Misplaced Pages

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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.

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 ${\displaystyle \Sigma {9/10^{n}}}$ 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 Misplaced Pages 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 , 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

${\displaystyle r=\sum _{i=0}^{\infty }{\frac {a_{i}}{10^{i}}}}$

where ${\displaystyle a_{0}}$ is a nonnegative integer, and ${\displaystyle a_{1},a_{2},\dots }$ are integers satisfying ${\displaystyle 0\leq a_{i}\leq 9}$.

Hence, the harmonic series is not a decimal expansion, because the individual terms cannot be written in the form ${\displaystyle a_{n}/10^{n}}$, where ${\displaystyle a_{n}}$ 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 ${\displaystyle \epsilon }$. What is the decimal representation of ${\displaystyle \epsilon }$? And why is it that ${\displaystyle \epsilon }$ 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 Misplaced Pages 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 ${\displaystyle 1-0.999...=\epsilon \neq 0}$, 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)

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, Misplaced Pages will 'fail'. Hail Knol!! Knowledgeable individuals will be able to post without fools like you trampling all over their pearls. No one on Misplaced Pages 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)

Absolute Proof that 0.999... IS NOT EQUAL TO 1

A math-wiki:

You are wrong about infinite representation. The ancient Greeks *never* represented 1/3 as 0.333...

Now let me refute your argument regarding the transitive property:

A) 1=3(1/3) B) 3(1/3) = 3(0.333...) C) 3(0.333...) =0.999...

The problem here occurs in B) where you assume that 3 * 0.333... = 3 * 1/3. You have 'not proved' anything. On the contrary, you have only proved that you are assuming 1/3 = 0.333... which it is obviously 'not equal'. 0.333... is an 'indeterminate' number because it cannot be represented finitely in base 10. 0.333... is not equal to 1/3 so you cannot say that 1/3 is a finite representation. An indeterminate number is a number that is repeating in any radix system.

You mention the division algorithm but do you know what it says? The division algorithm has 'nothing' to do with radix systems. Let me state it for you - If a and b are two given integers, there are unique integers q and r such that a=bq + r and 0<= r < b. Now I will use the division algorithm to prove that you are WRONG once and for all. Suppose a = 1 and b = 0.999..., then what are the integers q and r such that the above is true?

Suppose that q = 0 and r = 1. Then the division algorithm fails because it makes a false statement i.e. q and r can both be 0 and 1 respectively. For example: 1 = (0.999...)(1)+0 and 1 = (0.999...)(0)+1. So by default, the division algorithm is proved false if 0.999... is equal to 1. And you can say a lot of things about the Greeks, but they would not make a stupid mistake like this. They were the smartest race in ancient times.

Now suppose that you try to tell me that either form is acceptable in the case of 0 and 1, you will still end up with a terrible contradiction:

1=(0.999...)(0)+1. The second form in this case implies that 1 < 0.999... since r < b. But how can 1 be less than 0.999... ??!!

I want the world to know what f..g idiots run this site. You are a bunch of retarded, conceited fools who I look upon with the utmost hatred and condemnation. Your site is absolute crap because it is run by fools like Michael Hardy, Kmrsq, Melchoir, Meni (idiot) Rosenfeld, et al. DIE WIKIPEDIA!!!! HAIL KNOLL!!!! 98.195.24.26 (talk) 12:00, 26 December 2007 (UTC)