Quote (khoi @ May 9 2010 04:54am)
Here's the proof, close this thread.
Why 0.999... is not equal to 1.
We know that 1 is a rational number, that is, it can be expressed as a/b. Now if 0.999... is equal to 1, it too must be a rational number. This means that there exists c/d such that c/d = 0.999... However, no such c and d exist. So 0.999... is not a rational number. Therefore 0.999... cannot be equal to 1.
Try c=1, d=1. You can't assume it doesn't equal 1 as part of your proof that it doesn't equal 1 - that's circular reasoning. --Tango (talk) 15:37, 30 March 2008 (UTC)
Um, sorry, no. You can't use c=1 and d=1 because then you are assuming that 0.999... is equal to 1. We know that 1/1 = 1. What we have to show is that 1/1 = 0.999... Where do you see 'circular reasoning'?
I'm not assuming it, I can prove it. You're trying to show that they are not equal by assuming they are not equal - that is circular reasoning. Your statement that there is no such c and d is implicitly an assumption it doesn't equal 1. --Tango (talk) 15:03, 8 April 2008 (UTC)
In fact, every repeating decimal can be produced from dividing one integer by another. Thus, all repeating decimal values are, fundamentally, rational numbers. 0.9~ is no exception. If you want to find a ratio from a repeating decimal, you should check out Repeating decimal - it has several useful methods. Gustave the Steel (talk) 15:47, 30 March 2008 (UTC)
You think? So show me two integers whose quotient is 0.999... ?
I suggest 3/3. We know that 1/3 = .333...: thus 3/3 = 3 * 1/3 = 3 * .333... = .999...
Clearly .999...=9/10+9/100+9/1000+... Let S(n) be the partial sum 9/10+...9/(10^n). It is also clear that S(n)= (10^n-1)/(10^n). Therefore the limit as n goes to infinity of S(n) is equal to the limit as n goes to infinity of (10^n-1)/(10^n)=1/1. Since .999... is equal to the limit as n goes to infinity of S(n), we have directly shown that .999... is equal to 1/1.
Going by the information from 0.999... and Repeating decimal, and the sources cited within, I can say with the utmost confidence that, for any non-zero real number x, x/x equals 0.999... Gustave the Steel (talk) 03:17, 8 April 2008 (UTC)
You can say what you like - it is still nonsense. Just because you say it or one of your wiki articles says it, do you think that it is true? Again, show me two integers whose quotient is 0.999...? If you cannot, then I will cease to respond to your posts. Sorry. 98.199.111.222 (talk) 19:03, 14 April 2008 (UTC)
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)
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.
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, . 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 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)
[edit]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.
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 Rm 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)
[edit]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 goes to 1 as n goes to infinity.
A: That equals , 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 . 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)
I am shocked how many false proofs are found here. I was dozing off in all my real analysis lectures, I read this article and I got it. I am astounded. :-O. --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'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: . 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)
-wiki
Care to elaborate on your thoughts after reading all of this? Or are you going to try and get your locked/scammer tag removed?