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

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.

1. Using simple logic, .99999999... Does not equal 1. Many have "proven" that 1 equals .999..., but use simple knowledge.

1.0 0.999... Ones Place Value: (1).0 (0).999... The values in the ones places are 1 and 0. 1 > 0. That's a known fact.

2. You cannot multiply a number by an infinite number. You cannot multiply something by .888..., you cannot multiply something by .777, nor any other infinite number. Using continuity, multiplying something by .999... would not work.

Definitions:

• $0.999\ldots =0.{\dot {9}}$, in other words it is the recurring decimal represented by an infinite number of nines following the decimal point.
• 1 represents a single entity, the unit of counting or measurement (taken from 1 (number)).

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.

Proof That 1=0.999 Using An Infinite Series

The number 0.9 repeating can also be expressed as an infinite geometric series, the sum from 2 to infinity of 9 * ((1/10)^(n-1)). Using the general form of a geometric series (a * r^n-1), we are able to say that in the given geometric series, a = 9 and r = 1/10. Because |r| < 1, the geometric series can be assumed to converge. The exact value of the series can be found using the following equation: First Term of Series / (1-r). For the series representing 0.999, this equation gives (9/10)/(9/10), which is equal to 1.

Algebra proof

(Moved from above to ease editing)

[[I must contest your explanation below here. You are misinterpreting the Multiplicative Identity Law. The Law states that for any number x, 1*x = x. The c in the equation you are manipulating is the constant "1", and the equation only applies if 1 is in the equation (since 1 is the multiplicative identity in the real number system). The point of the equation is to replace x with any real number, and see that the equation is the same. Now, if you replace 1 with any other number, you will get an equation that is not equal (with the exceptions of x=1 and x=0). The point of the proof 10c-c=9 is derived from 9c=9, where our actual variable x=9, and our constant c is the identity. If .9999... truly equals 1, then it too is a multiplicative identity. The equation holds for c=1 and c=.9999..., so the number .999... is indeed equal to 1. Additionally, you erroneously stated that 9*.9999... is equal to 9.9999..., which it is actually equal to 8.9999...=8+.9999...=9.]]

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."

This is not true because if c were a variably then it may be replaced by any number. For exaple, if c=5 then 10(5) -5 would equal 9, which it clearly does not. That leaves c to be a constant, which in this case c will aways equal .9 infinity. Although 10(.9999...)-(.9999...) does equal 9, the second part of the equation does not equal 9. When you substitute c is the equation 9 *(.9999...)= 9.9999... and not 9.

^^ The above proof is invariably flawed as in stating that c = 0.999... then it can't be replaced by any number. Just because it's a letter doesn't mean it can represent any number lolz. Saying "let c = any real number" means that it's a true variable, i.e. it can represent any number. In this case, c can only equal 0.999... In short, lrn2math, you're not good at it.

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)

No, I think you're the one confused here, the problem is that this is a countably infinite set, you absolutely have a right to count the number of digits. If the infinities are not the same (which they clearly are not), then this is a valid point. - M Braithwaite

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

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I must disagree on this, as 1 divided by 3 is one third, and a third is written by 0.333....., then that times three is 0.999......! and one third times three is 1!

However, 1/3 != .33..., it is merely a near-perfect approximation of the fraction 1/3, as 1/3 cannot be perfectly expressed as a decimal. So even though 1/3 * 3 = 1, .33... * 3 does not.

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Must disagree that you disagree, 3 times 0.333... is 1 and not 0.999..., it's true that 0.333*3 equals 0.999 but 0.333...*3 is in fact 1 because its endless.

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.333...=1/3 is indeed provable by using the least upper bound property in ℝ, one can find the exact value. Perhaps in another number system not in or based of ℝ could you argue otherwise...

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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?

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The curve is asymptotic, that means it approaches 1 as you go off into infinity. It does not mean that it ever reaches 1. Its limit is 1, not its value at infinity.

A limit of a function need not equal the function at ANY point. This is I believe the entire issue of this problem. If you have an assumed limit in a series, then 0.9999... = 1. In mathematics we should never have an assumed limited to a definition that has not been strictly assigned. So if you look at the next two functions you will see the difference.

(A) = SUMMATION (n=1 → Infinity) of (9 * 0.1^n) = 0.999...

(B) = limit (k → infinity) SUMMATION (n=1 → k) of (9 * 0.1^n) = 1

My basic argument against 0.999... = 1 is that from the above (A) != (B). {-JLA-}

Finally, another carbon unit that thinks the way I do. Yes, the crux of the problem is Euler and his wrong ideas in this regard. http://johngabrie1.wix.com/newcalculus

I also think that many mathematicians do not understand what is a limit. For example, to write Lim (x->+inf) 1/x =0 means:

[1] The least value that 1/x cannot attain as x increases without bound is 0. [2] It is incorrect to say "as x approaches infinity" because a) Infinity does not exist (it is an ill-defined concept) and b) It cannot be approached - ever. [3] To use the word "limit" almost implies that 1/x can actually attain the value which in this case is false. In fact, limit should be called greatest lower bound in this case. If for example one had Lim (x->2) x^2 = 4, then the limit is not an upper or lower bound and it can be attained. [4] Limit theory has no place in calculus and should be removed completely. A good place to dump it is in the topic of set theory.

173.228.7.80 (talk) 00:07, 7 March 2013 (UTC)

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The problem with arguments above is that c=0.999... is not a clearly defined number. Infinity is not a set number. (Inifity - Infinity) != 0. That being said. 0.999... - another 0.999... is an undefined value because the two infinities in question have no rule to make them equal. Therefore 0.999... - 0.999... can not be proven to equal 0 since they do not have the same number of decimal places. {-JLA-}

Alright sorry guys but I do agree that .999 (repeating) = 1 and I will be furthering the explenation into why I belive such. With this method.

The concept of infinty has been used for a while and it does truly exist. Maybe we don't know many things that are infinite but welcome to math where we are talking about a very nice subject in which infinity has a place to live.

Ok now that that's away

.999...= x. x can NOT BE EQUAL TO ANYHING ELSE BUT .999 do you know why? because I just define x to be equal to .999.... So adding any other number would break the constraint I placed it. That is the only possible solution. X represents a number in which it's value may not be known but it doesn't have to be equal to any number in the number spectrum. It is not a paramater and that's why I personally use X to make it less confusing. We are aloud to make x be define by making it equate to something

.999.. (multiplied by ten) = 9.999.....

x (multiplied by ten) = 10x

I think we can agree with x being equal to ten x and that is not a problem. So for the one above the argument often is that a 0 will be added to the end and shift everything to the decimal place. BUt let's not forget that we know there is a infinite amount of 9's. So let's say we add a 0 at the end of an infinite amount of nines. That would make sense but let's not forget it is placed after the infinite amount of 9's.

So .999(after infinite amount of 9's) 0 = 10x

OK now let's get to the interesting part

9.999 (after infinate amount of 9's ) 0 = 10x

- .999 (infinte amount of 9's) = X

Let's agree now that we can add a infinite amount of zeros after the infinite amount of 9's which is always assumed that we can add as many zeros after wards in the decimal place and it does not affect the number .0010 = .001

So that means this can be rediscribed as

9.999 (after infinate amount of 9's) 0 = 10x

- .999 (after infinate amount of 9's) 0 = x _______________________________________________

9 = 9x

Now because of this we can do a simple subtraction to get this result. And of course from here.

1 = x

Ok so now you might say wtf, I still can't believe it because they are not the same number. That was the reaction I had. Try to find a fraction, which you know is a real and defined number and make that fraction equal to .999.... If you can achieve this place it on here and that would be a great way to disprove it. To me .999 infinite is impossible to get from a single fraction because it would be equal to the number 1 or else.

For the argument 1/3 not equal to .333...well exactly there is no real fraction that can be equivalent to that in the same way .999 cannot be achieve and therefore really is a mathematical equal to equal to one. Don't forget this is a single explanation and further proof is needed to prove it because it does not fill in all the loopholes. But as for this argument, it does prove what it needs

Gabzo

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For anyone who asserts .9... ≠ 1 because .9... never reaches 1 (being asymptotic), I would like this person to explain Zeno's paradox. The paradox and variants usually go something like this: Suppose you have a race between Achilles and a tortoise. Since Achilles is clearly faster than the tortoise, the tortoise is allowed a 100 meter head start. Let us denote the positions of the racers Achilles and the tortoise as "p0" and "p1" respectively. When the race starts, Achilles sprints and reaches the tortoise's initial position 100 meters ahead, but the tortoise has moved 10 meters since the start of the race, meaning Achilles is now at "p1", and the tortoise is now at a new position, which shall follow the trend for my system of labeling, and be denoted "p2". Therefore, the tortoise is now 10 meters ahead of Achilles still. (Note that means Achilles has closed 90% of the distance the tortoise had.) Therefore, Achilles sprints again, this time 10 meters, to the tortoise's new position. The tortoise has moved again though, this time 1 meter ahead of his position at "p2", and is now at "p3". (Note that means Achilles has closed 99% of the distance this time. This means percentage of the distance is approaching 99.9...%, or just .9... in decimal form.) Dividing this into an infinite number of steps, does Achilles ever catch up (or, in other words, achieve 100% of the distance between himself and the tortoise)? If you can justify that .9... never reaches 1, then Achilles should indeed never reach the tortoise, correct? But it is ludicrous to suggest that, as we can test the scenario and prove that with the given parameters, a runner could overcome a tortoise moving at the speeds described. So where does that leave infinity? Does the limit as n approaches infinity for the sum of n=1 to infinity for the sequence .9+.09+.009+...+(1/10^(n-1)*.9)=1, or is Achilles destined to never catch up to our devious tortoise who planned this paradox from the beginning?

--sly625

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The Achilles "paradox" would only work if Achilles is either slowing down or stopping at every new p1. If both Achilles and the tortoise occupy an infinitesimal space and we count their distances as line segments, the distance from origin of the tortoise will be 1 and the distance from origin of Achilles would be some value 1*x . Right before surpassing the tortoise, their distances from origin have to be equal. That is 1 = 1*x We can look at this by taking apart 0.999... 1-(1*0.9)=0.1 ans-(1*0.09)=0.01 ans-(1*0.009)=0.001 If this happens forever there is some 0...1 that is missing at all times meaning Achilles never reaches the tortoise.

--homeslice

I'm afraid your understanding of "forever" is flawed. There is only ever some 0.0...1 "missing" if there was an end. Because this is infinity, the end is never reached, ie. the value missing tends to 0 and will never be greater than 0. JaeDyWolf ~ Baka-San (talk) 10:04, 6 May 2014 (UTC)

Contesting the "Digit Manipulation Proof" that 1 = 0.999...

A proof that has been put forward for 1 = 0.999, as follows:

$x=0.999...$

$10x=9.999...$

$10x-x=9.999...-0.999...$

$9x=9$

$x=1$

In the 4th line of the proof we see that from the RHS: 9.999...-0.999... = 9. This is not the case. While it is true that the number of 9s after the decimal point is infinite, it is neither sufficient or trivial to assume this step. For example:

Take an infinitely long integer consisting exclusively of 9s, call this number A. Take second infinitely long integer also consisting exclusively of 9s and call this number B.

A - B = 0

If and only if the number of digits in A and B are the same. Since x has been multiplied by 10, its infinite number of 9s after the decimal point is one less than in x by definition. It is not sufficient either to say that there are an infinite number of 9s after the decimal place therefore it does not matter, irrespective of whether the repetition is finite or not, the number of digits does need to be the same, and as Mathematicians know, not all infinities are the same. Therefore the 4th line should say:

That is only true for a finite series.

Say you have two sets, A and B. If A = {0, 1, 2...} and B = {1, 2, 3...) (both infinity), does that mean A has more digits than B simply because it has a zero? No, both sets are equal.

9x = 9 - d

Where d is a infinitesimally small number which is strictly non-zero but limited by zero.

0 is an infinitesimal, and the only one to exist in the real number system

(X=.99999 and X= 1?) how does that work?

You say the number of digits in A and B would not be the same if you multiply x by 10. I'm not entirely sure here, but wouldn't saying that multiplying a number by 10 (or 100, 1000 etc.) moves the decimal place one to the right be a simplified way of expressing the final answer? Doesn't 10 * x = x + x + x + x + x + x + x + x + x + x or x added together 10 times? Therefore, if you add .999... to .999... 10 times, shouldn't you maintain the same number of decimal places? I feel like I'm probably wrong but I'm not seeing where.

...and all this time I thought wikipedia had all the answers...

-> 1 cannot equal 0.999... because although 0.999... is an infinite it will never reach the value 1. 1 and 0.999... will always get closer to the same value the further towards infinity but they will NEVER be the same value. The separation between 0.999... and 1 is infinitely small but it is not 0.

0.99... is 1. 1/9 = .11... 2/9 = .22... 3/9 = .33... 4/9 = .44... 5/9 = .55... 6/9 = .66... 7/9 = .77... 8/9 = .88... Therefore 9/9 = .99... but wait 9/9 is also 1

Ignoring the fact you've just said 9/9=0.999...=1, what you're describing is just the nature of infinity at work when it's used in equations; in real life nothing can continue for infinity but the theory states that if we COULD make something infinitely small then it would be equal to 0. JaeDyWolf ~ Baka-San (talk) 16:02, 6 March 2014 (UTC)

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).

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)

If your proof is correct, then everybody who uses irrational numbers is apparently hallucinating. And if duplicate representation is not allowed, what about the fact that 2.5 = 2.50 = 2.500 = ...? Double sharp (talk) 13:45, 27 February 2012 (UTC)

They are indeed hallucinating. Irrational numbers do not exist, only incommensurable magnitudes. 2.5 = 2.50 = 2.500 = ... are NOT different representations, they are different shorthand notations of the one and only representation of 2.5 in base 10, that is, ...00002.50000... 173.228.7.80 (talk) 00:04, 7 March 2013 (UTC)

I have to agree with Conman on this one. In decimal expansions, all decimals extend infinitely. In the cases of decimals that terminate, the recurring digit is 0. So Conman is currect here, the radix system does represent numbers uniquely.

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)

For another counterexample, 1.000 and 1 are different representations of the same number.

If you accept that 0.9~ is an infinite sequence and every DP is a factor of 10 less than the one before then 0.9~ = 1 - (0.1 * 10^(-infinity)). The 10c-c proof appears to be a bodge on infinity - 1 = infinity? In the end you have to face facts that 1/n where n = baseSystem-1 = 0.1~ = BANG. It is just a HUGE bodge to solve a problem. 1/3 in base 3 = 0.1 but move to another base and you have the same problem again, it's bodge time again. Base 10 is a great system but why don't people just acept a bodge is a bodge instead of wrap it up in garbage like the limit of 0.9~=1. Does not matter there are an infinite number of 9s that follow 0. there can never be enough to actually converge. It gets "infinitly close but NEVER gets there. You are always 0.1 * 10^(-infinity) away :)

That's a common misconception. 0.9... isn't an infinite sequence, it's the limit of an infinite sequence. That is, it's what you get at the end of the sequence after infinitely many steps. It's a fixed number, not a process, so "never gets there" makes no sense. --Tango (talk) 13:06, 8 October 2008 (UTC)

Ok, so 0.9~ is the result of the infinite steps caused by 1/3. And you say the "limit" is 1 because while there is a difference of 0.1 * 10^(-infinity) between 1 and 0.9~ as that value is impossible to describe or define because in reality 0.1 * 10^(-infinity) is not a valid calculation the difference is treated as 0? Because 1/3*3=1 is fact the 1=0.9~ is in reality a "render" issue of trying to represent 1/3 in base 10, or any 1/n type calculation where n=base-1 ?

The difference between two real numbers needs to be a real number (otherwise the real numbers are pretty useless for describing the real world). That would mean that 10^(-infinity) would need to be a real number. It's clear that n*10^(-infinity)<1 for any integer n, in other words, whatever you multiply it by, you'll still have a really small number. That's the definition of an "infinitesimal" number, however there are no infinitesimal real numbers other than zero (that's the Archimedean property), so it must be zero. If two numbers differ by zero, then they are equal. (The Archimedean property is rather technical, but it does follow directly from the standard definition of the real numbers [which is, again, rather technical!].) --Tango (talk) 10:20, 9 October 2008 (UTC)

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.

Assuming nothing about 0.999... then you must work in real numbers (under the completion of Cauchy sequences, where 0.999... is {0, 0.9, 0.99, 0.999, ...}), not rational numbers, then you can conclude that 0.999... is a rational.

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'?

Try using inefficient long division, and the .999... pops right out.

1/1 = .9 R .1

.1/1 = .09 R .01

.01/1 = .009 R .001

On to infinity gives 1/1 = .999...

But in any case, proving that .999... = 1 and 1 = 1/1, and therefore .999...=1/1 isn't circular logic. It's just the transitive property of equality. 68.12.180.246 (talk)

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)

How about c=9 and d=9. If c = 1 and d = 9 then c/d = 0.(1). Therefore if c = 9*c then 9*c/d = 0.(1)*9 = 0.(9). From this we have 0.(9) = 9/9 = 1.

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... ?

How about 0.999.../1? The numerator does not need to be an integer. Do you believe that pi has an infinite number of decimals? Since pi is a real number, 0.999... is also a real number. So 0.999.../1 is a valid fraction for 0.999...

I suggest 3/3. We know that 1/3 = .333...: thus 3/3 = 3 * 1/3 = 3 * .333... = .999...

(But 1/3 = .333... is as much at issue as 1 = .999...Craig Erb (talk) 05:45, 13 September 2010 (UTC))

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)

Do you accept that 10/9 == 1.111... ? --143.50.212.229 (talk) 23:33, 15 February 2009 (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)

Cometflash (talk) 20:37, 31 October 2010 (UTC):So UTC what you are saying is that, a number that doesn't have a last number, will be determined if is odd or even, by having a number that is supposedly = to another form of number, in this case 2? Could you link me to such of rule? You state 1.999... is a even number with such assurance, so I figure you might have a link for me. Is hard to imagine a number that has not a single even number in them, and still be even. I'm not a math genius, no even close, so please point to my flaw. Imagine a object that is expending, and you accept to be 0.999... cm all around, and you have another object that is not expanding and is 1cm all around. Imagine being a circle for example. So you have two circles, one is expanding, and the other is not. By the logic that 0.999...=1. Then a circle that is expanding= the circle that is not expanding, if the logic is applied to the object.Cometflash (talk) 20:37, 31 October 2010 (UTC)

Why would anyone accept that an expanding object has a fixed size? 0.999... is not an ever-expanding object, in the exact same way that no number is an ever-expanding object. Let me demonstrate using your analogy: a circle whose area starts at 0.9, then expands to 0.99, then 0.999, and so forth will never have an area equal to 1 - and it will also never have an area equal to 0.999...

The main flaw in your reasoning is that you think some numbers expand, and others don't; you weren't taught this in any class, and you'll never find a mathematician who claims this, but you think it all the same. Numbers are just finite quantities; the values they represent are already all there, all at once. Gustave the Steel (talk) 23:09, 31 October 2010 (UTC)

I see you target only one of my question, but I'm ok with that, lets just work on that one for now. 0.999... is not a fixed size, 0.999... is infinite from the way I see. If .999...=1, could I say I have .999... of an apple, if I have 1 apple? The thing is, if I have 0.999... of an apple, that apple should be infinite to me, since 0.999... is an infinite number. However if I have 1 apple, I can eat the whole apple and have nothing, but how can I eat 0.999..., if never ends? Is almost as like, I cannot substitute an apple with the number 0.999..., but I can substitute an apple with the number 1. If 1=0.999..., why can't I be able to use either? Here is a few more questions, can a expending object be measure it? Does any number has a number right before, or after the number being used? Like is there a number that comes just before, or just after 0.999...? If yes, which numbers are they, if not, can numbers really be counted? Cometflash (talk)

Before comparing .999... and 1 it may be helpful to clarify how you perceive .999... exactly: (1) would you agree that the sequence .9, .99, .999, ... gets closer and closer to 1 in such a way as to get arbitrarily close to it? (2) do you think of .999... as the infinitieth term in the sequence above? (3) if you feel that the difference 1-.999... is nonzero, how would you describe it, as an infinitesimal or perhaps something else? Tkuvho (talk) 19:46, 2 November 2010 (UTC)

I'll try my best in answering your questions, I would be happy if someone would do the same to the questions I asked, even if is to say the question cannot be answer it. Also like to note, I don't think things are as simple as a yes or no, right or wrong, and so on. (1)"Yes" (2)"No" (3) I don't know if what I'll write will give you a idea of the answer, will probable just give you more questions of my understanding, but here is for the hope. I see each number as being individuals, therefor, I don't see 0.999... being 1. I believe every logic has a flaw, even know I don't see a flaw in every logic presented to me. To some, if not all infinite numbers, are an approximation of a number you are trying to find, and not the number you want, like 1 divided by 3 equals 0.333..., that alone proves a flaw in math to me(I bet I'll be ridiculousied by this ( I know the word rediculousied probable does not exist, but I think you got the point)). I also think any number fallow by . should be called a fraction of a number and not a number, because to me that is what it is. My views are not final, as the concept of math is very complex. I don't think many people if any can say they understand math fully, otherwise neither this or the .999...=1 entry would be needed.Cometflash 22:21, 2 November 2010 (UTC)

Thanks for your comments. Concerning your answer to (1), keep in mind the following important point: the convergence of the sequence .9, .99, .999, ... to the real number 1 is all that mathematicians mean when they claim that .999... equals 1. That's how they define .999... whereas you appear to view it as a process. Thus there is no real argument here, or rather the argument is not about facts but rather about definitions. This is what makes it so thorny, as you point out. Tkuvho (talk) 03:06, 3 November 2010 (UTC)

If you ask multiple questions at once, you run the risk of having some of them ignored. In any case, I only answered the one that is at the root of your inability to equate 0.999... and 1. After all, once you realize that, 1.999... is obviously even. I will now similarly ignore some of the new questions you raised, while answering those that I feel are most important for your understanding. The point I will address now is your objection to 1/3 = 0.333... To begin with, I will introduce you to number bases.

You have taken for granted all your life that the numbers go from 0 to 9, then 10 to 19, and so forth. Counting up from zero, you always assume that a new digit must be added after the tenth number. Thus, we say that we count in base 10. However, other bases are also possible. Base 8 is sometimes used, and it goes from 0-7, then 10-17, - note that the number 11 in base 8 is the number 9 in the usual (base 10) system. Computers use base 2; they count 0, 1, 10, 11, 100, 101... and 101 in base 2 is what you would call 6.

The point? Apart from the number bases, let me remind you that every (non-integer) fraction may be represented using decimal values. Have you ever wondered why some fractions have finite representations, where others have infinite ones? You are about to learn:

In base 10, only those fractions whose denominator (when reduced) may be expressed as (2^x)(5^y) will have finite decimals. 1/5, 3/100, 21/3000 (after reduction), 1/4, and many others. Every denominator which (when reduced) has a factor of 3, 7, 11, or some other prime besides 2 or 5 will yield a repeating decimal.

With me so far?

THIS IS ONLY TRUE IN BASE TEN!

If our society used base 8, only denominators of the form 2^x would yield terminating decimals. 1/5, for example, would be 0.1463146314...

Some more: In base 3, 1/2 turns into 0.111... and 1/3 is just 0.1. In base 6, 1/2 is 0.3, 1/9 is 0.04, and 1/5 is 0.111...

You must now admit that finite decimals and infinitely repeating decimals both represent rational numbers, and the only difference between the two is a result of the number base we choose to employ. Admitting such, you will then see that repeating decimals are not "infinite" in the sense of being unequal to terminating numbers. This is why 1/3 is exactly equal to 0.333..., and why even numbers with unending representations cannot be said to grow, move, or expand. Gustave the Steel (talk) 03:34, 3 November 2010 (UTC)

Separate thread

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)

Thanks for that- Daan

27 February 2009 Here is an idea. First we look at (1/∞). Although the limit of this is equal to 0, since there is a 1 in the numerator, the value will never reach 0. Now we take a number between 0 and 1, say .9, and raise it to (1/∞). This will approach 1, but like before, it will never reach it. It will be infinitely close to 1. This is to say that as we go to infinity this will equal .9999 repeating. Now if .9999 repeating is equal to 1, than we can set up this equation.

(.9)^(1/∞)=1

Now, if we raise both sides to ∞, then we will discover a flaw in this argument.

((.9)^(1/∞)^∞)=1^∞

Our (1/∞)^∞ on the left side of the equation will cancel out, giving us .9. 1 raised to anything will always be one. Hence:

.9=1

This is false so possibly .9999 repeating is not equal to 1.

Eric

In what way is "1/∞" a function? --Tango (talk) 23:45, 27 February 2009 (UTC)

Sorry, it is not a function, I was poorly referring back to (1/∞). None of the above are functions. I will correct it.

(1/∞) does not have a limit. It is, by definition, the limit of 1/x as x approaches ∞. It should be obvious to see that this is 0. Let me repeat: (1/∞) = 0. Thus, 0.9 raised to (1/∞) is equal to 1. --Taeshadow (talk) 13:06, 22 May 2009 (UTC)

Although I agree that 1/∞ is approximately zero, I hotly dispute that it is actually 0. The limit (x->∞) (1/x) = 0, however 1/x is not zero for any number finite or otherwise.

A limit is not a value, you sure you can show that the limit of 1/x is 0. But I challange you to find me a value for which 1/x actually is 0...

---

The logic for the proofs that 1 = .999... are flawed. ((.999...)*10-.999...)/9=1 is flawed. After multiplying by 10 you have 1 less 9 after the decimal place. Each 9 in the above number can be represented by a counting number. This is equivalent to saying 2 consecutive counting numbers are equal which we know is not true, for example 1<>2, We have a contradiction

Actually infinity-1=infinity . Imagine a number line. For any point on the number line, there will be an infinite amount of points on each side of the point. Even though there may be a difference between the values of the actual points, infinity will be unaffected.--68.239.43.86 (talk) 21:41, 8 June 2009 (UTC)

---

Define: .999... = sum(9/10^i) from 1 to infinity Proof: for n=1: sum(9/10^i) from 1 to 1 =.9=1-1/10^1 < 1 . Assume sum(9/10^i) from 1 to n = 1 - (1/10^n) < 1 Then sum(9/10^i) from 1 to n+1 = 9/10^(n+1) + sum(9/10^i) from 1 to n = 9/10^(n+1) + 1 - 1/10^n = 1 + 9/10^(n+1) - 10/10^(n+1) = 1 - 1/10^(n+1) < 1 so by mathematical induction .999... is clearly less than 1

You may have one less nine, but since you had infinitely many nines to start with, you still have the same number left. Infinity-1=infinity. (That's pretty much what "infinity" means!) --Tango (talk) 13:40, 16 May 2009 (UTC)

I agree that if you subtract a finite number from an infinte number you will get another infinite number, however it will be a different infinite number. Infinity is a class of numbers rather than a number itself. Infinite numbers do not follow the same rules as the real number system. For example, does infinity-infinity=0 or does it equal infinity or some finite number. In our case it equals 1 since we have 2 consecutive counting numbers. There are also different classes of infinite numbers. In real analysis it has been shown there are more real numbers than counting numbers. The difficulty in solving problems that deal with infinite numbers is that it is usually impossible to identify the particular infinite number you are dealing with. It would be more interesting to determine what rules the infinite numbers do follow and a way to identify which infinite number we are referring to so we can use them correctly in the future.

Actually, an infinite value minus (or plus, or multiplied by, or divided by) a finite value yields the same infinite value.

There are only two classes of infinite values - those which are countably infinite, and those which are uncountably infinite. The difference is that countably infinite ranges may be counted according to some scheme; for example, the set of positive integers is clearly countably infinite, since a scheme of adding 1 to the previous value can obtain any arbitrary positive integer. In fact, the set of all rational numbers is also countably infinite. In contrast, the set of irrational numbers between 0 and 1 is uncountable; no scheme exists which could, for every irrational n, be guaranteed to reach n. The number of 9's in 0.999... is countably infinite, as each nine may be mapped to an integer.

You are right that infinity-infinity has no certain result. It is one of the indeterminate situations - its value depends on context, and can be shown to result in any desired value (including infinity). However, that has no bearing on the discussion of 0.999... Gustave the Steel (talk) 20:36, 17 May 2009 (UTC)

Countably infinite isn't a class of infinities, it is a single infinity, $\aleph _{0}$. The infinities we are are talking about here are cardinal numbers (number used for counting, as opposed to ordinal numbers which are used for ordering). Any cardinal bigger than $\aleph _{0}$ is uncountable, that includes the cardinality of the continuum and infinitely many (don't ask me which infinity!) other cardinals. --Tango (talk) 20:53, 17 May 2009 (UTC)

---

Proving that 1 does not equal .9.. using the definition of number sets:

[1,0] is a closed set. In this set all real numbers between 1 and 0 are included.

(1,0] is an open set. In this set 1 is not included, but .999.... is.

Error here. Circular reasoning by stating that since .999.... is not 1. If it is 1, then obviously .999... is NOT in the set.

Error, (1,0] is not open, the value 0 fails the definition of open (assuming a metric topology, which is what most of us think about when we say open set of real numbers).

How can two numbers that are equal not be in the same set? It is because they are not.

Here is why:

A set S in Rm is open if for each xεS:

Ǝex>0:Bex(x)cS

All .99… does is make ex ¬ infinitely small, but it still is bigger than 0.

Therefor these two numbers are not equal.

(sorry for the poor symbolism, I do not know how to make them look right on Wiki)

I believe this definition was proven by Gauss if I am not mistaken.

- Stephen

[1,0] and (1,0] are both empty, I think you mean [0,1] and [0,1). The former is closed, but the later is not open - it is "half-open", if you like, but topologically it is just neither open, nor closed (there is no requirement that all sets fall into one or other category). But that's just terminology. 0.999... is not in [0,1), since it isn't less than 1 (it is equal to 1). A set in R^m is open if it contains an open ball around every point, a ball must have finite positive radius (ie. non-infinite and non-zero - there is no such thing as an infinitesimal in the real numbers, see Archimedean property). Can you give an example of an open ball around 0.999... that is contained in [0,1)? --Tango (talk) 15:09, 30 May 2009 (UTC)

Can you please define the last number in the set [0,1)? — Preceding unsigned comment added by 71.240.228.52 (talk)

There isn't one. It contains all positive real numbers strictly less than one, they can be arbitrarily close to 1. --Tango (talk) 17:21, 30 May 2009 (UTC)

If it exists, then it has to have a definiton right? — Preceding unsigned comment added by 71.240.228.52 (talk)

If what exists then what has to have a definition? (Please sign your comments by adding ~~~~ to the end of your posts.) Maelin (Talk | Contribs) 02:10, 3 June 2009 (UTC)

The last number in [0,1) doesn't exist, there is no such thing. We can prove that by contradiction: Assume there is a last number and call it L. Since L is in the set it must be strictly less than 1, so 1-L is a positive number. Therefore (1-L)/2 is a positive number. L+(1-L)/2 is therefore larger than L, but is still less than 1, but that contradicts L being the last number in the set. Therefore, there is no last number in the set. --Tango (talk) 02:18, 3 June 2009 (UTC)

Places for Questions

because people like me who see the proof, understand the proofs, but still have questions. (please answer to help with a better understanding)

It seems to me that all of these calculations and proof revlove around the asssumption that 1/infinty = 0? Is this not ture?

Is 1/3 = .33..? If 1/3 assumes with in it the remainder to an infinite ammount of positions which is why .3.. + .3.. + .3.. = .99.. and 1/3 + 1/3 + 1/3 = 1? I am not sure how you can do any mathmatical calculation on an infintly repeating number, becuse your calculations never end.

.4 +.3 +.3 =1, .34 +.33 +.33 =1, .334 +.333 +.333 = 1 .3..4 +.3..3 +.3..3= 1. so 1/3 + 1/3 + 1/3 assumes in the remainder right?

How would you define the last number in the set [0,1)?

The limit of 1/x as x tends to infinity is 0 (which is sometimes written as 1/infinity=0, but it is best to avoid that kind of notation unless you know what you're doing as it doesn't always work as you would expect). That isn't an assumption, it is a simple result of the definition of the real numbers and of limits. You can't always do calculations with infinite series because, as you say, the calculation doesn't end. You need to know that when you jump to infinitely many terms everything behaves as you would expect based on the finite approximations you can actually calculate. That isn't always the case, but it is the case if the series in question converge, which 0.333... and 0.999... both do (as do any other decimal expansions). --Tango (talk) 17:27, 30 May 2009 (UTC)

No wonder you can't write this in an exam:

Q: Prove the that ${\frac {n}{n+n^{2}}}$ goes to 1 as n goes to infinity.

A: That equals $1+{\frac {1}{n}}$, now let n = infinity to get 1 + 0 = 1. ²³¹₉₁Pa (chat me!) 02:37, 18 December 2009 (UTC)

That has nothing to do with non-existent limits, it's just bad algebra... the method is sound, but it should be ${\frac {1}{1+n}}$. You them substitute in infinity and you get zero. You are implicitly using the continuity of division and addition, but that's obvious enough not to write down in my book. --Tango (talk) 00:11, 6 January 2010 (UTC)

Yes, I formatted the equation wrongly...I can't remember exactly what I meant, but its main point was substituting in infinity for n. Double sharp (talk) 14:11, 27 February 2012 (UTC)

I am shocked at how many false proofs are found here. --116.14.72.74 (talk) 10:48, 25 July 2009 (UTC)

if any of you had any sort of real intelligence you would see that not only does .999... not equal to 1, but that there are no such things as fractions of any sort in reality, when you cut something into thirds you dont have 3 pieces that are equivalent, you have 3 different pieces of that item. it is not possible to weigh or measure one third of anything being that every single piece is going to be different on molecular, atomic, and subatomic levels. this proves that if there is such a thing as .9999..... then it would have to be finite, and that it would be less than 1, and in all honesty outside of a virtual simulation, it is impossible to have a multiple equal fraction of anything, whether it be 1/4 or 1/3. so until we are capable of finding the smallest number possible, and be able to cut it perfectly equal, then fractions are nothing more than a theory. it is not at all possible to create more than one of anything that is perfectly identical, therefore the truth of the matter is that in the three dimensional world in which we currently live there are not more than one of anything. end discussion -mentalninja

Maths isn't supposed to be directly related to the real world, it is an entirely abstract subject that happens to be useful for modelling aspects of the real world. --Tango (talk) 11:29, 15 August 2009 (UTC)

Suppose you have a bar of pure gold, containing six million atoms. Through the use of technology, you separate the hunk into three equal pieces, each containing exactly two million gold atoms. I would argue that each piece constitutes 0.333... of the original bar. Care to prove me wrong? Gustave the Steel (talk) 15:04, 15 August 2009 (UTC)

there currently is no technology to divide them perfectly first of all, and secondly even if you had those six million atoms divided down into two million a piece, the weight of each individual atom cannot be exactly identical unless you can prove to me that all atoms of the same element are exactly identical in weight, volume, mass... ect. but since we don't really even know if atoms are truly the smallest thing out there how can you base it? hell for all we know, our own universe could just be another atom to some molecule of something completely insignificant to it's own universe. -mentalninja

Of course. However, gold may be measured in atoms - whether or not they have varying amounts of subatomic particles is irrelevant to the amount of gold in each piece. Gustave the Steel (talk) 07:12, 17 August 2009 (UTC)

I have three dollars in my pocket. You have one dollar. You have exactly one-third the amount of money I do.—Chowbok ☠ 21:54, 16 June 2010 (UTC)

I'm a humanities major, not a math major, but it seems to me that one of the key challenges is to find a number which is the difference between 1 and .999... if they are not equal. Wouldn't such a number would be the remainder, or .1^infinity. If it is true that for every possible non-infinity iteration of .9, .99, .999, etc., there is a corresponding iteration of .1^n that comprises the remainder, then why isn't .1^infinity the remainder for .999... It keeps being pointed out that .999... is a fixed number, not a process. Well, then why isn't .1^infinity? They both utilize the same concept of infinity, no? Or would math professionals do the exact same thing to .1^infinity that they would to .999... and simply identify it by its limit (i.e., by its standard part, in this case 0)? It also seems to me that this argument is partly due to general confusion/lack of understanding about the inherent problems with the concept of infinity, i.e. Zeno's Paradox: If a single distance is comprised of infinitely smaller, discrete distances, then how can motion be possible at all? E.G. To go from Point A to Point B, with a total distance D, you first travel d/2, or half the distance. But to travel d/2 you must first travel (d/2)/2 or 1/4 the original distance; continued ad infinitum. Thus, to go between any two spaces, you must cover an infinite number of distances. And even if you solve it mathematically using examples of motion from experience, it still doesn't answer the base question of how motion is possible to begin with. You are begging the question. Likewise, if a limit is the infinite approach toward a number, how can the limit ever be reached? Its almost equivalent to saying two parallel lines intersect (in Euclidean geometry), or that an object could attain absolute zero temperature (and still be an object). It seems to be ruled out by the very definition of the thing itself. I am a fan of the scientific method, so I would be gladly humbled if someone could explain why this is wrong. 66.229.236.236 (talk) 05:15, 10 December 2009 (UTC)

.1^infinity would, indeed, be defined as a limit: $\lim _{n\rightarrow \infty }10^{-n}$. That limit is zero. The difference between 1 and 0.999... is zero, so they are equal. --Tango (talk) 17:16, 11 December 2009 (UTC)

I have to dispute this claim: the difference between 1 and 0.999.. is zero just because the limit is zero. Calculus is about the infinite summation of infinitesimal values. If Mathematicians wrote off these infinitesimal values as zero, integral calculus would not work. Just to clear up any confusion about this, the limit of a function is a value which cannot be crossed. Furthermore, in general, if a value is converging towards this limit, it's never actually going to reach it, even at infinity.

Here is a challenge: Find me the greatest value of x such that x is less than one. In other words, what is the highest possible value less than one. .999...8 ??? but if that is true then it is also possible for a .000...1 to exist.

There is no such greatest value. Just because you can describe something doesn't mean it exists - not in maths, anyway. .999...8 is just as nonsensical as .000...1. --Tango (talk) 00:13, 6 January 2010 (UTC)

Responses to 82.45.18.63

Because the person on the abovementioned IP posted in about 6 places yesterday (btw, please sign your posts on this page to make it easier to see who said what) I'm going to try to explain the main issues all here.

First, on the concept of infinity and infinitesimals and how they relate to the real numbers: When working with real numbers and functions of real numbers, infinity does not exist. As a symbol, it is convenient notation to describe something that grows without bound or end, but there is no point on the number line you can point to and say "that's infinity". Similarly, the reciprocal of infinity is not a real number either - however, using it as a notation of something growing without bound, then if something "goes to infinity" its reciprocal "goes to zero". So your comment that "1/x is not zero for any number finite or otherwise" is true to an extent, except that the "or otherwise" part is redundant when we're talking about real numbers.

As for the comment that "Calculus is about the infinite summation of infinitesimal values. If Mathematicians wrote off these infinitesimal values as zero, integral calculus would not work", this is not true. Again, calculus tends to use infinitesimal notation, thanks to Leibniz among others, but these days the formal definitions of calculus refer instead to limits or measures. Certainly we sometimes manipulate the dy or dx in a notation as though it were an actual number and get the right answer, but this is a fudge and not strictly correct. In the real numbers, the Archimedean property holds, meaning that there are no non-zero infinitesimals.

A result of these two facts is that (a) we shouldn't say that 0.999... has infinity decimal places, but rather that there are an infinite number of them, or even better that the 9s continue without end - and as a result, there is no "last" one that can get moved out of place when you multiply by 10 in the digit manipulation proof (however please note that this is by far one of the least rigorous proofs, and is meant to be more of a demonstration of how it works than a formal proof in itself) and (b) when we're talking about real numbers, we can show that 1 - 0.999... is infinitesimal, and as a direct result must be zero (since, as I said, there are no non-zero infinitesimals in the real numbers). It's a difficult fact, and a non-intuitive one, but that's why we have to build the really complicated rigorous proofs to show it.

Also, a side note, "Just to clear up any confusion about this, the limit of a function is a value which cannot be crossed. Furthermore, in general, if a value is converging towards this limit, it's never actually going to reach it, even at infinity." is also not entirely true. The formal definition of a limit (if I try to put it in words, rather than going all epsilon-delta on you) is that by coming arbitrarily close to a given point, a function comes arbitrarily close to a particular value, i.e. the limit of the function at that point. It is entirely possible for a function to (a) take on the value at its limit (that is, in fact, the definition of a continuous function) and (b) to take on a limit value at other points - for example, y = x sin(1/x) takes the value 0 infinitely many times, and its limit as x approaches 0 is also 0. I can go further than that and say that in any open neighbourhood of x containing 0, y takes the value 0 infinitely many times. The same is true of y = 0, but I think the oscillating function demonstrates the point a bit better. Confusing Manifestation(Say hi!) 00:10, 4 May 2010 (UTC)

People, the argument is nonsense. You're using two different number systems that are inherently incompatible. Rational numbers represent a quantitized number line. Irrational numbers represent an infinitely divisble number line. When you mix the two, you end up with Xeno's paradox and other nonsensical results, such as .9~=1. It's not that .9~=1 is false, but rather it is an impossible statement. It's like being given the question: "does Zerdok = 1?" without having any idea what Zerdok is. Or consider this analogy: Let the rational number "1" be a unit of time (sec) and ".9~" be a measure of distance (m). If I travel .9~m/1sec, what is my velocity? The answer: undefined. .9~ represent an infinitely divisble (irrational) distance. "1" represents a quantitized (rational) time. One cannot cover an infinite (irrational) number of distance intervals in a rational (lesser) number of temporal intervals. When one converts between rational and irrational, certain things are lost in translation. They are not the same language, so there is no strict equivalency.

Why hasn't anyone thought of this?

If 1 = 0.999... then since 1 - 1/infinity = 0.999..., 1/infinity must be proven to equal zero. Zero, when multiplied by anything, must equal itself (that is, zero). though 22 multiplied by 1/infinity equals 1/infinity (because 22/infinity equals 1/infinity), infinity multiplied by 1/infinity yields one. Zero multiplied by infinity yields zero, not one, so 0 does not equal 1/infinity so 1 does not equal 0.999...

"infinity multiplied by 1/infinity yields one." Not true.

"Zero multiplied by infinity yields zero" Also not true.

I am sorry that i contest this point. Zero times anything is zero; although infinity may not be treated as number, it is not exempt from the laws of mathematics. I understand that limits of infinitesimal things can be zero, but a limit is a boundary and it shows a total lack of understanding of the concept of a limit as a boundary. Furthermore zero multiplied by anything is zero is a fundamental axiom of mathematics and is not open for dispute. So I am sorry, but in my view, this assertion is correct. Zero multiplied by infinity is zero. If you are in doubt graph y = 0*x, or better still prove that it's the case by induction, however don't muddy the water with your erroneous concept of limits.

"Zero multiplied by infinity yields [...] not one" ALSO not true.

All of the situations you described are indeterminate forms. Read up on those. Gustave the Steel (talk) 20:09, 1 June 2010 (UTC)

My point remains. There is no decent reason to consider 1/infinity equal to zero.

I was trying to access the talk by Gustave the Steel but I could not. Where else could I find a talk on whether 1/infinity is 0 or not?

You may begin with .9, which does not equal one. .99 makes you ten times closer, but does not make it equal to one. The same goes for .999. You can carry the operation of adding 9/10^x forever, but even though you will always get closer to one, you will never reach one. There will eventually (for the sake of argument, infinity has been traversed) be an infintesimal difference (1/infinity) but there is no reason to discount that. Elementary particles are infintesimal, but they make up our universe.

Actually, there is a very good reason to consider 1/infinity equal to zero. Mainly, that the limit of 1/x as x approaches infinity is 0 - and 1/infinity is not meaningful in any other context.

Also: your last paragraph doesn't apply to the discussion at hand, so I won't be replying to it. Start a new topic to raise new arguments. Gustave the Steel (talk) 21:21, 1 June 2010 (UTC)

It does very much apply to the discussion. If infintesimal equals zero, then elementary particles are irrelevant/nonexistent and this is an imaginary discussion.

"Infinitesimal" doesn't mean what you think it means. It simply means "too small to be measured"; thus, infinitesimals occur quite frequently in real life due to our measuring devices, and they occur nowhere in our number system. Since 1/infinity equals 0, you cannot say that 1/infinity is the same thing as infinitesimal.

You might be interested in learning about some other number systems which include infinitesimal numbers, such as the hyperreal system. (Note that 0.999...;...999... still equals 1 in that system.) Gustave the Steel (talk) 21:50, 2 June 2010 (UTC)

How are elementary particles infinitesimal? The mass of elementary particles have been determined and are quantified. Yongrenjie (talk) 09:50, 15 October 2010 (UTC)

Another attempt

Assume that for any x, x/x = 1. Assume also that for z=x/y, if x is greater than y then z > 1, and if x is less than y then z < 1.

To find z=0.999... we can construct x=(10^n)-1 and y=(10^n) assuming n=∞. This shows rather plainly that x will always be less than y, and according to the above assumptions, z < 1.

You are basing your proof on the assumption that ∞ - 1 < ∞, which is false anyway because ∞ - 1 = ∞. Yongrenjie (talk) 09:40, 15 October 2010 (UTC)

It is your assumption that ∞-1=∞. Notice that his example involved (10^∞)-1 and 10^∞. It's clear to see that 10^∞ would be a 1 followed by an infinite amount of 0s. In the case of (10^∞)-1, we would have an infinite amount of 9s. It's clear to see that they're not the same thing. ~~

Yes you can say that ∞ - 1 = ∞, however Cantor showed that infinities differ in size and it is absolutely crucial to pair of elements within the infinity to guarantee that you have infinities of the same size. In this case ∞ - 1 = ∞ is merely a technical tautology of saying that two infinities of different sizes are both infinity, so therefore they must have the same number of elements. This violates all of Cantors mathematics, it violates the priciples of calculus, it also violates the rules of Mathematical Analysis.

Argument from Logic

This is why 1 = .999~ is absolutely logical. If the two weren't equal, some value would exist that could be added to .999~ to make it equal to 1. That value would be equivalent to .0 repeated with a 1 stuck on the end (also an infinitesimal). By definition, infinity is an abstract concept describing something without any limit; in other words, it never ends. So .0 repeated with a 1 at the end is equal to 0. And therefore .999~ + 0 = 1

Another way to think about this proof: 1 - .999~ is infinitesimal and therefore 1 = .999~ (Andrew F)

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If A=A then A cannot equal not A.

((Alright I'm jumping in here to explain something. Start of gabzo's insert)

we are not saying that both numbers have different values, we are arguing that they both represent the same number in the number spectrum. In other words we are saying .999... is the same number as 1.

Gabzo ((I'm done for this part. End of Gabzo's insert.))

Taking 1 for A gives the result of 1=1 which is true by identity.

Therefore, 1 described as a decimal is 1.0~, as state elsewhere above when discussing terminating decimals. This, however, does not create an equation where 1.0~=0.9~ is true, though it appears to be if taken as an approximation and the remaining 0.0~1 is ignored. The problem here, however, is in ignoring the 0.0~1 which, however small it is, is still a real number and forms the base of the set, being the first number in the set larger than 0, while 0.9~ is the last number in the set smaller than one.

To clarify by example, taking those integers larger than 0 and smaller than 10, the largest plus the smallest equals the upper delimiter, the second largest plus the second smallest also equal this same value, so that 9+1=10 and 8+2=10, therefore it follows not that 0.9~=1.0~ but that 0.9~+0.0~1=1.0.

Jwlockhart (talk) 05:02, 17 February 2011 (UTC)

I just want to say that I think you are correct about everything in this. My point is that the proof that 1/3 = .333~ .333~ * 3 = .999~ and 1/3 * 3 = 1 is flawed because they have it backwards. They are assuming that 1/3 = .333~, while .333~ is just the most accurate representation of 1/3 using the base 10. The 1/3 = .333~ proof does nothing to prove that .999~ = 1, it in fact disproves that 1/3 = .333~. Callmenar

(Start: Writing by Hypocrism) I don't know if this is the right section but it reminded me of a "disproof of a disproof" I wanted to test out. Apologies for my logic symbols - I do not know how to make all of them on a keyboard. I have used != as not equals, € as member of, (R+) as positive real numbers. (NOTEX) designates a conclusion I want to mark to use later in the disproof.

Assumption 1 (A1): 0.999...!=1

Assumption 2 (A2): ∄ q€(R+): 0.999...<q<1

From A1 ⇒ ∃a€(R+): 0.999...+a=1 (NOTE 1)

∀a€(R+) , (a/2)€(R+) and MOD(a/2)<MOD(a) ∴ 0.999...+(a/2)<1 , 0.999...+(a/2)€(R+)

Let 0.999...+(a/2) = b : 0.999...<b<1

This contradicts A2 ∴ either b=0.999..., OR b=1, OR A2 is false.

Possibility 1: b=0.999...

If b=0.999... ⇒ 0.999... + (a/2) = 0.999... ⇒ a/2 = 0 ⇒ a = 0

If a=0, 0.999...+a=1 ⇒ 0.999...+0=1 ⇒ 0.999...=1

Possibility 2: b=1

If b=1, 0.999...+(a/2)= 1

But (NOTE 1) ⇒ 0.999 + a = 1

⇒ (a/2)=a ⇒ a=0 or a=(infinity)

If a=0 0.999...+0=1 ⇒ 0.999...=1

If a=infinity 0.999...+(infinity)=1 and this is not a possibility.

Possibility 3: A2 is false [section needs improvement]

∴ ∃ q€(R+): 0.999...<q<1

This q must be represented numerically but there is no numerical possibility greater than 0.999... and less than 1.000... ⇒ ∄ q€(R+): 0.999...<q<1 ⇒ (A2 is correct)

Possibility 1, possibility 2, and possibility 3 are all impossible ⇒ (A1) is false ⇒ 0.999... = 1 Q.E.D.

[[OLD If (A2) is false ⇒ ∃ y€(R+) : 0.999...<y<1 , Then by repetition of logic ∃ an infinite number of g€(R+) : 0.999...<g<1. Let us consider a subset of them: a, a/2, a/4, a/8...

SUM(a,a/2,a/4,a/8...) is a finite geometric sequence. Sum to infinity = a/(1-0.5) = 2a

∴ for (a: 0.999...+a=1), implies a value of 2a between 0.999... and 1 due to a, a/2, a/4, a/8.. also in the interval.

0.999...+a = 1]]

(End: Writing by Hypocrism)

Reply to Writing by Hypocrism

Hypocrism, I would see your Assumption 2 as being false, at least in relation to the rest of what was stated in this section, namely that 0.9~ is the last and largest number in the set of all numbers smaller than one. Following from that assumption, which was given as a basis for this section, 0.9~<q<1 is by necessity false due to its contradicting a preceding assumption. Further, in your Note 1 reference in Possibility 2 you state that 0.9~+A=1 yet go on to suggest that A must either equal 0 or be infinity. There would seem to be a third possibility that you overlooked, that A is infinitesimal, that is it is properly 1/∞ or as I notated it above 0.0~1. Thus, if you take A as 1/∞ then A/2 would be (1/∞)÷2 which would still be infinitesimal.

Further, since, as we restated above, 0.9~ is the largest number smaller than one, then it is smaller than one by an infinitesimal amount, that is, it is 1/∞ smaller, or 0.9~=1-1/∞ . This would leave you with A1 true and A2 false, as there can be no number between 1 and 1-1/∞ due to there being no smaller quantity to remove than one that is infinitesimally so. --Jwlockhart (talk) 07:34, 15 May 2011 (UTC)

(Start: Writing by Rustebucket) From what I have seen the argument that 0.999... is equal to 1 is based on the idea that a value that is infinitely small is equal to zero. Infinitely small and zero are two very different things. (End: Writing by Rustebucket)

"1/3" is similar to the idea of exponential decay. Exponential decay will infinitely move towards zero but will never reach zero. The term 1/3 is a one dimensional representation of exponential decay the number "0.333..." will infinitely try to reach 1/3 but by the same token never will. Unfortunately this means that any number with a repeating decimal is absolutely an irrational number. The decimal form of "1/3" is no more usable than the decimal form of "pi" or "e". 1/3 is the only correct way to describe "1(1/3)" and the only way to end the decimal is with it's proper form. *0.33⅓ or 0.3333⅓ or 0.33333333⅓* All of these are correct values for "1/3" but all require "1/3" to define. No matter how far out you take the number the only way to end it and get a proper value for 1/3 is to use 1/3 and that means "1/3" is the simplified version. The decimal form of 1/3 is described with 0.333-repeating infinitely. However infinity is just a mathematical concept and not actually a number. So to describe 1/3 as "0.333..." is just a "concept". --tyty8383 Oct 30, 2011

The Nature of Infinity

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One problem not as clearly addressed in this argument is the nature of infinity. There are two primary definitions being discussed. Infinity meaning to step off into forever, and infinity meaning a complete inifinitum.

(1) If infinity means to step off into forever, then 0.999... does not ever equal 1. Any asymptotic function does not ever reach its limit, it forever approaches it. It has a limit, but it never reaches it if you are stepping off into forever. With this definition infinity is something that can be approached but never attained.

(2) If infinity means a complete infinitum the idea changes. An example of this is the X-Y plane. If you can fathom this as simply existing, instead of stepping off into forever, then that is a complete infinitum. If infinity means this, then it is possible that 0.999... = 1. With this definition infinity is actually attained.

With that being said, most mathematics treats infinity as (1) above. Set theory and Real analysis use the stepping off into forever for most of the proofs. My problem with 0.999... = 1 is this inconsistancy with using infinity to mean what you want it to for the moment to prove a point. As they say, "You can't have your cake and eat it too." So if we stick with the most common definition of infinity, (1) above, then I argue that 0.999... != 1. {- JLA-}

You're conflating two things. 0.999... isn't a function or a process, it's a number. While it certainly is a limit, all numbers are limits. In particular, it does not refer to the process of taking 0.9 and adding another 9 to the end of the decimal repeatedly, which indeed never reaches 0.999.... 0.999... truly has an infinite number of digits, as there is no last 9, and indeed equals 1: it is the limit of the above process, which as you said never reaches 0.999... = 1. Double sharp (talk) 08:10, 19 June 2014 (UTC)

X approaching 1

How does that work when evaluating the function (1/(1-x)) as x approaches 1? The equation has a legit value for every value of X as infinitely close to 1 as we want as long as it's not exactly 1. That would seem to indicate there's two separate categories, "infinitely close to 1" and "exactly 1", which generate different answers. JeramieHicks (talk) 23:50, 16 November 2011 (UTC)

Contradiction

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Let us assume that 1 is not equal to 0.999... There should be a number in between them. What is that number?

Hello. My name is David, and let me just say that you cannot "prove" 0.999 repeating =1 using a mathematical formula. To do that you would need the complete number, and that is simply impossible. The fact is that 0.999 repeating will never reach 1, with every digit, it will get 90% closer, that gap just keep shrinking and shrinking infinitely (which is why it is repeated infinitely.)

Mathematical validity of infinitesimal

The whole argument over 0.999~ = 1 is really an argument over whether infinitesimal is even a valid mathematical construct. Many mathematicians give many proofs to show that infinitesimal = 0 in an effort to show that infinitesimal is not a valid construct. Today i give you proof that infinitesimal is a valid mathematical construct.

$Round(0.5)=1$

$Round(0.5-{\frac {1}{\infty }})=0$

$Round(0.5-{\frac {1}{\infty }})!=Round(0.5)$

If infinitesimal = 0, then rounding infinitesimally less than 0.5 would equal 1 just as rounding 0.5 does. Instead it equals 0. Now, I don't disagree that in a quantized universe infinitesimal has no useful meaning to the finite things within it. But is does have meaning on the infinite scale. Looking at everything from an infinite scale, anything finite is infinitesimal by that scale. It is difficult for us to comprehend infinite because it is so much more than us. But similarly, and oddly, it is even more difficult for us to comprehend infinitesimal, because in some sense we are already at infinitesimal. We exist as collections of infinitesimal parts of a collection of infinities that make up our universe. To us, these infinitesimals seem finite and real, but to the universe, all the infinitesimal pieces of matter might as well not exist is is so small. And yet despite making up an infinitesimal fraction of the universe each, there are an infinite number of particles. We (By "we", i mean the matter. Not just humans or even just living things.) make up a measurable portion of the infinite universe. So for my final proof, with empirical evidence (the fact that the universe is neither solid matter nor total void):

${\frac {1}{\infty }}!=0$

${\frac {1}{\infty }}\infty !=1$

Infinity and infinitesimal are not the smallest and largest values in a set. They are only loosely related in being at opposite ends of a set of sets, with all the finite sets existing in between. It is not truly possible to prove or disprove either concept using a finite set as your point of view, only approximate it.

98.163.21.16 (talk) 01:59, 11 February 2012 (UTC)Corscaria

1 - 0.5 = 0.5

0.999... - 0.5 = 0.4999...

Round to the nearest integer:

0.5 ==> 1

0.49... ==> 0 (just in case people were wondering, that is primary school maths)

the argument that 0.4999... also rounds to 1 because it IS 0.5 is not valid, as there you are assuming it to be true, and saying therefor it is true...

I hope everyone agrees that you can't have something that rounds to BOTH 1 and 0, therefore 1 ≠ 0.999...

I've just noticed this one. Your whole argument is based upon the idea that 0.5 rounded is mathematically equal to 1. This is flawed logic and I recommend you read the section on tie-breaking in the "Rounding" article. Rounding up when there is an equal distance between the two potential points is simply the most common choice made when encountered with the arbitrary decision. 0.5 DOES round to both 1 and 0, it's just commonly chosen to round to 1.

I also don't understand why saying 0.49... = 0.5 is wrong just because I'm assuming it to be and therefore I'm just saying it to be. There are plenty of rules already which state that it is actually true; wouldn't using that arugment also point toward saying it isn't true it just saying it? JaeDyWolf ~ Baka-San (talk) 11:07, 29 May 2014 (UTC)

That argument is base-dependent, and relies on the radix being even. If we used base 15, now there's only one way to express 1/2, which is 0.7. That also rounds ambiguously: it could round to 1 or 0, and you must choose one. 0.5 and 0.49 are just two ways in decimal to express 1/2, which rounds ambiguously. Double sharp (talk) 08:12, 19 June 2014 (UTC)

Infinitely expanding numbers being equal to 1 which is finite?

BTW I do accept 0.999 = 1 (I have always believed this intuitively so was pleased to find it proved) but have the same niggles as others here at the sloppiness of the proofs.

I'm sure the following idea is wrong, but I like it:

If 0.999... is infinitely expanding it must be the number infinity. It does not matter how small the pieces are that you add, if you are adding them forever; you will achieve infinity!

~Aaron Wilde

This is actually not correct. There are plenty of series that "add forever" and don't go to infinity. They're called convergent series. Geometric_series is a good page to read about simple geometric series on. The first example it gives is the series 1/2 + 1/4 + 1/8 + 1/16... the terms go on forever, any individual term has a finite size, but the series converges to 1, not infinity :). Paychonaut (talk) 20:55, 4 December 2014 (UTC)

9.999... = 10 x .999... Sloppy?

If we are multiplying by ten by moving everything one column to the left (not everything except the last 9, everything!), then surely:

10 x 0.999 = 9.999...0

(the zero is in the new column we have created)

Rather than say that's wrong, let's go with it for a moment and have a look at what would then happen to the sum:

(9.999 − 0.999...0)

We get the result:

9.000...9

Divide that by 9 and instead of 1, we get:

1.000...1 !

In other words a number which is an infinitely small amount larger than 1. I find this more elegant and satisfying than the usual version of this proof, which pretends to ignore the extra column we have created.

One could say in conclusion that 0.999 can be thought of as being an infinitely small amount smaller than 1, and that mathematics cannot measure the difference!

Indeed, surely if 0.999... = 1 then 1.000...1 = 1 also, since it is the exact same amount greater than 1 as 0.999... is less than 1.

~Aaron Wilde

Well, decimal arithmetic on rational numbers with "post-infinite" digits is sort-of workable, once we recognize that no matter the context, all post-infinite digits "may as well" be zero (or any other finite sequence of digits, or ideally, are really identical to the repeating sequence that precedes them). So yes, in a sense,, 0.{9} = 1 = 1.{0}1. (The angle brackets mean "repeat digits within this forever.") We have to be very careful not to think that post-infinite digits can "mean" anything, though.

Here's one problem that happens if we grant such digits any significance: what is 0.{0}1 doubled? It would have to be 0.{0}2. Well, surely this can be multiplied by 5, right? This would result in 0.{0}10. But that would be identical to the original quantity! Unless, of course, we decide that the number of zeros in that number is "one fewer" than in 0.{0}2. This requires additional notation; let's say that 0.{_∞-10}1 means "a zero, followed by a decimal point, followed by infinity-minus-one zeroes, followed by a 1." Whew!

But what about other bases? Maybe our general rule should be "Everything we say about an infinite sequence of digits should be a simple extrapolation from finite sequences." So let's consider the binary number 0.{1}, which is "normally" exactly equal to 1. What's its equivalent in quaternary? 0.{3}, also usually equal to 1. But let's extrapolate from sequences:

• • 0.3_quaternary = 0.11_binary
  • 0.33_quaternary = 0.1111_binary
  • 0.333_quaternary = 0.111111_binary
  • In general, for every 3 in the quaternary number, there are two 1s in the binary equivalent.

So under our "new" arithmetic, the binary number 0.{_∞1} is equal to the quaternary number 0.{_∞/23} -- which is to that there are half as many 3s in the quaternary number as the binary one! By contrast, 0.{_∞3}_quaternary = 0.{_∞*21}_binary -- the number of 1s in the binary number is twice infinity. My head is spinning from my own nonsense. (Comparison of quaternary 0.{3} to octal 0.{7} is left as an exercise to the reader; it's a doozy for a couple reasons.)

All right, let's say one doesn't mind all that. Consider the type of person who feels that Hilbert's Hotel cannot accomodate anyone if it is full, because they can't just kick the infinitieth guest out, any more than you're allowed to kick the hundredth guest from a hotel with a hundred occupied rooms.. For this person, anything we say about infinities has to be true of finite quantities and vice versa. Fair enough. Well, here's a truly "un-swallowable" problem, I think. Let's go back to base 10. What is one-twelfth in decimal? Well, use of long division or a standard calculator suggests the answer to be 0.08{3}. Meanwhile, would you agree that one-fourth times one-third equals one-twelfth? Hopefully yes.

So what do we find to be the product of one-fourth and one-third in decimal? Well...

• • 0.3 * 0.25 = 0.075
  • 0.33 * 0.25 = 0.0825
  • 0.333 * 0.25 = 0.08325
  • 0.3333 * 0.25 = 0.083325

... and from this point on, each additional 3 on the left-hand side of the equation means another 3 on the right-hand side. Our "new arithmetic" would say that 0.{_∞3}*0.25=0.8{_∞-23}25. There's a "25" at the end and our new arithmetic says we can't ignore it. But it's nowhere to be found in the problem as solved by long division, which implies that "one-twelfth" has a different value than "one-fourth times one-third". Madness! This seems genuinely unsolvable without "resorting" to conventional (and correct) arithmetic, whereby the 25 at the end is thoroughly ignorable. I assume there are an infinite number of similar problems.

Altogether, the only resort a .999 "denier" has is to deny the utility of repeating decimals altogther. To that there's no real answer but... have fun! :) ± Lenoxus (" *** ") 00:42, 20 October 2012 (UTC)

This is so wrong, at the beginning you basically say that you can put a number at the end of infinity. The end of infinity. How can there be an end to infinity?

That's right, this 0.000....1 is a common failure, the number is meaningless in real number analysis as the 1 has no specific place after the decimal point. So the value the 1 gives to the number is meaningless (eg 0.000001, the 1 represents the value one millionth). Then you could go further and write 0.000....01, but then is this the same number as 0.000....1 ? By comparing 1 with 01, then the 01 number must clearly be smaller, But by looking at the number of 0's after the decimal point, they are exactly the same (countable infinity). So in another sense, they are equal. So that just shows its gibberish, and these numbers should be avoided as they are not real. ___________________________________________________________________________________________________________________________________________________________________

The number at the end of the decimal does not mean that the infinitely recurring decimal suddenly ceases its infinite recurrence. take for example 10^-∞. The 1 will be there the whole time. The decimal representation would be 0.0...1 , or 0.0repeating1. Again, the 1 will be there the whole time. The fact that there is something outside of the infinitely recurring decimals does not mean that the infinitely recurring decimals are finite, it just means that there is something else there.

Additionally, the real problem with the algebra proof is that it is assumed in 10x=9.9... , that the amount of 9s to the right of the decimal point in the product of 10 and x is equal to the amount of 9s in the initial X. The reality is that when you multiplied it by 10, you have essentially shifted one of the 9s over to the left of the decimal point. While the amount of 9s to the right of the decimal point may still be infinite, and therefore never end, there is a difference between it and the amount of 9s in the initial X. That can be looked at one of two ways: either...

A) There is one less 9, or B) the set of the amount of 9s in 9.9... is one 9 behind the set of the amount of 9s in x=.9...

By looking at the problem another way, it is clear to see that something is wrong with the algebra proof when we get x=1 at the end.

         x=0.9...
       10x=10 x .9...

Now we subtract x from both sides, which would really just yield us with..

        9x=9 * .9...
         x=.9...

Similar to how

         x=4
       10x=40     10 x 4
        9x=36      9 x 4
         x=4

         x=4
    10 x 4=40
     9 x 4=36
 9 x 4 ÷ 9=36÷9
         4=4

If the value of X magically changes throughout the course of the equation, something went wrong. Again, this has to do with the assumption that the amount of 9s to the right of the decimal point in 9.9... is the same as the amount in x=.9...

As you can see we got our original x, which means we did it right. Again, this has to do with the fact that the amount of 9s to the right of the decimal point in 9.9... is technically one 9 less than that of the amount in the initial x=.9...

Observe the following:

         .99999*10=9.9999 (notice how there is 1 less 9 to the right of the decimal point in the product than there is in the left-hand side of the equation)

By assuming that the amount of 9s to the right of the decimal point is the same as it was in the first place, we could solve and get 1 for .99999 (or .9 of any amount of 9s, for that matter)

                x=.99999
              10x=9.99999
            10x-x=9.99999-.99999
               9x=9
                x=1

But does .99999=1? No, it doesn't.

Let's try it with .9999999999

                x=.9999999999
              10x=9.9999999999
            10x-x=9.9999999999-.9999999999
               9x=9
                x=1

But does .9999999999=1? No, it doesn't.

By assuming that the amount of 9s to the right of the decimal in the product of 10x is the same as the amount of 9s in the initial X, we end up with .9 of a certain amount of 9s being equal to 1. However, when done properly, we get:

                x=.99999
              10x=9.9999
            10x-x=9.9999-.99999
               9x=8.99991
                x=.99999

Similarly,

                x=.9999999999
              10x=9.999999999
            10x-x=9.999999999-.9999999999
               9x=8.9999999991
                x=.9999999999

Similarly, the "algebra proof for .9...=1" is performed improperly. It should be performed as follows:

                x=.9...
              10x=9.9... (with notation to show that the amount of 9s to the right of the decimal point is now 1 less than it was in x=.9...)
                OR
              10x=9.9...0 (STILL with notation to show that the amount of 9s to the right of the decimal point is now 1 less than it was in x=.9...)
            10x-x=9.9... - .9... (where the amount of 9s to the right of the decimal point in 9.9... is 1 less than it is in .9...)
               9x=8.9...1 (STILL where the amount of 9s to the right of the decimal point in the product of 9x is 1 less than it was in x=.9...)
                x=.9...

Voila.

It is argued that since the 9s in x=.9... and 9.9... are both infinite, that they are essentially the same value, however this is not true. When you multiply by 10 you are essentially shifting one of the 9s to the left of the decimal point. This is the case whenever you multiply .9 with any amount of 9s, no matter how many, finite or infinite, by 10.

WRONG WRONG WRONG!!!

This logic really annoys me, I've seen it several times now. You start with x = 0.999..., and then by whatever algebra you want, you end up with x = 0.999... Well you don't need to use any algebra to show that, x = x is axiomatically true. And secondly, just because x = x, that does not imply x ≠ y, it only implies x = x. Its a dirty trick used to fool the non-mathematically minded person, but is obviously invalid.

Using that logic, 5 = 5, therefore 5 ≠ 10/2

There are many proofs that 0.999... = 1, so if you don't believe it, you should try to refute all of them. And if can refute them, then you have made many errors, as they are correct proofs, and you should check for your mistakes.

0.999... * 10 = 9.999...

At first look seems there is no problem. But there is. Both nines after zero in 0.999... and nines after nine in 9.999... have an infinite count. But that count is not equal in both cases. It is hard to imagine how that causes problems, but it does. And I will try to make it visible.

0.999... = 9/10 + 9/100 + 9/1000 + ...

let's multiply both sides by ten:

9.999... = 9 + 9/10 + 9/100 + ...

Now let's subtract the first expression from the second:

9.999... - 0.999... = 9 + 9/10 - 9/10 + 9/100 - 9/100 + ...

9 = 9

Seems like it's ok?

Now let's try another series:

1 + 10 + 100 + 1000 + ... = x

We multiply both sides by ten:

10 + 100 + 1000 + 10000 + ... = 10*x

Subtract first from second:

-1 + 10 - 10 + 100 - 100 + ... = 10*x - x

-1 = 9*x

x = -1/9

Now we have an equation which most people will doubt. Actually I'm writing something well known. There are topics in wikipedia like 1 + 2 + 4 + 8 + ... = -1 and like 1 + 2 + 3 + 4 + ... = -1/12. They are all based on the same principal like 0.999... *10 -0.999... = 9. The only difference is whether the series are converging or diverging. But if we allow subtracting in one case, we have to allow it in all the other too. Or better not allow it at all. 1+10+100+1000+... will never equal -1/9, though -1/9 is the limit of sum(10^n) as n->-∞.

If we have a number like 0.999999 we will have:

0.999999*10=9.99999

9.99999-0.999999=8.999991

When we multiply by ten, the nines after the decimal get one less. It is the same when the nines are infinitely many - one of the nines after the decimal goes before it. The same with the diverging series. But there the mistake is obvious. We just subtract series with different lenghts (no matter the lenghts are infinite, what does matter is they are different).--77.85.46.84 (talk) 19:06, 26 November 2014 (UTC)

Subtracting 0.999... from 1.000...

Am I the first to try this? I've not seen this simple test anywhere.

(1.000... − 0.999...)

you can start at the far right end in the traditional way, because you know what the final digits are of both numbers.

and you get the answer:

0.000...1

Does this number equal zero? The only way 0.000...1 can equal zero, is if it "never gets there", but this is precisely the description that all the mathematicians disregard and ridicule, stating that, like the final 9 in 0.999... it does not need to "get there" but is simply there. So is the final 1 of 0.000...1 there or not?

~Aaron Wilde

Hey Aaron,

To answer your question, no, that final 1 is not there. This isn't a mathematically rigorous argument but think about it this way. If you have an infinite number of zeroes how can you have something AFTER them? Every time you'd go to write down your 1 you'd have to consider another zero.

This is like asking whether 0.00000... (recurring) is the same as zero. In the real numbers it is, though it might not be in the Hyperreal numbers. Dbfirs 17:03, 26 October 2012 (UTC)

Even if you don't agree with the notation of 0.000...1, there is no denying that 1 - 0.999... = something <Karlww _{(contribs|talk)} 04:34, 29 October 2014 (UTC)

That something is 0. Paychonaut (talk) 20:58, 4 December 2014 (UTC)

You don't have to write an infinite amount of zeroes, you only have to put the repetend symbol above the 0 to represent the fact that the 0s are infinitely repeating. The 1 is there, has been there the whole time, and will continue to be there the whole time. It never goes away. To answer the question posed by "aaron wilde", yes, the final 1 is there. But the word 'final' in this case might not be appropriate, that is, it might be more appropriate to say "the 1 in the far-right of the decimal". People seem to misconstrue the concept of a '1 at the end of the number' to mean that there is an end to the amount of 0s in between it and the decimal point.

If we try to subtract 0.999... from 1 we will never complete the process. 1-0.9=0.1; 0.1-0.09=0.01; 0.01-0.009=0.001 and so on, we'll carry the one forever. There is no need to try to define all this with numbers like 0.000...1. We cannot finish the subtraction and that means there is a difference between 1 and 0.999 and it is impossible to express that difference in decimal form. The same is with numbers like 0.333... The infinitely repeating numbers only tells us it is impossible to fully represent a given fraction in a decimal form. There always exist a difference which will remain unexpressed. Same is with the irrational numbers but there we don't have fractions but something else like sum of series for example. It is the same principle manifested. Infinitiness of the numbers only shows their inability to fully express a given idea.--95.43.65.53 (talk) 14:53, 18 October 2014 (UTC)

Why the Common Objections To A Number Like .000...01 Are Weak

Some, such as the user above, argue that 1 - .999... cannot equal .000...01 because such a number is impossible. Why would that be true? Infinite numbers are not like infinite spaces. To assume as much is a logical fallacy called "reification", where one treats an idea like a concrete, physical object.

We all agree that there are an infinite number of real numbers between 0 and 1. If you listed all of these numbers from least to greatest, you'd find an infinite count of real numbers that don't go below 0 or above 1. This is similar to an infinite series stuck between a decimal point and a final digit. You may think the difference is subtle when we discuss numbers rather than digits, so let's convert all of the real numbers between 0 and 1 to digits (while they are still in order from least to greatest) by simply taking the first digit of each number and constructing a new real number out of them. It would read like 01234567891, where every digit is infinitely repeating except for that final 1. Why is this number impossible? It's an idea, not bound by the constructs of how we view infinity working in the real world.

Another example of how numbers are different (as ideas rather than objects) is that we can have different sizes of infinity. There are an infinite number of real numbers between .47 and .48, but there are even more of them between .46 and .48. Again, if we make a number out of the first digits of all of these numbers, you'd have two sets of .444... where one has roughly twice as many 4's as the other.

For any finite length of .999..., we know that subtracting it from 1 will never give us anything but a decimal point, a series of zeroes, and a final 1. Why would we expect that to change just because it becomes an infinite length of 9's? We shouldn't. Through inductive reasoning, we know that this pattern has every reason to continue, and our fear of an infinite series with a different final digit shouldn't stop us from accepting the obvious answer.

On the other hand, if someone has a method of subtracting .999... from 1 so that it results in a 0, I'd love to see it. It always seems that those who argue that it isn't .000...01 never seem to demonstrate an alternative method that gives them the result they posit.

Supernova 20:12, 26 December 2012

If 1 - 0.999... = 0.000...01, then 0.999... + 0.000...01 = 1. But 0.999...91 = 0.999... + 0.000...01. If 0.000...01 is greater than 0, we can further conclude that 1 + 0.000...01 > 1. Using 1 = 0.999...91, we get 0.999...91 + 0.000...01 = 0.999...92 > 1.

A load of nonsense, supernova is obssessed by this, but is completely wrong. He says there are more digits between .46 and .48 than .47 and .48. Absolute rubbish, they both have the same cardinality, (c), which is the same as the cardinality of the whole real number line. Then he talks about 0.000...01, the number is completely meaningless. For example, say x = 0.000...01 and y = 0.000...001 (the 1 in x gets replaced by 01) Then they both have countably infinitely many 0's in, so that they are the same number. Please just disregard what this guy says, and observe that 0.9999... = 1

0.999... interpreted as number, is an ill-defined concept.

Unless one thinks of the limit of the series 9/10 + 9/100 + 9/1000 + ..., then to treat 0.999... as a number is completely erroneous. http://thenewcalculus.weebly.com/uploads/5/6/7/4/5674177/proof_that_0.999_not_equal_1.pdf

101.16.150.147 (talk) 11:54, 16 January 2013 (UTC)

Logic and Mathmatical Induction

So, this is a proof against the .999... = 1 proof. It has to do with the identity of infinity, and that infinity is not always equal to infinity (no matter what they tell you). It's sort of a bad "induction," because I couldn't think of how to represent it properly inductively. If someone knows how, please feel free to do so.

It goes like this: BASE CASE:

Say x = .99, then 10x = 9.9, thus, you have essentially shifted the decimal place one to the right. Now, note that 10x -1x = 9.9 - .99 = (9 - .09)

INDUCTIVE STEP (not really, but close enough):

Essentially, let’s test x = .999: 10x = 9.99, 10x – 1x = 9.99 - .999 = (9 - .009) = 8.991. So, we see that as the # of 9’s in x approaches infinity, 10x – 1x gets infinitely close to 9, but still has that pesky 1 at the end. That is the key to this whole problem. In essence, we find that 10x -1x = (9 - .000000000000000…9) when we have x = .999… This may take a while to wrap your head around, but given that true reasoning, we find that:

9x = 9 - .00000000000…9,

So x = 1 - .000000000000…1 = .9999999999999… , which is NOT = 1 and the laws of mathematics in the universe is preserved.

~Travis7

   I just wanted to say: In an isosceles right triangle when its base&height=1 it has a hypotenuse of sqrt 2.
   what is the value of sqrt 2?
   its decimal continues forever?
   Meaning that the hypotenuse never completes the triangle?
   yet it has to...
   So it is but it isn't?
   to say the hypotenuse doesn't have an exact number to me...is pretty much saying 0.999... = 1
   It does but it doesn't at the same time?... SilverTongue92 (talk) 10:11, 3 October 2013 (UTC)

You know the arrow never hitting the tree Paradox? Well we could use it here.

0.9 does not equal 1 0.99 gets you closer, but not at 1 0.999 still gets you closer 0.9999 we could do this forever.

Adding a nine one at a time gets you closer and closer to one, but you NEVER GET THERE. Just like the halfway paradox.

Why .999... does equal 1

.000...1 is simply not a "real number" (similar to i)

The very idea of 0.000...1 is flawed because you cannot have a number at the end of a repeating decimal. This is because whatever number is at the end would function as a regular integer, which is impossible on the right side of the decimal point.

For example. According to this method of thinking .999...8 (which is not real) would equal 9.999...80 when multiplied by 10. A decimal with a 0 on the end is no larger than a decimal without the 0 (ie. .4 is equal to .40) Therefore 9.999...80 - .999...8 = 9, making .999...8 = 1, too.

Proof:
x = .999...8
10x = 9.999...80
9x = 9
x = 1
.999...8 = 1

To say that that the 9.999...80 has one less 9 in it than .999...8 is to consider that "infinity - 1" is an actual thing. By doing so you would be required to accept that infinity is... well, "finite."

Nonsense, you started using numbers that don't exist, so your proof is nonsense.

_____________________________________________________________________________________________________________________________________________________________________

Not to sound too offensive but the fact that you think .9...8 * 10 would equal .9....80 just shows your math skills are not even up to par. It would be 9.(9repeating)8 with one less 9 in the infinite string.

Additionally, why would the number at the 'end' of .0...1 function as an integer when it's a fraction of 1? .00001 does not function as an integer, .0000000000000001, does not function as an integer, why would .0...1?

Also you have done the math wrong again.

x=.9...8 10x=9.9...8, with one less 9. this can be broken down into 9+ (.9...-(.0...9*10)+(.0...8*10) Then you will find that 10x-x is actually.. 9x=8.9...82, with one less 9 than the initial x. this can be broken down into 8+(.9...-(.0...9*10)+(.0...8*10)+(.0...2) then dividing by 9 yields exactly what we started with, x=.9...8

Having an infinite amount of digits does not alter certain patterns that said numbers undergo. TO assume that there is not one less 9 when you multiply by 10 is downright foolish and ignorant.

"To say that that the 9.999...80 has one less 9 in it than .999...8 is to consider that "infinity - 1" is an actual thing. By doing so you would be required to accept that infinity is... well, "finite.""

The reason why "infinity-1" is not actually a "thing" is because "infinity" does not actually represent a particular value. The whole concept of using ∞ to mean "infinity", as if it were a value, as opposed to using ∞ to mean "an infinite value" is mathematically flawed, and essentially makes no sense in the first place. An infinite value minus one is still an infinite value, yes. But does that mean that the infinite value we have after subtracting 1 is the exact same infinite value that we started with? Absolutely not. Similar to how a value greater than 5 but less than 10 might minus one might still be a value greater than 5 but less than 10, but not be the exact same value we started with. i.e., if we were using a triangle to represent a number 5<triangle<10,sure triangle-1 might still =triangle. But does that mean that we have the same value? say, if triangle was 7? Absolutely not. The word infinite is just a descriptive, broad term, that does not tell us anything about how many of the number we have. It just means that we have an infinite amount. It is broad and descriptive, not particular and exact, just like using the triangle to represent a number greater than 5 but less than 10.

In other words, infinity-1 does not necessitate thinking of infinity as "finite" at all. On the contraire, it is the belief that "infinity-1=infinity" that leads to thinking of infinity as a finite value which cannot be changed or altered. Also, infinity-1=infinity implies that we can create something out of thin air by taking it from an infinity, or that we can completely remove something from existence by adding it to infinity. None of these make any sense, and if it were true, would require actual physical evidence from the universe. And if infinity-1 is not an actual thing, then it is because infinity is an unknown, unquantifiable value.

Oranges

If .000...1 were a real number, it would be continually changing number. But, we live in a world of constants.

EXAMPLE Three people cut oranges into equal thirds.
One person is decent and cuts his orange into .333, .333, and .334 slices.
A more skilled person cuts his orange into .3333332, .3333332, and .3333336.
The perfect cut would be to cut it into fourths .333..., .333..., .333..., and .000...1?

The final method, while true in theory, would be impossible to perform however because any cut would be too large for .000...1 as the number is constantly smaller. This only makes, because it is equal to zero.

______________________________________________________________________________________________________________________________________________________________________

mate, cutting an orange into fourths would be .25, .25, .25 and .25. Not .3.. .3... .3.. and .0...1.

and constantly smaller does not equal 0. Either something can be infinitely small or it cannot. If it can, it still exists. And here's why. If it cannot be infinitely small, that means that there is a finite limit to how small something can be, which means that there is a finite limit to, say, the amount of zeroes that can come in between the decimal point and 1, meaning we can not have .0(infinite zeroes)1, but can have .0(finite zeroes)1. If this is true, then .9repeating wouldn't even exist. Since .9repeating can be viewed as the infinite geometric series of the form Σ9/10^n n=1 to infinity, and every term past .9 is of the form .0(certain amount of zeroes)1, it must follow that we can have an infinite amount of zeroes in between the decimal point and the 9. If we couldn't, then there would be a finite limit to the amount of zeroes that could come in between the decimal point and the 9. If this were the case, then we couldn't even have .9repeating, because eventually we couldn't add more terms to the series. But the series was supposed to be infinite.

So either .9...repeating and .0...1 are real, or neither one are real. You cannot have one without the other.

Face the Truth

Repeating decimals were created to artificially represent something that cannot be written down. There is no such thing as .999..., because it equals 1.

_____________________________________________________________________________________________________________________________________________________________________

If there's no such thing as .9..., it sure as hell doesn't equal 1. How could something that doesn't exist equal something that does exist? Your statement is a complete logical fallacy and makes zero sense. "There is no such thing as .9... because it equals 1" is literally tantamount to saying "1=.9.., therefore there is no such thing as .9...". So 1 doesn't exist? Nice try but no cigar. We all know that 1 exists. And your claim about repeating decimals being created to "artificially" (whatever that means) represent something that cannot be written down is nonsense. That implies that .3... was invented in order to have a decimal form of 1/3. Why would that be done?? Considering that we could just write 1/3, it makes no sense to make up an infinitely recurring decimal. There is no advantage to writing something such as .3..., or .3 with a bar over the 3, over writing something such as 1/3, be it on a computer or on paper, whiteboard, blackboard, etc. _____________________________________________________________________________________________________________________________________________________________________ Allow me to counter with:

Imagine one comlete circle. In case you cannot do this, here is a circle: O
What does 75% a circle look like? It looks something like this: C
What does 99.999...% of a circle look like? It is impossible to draw because it would be continually growing until it reach the size of 1 (which it never would, but how does one draw a continually growing shape any way...)

Who says 99.999...% has to reach 100%? That is clearly begging the question. <Karlww _{(contribs|talk)} 07:21, 29 October 2014 (UTC)

The truly simple proof

There are a lot of "simple" proofs on this page which ultimately result in people getting bogged down in the definitions of infinity and so on. But there is a simple mental exercise which will allow any non-believer to understand, and it goes:

• What's 1 - 0.9? 0.01
• What's 1 - 0.99? 0.001
• What's 1 - 0.999? 0.0001

And so on. It does not matter how many times you append a 9, the 01 will always be there. Waiting for you. Hide yo wife. <Karlww _{(contribs|talk)} 07:06, 29 October 2014 (UTC)

Using your logic here, 1 - 0.99... would be 0.00... because the "01" you're looking for at the end will never be reached. JaeDyWolf ~ Baka-San (talk) 14:32, 11 November 2014 (UTC)

Yes of course you will never reach the 01, which is why it's 0.00..., but that little piece of leftover will always be there, which is why it's not 0. <Karlww _{(contribs|talk)} 18:52, 24 November 2014 (UTC)

No, there is no "leftover" piece. 0.000.... goes on forever. There is no one at the end, there will never be a one at the end by the time the universe crunches or expands into its death, there will never be a one at the end even if you wrote zeroes every second for an infinite amount of time. There is no end to a repeating decimal. 98.255.135.163 (talk) 21:21, 25 November 2014 (UTC)

If 1 - 0.999... = 0, then 0.999... + 0 = 1. The fact is we can not express the difference using decimal notation. But the difference may still exist.--77.85.46.84 (talk) 17:30, 26 November 2014 (UTC)

The semantic proof

The definition of infinity is something that has no end. The definition of a repeating decimal is one that repeats infinitely. Therefore it must not end. 1 has an end in other words it is not infinite. Infinite cannot be the same as non-infinite. Take the example of the universe. If the universe were to slow down in its growth this would mean it was growing closer to something that it cannot react. In this case it would still be growing always growing, yet never reaching. If.9 repeating were to equal 1 .9 repeating would not be infinite and therefore not repeating. So therefore .9 repeating != 1. Koibeast (talk) 15:12, 5 November 2014 (UTC)

You're confusing what an "infinite number" is. 0.99... is not an infinite number; it's a recurring decimal. An infinite number is a number which is infinitely large. Recurring decimals are only related to infinity because of the definition of what a recurring decimal is. 0.33... is finite because it isn't infinitely large and the same goes for 0.99... JaeDyWolf ~ Baka-San (talk) 14:59, 11 November 2014 (UTC)

"If the universe were to slow down in its growth this would mean it was growing closer to something that it cannot react. In this case it would still be growing always growing, yet never reaching." That's like saying that if someone pushes the brake pedal while driving a car, the car will never stop moving forward. Here's a hint - it does. 98.255.135.163 (talk) 21:36, 25 November 2014 (UTC)

Infinitesimals

0.999... > 1 by 1/. It is less than one by an infinitesimal. Commonly, it is the same thing. For every value you claim to be equal to 0.999 and is not one, I can find one closer to one by an infinite amount. It is not equal, but only by an infinitesimal, which you may choose to interpret in any way. JamesJNHu (talk) 01:04, 3 December 2014 (UTC)