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Topic: 0.999. (Read 125651 times) 

Sir Col
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Re: 0.999.
« Reply #275 on: Nov 15^{th}, 2003, 5:47pm » 

In England, the term fraction is used to strictly mean the ratio of two numbers. More recently, we have been using the term in relation to noninteger parts of decimals, but I can't imagine many children in this country knowing what a decimal fraction was. They would probably think you were talking about something like 1.5/4.5. They accept that all fractions will be terminating or recurring decimals. The process of division confirms this: in the event that there is a remainder, the variations is limited to the range, 0 to d–1 (d represents the value of the denominator); hence it will either terminate (remainder 0), or will produce a previously 'seen' remainder and become locked in a never ending cycle. The question they then asked was, how can you be sure that all recurring decimals necessarily have a fraction equivalent? It's one of those questions that you think to yourself, come on, it's obvious. But I believe that this gives an insight into the difficulty that some people have with understanding that 0.999...=1. I suspect that they have failed to grasp that all recurring decimals can be written as fractions, and if 0.999... is not equal to 1, what fraction is it equal to?


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Icarus
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Re: 0.999.
« Reply #276 on: Nov 15^{th}, 2003, 8:00pm » 

Actually, when you say "decimal fraction", something like "1.5/4.5" is what I think of first as well. I am not very fond of that terminology. It is misleading. I think the point of TimMann's question is more one of: do they accept that integers are fractions, or by "fraction" do they mean only noninteger rational numbers (expressed as integer ratios). I am sure that those who fail to realize 0.999... = 1 do not understand about repeating decimals representing fractions. The procedure for finding the fraction for a repeating decimal produces the "10 times" proof that 0.999... = 1. So if they knew of and understood this relationship, they would have known the truth already. But by the same token, those that the "10 times" proof fails to convert will have a hard time accepting the whole relationship. Most of them also have significant issues with understanding what an infinite decimal even means. Witness the various "approximation" misconceptions that have been argued. It is only after someone gets a good handle on this idea that they have any real hope of understanding the connection between repeating decimals and fractions. Personally, I put a lot of the blame to the way irrational numbers are introduced in the first place  not that I have any ideas how to do it better. Counting is something that kids learn as toddlers, so when they enter school, they are already familiar with the idea of the natural numbers (not the name, or the properties, but with the numbers themselves), and zero  if not already familiar  is an easy addon to this. Next we introduce fractions, and the motivation behind them is easy to see. Of course, we can provide them with plenty of "real world" examples  usually culinary . So, while they hate the complexities of dealing with them, the concept of rationals is easily digested. But then we introduce two other types of numbers: negatives and irrationals. Negatives are confusing, but we still can provide a few, mostly monetary or temperaturerelated, "real world" examples. They are a bit harder to swallow, and the rules sometimes don't seem to make sense (why is (1)(1) = 1?), but students understand the idea. Irrationals are harder. There are no definite realworld examples. We encounter them as geometric abstractions only. Physical measurement is not capable of definitely producing an irrational length  that would require infinite accuracy. So how do you get the idea of these numbers across to students? The most obvious way is to introduce them as nonrepeating infinite decimals! The problem is, this gives the students the concept that real numbers consist of decimal representations; exactly the misconception demonstrated by Kozo and others in their creation of notations such as 1.000...([infty] 0s)...1 and demanding that these are  in and of themselves  real numbers not represented by ordinary decimal notations. The more I think about it  the more I believe this is the core of why some people are troubled at the idea that 0.999... = 1. If your understanding of real numbers is that they are decimal notations, it does not make sense to say that two obviously different notations are actually the same! We need to make them understand that the real numbers are a different set of objects, and that the notations are just our way of giving names to this other set of objects. Then they will realize there is nothing weird about a number having two names. (I think I may add this argument to my summary thread. Perhaps it will help clear things up for those who are still confused, and not educated enough yet to follow the calculus stuff.) Perhaps the answer is to emphasize the number line, and introduce irrationals as points on the line not represented by fractions. The existence of such points can be demonstrated with an isosceles right triangle and the standard [surd]2 is not rational proof. But it is admittedly much easier to show the existence of nonrepeating decimals (I still remember being shown 0.1010010001... for this very purpose), and much more easily comprehended. Maybe introducing irrationals as decimal expressions is best. But somewhere along the way, a concerted effort should be made to bring the student past this concept to a more fundamentally sound one. I don't think most teachers ever do this. In fact, I suspect a majority of primary (K12 in the US) math teachers have never really moved past this concept themselves.


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Sir Col
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Re: 0.999.
« Reply #277 on: Nov 16^{th}, 2003, 2:57am » 

I missed that part of TimMann's question: yes, they are quite happy with the idea that an integer can be a fraction. They know that one fraction divided by another fraction gives a fraction; ask them what 1/2 divided by 1/4 is and they'd say, 2. I agree with your points, Icarus, which is why I always spend quality time getting them to first understand that all fractions, when expressed as decimals, will either terminate or recur. As I mentioned, I then encourage them to think of an example of a decimal that neither terminates nor recurs. After a bit of thought they begin to realise that there are literally a countless list of such examples. Having previously spent time looking at sequences they can present examples such as: 0.12345678910111213..., 0.369121518..., or for that matter any type of nonrepreating sequence (they're probably happier with arithmetic examples, as they know there are infinitely many varieties of these). In fact, we could append an infinite string of random digits. After this, they appreciate that there are decimals that exist, which obviously didn't come from fractions (they don't repeat or terminate). I believe that this gives them a complete understanding of the nature of irrational numbers; it also has prepared them to appreciate transcendentals in the future.

« Last Edit: Nov 16^{th}, 2003, 3:38am by Sir Col » 
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Icarus
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Re: 0.999.
« Reply #278 on: Nov 16^{th}, 2003, 12:14pm » 

A "complete understanding" is saying far too much, as I am more than well aquainted with these facts, but lack a complete understanding of irrationals! (But I know what you meant.) However, this still does not address the problem I refered to: the misconception that decimals are what real numbers are, instead of only being names for real numbers. I trust that you do a good job of getting this across. Some other things you have said indicate that you are giving your students an excellent grounding in mathematics. Unforturnately, many teachers fail to do so. Many have never received such a grounding themselves, and therefore are unable to give such to their students. For instance, consider Alex Wright's reproduction of a "proof" by his teacher. I hope that this was nothing more than Alex misunderstanding what was the proof, and what was "scratchwork", but I fear  because I have seen it far too often  that his teacher actually did produce this as "proof" that 9.999... = 10. (Some other posts by Alex indicate a better understanding developing, so I am more hopeful now than I was at the time.) Similarly, I am hoping that matt's post at the top of the previous page was a result of him totally misunderstanding what his calc teacher was saying, rather than having a calc teacher who doesn't understand the basic calculus involved here. But again  I fear the worst.


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Sir Col
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Re: 0.999.
« Reply #279 on: Nov 16^{th}, 2003, 2:25pm » 

I always wondered how young Alex was doing; it's nice to see he remembered something I taught him. I like the idea of teaching that fractions, decimals, and percentages, are just different ways of describing numbers. Inevitably, students are well grounding in basic fraction arithmetic, and simple percentage and decimal calulations, when they arrive at my school, but whenever I review these topics, I usually ask the question, "What IS a fraction?" I am pleased if they leave my lessons with an understanding that a fraction is simply one number divided by another, and we usually use fractions when two numbers do not divide evenly. Teaching the value of fractions in this way encourages children to stop writing rubbish like, x=0.1428571429, when solving the equation, 7x=1. I normally tell them that fractions were invented by mathematicians that kept forgetting to buy new batteries for their calculators! So although I normally get children to question what fractions/percentages actually are, I must say that I hadn't thought of doing the same thing with decimals. Upon your suggestion, I believe there may be some value in this and will try it out; that is, getting the students to explicity understand that decimals are just another way of describing a position on the number line. It may also help prepare those that continue developing their mathematics beyond the required, to appreciate different bases, as they will realise that decimal notation is just one arbitrary choice of many.


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jay c
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Re: 0.999.
« Reply #280 on: Feb 13^{th}, 2004, 7:15am » 

You can prove that 0.999 repeated is equal to 1: let x=0.9999 repeated multiply each side of the formula by 10... therefore you get 10x = 9.9999 repeated now subtract x from each side of the equation therefore you get 9x = 9 now divide each side by 9 boom baby.... x = 1 therefore 0.9999 repeated = 1


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Sameer
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Re: 0.999.
« Reply #281 on: Feb 13^{th}, 2004, 7:37am » 

For infinite series the equal to sign disappears and becomes a limit when you use multiplication .. so the step where u equated it to x actually is a limit ... as for e.g. consider S = 1+2+4+8 .... 2S=2+4+8+16 ... 2S+1=1+2+4+8+16... 2S+1=S S=1 By your method sum of 2^n comes to 1. Looks like a computer error


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ThudnBlunder
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Re: 0.999.
« Reply #282 on: Feb 13^{th}, 2004, 7:40am » 

Quote:You can prove that 0.999 repeated is equal to 1... 
 ...by repeating the same method as the very first post in this thread? Why not read this?

« Last Edit: Feb 13^{th}, 2004, 8:12pm by ThudnBlunder » 
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Icarus
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Re: 0.999.
« Reply #283 on: Feb 13^{th}, 2004, 9:06pm » 

on Feb 13^{th}, 2004, 7:37am, Sameer wrote:For infinite series the equal to sign disappears and becomes a limit when you use multiplication .. so the step where u equated it to x actually is a limit ... as for e.g. consider S = 1+2+4+8 .... 2S=2+4+8+16 ... 2S+1=1+2+4+8+16... 2S+1=S S=1 By your method sum of 2^n comes to 1. Looks like a computer error 
 Sorry, Sameer, but this is pretty much entirely wrong. Even if you demand that 0.999... be interpreted as an infinite series (there are other approaches to defining it), it does not make the "equal to sign disappear and become a limit". The = sign remains. It is 0.999... itself that is replaced with the limit of the sequence of partial sums. But, the limit of the sequence of partial sums IS a number. One need only show that this sequence converges. The limit does NOT represent the series. It represents the particular number the series converges to. So the only thing necessary to justify jay c's equation is that the series does converge. You might argue that jay c should have shown this, but I disagree. It is understood as a given in this problem that 0.999... represents an actual number. The problem is only the question of which number it represents. jay c's argument is entirely valid, although far from new to this thread. YOUR sum on the other hand fails to meet the condition of convergence. Whereas jay c's statement x = 0.999... is valid because 0+.9+.09+.009+... represents an actual number, your statement S=1+2+4+8+... is not valid, because there is no such number as 1+2+4+8+... . You could say that this sum = [infty], and that would be correct. But [infty] does not obey the same laws of addition and multiplication that real numbers do, so the calculation you followed with is invalid either way. For completeness, a proof that all decimal expressions (to any base) converge. It requires the completeness property of real numbers. This basic property has many forms, but the one I use here is "An increasing sequence which is bounded above must converge". Let {d_{n}} be a sequence of integers with d_{n} [in] {0, 1, 2, ..., b1} for some fixed integer b > 1. Each sum [sum]_{n=1}^{N} d_{n}b^{n} [le] [sum]_{n=1}^{N} (b1) b^{n} = 1  b^{N} < 1. So the partial sums are all bounded. They also clearly form an increasing sequence, since each adds a nonnegative number, d_{n}b^{n}, to the previous sum. Since the sequence of partial sums is increasing and bounded above, it must converge. Hence any decimal expression to any fixed base represents a actual real number.


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Sameer
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Re: 0.999.
« Reply #284 on: Feb 16^{th}, 2004, 10:59am » 

My bad of comparing a convergent problem with a divergent one... But Icarus I would have to argue that even though the number is bounded by an actual number it doesn't meant it is 'equal to' that number As for e.g. if and only if M <= S_{n}<=M then S_{n} = M Unless otherwise it always 'tends to' that number. So in our original problem .999.... tends to 1 and hence the 'limit' is 'equal to' 1 and not the number. My question for you would be then is it safe to assume that for a convergent problem 'limit' is same as 'equality'?


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towr
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Re: 0.999.
« Reply #285 on: Feb 16^{th}, 2004, 11:16am » 

Quote:Unless otherwise it always 'tends to' that number. So in our original problem .999.... tends to 1 and hence the 'limit' is 'equal to' 1 and not the number. 
 0.999... = [sum] _{k=1}[supinfty] 9*10^{k} = lim n[to][infty] [sum] _{k=1}^{n} 9*10^{k} so it doesn't tend to 1, it is 1 Quote:My question for you would be then is it safe to assume that for a convergent problem 'limit' is same as 'equality'? 
 no, but [sum] _{k=1}[supinfty] f(k) [equiv] lim n[to][infty] [sum] _{k=1}^{n} f(k)

« Last Edit: Feb 16^{th}, 2004, 11:20am by towr » 
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Sameer
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Re: 0.999.
« Reply #286 on: Feb 16^{th}, 2004, 3:45pm » 

on Feb 16^{th}, 2004, 11:16am, towr wrote: 0.999... = [sum] _{k=1}[supinfty] 9*10^{k} = lim n[to][infty] [sum] _{k=1}^{n} 9*10^{k} so it doesn't tend to 1, it is 1 
 Ok someone needs to refresh the fundamentals of limits to me... Now you just said that limit is equal to 1. How can I say that hence the number .999... is equal to 1 ... Ultimately isnt the limit defined within a open interval neighborhood with some epsilon > 0. Isn't limit another way of saying what is this value closest to and tends to converge to? *sigh* I think I should go back to school...


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Icarus
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Re: 0.999.
« Reply #287 on: Feb 16^{th}, 2004, 3:51pm » 

Sameer, please follow the link in the header to my "Misconceptions" post, and read #1 & #2. You seem to be holding them. I shall try to make the point clearer: A limit is not a set of approximations as you seem to believe. The limit is a particular number  it is the exact number that is being approximated. 0.999... does not represent 0.999 or 0.9999 or 0.999...9 for any finite number of 9s. It does not represent them individually or collectively. What it represents; what it means as the limit of the sequence of the finite expressions; is the particular number that the finite expressions are approaching. As such, 0.999... is exactly 1, no more, no less. And I did not argue that the sequence {0.9, 0.99, 0.999, ...} being bounded by 1 meant that its limit was 1 either. All my argument showed  all it was supposed to show  is simply that the sequence converges  without regard to what it converges to. This demonstrates that 0.999... is an actual real number. Which is the only way in which jay c's proof falls short of being complete. (Okay, not quite  true completeness also requires showing that multiplying by 10 and subtracting decimals really does work the way we all know it does). As I said in the previous post, all of this I consider "given". Not something that jay c should be required to show. Hence his proof is quite accurate.


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Icarus
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Re: 0.999.
« Reply #288 on: Feb 16^{th}, 2004, 4:20pm » 

Apparently you posted while I was creating mine. My description of what I thought you were saying was based on your earlier posts, not this latest one. The basics of my reply remain the same, though. The limit is not an interval. The limit is the particular number L for which  a_{n}  L  < [epsilon] can be made to hold (by increasing the minimum value of n) for ANY [epsilon] > 0. L is a single number. It does not depend on [epsilon] or the value N for which the inequality holds for n [ge] N. (N depends on [epsilon], but L must be the same for every [epsilon].) So 0.999... does not represent an interval about 1. It represents 1 itself.


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Sameer
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Re: 0.999.
« Reply #289 on: Feb 17^{th}, 2004, 6:34am » 

hmmm Icarus before I read that... can you explain me following: Consider three functions f(x) = 4 g(x) = x+2 h(x) = (x^24)/(x2) If you take limits of all three functions as x>2 we have the answer of 4. Does that mean all the functions have a value of 4 at x=2?


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towr
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Re: 0.999.
« Reply #290 on: Feb 17^{th}, 2004, 7:00am » 

h(2) is not defined, so no If h was a continuous function, then it would be true. I'm not quite sure how this works for discrete maths though (like series) For a discrete function lim n>k (k < [infty]) is generally undefined, as the minimum [epsilon] you can find a [delta] for is max( f(k+1)f(k) , f(k1)f(k) ), unless I'm mistaken (so you can't find a [delta] for every [epsilon] > 0, which is necessary for the limit to exist) If there's convergence you don't face that problem with a limit from n > [infty]

« Last Edit: Feb 17^{th}, 2004, 7:33am by towr » 
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Sameer
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Re: 0.999.
« Reply #291 on: Feb 17^{th}, 2004, 7:51am » 

Hmm so I am gonna quote Icarus from the misconception post Quote: Note that by the definition, the limit is not any of the or all of them, or some "process". The limit is the number which the sequence elements approximate. 
 Doesn't this imply that if Lim_{n > inf} a_{n}=L then a_{n} approximates L and is not equal to L. Going by this then how can we say .9999.. = 1 isnt is the limit that is 1 which is by definition approximation of the series.


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towr
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Re: 0.999.
« Reply #292 on: Feb 17^{th}, 2004, 8:11am » 

on Feb 17^{th}, 2004, 7:51am, Sameer wrote:Doesn't this imply that if Lim_{n > inf} a_{n}=L then a_{n} approximates L and is not equal to L. 
 yes, for any n < [infty] (just to be clear, a_{n} is not necessarily L, but it may be) But 0.999... isn't any of these a_{n}, it represents an infinite number of 9's after the decimal point. So it is_{(identical to)} lim n > [infty] a_{n} (rather than is_{(has the same value as)} ).

« Last Edit: Feb 17^{th}, 2004, 8:16am by towr » 
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Sameer
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Re: 0.999.
« Reply #293 on: Feb 17^{th}, 2004, 8:23am » 

on Feb 17^{th}, 2004, 8:11am, towr wrote: yes, for any n < [infty] (just to be clear, a_{n} is not necessarily L, but it may be) But 0.999... isn't any of these a_{n}, it represents an infinite number of 9's after the decimal point. So it is_{(identical to)} lim n > [infty] a_{n} (rather than is_{(has the same value as)} ). 
 awesome towr... thanks for clearing the cloud in my head!!! . I learn so much everyday!!!


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TenaliRaman
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Re: 0.999.
« Reply #294 on: Feb 18^{th}, 2004, 9:18pm » 

I am afraid the biggest question still looms at large, "why isn't this thread locked yet?" Isn't 12 pages of sarcastic remarks, mathematical buffoonism and some really insightful mathematical posts enough for a thread?


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towr
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Re: 0.999.
« Reply #295 on: Feb 19^{th}, 2004, 1:30am » 

on Feb 18^{th}, 2004, 9:18pm, TenaliRaman wrote:"why isn't this thread locked yet?" 
 Why ought it be? I suppose the thread is long, but is that good enough of a reason?


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TenaliRaman
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Re: 0.999.
« Reply #296 on: Feb 19^{th}, 2004, 6:42am » 

Fooey comes into this riddle forum. He goes through the site and sees 0.999....=1 and he says to himself "hey i know this one". He clicks the thread and says "whoa 12 pages i am not going to read through that" (obviously he also missed icarus' compendium of proofs and misconceptions". Then he posts Fooey : so the proof of the above is ..... <blah blah> Uberpuzzler X : That proof is already done check here here and here. ============================================= Now copy paste the above several times and we would have reproduced atleast 25% of the discussion that has been going on in this thread. My point of view : Waste of space.


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towr
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Re: 0.999.
« Reply #297 on: Feb 19^{th}, 2004, 6:46am » 

If this thread were locked, they would probably just make new threads. Which isn't any better..


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Sameer
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Re: 0.999.
« Reply #298 on: Feb 19^{th}, 2004, 6:49am » 

Who cares about the minute amount of space occupied by this thread when there is incredible amount of knowledge flowing...


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TenaliRaman
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Re: 0.999.
« Reply #299 on: Feb 19^{th}, 2004, 7:23am » 

towr, yes i think you got a point there. Sameer, my waste of space comment was actually a sarcastic remark.But its explanation isn't important at this moment.


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