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

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This is an attempt to provide those who dispute the accuracy of Proof that 0.999... equals 1 with a space in which to give their own careful, meticulous proofs to the contrary.

Contents 1 Definitions: 2 Rules: 3 Algebra proof 4 What needs to be proved? 5 Why 0.999... is not equal to 1. 6 Simple Proof that 0.999... is not equal to 1

[edit] Definitions:

• , in other words it is the recurring decimal represented by an infinite number of nines following the decimal point.

[edit] Rules:

• By contributing to this page, you agree to abide by the rules.
• I (Confusing Manifestation) may introduce additional rules or definitions as necessary. If you do not agree with a new definition, then give your alternative definition in your proof along with justification why it is a valid definition. If you do not agree with a rule, take it up on the talk page. I will attempt to make the rules fair, with the aim of letting you provide your arguments.
• Your proof must hold up to the same scrutiny that you attempt to apply to the proofs on the original page. Thus, the following apply:

• You cannot claim anything as "trivial" or "obvious" unless you are prepared to provide further explanation when questioned.
• You must give careful definitions of anything not defined in the above Definitions section. You must be ready to justify these definitions as well.

• Wikipedia policy such as no personal attacks applies as usual. This is a page for cogent proofs, not insults and jibes.

[edit] Algebra proof

(Moved from above to ease editing)

Another kind of proof adapts to any repeating decimal. When a fraction in decimal notation is multiplied by 10, the digits do not change but the decimal separator moves one place to the right. Thus 10 × 0.9999… equals 9.9999…, which is 9 more than the original number. To see this, consider that subtracting 0.9999... from 9.9999… can proceed digit by digit; the result is 9 − 9, which is 0, in each of the digits after the decimal separator. But trailing zeros do not change a number, so the difference is exactly 9. The final step uses algebra. Let the decimal number in question, 0.9999…, be called c. Then 10c − c = 9. This is the same as 9c = 9. Dividing both sides by 9 completes the proof: c = 1."

As promising as this theory may seem, I have spotted what I believe to be a flaw in the calculations. 10c - c should be equal to 9 following the evidence given above. However, since 10c is c multiplied by 10, there will be one less decimal place than c. This means that 10c - c is not equal to 9, but 8.999.......1 and so 9c is not equal to 9 and c is not equal to 1. —The preceding unsigned comment was added by 166.87.255.133 ([[User talk:|talk]] • contribs) 06:33, 1 August 2006 (UTC)

A good try, but it falls flat because there are an infinite number of decimal places and one less than an infinite number is still infinite. No matter how far you go, you can never reach the end of that decimal expansion to include the extra 1 that you think should be there. This is one of the simpler proofs that, by necessity, has some flaws, but they are flaws that can be fixed by application of more rigorous definitions and calculations than the simplicity of the proof allows. (In fact, if you tighten the proof, you end up with something resembling the limit / infinite series proofs.) Confusing Manifestation 10:54, 1 August 2006 (UTC)

Yeah, I agree with ConMan on that one, whether the infinite number is 8.9999999..1 or 9.99999.., the two sides of the algebraic equation remain equal, but c still doesn't equal 1. My algebra is very rusty, that said, the only problem I see is with the algebraic equation as provided by ConMan, where Step 3 is written as 9c = 9. This is incorrect. It should be 9 = 9. Step 1 is: (10 * c) - c = 9.999.. - c; So we process (10 * c), resulting in 9.999... and then removed c from both sides of the equation, leaving 9 = 9, not 9c = 9. Well, you can do this with any variable.
10x = 40.
10x - x = 40 - x.
36 = 36.
36/36 = 36/36.
x doesn't equal 1, x = 4.
Nor is c equal to 1, c = .999...
lets say that x = 0.999...

multiply both by 10 10x = 9.999...

take the first from the second 9x = 9

and divide by 9! x = 1

There fore, 1 = x = 0.999... If you want further proof that .999... != 1, try getting 1 to equal .999... hah = )
This was great fun though, thanks = )

As much as I'd like to disprove this, I'm afraid the above method isn't entirely correct. The algebra is flawed. If you have 10c = 9.9999 and you take away a c (10c - c = 9.9999 - .9999) you are left with 9c. Not 9. Just as if you had 10c and took away 4c you would have 6c and not 6. In order to get 9 from 9c you would have to divide by c, not subtract (10c / c = 9.999 / c which would be 10 = about 10 although not EXACTLY 10 since in order for a divided number to equal itself the number needs to be 1 which .999 is not) Throktar

Quite possibly the biggest debaser is the fact that .9~ is a smaller number than 1.

As you have functions that gradually curve toward a value (say 1), this function will eventually hit every possible value along .9999 repeating. As we know the curve will NEVER hit 1, then .9999 repeating is by necessity less than 1.

The curve is asymptotic, it reaches 1 at infinity. Since the .999 sequence is infinite, it is equal to 1.

Moreover, as you have a square on a plain, with sides of .9999 repeating units on a side. Will a square with sides of 1 fit inside?

[edit] What needs to be proved?

There is no need for anyone to prove 0.999... < 1 because by definition of the decimal radix system, it is. The onus of proof rests on those who claim these two values are equal. All their proofs are false: starting with the most common (as in the above example) and moving to the most complex (as in the Archimedean property). If x = 0.999... then 10x is not well-defined. In fact it is not defined at all because our rules of arithmetic for multiplication apply to rational numbers only. We extend these rules to non-rational numbers by approximation. That is to say, we can never calculate the area of a circle exactly and can never find the exact length of the hypotenuse in a right angled triangle where the remaining sides are both of unit length. So, if multiplication is defined for rationals, we must therefore be able to express 0.999... as a rational number. This is impossible for there are no numbers a and b such that a/b = 0.999...

Rational numbers were at first invented to represent only quantities that are less than 1 but greater or equal to 0. Numbers were written in a finite representation format. Recurring numbers were not considered rational until centuries later when the limit concept was introduced. This definition was later reworked (or extended) so that we could perform mixed arithmetic a lot easier. For example arithmetic involving an originally defined rational number and a whole number, eg. 2 + 2/3 = 6/3 + 2/3 = 8/3 This is an exact calculation because it can be represented finitely.

Much later very stupid men decided that a number would be defined as a limit of a Cauchy sequence. The definition they invented is in fact circular because they use a limit to define a number. In fact, if they were even remotely intelligent, they would have stated that any number is a rational number or the limit of the sum of infinitely many rational numbers where a rational number is a/b with b not zero and a < b (Was part of original definition but later dropped).

Sorry to comment in the middle of your argument, but any Cauchy sequence is the limit of infinitely many rational numbers. Let (a_1,a_2,...) be a Cauchy sequence. Let b_1=a_1, b_2=a_2-a_1, b_3=a_3-a_2 etc. Then the limit of the sum from i=1 to n of b_i can be represented by the Cauchy sequence (b_1, b_1+b_2, b_1+b_2+b_3,...). But this sequence is the same as the original sequence (a_1,a_2,...). So every Cauchy sequence can be represented as the limit of the sum of infinitely many rational numbers. However the converse is not true, not every limit of a sum of rational numbers can be represented as a Cauchy sequence. This is obvious because many limits of sums of rational numbers are not even finite. If you were to define a real number as a limit of a sum of rational numbers, then you would get a set which behaves very differently from what everyone else refers to as the real numbers. For example, you would have, 1/2+1/2+1/2+... as a real number which is strictly greater than every rational number.

So as long as we stay with the same rational representation of numbers, our result will generally be true. eg. 1/3 + 1/3 + 1/3 = 3/3 = 1. However, when we try to use other representations such as radix or mixed rational/radix, we end up with anomalies. eg. 1/3 + 0.333.. + 0.3333... = 1/3 + 0.666.... = ? Oops, 0.666... is not finitely represented and thus cannot be used in our arithmetic except as an approximation. e.g. 1/3 + 66/100. We cannot say 0.666... = 2/3 because it is not. Sorry, duplicate representation is not allowed in any radix system. And no radix system is capable of representing all numbers.

Let's summarize the rules: In order for arithmetic to be performed, numbers must be finitely represented using the same representation. 2pi/3 + pi/3 = pi. 1 + sqrt(5) = 1 + 2.23 = 3.23

In light of the above it is evident that even debating the 0.999... and 1 equality is completely idiotic.

75.87.69.48 (talk) 03:20, 30 November 2007 (UTC)

[edit] Why 0.999... is not equal to 1.

The proofs used in the original Wiki article are incorrect because these are based on assumptions that our arithmetic deals with incomplete representations of numbers. The Wikipedians make several assumptions of which some are correct and some incorrect. For example, they start off 'proofs' by stating that a sum or difference is related in some way to a limiting value and then proceed to demonstrate a contradiction that supposedly proves their viewpoint. It is true that the limit of the series 9/10+9/100+... is in fact 1. It is also true that this sum is always less than 1 as can be easily verified by mathematical induction. A wikipedian will immediately claim that numbers have duplicate representations - this of course is absolute nonsense. The invention of radix systems came long after the foundations of arithmetic were laid. It is a fact that not all numbers can be represented using any radix system. This frightens wikipedians and academics alike. It's hard to see why though, because they cannot represent any irrational number exactly. Maurice Freches' theory of metric spaces added to the confusion with its identity of indiscernibles that states d(x,y)= 0 iff x = y. This identity needs some clarification: we must also have that x and y are represented in 'exactly' the same (finite) way. That is, they 'look' or 'appear' exactly the same. If there is no exact representation, then it must be understood explicitly that x and y are in fact represented the same way. This means that 0.999... and 1 are equal iff: d(1,0.999...) = 0 and Representation(1)=Representation(0.999...). Since the representations are different, i.e. 0.999... and 1 do not look alike in any way but quite different in fact, it follows automatically these cannot be equal. There is no place for duplicate representation in any radix system. This is a falacy that has been propagated by ignorant academics. As long as there are fools and Wikipedia is the viewpoint of its editorial staff, this grossly incorrect article will serve to misled many students. 159.140.254.10 (talk) 19:06, 27 November 2007 (UTC) cool

Why do the representations have to be the same? 0.5 and 1/2 are equal, but have different representations. Just because 1 and 0.999... are expressed in the same system doesn't mean they have to be expressed in the same way. You've just asserted that it's nonsense without giving any reasoning. --Tango (talk) 15:40, 30 March 2008 (UTC)

I actually think this argument is very interesting and illustrates a common misconception about the construction of the real numbers. I will discuss the construction using Cauchy sequences, but my points will apply to Dedekind cuts as well. As the original poster suggested, if two Cauchy sequences don't look exactly the same then they are not equal. Intuitively this is rather obvious and is in fact correct. However, the real numbers are not a set of Cauchy sequences, they are a set of equivalence classes of Cauchy sequences where two Cauchy sequences (a_n),(b_n) are defined to be equivalent if the limit as n goes to infinity of a_n-b_n is 0. We can show this is an equivalence relation, so there is no ambiguity in referring to an equivalence class by any one of its members. In this case, one equivalence class can have many representations because it can be represented by any Cauchy sequence that is an element of it. Also, I find your comment that "This is a falacy that has been propagated by ignorant academics," unnecessarily rude. How exactly did you come to the conclusion that "There is no place for duplicate representation in any radix system,"? --128.135.169.19 (talk) 04:47, 15 May 2008 (UTC)

[edit] Simple Proof that 0.999... is not equal to 1

We know that 1 is a rational number, that is, it can be expressed as a/b. Now if 0.999... is equal to 1, it too must be a rational number. This means that there exists c/d such that c/d = 0.999... However, no such c and d exist. So 0.999... is not a rational number. Therefore 0.999... cannot be equal to 1.

Try c=1, d=1. You can't assume it doesn't equal 1 as part of your proof that it doesn't equal 1 - that's circular reasoning. --Tango (talk) 15:37, 30 March 2008 (UTC)

Um, sorry, no. You can't use c=1 and d=1 because then you are assuming that 0.999... is equal to 1. We know that 1/1 = 1. What we have to show is that 1/1 = 0.999... Where do you see 'circular reasoning'?

I'm not assuming it, I can prove it. You're trying to show that they are not equal by assuming they are not equal - that is circular reasoning. Your statement that there is no such c and d is implicitly an assumption it doesn't equal 1. --Tango (talk) 15:03, 8 April 2008 (UTC)

In fact, every repeating decimal can be produced from dividing one integer by another. Thus, all repeating decimal values are, fundamentally, rational numbers. 0.9~ is no exception. If you want to find a ratio from a repeating decimal, you should check out Repeating decimal - it has several useful methods. Gustave the Steel (talk) 15:47, 30 March 2008 (UTC)

You think? So show me two integers whose quotient is 0.999... ?

Clearly .999...=9/10+9/100+9/1000+... Let S(n) be the partial sum 9/10+...9/(10^n). It is also clear that S(n)= (10^n-1)/(10^n). Therefore the limit as n goes to infinity of S(n) is equal to the limit as n goes to infinity of (10^n-1)/(10^n)=1/1. Since .999... is equal to the limit as n goes to infinity of S(n), we have directly shown that .999... is equal to 1/1.

Going by the information from 0.999... and Repeating decimal, and the sources cited within, I can say with the utmost confidence that, for any non-zero real number x, x/x equals 0.999... Gustave the Steel (talk) 03:17, 8 April 2008 (UTC)

You can say what you like - it is still nonsense. Just because you say it or one of your wiki articles says it, do you think that it is true? Again, show me two integers whose quotient is 0.999...? If you cannot, then I will cease to respond to your posts. Sorry. 98.199.111.222 (talk) 19:03, 14 April 2008 (UTC)

We have done - 1/1. If you don't accept that, give a proof that we're wrong - one that does not involve assuming we're wrong to start with. --Tango (talk) 19:35, 14 April 2008 (UTC)

Or to put it another way, I claim that x = 2 isn't the solution to x + 3 = 5, because there are no real solutions to the equation. And no, you can't just plug in x = 2 and show it works, you've got to show that there's another real solution. Confusing Manifestation(Say hi!) 00:09, 9 April 2008 (UTC)

This is not what I have done. You are confused. 98.199.111.222 (talk) 19:03, 14 April 2008 (UTC)

What is 0.333... x 3 then? I would say this is 0.999... 0.333... is 1/3. 1/3 (0.333...) x 3 = 1 (0.999...).

No, because 0.3333... is not equal to 1/3. —Preceding unsigned comment added by 98.199.111.222 (talk • contribs) 06:03, 15 April 2008

Yep, it even says that on 0.999..., and yet you'd be amazed at the number of people who argue that point (generally by trying to say that 0.333... and 0.999... are only "approximations"). Confusing Manifestation(Say hi!) 23:56, 13 April 2008 (UTC)

A quick note to 98.199.111.222 - first, please don't place your response above mine, and secondly please sign it with four tildes (~~~~), as otherwise it looks like I had said that. And thirdly, that's a fairly categorical statement given that there's fairly strong agreement that they are equal. Confusing Manifestation(Say hi!) 23:16, 14 April 2008 (UTC)

If you want to talk about 0.3~ and 1/3, I've got a great discussion already going on 0.999...'s arguments page. You should read it and join in there. Obviously, if you don't think 1/3 equals 0.3~, you'll never see that 0.9~ = 1. Be warned: I've set some ground rules that you might have trouble with. Gustave the Steel (talk) 23:22, 14 April 2008 (UTC)

Even though I am actually a 0.999...=1 zealot, out of interest, could it be said that if 0.999...=1 that, by extension, 1.999...= 2? (because 1+0.999... = 1+1). Would this mean that it could be said that because 2 is an even number and 1.999... is not that they must be unequal, hance showing a difference between 1 and 0.999...? - Daan

1.999... does indeed equal 2. Therefore, 1.999... is an even number. What is it that makes you think otherwise? The fact that its last digit isn't an even number? It doesn't have a last digit, so that standard test simply doesn't apply. --Tango (talk) 21:26, 25 May 2008 (UTC)

Ok, thanks for sorting that out... the reason I would have thought otherwise would be that despite the 9s carrying on forever, I would have thought they continued the pattern in which 1.9, 1.99, 1.999, 1.9999 are odd. But without a last digit then... -- Daan

Only integers can be odd or even - 1.9 isn't an integer, so is neither. It's not unusual for the limit of a sequence to have a property no member of the sequence has - it seems a little strange at first, but you get used to it! --Tango (talk) 12:32, 28 May 2008 (UTC)

In fact, here's a sequence where all of the members display an obvious property, but their limit doesn't - (0.9, 0.99, 0.999, 0.9999, ...) are all strictly less than 1, but since the limit point is 1, well ... Confusing Manifestation(Say hi!) 23:50, 28 May 2008 (UTC)