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Topic: 0.999. (Read 124728 times) 

Icarus
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Re: 0.999.
« Reply #225 on: Jun 11^{th}, 2003, 3:58pm » 

Hyperreals are an extension of the real numbers. That is, they form a set ^{*}R containing the real numbers as both an algebraic and a topological subspace. So every algebraic action on real numbers in the hyperreals provides the same outcome as in the reals, and every limit of realvalued functions that converges in the real numbers also converges in the hyperreals to the same limit. In particular, 0.999... = [sum]_{k=1}^{[supinfty]} 9*10^{k} = 1 in the hyperreals, just as it does in the reals.

« Last Edit: Aug 19^{th}, 2003, 6:28pm by Icarus » 
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Sir Col
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Re: 0.999.
« Reply #226 on: Jun 12^{th}, 2003, 4:57pm » 

Okay, I'm still reading the book, Elementary Calculus: An Infinitesimal Approach, but I'm getting happier with the idea of hyperreals. I must say, though, that the jury is still out on 0.999... being a real number. Is 0.999...=1e (where e is a nonzero infinitesimal)? In other words, is it infinitely close to 1, but not equal to 1? I am still getting my head around a rigorous approach (via infinitesimals) to the infinite sum tending towards a real number limit. I accept the proofs I've read that infinite partial sums of a converging geometric series are infinitely close to the limit, but is it equal? If the difference between the infinite sum and the limit is an infinitesimal and the infinitesimal is nonzero... More reading I thinks!


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Icarus
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Re: 0.999.
« Reply #227 on: Jun 12^{th}, 2003, 7:20pm » 

I admit my knowledge of hyperreals (^{*}[bbr]) has no great depth, but I have understood that they are an extension of the reals ([bbr])  both algebraically, and topologically. This means that [bbr] [subset] ^{*}[bbr], and when all the algebraic operations (+, , *, /) defined on ^{*}[bbr] are restricted to [bbr], they are the standard operations on real numbers. (Ie. if [smiley=a.gif] + [smiley=b.gif] = [smiley=c.gif] in [bbr], then [smiley=a.gif] + [smiley=b.gif] = [smiley=c.gif] in ^{*}[bbr] as well, etc). It also means that if, as a real function , [smiley=f.gif]([smiley=x.gif]) [to] [smiley=a.gif] [in] [bbr] as [smiley=x.gif] [to] [smiley=y.gif], then [smiley=f.gif]([smiley=x.gif]) [to] [smiley=a.gif] as [smiley=x.gif] [to] [smiley=y.gif] in ^{*}[bbr] as well. If these are true, then the answer to your question is that any limit of real numbers has a real value, not an infinitesimal one. To have infinitesimal limits, you would need infinitesimal values in the thing you are taking the limit of. I stand by my statement that 0.999... = 1 in the Hyperreals as well as in the reals. If it didn't, then quite frankly the hyperreals would be worthless for the study of problems in real (not hyperreal) analysis. This does not appear to be the case. But even if I am wrong, and the hyperreals induce a different topology on the Reals (the algebraic extension is immediate from the definition, so only the topology could be different), it still does not matter to the point of this thread. 0.999... is defined in the real numbers, according to the standard topology of the real numbers. The definition is 0.999... = Lim_{n[to][infty]} [sum][supn]_{k=1} 9[cdot]10^{k} All the algebraic operations are as defined in the reals. The limit is also as defined for real  not hyperreal numbers. That limit exists. It is unique, and it is 1. If it turns out that hyperreals induce a different topology on real numbers, so that the limit above converges to something else, it would not matter. It would only mean that in the hyperreals 0.999... would have a different definition than it does in the reals. The definition for REAL numbers, which is the only one of any concern here, is still = 1. I have finally been able to track down the topology on ^{*}[bbr]. Because ^{*}[bbr] is an ordered set (the order being an extension of the ordering of [bbr]), their "natural" topology is the "order topology"  which induces the order topology on their subset [bbr]. But the order topology on [bbr] is the standard topology, by which the limit above is taken. [therefore] What I said above was true: 0.999...=1 in the Hyperreals, as well as the Reals. A few notes on Topology: ([forall] = "for all"; [exists] = "there exists") A topology on a set [smiley=cx.gif] is a collection [calt] of subsets of [smiley=cx.gif] such that: (1) Both [emptyset], [smiley=cx.gif] [in] [calt]. (2) [forall] [smiley=ca.gif], [smiley=cb.gif] [in] [calt], [smiley=ca.gif][cap][smiley=cb.gif] [in] [calt]. (3) [forall] [calf] [subseteq] [calt], [cup][calf] = {[smiley=x.gif]  [exists] [smiley=ca.gif] [in] [calf] with [smiley=x.gif] [in] [smiley=ca.gif]} [in] [calt]. The sets in the topology are called "open". A set [smiley=ca.gif] is "closed" if its complement [smiley=cx.gif] [smiley=smallsetminus.gif] [smiley=ca.gif] is open. In any topology, the empty set and the whole space are both open and closed. Every set [smiley=cx.gif] has two simple topologies: The "trivial" topology: [calt] = {[emptyset], [smiley=cx.gif]} (only the empty set and [smiley=cx.gif] are open or closed), and the "discrete" topology: [calt] = { [smiley=ca.gif] [smiley=ca.gif] [subseteq] [smiley=cx.gif] } (every set is both open and closed). The standard topology of [bbr] can be defined by: [smiley=ca.gif] is open [bigleftrightarrow] [forall] [smiley=x.gif] [in] [smiley=ca.gif], [exists] [smiley=r.gif][smiley=subx.gif] > 0 such that [smiley=y.gif][smiley=x.gif] < [smiley=r.gif][smiley=subx.gif] [bigto] [smiley=y.gif] [in] [smiley=ca.gif]. (In other words, a set is open if for each of its points, the minimum distance to anything outside of the set is greater than zero.). Its not hard to see that an interval is open by this definition if and only if it is an "open interval". Any ordered set [smiley=cx.gif] can be given the "order topology": a set [smiley=ca.gif] is open [bigleftrightarrow] [forall] [smiley=x.gif] [in] [smiley=ca.gif], [exists] [smiley=a.gif], [smiley=b.gif] [in] [smiley=cx.gif] with [smiley=a.gif] < [smiley=x.gif] < [smiley=b.gif] such that ([smiley=a.gif], [smiley=b.gif]) [subseteq] [smiley=ca.gif] (ie, every [smiley=x.gif] [in] [smiley=ca.gif] lies within an open interval contained in [smiley=ca.gif].) The order topology on [bbr] turns out to be the same as the "metric topology" I defined above. If [smiley=cx.gif] and [smiley=cy.gif] are topological spaces (sets with specified topologies), and [smiley=f.gif]: [smiley=cx.gif] [to] [smiley=cy.gif] (that is, [smiley=f.gif] is a map, or function, from [smiley=cx.gif] into [smiley=cy.gif]) and if [smiley=a.gif] [in] [smiley=cx.gif], then the "limit of [smiley=f.gif] as [smiley=x.gif] approaches [smiley=a.gif]" is defined to be an element [smiley=b.gif] [in] [smiley=cy.gif] satisfying: For every open set [smiley=cb.gif] [subseteq] [smiley=cy.gif] containing [smiley=b.gif], there is an open set [smiley=ca.gif] [subseteq] [smiley=cx.gif] containing [smiley=a.gif] such that, if [smiley=x.gif] [in] [smiley=ca.gif] and [smiley=x.gif] [ne] [smiley=a.gif], then [smiley=f.gif]([smiley=x.gif]) [in] [smiley=cb.gif]. The large majority of the various types of limits encountered in mathematics can be considered special cases of this definition (you just have to find the right topologies on the right sets). Last, if [smiley=cx.gif] is a topological space and [smiley=cy.gif] [subseteq] [smiley=cx.gif], then the "subspace" topology on [smiley=cy.gif] is defined by declaring [smiley=ca.gif] [subseteq] [smiley=cy.gif] open [bigleftrightarrow] if [exists] [smiley=cb.gif] open in [smiley=cx.gif] such that [smiley=ca.gif] = [smiley=ca.gif] [cap] [smiley=cy.gif]. It is not hard to show that if [smiley=f.gif]: [smiley=cz.gif] [to] [smiley=cy.gif], and if [smiley=f.gif]([smiley=z.gif]) [to] [smiley=y.gif] [in] [smiley=cy.gif] as [smiley=z.gif] [to] [smiley=a.gif] [in] [smiley=cz.gif] according to the subspace topology on [smiley=cy.gif], then [smiley=f.gif]([smiley=z.gif]) [to] [smiley=y.gif] in [smiley=cx.gif] as well. Applying this last result to the real numbers [bbr], a topological supspace of the hyperreals ^{*}[bbr], is how I can be sure that 0.999... = 1 in the hyperreals as well as in the reals.

« Last Edit: Aug 19^{th}, 2003, 6:43pm by Icarus » 
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otter
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Re: 0.999.
« Reply #228 on: Jul 14^{th}, 2003, 5:31am » 

The Straight Dope has weighed in on the original subject matter. You can check out Cecil's answer at http://www.straightdope.com/columns/030711.html.

« Last Edit: Jul 14^{th}, 2003, 5:32am by otter » 
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wowbagger
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Re: 0.999.
« Reply #229 on: Jul 14^{th}, 2003, 6:18am » 

Following the link in otter's post, we read the following: Quote:We readily concede that .999~ gets infinitely close to 1to put it in mathematical terms, 1 is the sum of the converging infinite series .9 + .09 + .009 + . . . But, we protest, .999~ never quite reaches that limit. If at any step we halt the progression to infinity to take a sum, we find that we remain separated from 1 by some infinitesimal amount. 
 Now, this is at least misleading. Of course the finite sum (halting the progression to infinity) always differs from 1 by a finite amount, not an infinitesimal one  using the word in its mathematical sense. There seems to be a nonmathematical meaning as well (tiny, miniscule), but when discussing mathematical topics, one should avoid such ambiguities. Thanks a lot, otter, for having introduced the Straight Dope to me (in different thread), by the way!

« Last Edit: Jul 14^{th}, 2003, 6:19am by wowbagger » 
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S. Owen
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Re: 0.999.
« Reply #230 on: Jul 14^{th}, 2003, 8:56am » 

Indeed  ".999~ never quite reaches that limit" .999~ is that limit!


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Chewdog
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Re: 0.999.
« Reply #231 on: Jul 15^{th}, 2003, 5:13pm » 

All this talk about hyper reals are way out of my league but i remain convinced that .999.... = 1 becuase: On any calculator u can enter the following problems and get the following answers: 1/9 = .111...... 2/9 = .222...... 3/9 = .333...... 4/9 = .444...... 5/9 = .555..... and so on so when you get to 9/9 = 1, following the patern 9/9 must also equal .999..... and there fore 1 must = .999....


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wowbagger
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Re: 0.999.
« Reply #232 on: Jul 16^{th}, 2003, 5:03am » 

on Jul 15^{th}, 2003, 5:13pm, Chewdogscp wrote:i remain convinced that .999.... = 1 becuase: On any calculator u can enter the following problems and get the following answers: 
 Sorry, but I don't think that's a good argument. Calculators and computers are known to produce nonsensical results sometimes. Therefore, they should not be trusted blindly (I don't assume that you do) if their results are counterintuitive or contradict common sense. This is not an advertisement to heavily use common sense in mathematics, but to use critical thinking and strict logic. After all, this is one of the rare occasions where mathematicians can prove helpful!


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Icarus
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Re: 0.999.
« Reply #233 on: Jul 17^{th}, 2003, 5:16pm » 

"Proof by calculator" is seldom good. But the basics of one proof are to be found in the information. It is fairly simple to demonstrate mathematically that 1/9 = 0.111...
 9*(0.111...) = 0.999...
 9*(1/9) = 1
From which it follows (again) that 1 = 0.999...


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aero_guy
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Re: 0.999.
« Reply #234 on: Jul 22^{nd}, 2003, 7:04am » 

You guys do realize that EVERY ONE of these arguements has been gone over several times in this thread. There have been some very eloquent and exhaustive refutations of all of the "anti equals" arguements. I suggest taking some time and reading through it.


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Sir Col
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Re: 0.999.
« Reply #235 on: Jul 23^{rd}, 2003, 8:58am » 

How about this one... ? Let x=0.999... 10x=9.999... 10x–x=9x=9.999...–0.999...=9 Therefore x=1. QED In fact, if I present 0.999... of an argument, is that the same as presenting a complete argument? Seriously, I think aero_guy is right. I, like everyone else here, am keen to see a new slant on this old chestnut, but it does get a little tedious when someone represents well known 'textbook proofs'.


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Icarus
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Re: 0.999.
« Reply #236 on: Jul 23^{rd}, 2003, 4:31pm » 

Calm down! Chewdogscp was simply putting his two cents in, explaining how he understood the truth here. Everyone else has had a say, he deserves one too. Wowbagger felt it important (and I agree) to point out the problems with "proof by calculator" when applied to nonterminating decimals. But since he had pointed it out, I felt it also reasonable to explain why Chewdogscp's proof works anyway. None of us suggested in any way that this was new. If you're not interested in a conversation, move on  don't complain that those involved shouldn't be having it!


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Sir Col
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Re: 0.999.
« Reply #237 on: Jul 23^{rd}, 2003, 6:15pm » 

I don't think that anyone was referring especially to Chewdogscp's post, Icarus; it's more the fault of the thread, as it has become so long now that it is not practical to spend time reading through it. That was precisely why I chose (with intended humour) to quote the essence of the first proof presented in the thread. I'm sorry if I sounded critical of Chewdogscp's post – that was not the intention. Anyway, I would suggest that, of all the arguments/proofs, the continued addition of 1/9 (Chewdogscp's method) is perhaps the most appealing to intuitition and is a refreshingly efficient demonstration of the principle.


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mook
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Re: 0.999.
« Reply #238 on: Jul 23^{rd}, 2003, 6:31pm » 

if x/y=z then z*y=x if x=1, y=9, z=0.111.... then 0.111....*9=1 not 0.999....


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Sir Col
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Re: 0.999.
« Reply #239 on: Jul 23^{rd}, 2003, 6:46pm » 

What Icarus was demonstrating was the following: The mechanical process of division demonstrates that the decimal ratio of 1/9 is 0.111.. Now we can see that 0.111...+0.111...=0.222..., or 1/9+1/9=2/9. Similarly 0.222...+0.111...=0.333..., or 2/9+1/9=3/9, and so on. Leading to 0.888...+0.111...=0.999..., or 8/9+1/9=9/9.


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mook
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Re: 0.999.
« Reply #240 on: Jul 23^{rd}, 2003, 6:55pm » 

if 1/9=0.111.... and 9*1/9=1 then 0.111... 0.111... 0.111... 0.111... 0.111... 0.111... 0.111... 0.111... +0.111...  1 not 0.999...


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Icarus
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Re: 0.999.
« Reply #241 on: Jul 23^{rd}, 2003, 7:02pm » 

Mook  did no one ever teach you the rules for adding and multiplying decimals? Follow those rules and guess what! 9*(0.111...) = 0.999... and 0.111... + 0.111... + (7 more copies) = 0.999... But they also equal 1! GEE what a great discovery! 0.999... = 1 !


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wowbagger
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Re: 0.999.
« Reply #242 on: Jul 28^{th}, 2003, 11:55am » 

on Jul 23^{rd}, 2003, 6:15pm, Sir Col wrote:I don't think that anyone was referring especially to Chewdogscp's post, Icarus; 
 So why do you think I quoted that post?


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Sir Col
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Re: 0.999.
« Reply #243 on: Aug 1^{st}, 2003, 2:02pm » 

You're a bad man, wowbagger! Hey, here's an interesting question... 0.9+0.9=1.8 0.99+0.99=1.98 0.999+0.999=1.998 That is, the result is always two 'parts' below 2. So what does 0.999...+0.999... equal?


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Icarus
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Re: 0.999.
« Reply #244 on: Aug 1^{st}, 2003, 2:13pm » 

Ouch! After all you've said about wanting something new, you throw this out  in this thread where we all know what someone's going to make of it. And you call Wowbagger bad!


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Sir Col
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Re: 0.999.
« Reply #245 on: Aug 3^{rd}, 2003, 4:39pm » 

I was just thinking about the mechanical process of division and, in particular, the similarity with 1/9 and 0.999.../9. It led to this interesting observation: 1=0.^{1}0 (in other words, 0 units and 10 tenths) Therefore, 1=0.9^{1}0 (9 tenths and 10 hundredths) Continuing this process, 1=0.99^{1}0=..., leading to the result 1=0.999...


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Icarus
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Re: 0.999.
« Reply #246 on: Aug 3^{rd}, 2003, 6:08pm » 

Nice. I doubt it would convince anyone who rejects some of the other demonstrations above, as they will demand that it shows that 1 = 0.999... + 10 [infty]^{th}s, or something like that. But it's still very nice. (For those of you who are asking "well, why doesn't it show that 1 = 0.999... + 10 [infty]^{th}s" (and anyone who has read through this thread knows that there are some who will ask that), the answer is: because there is no such thing as "infinitieths". If N is allowed to become infinitely large in 1/10^{N}, then the value becomes infinitely small. There is only one infinitely small real number: 0. So indeed 1 = 0.999... + 10 [infty]^{th}s, but the latter term is just 0, which leaves 1 = 0.999...)

« Last Edit: Aug 16^{th}, 2003, 9:40am by Icarus » 
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Re: 0.999.
« Reply #247 on: Aug 4^{th}, 2003, 3:37am » 

Absolutely. I think the problem that people have with understanding the result, 1=0.999..., is trying to make sense of it in a finite way. In exactly the same way that the infinite geometric sum, 0.999...=0.9+0.09+0.009+...=1, demonstrates, the result above holds because the process continues ad infinitum, not towards some finite position.


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Icarus
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Re: 0.999.
« Reply #248 on: Aug 31^{st}, 2003, 10:26pm » 

A compendium of the various correct demonstrations found here. I will provide links only to the first properly stated post showing each method. First though, I would like to address this question: "How is it even possible that two such obviously different things as 0.999... and 1 could be the same?" I have come to the conclusion that much of the confusion on this issue arises from a misconception about the nature of real numbers. Irrational numbers are usually introduced to students as decimal notations, such as 0.1010010001... . Unfortunately, the student often is never moved away from this idea to a more accurate definition. So: Real numbers are not decimals. Perhaps the easiest way conceptually to describe what a real number is would be to say that the positive real numbers consist of all possible distances between two points (in units of a fixed reference distance, called 1). To this set we add in opposites for each member except 0 to get the full set of Reals. Decimal notation is only a means we use to put names to the real numbers. Just as some people go by more than one name, so also it is possible for some numbers to have more than one name. [ All real numbers have only one decimal notation except for those with a terminating decimal. These have two  one terminating (ending in repeating 0s), the other ending in repeating 9s, with the last non9 digit one less than in the terminating form. ] Arguments depending on the reader knowing how to add/multiply decimals 1) Times 10 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 1/3 = 0.333... 1/3 = 0.333... + 1/3 = 0.333... 3/3 = 0.999... 1 = 0.999... 3) Take the Difference 1.000... 0.999... 0.000... = 0 Since their difference is 0, they are the same. 4) 9*0.111... 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 + ... 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(( 110[supminus][smiley=supcn.gif][supminus][sup1])/(110[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) Epsilondelta 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],  (110[supminus][supn])  [smiley=cl.gif]  < [epsilon]. (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] log_{10} [epsilon]. Then for [smiley=n.gif] > [smiley=cn.gif],  (110[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 I'm not going to try and summarize the first contribution I made to this thread. If your suffering from insomnia, or want to experience all the nasty effects of getting drunk without any of the fun, follow the link and read it. This isn't a calculusbased proof. It is actually a "what calculus is, but from before calculus was invented" proof. 8 ) 1=0.999...9^{1}0 For all [smiley=n.gif], 1 = 0.99...([smiley=n.gif] 9s)...9^{1}, 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 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 nonexistant 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.

« Last Edit: Nov 15^{th}, 2003, 8:36pm by Icarus » 
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Icarus
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Boldly going where even angels fear to tread.
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Re: 0.999.
« Reply #249 on: Sep 1^{st}, 2003, 12:56pm » 

Misconceptions about Decimals and Infinity found in this thread 1) Infinite Decimals are Approximations This argument says that, for example, 1/3 [ne] 0.333... because the righthand 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 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 "shorthand" 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 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 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 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 , I gave a short overview of the three types of infinities most commonly encountered in mathematics. 6) Infinity is an actual number (The linked post does not state infinity is a number like some other posts 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", "padic 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 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 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 While noone 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 welldefined real number  without bothering to define what this new variation actually means. The post linked introduces a new digit, "#", to base10 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 topologicallycomplete 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 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 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! I think it's clear who the real idiot is with these posters!

« Last Edit: Nov 1^{st}, 2003, 5:33pm by Icarus » 
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"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



