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Title: 0.999... Post by Icarus on May 1st, 2004, 7:19pm The original 0.999. thread (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564) has grown so large that it is difficult for those without broadband to reply to it. For this reason, I have locked that thread and created this one to continue the discussion. In this post, I will start off by addressing, for reference, three issues related to this topic. The second and third posts are summaries of the original thread that I had created, and reprise here. What are Real numbers? 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. There are many ways of understanding the real numbers conceptually. Probably the most familiar is that the real numbers represent all possible directed distances between points ("directed" means we have both positive and negative distances). Mathematically, the Real numbers are defined to be "the smallest topologically complete ordered field". To understand this phrase, we must start at the right and work left: A Field is a triple ([smiley=cf.gif], +, [cdot]), where [smiley=cf.gif] is a set, and + and [cdot] are binary operations on [smiley=cf.gif] which satisfy the following axioms: Commutivity: For all [smiley=x.gif], [smiley=y.gif] [in] [smiley=cf.gif]; [smiley=x.gif] + [smiley=y.gif] = [smiley=y.gif] + [smiley=x.gif] and [smiley=x.gif][cdot][smiley=y.gif] = [smiley=y.gif][cdot][smiley=x.gif]. Associativity: For all [smiley=x.gif], [smiley=y.gif], [smiley=z.gif] [in] [smiley=cf.gif]; ([smiley=x.gif] + [smiley=y.gif]) + [smiley=z.gif] = [smiley=x.gif] + ([smiley=y.gif] + [smiley=z.gif]) and ([smiley=x.gif][cdot][smiley=y.gif])[cdot][smiley=z.gif] = [smiley=x.gif][cdot]([smiley=y.gif][cdot][smiley=z.gif]). Identity: There exists 0, 1 [in] [smiley=cf.gif] such that for all [smiley=x.gif] [in] [smiley=cf.gif], 0 + [smiley=x.gif] = [smiley=x.gif] and 1[cdot][smiley=x.gif] = [smiley=x.gif]. Inverse: For each [smiley=x.gif] [in] [smiley=cf.gif], there is a [smiley=y.gif] [in] [smiley=cf.gif] such that [smiley=x.gif] + [smiley=y.gif] = 0, and if [smiley=x.gif] [ne] 0, then there is a [smiley=z.gif] [in] [smiley=cf.gif] such that [smiley=x.gif][cdot][smiley=z.gif] = 1. Distributivity For all [smiley=x.gif], [smiley=y.gif], [smiley=z.gif] [in] [smiley=cf.gif]; ([smiley=x.gif] + [smiley=y.gif])[cdot][smiley=z.gif] = ([smiley=x.gif][cdot][smiley=z.gif]) + ([smiley=y.gif][cdot][smiley=z.gif]). Fields are usually denoted by their set, suppressing explicit mention of the two operations. An Ordered Field is a pair ([smiley=cf.gif], [le]), where [smiley=cf.gif] is a field, and [le] is a binary relation satisfying the following axioms: Reflexivity: For all [smiley=x.gif] [in] [smiley=cf.gif], [smiley=x.gif] [le] [smiley=x.gif]. Antisymmetry: For all [smiley=x.gif], [smiley=y.gif] [in] [smiley=cf.gif], if [smiley=x.gif] [le] [smiley=y.gif] and [smiley=y.gif] [le] [smiley=x.gif], then [smiley=x.gif] = [smiley=y.gif]. Transitivity: For all [smiley=x.gif], [smiley=y.gif], [smiley=z.gif] [in] [smiley=cf.gif], if [smiley=x.gif] [le] [smiley=y.gif] and [smiley=y.gif] [le] [smiley=z.gif], then [smiley=x.gif] [le] [smiley=z.gif]. Additivity: For all [smiley=x.gif], [smiley=y.gif], [smiley=z.gif] [in] [smiley=cf.gif], if [smiley=x.gif] [le] [smiley=y.gif], then [smiley=x.gif] + [smiley=z.gif] [le] [smiley=y.gif] + [smiley=z.gif]. Multiplicativity: For all [smiley=x.gif], [smiley=y.gif], [smiley=z.gif] [in] [smiley=cf.gif], if [smiley=x.gif] [le] [smiley=y.gif] and 0 [le] [smiley=z.gif] and 0 [ne] [smiley=z.gif], then [smiley=x.gif][cdot][smiley=z.gif] [le] [smiley=y.gif][cdot][smiley=z.gif]. Again, explicit mention of the ordering is suppressed. An ordered field [smiley=cf.gif] is Topologically Complete if it satisfies the supremum property. Before giving it, a few definitions are needed: A set [smiley=ca.gif] [subseteq] [smiley=cf.gif] is bounded above if there is a [smiley=b.gif] [in] [smiley=cf.gif] such that for all [smiley=x.gif] [in] [smiley=ca.gif], [smiley=x.gif] [le] [smiley=b.gif]. In this case, [smiley=b.gif] is called an upper bound for [smiley=ca.gif]. A supremum, or least upper bound, of [smiley=ca.gif] is an upper bound [smiley=s.gif] of [smiley=ca.gif] such that for all upper bounds [smiley=b.gif] of [smiley=ca.gif], [smiley=s.gif] [le] [smiley=b.gif]. [smiley=cf.gif] satisfies the supremum property if every subset of [smiley=cf.gif] which is bounded above has a supremum. (This is where the real numbers differ from the rationals. Every property mentioned before this one is also satified by the set of all rational numbers. But the rational numbers are not topologically complete. For example, the set { p/q | p, q [in] [bbz], p2 < 2q2 } is bounded, but has no rational supremum.) Finally, we come to the word "smallest". It can be shown that the intersection of topologically complete ordered fields is also a topologically complete ordered field. (Sort of - I'm glossing over a few things with that statement.) The intersection of all topologically complete ordered fields is the smallest such field, which we call the Real numbers. Some common set notations: Real numbers: [bbr] Natural numbers: [bbn] = {1, 1+1, 1+1+1, ...} (the set of all real numbers obtainable by starting with 1 and repeatedly adding 1). Whole numbers: [smiley=bbw.gif] = [bbn] [cup] {0} Integers: [bbz] = { x | [pm]x [in] [bbn] or x = 0 } Rational numbers: [bbq] = { p/q | p, q [in] [bbz] and q [ne] 0 } Complex numbers: [bbc] = { a + bi | a, b [in] [bbr] and i2 = -1 } The definition of a limit. Many confusions about 0.999... centered on a failure to understand the idea of a limit. More is said about this in the Misconceptions post below. For reference, I give Cauchy's definition of the limit of a sequence. A sequence is a function [smiley=a.gif] from the natural numbers [bbn] into [bbr]. The value of [smiley=a.gif] for a number [smiley=n.gif] is usually denoted [smiley=a.gif][subn] and the sequence by {[smiley=a.gif][subn]}. The limit of a sequence {[smiley=a.gif][subn]} is a number [smiley=cl.gif] such that for any [epsilon] > 0, there is a natural number [smiley=cn.gif] such that if [smiley=n.gif] > [smiley=cn.gif], then The limit [smiley=cl.gif] is usually denoted by "lim [smiley=a.gif][subn]" or "lim[subn] [smiley=a.gif][subn]" or "lim[subn][to][subinfty] [smiley=a.gif][subn]". Decimal Notation Decimal notation is a means of identifying particular real numbers. In brief, it is defined as follows: Define 2=1+1; 3=1+2; 4=1+3; 5=1+4; 6=1+5; 7=1+6; 8=1+7; 9=1+8. A decimal expression is a function [smiley=d.gif] from [bbz] into the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} with the property that for some [smiley=cn.gif] [in] [bbz], for every [smiley=n.gif] [ge] [smiley=cn.gif], [smiley=d.gif][subn] = 0. The decimal expression is denoted by adjoining the digits: Where [smiley=cn.gif] is the larger of 0 or the largest index of [smiley=d.gif] for which [smiley=d.gif] > 0. Each decimal expression is assigned a real number as its value: For each [smiley=i.gif] [in] [bbn], let Then the value assigned to the decimal expression is lim[subi] [smiley=cd.gif][subi]. Note that by definition, every decimal expression represents a single number, but that there is nothing in this definition to say that two decimal expressions cannot have the same value. In the Misconceptions post below is a proof that every real number is the value of some decimal expression. |
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Title: Re: 0.999... Post by Icarus on May 1st, 2004, 7:19pm Arguments depending on the reader knowing how to add/multiply decimals 1) Times 10 (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564#0) Let [smiley=x.gif] = 0.999... 10[smiley=x.gif] = 9.999... - [smiley=x.gif] = -0.999... 9[smiley=x.gif] = 9.000... = 9 [smiley=x.gif] = 9/9 = 1. 2) Thirds (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564#2) 1/3 = 0.333... 1/3 = 0.333... + 1/3 = 0.333... 3/3 = 0.999... 1 = 0.999... 3) Take the Difference (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564#3) 1.000... -0.999... 0.000... = 0 Since their difference is 0, they are the same. 4) 9*0.111... (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=225#231) 1/9 = 0.111... 1 = 9/9 = 9*(1/9) = 9*(0.111...) = 0.999... (The linked text has the flaw of depending on "calculator expressions", however this can be cleared up into a true proof.) Arguments based on calculus 5) 0.999... = 9/10 + 9/100 + 9/1000 + ... (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=25#28) By definition 0.999... = [sum][subn] 9*10[supminus][supn] (1 [le] [smiley=n.gif] [smiley=lt.gif] [infty] ) That is, 0.999... = Lim[smiley=subcn.gif][to][subinfty] [sum][subn]=1[smiley=supcn.gif] 9*10[supminus][supn] = Lim[smiley=subcn.gif][to][subinfty] 9(( 1-10[supminus][smiley=supcn.gif][supminus][sup1])/(1-10[supminus][sup1]) )-1) = Lim[smiley=subcn.gif][to][subinfty] 10 - 10[supminus][smiley=supcn.gif] - 9 = 10 - 0 - 9 = 1 While some posters discussed this before the linked post, this post was the first to actually lay out the argument clearly (except when the poster got confused and said that the definition he had just given could then be proved using Analysis. Since it was a definition, it is not subject to proof.) 6) Epsilon-delta (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=50#67) 0.999... = lim[subn][to][subinfty] 0.999...9 ([smiley=n.gif] 9s) = lim[subn][to][subinfty] 1 - 10[supminus][supn] The definition of this limit is: The limit is the number [smiley=cl.gif] such that for every [epsilon]>0, there is an [smiley=cn.gif] such that for all [smiley=n.gif] > [smiley=cn.gif], (That [smiley=cl.gif] is unique is easily proven from the definition.) So let [epsilon] be an arbitrary number > 0. Choose [smiley=cn.gif] [ge] -log10 [epsilon]. Then for [smiley=n.gif] > [smiley=cn.gif], | (1-10[supminus][supn]) - 1 | = 10[supminus][supn] < 10[supminus][smiley=supcn.gif] < [epsilon]. Hence 0.999... = 1. Pietro K.C. has a similar proof earlier to the one linked. But it was so wrapped up in refutations of other arguments that I choose to link to James Fingas' instead. 7) Method Of Exhaustion (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=75#83) Since every finite stretch of 9s, 0.999...9, is < 1 and these get arbitrarily close to 0.999..., we must have 0.999... [le] 1. But let x < 1, then 1-x > 0 and by the Archimidean principle, there must be an n such that 10-n < 1-x. Therefore x < 1 - 10-n = 0.999...9 (n 9s) < 0.999... . So 0.999... > x for all x < 1. The only possibility left is: 0.999... = 1. This isn't a calculus-based proof. It is actually a "what calculus is, but from before calculus was invented" proof. 8 ) 1=0.999...910 (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=225#245) For all [smiley=n.gif], 1 = 0.99...([smiley=n.gif] 9s)...91, where the notation means that after all the 9s we have a "10" for the last digit (ie, we are extending base 10 notation to include a "10" digit). Letting [smiley=n.gif] [to] [infty] leaves 1= 0.999... This works, but unfortunately it requires even more of an understanding of infiniteness than the other methods. Other proofs 9) Nothing Between (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564#6) Suppose 0.999... [ne] 1. Let [smiley=x.gif] = (1+0.999...)/2. What number is it? By our understanding of decimals it must have a decimal expansion with digits >9. Since there are no such digits in base 10, there cannot be such an [smiley=x.gif]. And so 0.999... = 1. My thoughts: Kozo's replies to this argument were very salient, even though they were based on non-existant numbers. This argument depends heavily on a deep understanding of the nature of decimal notation that those who expounded it never explained. In order to truly demonstrate that 1 = 0.999... using this idea, you must also prove that every real number has a decimal expansion, and prove that 0.999... < [smiley=x.gif] < 1 requires the digits of [smiley=x.gif] to be greater than those of 0.999... To do this properly is a royal mess. |
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Title: Re: 0.999... Post by Icarus on May 1st, 2004, 7:20pm 1) Infinite Decimals are Approximations (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=0#4) This argument says that, for example, 1/3 [ne] 0.333... because the right-hand side is only an approximation. Not so. The definition of the expression 0.333... is 0.333... = Lim[subn][smiley=subto.gif][subinfty] [sum][smiley=subk.gif][smiley=subeq.gif][sub1][supn] 3*10[supminus][smiley=supk.gif] = Lim[subn][smiley=subto.gif][subinfty] (0.3 + 0.03 + ... + 3*10[supminus][supn]) The value of the limit is exactly 1/3. This argument is closely tied to the next two: 2) Limits are Approximations (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=25#49) There are several posts with this misconception. The one linked has both it, and a variant of the "infinite process" misconception, which is described below. According to this view, limits are nothing more than a "short-hand" for describing approximation schemes. I believe this idea comes from the descriptions used by math teachers to first introduce the idea of limits. Unfortunately, the student never moved beyond these original incomplete conceptions. The basic definition of the limit of a sequence (the particular type of limit needed here) is: The limit of {[smiley=a.gif][subn]} as [smiley=n.gif] goes to infinity, written as "lim[subn][smiley=subto.gif][subinfty] [smiley=a.gif][subn]", is the real number [smiley=cl.gif] which satisfies the following: For every [epsilon] [smiley=gt.gif] 0, there is an [smiley=cn.gif] [in] [bbn] such that for all [smiley=n.gif] [smiley=gt.gif] [smiley=cn.gif], [smiley=vert.gif] [smiley=a.gif][subn] [smiley=minus.gif] [smiley=cl.gif] [smiley=vert.gif] [smiley=lt.gif] [epsilon]. Note that by the definition, the limit is not any of the [smiley=a.gif][subn] or all of them, or some "process". The limit is the number [smiley=cl.gif] which the sequence elements [smiley=a.gif][subn] approximate. 3) Decimals and Limits are Processes (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=0#22) The linked post is the first I could find that treats 0.999... as being a process of constantly adding more 9s. In the linked post for (2), limits themselves are also described as processes. From the definitions for decimals and limits provided, it is evident that decimal expressions such as 0.999... and that limits are defined to be particular numbers, not processes. 4) Infinite decimals and limits are the result of an infinite process (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=25#30) This misconception is very common, and is evident in many posts on both sides. The truth is rather difficult to explain. Mathematics, and particularly these two situations, contains a concept of infinite processes, but not the actuality of them. This is why we have such notational conventions as the ellipsis (...) and the overline for repeating decimals that is not available in this forum. Conceptually, the ellipsis means that the digits go on forever. In reality, they can't, and in any actual mathematics, they don't - they always end with an ellipsis or similar notational convention. Since mathematics has no means to actually carry out any infinite processes, we use other means to define and/or calculate the outcomes that such processes would have if they actually could be performed. The most basic idea we use for this is: if for all potential outcomes of an "infinite process" but one, you can show (by finite means) that the outcome cannot be the result of the process, then the one remaining potential outcome is the result of the process. This is the basis for the definition of limit (which is due to Cauchy, by the way) that I gave earlier. In the concept of limits, they involve an infinite process of improving approximation - with the idea that if only you could continue the approximation on indefinitely, the result would be the actual value rather than any approximation. The actuality of limits is that we show for any particular limit that there is exactly one number [smiley=cl.gif] that can be the result of infinite continuation of the approximating process. We define the limit to be that number. The same thing goes for the manipulation of decimals. Conceptually, 0.333... + 0.333... requires an infinite number of additions of the form 3+3 = 6. In actuality, it is evident that all the additions are 3+3=6, so the sum must be a number whose decimal expansion is 0.666.... Using the definitions of decimals and limits above (along with certain properties of the real numbers), we can also show that there is exactly one real number with this decimal expansion. All of this is done in a finite number of steps in the actual mathematics. It is only in concept that it involves an infinite number of steps. 5) Infinity is only a concept, not an actual number (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=25#37) This misconception - which is almost universal among those who understand the previous point - takes the previous point too far: the belief is that infinity has no place in actual mathematics. This is not true. Infinite processes have no place in mathematics other than as a concept. Infinities themselves are well within the domain of mathematical exploration. They are nothing more than a matter of definition. In this post (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=125#135), I gave a short overview of the three types of infinities most commonly encountered in mathematics. 6) Infinity is an actual number (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=25#46) (The linked post does not state infinity is a number like some other posts (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=75#93) do, but it the first to attempt treating infinity as a number without any understanding of what it requires.) Why am I including the opposite side of the argument above as another misconception? Because those who argued this side did not understand what they were saying either. The problem here is the word "number". Oddly enough, the word "number" has NO MATHEMATICAL MEANING. In mathematics, we have "Natural numbers", "Cardinal numbers", "Ordinal numbers", "Rational numbers", "Real numbers", "Algebraic numbers", "Complex Numbers", "Hamiltonian numbers (quaternions)", "Cayley numbers", "Hyperreal numbers", "p-adic numbers", "Surreal numbers", and a vast host of other types of numbers someone has defined here, there, or yonder. But we do not have a meaning for "number" in and of itself. Many who have argued in this thread that "infinity is a number" both had no workable concept of what they meant by "infinity", and generally tried to treat infinity as if it were a REAL number (ie. part of the set of Real numbers). There is no infinity in the real numbers. To get infinities, you must expand the real numbers (to the Extended Reals, the Hyperreals, the Surreals, the Long Line, or by other means). 7) There are numbers without decimal representations (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=0#7) I don't know if Kozo actually thought so when he made the post, or was only trying to point out the weakness of the previous "Pro" argument (and it does have serious weaknesses, though it is salvagable), but just for clarity's sake: Every real number has at least one decimal expression. Proof: There is for any real number [smiley=x.gif] an integer [smiley=cn.gif] such that 10[smiley=supcn.gif][supplus][sup1] > [smiley=x.gif] [ge] 10[smiley=supcn.gif]. For every integer [smiley=n.gif] [le] [smiley=cn.gif] we can define the two numbers [smiley=a.gif][subn] and [smiley=d.gif][subn] inductively by: [smiley=a.gif][smiley=subcn.gif][subplus][sub1] = [smiley=x.gif]. For all integers [smiley=n.gif] [le] [smiley=cn.gif], [smiley=d.gif][subn] = [lfloor][smiley=a.gif][subn][subplus][sub1]/10[supn][rfloor] ( [lfloor] [rfloor] is the floor function - the greatest integer [le] the contents of the brackets) [smiley=a.gif][subn] = [smiley=a.gif][subn][subplus][sub1] - [smiley=d.gif][subn]*10[supn] It is not hard to show that [smiley=x.gif] = [sum][subn][smiley=subeq.gif][subminus][subinfty][smiley=supcn.gif] [smiley=d.gif][subn]*10[supn], so [smiley=d.gif][smiley=subcn.gif] ... [smiley=d.gif][sub0].[smiley=d.gif]-[sub1] ... provides a decimal representation for [smiley=x.gif]. 8 ) There is a least number greater than or greatest number less than a given real number (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=0#10) If [smiley=x.gif] and [smiley=y.gif] are any real numbers, then ([smiley=x.gif] + [smiley=y.gif])/2 is one of infinitely many real numbers lying strictly between [smiley=x.gif] and [smiley=y.gif]. Thus there is no such thing as the least real number greater than, or greatest real number less than, [smiley=x.gif]. 9) Real numbers consist of decimal representations (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=0#7) While no-one has expressed this view explicitly, it is implicit in many of the arguments made (and not just by Kozo, or just by "Con" posters). The crux of these arguments is to introduce a new variation of decimal notation, and then to talk about the new decimal representation as if it were a well-defined real number - without bothering to define what this new variation actually means. The post linked introduces a new digit, "#", to base-10 decimal notation. Other posts introduce notations with "tranfinite" decimal places (i.e. decimal places with an infinite number of other decimals preceding them). In all of these posts, the poster never bothers to actually tie these notations to any real number. They are simply thrown out, and it is assumed that they somehow represent real numbers. Much of the argument that follows comes from other posters trying to put a definition to the new notation only to be told "no - that's not what it means" (but still without any attempt on the originator's part to provide a meaning). Real numbers have an existance entirely separate from any means of denoting them. The simplest concise definition of the Reals is "the smallest topologically-complete ordered field". (Of course all of these terms need their own definition before this one is meaningful.) Decimal notation is a defined means of denoting the members of this field. If you introduce new notations for real numbers, then YOU must provide the definitions of the new notations also. 10) There is no such thing as transfinite digits (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=0#24) This misconception is that such expressions as "1.000...01", where the ellipsis represents infinitely many zeros, are not only undefined, but undefinable. They are undefined, unless the person introducing them also bothers to define them. But they are not senseless. You can define them. For instance, you could define 1.00...01 as representing the real number pair (1, 0.01). Such expressions could represent members of the set [bbr] [times] [0,1), where the digits of the second term are considered to follow every digit of the first (terminating decimals are filled out with 0s). 11) 0.999... is a hyperreal number, not real (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=200#220) This sounds apealling, but only because only us severe math wonks have ever even heard of hyperreals. The idea is that hyperreal numbers extend the reals, including in "infinitely small" numbers. That's it! we say: 1 - 0.999... is one of these infinitely small numbers, rather than 0! Sorry, but it ain't so. By definition 0.999... = limit[subn] {0.999...([smiley=n.gif] 9s)}. Each element of the sequence is a real number. By the definition of limits, the limit of a sequence of real numbers is a real number. So 0.999... is real, not hyperreal. No matter what larger set of numbers we decide to work in, decimal notation is defined for real numbers, and it is in the real numbers alone that this matter is settled. 12) ANYONE WHO DISAGREES WITH ME IS AN IDIOT! (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=100#111) I think it's clear who the real idiot is with these posters! ::) |
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Title: Re: 0.999... Post by Typhoon on May 4th, 2004, 7:28pm I cant explain anything scientific or complicated like others that are replying to this post, but i will say something that my math teacher told me. If I am standing 1 foot away from a wall and i walk 9/10 of the space between me and the wall, and from there move another 9/10 closer to the wall again and again and again. Theoreticaly, you will never reach the wall. You might be up against the wall, but you will never get there by moving a fraction of the space left. |
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Title: Re: 0.999... Post by towr on May 5th, 2004, 1:05am Your teacher must not be an engineer ;) And you should always be sceptical at what teachers say. It is true that for any finite number of steps you don't reach the wall, but if you take the limit to infinity then it no longer is a finite number of steps. |
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Title: Re: 0.999... Post by Icarus on May 5th, 2004, 3:30pm Exactly. All of your positions at which you are stopping correspond to .9 or .99 or .999 or .999...9 for some finite number of nines. No one is claiming that any of these equal 1. What is equal to 1 is 0.999..., with an infinite number of 9s. This does not correspond to any of the steps in your approach to the wall, but is beyond all of them. Think about it: if you mark any spot short of the wall, your steps will eventually get past it. So, since 0.999... is beyond all of your steps, it also must be beyond all possible marks short of the wall. The only place it can be at is at the wall itself. 0.999... has to be greater than every number less than 1. Since it is also [le] 1, that leaves no value but 1 for it to equal. |
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Title: Re: 0.999... Post by Guest on Jul 14th, 2004, 4:13am "ANYONE WHO DISAGREES WITH ME IS AN IDIOT!" Damn right. Good to see you are educating the peasants... |
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Title: Re: 0.999... Post by EZ_Lonny on Aug 11th, 2004, 5:37am Dear Icarus, On your explanation about Ordered Fields (F,=<), I think you made a small mistake that could lead to complete nonsense in the rest of your statement. You say: For all x,y,z in F; if x=< y and 0<>z, then x.z =< y.z Did you think of the possibility that x is negative and z is negative and y is positive? So x.z is positive and y.z is negative. Excuse me if I'm wrong but then x.z > y.z. I havent checked on other flaws in your statement, bus such easy to mistakes may imply more to be made in apparently harder assumptions. |
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Title: Re: 0.999... Post by towr on Aug 11th, 2004, 9:22am on 08/11/04 at 05:37:56, EZ_Lonny wrote:
on 05/01/04 at 19:19:11, Icarus wrote:
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Title: Re: 0.999... Post by Icarus on Aug 11th, 2004, 6:50pm Nothing I gave in that post is original to me, nor is any of it an assumption - It is a definition. I.e. I am not making assumptions about what the real numbers are, I am telling you what the meaning of "real number" is. This is where they come from. The particular problem you find comes about, as towr has pointed out, because you have inadvertently dropped part of the defining statement. As it is, this is a well-known property of inequalities: when you multiply both sides of an inequality by the same positive number, the inequality is preserved. That is all this statement says. As an aside: why did I say (0 [le] z and 0 [ne] z), instead of simply saying (0 < z)? Because I had not defined the symbol "<" yet, and in a development such as this, you don't use anything until it has been defined. |
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Title: Re: 0.999... Post by EZ_Lonny on Aug 12th, 2004, 8:46am My mistake |
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Title: Re: 0.999... Post by StatusX on Aug 23rd, 2004, 10:22am To those who say you'll NEVER reach one (or never reach the wall): isn't saying "never" just another way of saying "it will take an infinitely long time"? Of course, you have to combine this with the fact that the terms in the series, (you) get closer and closer to 1 (the wall), otherwise the limit could be anything (anywhere), but it is combining these two concepts that lets you understand an infinite limit. |
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Title: Re: 0.999... Post by sam_i_am on Sep 25th, 2004, 8:36am In the movie dumb & dumber, jim carrey asks the girl in the bar about the chances he has to go on a date with her. she replies ' one in a million'. carrey counters, 'atleast i have a chance' and in the end he gets the girl. looking logically at this situation we can conclude that 0.9999..<1 otherwise 1-0.9999..=0 which would definitlely contradict the ending to the movie. ;D |
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Title: Re: 0.999... Post by rmsgrey on Sep 25th, 2004, 9:57am on 09/25/04 at 08:36:04, sam_i_am wrote:
The movie has nothing to do with it. One in a million (1-0.999999=0.000001) is definitely larger than 0, but 0.999999 is definitely smaller than 0.999... |
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Title: Re: 0.999... Post by llamario on Oct 17th, 2004, 4:12pm Wow, that "Dumb and Dumber" analogy could be the worst argument I ever heard. It's just wrong on so many levels. 1) Carrey's reply was "So you're telling me I've got a chance!" 2) Carrey did not get the girl in the end, she was re-united with her husband. (He didn't even get the Hawiian Tropic girls...) 3) 1 minus .999... does equal zero, because there are no real numbers in between them. (See first post in this thread) 4) You are (erroneously) basing your answer on the philosphy of a movie character whose title clearly implies that he is dumb at best, and possibly even dumber than that. |
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Title: Re: 0.999... Post by Icarus on Oct 18th, 2004, 2:55pm llamario, have considered the possibility that sam_i_am was, just possibly, less than completely serious? |
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Title: Re: 0.999... Post by Sputnik on Oct 21st, 2004, 2:29pm Sorry to burst your bubble, folks, but if (infinity+1) has no meaning then (0.999... * 10) also has no meaning. Infinitely recurring values cannot be the subject of normal mathematical procedures any more than infinity itself can. 0.999... does not equal 1 It equals 0.999... and no other value. It is functionally equivalent but NOT equal to 1. |
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Title: Re: 0.999... Post by Icarus on Oct 21st, 2004, 6:41pm on 10/21/04 at 14:29:02, Sputnik wrote:
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Title: Re: 0.999... Post by Emilie on Nov 5th, 2004, 2:21am " 3) Take the Difference 1.000... -0.999... 0.000... = 0 Since their difference is 0, they are the same. " -------------------------- I thought that 1.000... -0.999... 0.000... = 0.000...1 <> 0.000...0, only approximately |
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Title: Re: 0.999... Post by Sir Col on Nov 5th, 2004, 4:49pm 1-0.9=0.1 1-0.99=0.01 1-0.999=0.001 If the string of 9's in 0.999... terminates, then the difference, 0.000..., will also terminate with 1. However, as 0.999... is an infinitely recurring decimal, the difference, 0.000..., is also an infinitely recurring decimal, so there is no 1 "at the end". |
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Title: Re: 0.999... Post by Icarus on Nov 9th, 2004, 5:30pm This is an example of why I put this one under the heading "Arguments depending on the reader knowing how to add/multiply decimals". (That is not intended as a put-down. The point here is a bit intricate, and there is no shame in never having come across it before.) Before you can trust the argument, you have to know what it is actually saying. Note that in the notations above the 3 dots represents the digits repeating to infinity. As such, there is no such decimal notation as "0.000...1", for it would require an infinite number of 0s to occur before the 1. If you check the definition of decimal notation given above, you will see that it does not include such transinfinite decimals. (It is possible to define such notations, but when you then attempt to attach these new expressions to real numbers, the only value that makes sense for 0.000...1 is 0.) When you manipulate decimals, once you've pushed the calculation "to infinity", there is no more that comes after. If you are not already aware of this, or do not understand it, then this argument is not for you. To make it more exact, I would have to bring in calculus concepts, and simpler versions of calculus proofs are already given. |
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Title: Re: 0.999... Post by Three Hands on Nov 10th, 2004, 5:19am OK, this may have already been stated, but I figured that it probably counts as a reasonable, not-very-mathematical way of looking at this. If 0.999... is the same as 1, then it presumably means that there can be no difference of 0.000...1. So, if you subtract 0.000...1 from 0.999... (resulting in 0.999...8 ), then you still have 1. This process could then be repeated, eventually resulting in 0=1 Presumably people do not want this to be the case. Hence I would argue that the difference of 0.000...1 between 0.999 and 1 is important in that respect, and that 0.999 is less than 1 But then, someone has probably already pointed this out, and the discussion suggested that they are some ignorant person who doesn't fully understand the intricacies of infinity, etc. But, hey, I never claimed to be a mathemawhatsit... [e] damn smileys... [/e] |
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Title: Re: 0.999... Post by Grimbal on Nov 10th, 2004, 5:53am The whole problem comes from the intuitive (but naive) view that a number is defined by its digits. That it is a set of digits. It works for the integers, after all. 1.000... is not 0.999... because 1 is not 0 and 0 is not 9. |
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Title: Re: 0.999... Post by rmsgrey on Nov 10th, 2004, 6:02am So what exactly is 0.999...0 - 0.000...1 ? And is 0.999... the same as 0.999...9 ? And what is (0.999... + 1)/2 ? From the definition of Real numbers I was taught (the minimal closure under Cauchy limits of the Rationals) you get that two numbers are the same if, for any finite (rational) value greater than zero, the difference between them is smaller than that. |
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Title: Re: 0.999... Post by Grimbal on Nov 10th, 2004, 6:37am on 11/10/04 at 06:02:05, rmsgrey wrote:
And what exactly are 0.999...0, 0.999...9, 0.000...1 ? |
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Title: Re: 0.999... Post by Icarus on Nov 11th, 2004, 3:48pm on 11/10/04 at 05:19:46, Three Hands wrote:
If you will note in my post right above yours, I state Quote:
Your whole argument simply ignores this problem, pretending that 0.000...1 exists! Let me try to be clear here: THERE IS NO SUCH REAL NUMBER AS 0.000...1 OR 0.999...9 OR 0.999...8! The definition of decimal notation simply doesn't extend to these transinfinite decimal positions. If you want to talk about such "numbers", YOU have to define them first. And even if you do define them, they have no bearing on the question of whether 0.999... = 1, as these two notations already have definitions, which are not changed by defining additional ones. By those standard definitions, both refer to the same real number. Quote:
Actually, this process cannot be carried out even once (unless you can come up with some sort of reasonable definition for your new notations - and even then, it is NOT going to be repeatable in such a way as to get to 0). You seem to think that 0.999... has some "last 9" that we don't bother to write down. This is most definitely not the case. 0.999... has an infinite number of 9s - one nine in order for each of the natural numbers. A last "9" would mean the same as a "highest natural number". But there is not a highest natural number. If you give me a proposed highest, I can always beat it by adding 1. So also 0.999... does not have a last 9 for us to decrement to get 0.999...8. Quote:
Ignorance itself is not a problem. Only when ignorance is willful is it something to be ashamed of. To believe wrong concepts is a common condition of all people. To hold to these concepts without bothering to examine the arguments against them (when such arguments have been presented), or to reject such arguments simply because you don't understand them, is unexcusable. And to presume you know the tenor of a conversation you haven't have bothered to follow is rude. |
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Title: Re: 0.999... Post by Three Hands on Nov 12th, 2004, 4:50am OK, fair cop, I should have thought through what I was implying a bit more before posting - especially given philosophers seem to enjoy questioning existence so much to begin with :-/ . I'll try and avoid making stupid comments in future (but almost certainly fail...) |
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Title: Re: 0.999... Post by h7 on Nov 13th, 2004, 6:08pm -- The idea is that hyperreal numbers extend the reals, including in "infinitely small" numbers. That's it! we say: 1 - 0.999... is one of these infinitely small numbers, rather than 0! If they're hyperreals between 0.999 and 1 does that mean 0.999 still equals 1? (I don't know how to express hyperreals or complex notions so please bear with me) Suppose we add a new set of numbers to extend the decimals in which (to show the new set of numbers I'll add a colon) : 0.9 : 9 which does not equal 0.99 suppose :1 + :9 = 0.00...1 the numbers after the colon will be something that is smaller then 0.000....1 (And by this I mean the smallest thing we can represent using our number system) Something similar of how time works. 1:42:51:75:35 letting 1 be intergers, 42 to be decimals, 51 to be these new sets of smaller numbers, and so on and so on. So we'll have a number like 0.999... : # which is smaller then 1 but greater then 0.999 # = any natural number I'm really getting a headache thinking about such infinitely small things... like this. I don't think I made my post clearly enough, so if anyone has any questions (or more probable corrections/retorts in which proves that this is all babble and nonsense), please post. |
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Title: Re: 0.999... Post by Sir Col on Nov 14th, 2004, 3:02am There is nothing wrong with asking genuine and original questions. But, as you said, you can always expect to be challenged. This is the arena of mathematics: everything we commit outwardly is open to refutation, including the points I make in this post. It was only during the 19th and 20th century when mathematicians were evicted from their ivory towers of certainties that we began to make such significant advancement. Now, any credible mathematician would not refuse to have premises questioned. Having challenged the perfect harmony of balance, where 0.999...=1, with your question, I would ask you to clearly define this "new set of numbers to extend the decimals". For example, I could say that x is a number that exists beyond the normal decimals, but resides between 0.999... and 1. Therefore 0.999...[ne]1. The argument fails because x cannot be defined in terms of the decimals, hence it cannot validly be used to draw conclusions based on the decimals. To say that :1+:9=0.000...1 doesn't help, because 0.000...1 is not defined in the normal decimal system anyway. Can you express your new system in terms of standard decimal notation? Or can you show how normal decimals operate in your extended system? If not, it makes no sense to combine the two distinct systems in your argument. |
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Title: Re: 0.999... Post by Grimbal on Nov 14th, 2004, 2:21pm Maybe that is what nonstandard analysis is all about? http://mathworld.wolfram.com/NonstandardAnalysis.html |
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Title: Re: 0.999... Post by Sir Col on Nov 14th, 2004, 2:58pm Hyperreals are an extension of the real number system and deal with infinitesimals. A positive hyperreal is defined as being greater than zero but less than every positive real number. Consequently, the decimal system, which describes real numbers, simply cannot describe the complete set of hyperreals. The only infinitesimal that real numbers can describe is zero. If you want to learn more about hyperreals, I can strongly recommend, Elementary Calculus: An Approach Using Infinitesimals, written by H. Jerome Keisler. It deals comprehensively with the entire system and you can download it from here: http://www.math.wisc.edu/%7Ekeisler/calc.html |
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Title: Re: 0.999... Post by Icarus on Nov 15th, 2004, 8:10pm on 11/12/04 at 04:50:48, Three Hands wrote:
That's alright. I wouldn't want you raising the bar higher than I could meet anyway! ;) on 11/13/04 at 18:08:16, h7 wrote:
(I added the dots to the quote, since without them you just have a trivially obvious fact: 0.999 < 1. This may seem pointless since we both know what you meant, but there are some who read this thread who have a hard time differentiating between 0.999 and 0.999.... I'd rather not make it harder for them by confusing the notation even more.) Since the hyperreals extend the real numbers, 0.999... is defined in the hyperreals exactly as it is in the Reals. I.e. 0.999... = 1 as hyperreal numbers, too. Therefore, there are no hyperreal numbers between 0.999... and 1, as no other number can come between 1 and itself. In order to have a number between 0.999... and 1, you would have to redefine what 0.999... means in the hyperreals. But even if you do redefine it, it has no bearing on the question of this thread, which is what 0.999... equals as a real number under the ordinary definition. Quote:
You can define 0.999...:# as a hyperreal much as you have suggested. The key is: during addition or multiplication, when carries reach the colon they are discarded, as they are considered "infinitely smaller" than any of the digits to the left of the colon. But this changes nothing. You simply have that 0.999...:# = 1:# > 1. Your assumption that anything you tag onto the end of 0.999... is going to result in something less than 1 is false. |
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Title: Re: 0.999... Post by rmsgrey on Nov 22nd, 2004, 6:35am Just to annoy Icarus... 0.999...<1[.000...] provided you use a lexicographic ordering. One reason so many people have difficulty with 0.999...=1 is that, except for the very rare cases where you compare the two representations of a terminating decimal, the lexicographic order (padded with leading 0s to align the decimal points) is the same as the usual order on the reals, and generally people judge the order of two numbers first by looking at the number of digits (to the left of the decimal point) or number of leading 0's (to the right of the decimal point) or by comparing the exponent of 10 (in the same way), then, in case of a tie, looking at the lexicographic order. This is fine except when (padded) lexicographic order differs from the usual order on the reals... |
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Title: Re: 0.999... Post by Icarus on Nov 22nd, 2004, 4:45pm As abstract decimal expressions 0.999... < 1.000... lexigraphically. But not as real numbers, since they are in fact the same real number. It is not quite accurate to say that the real numbers do not follow lexigraphical order. After all, if x.xxx... [le] y.yyy... lexigraphically as abstract decimal expressions, then x.xxx... [le] y.yyy... as real numbers as well. The only difference is that sometimes x.xxx... < y.yyy... lexigraphically, but x.xxx... = y.yyy... as real numbers. The general rule is, x.xxx... and y.yyy... represent the same real number if and only if they are adjacent lexigraphically (i.e. no other decimal expression lies between them). So another way of defining the real numbers (not a very natural one) is to take the set of all decimal expressions, and mod out the relation (x is adjacent to y). |
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Title: Re: 0.999... Post by THUDandBLUNDER on Dec 17th, 2004, 9:02am Consider the following (generally accepted) 'proof' that 0.9999999... = 1 Let 0.9999999... = x Multiply both sides by 10 9.999999... = 10x Therefore 9x = 9 x = 1 But 1) We cannot assume that subtraction applies to infinite decimals in the same way that we know it applies to finite decimals. (Hence the original question of whether 0.9999999... = 1). 2) Point 1) above implies that we cannot assume that addition applies to infinite decimals in the same the same way that we know it applies to finite decimals. 3) Given that multiplication is no more than repeated addition, we cannot therefore assume that multipication applies to infinite decimals in the same the same way that we know it applies to finite decimals. 4) Therefore the above 'proof' is begging the question (assuming to be true that which is to be proved). 5) Therefore the above 'proof' is bogus. Given that a proof is either valid or it isn't, why is the above 'proof' so often quoted as being acceptable? (Even in this thread.) :o |
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Title: Re: 0.999... Post by towr on Dec 17th, 2004, 11:04am Quote:
Or would you want to object to 10*1/3-1/3=3 just because someone wrote it as 3.333... - 0.333... = 3 ? We could always write 0.999.. as [sum]i=1[supinfty]9*10-i, but it's still the same thing. 10*[sum]i=1[supinfty]9*10-i = [sum]i=1[supinfty]9*101-i = [sum]i=0[supinfty]9*10-i [sum]i=0[supinfty]9*10-i - [sum]i=1[supinfty]9*10-i = 9 * 10-0=9 so [sum]i=1[supinfty]9*10-i = 1 |
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Title: Re: 0.999... Post by THUDandBLUNDER on Dec 17th, 2004, 12:04pm Quote:
1) Are you saying that 'infinite decimals' are well-defined. Therefore 0.9999999... = 1? and/or 2) Are you claiming that the proffered 'proof' that 0.9999999... = 1 is acceptable? |
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Title: Re: 0.999... Post by Sir Col on Dec 17th, 2004, 1:36pm I don't think anyone with less than a graduate level understanding of mathematics, in presenting that particular proof, knows what is really going on. However, what makes it such a popular proof is that it appeals to the intuition and confirms what is mathematically true, albeit with somewhat naive assumptions; that is, the heart of the proof requires a sophisticated appreciation of arithmetic with infinite decimals. Although a novice could begin to grasp what would happen if you added 0.111... to 0.222..., they would be hard pushed to explain what happens with the "carry" when 0.666... is added to 0.444... After all, their understanding of adding is aligning numbers and adding from right to left. So where do you start with infinite sums? Obviously to mathematicians, arithmetic involving infinite decimals is clearly defined. Even if the definition circumvents the question of what is actually going on. Which human can honestly grasp the infinite or the infinitesimal. As I say to my students, "Quite literally, the infinite is too big to understand and the infinitesimal is too small." The real question is, does presenting a proof, which may make use of a higher concept, become invalid if the writer hasn't yet grasped the higher concept? For example, some of my students could prove a series of neat number theory results that make use of the fundamental theorem of arithmetic. Just because none of them have seen a proof for the FTA, does that make their proofs any less valid? There is an excellent book by A. Gardiner called, Understanding Infinity: The Mathematics of Infinite Processes, which asks the big question, "Why does calculus work?" It has chapters dedicated to the exploration of real numbers, infinite decimals, recurring decimals, and even recurring nines! http://store.yahoo.com/doverpublications/048642538x.html |
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Title: Re: 0.999... Post by Icarus on Dec 17th, 2004, 6:28pm on 12/17/04 at 12:04:53, THUDandBLUNDER wrote:
I think he is saying both, and if he isn't, then I am. |
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Title: Re: 0.999... Post by Sir Col on Dec 18th, 2004, 3:24am On a more general note... I am reminded of when I extended one of my classes work on arithmetic into different bases and whilst working in base 2 they explored decimal fractions. Having previously looked at recurring decimals in base 10 and the result, 0.999...=1, the class were a little divided on whether or not they "believed" it. You can imagine their reaction when they converted 0.111...2 into base 10; that is, 0.111...2=1/2+1/4+1/8+...=1. I suspect that I won across most of the remainging stubborn doubters; at least those who are bright enough to appreciate the significance of the digit (b-1) in base b. |
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Title: Re: 0.999... Post by towr on Dec 20th, 2004, 12:57am on 12/17/04 at 12:04:53, THUDandBLUNDER wrote:
To be honest, I probably wouldn't have if Icarus hadn't convinced me of it earlier in the discussion :P |
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Title: Re: 0.999... Post by kellys on Dec 28th, 2004, 5:38pm Well, some people might believe that 1/3=0.333... because they can work it out using long division (after "giving up" and "getting the picture"). So maybe they'll believe that 1=0.999... after using long division to divide 2 by 2, say. Since 2/2=1, they could write, 0.99.. If they can't comprehend that, then they should probably be questioning the fact that 1/3=0.333... as well. |
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Title: Re: 0.999... Post by igabo jj on Jan 12th, 2005, 1:32pm Sorry Icarus, my math needs some help. For your first proof in calculus, could you explain how you got the explicit equation from the geometric series? I'm a bit lost as to why the numerator is 1 - 10^(-N - 1). I'm not sure where the -1 in the exponent comes from. |
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Title: Re: 0.999... Post by Icarus on Jan 12th, 2005, 3:18pm Let y = [sum][subn]=1[smiley=supcn.gif] x[supn] = x + x[sup2] + x[sup3] + ... + x[smiley=supcn.gif] 1 + y = 1 + x + ... + x[smiley=supcn.gif] (1 - x)(1 + y) = 1 + x + x[sup2] + ... + x[smiley=supcn.gif] -x - x[sup2] - ... - x[smiley=supcn.gif] - x[smiley=supcn.gif][supplus][sup1] = 1 - x[smiley=supcn.gif][supplus][sup1] So y = (1 - x[smiley=supcn.gif][supplus][sup1])/(1 - x) - 1 Substitute x = 10[supminus][sup1], and you have the result. |
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Title: Re: 0.999... Post by Qrimson Fury on Jan 20th, 2005, 8:37am If only the question would have been worded like the Title, then it would actually be a riddle, i.e.: Which is true? .999...<1 .999...=1 .999...>1 The answer would obviously be: .999...>1 since 0.999.999.999.999 and so on > 1 Of course the leading "0." gets dropped just like if you wrote 015. |
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Title: Re: 0.999... Post by rmsgrey on Jan 20th, 2005, 9:04am on 01/20/05 at 08:37:41, Qrimson Fury wrote:
The convention usualy used on this forum is for '.' to denote a decimal point rather than a thousands separator - in those cases where a thousands separator is used, it's usually a ',' so: 9999999.9999 or 9,999,999.9999 |
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Title: Re: 0.999... Post by Qrimson Fury on Jan 21st, 2005, 1:37am "." may not be the commonly used convention for separating thousands, but it is an accepted convention. Am I missing something? Where exactly does the "riddle" state that we are dealing with Real numbers? I would say that (if my above solution is accepted) all three possibilities have been proven. .999... = 1 was proven several ways, and I believe the following is an agreement by Icarus that .999... < 1 on 11/22/04 at 16:45:53, Icarus wrote:
A couple questions: 1) Can you use the 1.000... -0.999... 0.000... = 0 proof for this since, when subtracting, don't you have to start at the right-most digit? How can you start at the right-most digit when there isn't one? 2) Doesn't the 1=0.999...91 go against what is given in the question: 0.999... has infinitely many 9s. Doesn't this mean that EVERY digit after the decimal point is a 9? 91 isn't a 9. Therefore it doesn't fit the question. 3) Using the 1=0.999...91 proof, does this mean that there can be a "last" digit? Thus, looking at only the last digit of .999... which, if my 2nd question is correct, would be a 9, using the 10x proof, when you multiply .999... by 10, wouldn't the last 9 shift right one place and you would be subtracting the last 9 of x from a 0 in 10x 10x = 9.999...90 -x = -0.999...99 9x = 8.999...1? |
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Title: Re: 0.999... Post by Icarus on Jan 21st, 2005, 6:02pm on 01/21/05 at 01:37:28, Qrimson Fury wrote:
Where you are, maybe. In the USA, it is not accepted at all. Quote:
Because it never occurred to William that the riddle could be given any other interpretation than as real numbers. The convention concerning decimal notation representing real numbers is so universal that it is necessarily assumed unless specifically said otherwise. I.e., in order for 0.999... to mean anything other than a real number, you must actually say that it can. Since William did not, it refers to Real numbers. Quote:
If you are allowed to come up with any meaning for 0.999..., then you can prove anything you want about it. But there is no point. Your result is meaningless except in the particular context of the definition you used, and the only context of only import is for 0.999... is the real numbers. In the example you gave, I specifically pointed out the context: Lexigraphical sequences of ordered characters. This context tells me nothing about the context of the question itself: decimal expressions for real numbers. Which was the point I was making in that post. Quote:
No, you do not have to start at the right-most digit. You can start at the left digit, and move right. Occasionally, you have to back up a bit to borrow, but whenever you have a non-zero digit in the result, you will never have to go back any further than it again, so you are still guaranteed to have unchanging digits eventually in any spot. Quote:
In general, an ellipsis, "...", indicates a continuance of whatever precedes it. Usually, this continuance is finite in length. Only in mathematics does the possibility of an infinite continuance come up. But even in mathematics, the ellipsis does not always mean an infinite continuance. Usually the meaning is determined through context. In the context of decimal expressions, an expression of the form 0.999... usually means infinite continuation. But an expression of the form 0.999...9 does not, because in any decimal, any digit has only a finite number of predecessors. In this case the ellipsis represents only a finite continuation. This is why, when discussing the concept of infinitely many digits before another digit in earlier posts, I was always careful to establish the context of infinite continuation when giving these expressions (which are NOT decimal expressions). Unless specifically said otherwise, 0.999...9, and similar expressions, represent only a finite number of digits, while 0.999... represents an infinite number of digits. |
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Title: Re: 0.999... Post by Qrimson Fury on Jan 22nd, 2005, 1:07am Since this is a site for riddles, wouldn't it be better to word the problem along the lines of finding a case where you can prove all three possibilities? As it is written, it's just a simple mathematical proof, and one that I would think many who frequent this site saw back in grade school when first taught the 10nx method for finding the fractional equivalent of a repeating decimal. My 999.999.999... answer was actually in response to someone in the original thread that challenged someone to come up with showing .999...>1. My answer was the best I could come up with. I do live in the USA. I saw this notation in an old Mathematical Recreations book several years ago, but unfortunately lost it. I did a web search under 1.000.000, and it looks like it might be a common convention in Europe. I've never been to Europe, so I have no first-hand knowledge whether this is the case. I don't believe I made up a new meaning for .999... in order to prove anything. |
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Title: Re: 0.999... Post by Sir Col on Jan 22nd, 2005, 3:56am Please don't feel got at, Qrimson Fury. Your reinterpretation was a nice idea, and you're absolutely correct about the comma convention. It is only in English speaking countries that the dot is used as a decimal point and the comma is used as a separator. Certainly in continental Europe, and pretty much the rest of the world, the dot is used as a separator and the comma is used as a decimal point. You can check these out by going to Control Panel, Regional and Language Options. If you change your location you will see examples of how money, time, date, and number is written in the box below. I found this interesting page: http://www.geocities.com/Broadway/1906/cultr14.html One thing I am not sure about, and I would really like this to be confirmed or refuted, but I believe that they use a semicolon in coordinates. |
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Title: Re: 0.999... Post by Vincent Lascaux on Feb 4th, 2005, 4:03pm Hi, I'm from France. We actually use semi colon to separate numbers in coordinate, sets... For example, we would say that the coordinate of a point are (0,5;0,6), where you would write (0.5,0.6). Actually, I'm so used to see (0.5,0.6) on computers that its odd to see (0,5;0,6), but on paper it looks ok :) |
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Title: Re: 0.999... Post by Jetru on Mar 29th, 2005, 5:44am Wow, great discussion. .999... IS 1 .999... IS THE SAME AS 1 .999... = 1 Alright tell me a number between .999... and 1. Oops. No such number. You can't say .999... with an EXTRA 9 because .999... already has infinite nines. Even if you put an extra nine, it's the same number. 1-.9 =.1 1-.99=0.01 therefore, 1-.99... =0.0000...1 there is NO such number as '0.0000...1'. You DO NOT even get to the terminating 1, because there are INFINITE 0's. Therefore, 0.0000...1 IS 0. Anyway, .999... is NOT a real number, because there are infinite nines.Infinite itself is not 'real'. |
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Title: Re: 0.999... Post by towr on Mar 29th, 2005, 6:44am on 03/29/05 at 05:44:10, Jetru wrote:
Every number equal to a real number is real. Altough you might maintain that nothing is real, and thus not 1 nor 0.999... nor anything. But aside from that it is, and infinity is just as real as anything else, especially in mathematics. Heck, even imaginary numbers are real in mathematics ;) |
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Title: Re: 0.999... Post by Grimbal on Mar 30th, 2005, 3:56am on 03/29/05 at 05:44:10, Jetru wrote:
But then, where did the extra 1 go? It's not because it is too far to see that it does not exist. ::) PS: it's a joke. Of course the diff becomes infinitely small, which is zero. |
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Title: Re: 0.999... Post by Sjoerd Job Postmus on Mar 30th, 2005, 4:26am If I were to describe this as an infinite sum, I'd use: http://tcw2.ppsw.rug.nl/cgi-bin/cgiwrap/towr/FORMULA/img.cgi?img=e7211819e95d0f671591dd2f30921ad6 Probably not too much help... Just in case it might give you something to work with. You've probably thought about it already, I guess... anyway |
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Title: Re: 0.999... Post by Icarus on Mar 30th, 2005, 2:56pm It's in the proofs post at the start (with 10-n instead of 0.1n). But you can't read it right now because the mathematical symbolry is currently disabled. Hopefully William will be able to get the symbolry back up sometime soon. |
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Title: Re: 0.999... Post by Jetru on Mar 31st, 2005, 12:58am I think this thread should be locked. The first few posts make it clear that .999...=1. If people don't understand it, then they should learn maths, then think about it again. |
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Title: Re: 0.999... Post by The_Godfather on Mar 31st, 2005, 2:42pm Just a question, but if there is a contradiction or problem in math with many solutions based on opinion, views, or moral standpoints, what makes that a riddle? |
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Title: Re: 0.999... Post by Icarus on Mar 31st, 2005, 6:35pm 1) There is no contradiction here. 2) There is only one solution. Its sole truth is unaffected by opinion, view, or moral standpoints. It is a direct consequence of the definition. 3) It is a truth that many people have trouble finding or understanding, because their past experience has led them to make false assumptions. 4) Point (3) is the most common basis for riddles. Most "riddles" rely on leading the person to make false assumptions. All those riddles you've recently posted in the easy forumare like this. 5) This forum and site does not restrict itself to "riddles" as you might think of them, but allows more general problems with some aspect that makes them interesting. |
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Title: Re: 0.999... Post by ricecakeboy on Apr 27th, 2005, 12:19am why is everyone arguing over this ... this is the way u proof repeating decimals anyways its like arguing over why 0.333.... = 1/3 0.666... = 1/6 0.999... = 1/1 i dont see why 0.999... is so special to have over 20 pages http://en.wikipedia.org/wiki/Recurring_decimal |
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Title: Re: 0.999... Post by towr on Apr 27th, 2005, 1:46am You're right, it's terribly easy. Yet many people simply can't believe it, regardless of the number of excellent proofs they're confronted with. (btw you mistakenly typed , 0.666... = 1/6, instead of 2/3 ) |
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Title: Re: 0.999... Post by Deedlit on Apr 27th, 2005, 8:00pm It was pointed out in a math FAQ somewhere that the "trivial" proofs of .99999... = 1 hid some important issues - namely, what an infinite decimal really means, and how you can justify using the same operations and theorems on them as you do on finite decimals. Of course, we really mean that .a1a2... = [sum] ai 10-i, where an infinite sum is defined as the limit of partial sums, etc. Then, showing that the things we do with infinite decimals are justified requires a bit of work. Since showing 1/3 = [sum] 3 * 10-i isn't any easier than showing 1 = [sum] 9 * 10-i, the above proof doesn't really shorten anything. Similarly for the 10x - x = 9 approach - you're subtracting two infinite sums, and you need to know that's justified when the two sums are absolutely convergent. So the easiest rigorous proof is probably just straghtforwardly showing the limit is 1. On the other hand, it's easier to give the naive proofs to doubters, if it will shut them up! Odds are they will tune out if you try to explain limits to them. |
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Title: Re: 0.999... Post by SMQ on Apr 28th, 2005, 5:32am on 04/27/05 at 20:00:05, Deedlit wrote:
Maybe not in full rigor, but demonstrating the reverse -- that 1/3 = [sum]3*10-i -- is relatively simple through long division. In fact, couldn't you construct a rigorous inductive proof based on the long division algorithm that, while less elegant than the delta proof given somewhere above, would be intuitively easier to grasp by someone unfamiliar with infinite series? --SMQ |
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Title: Re: 0.999... Post by Icarus on Apr 28th, 2005, 6:37pm It might be easier to understand, but it still would require the same delta style argument in the end. On of the proofs given in my summary (suggested by Sir Col) is essentially exactly this sort of long-division based proof. But there is a conceptual leap between saying that 1 = 3(.33...33) + .00...01 and saying that 1 = 3(.333...) that most people who are bothered by this problem are unable to overcome. |
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Title: Re: 0.999... Post by Deedlit on Apr 28th, 2005, 8:04pm As Icarus says, technically you need the epsilon-delta proof as that's how limits are defined; still, it's commonplace to just assert sequences go to zero when they obviously do, so perhaps we can be a little casual here. But it seems that the rigorous justification of the long division will justify 1/1 = .99999... just as easily (instead of requiring the remainders to be less than 10-k, make them less than or equal to 10-k). So there's still no advantage to proving 1/3 = .333... first. Of course, a lot of people will probably see one as rock solid and the other as suspicious, so again it may well be a more convincing argument. |
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Title: Re: 0.999... Post by Icarus on Apr 29th, 2005, 3:11pm Unfortunately, the problem people have had is exactly where you are suggesting we be casual. These people either do not accept that the sequence goes to zero (because they do not understand the concept of a sequence, having never been exposed to it before). Or else, they do not understand the relationship between the various finite decimals 0.999...9 and the infinite 0.999..., so the fact that the sequence (1-0.999...9) goes to zero is immaterial to the value of 0.999... to their mind. This is why you can't really convince someone of this until they have a good understanding of the notation. And once they do have it, the value of 0.999... is trivial. |
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Title: Re: 0.999... Post by SMQ on Apr 30th, 2005, 7:21am Which is why I thought going through 1/3 = .333... might be useful: Most people, because of their experience with long division, seem ready to accept that 1/3 = .333... with a never-ending series of 3's. Many people can, from there, just multiply both sides by 3 to get 3/3 = 1 = .999... , but these people aren't the one's we're concerned with (and that step did indeed hide an important detail of the multiplication) I was thinking that with a little reflection, at least some of the remaining people might be able to agree with the mathematical conjecture that 1/3 = .333... = .3 + .03 + .003 + ... = sum from i = 1 to infinity of 3*10-i. Now, strictly mathematically, 3*1/3 = 3/3 = 1 = 3*(sum from i = 1 to [infty] of 3*10-i) = sum from i = 1 to [infty] of 3*3*10-i = sum from i = 1 to [infty] of 9*10-i = .999... But maybe those who would accept that line of reasoning are already convinced by other demonstrations. --SMQ |
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Title: Re: 0.999... Post by Whiteshadows on Oct 2nd, 2005, 6:28pm you people will believe anything these days won't you? you are all narrow minded and anything that is put on your plate, you will eat it. even if it is poop which is everyone who believes .9999... is equal to 1. so you narrowminded guys think this.... 1/3=.3333... 2/3=.6666... 3/3=.9999... and since 3/3=1, therefore 1=.99999... that is the most ridiculous bull crap i have ever heard. we use .3333 etc. as a decimal because there is no real decimal to describe thirds. it is impossible to describe thirds in decimals becease 10/3 just doesn't end. saying .9999 is equal to 1 is like saying 9=10 or 99=100 or 1000=999. IT JUST DOESN'T WORK THAT WAY YOU IDIOTS. you guys are freaking stupid to believe that .9999=1. it doesn't. 1 is only equal to .9999999 if you add .0000001 to it. which we are not so sal;fjdsl;fjkdsa;flkj. man you guys are so stupid that if your brains were dynamite then there wouldn't be enough to even blow your nose. that is if you think .9999... is equal to 1. phew! i am glad i got that out of my system. good luck trying to deny me. you can go ahead and break a couple of sweats trying to disprove me if you want but you might as well give up. Blackshadows out. |
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Title: Re: 0.999... Post by Icarus on Oct 2nd, 2005, 7:51pm From the Misconceptions list: on 05/01/04 at 19:20:52, Icarus wrote:
You know, it really funny when someone like you comes around pronouncing how smart you are and insulting every one else, while at the same time spouting utter nonsense in your actual reply. At least the guy whose post I linked to as an example knew what he was talking about. You are completely clueless! For the record, since it is not clear that you even know what it is we are talking about: By 0.999..., we do not mean 0.999 or 0.999999 or the like. The three dots indicate that the sequence of 9s continues forever. This is more often denoted by placing a bar over the final 9, but that convention is not available in this forum. So we use an ellipsis (the three dots) instead to indicate the same thing. Because the 9s continue on forever, the behavior of 0.999... differs significantly from that of 0.999 or 0.999999 or any finite number of decimal places. The difference is why 0.999... = 1 while all the finite versions fall short. I am not going to try and prove it to you. If William ever gets the math symbolry repaired, then you can read several proofs in the second post of this thread. But you would still reject out of hand the ones simple enough for you to have some hope of following (as you already have one of them). So instead, if you are still in school, I strongly suggest asking your math teacher. If you are not in school, I suggest calling the math department of your nearest university. Ask them, and they will tell you that 0.999... = 1. |
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Title: Re: 0.999... Post by Grimbal on Oct 3rd, 2005, 1:00am on 10/02/05 at 18:28:58, Whiteshadows wrote:
lol! The kind of post that tells nothing about the subject at hand, but a lot about the poster. |
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Title: Re: 0.999... Post by Whiteshadows on Oct 5th, 2005, 9:08pm :)Please explain to me where i said that i was a whole lot smarter than you? that i am positive i didn't say. and i was just not caring enough for the triple period because there is no such thing as an infinite number of nines. something always ends somewhere, except for pi. who taught you math anyway icarus? your mom? your two year old sister? ::) your threats are so scary i almost peed my pants. or died from laughter of the stupidity of your comments. i will let that tingle in your brain MR. Always right. and, since i am done ridiculing you, it doesn't even matter how many nines there are, that just means the difference is that much smaller even though it may be an extremely small difference, there is still a difference. and for grimbal, what in the world does me saying,"whatever" have to do with anything? ??? just wondering is all. blackshadows out. |
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Title: Re: 0.999... Post by towr on Oct 6th, 2005, 4:16am on 10/05/05 at 21:08:13, Whiteshadows wrote:
Quote:
Rather peculiar if there was only one such number wouldn't it? There are in fact infinitely many (and not just ones you get by taking the average of pi with 1,2,3,.. either; try sqrt(2) for example) Quote:
Quote:
If 0.999... is smaller than 1, then the average of 0.999... and 1 must lie in between the two, and be distinct from both. I'll personally nominate you for the nobel prize for mathematics if you tell me what number that is :P |
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Title: Re: 0.999... Post by BNC on Oct 6th, 2005, 7:01am on 10/05/05 at 21:08:13, Whiteshadows wrote:
;D LOL... Asking this of Icarus of all people... ;D |
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Title: Re: 0.999... Post by Icarus on Oct 6th, 2005, 3:34pm on 10/05/05 at 21:08:13, Whiteshadows wrote:
Okay! How about here: on 10/02/05 at 18:28:58, Whiteshadows wrote:
on 10/05/05 at 21:08:13, Whiteshadows wrote:
As towr has indicated, didn't it ever occur to you that pi might not be such a special number (at least in that aspect)? It is quite true that you cannot write down an infinite number of digits, not even for pi. However the concept of decimal expressions containing an infinite number of digits has long been a part of mathematics. Indeed, mathematicians generally consider all decimal expressions to contain an infinite number of digits, even if they are not written down (in the case of terminating decimals, they are considered to end in repeating zeros). For repeating decimals we even have a notation that denotes exactly this: overlining the block of digits that repeats indefinitely. These expressions are well-defined, meaning that there is a unique real number associated with each (but this does not mean that each real number has a unique decimal expression ;)). on 10/05/05 at 21:08:13, Whiteshadows wrote:
I earned my PhD in mathematics from Wichita State University in 1991. on 10/05/05 at 21:08:13, Whiteshadows wrote:
I am sorry to hear about your problems with incontinence, but you will have to point out where I made any threats. Rereading my post, I find nothing that even slightly resembles a threat. Unless you count my telling you that if you would just ask anyone you know who, unlike you, is educated in mathematics, they would confirm that 0.999... = 1. I strongly suggest you try it, just as soon as you put on a fresh diaper. on 10/05/05 at 21:08:13, Whiteshadows wrote:
Not always, but definitely so here. on 10/05/05 at 21:08:13, Whiteshadows wrote:
And what non-negative numbers are so small that they are less than any positive number? Let x = 1 - 0.999..., and you can show that for any h > 0, h is also greater than x. In particular, let n be an integer > 1/h (that such an n exists is guaranteed by the Archimedean principle). Then 10n > n > 1/h, so h > 1/10n = 10-n. Therefore 1 - 10-n > 1 - h. But 1 - 10n = 0.999...(n-1 9s total)...9998 < 0.999...(infinite 9s)... . So 0.999... > 1 - 10-n > 1 - h. This gives us that x = 1 - 0.999... < 10-n < h So the difference is between 1 and 0.999... is smaller than any number above zero. It cannot be anything other than zero itself. That is one of the many proofs that 1 = 0.999... |
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Title: Re: 0.999... Post by rmsgrey on Oct 7th, 2005, 12:52pm on 10/02/05 at 18:28:58, Whiteshadows wrote:
I deny you. No that that's out of the way, could I just ask which part of your post is meant to constitute a proof of anything for us to try and disprove? You make a number of unsupported assertions, such as: on 10/02/05 at 18:28:58, Whiteshadows wrote:
but never offer an actual proof. If necessary, Icarus could produce an answer to any question of "why is that assertion true" for any of his assertions in proving that 0.999...=1 - ultimately, being reduced to "because that's just the way we choose to assume things to be" - at which point, you're free to take exception with any of the axioms that reflect several thousand years of mathematical thought (in fact, non-Euclidean geometry requires you to discard an axiom that was used for geometry for a couple of thousand years) or choose to replace any of the definitions or terminology or notation used that you don't like, though you should really explain where you choose to use non-standard terms, and then you shouldn't be surprised if your results don't apply to the standard terminology... Can you provide any proof that "it just doesn't work that way" or are we to take it as an additional axiom - in which case I should point out that it's inconsistent with the usual sets of axioms (and would enable us to prove that you are the Pope) |
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Title: Re: 0.999... Post by Sir Col on Nov 29th, 2005, 4:44pm (I hope I'm not wasting my time with this post, but it may help identify your misconceptions, Whiteshadows) May I ask what you think 0.999... is equal to? Either it (i) exists, or (ii) it does not exist. (i) If it exists then it is either equal to one, less than one, or more than one. I am sure you would agree it is not greater than one, and as you don't believe that it is equal to one, you must believe it is less than one. Right? In which case you probably believe that it is the "last" number before one. Right? What happens if you take two numbers, x and y, and calculate (x+y)/2? That's right it gives a number in-between. So what does (0.999...+1)/2 equal? There certainly cannot be a number in-between, so I guess the only resolution would be to say that the numbers are equal. However, let us consider the other possibility... (ii) If it does not exist we must be saying that no infinitely recurring decimals exist. Right? Let x = 3/10 + 3/100 + 3/1000 + ... = 0.333... 10x = 3 + 3/10 + 3/100 + ... = 3 + x Hence 10x-x = 9x = 3, so x = 3/9 = 1/3. If we ever terminate that series, we have a problem... If y = 3/10 + 3/100 + 3/1000 10y = 3 + 3/10 + 3/100 10y-y = 9y = 3-3/1000 = 3000/1000-3/1000 = 2997/1000 = 2.997, so x = 2.997/9 = 0.333 So what is 0.333... ? Of course we can use the same method for 0.999... z = 9/10 + 9/100 + 9/1000 + ... = 0.999... 10z = 9 + 9/10 + 9/100 + ... = 9 + z 10z-z = 9z = 9, so z = 9/9 = 1 But if you deny the existence of 0.999... = 9/9 = 1 then I guess you must be denying the existence of all other recurring decimals. Right? |
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Title: Re: 0.999... Post by groll on Dec 12th, 2005, 7:50am I'm having slight problems with one of the equations commonly used in this discussion. I hope some one here can take their time and explain my errors. x = 0.999... 10x = 9.999... 10x - x = 9.999... - 0.999... We assume that we have n decimals Let a = 0.999.. (with n decimals) b = 10 * a = 9.999... (with n-1 decimals) c = b - 9 = 0.999... (with n - 1 decimals) Now subtract c from a and you will have a number greater than 0. Now even as we let n travel towards infinity we will have a difference, you can allways find a value between a and c. This would lead to 10x - x = 9.999... - 0.999... 9x = 8.999...1 (since there will allways be one more nine in the second number) x = 8.999...1 / 9 x < 1 ================================== I assume that this might have to do with how we're allowed to work with infinity. If I let "m = infinity" would then "m - 1 = m" ? If you state that you have an infinite number of decimals (say 9's) would I be able to add one more? If there are m natural numbers where m is an infinite number, would it then be correct to state that there exists a number m + 1? ================================== Thanks in advance /Dan |
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Title: Re: 0.999... Post by towr on Dec 12th, 2005, 8:34am on 12/12/05 at 07:50:39, groll wrote:
Quote:
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take for example an ordered set of numbers, {1,2,3,4,5, ..} it contains infinitely many numbers, but you can add a new one, that's not in it. {0,1,2,3,4,5, ..} So you can add to infinity. But in fact, there are also equally many numbers in both sets. For each number X in the first set, (X-1) is in the second one. And conversely, for each number Y in the second set, (Y+1) is in the first set. |
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Title: Re: 0.999... Post by Marvin on Dec 12th, 2005, 1:42pm on 12/12/05 at 07:50:39, groll wrote:
This is correct with finite number of decimals. For n (finite) number of decimals, you can add or subtract one and get n-1 < n < n+1. on 12/12/05 at 07:50:39, groll wrote:
Infinity is greater than any number, so if you add one decimal, you won't get "one more than infinity" decimals, the same with subtracting. infinity + x = infinity for any real number x (including negative numbers - subtracting). With infinite number of decimals, there would be no last n'th decimal (and no n-1'th before it). When we talk about some last digit, then it must be with some finite number of digits - which is less than infinity. With n, finite number of decimals, a - c = 9*10^-n When n -> infinity, the limit of the series equals 0. The limit means what would a - c be if n = infinity (it shouldn't be written this way, because no natural number equals infinity). |
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Title: Re: 0.999... Post by Lord_of_math on Feb 22nd, 2006, 2:59pm X=.999999.... This is my proof, same as icarus, but hopefully simpler 10x=9.999999 I multiplied by 10 10x-9=.999999 subtracted 9 from both sides 10x-9=x Substituted from first step 9x-9=0 subtracted x from both sides 9x=9 added 9 to both sides x=1 divided both sides by 9 By the way, I am 15, and i did this with the use of math i learned in like 6th and 7th grade and http://mathworld.wolfram.com/ is an excellent math cite... look up aleph, it discusses different sizes of infinity // I accidentally deleted the post following this one trying to take care of the tripple post following that one. So I decided to reinsert it here. Hence the following is a repost of Icarus reply. --towr posted Feb 23rd, 2006, 4:07am by Icarus That is exactly the first of the proofs I listed, though it is hard to tell now, because I used the math symbolry we had before William updated the site to use a newer and more efficient version of YaBB. Since the symbolry was an add-in that he had provided, the new bulletin board did not have it, and he has never gotten around to adding it back in. So now all you see is the codes for the symbols. I can't even fix it because the upper limit for post sizes was decreased significantly, so I am not allowed to save the fixed post, as it is too long. |
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Title: Re: 0.999... Post by said_yam on Apr 29th, 2006, 1:48pm There is a new theory that may end the debate about 1=0.999... the whole theory can be found in this link http://www.geocities.com/humood_theory/index.htm here is a Quote about cracking the famouse proof: Proof (A) x=0.999... 10x= 9.999... 10x-x=9.999... - 0.999...=9 9x=9 x=1 Quote: -------------------------------------------------------------- I will start by cracking this proof and show it is in fact a Paradox not a proof. Let x = 0.999... 10x = 9x + x = [ 9* (0.999...) ] + 0.999... [ I expressed 10x as (9x +x)] 10x - x = ( 9x + x ) - x = [ 9* (0.999...) ] + 0.999... - 0.999... [Positive 0.999... goes with Negative 0.999...] 9x = 9* (0.999...) x = 9* ( 0.999...) / 9 = 1 * 0.999... x =0.999 We did not end with x=1 In fact what made it looks like a proof was the mistake we did in this step: 10 x - x = 9.999... - 0.999... = 9 This step assumes the value of infinite 9s after the decimal point in ( 10x ) equals the value of infinite 9s in (x) The infinite decimals in x = 0.999... have different behavior than the infinite 9s in 10x =9.999... Because 9s in x= 0.999... Go to infinity faster than 9s in 10x=9.999... (10 x ) is always one decimal behind (x) When x = 0.99 then 10x=9.9 When x = 0.9999 then 10x=9.999 When x = 0.999999999 then 10x=9.99999999 Value of Infinite decimals in (x) - Value of infinite decimals in (10x) = Lim x-> 8 9 / (10^x) So the difference between the two values equals an infinitely small fraction which is Lim x-> 8 9 / (10^x) This means in Proof (A) when we say: 10 x - x = 9.999... - 0.999... = 9 We are already Assuming that a number minus an infinitely small fraction is the number itself. Which means we are already assuming that Lim x-> 8 1 - ( 1/x ) = 0.999... = 1 So, in Proof (A) we already assumed 1= 0.999... Then no wonder we finally ended with our assumption, which make it not a Proof at all. You cannot prove any statement by assuming it to be true from the beginning. But when we explain (10 x ) in a very fundamental method no body can argue with, we will say: 10x =0.999... + 0.999... + 0.999... + 0.999... +0.999... +0.999... +0.999... +0.999... +0.999... +0.999... And when we subtract x = 0.999... from it : 10x - x =0.999... + 0.999... + 0.999... + 0.999... +0.999... +0.999... +0.999... +0.999... +0.999... + 0.999... - 0.999...= 0.999... * 9 The infinite 9s in x after the decimal is gone with only one of those 0.999... Not all of them as we did in Proof (A). ------------------------------------------------------------------ All the theory and its results can be seen in the link: http://www.geocities.com/humood_theory/index.htm For comments please write it here in this forum or send it to the said_...@yahoo.com |
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Title: Re: 0.999... Post by Grimbal on Apr 29th, 2006, 6:17pm The road to enlightenment is long and difficult. |
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Title: Re: 0.999... Post by Icarus on Apr 30th, 2006, 1:05pm Hoo boy. That is almost as bad as the "disproof" of innumerability of the reals that T&B referenced awhile back (although T&B wasn't promoting that one, just bringing it to our attention). said_yam, all that this shows is that you do not understand the mathematical meaning of the terms "limit", "proof", and "real number". There are so many things wrong with this, it is impossible to address them in the size limits of an individual post, so I will start off with your supposed "cracking" of the 10x-x proof. One thing you will note about my listing of this proof is that I put it under the heading "Argument depending on the reader knowing how to add/multiply decimals". That is, the proof expects the person reading to already know that 9 + 0.999... = 9.999.... If you do not accept this, then you are not going to accept the proof as stated. This in itself is not a weakness in the argument. Most proofs assume that the reader is aquainted with certain previous results, and does not try to prove those results again. If we did not do this, but proved everything directly from the axioms and definitions, our proofs would quickly become too long and unwieldly to follow. In Nicholas Bourbaki's Elements of Set Theory, he estimates that if you were to write out the expression he uses to define the number "1" in the basic symbols of his theory, it would require better than 22,000 characters. So what about your "crack"? You make a big deal of doing this manipulation and ending up with x = 0.999... again. In fact, you could have gotten to this point much easier! Just stop after the initial line "Let x = 0.999...", and WOW! you get x = 0.999... right there! That you end up with 0.999... instead of 1 would only indicate a problem with the 10x - x proof if 0.999... is NOT equal to 1. If 0.999... = 1, then it is only to be expected that sometimes as you do the arithmetic, you get 0.999..., and sometimes you get 1. Your crack is not a crack at all, but is an expected consequence of the proof being true. You only see it as a problem, because you are assuming at the out set that the result is false. But as I said, if you do not accept that 9 + 0.999... = 9.999..., then this proof is not for you. First, you will need to see why this is true. This requires a delving of the definitions of real numbers, decimal expressions, and limits. I have posted just such a delving in the first post of this thread. Unfortunately, changes in the bulletin board software since then have rendered the mathematical symbolry I used illegible. And the new post size limits have prevented me from being able to go in and fix the problem (you can't post anything that long anymore, nor can you modify a post that long without truncating it to the new limits - which I run into about a quarter of the way through). |
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Title: Re: 0.999... Post by Icarus on Apr 30th, 2006, 2:23pm Let's look at the opening "calculation" of your Humood Theory page, shall we? You start off with this calculation: Limx->oo (1 - 1/x) = 1 (which you are attempting to disprove). So, you say, Limx->oo (1 - 1/x)x = Limx->oo 1x 1/e = 1 You claim that this disproves that Limx->oo (1 - 1/x) = 1. I claim that is disproves the idea that whenever lim f(x) = A, then the lim g(x, f(x)) = lim g(x, A), regardless of the function g. That is the "rule" you used to go from Limx->oo (1 - 1/x) = 1 to Limx->oo (1 - 1/x)x = Limx->oo 1x. But in fact, it is not true in general and exponentiation is one of many exceptions. Even if g had no direct dependence on x, it would be true only if g were continuous at A. With g depending on x itself, more stringent conditions need to be applied before you can move limits around like that. That limx->oo (1 - 1/x) = 1 can be proven easily from the definition of a limit. By definition, this holds if we can show that for any h > 0, we can find a number M (depending on h), such that if x > M, then | (1 - 1/x) - 1 | < h. So, let h > 0 be an arbitrary number. Choose M = 1/h. If x > M = 1/h, then h > 1/x = | - 1/x | = | (1 - 1/x) - 1|. Which is exactly what was required. So, by the definition of "limit", limx->oo (1 - 1/x) = 1. The problem was with your next step, not here. _________________________________________________ Let's state some definitions. Size limitations again prevent me from going into detail, but this is the basics: Real numbers: The minimal topologically complete ordered field. In particular, the real numbers are a set which contains two particular elements, which we label "0" and "1", and call the additive and multiplicative identities, respectively. It also has two operations, addition and multiplication, and an ordering "<" which satisfy: (1) commutivity: a+b = b+a, ab = ba (2) associativity: (a+b)+c = a+(b+c), (ab)c = a(bc) (3) identity: a+0 = a, a1 = a (4) inverse: for all x, there is y such that x + y = 0. For all x != 0, there is z such that xz=1. (5) distributivity: a(b+c) = ab + ac (6) order: if a>b, then a+c > b+c. If a>b and c>0, then ac > bc. (7) completeness: if S is a non-empty set of real numbers such that if x is in S, and y < x, then y is in S, and if for all x in S, x < a for some fixed a, then there is a real number p such that if x < p, then x is in S, and if x > p, the x is not in S. (p may or may not be in S, and is called the "supremum" of S). The real numbers are also minimal, in the sense that no subset of them closed under addition and multiplication has all of the properties above. There are many ways of building a set with those properties. If you build such a set, we call it a "model" of the real numbers. Any two models of the real numbers can be put in 1-to-1 correspondence with each other in a unique fashion that preserves addition, multiplication and order. As far as real number behavior is concerned, all models behave exactly the same (they have to, because of the correspondence). For this reason, we do not specify any particular structure or way of building the real numbers. Merely that they must meet the requirements mentioned above. Limits (I will restrict this to limits at infinity) are defined, per Cauchy, as follows: limx->oo f(x) = L if and only if, for every h >0, there is an M such that if x > M, then | f(x) - L | < h. All of the rules for taking limits that you have been taught can be derived from that definition. This includes the fact that limits are unique (so, since limx->oo (1-1/x) = 1, as I already proved, and since, as you yourself have admitted, limx->oo (1-1/x) = 0.999..., we are left with only one conclusion: 1 = 0.999...). Infinite Decimals: the decimal expression dndn-1...d0.d-1d-2... by definition represents the real number D, where D = limK->oo dn...d0.d-1...d-K One thing to note about this definition (which is also true of the definition of an infinite series), is that it does NOT require an infinite amount of summing to define the unique value D that the decimal expression represents. Though mathematics deals all the time with concepts such as "infinite sums" or other infinite processes, the meaning of such things is always defined by some finite means. |
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Title: Re: 0.999... Post by fireashwinter on May 19th, 2006, 9:46am I'm responding to item 9 in Icarus' post on common misconceptions. Actually, I'm responding to the various discussion related to the interposing of additional numbers. I don't doubt that the various informed posters are aware of all of this. Perhaps my statements are even implied by theirs. I'm sorry about that-- I am not a mathemetician, engineer, or the like. I should start off by affirming that I agree that .999... = 1. However, the "disproof" I've seen of interposing numbers are not 100% convincing (I admit I have not looked at every single one). The most common "disproof" is that if there is a number between 9 and 10, designated #, then since #=.95, .###... would be greater than 1, so there is no such number .###... that lies between .999... and 1. That proof is disturbing to me, because I think that the suggestion "#" can be interpreted differently. In fact, I think that it should be interpreted differently, rather than assumed to have the most absurd possible meaning. (Even if it was intended as most absurd, it is more interesting to treat the intent as the most legitimate.) Obviously, I do not mean to propose that "#" is something like 9.000...0001, because that would be the same as proposing that #=9. What I propose is that # is the addition of an actual digit, and that as such there is not the correspondence that #=9.5. The addition of # creates an 11-digit system, aka "ak." Our decimal system can be thought of as: x/n +y/n2 +... where the numerators are less than n, and n is the number of digits. Also, the largest numerator we will say is n-1, because 9 = 10-1. (Note, I understand that you could express things differently, but with the same meaning, if you wanted to.) Let's keep the same idea with "ak," so that So, .kk... = .k/n+.k/n2... Since k is the largest numerator, and n (the number of digits) is 11, we can make an analogy between k/n + k/n2+... in the "ak" system and our own 10-digit system. Having k/n in "ak" is not quite the same as having 9/10, or 9.5/10. More reasonably, having k/n is like having 10/11. So a good analogy is that .kk... corresponds to the 0-9 decimal system as: 10/11 + 10/121 + 10/1331 ... Which you can show mechanically seems that it will be .999... In fact, for any system of n digits (n > 1, n is real), the decimal represented as .xxx... when x=n-1 will correspond to .999... This can be better shown using the fact that a converging infinite series has the sum of: a/(1-r), which you can look up in a Calculus book or website if you don't believe it. series is [(n-1)/n]/[1-(1/n)] = 1 (sorry about using "n" in this potentially confusing way) To me, this shows that the idea that you can propose any decimal system based on n>1 digits is not proof that there is a number between .999... and 1, because your number will actually correspond to .999..., or 1 revolution of the system. Actually, the bit where I discuss a/(1-r) is good enough all by itself, but might not be understood alone. In any system of that like, .xxx... is going to analogize to .999... of that "system," or as I like to think of it .999... of that mode. At that point, the # argument seems reduced to wondering whether we are actually looking at a 0-9 system, or some other system which arbitrarily uses "9." There is also the issue of whether digits have essential correspondence. However, we are discussing within a system or systems where digits have constant interval relationships. If you don't accept that, then arguments against .999... =1 challenge the meanings of both "1" and ".999...," and constitute neither immediate proof that the two are unequal, nor that another number can intervene. |
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Title: Re: 0.999... Post by fireashwinter on May 19th, 2006, 9:59am Quote:
Sorry, 0<abs(r)<1. Oops. It's the case, though. |
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Title: Re: 0.999... Post by Icarus on May 19th, 2006, 3:28pm on 05/19/06 at 09:46:08, fireashwinter wrote:
That is the sum of a power series with constant coefficient: a + ar + ar2 + ar3 + ar4 + ... = a/(1-r) (Which converges if and only |r| < 1. Note that it does converge for r = 0.) There are many, many convergent infinite series that are not of this form, and their sums are not given by that expression. As for the rest, what you are describing is simply a base-11 decimal system instead of our normal base 10. The real numbers can be expressed in decimal systems for any base n > 1. In fact you can define a "base-n" decimal system for any real number n with |n| > 1 (even bases -1 and 1 are workable with some special rules). The digits are the integers d with 0 <= d < |n|. However, if n is not an integer, every real number has multiple representations. If n is integer, then only rational numbers r such that nkr is an integer have multiple representations - exactly two: one terminating (ending in repeated zeros), and one ending repeated (n-1)s. However, the "con" arguments I was referring to were not talking about a base-11 counting system. Rather, they were trying to introduce another digit # to the base-10 decimal system, where 9 < # < 10. A base-11 system does not in any way support their position, and they knew it. They needed a base-10 system with an additional digit. How did they define their digit # or decimal expressions involving it, such at 0.###... ? They didn't! They simply assumed that such expressions were meaningful even though they were not defined. And whenever someone tried to address their arguments by supplying a meaning to their meaningless expressions, they simply declared that is not what they meant. And that, above all else, is why the argument was simply hot air. |
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Title: Re: 0.999... Post by fireashwinter on May 21st, 2006, 6:49pm Thanks. |
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Title: Re: 0.999... Post by GeMsToNe on Aug 20th, 2006, 9:21pm I just noticed... 1.000 - 0.999 = 0.0001 (in which 10 - 9 is equal to one, because you cancel the next zero to nine) Which means... 1.000... - 0.999... = 0.000...1 And isn't... 1/3 = .3333... 2/3 = .6666... 3/3 = 1 Because 3 divided by 3 is equal to one. I don't know why I even posted... blah, I'm not so good with math. I just noticed some stuff, that's all... =D |
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Title: Re: 0.999... Post by towr on Aug 21st, 2006, 2:13am Or you could conclude 0.000...1 (if it means anything at all) must be 0 Because 0.999... is obviously thrice 0.333... If the latter is 1/3rd, then the former must be 1 |
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Title: Re: 0.999... Post by GeMsToNe on Aug 21st, 2006, 7:51am Oh, I get it now. Thanks towr ^-^ |
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Title: Re: 0.999... Post by Icarus on Aug 21st, 2006, 4:38pm The normal definition for decimals does not define "decimal expressions" such as 0.000...1, where the ... represents an infinite number of zeros (or any other digit). For any defined decimal expression, each digit has only a finite number of digits to its left (it can be infinite on the right, but it has to be finite on the left). But the "1" at the end of this one has an infinite number of digits to its left. So the standard definition does not include 0.000...1. But as towr has indicated, you can define this expression yourself. But the only way to do so and be consistent with the rules for adding and multiplying decimal expressions is to have 0.000...1 = 0, exactly because of the calculation you've shown. |
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Title: Re: 0.999... Post by A-man on Dec 14th, 2006, 8:32pm Sorry to continue to beat a dead horse but I have always thought of 0.999... as the greatest number that is less than 1. If not, how would you represent the greatest number less than 1? |
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Title: Re: 0.999... Post by Michael_Dagg on Dec 14th, 2006, 9:06pm on 12/14/06 at 20:32:31, A-man wrote:
max{x: x < 1} |
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Title: Re: 0.999... Post by Eigenray on Dec 14th, 2006, 9:14pm on 12/14/06 at 20:32:31, A-man wrote:
There isn't one: for any number x < 1, there's a larger number less than 1, such as (1+x)/2: x = (x+x)/2 < (1+x)/2 < (1+1)/2 = 1. |
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Title: Re: 0.999... Post by A-man on Dec 15th, 2006, 8:35am Last two posts seem to contradict each other. Or am I reading them wrong? |
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Title: Re: 0.999... Post by Marvin on Dec 15th, 2006, 9:00am They do contradict, but Eigenray is right. The greatest number less than 1 would be max{x: x < 1} if there were such a number. But this set has no maximum, for the reasons Eigenray wrote. |
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Title: Re: 0.999... Post by A-man on Dec 15th, 2006, 11:29am OK. So is the issue that there is no way to represent the greatest number less than 1, or that there is no greatest number less than 1? |
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Title: Re: 0.999... Post by Marvin on Dec 15th, 2006, 11:40am Probably I was not clear enough, sorry. There is no greatest number smaller than 1. It is a property of the real numbers: If x<>y then there are infinitely many numbers between them. So for any x, if x<1, there are always numbers larger than x and still smaller than 1. |
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Title: Re: 0.999... Post by Icarus on Dec 15th, 2006, 4:21pm More to the point: for any two distinct numbers x and y, x < (x+y)/2 < y. So, if M were the largest number less than 1, we would have a contradiction: M < (1+M)/2 < 1, so M is not the largest number less than 1 after all. Michael_Dagg (who is a mathematician, by the way) was answering tongue-in-cheek to your question. Your question assumed that such a number existed, so he simply translated the phrase "greatest number less than 1" into a mathematical expression. Being able to write an expression that means "greatest number less than 1" does not mean that such a value actually exists. For instance, Max{x : x>= 2 and x <= 1} clearly fails to exist, since no number is a member of that set. |
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Title: Re: 0.999... Post by Michael_Dagg on Dec 15th, 2006, 5:18pm on 12/15/06 at 08:35:27, A-man wrote:
No they don't (why?). |
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Title: Re: 0.999... Post by Marvin on Dec 16th, 2006, 7:03am 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 A-man, I think. |
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Title: Re: 0.999... Post by rmsgrey on Dec 16th, 2006, 12:50pm 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) |
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Title: Re: 0.999... Post by Michael_Dagg on Dec 16th, 2006, 3:47pm sup{x: x < 1) exists and is 1 = sup{x: x< 1}. You need max so-to-speak 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. |
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Title: Re: 0.999... Post by Marvin on Dec 17th, 2006, 5:45am 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..(n-times)..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..(n-1-times)..9899... < 0.99..(n-times)..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..(n-times)..9 < 0.99..(n+1-times)..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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Title: Re: 0.999... Post by UNKNOWN on Feb 5th, 2007, 7:10pm 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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Title: Re: 0.999... Post by UNKNOWN on Feb 5th, 2007, 7:41pm 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 small-marked 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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Title: Re: 0.999... Post by Whiskey Tango Foxtrot on Feb 5th, 2007, 8:03pm 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... [hide]I am infinitively upset by your post[/hide] >:( |
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Title: Re: 0.999... Post by BNC on Feb 6th, 2007, 12:23am on 02/05/07 at 19:10:21, UNKNOWN wrote:
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 <a-hmm> infinitively more so at math-related issues. |
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Title: Re: 0.999... Post by Three Hands on Feb 6th, 2007, 1:37am on 02/05/07 at 19:10:21, UNKNOWN wrote:
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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Title: Re: 0.999... Post by Icarus on Feb 6th, 2007, 6:27pm 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 (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_suggestions;action=display;num=1035249563;start=25#49)). 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 no-brainer. Let's look at 0.999.... The definition of this expression is: 0.999... = 9/10 + 9/102 + 9/103 + ... 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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/leqslant.gif 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 0.999---9 (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 0.999---9 and 1 is 0.000---1 = 1/10n 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 1-y > 0, and I can choose n large enough so that 1/10n < 1-y. Turning this around gives y < 1 - 1/10n = 0.999---9. So x=1 satisfies property (2). Since 0.999---9 < 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 + 9/102 + 9/103 + ... . 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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Title: Re: 0.999... Post by UNKNOWN on Feb 7th, 2007, 5:25pm you know whats weird? why this thread has 110 replies and only 189 views when smaller threads have less ??? |
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Title: Re: 0.999... Post by Icarus on Feb 7th, 2007, 6:25pm 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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Title: Re: 0.999... Post by UNKNOWN on Feb 7th, 2007, 7:25pm Oh, I never knew that! Dang Nabbit!!! am i saying that correctly? |
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Title: Re: 0.999... Post by Icarus on Feb 8th, 2007, 4:31pm Gol Dang it!, its "Dag Nabbit!", ya crazy young whippersnapper! |
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Title: Re: 0.999... Post by CowsRUs on Feb 8th, 2007, 4:45pm Ima put you old Wheezers into your PLACE!!! ;D ;D ;D |
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Title: Re: 0.999... Post by WakeTFU on Feb 10th, 2007, 5:44pm 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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Title: Re: 0.999... Post by Icarus on Feb 11th, 2007, 12:19pm 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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Title: Re: 0.999... Post by WakeTFU on Feb 15th, 2007, 9:56pm " For them, to say that 0.999... = 1 is patently ridiculous" -Icarus No it isn't, is all I'm saying :P (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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Title: Re: 1 Post by Grimbal on Feb 16th, 2007, 5:08am Hey! why isn't this thread simply called "1"? ;) |
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Title: Re: 1 Post by rmsgrey on Feb 16th, 2007, 10:59am on 02/16/07 at 05:08:00, Grimbal wrote:
Notational convention :P |
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Title: Re: 0.999... Post by Icarus on Feb 16th, 2007, 4:41pm on 02/15/07 at 21:56:49, WakeTFU wrote:
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 05/01/04 at 19:19:11, Icarus wrote:
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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Title: Re: 0.999... Post by Icarus on Feb 16th, 2007, 8:16pm 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 Russell-Whitehead collections rather than Zermelo-Frankel 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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cup.gif B = S (ii) A http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cap.gif B = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/emptyset.gif (iii) for all a http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif A and b http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif B, a < b. If A, B are a cut of S, define {A|B} 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: {a1, a2, ... | b1, b2, ...}. So, for instance {0,1 | 2,3} would represent the pair ({0,1},{2,3}). (*) We start with the empty set http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/emptyset.gif as our un-indexed base. (0) Next we define 0 = {|} and define the set S0 = {0}. (1) S0 has exactly 2 cuts: Define -1 = {|0}, 1 = {0|}. Define S1 = {-1, 0, 1}. (2) S1 has 4 cuts: -2 = {|-1,0,1}, -1/2 = {-1|0,1}, 1/2 = {-1,0|1}, 2 = {-1,0,1|}. Define S2 = {-2, -1, -1/2, 0, 1/2, 1, 2}. ... Suppose that Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subalpha.gif has been defined for all http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif for some ordinal http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif, such that if http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif, then Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subalpha.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subset.gif Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subgamma.gif. Further suppose that all the Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subalpha.gif are ordered with compatible orders (that is if x, y http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subalpha.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subset.gif Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subgamma.gif, and x < y in Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subalpha.gif, then x < y in Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subgamma.gif as well). Define O = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bigcup.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subalpha.gif Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subalpha.gif. O is ordered by the common ordering inherited from the Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subalpha.gif. This ordering determines cuts in O. Define N = { {A|B} : {A|B} is a cut of O}, and define Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subbeta.gif = O http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cup.gif N. We must still define the ordering on Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subbeta.gif. If x,y are in O, then we inherit the ordering of O. If x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif O and y http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif N, Let y = {C|D}. Since y is a cut of O, x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif C or x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif D. If x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif C, define x < y. If x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif D, we define y < x. Lastly, if x, y http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif N, then x = {A|B} and y = {C|D}. By the nature of cuts, either A http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subseteq.gif C or C http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subseteq.gif A. In the former case, define x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif y, and in the latter case, y http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif x. This completes the induction. |
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Title: Re: 0.999... Post by Icarus on Feb 16th, 2007, 9:00pm 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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif such that x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subalpha.gif. 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/2m. The entire rest of the real numbers have http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif (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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif A and b http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif B, a < b. Then for any pair of sets A, B of surreal numbers, define {A|B} to be the oldest surreal number x such that for all a http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif A, b http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif B, a < x < b (by the definition of surreal numbers, such an x must exist and is unique). (4) Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subomega.gif is not http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbr.gif, however. If we define Sf = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bigcup.gifn=0http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supinfty.gif Sn, then some elements of Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subomega.gif not in http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbr.gif are: -http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif = {|Sf}, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif= {Sf|}. Also, consider the somewhat troublesome numbers [x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif 0} | {x > 0} } and { {x < 1} | {x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 1]. That is, in Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subomega.gif, there IS a smallest number greater than 0, and a largest number less than 1! (However, the surreals do not end with Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subomega.gif - 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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/emptyset.gif, both operators are defined "vacuously". If +, * have been defined for all surreals with birthday < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif , let x, y http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif. Then x = {A|B}, y={C|D}. [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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cup.gif y+A | x+D http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cup.gif y+B}.[/edit] |
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Title: Re: 0.999... Post by SMQ on Feb 17th, 2007, 5:16am What I think is "fun" is that under those definitions, expressions like http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif - 1, (technically http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif - 1) and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif are meaningful and useful numbers! Now here's a question I haven't found an answer to: the Wikipedia article (http://en.wikipedia.org/wiki/Surreal_number) 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 + Bhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/imath.gif where A and B are surreal numbers and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/imath.gif is defined to be http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif-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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Title: Re: 0.999... Post by Icarus on Feb 17th, 2007, 7:26am on 02/17/07 at 05:16:45, SMQ wrote:
I don't have a definitive answer for this one either, but I doubt it. If we consider an ordered field to be one dimensional, then the surreal complex numbers are 2 dimensional. I suspect that there should be algebraically closed fields that are not 2 dimensional. In fact, I doubt that the p-adics are embeddable in the surreals. on 02/17/07 at 05:16:45, SMQ wrote:
Of course you know, but just so the record is clear for others: In any extension of the real numbers, 0.999... = 1. Otherwise, it couldn't be an extension of the reals. |
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Title: Re: 0.999... Post by Icarus on Feb 17th, 2007, 7:43am Reading that Wikipedia article, I notice that they define the set Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subomega.gif differently than I do. My version has the S for each ordinal including cuts of all older surreals. But they define Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subomega.gif to simply be the collection of all older surreals, with the cuts added in only for http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif + 1. As a result, all non-dyadic real numbers have birthday http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif + 1 for them, but only http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif for me. In fact, no surreal has http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif for a birthday in their scheme, since their Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subomega.gif is simply a compilation of all that was defined before. Personally, I prefer my approach (of course :P). Theirs treats limit ordinals differently than others. Their induction has one method to construct Shttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subalpha.gif if http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif is a limit ordinal, and another method when it is not. My approach has a single method that works no matter type of ordinal we have. Further, every step in my construction introduces new numbers. Limit ordinals do not introduce anything new in their construction. I'm curious now if their approach is "standard", or did whoever wrote up the wikipedia article misunderstand and introduce an unneeded variation? |
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Title: Re: 0.999... Post by kiochi on Mar 21st, 2007, 9:29pm Quote:
I don't think it really makes sense to talk about the "largest" ordered field if we allow it to be a proper class. For several reasons: First, any field ought to be a set, as is every mathematical object. Heck, in a math class, even logical symbols like "=" are formally considered to be sets. The "size" of any mathematical object is the cardinality of it, namely the smallest cardinal that can be put into bijective correspondence with it. Now if we allow a proper class to have a cardinality, then we'll run into something like the Burali-Forti paradox (basically we just contradict the foundation axiom). Of course we still talk about things that aren't sets, like the collection of all ordinals, "On," but we don't compare these constructs to sets. We can use this construct as a way to do transfinite induction (as Icarus did when defining the surreal numbers). Formally, though these aren't sets, they are basically just logical formulae, e.g. "x is an ordinal," which we can talk about and manipulate only metamathematically. EDIT: To distinguish a set from a class, note that we are only allowed to construct sets from other sets. e.g. we can take unions of sets, power-sets of sets, as allowed by ZFC. Also, by the replacement schema, we can define a set as a collection with a certain property, but even this requires a set, e.g. given a set X we can form the set {x in X : some property of x holds} (often the "in X" part is left out when the set X is obvious or implicit). Thus we can't write something like: {x: x is an ordinal} because there is no set from which x is to be chosen. |
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Title: Re: 0.999... Post by Icarus on Mar 22nd, 2007, 6:03pm In ZFC, this is true, as ZFC does not consider the existence of collections other than sets. There are other Set theories that do, however. Perhaps the simplest example would be NBG, or Von Neumann-Bernays-Godel set theory (http://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Bernays%E2%80%93G%C3%B6del_set_theory). This is a conservative extension of ZFC that allows proper classes which cannot themselves be part of any collection. Such classes are actual objects of the theory, not just logical formulas. In NBG, {x : x is an ordinal} makes perfect sense. It is just a proper class, not a set. Most of the time when mathematicians deal with things like surreals and ordinals and categories that cannot be contained within a set, they will use NBG as their set theory. But in many ways, it has uncomfortable limitations. You cannot build constructions on proper classes, because they cannot be the element of anything. So you cannot speak of "the topology of the surreals" like you can "the topology of the reals", because open subsets of the surreals are all proper classes, and so by NBG cannot be collected together in a larger object, the topology itself. But NBG is not the only alternative. For example, Russell and Whitehead built an entire pantheon of higher and higher tiers of collections, each of which cannot be the elements of lower tier collections, can be gathered in collections of the same tier in most, but not all, useful fashions (powers, unions, etc), and can always be gathered in higher tier collections. (If that last sentence was confusing to you, then you have a true taste of RW - an extremely influential set of books, but convoluted to an extraordinary level. I'm not sure that anyone other than Russell and Whitehead themselves ever truly understood it.) A much more modest approach gives us all we need for just about anything people want to do: use a Grothendieck Universe. Start with ZFC (or NBG) and add the following axiom: There is a class U which 1) contains the empty set and all "urelements" (i.e., anything that isn't a collection). 2) for x, y http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif U, {x, y} http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif U. 3) for all indexed collections {Sx}xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gifA with A http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gifU, {Sx} http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subseteq.gif U, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bigcup.gifxhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gifA Sx http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif U. 4) for all A http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gifU, A http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subseteq.gif U 5) for all A http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gifU, P(A) = {x : x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subseteq.gifA} http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gifU. 6) U contains an infinite collection. The set U is called a Gronthendieck universe (actually 1-5 make it that, the infinite element is necessary for much of mathematics, however). You can restrict the word "Set" to mean an element of U (and not an urelement, if your theory has them), and the word "class" to mean things not in U. At first, this seems like we are restricting our sets to just a portion of what we should have under ZFC. This is not so, though. If we throw out everything not in U (which includes U itself, because ZFC does not allow collections to contain themselves), what we have left is the complete ZFC theory!. Every axiom of ZFC is satisfied by the elements of U alone, without anything more. Further, The elements of U do not add anything to ZFC. If you can prove something about the elements of U that does not involve U itself, then you can prove the same thing in ZFC. Effectively, U is the universe of ZFC, or of NBG. But unlike NBG, you can construct collections that include U as a member. Rather than thinking of U as a restriction of ZFC, the various classes built on U should be considered additions to ZFC. All of the various definitions made with sets also apply to classes in ZFC + U. The Surreals, ordinals, and cardinals are all actual classes, and you can build upon them to create more classes. The topology of the Surreals is a well-defined object as is its dedekind completion, and the topology of its dedekind completion (an idea impossible to express in NBG). Yet the "class" {A : A http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/notin.gif A} does not exist, any more than it does in ZFC. The downside of ZFC + U is that sethood is not a matter of size, as we are used to thinking. Even the singleton {U} is a proper class, not a set. One other thing: since we can make these constructions, the temptation is to extend the surreals, ordinals, and other inductive constructions outside of U to classes. This is possible, but it isn't profitable. Since all our classes can be considered ZFC sets (after all, our theory is just ZFC with a special "set" singled out), and since every ZFC set can be considered an element of U, everything we can do by extending Surreals et al to classes, we've already can do within U. |
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Title: Re: 0.999... Post by kiochi on Mar 27th, 2007, 11:50am Ah yes, I remember reading a set theory book once that used NBG. I don't recall exactly, but it seemed to me that it was just an attempt to expose what people who use ZFC already know: you can form collections, these are called classes. Some classes are called sets. Then you can formally talk about the distinction between sets and classes; still, in ZFC you can talk about classes as well, like the universe V, or the class of all ordinals, and these distinctions are still formal. I guess I just don't see what the advantage of NBG is. It would still be impossible to have the same notion of size for classes and sets, so you could develop a new notion of size for classes, but this you could do equally well within ZFC as NBG. What's a urelement? If I understand what you are saying, a urelement is "not a collection," it contains nothing. If we're in ZFC (or NBG, I think), then, extensionality says that every "urelement" is 0, but perhaps you aren't using that kind of an axiom? This U is also new to me, probably becasue I haven't read much about this stuff, but it seems to me to be a perfectly reasonable universe for set theory. You called U a set, though, and for U to be a model of the theory of ZFC, it would have to "think" that it's universe, U, is not a set, right? How does U have to be? It seems to me that U could be very large, e.g. U=V seems reasonable, but then I think U=V_i for some limit ordinal i might also work (in which case, if we took the usual universe V and ZFC, we would say U is a set in the usual model of ZFC). You say a "downside to ZFC + U is that sethood is not a matter of size." I'm really not sure what you mean by this. What is ZFC + U? Are you just taking a model (U, \in) of ZFC or is U some axiom involving U? I'm not sure what you mean by "sethood" being a "matter of size," but if we have two models of the same theory, then sethood in either of them will be the same (up to isomorphism). You seem to be astounded that this U can be a model of the entire theory of ZFC. It reminds me of a logic book I read in which they proved using things called skolem functions that there exists a countable model of ZFC. In this book they called it a "paradox," but I didn't think it was. Sure, this structure's universe is only countable, but it still "thinks" it has uncountable elements, which is perfectly fine, it's just that, within the structure, you can't find any bijection from the "countable" elements to the "uncountable" ones (here the quotes indicate that the terms are relativized to the countable model). |
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Title: Re: 0.999... Post by Icarus on Mar 27th, 2007, 5:56pm A urelement is any object in a set theory that is not a set or class or other sort of collection. For instance, in the old "chinese proverb" when you see a horse and a cow, you are actually seeing three things: a horse, a cow, and the set consisting of the horse and the cow, the horse and cow are urelements of the set theory. (Of course, mathematicians count three more things that the proverb misses...) ZFC does not contain urelements. Nor does NBG or most other set theories. However, you can add axioms and symbols to all of these that would constitute the introduction of urelements. The only change you would need to make is that the axiom of extent only applies to collections (a restriction not needed in the unaltered theory, as everything is a collection there). In fact, the axiom of extent essentially is the definition of the concept of a "collection" - specifying that its identity is completely determined by its elements. Urelements would have their own axioms specifying when you can claim they are equal. If you choose to add a Grothendieck universe to a set theory that has urelements, then you want all urelements to be a part of your universe. You cannot set U = V in ZFC, as there is no such object as V in ZFC (as you yourself have said, the universe of ZFC and all proper classes are not actual objects in it, but formal expressions that actually represent logical properties). The whole point of U is that it is an actual object that can be used in other collections and constructions. V does exist as an actual object in NBG. This is one of the reasons this theory was developed: to have a theory where the things you are pretending are allowable in ZFC when you talk about proper classes actually are allowable. Another advantage of NBG is that is "finitely axiomable". Most set theories, including ZFC, rely upon metamathematical expressions, called axiomatic schema, that are essentially rules that allow you to create an infinite number of actual axioms within the theory itself. NGB, however, can be completely expressed with a finite number of axioms within the theory itself. However, even though the total universe V exists within NGB, it would be pointless to set U = V. Again, the point of U is to be able to construct with it, as it can be an element of collections. In NBG, proper classes, including V, cannot occur on the left side of the http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif sign. For Grothendieck universes, you need more than just a limit ordinal, what you need is a strongly inaccessible cardinal (http://en.wikipedia.org/wiki/Strongly_inaccessible_cardinal). This a cardinal http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif such that for all http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif, 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supbeta.gif < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif, and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bigcup.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subgamma.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sublt.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subbeta.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif. The existance of a Grothendieck universe is equivalent to the existance of a strongly inaccessible cardinal (neither can be proven or disproven within the confines of ZFC - which is why a Grothendieck universe must be introduced with an axiom). Yes, when I made the previous post, I accidentally called U a set. I noticed this a couple days ago and changed it to "class", because I want "set" to mean "a collection which is an element of U". In ZFC + U (by which I thought it would be obvious from context means "ZFC with a Grothendieck Universe"), the objects are all "classes", while "set" means a class in U. By this definition, U itself is not a set (in ZFC, no class can be an element of itself). Concerning size and sethood: In ZFC, any property that is satisfied by only a small number of elements forms a set. In NBG (at least in the version with the axiom of choice), this concept is more explicit: every property determines a class. That class is a set if and only if its size is strictly less than that of V, the set universe. This concept does not hold true in ZFC+U. U cannot be a set. But if A http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif U, then A http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subset.gif U as well. Therefore {U} is not a set, even though it has only one element. If it were a set, then we would have U http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif U, which is impossible (the axiom of regularity prevents it). I am not "astounded" that U alone is a model for all of ZFC. (At one time, I would have been, but I have seen this sort of thing far too often to be amazed by it anymore.) My comments about this had a different purpose entirely. Effectively, what U does is chose a particular set in ZFC and limit all sets to it. If I define ordinals or surreals in ZFC, then I define them for all sizes. Nothing stops me until I run into the immovable barrier of logical consistency. However, if I define them in ZFC+U, I choose a particular collection U in ZFC's universe of discourse, and limit my construction to keep it inside U. The natural supposition here is that I don't get the "full surreals" or "full ordinals" this way. I could continue my constructions beyond U and get more ordinals, more surreals. Hence I lose out on these larger objects by limiting myself to U. What I was trying to show is that this supposition is false. Since U is a model for ZFC, everything I can construct in ZFC+U that does not depend intrinsically on U itself, I can also construct inside of U. My point here is I don't have to worry about missing anything by restricting myself to U. Instead of considering U to be just a portion of ZFC, I can consider U to be all of ZFC, and everything outside of U to be added on to ZFC. |
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Title: Re: 0.999... Post by kiochi on Mar 28th, 2007, 10:18am Thanks, That was really interesting to read and it helped me understand to a large extent what you are talking about. It makes sense to me that you would need a strong inaccesible cardinal to have a set like your U, because you don't want to be able to "get out of U" using your sets in U, but why doesn't omega work? For every finite n<omega certainly 2^n<omega, and so on. And finally one last question, because I am still a little confused about this ZFC+U. The advantage of U seems to be that it's a set in ZFC (that is, maybe, if we assume some large cardinal exists). But then when you're in ZFC+U you say you can sort of restrict sethood to mean "in U" and then do ZFC entirely within U (or am I totally mising the point?). But then U ceases to be a set, at least as far as the this new model is concerned. So what's going on here? Do you have two models of ZFC at the same time, one relativized so that U is the universe? I will stop speculating because I know you will explain it clearly.. Thanks |
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Title: Re: 0.999... Post by Icarus on Mar 28th, 2007, 7:17pm Well, I'll try to explain it clearly, but given that I've failed twice now... :P First of all, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif is a strongly inaccessible cardinal, and it does work for the "standard definition" of Grothendieck universe, which does not include my axiom (6). But I want U to be a model for ZFC in its own right. Since ZFC includes the axiom of infinity, U has to contain infinite sets for it to work. For this reason, I added axiom (6): U contains an infinite set. We add U to ZFC by adding the axioms (1)-(6) to the axioms of ZFC. This is equivalent to assuming the existance of a strongly inaccessible cardinal > http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif. (i.e., adding an axiom that says such a cardinal exists). What I am doing with U is that I use it break the sets in ZFC into two groups: those sets within U, I still call "sets". Those sets outside U, I no longer refer to as "sets", but only as "classes". Then I argue that this restriction of sethood actually does not cost me anything, as U alone is a model for the full universe of ZFC. The point is to establish a theory of classes that allows me to have classes contained within other classes. To see the utility, compare the three theories:
For example, consider the order topology of the Surreals. Every interval in the surreals contains a subclass that is order-isomorphic to the ordinals. Therefore every non-empty open class of surreals is not a set.
I hope that gives you an idea of why one would want to use a Grothendieck universe. It makes dealing with collections too large to be sets no harder than dealing with sets themselves. If I use a Grothendieck universe, I have no need to worry about the difference between "fields" and "big fields" - the latter being a field that is a class instead of a set. If I don't use ZFC+U, or some other set theory that allows building on classes, then I have to define the concept of a "big field" somewhat differently than that of a field (the basic concepts are the same - but the formalism has to be adapted to not refer to sets). Of all the set theories that allow you to build on classes, ZFC+U strikes me as by far the simplest. Yet it is more than sufficient for anything I'm likely to need. |
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Title: Re: 0.999... Post by Product on Apr 26th, 2007, 12:37pm I will start off this post admitting that much of the math and mathematical theory in this thread is above me. I post not to debate or offer my own solution, but to have the flaws in my thinking pointed out to me. I've always stayed out of the whole debate because I'm not a number theory wizard. I do teach elementary school though, and while teaching a little about probability the other day, a thought occurred to me. If you flip a fair, two headed coin where the odds are exactly 1:1 that it will land heads, and you plan to flip it an infinite number of times, the probability that it will land heads at least once is .999~ but it is not 1 because the coin never has to come up heads. Theoretically it could land tails every time out into infinity, regardless of the realistically non existant chance that it would even happen out to 100. So that's the thought I had. I'm sure there are flaws and I just want to see where my reasoning is wrong. Is it that probability can't accurately be measured in decimals? Am I comparing wrong types of numbers? My best guess is that the .999~ probability comes from 1-1/(infinity), and you can't divide by infinity, so the whole idea of .999~ being a probability is silly anyway. Thanks for any discussion or answers. |
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Title: Re: 0.999... Post by Sir Col on Apr 26th, 2007, 3:33pm I'll have a go at addressing your query. It seems that the problem comes down to the difference between infinite and finite. To make a better comparison with 0.999..., consider the following experiment. A bag contains nine red discs and one blue disc. A disc is taken at random and its colour is noted. The disc is returned and the experiment is repeated N times. Therefore P(R=0) = 1/10N. Hence P(R>0) = 1-1/10N. N=1: P(R>0) = 0.9 N=2: P(R>0) = 0.99 N=3: P(R>0) = 0.999 et cetera. Now for any finite number of times the experiment is repeated, N, the resulting probability will be zero point followed by a finite string of N nines. If the experiment is repeated forever, then we move from a finite set of possible outcomes: 0 reds, 1 red, 2 reds, 3 reds, ..., N reds, to an infinite (continuous) set; there is nothing in-between. So the probability of getting no reds is zero by definition. In fact, P(R=0) = P(R=1) = P(R=2) = ... = P(R=N) = 0; for any continuous data, P(R=r) = 0. Hence P(R>0) = P(R>=0) = 1. The problem is that we try to extend the finite cases as N increases to the realm of the infinite as if the infinite case is the next one after the "last" finite case. The reality in the context of probability is that we are dealing with an either/or situation. We have a finite number of cases (discrete), in which case there is a finite string of nines; or we have the infinite case (continuous), in which case we have unity (or zero). |
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Title: Re: 0.999... Post by Icarus on Apr 26th, 2007, 4:18pm An insightful question, but what it points out is the limitations of the concept of probability, and not of the fact that 0.999~ = 1. What you have to understand is that when there are an infinite number of possible outcomes, saying that a particular outcome has probability 0 does not mean that the outcome cannot occur. And saying that the outcome ("event", in the language of probability theory) has probability 1 does not guarantee that the event will occur. For example: consider the case of picking a real number between 0 and 1 at random (i.e., with uniform probability). For any two numbers a and b with 0 <= a < b <=1, the probability that the number x picked lies between a and b (a <= x <= b), is b - a. So if I choose any particular number N and ask what is the probability that N will be picked, the answer must be exactly 0: If P(N) > 0, then I could note that N lies between (N-P(N)/4) and (N+P(N)/4), but the probability that the number picked is between (N-P(N)/4) and (N+P(N)/4) is only P(N)/2, which cannot be. Therefore we must have P(N) = 0. This holds for every number N between 0 and 1. Each individually has exactly 0 probability of being picked. So every time a pick is made, it was made despite the probability being 0. Note that the argument does not allow for P(N) to be some number "just slightly above zero" in any fashion at all. No matter how close P(N) is to 0, if it is not actually 0, it leads to the conclusion that N is more likely to be picked itself than any number is out of a set to which N belongs. [edit] This is what happens when you take a break in the middle of composing a post! Someone slips in ahead of you![/edit] |
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Title: Re: 0.999... Post by Grimbal on Apr 27th, 2007, 1:30am This reminds me of the following paradox. Ask what proportion of all integers can be written in decimal without the digit 3? For up to n-digit numbers, the proportion is is (9/10)n. At the limit for n->inf, it is 0. The conclusion is that 0% of all integers can be written without the digit 3. Does it mean there are none? Of course not, I can think of plenty of them. Does it mean, then, that the limit isn't really 0? No, limn->inf (0.9)n is exactly 0. The problem comes from the fact that you cannot compare the size of infinite sets in this way. |
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Title: Re: 0.999... Post by TruthlessHero on Jun 8th, 2007, 8:15am Is there any way you could edit the original posts so they didn't have the smiley stuff? Not sure if since it's so old that it wouldn't work? |
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Title: Re: 0.999... Post by towr on Jun 8th, 2007, 1:45pm on 06/08/07 at 08:15:31, TruthlessHero wrote:
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Title: Re: 0.999... Post by javacodeman on Jul 6th, 2007, 11:08am Mentor a guest wrote in the original Thread (#2) Quote:
I had to write that this is incorrect as infinity is not a number to be used that way. It is correct to say, lim (x -> infinity) 1/x = 0. while, lim (x -> +0) 1/x = infinity lim (x -> -0) 1/x = -infinity java |
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Title: Re: 0.999... Post by srn347 on Aug 29th, 2007, 12:38pm Four words: geometric series. |
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Title: Re: 0.999... Post by CHIMELA on Nov 22nd, 2007, 5:34pm I remember our teacher told our class about how you take an infinite amount of half-steps whenever you walk to any given place. For example, if you walk towards a stool, first you must walk half the distance to that stool. But you must walk half the distance of that half. And the half of that half. Thus, you walk an infinite amount of half-steps. I've been wondering something, though. (And keep in mind I'm still in secondary education.) If 1=0.999... and you multiply both sides by 2, would you get 2=1.999...8, or is that not allowed because you can't have an infinite amount of decimals and then have a finite number at the end, so you'd have to simplify 0.999... to 1 first? |
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Title: Re: 0.999... Post by ThudanBlunder on Nov 22nd, 2007, 5:49pm on 11/22/07 at 17:34:11, CHIMELA wrote:
Yes, see this (http://mathworld.wolfram.com/ZenosParadoxes.html)link. on 11/22/07 at 17:34:11, CHIMELA wrote:
That is correct. |
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Title: Re: 0.999... Post by SMQ on Mar 28th, 2012, 6:41am The ever-amazing Vi Hart (http://www.youtube.com/user/Vihart) weighs in: youtube.com/watch?v=TINfzxSnnIE (http://www.youtube.com/watch?v=TINfzxSnnIE) --SMQ |
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Title: Re: 0.999... Post by steamypi on Aug 5th, 2012, 1:59pm What I would say is: First think of this separately, 1/3 = .333333(infinite # of 3's) OK, 1/3 + 1/3 + 1/3 =3/3 3/3 = 1 RIGHT? YES SO THEN, 1/3 = .333333(infinite # of 3's) 1/3 = .333333(infinite # of 3's) +1/3 = .333333(infinite # of 3's) ______ 3/3 = 1 So when you say .3333(infinite # of 3's) +.3333(infinite # of 3's) +.3333(infinite # of 3's)=.999999(infinite # of 9's), that's not true because it really equals 1 instead of 999999(infinite # of 9's). |
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Title: Re: 0.999... Post by steamypi on Aug 5th, 2012, 2:45pm And what you must remember is that .333333(infinite # of 3's) is actually 1/3, although many people treat .333333(infinite # of 3's) as something you can add like .333+.333+.333=.999. Using these numbers that continue infinitely is like rounding a number. ex. Say you add like this: Here's the equation, .7845 +.4839 _______ But these numbers are too long for you so you round. .7845~.785 and .4839~.484 Now we have: .785 +.484 ________ Add it up and you get 1.269 But when you round like that, the solution is not the actual solution, nor are the numbers that were added together. The actual solution, using the original numbers, is 1.2684 So when you say that .999(infinite # of 9's) is equal to 1, it is like saying, using the example, that 1.269 equals 1.2684. |
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Title: Re: 0.999... Post by towr on Aug 5th, 2012, 10:09pm on 08/05/12 at 14:45:25, steamypi wrote:
Also 0.999... = sum i=1..inf 9*10-i = 3 * sum i=1..inf 3*10-i = 3 * 0.333... So it 3 * 1/3 is 0.999... and it is equal to 1. 0.999... = sum i=1..inf 9*10-i = lim i->inf 1 -1/10i = 1, or actually read the thread to find a hundred other ways to conclude the same simple truth. |
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Title: Re: 0.999... Post by marsh8472 on Sep 15th, 2012, 1:05am There's a good chance this has already been brought up but there's an axiom called the Archimedean property for real numbers which states there's no such thing as an infinitesimal number or infinitely large number. Arguing that 0.99999... < 1 violates that property. I'm sure it's possible to use a different system of math that assumes the Archimedean property does not apply to real numbers though. But this is just an argument in circles otherwise. |
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Title: Re: 0.999... Post by Edward_E on Nov 14th, 2012, 3:59pm I like thinking of this problem a little differently. 1 / 1111 = 0.000900090009... 1 / 111 = 0.009009009... 1 / 11 = 0.09090909... Based on this pattern one would assume that. 1/1 = 0.999... So I believe that. 1 = 0.999... This inverse of numbers can be found also in 1/2222 = 0.0004500450045... 1/222 = 0.0045045045... 1/22 = 0.045454545... 1/2 = 0.4999... = 0.5 |
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Title: Re: 0.999... Post by Pavel on Feb 25th, 2013, 10:41pm amuse we have the secession of real numbers, S = {a(i) such that a(1)=0.9 and a(n+1)=a(n)+9*10^-(n+1):n=1,2,3,...} So for every a(i) of S there must exist some number number [Epsilon](a(n))>0 such that for any n=1,2,3,... 1-[Epsilon]=a(n) thus 1=a(n)+[Epsilon] since [Epsilon] is a positive real number 1>a(n) and 1>a(i) for all a(i) belonging to S therefore .999...<1 ..... thoughts: .... .999..... is a concept not a number and its unfair to compare it to the number 1 and not the concept of 1 (thus most proofs above). the discussion i believe is not about limits or convergence but one of a number that looks like this: 0.9999999999999999999999999999 but there is always one bigger but with the trait that it MUST start 0.9..... , and so never quiet 1. the series S contains infinitely many elements and it does converge to 1 but there isn't a single number in there like 1. |
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Title: Re: 0.999... Post by Grimbal on Feb 26th, 2013, 1:22am on 02/25/13 at 22:41:33, Pavel wrote:
Epsilon is not a positive real number. In this case, Epsilon is a function of a(n). You should write Epsilon[n]. on 02/25/13 at 22:41:33, Pavel wrote:
You cannot draw that conclusion. You can have a(i)<1 and lim a(i)=1. on 02/25/13 at 22:41:33, Pavel wrote:
I disagree. 0.999... is a number. Its value is sum k=1,2,... (9*10-k). |
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Title: Re: 0.999... Post by rmsgrey on Feb 26th, 2013, 4:11am on 02/25/13 at 22:41:33, Pavel wrote:
We also have: .999...>a(i) for all (finite) i therefore? .999...<.999... ??? Something is wrong with the logic there. |
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Title: Re: 0.999... Post by peoplepower on Feb 26th, 2013, 4:44am Something is wrong with the logic elsewhere too. on 02/25/13 at 22:41:33, Pavel wrote:
There are multiple meanings that can be and often are attached to the single symbol 1. The meanings of 1 as a natural number, rational number, and real number are all different even though in some sense it has the same value, which is grounds for confusion. When we are asked to compare the real number 0.999... with 1, we need to ask which meaning fits best from the context we are given. Of course, one is to choose the real number 1. Quote:
The assumption inherent in the problem is that we choose to work in some order-completion the rational numbers (like the real numbers). Thus, by definition really, the supremum of S exists taking the value 0.999... Quote:
We are working in an ordered field. Likeness is based on distance rather than some properties of the decimal representation of the number. |
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Title: Re: 0.999... Post by riddler358 on May 4th, 2016, 11:15pm note: i didn't read all of the answers if we agree that 0,(3) = 1/3 and we agree that 3 * 0,(3) = 0,(9) we substitute and get 0,(9) = 3 * 1/3 then we probably should conclude that 0,(9) = 1 |
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Title: Re: 0.999... Post by Grimbal on May 25th, 2016, 8:34am And that's perfectly correct. The problem most people face, I think, is that they have the intuition that if two numbers are written differently, then they must be different. |
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Title: Re: 0.999... Post by towr on May 25th, 2016, 8:53am one != 1 ;) |
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Title: Re: 0.999... Post by Grimbal on Sep 28th, 2016, 8:39am one != 1 by commutativity: 1=one ! |
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Title: Re: 0.999... Post by towr on Sep 28th, 2016, 8:51am I wonder if there's a programming language with ! as factorial operator and where it binds stronger than than unequal (or just doesn't have != as unequal operator, and uses = as equality operator) $: 1!=1 > True |
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