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Topic: 0.999... (Read 9750 times) 

Michael Dagg
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Re: 0.999...
« Reply #100 on: Dec 15^{th}, 2006, 5:18pm » 
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on Dec 15^{th}, 2006, 8:35am, Aman wrote:Last two posts seem to contradict each other. 
 No they don't (why?).


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Marvin
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Re: 0.999...
« Reply #101 on: Dec 16^{th}, 2006, 7:03am » 
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Yup, I was wrong, they don't contradict Michael_Dagg wrote the representation of the number and Eigenray showed that such a number does not exist. Both correct. It was only confusing for Aman, I think.


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rmsgrey
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Re: 0.999...
« Reply #102 on: Dec 16^{th}, 2006, 12:50pm » 
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Of course Sup{x: x<1} does exist and is 0.999... (the supremum of a subset being its least upper bound within the parent set)

« Last Edit: Dec 16^{th}, 2006, 12:51pm by rmsgrey » 
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Michael Dagg
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Re: 0.999...
« Reply #103 on: Dec 16^{th}, 2006, 3:47pm » 
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sup{x: x < 1) exists and is 1 = sup{x: x< 1}. You need max sotospeak in this context. This idea is about context if you ask me. You can say that there is no largest number less than 1 by using a set of no elements or some set upon which you assert some kind of metric that does not exist. Makes more sense to say max{x: x < 1} than say something like {x: x = 1/0} or something like that. But, my choice one of many. Anything wrong with this sqrt(2) = sup{x in Q: x^2 < 2}. The sqrt(2) is not even is in that set.

« Last Edit: Dec 16^{th}, 2006, 4:32pm by Michael Dagg » 
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Marvin
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Re: 0.999...
« Reply #104 on: Dec 17^{th}, 2006, 5:45am » 
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This gave me an idea of a proof that 0.999... = 1 Let S={x: x<1}. We will show that 0.999...=sup(S). We need to show that 0.999...>=x for all x from S Let epsilon = 10^n, for an arbitrary natural n x= 1  epsilon = 0.99..(ntimes)..9 < 0.999... Now we want to show that for any epsilon>0, there is x in S such that x > 0.999...  epsilon. We will use the same epsilon... 0.999...  epsilon = 0.99..(n1times)..9899... < 0.99..(ntimes)..9 which lies in S (it is < 1) It is obvious that sup(S)=1 : 1 is greater than any x from S trivially (it results from the definition of S) 1  epsilon = 0.99..(ntimes)..9 < 0.99..(n+1times)..9 which lies in S ( it is < 1) From the uniqueness of a supremum: 0.999... = 1 I hope I was formally correct...


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CowsRUs
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Re: 0.999...
« Reply #105 on: Feb 5^{th}, 2007, 7:10pm » 
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well i may be contradicting Icarus, and i may be repeating sum1 else i dont know at all... Icarus [smiley...] sort of lost me there... but i think he said this was true or not true 1.000...01 because if it was true then this whole argument is a waste: .99999.... *10= .999999.....90 .999999... would equal 8.99999999.....91 u get my point? as for the fractions 1/3 is not equal to .333... meaning. .333..... is short by a tiny bit to 1/3 unless u mean .333... as really 1/3 then 3(.333...) would equal 1 i may be repeating as i said but who really reads through 5 pages of stuff anywayz?


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CowsRUs
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Re: 0.999...
« Reply #106 on: Feb 5^{th}, 2007, 7:41pm » 
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Omg, the subject field wasnt filled out so i have to retype everything... im going to try to summarize it if there are numbers after infinitives then there can be infinitives after infinitives(why not) if u add .999... to 1 and divide that by two. then u would get .999...945, and then u would do that infinitive times, therefore making it .999...999... infinitively*infinitively FOREVER. .999...999...999... for ... times. thats why infinitives are difficult to work with... im probably wrong so explain to me why there cannot be numbers after an infinitive because however infinitively small it is, it will be still the difference between a difference of infinitively smallness. That is if we could be infinitively small then if we stand by an infinitively smallmarked ruler than we would see the difference between .9999999...91 and .999...92 Prove me wrong... that also brings me to the meaning of forever, and now that word confuses me since it involves an infinitive...


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Whiskey Tango Foxtrot
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Re: 0.999...
« Reply #107 on: Feb 5^{th}, 2007, 8:03pm » 
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Unknown. Firstly, we are not discussing grammar, so infinitives are definitely out. Please don't bring them up again. Secondly, you demanded that we prove you wrong. Fortunately, several have already done so for us. If you had read the above posts, you would see that you have made something of a classic mistake in your reasoning. To summarize it, 0.999...995, were it to actually make any sense, would still be less than 0.999... I am infinitively upset by your post

« Last Edit: Feb 5^{th}, 2007, 8:26pm by Whiskey Tango Foxtrot » 
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BNC
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Re: 0.999...
« Reply #108 on: Feb 6^{th}, 2007, 12:23am » 
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on Feb 5^{th}, 2007, 7:10pm, UNKNOWN wrote:<snip> who really reads through 5 pages of stuff anywayz? 
 People who want to understand? People who want to avoid repeating common errors? People who respect other people's time and effort in posting those 5 pages? Short answer: number like 0.999....91 are meaningless. You never "reach" the last spot to write the "1". PS: Contradicting Icarus is neer wise. And <ahmm> infinitively more so at mathrelated issues.


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Three Hands
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Re: 0.999...
« Reply #109 on: Feb 6^{th}, 2007, 1:37am » 
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on Feb 5^{th}, 2007, 7:10pm, UNKNOWN wrote: i may be repeating as i said but who really reads through 5 pages of stuff anywayz? 
 As one who made a similar argument towards the bottom of page one (I was young and foolish then...), I suggest you look at the top of page 2 for a suitable response... (Granted, this is repeating much of what others have already said, but much of this topic has ended up repeating what others have said...)


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Icarus
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Re: 0.999...
« Reply #110 on: Feb 6^{th}, 2007, 6:27pm » 
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Unknown  I am sorry for the somewhat hostile reception, but it really is hard to keep your temper when you have to read through a badly written post several times trying to make any sense out of it. While spelling and grammar mistakes happen, it really is a good idea to make an attempt at clear communication. As for the issue. It is possible to define notations such as 0.999...95 etc (where here the ... are meant to represent an infinite number of digits). However, simply having such notations does NOT mean they are actual real numbers. The real numbers are NOT decimal notations. They have an existence entirely apart from how we write them down. They are completely and uniquely determined by certain properties they possess. In particular, the real numbers are the minimal topologically complete ordered field (a phrase I defined in that first post  though it is now unreadable unless you use the tricks described in towr's post here). Decimal notation is not how we define Real numbers. It is simply a scheme for naming them. In ordinary decimal notation, there are NO decimal digits in infinite positions. Every digit has only a finite number of digits to its left (and an infinite number to its right  terminating decimals are those for which the infinite number of digits are all 0) . There is no need for digits in infinite positions, as the ordinary decimal notation yields enough names so that every real number has one. In fact  as this riddle shows  some real numbers actually have two names! If you decide to introduce decimal notations with with an infinite number of digits to the left of some digits, you are quite free to do so. But there is a caveat: You have to figure out what numbers they are supposed to represent. You need to define which number is to be represented by our new notation. This requires assigning even more names to our already named real numbers, or else introducing a new set of numbers to be represented by your new notations. In either case, your result does not in any gainsay 0.999... = 1, since these are already defined by ordinary decimal notation to be ordinary real numbers, and in that system, the two notations in fact represent the same number. In order to understand the arguments you have referenced, you have to understand decimal notation. Of course, once you actually understand what infinite decimals mean, this whole problem becomes a nobrainer. Let's look at 0.999.... The definition of this expression is: 0.999... = 9/10 + 9/10^{2} + 9/10^{3} + ... Now the question comes up: How do you sum up an infinite number of terms? The answer is, you can't. We have to find a means of defining the value that doesn't require that we actually add an infinite number of times. The means of doing this has been known since the time of the Greeks. We can't add up an infinite number of terms, but we can show that there is exactly one number x with the following properties: (1) No matter how many terms of the sum we add up, the total will always be x. (2) For any real number y < x, we can always add up enough of the terms to get a total > y. We define the value of the infinite sum to the unique x for which this is true. That value is then also the real number named by the decimal expression. (How do we know the number is unique? That is built into the properties: if z is a second number with the same properties, then either z < x or x < z. But if z < x, then property (1) for z contradicts property (2) for x, and vice versa if x < z. Hence there can be only one number which satisfies both properties). Now let's try to compare 1 to 0.999.... Summing up any finite number of terms is going to give you a number 0.9999 (where I am using  to represent a finite number of additional 9s). The meaning of this finite decimal notation is clear to everyone, I hope. The difference between 0.9999 and 1 is 0.0001 = 1/10^{n} for some value of n. By choosing enough terms, I can make this expression as small as I like. In particular, if y < 1, then 1y > 0, and I can choose n large enough so that 1/10^{n} < 1y. Turning this around gives y < 1  1/10^{n} = 0.9999. So x=1 satisfies property (2). Since 0.9999 < 1, regardless of how many 9 are there (remember,  means only a finite number), x=1 also satisfies property (1). So 1 is the unique number satisfying both property (1) and property (2) for the sum 9/10 + 9/10^{2} + 9/10^{3} + ... . By definition, it is the value of that sum. By definition as well, the value of that sum is the value of 0.999... That is, 0.999... = 1. You are mistaken as well in your claim that 0.333... is a little less than 1/3. By the definition, it is exactly 1/3.


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CowsRUs
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Re: 0.999...
« Reply #111 on: Feb 7^{th}, 2007, 5:25pm » 
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you know whats weird? why this thread has 110 replies and only 189 views when smaller threads have less


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Icarus
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Re: 0.999...
« Reply #112 on: Feb 7^{th}, 2007, 6:25pm » 
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I would guess that the count was corrupted when the server crashed a couple months ago. They eventually recovered most of the data, but every so often we come across something that got lost. Worst perhaps was that Alien's membership disappeared and he had to start again as Iceman. At least William was able to restore his post count, so he didn't have to wait to become an uberpuzzler all over again.


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CowsRUs
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Re: 0.999...
« Reply #113 on: Feb 7^{th}, 2007, 7:25pm » 
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Oh, I never knew that! Dang Nabbit!!! am i saying that correctly?


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Icarus
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Re: 0.999...
« Reply #114 on: Feb 8^{th}, 2007, 4:31pm » 
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Gol Dang it!, its "Dag Nabbit!", ya crazy young whippersnapper!


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CowsRUs
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Re: 0.999...
« Reply #115 on: Feb 8^{th}, 2007, 4:45pm » 
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Ima put you old Wheezers into your PLACE!!!


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WakeTFU
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Re: 0.999...
« Reply #116 on: Feb 10^{th}, 2007, 5:44pm » 
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For them, to say that 0.999... = 1 is patently ridiculous, since they quite evidently look nothing alike. But in fact, decimal expressions are not the numbers themselves, but are names for those numbers. They are labels which we have attached to underlying ideas to allow us to easily discuss them. 0.999... and 1 are two different names we use for the same concept Yeah..you contradict yourtself, in case no ones had pointed that out yet...If they are different names for the same thing then it's not rediculous to write .999...=1, because it does. If they aren't equal I double dog dare you to subtract something from one to get .999... (if they're different numbers you will ALWAYS beable to subtract from one to get the other, and you can't) Your best guess would be .000...1 which, if you understand the way infinity works, that number doesn't exist. Not in any usable form at least. THat number is entirely useless when doing math (except in this case, LOL) because it's just a zero. .999...=1


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Icarus
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Re: 0.999...
« Reply #117 on: Feb 11^{th}, 2007, 12:19pm » 
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Would you please be so kind as to point out what I have ever said that contradicts that? And if I somehow have contradicted myself, why are you arguing the same thing yourself?


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WakeTFU
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Re: 0.999...
« Reply #118 on: Feb 15^{th}, 2007, 9:56pm » 
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" For them, to say that 0.999... = 1 is patently ridiculous" Icarus No it isn't, is all I'm saying (ridiculous that is, it's not ridiculous) heh, moving on... You guys are all really smart...I hope your jobs make use of y'all...


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Grimbal
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Re: 1
« Reply #119 on: Feb 16^{th}, 2007, 5:08am » 
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Hey! why isn't this thread simply called "1"?


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rmsgrey
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Re: 1
« Reply #120 on: Feb 16^{th}, 2007, 10:59am » 
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on Feb 16^{th}, 2007, 5:08am, Grimbal wrote:Hey! why isn't this thread simply called "1"? 
 Notational convention


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Icarus
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Re: 0.999...
« Reply #121 on: Feb 16^{th}, 2007, 4:41pm » 
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on Feb 15^{th}, 2007, 9:56pm, WakeTFU wrote:" For them, to say that 0.999... = 1 is patently ridiculous" Icarus No it isn't, is all I'm saying 
 You know, if you take something completely out of its context, then you really cannot expect it to make sense. Why don't we just look at this quote where it belongs: on May 1^{st}, 2004, 7:19pm, Icarus wrote:One of the problems that people have with the question of the meaning of 0.999... is a misunderstanding of what real numbers themselves are. In particular, many mistake the decimal expressions we write out as being the numbers themselves. For them, to say that 0.999... = 1 is patently ridiculous, since they quite evidently look nothing alike. But in fact, decimal expressions are not the numbers themselves, but are names for those numbers. They are labels which we have attached to underlying ideas to allow us to easily discuss them. 0.999... and 1 are two different names we use for the same concept. 
 If you will read that carefully this time, you will notice that I am describing a misconception that is the main cause of people having trouble with this. I never said myself that 0.999... = 1 is ridiculous. Instead I am pointing out that people who have this misconception believe that it is. Then I immediately point out what that misconception is! There is no contradiction in this.


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Icarus
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Re: 0.999...
« Reply #122 on: Feb 16^{th}, 2007, 8:16pm » 
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Thought I'd play around with SMQ's new math symbols tool, and give a short intro to surreal numbers (a concept I'm fairly new to myself). Most of what is here is standard to the subject. A few aspects are my own interpretation (that is, I didn't get it from any other source  someone else may have done the same thing before me though). In particular, my sources defined surreal numbers with looser rules on the sets used, and as a result end up with many equivalent representations for the same number. The numbers themselves were defined as equivalence classes on the representations. But I saw that there was a "privileged" member in each class, and it made more sense to me to define the surreal number to be this member. Surreal Numbers Surreal numbers are amongst those "collections" that cannot be held within an ordinary set without introducing contradictions (cardinal numbers and ordinal numbers likewise suffer this problem). So instead of defining a set that defines all of them, we must define them in smaller increments. The standard method is to define them inductively using an ever increasing collection of sets. If we are going to use induction, then we need a well ordered collection to induct on. For this, we choose the ordinals (which is why the surreals require RussellWhitehead collections rather than ZermeloFrankel sets). This may seem like cheating, since the ordinals already have 0,1,2,... defined, and we are going to want to get these same numbers out of our Surreal construction. However, for our purposes the only property of the ordinals needed is the order. So we don't actually think of ordinal 0,1,2,... etc as numbers. To build the Surreal numbers, we need the Cut operator: { ...  ... }. If S is a totally ordered set, a cut of S is a pair of subsets A,B of S such that: (i) A B = S (ii) A B = (iii) for all a A and b B, a < b. If A, B are a cut of S, define {AB} to be the pair (A, B). A common notational convention used in Surreal numbers is to replace the sets A and B with lists of their members: {a_{1}, a_{2}, ...  b_{1}, b_{2}, ...}. So, for instance {0,1  2,3} would represent the pair ({0,1},{2,3}). (*) We start with the empty set as our unindexed base. (0) Next we define 0 = {} and define the set S_{0} = {0}. (1) S_{0} has exactly 2 cuts: Define 1 = {0}, 1 = {0}. Define S_{1} = {1, 0, 1}. (2) S_{1} has 4 cuts: 2 = {1,0,1}, 1/2 = {10,1}, 1/2 = {1,01}, 2 = {1,0,1}. Define S_{2} = {2, 1, 1/2, 0, 1/2, 1, 2}. ... Suppose that S_{} has been defined for all < for some ordinal , such that if < < , then S_{} S_{}. Further suppose that all the S_{} are ordered with compatible orders (that is if x, y S_{} S_{}, and x < y in S_{}, then x < y in S_{} as well). Define O = _{} S_{}. O is ordered by the common ordering inherited from the S_{}. This ordering determines cuts in O. Define N = { {AB} : {AB} is a cut of O}, and define S_{} = O N. We must still define the ordering on S_{}. If x,y are in O, then we inherit the ordering of O. If x O and y N, Let y = {CD}. Since y is a cut of O, x C or x D. If x C, define x < y. If x D, we define y < x. Lastly, if x, y N, then x = {AB} and y = {CD}. By the nature of cuts, either A C or C A. In the former case, define x y, and in the latter case, y x. This completes the induction.


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Icarus
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Re: 0.999...
« Reply #123 on: Feb 16^{th}, 2007, 9:00pm » 
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Continuing... Some comments: (1) In addition to the normal ordering of numbers that develops with this, we also get another relationship: "age". Some numbers are introduced earlier in the process than others. The "birthday" of any particular surreal number x is the lowest ordinal such that x S_{}. If the birthday of x is before the birthday of y, then x is said to be "older than" y. If they have the same birthday, then x is "the same age as" y. (2) The only numbers with finite birthdays are the integers and fractions of the form n/2^{m}. The entire rest of the real numbers have (the first infinite ordinal) as their birthday. (3) The age operator allows us to expand the definition of the cut operator: Define "A < B" to mean that for all a A and b B, a < b. Then for any pair of sets A, B of surreal numbers, define {AB} to be the oldest surreal number x such that for all a A, b B, a < x < b (by the definition of surreal numbers, such an x must exist and is unique). (4) S_{} is not , however. If we define S_{f} = _{n=0}^{} S_{n}, then some elements of S_{} not in are:  = {S_{f}}, = {S_{f}}. Also, consider the somewhat troublesome numbers {{x 0}  {x > 0} } and { {x < 1}  {x 1}}. That is, in S_{}, there IS a smallest number greater than 0, and a largest number less than 1! (However, the surreals do not end with S_{}  the next iteration drops more numbers in between these.) No collection deserves the title "numbers", however, unless you can add and multiply them. For surreals, these are defined inductively: On our base set , both operators are defined "vacuously". If +, * have been defined for all surreals with birthday < , let x, y S_{}. Then x = {AB}, y={CD}. It is easy to see that A + C < B + D. Hence by (3) above we can define x + y = {A+CB+D}. Unfortunately, the corresponding expression for multiplication is actually quite nasty (because A*C is not less than B*D, finding the right sets is much more complicated). [edit]My "definition" for addition was written from a poor memory. It breaks down in certain instances. The official version is x + y = {x+C y+A  x+D y+B}.[/edit]

« Last Edit: Feb 25^{th}, 2007, 7:57pm by Icarus » 
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SMQ
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Re: 0.999...
« Reply #124 on: Feb 17^{th}, 2007, 5:16am » 
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What I think is "fun" is that under those definitions, expressions like  1, (technically  1) and are meaningful and useful numbers! Now here's a question I haven't found an answer to: the Wikipedia article states (witout proof, although I believe it to be correct) that where the reals are the "smallest topologically complete ordered field", the surreals are the largest ordered field (if we allow fields to be defined over proper classes/collections as well as sets); does that make the complex surreals (A + B where A and B are surreal numbers and is defined to be 1 as usual) the largest algebraicly closed field? I.e. can all algebraically closed fields be embedded in the complex surreals? And does 0.999... still equal 1? SMQ


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