As for the limit of a sequence determining an number - that's how the reals are defined, as the limits of Cauchy sequences of rationals. Something is a real number if and only if it is the limit of a Cauchy sequence of rational numbers. 0.999... is the limit of (0.9,0.99,0.999,...), pi is the limit of various rational sequences (including simply (3,3.1,3.14,3.141,...) although that's impossible to define all the terms of, since pi is irrational, so it's not a particularly good choice, there are plenty of choices with clear patterns though). If you can't write down a number that is between 0.999... and 1 in a simple way, just give me the sequence that tends to it, if it's a real number, it must have such a sequence. --Tango (talk) 17:32, 18 December 2007 (UTC)

[edit] No indeterminates in the reals?

Of course there are indeterminates in the real numbers - can you provide a complete representation in *any* number system for those numbers we call irrational? The answer is no. Even if it were possible for 0.999... to equal 1, it breaks the decimal system as it was intended since its invention. Do you agree that not all real numbers can be represented in a radix system? If yes, why would you insist on having a cauchy sequence define a number? Just because you can list the first few terms says nothing about the limit. For example, you can provide a Cauchy sequence that 'represents' pi but you cannot find its limit. There are several contradictions in this branch of 'analysis' - All Cauchy sequences must have limits but you cannot find the actual limit of 'most' Cauchy sequences that represent the majority of real numbers you cannot represent completely. I am not calling you an idiot but I am glad that you think you are an idiot. It's the first step to enlightment. Am I logamath? Nope. Not logamath but I am the so-called 'troll' most of your wiki math gods hate. I have the most extreme disdain for Michael Hardy, Krmsq(close runner up bum), Melchoir and small fry like Meni Rosenfield in this order. The rest of you are all pawns who have to support their 'view' of the math-world.

In the dark ages, we had the leaders of the church who supported only their conception of the universe until someone smart came along and proved them to be all idiots who were hindering the progress of mankind. Likewise, Wikipedia will 'fail'. Hail Knol!! Knowledgeable individuals will be able to post without fools like you trampling all over their pearls. No one on Wikipedia is qualified to edit my work. You first have to know what I know and then you can speak to me on my level. As arrogant as this sounds, Hardy and company are bishops of the math church whose founders were exceptional fools like Cauchy, Weierstrass and the rest who I don't care to mention. 98.195.24.26 (talk) 14:22, 18 December 2007 (UTC)

You should consider returning to using logical arguments rather than heresay, it makes your position more justifiable. I honest do not agree with your position on 0.999...=1 it is definitely true and the best way I can thing to prove it is as follows. First of when the decimal system was created it's rules never forbade the exist of an infinite number of digits, it would never be useful to have an infinite number of digits in application but theoretical work shows that it occurs as is the case with 1/3=0.333... which if you don't believe that yet bear with me as I shall hopefully make it clear why that is. The long division algorithm existed before decimals but instead of calculating the decimal they usually used mixed fractions like 3 2/5 or in it's improper form 17/5 the other way of notating this is with the remainder, which would 3r2 (given that 17/5= 17 divided by five) read 3 remainder 2 if you were to divide 17 by 5 via long division you would observe that 3*5=15 and 4*5=20 so you would write 3 in the ones place and then take 17 and subtract 15 17-15=2 since we don't have a zero as our remainder then 17 is not divided completely by 5 yet, either we can leave the remainder to attest to this or using base 10 decimals we can finish the division. This is done by simply moving to the next place value and continuing our division operation. so we have 20 now being divided by five, so we get 4*5=20 thus our answer for 17/5=3.4 because our remainder is zero, this algorithm always works for division of one integer by another (do what me to prove that as well??? I will if you ask). So lets use this algorithm on 1/3, we 0 for the ones place, and thus next is 10/3, 3*3=9 is means our partial answer is 0.3 and our corrisponding remainder is 1 since 10-9=1, now for the next digit 10/3 (hey that looks familiar!!), 3*3=9, 10-9=1 and looky here it repeats, the remainder is once again and 1 and will always be 1 so we can never finish working this out the long way, but there is a very clear pattern here 0.3, 0.33, 0.333,... (... means this pattern continues with ceasation) so with the knowledge of this pattern we can immediately jump to the final answer, since it is logically obvious that this pattern never stops as there is no "anomaly" to change the remainder ever. We notate this answer as 0.333... in order to avoid having to spend an eternity writing it. (The ... notations means in this context continues without ceasation and it a standard in mathematics, that is to say those of us how work with math use that symbol to communicate the idea aforementioned) so now we have the equality 1/3=0.333... using basic arithmetic properties we can show that 3(1/3)=1, 3=3/1 since any integer divided by one is itself an integer since 1 is the Multiplicative Identity. so 3(1/3)=(3/1)(1/3) the 3's cancel, leaving 1/1 which reduces to 1 since by the multiplicative identity any integer divide by one is itself. So 1=3(1/3) and we also already showed that 1/3=0.333... so we can substitute 0.333... for 1/3 thus gives 1=3(1/3)=3(0.333...) and now we can use the distributive property (works for number as well as variables, and I can prove if you that insist too) that gives 1=3(1/3)=3(0.333...)=0.999... and now by the Transitive Property of Equality we have 1=0.999... I think is about as Ironclad a proof as can be constructed anything more fundamental will be dealing with axiomizations which are a necessity for any productive results. A math-wiki (talk) 02:53, 20 December 2007 (UTC)

If something cannot be determined, it cannot be a member of a set, thus there are no indeterminates in the reals. You are clearly using a non-standard definition of "indeterminate", so please define it. I can't understand your point until you do. --Tango (talk) 18:05, 21 December 2007 (UTC)

[edit] side point on rational vs. irrational

A lot of these proofs seem to rely on the assertion that 0.999... is irrational, since it never terminates. What, then, of 0.142857142857142857... ? Is that rational or irrational? —Steve Summit (talk) 21:21, 2 March 2008 (UTC)

A nonterminating, repeating decimal is still rational. 0.142857... = 1/7, a rational number. Irrational numbers are characterised by nonrepeating decimal expansions, such as 3.14159265358979..., and 2.71828182874... Confusing Manifestation(Say hi!) 21:48, 2 March 2008 (UTC)
I know that, and I know you know that; I wanted to hear from the people claiming that 0.999... is irrational! —Steve Summit (talk) 23:14, 2 March 2008 (UTC)

[edit] Proof that 0.999... < 1 according to Archimedean Axiom

    1>1/x            (i)
    0.999...>1/x     (ii)
 => 1-0.999...>0     (iii)

We can find infinitely many x that satisfy both inequalities. Since the same x always satisfies both inequalities, we can subtract (ii) from (i) giving (iii). From (iii) we can be certain that the numbers 0.999... and 1 are not the same because their difference is not 0. —Preceding unsigned comment added by 98.199.111.222 (talk) 14:07, 8 March 2008 (UTC)

That's not how inequalities work. You have to subtract the same thing from both sides for the inequality to still hold, not one thing from each side of another inequality. If your manipulation was valid we could use 2>1 and 3>1 to prove 2-3>0, which is obviously false. --Tango (talk) 14:46, 8 March 2008 (UTC)
All of (i), (ii) and (iii) contain positive quantities so that the manipulation is true provided the lhs of (i) is greater than the lhs of (ii). If this is not the case, then we have to assume that either they are equal or the lhs of (ii) is less than the lhs of (i). They cannot be equal because then we would have 0>0 which is obviously false. Therefore the only possible conclusion is that 0.999... must be less than 1. In your example, 3>1 and 2>1 implies that 1>0 which is correct. If this is not how inequalities work, then all the proofs provided in your archive are false because these proofs start by supposing that 1-0.999.. > 0 and then reach a conclusion based on an assumption that 0>0 which these assert to be a contradiction that is in reality a false conclusion. Generally, if a>x and b>x, then provided a>b with a, b > 0, then it is true that a-b>0.
Um ... no. Firstly, in your "generally" - a>b => a-b>0, all the rest of the conditions are completely irrelevant. Secondly, by stating that "the manipulation is true provided the lhs of (i) is greater than the lhs of (ii)" you are, in fact, affirming the consequent - in other words, you're saying "I can do this manipulation only if 1 > 0.999..., and when I do it I get the result that 1 > 0.999... so therefore the manipulation was valid".
The fact that you state that 1 = 0.999..., combined with your manipulation, leads you to 0>0, in fact demonstrates why your manipulation doesn't work unless you already know something about the relationship between the two left hand sides. If we replace everything with letters, then maybe you can see what the problem is:
    a > c     (i)
    b > c     (ii)

=> a - b > 0 ?

In fact, unless you "know" the relationship between a and b beforehand, you can draw no conclusion from (i) and (ii). And if you *do* know the relationship, then you don't need (i) and (ii) to get it. Confusing Manifestation(Say hi!) 23:28, 10 March 2008 (UTC)

Um...Yes, I would think. Of course I assume that 1 > 0.999... (by definition of the decimal system). It would be preposterous to assume that 0.999... > 1, would it not? See Gustav Steele's post below which is not correct but you get the idea. Now if I assumed these are equal, I would be stating that 0>0 which is obviously false. So if I cannot assume these are equal, then one must be less than the other - Yeah?
"if I assumed these are equal, I would be stating that 0>0 which is obviously false". To me that doesn't necessarily prove that they're not equal. That just proves that there's a problem with how you get from (i) and (ii) to (iii). For example:
   1.5 > 1 (i)
   3/2 > 1 (ii)
   1.5 - 3/2 > 0 (iii)
   0 > 0
Steps (i) and (ii) are clearly correct. And I used the same logic as you to get from there to (iii). But then I got 0>0, which is obviously false. So does that mean that 1.5 != 3/2? Or that I used faulty logic? --Maelwys (talk) 14:57, 24 March 2008 (UTC)
Just for fun:
Proof that 0.999... > 1 "according to Archimedean Axiom"
    0.999...>1/x     (i)
    1>1/x            (ii)
 => 0.999...-1>0     (iii)
Wow! That's awesome. 0.9~ is both greater than and less than one! Gustave the Steel (talk) 19:29, 18 March 2008 (UTC)
(Reply to the anon's comments, bumped down for chronology) Like I said, by assuming that 1 > 0.999..., you are already excluding the possibility that 1 = 0.999... - of course you could have meant 1 >= 0.999..., in which case your whole argument falls down (since the possibility of equality eliminates any contradiction). It will take me a little digging, but I seem to recall somewhere in the archives of Talk:0.999.../Arguments that I actually proved, under reasonable assumptions, that 0.999... >= 1 (I think this was without assuming that "limit of partial sums = infinite sums" since that seemed to be a sticking point for some)! Funnily enough, without resorting to any of the existing proofs, it was harder to disprove the possibility 0.999... > 1 (and no, "it's obvious" doesn't cut it, since "it's obvious" would have you stating 1 > 0.999... and I'd already shown that wasn't the case). Confusing Manifestation(Say hi!) 01:22, 26 March 2008 (UTC)
Like I said, I assumed 1 > 0.999... because it makes no sense for it to be equal and certainly not less. It is 'obvious' that exactly one of three statements is true: (a) 1 > 0.999... (b) 1 = 0.999... or (c) 1 < 0.999... (c) is absurd by the structure of the decimal system. (b) leads to a contradiction - 0>0. Therefore the only possible answer is (a). Just as I assume that 1 > 0.999..., the only way you can justify your argument is to assume that 1 and 0.999... are equal. You can only do this if you are comparing the limit of the partial sums of 0.999... which is 1. However, as not all mathematicians agree that the sum of an infinite convergent series is its limit, you cannot assume that 0.999... = 1. Furthermore there are mathematicians who do not accept duplicate representation in any radix system. If the numbers do not look exactly the same (equivalent fractions aside), then they are not the same quantity. Lastly, we compare decimal numbers using the radix position system, not the limit which is indetermine in most cases. For example, you cannot find the limit of all irrational numbers - you can only estimate it to a finite number of places. 98.199.111.222 (talk) 11:49, 27 March 2008 (UTC)
Actually, (b) doesn't lead to a contradiction. You only got the result you wanted because you already assumed (a) was true. It is illogical to say "I can prove statement X! Here's how: assume X to be true..." If you want to make a real, formal proof, you need to start with something already known to be true (i.e., not in dispute) and show that your claim follows logically from it. Alternatively, you could start by assuming the opposite of what you intend to prove, and show that it leads to a contradiction. You have, so far, done neither of these. Gustave the Steel (talk) 15:24, 27 March 2008 (UTC)
Actually, I am sorry, but (b) does lead to a contradiction. 0.999... must either be equal or less than 1. If you assume that it is less, then the inequality is true. Any other assumption renders the inequality untrue. In any event, the only way one can accept these two numbers as being equal, is if one accepts that the limit of a converging infinite series is its sum. This being the case, most real numbers remain undefined. —Preceding unsigned comment added by 98.199.111.222 (talk) 18:56, 28 March 2008 (UTC)
You'll need to show how (b) leads to a contradiction. Your proof above assumed (a), not (b). Gustave the Steel (talk) 19:21, 28 March 2008 (UTC)
You are correct that most real numbers are not definable, but 0.999... is definable, so that's not a problem. --Tango (talk) 15:39, 29 March 2008 (UTC)
Could you name some of these mathematicians that dispute these basic facts of real analysis? --Tango (talk) 15:39, 29 March 2008 (UTC)
You can search for these yourself. Two mathematicians are Dr. Escultura and Nrich. Although I do not endorse all they write, they happen to have similar views on this topic. —Preceding unsigned comment added by 98.199.111.222 (talk) 19:57, 29 March 2008 (UTC)

[edit] If 0.999... is rational, ...

If 0.999... is a rational number, then you must be able to express it as a/b. You cannot say that 0.999... = 1/1 because then you have already accepted that these values are the same. 1 = 1/1. Now if you can find an a and b such that a/b = 0.999..., then you can start to argue about the equality of these numbers. Till then, your so-called 'proofs' (they are nonsense) rely on the definition that you claim for real numbers, that is, a real number is equal to the limit of a convergent sequence/series. Of course this definition is also flawed,because in order to arrive at a Cauchy sequence, you must first be able to calculate partial sums using an appropriate series. For most real numbers you cannot find the limit of such a series; you can only find an approximate value accurate to a certain number of places in a given radix system. —Preceding unsigned comment added by 98.199.111.222 (talk) 19:07, 28 March 2008 (UTC)

0.9~ is x/x for any non-zero x. We can say with confidence that 0.9~ = 1/1 because our other proofs, which do not rely on the rationality of 0.9~, show that it has the same value as one. The simplest explanation is that for two real numbers to be different, there must exist an infinite amount of values in between them - yet of all the infinite numbers that MUST exist between 0.9~ and 1 if they're unequal, the inequality believers have yet to show the existence of a single such value. Gustave the Steel (talk) 19:26, 28 March 2008 (UTC)
Unless you can show that 0.999... is a rational number, your so-called other 'proofs' are false.

You claim that the inequality believers have yet to show the existence of a number between 0.999... and 1, yet your archive shows how you can find infinitely many numbers between 0.999... and 1. Also, wikipedia claims infintesimals exist, but you have yet to show the existence of any infinitesimal. So, unless you can prove that 0.999... is a rational, your 'proofs' mean nothing. —Preceding unsigned comment added by 98.199.111.222 (talk) 19:49, 29 March 2008 (UTC)

Infinitesimals exist, but there aren't any in the real numbers - does Wikipedia claim otherwise somewhere? If so, it needs fixing. Where in the archive does it show infinitely many numbers between 0.999... and 1? --Tango (talk) 21:47, 29 March 2008 (UTC)
Yes, if you or anyone could link to something showing some of the numbers between 0.9~ and 1, I'd appreciate it.
And, actually, if you look at the main article, some of the proofs there do not rely on the rationality of 0.9~. The digit manipulation proof, for example, doesn't assume rationality. Gustave the Steel (talk) 22:01, 29 March 2008 (UTC)
The rigorous definition of the real numbers as the completion of a rationals is rather technical, and probably beyond the scope of this talk page. I assure you, it does work, though. I suggest you find a decent introductory textbook into real analysis and work through it until the validity of the construction is proven. I think I may see where you are getting confused, though - the Cauchy sequences that construct the reals are sequences of rational numbers, not real numbers, so the calculations required are just arithmetic on rationals, so doesn't require any prior knowledge of reals. --Tango (talk) 15:35, 29 March 2008 (UTC)
A Cauchy sequence defines a real number strictly in terms of the partial sums of rational numbers. The limit of these partial sums is claimed to be a real number. This being the case, you cannot find the limit of most real numbers. Example: pi, e, or any irrational number. Arguing that you can compute it to any desired degree of accuracy does not mean anything because you can never represent it completely, only approximately. Thus, whatever the so-called real number is, it is always (without exception) an approximation representation given by rational numbers. So do you still think real analysis is rigorous? Also, please do not tell me I am getting 'confused'. This is condescending and has no place in civil discussion. Address the topic. If you cannot do this, then allow others the opportunity to do so.
I think you have your terminology a little confused. A partial sum is related to a series, not a sequence (a sequence is something like 1, 2, 3..., a series is something like 1+2+3+...), you can make a sequence out of the partial sums of a series, but you can define sequences without reference to a series. You can define certain irrational numbers (although, most you can't) exactly, for example e = \sum_{n = 0}^\infty \frac{1}{n!}. Yes, it takes an infinite number of terms, but it is exact. --Tango (talk) 21:47, 29 March 2008 (UTC)
You think I have my terminology confused? A partial sum is related to a sequence. Let me enlighten you. Suppose you are calculating a Cauchy sequence for pi. The first term is 3, the second term is 3.1 or 31/10, the third term is 314/100, etc. Now please dare to tell me that 3, 31/10 and 314/100 are not partial sums. The rest of your paragraph contradicts itself and if I were confused, I would be even more confused. —Preceding unsigned comment added by 98.199.111.222 (talk) 13:30, 30 March 2008 (UTC)
They are partial sums. As I said, you can form a sequence out of partial sums, but you don't have to. (4,2,3.2,3.0,3.15,3.13,3.142,3.140,3.1416,3.1414,...) is a Cauchy sequence tending to pi, but isn't constructed from a series (you could construct a series from it if you wanted, I suppose). When talking about sequences, you shouldn't use the terminology from series - they are closely related concepts, but they aren't the same. --Tango (talk) 15:45, 30 March 2008 (UTC)
You just contradicted yourself. Go back and read what you wrote about partial sums. As for your example, it is not a Cauchy sequence that is formed from the partial sums of a convergent series. You shouldn't talk about these concepts unless you understand them. I suggest you study the terms sequence and series before you make any further edits. 98.199.111.222 (talk) 00:39, 8 April 2008 (UTC)
It is a Cauchy Sequence, it isn't formed from the partial sums of a convergent series. Considering that's what I said I was giving an example of, that would mean I was successful. Thank you. --Tango (talk) 15:00, 8 April 2008 (UTC)
I think the other thing that you're getting confused is that while we cannot physically sit down and add together an infinite number of terms, we can still determine rules about how to manipulate these series, and once we've done so we can treat them as numbers to our hearts' content. If it really bugs you, just assume that "the infinite sum" is just shorthand for "the limit of the partial sums", a notation that is still completely consistent and rigourous. Confusing Manifestation(Say hi!) 02:18, 30 March 2008 (UTC)
And your point is? Please address the topic. Nothing in your paragraph addresses the topic. —Preceding unsigned comment added by 98.199.111.222 (talk) 13:33, 30 March 2008 (UTC)
His point is that your point doesn't actually make sense. --Tango (talk) 15:45, 30 March 2008 (UTC)
That's pretty close. My point was that you said (paraphrased) "You can't believe a decimal expansion because it's infinite and we can't do calculations with an infinite number of terms in finite time", but, frankly, we can, by setting appropriate rules. Confusing Manifestation(Say hi!) 22:59, 30 March 2008 (UTC)
Hmmm, I believe you both missed the 'point'. Frankly, you can't do calculations with any infinite representations unless they are approximations or you already know the limit of a convergent sequence and agree that this limit defines the sum. 98.199.111.222 (talk) 00:43, 8 April 2008 (UTC)
Repeating decimal begs to differ. Gustave the Steel (talk) 03:20, 8 April 2008 (UTC)
You don't have to know the limit, you just have to know the limit exists (ie. the sequence converges). If you define real numbers in terms of Cauchy sequences (which is one of the 2 standard ways), then the real number is simply defined to be the limit of the sequence, so it's impossible to know what it is in advance. --Tango (talk) 15:00, 8 April 2008 (UTC)
Repeating decimals are not rational numbers. The ancient Greeks did not consider repeating decimals or repeating 'anything'. This definition was accepted much later on. It is incorrect. Rational numbers are numbers that can be represented finitely, for example, a/b. However, when you attempt to represent these in certain radix systems, you end up with repeating representation that is not equivalent to the number you started with - in other words, repeating or non-repeating representation of a number means it is not rational. As for the limit, you have to know the limit. Why? Well, when you compare 0.999... and 1 and assume these are equal, you are comparing the limit of the partial sums of 0.999... and 1. But, when you compare any irrational number, you do not use the limit (because you don't know it!). How is it that you use two different methods to compare two sets of numbers that you claim are both real? I cannot continue this discussion with you because you are mentally inferior to me (please, this is not an insult, it is a fact. Hope you will not be offended). Take me to your leader and I will carry on the discussion. —Preceding unsigned comment added by 98.199.111.222 (talk) 18:58, 14 April 2008 (UTC)
0.999... can be represented in the form a/b as 1/1. By refusing to accept that without reason, you are making a circular argument. I don't know what you mean by "my leader", but any professional mathematician will tell you exactly the same thing. I don't understand your problem with their being two methods for achieving a result - it's commonplace throughout mathematics. Some methods work for some things, other methods work for other things, there is often an overlap. --Tango (talk) 19:34, 14 April 2008 (UTC)
Every repeating decimal can be produced by dividing one integer by another. The easiest example is 1/3 = 0.3~, of course. Unless you believe that not every rational number has a corresponding decimal value? In which case, rather than assert it from a position of mental superiority, you should provide some proof. Gustave the Steel (talk) 23:18, 14 April 2008 (UTC)
You divide 1 by 1 and you get 0.999... (if you want to - usually you just get 1, but you can do it in such a way as to get 0.999...) Of course every rational number has a corresponding decimal value, who has said otherwise? The decimal value of 0.999... is 0.999..., I don't understand what you're getting at. --Tango (talk) 23:30, 14 April 2008 (UTC) I think I misunderstood you there. Sorry. I'm still not sure I understand you, though. Who has said that there's a rational number without a decimal expansion? --Tango (talk) 23:32, 14 April 2008 (UTC)
I was replying to the anonymous guy, not to you. Gustave the Steel (talk) 00:09, 15 April 2008 (UTC)
Yes, that's what I realised (eventually!) - the anon doesn't seem to be claiming that there are rational numbers without decimal expansions, though. --Tango (talk) 00:11, 15 April 2008 (UTC)
He claims that repeating decimals are not rational numbers. It follows from that statement that 1/3 has no corresponding decimal value, since 0.3~ isn't rational on his planet. Gustave the Steel (talk) 00:16, 15 April 2008 (UTC)