

Title: 0.999. Post by mentor on Jul 26^{th}, 2002, 10:27am There's a method to remove reccurence form a decimal: FIrst, take the number and time by 10^(no of places before reccurence): 0.999. x 10 = 9.99. Now take the original number away from the answer: 9 If we repeat the same process with a variable x instead of the decimal, we get: 9x We now equate the two and solve for x, which gives us: 1 so, 0.999. = 1 

Title: Re: 0.999. Post by mentor on Jul 26^{th}, 2002, 11:40am An Addendum: it's a short step (left as an excercise to the reader) to show that 1/(ifinity) is 0. 

Title: Re: 0.999. Post by Frothingslosh on Jul 26^{th}, 2002, 12:52pm Since .9999... = 1, there are any number of ways to prove this. When I find someone who can't accept some of the classic proofs, I resolve the matter once and for all with the following: 1/3 = .333333...... 2/3 = .666666...... ____ ___________ 3/3 = .999999....... Just add each side together and it becomes obvious! 

Title: Re: 0.999. Post by Phezik on Jul 26^{th}, 2002, 4:10pm Another even easier way to think of this for those who won't accept the classic proof is 1  .9999... = 0.0000... If you take the difference between two numbers, and the difference is 0, then they have to be the same number. 

Title: Re: 0.999. Post by Kozo Morimoto on Jul 26^{th}, 2002, 5:16pm First off 1/3 is NOT 0.3333... that's just the decimal approximation. I don't understand the first solution. And 0.0000... is NOT 0, very close to zero, but not zero, so that equal sign in the middle is wrong. 1 is not equal to 0.999... I suppose you have to define what EQUAL is... I've heard that something with a probability of less than 1 out of 10^40 is mathematically classified as impossible, although it isn't 0. 

Title: Re: 0.999. Post by Rhaokarr on Jul 26^{th}, 2002, 9:10pm I want to see who would take up the argument that 0.999' > 1 ;) 

Title: Re: 0.999. Post by Davers on Jul 27^{th}, 2002, 3:32am A nonmath solution: What number is between .9999... and 1? See? There isn't one. Hence: .9999... = 1 ;D 

Title: Re: 0.999. Post by Kozo Morimoto on Jul 27^{th}, 2002, 6:32am "A nonmath solution: What number is between .9999... and 1? See? There isn't one. Hence: .9999... = 1" Of course there is, we just can't express it with the Arabic numeral that we use. Lets create a new digit # which is bigger than 9 but less than 10. Then .999999... < .####..... < 1 so there is a number between .999... and 1, but we just can't express it with the digits that we NORMALLY use. Just because we can't express it doesn't mean that it isn't there. 

Title: Re: 0.999. Post by NickH on Jul 27^{th}, 2002, 8:01am Suppose, for the sake of argument, that # = 9.5. Then, .## = .95 + .095, which is already greater than 1! In fact, for # = 9.5, .#####... = 19/18. Further, .#####... > 1, for any # > 9. Your argument that there exists a # such that 9 < # < 10 does not imply the existence of a number @ such that .999... < @ < 1. There is no such @. 

Title: Re: 0.999. Post by cnmne on Jul 27^{th}, 2002, 10:58am > First off 1/3 is NOT 0.3333... that's just the decimal approximation. that is kind of the point. 0.9999... is the decimal approximation of 3/3 = 1 

Title: Re: 0.999. Post by obtuse on Jul 27^{th}, 2002, 1:56pm I recall reading an article defining a counting system in which you could easily write infinitely large and infinitely small numbers. It incorporates ideas from Zeno's Paradox. I hope my explanation makes some sense. ^ = up arrow V = down arrow The first arrow signifies a change of 1. Every arrow in the same direction as the first counts as another increment of 1. ^^ = 2 VVV = 3 The first arrow to change direction meant a 'decrement' of 0.5, and every arrow thereafter was a 'decrement' of half the previous value. ^V = 1  0.5 = 0.5 ^VV^ = 1  0.5  0.25 + 0.125 = 0.375 Expressing a repeating pattern was done by adding a bar over the top, which I will exchange for an underline. Multiple bars were allowed, to allow embedded patterns and multiple infinities. ^ = infinity ^^ = infinity plus 1 (there IS a difference here!) ^VV^ = 1/3 (if I'm doing my math right) It defined an iota as ^V, meaning the smallest possible number above zero. Thus: 0.999... + 0.000... = 1 or ^^V^ + ^V = ^ I assume this isn't what the problem was looking for, but interesting regardless. 

Title: Re: 0.999. Post by domn on Jul 27^{th}, 2002, 4:35pm 10 x .9999... = 9.9999... 9.9999...  .9999... 9 9/(101) = 1 

Title: Re: 0.999. Post by Frothingslosh on Jul 27^{th}, 2002, 8:32pm >> First off 1/3 is NOT 0.3333... that's just the decimal approximation. > that is kind of the point. > 0.9999... is the decimal approximation of 3/3 = 1 You guys are dead wrong. 1/3 = .333....... exactly. The .... is an indication that the 3's continue on to infinity. This is not an aproximation, it is the exact value, represented by the dots to save time from writing an infinite number of 3's. But it is the exact answer, if you didn't learn that much math in grade school there isn't much point in further discussion. I suggest you both do a little research and try to come up with anything to support your position. 

Title: Re: 0.999. Post by jeremiahsmith on Jul 28^{th}, 2002, 12:46am Actually, .99999... is a geometric series... 9/10 + 9/100 + 9/1000, each term is 1/10 of the last. I remember a formula from my calculus class dealing with the sum of geometric series, and although it's much too late (or early...) for me to remember, I do remember that it showed that .9999... = 1. Maybe I'll go find that formula tomorrow. 

Title: Re: 0.999. Post by Kozo Morimoto on Jul 28^{th}, 2002, 6:37am "You guys are dead wrong. 1/3 = .333....... exactly. The .... is an indication that the 3's continue on to infinity." You are wrong. .333... is an aproximation of 1/3. That's why 1/3 is a rational number and thus can not be represented by decimal notation. So .333... is an approximation. Otherwise it wouldn't be a rational number. Rationals are a subset of reals and a superset of integers. "Suppose, for the sake of argument, that # = 9.5. Then, .## = .95 + .095," No, .###... = .###.... and can not be represented by normal decimal notation in my example. If I could I would need to use the new symbol/digit "#". It's still a number between 0.999... and 1 by definition. 

Title: Re: 0.999. Post by NickH on Jul 28^{th}, 2002, 8:44am "You are wrong. .333... is an aproximation of 1/3. That's why 1/3 is a rational number and thus can not be represented by decimal notation. So .333... is an approximation. Otherwise it wouldn't be a rational number. Rationals are a subset of reals and a superset of integers." I think you are confused by the concepts of "rational" and "representable in decimal notation." A rational number is a real number that can be represented as a/b, where a and b are integers, b != 0. Some rational numbers have a decimal expansion that terminates; example: .25. All other rationals have a decimal expansion that repeats indefinitely from some point on; example 1/11 = .09090909..., where '09' is the repeating group. Conversely, any number whose decimal expansion either terminates or repeats from some point on is rational. As jeremiahsmith says above of .999..., .333... is mathematical shorthand for the infinite geometric series: 3/10 + 3/100 + 3/1000 + ... The sum of such a series is given by a/(1r), where a is the first term and r is the common ratio, if r < 1. In this case we have a = 3/10, r = 1/10, yielding a sum of 1/3. This is the exact answer. It is what we get when we sum to infinity. Let me ask you this: if the sum to infinity of 3/10 + 3/100 + 3/1000 + ... is not equal to 1/3, what is it equal to? "No, .###... = .###.... and can not be represented by normal decimal notation in my example. If I could I would need to use the new symbol/digit "#". It's still a number between 0.999... and 1 by definition." I think you need to explain more clearly exactly what "#" is, what ".#" means, and what ".###..." means. Clearly # is not an integer, for there are no integers > 9 and < 10. Is # a real number? If so, then what is .#? What number base are you working in? Once you've explained the concepts you are using, attempt a proof that .999... < .###... < 1. 

Title: Re: 0.999. Post by srowen on Jul 28^{th}, 2002, 9:13am NickH is 99.9999...% right. By which I mean 100%. I agree that the problem here is that people are throwing around ".333..." like any other decimal. Indeed, it's a way of expressing the value 1/3 as the sum of an infinite geometric series. ".333.." just looks like an ordinary decimal, and that's the trick of this question. But it's an "infinitely long decimal" which is quite an imaginary thing to begin with. So .999... = 1. It's just a roundabout way of expressing "1". And no, there is no value between .999... and 1, because they are in fact equal! This business about a digit "#" equalling 9.5 is misguided. 

Title: Re: 0.999. Post by Gareth Pearce on Jul 28^{th}, 2002, 6:50pm 0.9999999... = 1 is simple but can you prove 0.9999... < 1 ? That is, can bother answers 1 and 2 be true at the same time... I dont think so, but i havent proven it. 

Title: Re: 0.999. Post by Kozo Morimoto on Jul 28^{th}, 2002, 9:49pm In my example, if # is half way between 9 and 10, then 0.# would be halfway between 0.9 and 1. So 0.## would be half way between 0.99 and 1 and so on. So 0.###... would be half way between 0.999... and 1 

Title: Re: 0.999. Post by jmlyle on Jul 28^{th}, 2002, 11:41pm Kozo, it's time to step away from this problem. Quote:
Imagine that there is a number that is half way between 8 and 10. Let's call this number "9." A little substitution: Quote:
And maybe we should also imagine that math works completely differently than it does. jmlyle 

Title: Re: 0.999. Post by NickH on Jul 29^{th}, 2002, 12:25am "In my example, if # is half way between 9 and 10, then 0.# would be halfway between 0.9 and 1. So 0.## would be half way between 0.99 and 1 and so on. So 0.###... would be half way between 0.999... and 1" Ah, I think I see what you mean by .##. You mean .995. Similarly, .### = .9995. Now, it's true that for any finite number of #'s, .9999...9 < .###...# < 1. However, when we sum to infinity, both .999... and .###... are equal to 1. 

Title: Re: 0.999. Post by anshil on Jul 29^{th}, 2002, 12:52am <i>if 9 is half way between 8 and 10, then, 0.9 would be halfway between 0.8 and 1. So 0.99 would be half way between 0.88 and 1 and so on. So 0.999... would be half way between 0.888... and 1.</i> I don't want to disappoint you, but take out the calculator and find out what's halfway between 0.88 and 1. Well it's (1  0.88) / 2 + 0.88 right? Well that's 0.94, not 0.99 I guess that is the pretty end of the # theory :D Don't forget what the decimal system is actually about. It's just a shortcut for writing like 0.99 is 0 * 10 ^ 0 + 9 * 1 * (10 ^ 1) + 9 * (10 ^ 2) You can't introduce a new decimal but leaving the base equal, of course you can create a new numbering system, with the base of 11, so we need a new digit in this system to represent 10, let it be #. So 10 decimal would #. 11 would be "10". 0.## would be written in the new system: 0 * 10 ^ 0 + # * 10 ^ 1 + # * 10 ^  2. Translated to the decimal system: 0 * (11 ^ 0) + 10 * (11 ^  1) + 10 * (11 ^  2) And thats equal to: 0.991736 Well we only have generated a new calculation system, but that still didn't create a new number per se. So in our new system we still can say: 0.########......... = 1 It's importand to understand that both systems are valid, and well equal, but you can't mix them Like using the number representation # in the decimal system. There are systems with a base other than ten which have their very own sex. Like of course the binary system (base 2), used in every computer, beside the decimal (base 10) the most important system today. The tertinary (base 3) almost not used, but having it's own very interesting behaviours. The hexadecimal system (base 16) used for easy representation of binary numbers. And of course a very old system, the system with base 12, forgot a name for it. Guess it's even older than the decimal, since 12 is devidable by 3, 4 and 6. While ten is only devideable by 2 and 5. Still our years are devided in 12 months, our days in 12 * 2 hours and so on. But I'm babbeling already :D Actually math works on Paper. Sometimes you have to believe in what your calculations proof and show. 

Title: Re: 0.999. Post by Kozo Morimoto on Jul 29^{th}, 2002, 6:40am OK, nobody seems to be getting the '#' besides me so I'll drop it... How about this. 0.9 is pretty close to 1, an approximation of 1. You add 0.09 to make it 0.99 and its even closer to 1 than 0.9 but still not 1. Repeat to infinity. You get ever closer and closer to 1 from the low side, but you never get there? Using the ideas given above, does it mean that 1.000 ... 001 with infinite zeroes between the two 1s make it equal to 1? So Does it mean that 0.999... = 1.000...001 ? How about 0.999...998 = 0.999... = 1.000...001? 

Title: Re: 0.999. Post by anshil on Jul 29^{th}, 2002, 7:15am "OK, nobody seems to be getting the '#' besides me so I'll drop it... " Well thats a very arrogant argument. Maybe all the people but you do "get" it, what you mean, but it just makes mathematically not much sense. " 0.9 is pretty close to 1, an approximation of 1. You add 0.09 to make it 0.99 and its even closer to 1 than 0.9 but still not 1. Repeat to infinity. You get ever closer and closer to 1 from the low side, but you never get there? " There you are, infinity and "near to infinity". Yes you get closer ever step you do, and as you are approaching infinity you are approaching 1 that is true. However when if you hit infinity you hit 1 also. Using the ideas given above, does it mean that 1.000 ... 001 with infinite zeroes between the two 1s make it equal to 1? No not, it's inifnitly small larger than one. So Does it mean that 0.999... = 1.000...001 ? No. 0.999999... .... 9998 would be smaller than 1, okay? But 0.9999....9999 _is_ 1. Like 1.00000......00000 is also 1. 

Title: Re: 0.999. Post by srowen on Jul 29^{th}, 2002, 11:33am on 07/29/02 at 06:40:49, Kozo Morimoto wrote:
Well, you "never get there" in the sense that there is no finite number of steps of this loop that gest you to 1. But "0.999..." is probably best viewed here as an expression of where you are headed to: 1. on 07/29/02 at 06:40:49, Kozo Morimoto wrote:
What does "1.000...0001" mean? It doesn't make sense to talk about a terminating decimal with a 1 after an infinite number of zeroes. Likewise for "0.999...998". 

Title: Re: 0.999. Post by anshil on Jul 29^{th}, 2002, 12:08pm Well srowen I agree that it doesn't make much sense on paper and can I be it can easly be proven to be wrong as soon I start to assert it to be true. But it does make a sense in imagination. Say x is the smallest rational number bigger than one. You can imagine it as 1.0000......00001, or x is the biggest rational number smaller than 1. That one you can imagine as 0.9999.....9998. And not 0.99999...... which is equal to one, and not smaller. 

Title: Re: 0.999. Post by srowen on Jul 29^{th}, 2002, 12:27pm on 07/29/02 at 12:08:10, anshil wrote:
There is no such thing as the "smallest rational number greater than 1". Call it x. What is (1+x)/2 then? 

Title: Re: 0.999. Post by anshil on Jul 29^{th}, 2002, 12:44pm Oops, you're right srowen. And I yet discovered another funny fact. Okay say x is the smalles rationonal number greater than one. So what would (1 + x) / 2 then be? Well I would say it would be still x. So what does the equation yield? (1 + x) / 2 = x > x = 1. Ha ha ha! :D 

Title: Re: 0.999. Post by aadash on Jul 29^{th}, 2002, 3:09pm Very interesting discussion... Here is what i found after search on google.com on the topic  http://mathforum.org/dr.math/faq/faq.0.9999.html Why does 0.9999... = 1 ? This answer is adapted from an entry in the sci.math Frequently Asked Questions file, which is Copyright (c) 1994 Hans de Vreught (hdev@cp.tn.tudelft.nl). The first thing to realize about the system of notation that we use (decimal notation) is that things like the number 357.9 really mean "3*100 + 5*10 + 7*1 + 9/10". So whenever you write a number in decimal notation and it has more than one digit, you're really implying a sum. So in modern mathematics, the string of symbols 0.9999... = 1 is understood to mean "the infinite sum 9/10 + 9/100 + 9/1000 + ...". This in turn is shorthand for "the limit of the sequence of numbers 9/10, 9/10 + 9/100, 9/10 + 9/100 + 9/1000, ...." One can show that this limit is 9/10 + 9/100 + 9/1000 ... using Analysis, and a proof really isn't all that hard (we all believe it intuitively anyway); a reference can be found in any of the Analysis texts referenced at the end of this message. Then all we have left to do is show that this sum really does equal 1: Proof: 0.9999... = Sum 9/10^n (n=1 > Infinity) = lim sum 9/10^n (m > Infinity) (n=1 > m) = lim .9(110^(m+1))/(11/10) (m > Infinity) = lim .9(110^(m+1))/(9/10) (m > Infinity) = .9/(9/10) = 1 Not formal enough? In that case you need to go back to the construction of the number system. After you have constructed the reals (Cauchy sequences are well suited for this case, see [Shapiro75]), you can indeed verify that the preceding proof correctly shows lim_(m > oo) sum_(n = 1)^m (9)/(10^n) = 1 0.9999... = 1 Thus x = 0.9999... 10x = 9.9999... 10x  x = 9.9999...  0.9999... 9x = 9 x = 1. Another informal argument is to notice that all periodic numbers such as 0.9999... = 9/9 = 1 are equal to the digits in the period divided by as many nines as there are in the period. Applying the same argument to 0.46464646... gives us = 46/99. References R.V. Churchill and J.W. Brown. Complex Variables and Applications. 0.9999... = 1 ed., McGrawHill, 1990. E. Hewitt and K. Stromberg. Real and Abstract Analysis. SpringerVerlag, Berlin, 1965. W. Rudin. Principles of Mathematical Analysis. McGrawHill, 1976. L. Shapiro. Introduction to Abstract Algebra. McGrawHill, 1975.  

Title: Re: 0.999. Post by drdedos on Jul 29^{th}, 2002, 3:36pm Kozi Morimoto would not win a race against the Tortoise. Kozi says "0.9 is pretty close to 1, an approximation of 1. You add 0.09 to make it 0.99 and its even closer to 1 than 0.9 but still not 1. Repeat to infinity. You get ever closer and closer to 1 from the low side, but you never get there?" Kozi, say you are asked to run across a football field. Before you get across the football field, you have to get 0.9 of the way across, right? Now before you get across the remaining 0.1, you have to get across 0.9 of THAT first, right? In other words, you have to first get to 0.99 of the football field? Likewise, you have to get 0.999 of the way across the football field before you get all the way across, right? And so on and so on... Using the same kind of logic you used, quoted above, you will never make it across the football field. Sorry to hear that. BTW, the reference to the Tortoise is from a variant of Zeno's Paradox written by Lewis Carroll. 

Title: Re: 0.999. Post by Kozo Morimoto on Jul 29^{th}, 2002, 4:17pm Yeah, that's right! I was thinking about using the turtle/hare paradox to explain MY side of the arguement, but you beat me to it! If this isn't the case, does that mean that drdedos actually get to the finish line following the rules? I don't think so. I apologise if my earlier statement implied that I was arrogant. I didn't intend that at all  I was trying to express how my English language skill was too ineloquent to express my thoughts in writing. The fact that this wasn't apparent also indicates that my English skill is well below par. Back to .999.... You start with 0.8. Divide by 10 and add 0.9. What do you get? 0.98. Divide that by 10 and add 0.9. What do you get? 0.998. If you do this infinite amount of time, don't you end up with 0.999...998? with infinite count of 9s between the 0 and the 8? Is this number equal to 0.999...? Along another direction, if 0.999... is equal to 1, doesn't this imply a nonhomogeneous (lumpy) R space? I'd assume that this effect only happens when 9 changes to 10 in decimal representation ie 1.999... = 2 or 0.4999... = .5 or 7.86999... = 7.87 so it only happens at certain places in R and not others? 

Title: Re: 0.999. Post by srowen on Jul 29^{th}, 2002, 8:29pm Kozo deserves credit for making everyone think very carefully about their arguments, and for refusing to accept anything less than perfect arguments! This is an excellent discussion. on 07/29/02 at 16:17:37, Kozo Morimoto wrote:
I think that we must not confuse "doing something many, many times" and "doing something an infinite number of times." Again, there is no such thing as "0.999...998" with an infinite number of nines. If this is a terminating decimal, then there must be a finite number of nines preceding the eight. If not, then it makes no sense to write it like it is a terminating decimal. This is an illusion, a subtle abuse of notation. If you are using "0.999...998" as a sort of shorthand to express the limit of your process, then yes, it has a value, and it is 1, and is equal to "0.999...". But it's important to realize that 1 is the limit of this process, the point to which it is heading, but never reaches in a finite number of steps. To write it as if it is a terminating decimal like "0.999...998" is only deceptive. 

Title: Re: 0.999. Post by NickH on Jul 30^{th}, 2002, 12:56am "Along another direction, if 0.999... is equal to 1, doesn't this imply a nonhomogeneous (lumpy) R space? I'd assume that this effect only happens when 9 changes to 10 in decimal representation ie 1.999... = 2 or 0.4999... = .5 or 7.86999... = 7.87 so it only happens at certain places in R and not others?" That's a good question, and the answer is no, this does not imply lumpy R. It's true that any number that can be represented as a terminating decimal has two decimal expansions, as in the examples you have given. But these are just two different ways of writing the same number. There is no lumpiness in the underlying reals. Note also that the same thing happens in other number bases, and sometimes for different numbers. For example, decimal 1/3 can be written in base 3 as either 0.1 or 0.0222... However, 1/3 has only the repeating representation 0.333..., in base 10. Ordinarily it does not matter that there are two ways of writing certain decimal numbers. However, in Cantor's famous diagonal proof of the uncountability of the reals, we must choose one or the other representation. See: http://c2.com/cgi/wiki?CantorsProof 

Title: Re: 0.999. Post by Kozo Morimoto on Jul 30^{th}, 2002, 1:52am Thanks for the counter arguments, its making it a lot more digestible. So the number I get by divide by 10 and 0.9, eg 0.9998 and ever increasing number of 9s, if there were really really large number of 9s (say large n but not "quite inifinity"), its still less than 0.999.. with n+1 9s? As soon as they both 'reach' infinite number of 9s, they are all both considered to be 1? So can you start with any finite number and keep dividing it by 10 and add 0.9 to it, and if you repeat the process to infinity you get 1? So 'inifity' is a really special situation? Are there any good newbie article/paper on infinity and its applications/consequences on the web? I'm very much interested in doing some personal research into it. I've learned quite a bit on this thread, not so much on solving riddles, but still, learned a lot. 

Title: Re: 0.999. Post by Willis on Jul 30^{th}, 2002, 5:15am First, let 0.999... be denoted by x. Observe that 10x = 9.999... Observe that 10x  9 = 0.999... = x, and so 10x  9 = x. Thus, 10x  x = 9, 9x = 9, and x = 1. But x = 0.999..., and so 0.999... = 1. 

Title: Don't think of it as "reaching infinity" Post by srowen on Jul 30^{th}, 2002, 6:07am Yeah, infinity is a special situation in the sense that it's not a number, not even a big one, so you can't think of it in the same terms. Thinking of it as "reaching infinity" eventually is misleading. Theses processes will never reach infinity and never equal 1. However, 0.999... = 1. This is not saying that some large number of 9's equals 1, but is a statement about where their sum is heading. I think it's misleading to think of it as "where you get after an infinite number of 9s". When you toss in infinity you are speaking about the limits, the imaginary endpoint that you never quite reach. on 07/30/02 at 01:52:55, Kozo Morimoto wrote:


Title: Re: 0.999. Post by Kozo Morimoto on Jul 31^{st}, 2002, 2:47am Hope people don't mind me ranting on... Talking about inifinity and bounds (ie 0.999...998), as a theoretical entity, is it possible to be holding an end of an infinitely long string in each hand? Like, imagine like holding a skipping rope, but the middle of the rope gets further and further away until its infinitely far away, like 0.999...998. Or at infinity, do you actually hold 2 ropes/strings of infinite length instead of 1? 

Title: Re: 0.999. Post by bartleby on Jul 31^{st}, 2002, 6:30am Hi Kozo! Yes, you are on the right track if you listen to srowen. The whole point to the "Does .999.... really EQUAL 1?" puzzle is really "Do you understand what 'infinity' means?" Every time you bring up an argument like "What if I had a number like .9999....998", you are demonstrating that you don't quite grasp the concept of infinity... Very few people do, I'm not saying I do either. You're probably a math Ph.D. candidate if you do. But you have to think about infinity not as a number, but rather as a concept, or even better as a process. You can use the concept of infinity to basically say, "Well, if I do something, and I do it again, and again, and again, and it's moving toward (but not quite reaching) a goal, (the concept of a limit), then I can project that if it were to happen an infinite number of times, that goal would be exactly reached." But infinity isn't a number... you can prove this by saying (with the example of .99999.... = .9 + .09 + .009 + ....) that if infinity is a number, then let's say x = infinity, and if I do this process x times, I am still 1.1*10^{x} away, and by the way I could still do it one more time, couldn't I? So the idea of saying "x equals infinity" is meaningless. This destroys the concept of ".999....98" because exactly HOW MANY 9's are there in there? And why can't you have one more? One thing I was waiting for you to say was to refute the proof: x = .999... , 10x = 9.999..., 9x = 9, x = 1 by saying "Well, if x = .999.... then 10x should be 9.9999....990, right?" And as shown above, that doesn't make sense to try and say that! 

Title: Re: 0.999. Post by Drake on Aug 2^{nd}, 2002, 3:02pm When I first looked at this, I thought it was dependent on the level of precision as to whether 1 >= .999..., but after thinking about it it makes perfect sense because the precision is infinite as defined by the question. As a test, try this on a un*x machine: bc l (1/3)+(2/3) The obvious answer is 1, but it yields .999... 

Title: Re: 0.999. Post by mook on Aug 3^{rd}, 2002, 12:07pm i'm no math major, but you've provided some pretty simple proofs which look correct on paper. However, I think that the interesting discussion is on the semantics of infinity. If you apply no math, you logically know that .999........ is less than 1, as it will repeat infintely as it approaches 1, never equalling it. First, let 0.999... be denoted by x. Observe that 10x = 9.999... Observe that 10x  9 = 0.999... = x, and so 10x  9 = x. Thus, 10x  x = 9, 9x = 9, and x = 1. But x = 0.999..., and so 0.999... = 1. just to get you thinking.... when you multiply .99999... times 10, how can you possibly ever get as answer. here's some grade school math for you to ponder: 99 x10  00 +99  990 if you try to solve .9999... x10  you will NEVER find the answer because you will start by multiplying 0 times 9 an infinte number of times; it may always be zero, but you will keep writing zeros down for infinity. 

Title: Re: 0.999. Post by Charles miller on Aug 4^{th}, 2002, 7:24pm I don't understand, even if .999... continues on indefinately, how it could ever = 0. It's length only gets it closer to approximating 1 but can never = one. An infinite number of 9's doesn't change this property. .999... 

Title: Re: 0.999. Post by srowen on Aug 4^{th}, 2002, 8:36pm Indeed, 0.999...9 does not equal 1 no matter how many 9s you add. It seems logical to then say that adding a infinite number of 9s still does not get you to 1, and so 0.999... < 1. However it's deceptive to talk about adding "an infinite number of 9s" because you can't. "0.999... = 1" does not say that you get to 1 after adding an infinite number of 9s. It is only a statement about what this process is getting infinitely close to, but never reaching of course. 

Title: Re: 0.999. Post by OMG on Aug 5^{th}, 2002, 4:11pm :/ 0.9999 approaches 1, never reaches it, but keep trying... I say 0.99999999... > 1.000... because if at some point in time .9999... = 1.000... intersect then there must be a point at which they intersected previously, where 0.99999... > 1.0000... life isn't Cartesian, it is not parallel, they must intersect more than once; think preBigBang math is only a concept, the current universe is reality and only temporary 

Title: Re: 0.999. Post by S. Owen on Aug 5^{th}, 2002, 6:50pm Oh man, that sounds just profound enough to be true... 

Title: Re: 0.999... Post by otter on Aug 6^{th}, 2002, 7:39pm Wow! The last time I dealt with anything even remotely resembling this was over 30 years ago. It seems to me that the entire problem rests with being able to grasp the concept of infinity. Not "very large numbers", but infinity. In fact, the statement .999... = 1 is *only* true as the number of nines becomes infinite. For any finite number of nines, the statement becomes false. Just as an exercise, though, does the logic change if we ask does .999... = 1.000...? Otter 

Title: Re: 0.999. Post by Paul Hsieh on Aug 6^{th}, 2002, 8:12pm Look people  (1a^{n}) = (1a) * (1+a+a^{2}+a^{3}+ ... +a^{n1}) So, 0.999999... = 9/10 * (1.11111...) Now putting 1/10 in the formula above we see that: (1(.1)^{n}) = (10.1) * (1+.1+.01+ ... +(.000...01)) Taking limit of n to infinity we get: (1(.1)^{inf}) = (0.9) * (1.11111...) But 0 < (.1)^{inf} < eps, for all eps > 0, therefore (.1)^{inf} = 0. Which means 1/0.9 = 1.1111... So, 0.999999... = 9/10 * (1.11111...) = 9/10 * 1/0.9 = 9/9 = 1. Got it?  I think the basic trouble people have is that they see a notation like: 0.9999... and they don't really have a good concept of what that even means. Well let me tell you. It means: 0.9 + 0.09 + 0.009 + 0.0009 + ... and nothing else. If you cannot sum that up and get the answer 1.0, then that means you cannot sum an infinite series in general, and that you should not be engaging in discussions of mathematics that go beyond high school. Zeno's paradox is in the same category. Let me be clear. This is just an infinite sum. Its a very straightforward thing. Either you can do it, or you can't. And the answer is 1. That's all. 

Title: Re: 0.999. Post by na on Aug 8^{th}, 2002, 10:03am I think the following is wrong, (1a^{n}) = (1a) * (1+a+a^{2}+a^{3}+ ... +a^{n1}) So, 0.999999... = 9/10 * (1.11111...) It should be (1a^{n}) = (1a) * (1+a+a^{2}+a^{3}+ ... +a^{n}) if n is infinite (hint: how much is inf  1?) so, 0.999999... =x= 9/10 * (1.11111...) 

Title: Re: 0.999. Post by S. Owen on Aug 8^{th}, 2002, 11:30am on 08/08/02 at 10:03:00, na wrote:
In taking the limit as n goes to infinity, it doesn't matter whether we put in n1 for n or not. So there is no problem here involving infinity1 or anything... we are taking a limit. But more importantly, the change to: (1a^{n}) = (1a) *(1+a+a^{2}+a^{3}+ ... +a^{n}) simply makes this not a true statement (try n=2 for example). Did you mean: (1a^{n+1}) = (1a) *(1+a+a^{2}+a^{3}+ ... +a^{n}) Same thing as the original equation, and certainly no different in taking a limit. 

Title: Re: 0.999. Post by Christopher Delovino on Aug 8^{th}, 2002, 3:15pm Lots of good answers why .9999.... would equal 1 but so far everything here has been done analytically. Let's try to graphically prove it. (I'm assuming that the reader has access to a graphing calculator) Let's first look at y = x  1 This gives us a diagonal line with x = 1 on the xaxis (in coordinate form (1, 0) ). Now, I can stop here and say that you can just trace .9999... for x and that will eventually get to 0, but that's no fun. So let's modify the equation a bit by multiplying 1 using (x  1)/(x  1) giving us y = ((x  1)^2)/(x 1) Now this gives us basically the same graph as the previous equation with the exception that there's a gaping hole at x = 1 now (because a 0 in the denominator is a nono). Now if we calculate x = .9, y = 0.1 If we go further, we get (once again in coordinate format) (.9, 0.1) (.99, 0.01) (.999, 0.001) (.9999, 0.0001) ... ad infinitum Well, from that we can see that as x gets closer to 1, y gets closer to 0. If my memory of calculus serves me, if the limit at a point of x approaches a yvalue c from the right the left then the value of y is c. Thus answering Kozo's previous question about 1.00...01 being equal to .9999. In short, .9999 will never reach 1. BUT considering that infinity will never come along we just skip the exact value semantics and say that it is 1. That's basically what infinity represents, "I'm too lazy and no computer has enough processing power to prove this. We're close enough so let's just say that we're there." :P 

Title: Re: 0.999. Post by NoYes on Aug 10^{th}, 2002, 5:18pm Looks like some people didn't pay full attention in calculus. I skipped to the end after the first page and a half so I am sry if this was brought up in the half a page I didn't read :o. Kozo you are a winner :D! He is demonstrating that if you all fallow your ideas of how limits work you will fail. Near the end of page one anshil said: "Using the ideas given above, does it mean that 1.000 ... 001 with infinite zeroes between the two 1s make it equal to 1? No not, it's infinitely small larger than one." If you believe that you must believe that 0.999... != 1 Because 10.9999.... = 0.00000.....1 1.00000....11=0.00000....1 Kozo is showing the error of your interpretation of infinity and limit when he uses the .99999....8 = .9999... = 1 = 1.0000...1 example. As I think Olsen said, we can't reach infinity, well if we can't reach infinity there is no problem using infinity + 1 since both are unreachable, and for that matter infinity  1 is also unreachable. 1 1/(infinity) = .9999999...., that seemed to be already agreed upon. so 1 1/(infinity1) = .99999......8, right? Because 1/(infintiy1) would have to be the number that is greater then 1/infinity by the smallest unit. 11/(infinity) just shorthand for the limit as n> infinity of 11/n, for those of you that are being a little more staunch with your notation, but now we are into limits. If you say infinity doesn't exist or is unreachable then "The limit" doesn't exist! If you say infinity does exist or is reachable then the limit does exist....but you have some other problems, of the infinity +1 sort :D. So again, 0.999999.... != 1 if you say infinity is unreachable, because .9999999 will always be short of 1 by 0.0000...1. You can say the limit of 0.99999... = 1 because you know that .99999... < 1 and is getting closer the deeper you go into the ... or infinity. limit != =, in english, limit is not equality, all it means is you are getting close to it, you are never at it, because if you did get to it you would be at infinity and still have infinity + 1 to go. If you want to argue that you do reach it then I would face you with this problem: if .99999 = 1 when you are at infinity then at infinity + 1 do you have 1.0000....1? How about infinity + infinity? do you have 2 then? Almost forgot to talk about it, but someone brought up the tortoise and the hare (found most famously in GEB...not sure why no one mentioned that...). To say it will never cross the finish line is to assume you take the same amount of time on each step ( i.e. 0>.5, .5>.75, ect.). Maybe I have always misread it, but I have never been entertained by that "paradox." Because of course you will cross the finish line because as the distance being traveled reaches 0 the time to cross it reaches 0. Said the same thing many different ways, hopefully that will help get the message to many different people reading it in many different ways. NoYes 

Title: Re: 0.999. Post by pythagoras on Aug 12^{th}, 2002, 1:17pm infinity + 1? As said earlier, infinity needs to be thought of differently than numbers. It is generally accepted that ... = infinity  1 = infinity = infinity + 1 = ... I can't quite follow NoYes's logic, but when we "reach" the limit we have "reached" infinity. Neither of these can actually be done, and since infinity is really the same as infinity + 1, there is no real problem. A.M. There are bigger consequences of treating infinity as a number. Say S = 1 + 2 + 4 + 8 + ... S = 1 + 2(1 + 2 + 4 + ...) S = 1 + 2S S = 1 S = 1 

Title: Re: 0.999. Post by heywood on Aug 12^{th}, 2002, 1:40pm Heres another example/riddle dealing with infinite sums that I find quite amusing. Suppose you have a perfect superball that bounces half its height on every bounce. Disregard all friction and other sources of energy loss (i.e. if the ball starts at 1 meter then the next bounce it will reach EXACTLY 1/2 a meter.) Now the question is, how long does it take the ball to stop bouncing? This was a great problem from an early calculus class I had. 

Title: Re: 0.999. Post by HaPpY on Aug 12^{th}, 2002, 2:22pm as n approaches infinity where .999... n is the number of digits past the decimal the value approaches 1. therefore if infinity is "reached" the value becomes one. 

Title: Re: 0.999. Post by Paul Hsieh on Aug 12^{th}, 2002, 4:25pm Grumble ... 1. You cannot "set n = infinity" for the simple reason that "infinity" is not a number. Its a concept. 2. One cannot so easily just "take the limit as n goes to infinity" unless you have established each step very carefully. It turns out in this trivial case, there are no issues and doing so presents no problems, but it is in a sense a circular argument. I.e., can't do it until you've established a convergence, which is really part of what we are trying to do in the first place. I did not mean that: (1a^{n}) = (1a) * (1+a+a^{2}+...+a^{n1}) Implies that: 0.9999 ... = 9/10*(1.1111 ...) These are actually two seperate and independent facts that are just true on their face. I tie the two together in later steps in my analysis. 

Title: Re: 0.999. Post by Mongolian_Beef on Aug 13^{th}, 2002, 10:02pm I think the issue really is our system and the conversions that it entails. The point is that the way our system functions, as many of you seem to be arguing in its context, .9999999 repeating is equivalent to 1. Arguing over it also appears pointless because regardless of opinion the variance between .99999999 repeating and 1 is negligible in any real world context. 

Title: Re: 0.999. Post by chris mallinson on Aug 22^{nd}, 2002, 9:41pm It has been said here that .99999....... multiplied by 10 is 9.999999999.... .99999....... plus 9 is also 9.999999999.... Is there something to be said for the fact that the latter will theoretically approach the number 10 faster than the former? chris 

Title: Re: 0.999. Post by Jason Short on Aug 22^{nd}, 2002, 11:42pm The notation I've seen for 0.999... (infinitely repeating) is to use 0.9 with a line over the 9. Instead, I'll put the line under the 9 so it becomes 0.9. If we switched from base 10 to base 20, then we could call our digits {0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J}. Thus "J" is the digit "19". This may be comparable to the # digit spoken of earlier: the average of 0.9_{10} and 1 is 0.J_{20}. So the number 0.9_{10} could be compared to 0.J_{20}, and we would be tempted to think that the latter was larger. But we are implicitly assuming here that they are different; in actuality both are simply equal to 1. Here is a related topic: a paradox I once heard. Imagine a game of chance in which a coin is flipped indefinitely until eventually it lands tails. You then count the number of times it landed heads (n), and the person playing the game gets 2^{n} dollars. The question then becomes: how much money does a player expect to win playing this game? Logically, we would think this is ((1/2)*1+(1/4)*2+(1/8)*4+...) since there is (for instance) a 1/2 chance that heads will be flipped 0 times, and if this happens the player will win $1. But each term simplifies to 1/2, so the series becomes (1/2+1/2+1/2+...). The infinite series of 1/2's seems to sum to infinity: it seems as though the payoff for this game is infinite. Yet this seems rather rediculous; in fact we've assumed that there is an "expected" payoff. The actuality is that the expected payoff is undefined. My point is that infinity isn't just a rather large number, it is a concept that has undefined value. There is no such thing as "infinity plus one" (^^). You simply cannot treat infinity as a value. Thus it is meaningless (although seemingly clever) to consider the value 0.99, and claim that it is larger than 0.9. In this case, assuming the given decimal representation has a value (which is sort of implicitly true, since decimal notation is simply an abstraction we made up to represent values), the best we can do is treat it as an infinite series and reduce that series to find the value. This has been done several times elsewhere: the value is 1. By contrast, the value 9 is "infinite", and certainly does not have a value. It would be quite meaningless to compare it to, for instance, 8, and try to say that it is "larger" since 9>8. 9>8 does not prevent the fact that 88>9. 

Title: Re: 0.999. Post by wowbagger on Aug 24^{th}, 2002, 4:16pm on 07/30/02 at 01:52:55, Kozo Morimoto wrote:
As far as I followed the discussion, nobody replied to this specific question yet, so here's my proof that yes, you get 1: What you do is calculate a sequence of numbers generated by the recursion formula a_{n+1} = a_n / 10 + 0.9, starting with some arbitrary a_0 (e.g. 0.8) If you "repeat the process to infinity", you take the limit of both sides of the recursion equation as n goes to infinity. If the limit of the sequence a_n exists  let's call it A  you can substitute a_n and a_{n+1} by A, since being elements of the same sequence, they must have the same limit A. So we have A = A / 10 + 0.9 9 / 10 A = 0.9 A = 1 Note that this result doesn't depend on your initial value a_0, it works for all real a_0. If you start with a_0 > 1, you will approach the limit from above, however. In the same way one can show that a_{n+1} = (a_n + 2 / a_n) / 2 goes to sqrt(2), i.e. A*A = 2 in this case. Well, I don't know whether this is of much help as some will probably feel uncomfortable with the substitution of the limit A, but I didn't want to leave the question unanswered. Kozo: I really respect your being sceptical. BTW, I'm convinced (not to say I know) the infinite series does sum up to 1. The bit about the "#" was really weird, though. Contributions of yours to other threads I read are also stimulating  keep it up! 

Title: Re: 0.999. Post by pythagoras on Aug 25^{th}, 2002, 3:39pm on 08/22/02 at 23:42:57, Jason Short wrote:
Well, actually this has been argued against by people I know. It's kind of nonsensical to imagine that the 9's repeat to the right in this case, so say they repeat to the left. I'll denote something repeats to the left in parentheses like this: (9)9. Now consider (0)0  (0)1, that is: ...000000 ...000001 Let's try to do this like we don't know what a negative number is. Since 1 is greater than 0 we have to borrow a 1 from, uh, somewhere... (An imaginary 1 in front of all the zeroes?) to get: ...000000 ...000001  ...xxxxx9 and since we borrowed our theoretical subtraction will end up: ...000000 ...000001  ...999999 so (0)0  (0)1 = (9)9. That's odd. There is actually a concept called nadics that deals with this sort of thing, and this is just an example in the 10adics. 

Title: Re: 0.999. Post by justin m on Aug 28^{th}, 2002, 2:40pm just like the limit of [f(x+h)  f(x)] / [xh] as h>0 is a "pretty good approximation of the rate of change of f its not just "pretty good", if you evaluate the limit, this IS the rate of change of f, its a simplet hing called the derivative. just like the infinite geometric series 9/10 + 9/100 + ... is not just a "pretty good" approximation, when you evaluate the limit you get an exact number, 1 additionally, 1.0000...001 is mathmatical nonsense, because you cant terminate a decimal that, by definition with the "...", is nonterminating. perhaps a college math course or two will help you swallow your pride on 07/29/02 at 06:40:49, Kozo Morimoto wrote:


Title: Re: 0.999. Post by wowbagger on Aug 29^{th}, 2002, 2:27am on 08/28/02 at 14:40:05, justin m wrote:
Of course, the correct formula is f'(x) = lim_{h>0} ( f(x+h)  f(x) ) / h Perhaps people writing illiterate posts shouldn't encourage others to take courses? 

Title: Re: 0.999. Post by Kozo Morimoto on Aug 31^{st}, 2002, 11:32pm OK, here is another thought exercise  I haven't work it all the way thru but... Imagine a number system with 20 digits A to T, but still in base 10 so that A maps to 0, C maps to 1 etc until S maps to 9 and T maps to something between 9 and 10. (B maps to somewhere in between 0 and 1, D maps between 1 and 2 etc) So when you have 0.9 in normal system, that would be same as A.S, however, there is a number A.T which is bigger than 0.9 and less than 1.0. So 0.999... would map to A.SSS.... but there is a number A.TTT... which is bigger than 0.999... but less than 1.0? 

Title: Re: 0.999. Post by Pietro K.C. on Sep 1^{st}, 2002, 11:57pm Sorry to insist on this point, but it seems necessary for the argument I'm going to give regarding Kozo's question. The set of characters "0.999..." DOES NOT MEAN adding the numbers 0.9 + 0.09 + etc, it is not the result of some infinite process, nor any other complicated nonsense like that. For the simple reason that an infinite process (be it addition or whatever) does not yield a result. Srowen and aadash have explained it best. The set of characters "0.999..." means the limit that the sequence (0.9 ; 0.99 ; 0.999 ; etc) tends to, in the sense of Cauchy, which is the following: a sequence (a(n)) = (a1, a2, etc) tends to a limit L if, for every real number r, there exists a natural number N such that a(n)  L < r for all n > N. The definition of the particular sequence we are dealing with does involve addition, but the concept of a limit does not depend on it; the concept of infinite addition is inconsistent. Come on, what is the point in knowing complicated integrals if you don't have a grasp of the beautiful underlying structure? No one denies that no finitenumbered term of the sequence equals or exceeds 1, but to say that, or to attempt proofs just by adding the succesive terms, is to miss entirely the meaning of the expression. Now, all of you have excellent problemsolving skills, as can be seen by your posts. Surely no one wil have trouble proving that the number 1 is a limit of the sequence (0.9 ; 0.99 ; etc) using the definition above. Hint: the difference between 1 and the nth term of this sequence is 10 taken to the (n)th power. (As an exercise, if you haven't already done it, you may also prove that, if a sequence of real numbers has a limit, it is unique) Hence the question "0.999... = 1" is resolved, and the answer is affirmative. About Kozo's alternative number system: the fact that a number T is less than 10 does not mean that 0.TTT... is less than 1. For instance, 9.9. If we take 0.TT to mean T/10 + T/100, then 0.TT = 0.99 + 0.099 = 1.089. Since 0.TTT... > 0.TT, it must also be greater than 1.089, which is greater than 1. However, my spidersense (and the previous posts) tells me that this isn't enough of an argument for Kozo. He may argue that there is some number not expressible in our decimal system for which the properties 9 < T < 10 and 0.TTT... < 1 simultaneousy hold. I intend to show that no such number exists. Never mind the number system, and suppose that 9 < T < 10. I take it we can represent ANY number by the letter T. Now construct the sequence S = (0.T ; 0.TT ; etc) by means of sums with progressively more numerous terms, in the same manner as we constructed the infamous "0.999...". By the expression "0.TTT..." we shall then mean the limit of this sequence, if indeed it does approach a limit. To prove that the limit exists, we could resort to the axiom of completeness of the set of real numbers, or to the axiom of existence of a supremum of an increasing bounded sequence (which is equivalent, cf any introductory text on real analysis, of which there are many  if you would like the proof, email me). But, since it obviously suffices to show a number that satisfies the limit criterion, and this is a simple case, we shall do so. I say that T/9 is the limit of the sequence S. To see this, consider the difference T/9  (the nth term of S); a simple proof by induction shows that this difference is T/(9*10^n) for all natural n. Clearly, for any number T, this difference can be made as small as we please by increasing n; in Cauchy's words, for all r > 0, there is an N such that T/(9*10^n) is less than r for all n > N. Hence, T/9 satisfies the definition of a limit of the sequence S = (0.T ; 0.TT; etc). And therefore 0.TTT... = T/9. It is not a coincidence that 0.TTT... "=" T*0.111... = T*1/9 = T/9. The multiplicative property of limits can be used to rigorously remove the sarcastic quotation marks around the equal sign. Now, you may notice that T/9 > 1 for all T > 9, which was the original hypothesis. So no number "not expressible in our decimal number system" exists that satisfies both T > 9 and 0.TTT... < 1 at once. That is the basic flaw in the # argument. As a side point, I would like to remark that there exists no numbers that cannot be arbitrarely well approximated by decimal expansions (prove this); so any eventual 0.###... would have a representation involving only the digits 0 through 9. If we were to make it greater than 0.999... but less than 1, the first digit would have to be 9; but also the second; and so forth, so that a number greater than 0.999... and less than 1 cannot exist. But this type of argument alone cannot make up for the rigor of the preeding proof. I'm sorry if I wrote too much, it's just that I would like to clear this up once and for all, so that such smart people (I was impressed by many others of Kozo's ingenious posts) would not waste their valuable time on matters of definition. 

Title: Re: 0.999. Post by Xanthos on Sep 2^{nd}, 2002, 5:42pm What about using the sum to infinity? If we take 0.999.... as 0.9+0.09+0.009 etc. then it is a geometric sequence. Therefore if we use the sum to infinity formula it should give us what 0.999.... is equal to. S_{inf}= a/(1r) In this case the starting number (a) is 9 and the common ratio (r) is 0.1. Substitute and you get 0.9/0.9, which is equal to 1. I might have missed something and it might have been said before but I thought I'd weigh in with that. Xanthos 

Title: Re: 0.999. Post by Kozo Morimoto on Sep 2^{nd}, 2002, 7:49pm 0.T maps to 19/20 = 0.95 which is > 0.9 and < 1 0.TT maps to 19/20 + 19/400 = 0.9975 which is > 0.99 and < 1 0.TTT maps to 19/20 + 19/400 + 19/8000 = 0.999875 which is > 0.999 and < 1 so on. so wouldn't 0.999... < 0.TTT... < 1 ? 

Title: Re: 0.999. Post by NickH on Sep 2^{nd}, 2002, 11:27pm "so wouldn't 0.999... < 0.TTT... < 1 ?" No. In the limit, both are equal to 1. What you have written as 0.TTT... is simply the base 20 equivalent of 0.999.... We can also consider 0.111... (base 2), 0.222... (base 3),.... In all cases the sum to infinity of the geometrical progression is 1. Nick 

Title: Re: 0.999. Post by S. Owen on Sep 3^{rd}, 2002, 5:41am How about this, define two sequences: A_{i} = 0.999...9 (there are i 9s) B_{i} = 0.TTT...T (there are i Ts) So the sequences are: {0.9, 0.99, 0.999, ...} {0.T, 0.TT, 0.TTT, ...} It is true that A_{n} < B_{n} < 1 for any n. However, the limit of both sequences is still 1. 

Title: Re: 0.999. Post by James Fingas on Sep 3^{rd}, 2002, 11:44am This is a killer thread! Has anyone done epsilondelta proofs? This is the most basic (consistent) way of saying that a given limit exists and converges to the specified point. The way you use them in this case is to say: The series 0.9 + 0.09 + 0.009 + ... converges to 1 iff for every epsilon around 1, there exists an M so that by summing up the first M terms of the series, the sum is within epsilon of 1. Basically, an epsilondelta proof challenges you to find a differenceany differencebetween the series 0.9 + 0.09 + 0.009 + ... and the number 1. It says that any purported "difference" that you find will vanish after I sum enough terms. Since you cannot find a difference, then the numbers MUST be the same. To prove this, I show how to get M from epsilon. This leads to the following argument: 1) Assume that the series 0.9 + 0.09 + 0.009 + ... sums to a real number a. By the ratio test, the series converges to a single value. 2) Now we assume that a != 1. 3) This means that a1 has a nonzero value. Let's take half the absolute value of a1, and call it epsilon. 4) But if we take M = 1  log10( epsilon ), and sum the first M terms of the series, we will find that it is closer to 1 than epsilon. This value of epsilon is therefore rejected. Since this holds for any epsilon, there are NO NUMBERS between 1 and a. 5) But this contradicts our assertion that a1 != 0. Consequently, it must be true that a=1. If there is no difference between a and 1, then I argue that they must be the same number. 

Title: Re: 0.999. Post by Pietro K.C. on Sep 3^{rd}, 2002, 2:39pm Yes, I did do a "deltaepsilon" proof, even though there is no delta when dealing with sequences, but as I thought that epsilon was an ugly name for a variable, when written out whole, I used the letter r instead. And in keeping with the spirit of this site, I didn't spell out all the details, such as constructing the number N such that limit  sum up to the nth term < epsilon for all n > N. Maybe I do deserve not being read for posting such a huge text. But I repeat: arguments of the type "the series sums to..." will not convince Kozo, and he is right. These arguments are incosistent, and his rebuttals go to the heart of this incosistency. Kozo, I don't know if I misinterpreted your previous construction or if your last post is a new question,but anyways. You can apply the same type of findthedifferenceasafunctionofn proof I gave in my LARGE post to show that "0.TTT..." = limit (0.T ; 0.TT; etc) (now in base 20) = T/19. For T = 19, this limit is 1. Same as my proof for 0.999... in base 10. Let me give YOU a puzzle now. Let the sequences S1, S2 be defined by: S1(n) = 1  (1/n) S2(n) = 1  (2/n) and let the numbers A, B be their limits as n>oo, respectively. Would you say that A < B, just because each term in S1 is less than the corresponding term in S2? That is not keeping with the definition of a limit. The limit of a sequence need not be included in that sequence, as the above examples demonstrate. Clearly, A = lim S1 = 1, and B = lim S2 = 1, but 1 is not a term in either sequence. The same phenomenon occurs in the definition of 0.999... and 0.TTT...; these symbols indicate limits of sequences, and are not actually terms, and so the inequality does not have to hold. The key here is that "a(n) < b(n) for all n" does NOT imply that "lim a(n) < lim b(n)". Hope that helps. 

Title: Re: 0.999. Post by Kozo Morimoto on Sep 4^{th}, 2002, 5:21am I've come to the conclusion that this has come to a point of definition/symmantics. I googled and came up with: http://www.math.toronto.edu/mathnet/answers/infinity.html which I think covers this discussion to my satisfaction in an easily accessible form. The definition I was using is covered by point 1 on the referred website, and in this context, I was right. The definition that everyone else here was using is covered by point 2 on the referred website, and in that context, everyone was correct. In this definition, convergence and infinity is used interchangeably but I was treating them as 2 different 'things'. It certainly has been an interesting discussion though! 

Title: Re: 0.999. Post by Pietro K.C. on Sep 4^{th}, 2002, 12:08pm I looked at the site, but I still don't get how point #1 validates your point of view. ??? It says infinity doesn't exist as an element of a "number system", but very few of us proposed that it did. I really didn't understand what you said. ??? ??? ??? I still think that a string of symbols such as 0.999... requires the concept of convergence to make sense, and is meaningless in the "number system" context alone. 

Title: Kozo was right! Post by James Fingas on Sep 5^{th}, 2002, 11:43am It turns out that Kozo was right all along. There is a difference between 0.9999... and 1. At least the floor function says so ;D 

Title: Re: 0.999. Post by Dustin on Sep 6^{th}, 2002, 5:58pm .99999... < 1 if someone wrote .99999... on a piece of paper for there entire lives, and every last decended of theirs did the same thing, it would still never reach 1. Someone also earlier said that if 1/3 = .3333... 2/3 = .6666... then technically 3/3 would = .9999.... However since we know that 3/3 = 1 then .9999... must also = 1 This idea is flawed because of the fact that the representation of both 1/3 and 2/3 is also flawed. 1/3 does not = exactly .333... and 2/3 does not = exactly .666... The reason math labels it as that is because with out math system the closes representation we can get of 1/3 is .333... but that does not make it = to .333... and hence 3/3 is not = .999.... and therefor .999.... is not = to 1 

Title: Re: 0.999. Post by jeremiahsmith on Sep 6^{th}, 2002, 7:05pm on 09/06/02 at 17:58:47, Dustin wrote:
Did you miss the entire thread, somehow? Read > Comprehend > Post 

Title: Re: 0.999. Post by Dustin on Sep 6^{th}, 2002, 7:58pm I read the first group of posts and then skipped over the rest because of the fact that the same things were being said. 

Title: Re: 0.999. Post by Kozo Morimoto on Sep 6^{th}, 2002, 8:07pm Ok, how about this... You have a set A = {0.9, 0.99, 0.999, ...} which is inifinite in size. You also have a set W = {1, 2, 3, 4, 5...} which is the set of whole numbers and its size is also infinite. You can see that there is a 1to1 mapping between set A and set W, where number of 9s in an element of A is an element in W eg 0.9 > 1, 0.9999 > 4 etc etc So if 0.999.... = 1, then 1 must be an element in set A. If so, what does 1 in set A map to in set W? Since it doesn't, set A must no include 1, therefore 0.999... <> 1 

Title: Re: 0.999. Post by Pietro K.C. on Sep 6^{th}, 2002, 11:59pm No, you're still hanging on to the notion that a sequence's limit must also be a term in that sequence. You have successfully proved that 1 is not in set A, and then used the fact that 0.999... is in that set to conclude that they are different numbers. There is a tiny flaw, and to uncover it I now pose the same question: what number does 0.999... map to? Surely not infinity! You yourself referred us to a website that said infinity does not exist in the context of a "number system". No, let's work backwards. What is an arbitrary member of set W? A positive integer x. A FINITE positive integer x. What element of A is mapped onto that FINITE positive integer? The number 0.999...9, with exactly x 9's after the decimal point. And that is surely not equal to 0.999... repeated indefinitely. So 0.999... is not mapped onto any member of W, so it is itself not in A. At this point I would like to encourage anyone who would object to the above argument by saying that infinity IS included in the set W (the integers) to pick up a book on the subject, study it, and stop waving their intuitions around like they're foolproof methods of ascertaining mathematical truth. If the book is too boring, at least pick up one on the history of mathematics, and read about the nineteenth and twentieth centuries. :P Anyways. If I were to use your kind of reasoning, Kozo, I could easily construct an example where 1 IS included in the set A. For instance, construct the set B as: B = {1/1, 1/2, 1/3, etc}. There is also a onetoone mapping between A and B. If we assume that 0.999... is in A, we must also assume that 0 is in set B. The situation is exactly the same. If you tell me that 0 is not in B because, no matter how large n is, 1/n is not zero, then I reply that, no matter how many 9's we put after the decimal point, 0.999...9 (terminating) is not 0.999... (repeating). If, however, you accept that 0 is in B, suppose that 0.999... = 1, and that it is in A. What does 1 map to in B? Why, 0, of course! Because the mapping is onetoone, and any terminating decimal with n 9's maps to 1/n > 0. Voilà, the contradiction is gone. The point is, for the same reason you don't believe that "infinity" is in the integers, you mustn't assume that a sequence's limit is in that sequence. I gave examples of this in my previous posts. And 0.999... indicates precisely the limit of set (sequence) A. It makes no sense to argue about inequalities satisfied by elements of A, because 0.999... is NOT in A. Please, read the posts of S. Owen, Aadash and myself a bit more carefully  the main ideas are there. And, if you come up with another of these "thought experiments" (VERY ingenious  you're like a modern Zeno :)), try to solve them yourself using the concepts we presented you. Chances are, you'll succeed. P.S.  Dustin, don't you think that "infinitely good approximation" is something of an oxymoron? Maybe you should think a bit more about the subject, because it is not easy  it took mankind about 3000 years to formalize it properly. 

Title: Re: 0.999. Post by Kozo Morimoto on Sep 7^{th}, 2002, 12:52am But 0.999... is in set A, otherwise size of set A would be finite. How can size of set A be infinite, but still not include 0.999...? 

Title: Re: 0.999. Post by Pietro K.C. on Sep 7^{th}, 2002, 9:04am In exactly the same way that {1, 2, 3, etc} can be infinite in size without including anything like "infinity". I ask you again, if 0.999... is in A, what does it map to in W? With that question you went to the heart of the matter; a sequence (not a set, mind you) of real numbers like (a1, a2, a3, etc) is in reality a function f : N > R, from the natural numbers (positive integers) to the reals. When we list the terms of the sequence, what we are saying is that f(1) = a1, f(2) = a2, f(3) = a3, etc. Therefore, a number x is in the sequence IF AND ONLY IF there is some positive integer n such that f(n) = x. Since there is no positive integer n satisfying f(n) = 0 in my previously defined sequence B, then 0 must not be in B  the same goes for 0.999... and sequence A. The concept of infinity, and things like infinite sets whose elements are all finite, is very often counterintuitive. From a nonformal point of view, what makes the set of natural numbers infinite is that, given any one of them, you can always find a bigger number by adding 1 to it. You may argue that, if we imagine this process carried on indefinitely, as it must be, "infinity" will be included in the set of naturals. But it will not. By adding 1 to a number, you are simply showing that there is a contradiction in assuming that N is finite. The argument is an abbreviation of "suppose N is finite. Pick its largest element. Add 1 to it. The new element is a natural number, but it is larger than the others. Therefore it was not included among the finite list of numbers. But that list was arbitrary, so N must have an infinite number of elements". If you do not assume that N is finite, then you cannot necessarily pick its largest element  which is another thing the argument says, that there are arbitrarily large naturals. Do you see this? That, after you stop assuming that N is finite, you have nothing to assure you that addition CAN be carried on indefinitely? It cannot. BORING PROOF Addition, too, is a function, one that maps two naturals on to a third one. What a + b = c means is that +(a,b) = c, where + is a function. However, the value A = 1+ 1+ 1 + ... = +(1, +(1, +(1, ...) ) ) is not defined. Because, if defined, it is either a natural number, or it is not. And we stated that + takes pairs of naturals as arguments. So, if A is not a natural, A = +(1, A) is not defined. If, however, it is a natural, it must satisfy +(1, A) = +(0, A), so 1 = 0 (by the properties of function +). So A must not be a natural, and therefore not anything.  I think you can see how subtle this is. The modern point of view is that a set A is infinite if and only if it can be put in onetoone correspondence with a part of itself. For instance, the naturals. The function f(n) = n + 1 is a onetoone mapping from {1, 2, 3, etc} to {2, 3, 4, etc}, that is, from N to N  {1}. Hence N is infinite. Now I define set A to include ONLY numbers of the form 0.999...9, with FINITELY many 9's, and ALL such numbers. I am deliberately excluding 0.999... from A. If I define f(0.999...9) = 0.999...99, that is, if given a number in A I map it to another which has exactly one more decimal place, I establish a onetoone correspondence between A = {0.9, 0.99, etc} and {0.99, 0.999, etc} = A  {0.9}. So A is infinite too, and includes nothing but finitely representable numbers. I refer you to texts on modern set theory (read about Georg Cantor's work) for more examples on how infinite sets can be confusing. :) The following page has one of the most basic examples: http://users.pipeline.com.au/owen/infinity.html 

Title: Re: 0.999. Post by Pietro K.C. on Sep 7^{th}, 2002, 8:23pm Thought of something today... suppose set A is infinite, and does contain 0.999... . However many 9's there are, that is still just one element. If we remove it, set A will no longer contain 0.999..., but will still be infinite. How about that? ;) 

Title: Re: 0.999. Post by Jason on Oct 1^{st}, 2002, 3:24pm If .9999... = 1, then (.9999...)^(infinity) should equal (1)^ infinity. However, any number less than 1, no matter how close, will eventually approach 0, not 1. The number 1 raised to 'n' as n > infinity will always be an absolute 1. Therefore, .9999... < 1. 

Title: Re: 0.999. Post by S. Owen on Oct 1^{st}, 2002, 3:41pm on 10/01/02 at 15:24:17, Jason wrote:
It sounds like you are presupposing that 0.999... < 1! 

Title: Re: 0.999. Post by Pietro K.C. on Oct 1^{st}, 2002, 5:17pm Quote:
And how close IS 0.999... to 1? Or better yet, what does 0.999... mean? A few links: http://hades.ph.tn.tudelft.nl/Internal/PHServices/Documentation/MathWorld/math/math/r/r223.htm http://www.math.vanderbilt.edu/~schectex/commerrs/#Infinity  "OTHER COMMON CALCULUS ERRORS  Jumping to conclusions about infinity" link 

Title: Re: 0.999. Post by Icarus on Oct 10^{th}, 2002, 7:49pm Let's back up here. The first tool ever devised for handling when two numbers calculated in two different ways are equal was Eudoxus' Method Of Exhaustion. It works like this: suppose you want to know if x = A, but you don't have a way of expressing what x is exactly, only a means of approximating x as closely as you like. (Eudoxus invented it to handle such things as the ratio of the circumference of a circle to its diameter. But it applies here. We can approximate x=0.999... as closely as we want with a terminating sequence of 9s.) Eudoxus reasoned that if for every number B < A, you could show that x > B, then x could not be less than A, otherwise x would have to be less than itself!. Similarly, if for every number C > A, you can show that C > x, then x cannot be greater than A. If x is not less than A, or greater than A, then x IS A. So first, can we truly approximate x with a terminating sequence of 9s? Yes. I will not prove it here, but if you doubt, ask and I will supply one. So given any [epsilon]>0, there is a number y=0.99...9 (the dots represent a finite set of 9s here) so that xy < [epsilon] Second let C be an arbitrary number > 1. let [epsilon] = C1, so [epsilon]>0. Then there is a y=0.99...9 so that xy<C1. In particular, xy<C1. But 1>y, so x1<xy<C1 Add 1 to get x < C. Since C was arbitrary, x is strictly less than every number greater than 1. Third let B be an arbitrary number < 1. This time we will be more restrictive with [epsilon]: [epsilon]=(1B)/2. We must also note some thing else, just like we can pick a y so that xy<[epsilon], we can also pick a y=0.99...9 so that 1y<[epsilon]. (Again, if you don't see this, ask and I will supply the proof). Further, we can pick one y that will do for both (the more 9s you add, the closer you get to both x and to 1, so pick a y with enough 9s to satisfy both conditions). This gives: y  x < xy < [epsilon] 1  y < [epsilon] Adding gives (1  y) + (y  x) < 2[epsilon] = 2(1  B)/2 1  x < 1  B B < x Since B was arbitrary, x is strictly greater than every number less than 1. x=0.999... is greater than every number less than 1, and less than every number greater than 1. There is no other choice: it has to be 1. As if this post were not long enough already, I have some comments. Some of you will see this as an [epsilon]e[delta] proof in disguise. I answer, No: [epsilon][delta] proofs are Method of Exhaustion proofs in disguise. Eudoxus was first. Cauchy borrowed from him. A few might also see another proof in that 0.99...9 is shown (actually only stated, but it can be shown) to approach both x and 1 as 9s are added. I answer, Not really. That proof requires a proof of uniqueness of limits, which brings us back to the above. I offer this proof not because it is better, or more accurate than previous ones, but because it involves the least number of concepts of any I have seen. It basically depends only on the following: That 0.99... actually is a real number. (If anyone has questioned this, I missed it, but it is something that should be addressed to be completely thorough.) That all real numbers are comparable: i.e. given any real numbers a and b, one of the following must be true: a < b, a = b, a > b That < interacts with + and  according to the wellknown rules. And some basic properties of decimal notation, used in the parts I skipped. No, I'm not done yet: It is possible to define numbers that are greater than 0.99...9 but less than 1. It is done in NonStandard Analysis all the time. But these are not Real numbers. (I am using "Real" here as the name of a set of numbers, not as opposed to "fake" numbers. All numbers are only concepts, and no concept possesses greater reality than any other.) These new "infinitesimal" numbers obey most, but not all, of the standard laws of arithmetic. They cannot be expressed as decimal numbers, and they do not negate the proof above. But they do show the fallacy of simply stating that there can't be any numbers between 0.99... and 1. If you would like to learn more about Nonstandard analysis, infinite numbers, or nadic numbers (someone mentioned them above, or I might have gone on about them myself  they're cool!) look on google. For infinite numbers, I suggest searching for "Transfinite numbers". 

Title: Re: 0.999. Post by Jeremy on Oct 11^{th}, 2002, 11:44pm ... right... how about this. what does .999...  1 = ? Either A) 0 (in which case 1 = .999...) B) something other than 0... i'm thinking most people would say "0.00001", or "0.0000000001"... basically the number of 0's will equal the number of 9's that you took away from the original one. well it's easy to figure out the number of 0's with small values of .999, for example .9  1 = .01 .9999  1 = .00001 ok... well .999... by definition has an infinite 9's after it. well infinity by definition does not stop... it just goes on and on and on and on. if it were to ever stop it just wouldn't go on for infinity. s you'll never get to that one at the end... just lots and lots of 0's... so you'de never reach that ...01. Well so B can't be the case, so lets all go back to A. ok i suck for not using any math to try to prove this, and instead using intuition... but i think the best bet was .999... = x .999... * 10 = x * 10 10x = 9.999... 10x  x = 9.999...  .999... 9x = 9 x = 1 but you know what? everything i just said has already been said before. yeesh why is this topic so long? think we'll ever get to intinity posts? 

Title: Actually Post by FenderStratFatMan on Oct 16^{th}, 2002, 6:30am I really don't see were you are coming from saying that _____ _____ .99999 = 1 because if .99999 then there is no way that 1 = .99999. Obviously a number isn't = to another number unless it is written in a form of the number. And I sure as heck can't see how .99999 is any way of writing 1. 

Title: Re: 0.999. Post by TimMann on Oct 16^{th}, 2002, 9:21am By that reasoning 1/2 is not equal to 2/4 because it's not written in the same form. I.e., that's not an argument at all, just a random wrong statement. Hmm, I sound crabby. Got to stop posting before breakfast. 

Title: Re: Actually Post by jeremiahsmith on Oct 16^{th}, 2002, 10:05am on 10/16/02 at 06:30:17, FenderStratFatMan wrote:
Perhaps, then, you should consider reading the thread first, as my fellow boardmates have explained it very well, in about twenty different ways. I'm crabby, too. 

Title: Re: 0.999. Post by GRAND_ADMRL_THUORN on Oct 29^{th}, 2002, 8:30pm (MY SECOND POST, YAAAAAA! ;D) unfortunatly, it seems that just about everyone in here is wrong, im surprised most of you intelligent individuals didnt see the answer from the begining ??? .9999..... is GREATER than 1!!! :o no computer can show you .999.... actually written out you cant actually put the true .999.... on paper its completely impossible :/ but showing 1 is easy and takes up infinitly less space on paper or hard drive/ram ;) so in actuality .999.... is infinitly larger than 1 :) 1x=.999.... :D (lol, i really wanted to add something to this wonderful thread, even if it was only a joke ;D) 

Title: Re: 0.999. Post by Icarus on Oct 29^{th}, 2002, 9:09pm 0.99... 1 Which did you say was bigger? ;D 

Title: Re: 0.999. Post by Marc Nielsen on Nov 4^{th}, 2002, 9:41am How long can you guys keep arguing wether 0.999....=1??? I think that it was pretty clear in one of the first postings. And even if the 1/3+1/3+1/3 = 1 argument doesn't convince you, then there's a much easier way to mathematically prove it. how did you learn to 'round up' (I don't know if that is the English term  I'm not thinking of a ranch ;D) a decimal figure??? Notice that the figure 1 isn't 1,00000... Anyways... Only the last two statements are true. That's a fact. I don't know if it's mentioned anywhere in this thread  I'm not about to read all four pages ;D And I'm not going to reveal it right away... You try 'n figure it out (pun intended) regards Marc Nielsen 

Title: Ooops... Post by Marc Nielsen on Nov 4^{th}, 2002, 10:54am 'Why O Why, didn't I read all 4 pages :'( ' Sorry 'bout that  didn't see the last couple of posts... :[ regards Marc 

Title: Re: 0.999. Post by Guess Who on Nov 5^{th}, 2002, 10:37pm let n = a real number, lets not deal with imaginaries yet. if n = real number that is NOT 0 n / 0 would not exist ( n divided by 0 is undefined ) yet, any n / m where m is a number that APPROACHES 0 is positive / negative infinity by definition, as the only undefined rational expression is when the divisor is 0. as any number divided by a very small number becomes a very very large number, and if this number approaches 0, then n / m would approach positive or negative infinity. > means approach E means from the set of R means reals so {mm > 0, m E R} does NOT equal 0 because as a divisor, one produces [approaching infinity] and the other is undefined and since there are 2 different "quotients" (infinity and the crossed out 0 denoting undefinition), the divisors therefore cannot be the same number so if a number that approaches zero is not zero, how can a number that is approaching one be one? ??? to dispel the theory that 1/3 is 0.33333...., do 1 / 3 by long division, there will always be a remainder of 3 that doesnt belong anywhere. Sure, it is really small, and you can divide it into another decimal place, but THE REMAINDER IS ALWAYS THERE; THERE IS NO WAY OF ELIMINATING THIS REMAINDER COMPLETELY. :o It will always lag behind the 3 in the smallest place. The remainder is the essential difference between 1 / 3 and 0.33333333..... WHICH NEEDS A REMAINDER OF 0.000000........3 before it can = (1/3) , so 1/3 actually does NOT equal 0.33333......., as there will always be something missing one place further down 

Title: Re: 0.999. Post by Guess Who on Nov 5^{th}, 2002, 10:44pm oh, and get it through your heads: infinity DOES EXIST in the real world, so you cant ignore it and disregard the 1 or whatever digits that does happen after an " infinity " of 0's, as the 1 would be getting smaller more quickly than you can put 0's on. i mean, hey, theres an infinity of numbers between every single real number, so there is a concept of grasping beyond infinity. ( if 1 < n < 1.0001, then n can be an infinite number of numbers ) 

Title: Re: 0.999. Post by Pietro K.C. on Nov 6^{th}, 2002, 4:41am AAAAAAARGHHHHHH!!!! :P :P :P :P :P :P :P :P I really REALLY try to be nice to the best of my abilities in answering people's doubts and wrong conclusions, because I myself have made more than my fair share of mistakes. But because of the "get it through your heads" comment, which is VERY snobbish, I will go against custom this time. So my answer is this. I won't even try to argue with you, since you have clearly not read a SINGLE post of this thread, and would therefore not have the patience to read the lengthy exposition necessary to clarify your misconceptions. Further, you obviously have no idea about what an expression such as "0.999..." actually MEANS, and are basing your arguments on things which have nothing to do with the question. Enlightening though it was that you pointed out that "infinity exists" (oooh... how clever...), that sort of infinity has no bearing whatsoever on the matter at hand, and I must also disagree that it exists in the "real world" (which is even more irrelevant). Now if you REALLY want to argue about 0.999..., go read some of Kozo Morimoto's posts, which at least are interesting. So there you have it. A post of mine on just four hours' sleep and with a project due in half an hour. 

Title: Re: 0.999. Post by richi on Nov 6^{th}, 2002, 5:00am I must say I am surprised by the vast amounts of incorrect proof that ".999...(rep indef) = 1 .999...(rep indef) is a number that approaches 1 as the number of significant figures (9's) approaches infinity, but as the number of significant figures never reachs infinity, .999...(rep indef) never reaches 1. 

Title: Re: 0.999. Post by S. Owen on Nov 6^{th}, 2002, 5:41am on 11/06/02 at 05:00:48, richi wrote:
The point is that "0.999..." does not equal any decimal like "0.999...9" that has a finite number of significant digits. The notation can easily fool one into thinking about "0.999..." as just a special kind of "0.999...9", one with "infinite" digits, and thus something that must be less than 1. That's the point of the riddle  this seemingly reasonable reasoning is wrong. "0.999..." can only mean "the value that "0.999...9" approaches, but never reaches, as the number of digits grows to infinity;" this is 1. The true issue here is not what "0.999..." equals, but what it really means to begin with. Once that is clear  and I think the meaning of "0.999..." can only be what I reiterated above  the value of "0.999..." is unquestionably 1. 

Title: Re: 0.999. Post by richi on Nov 6^{th}, 2002, 5:35pm It seems to me that the point has been blatantly missed. To say "is equal to" is not an approximation. The induction is: as the number of significant figures of (lets call it x) increases, x approaches 1. Because, by definition, we can never reach infinity, x will never reach 1. 

Title: Re: 0.999. Post by S. Owen on Nov 6^{th}, 2002, 5:47pm "0.999..." is not an approximation of anything. It is, exactly, the limit as the number of significant figures increases, and you rightly agree that this limit is 1. To say that "0.999..." equals 1 is not to say that any number of significant figures ever gets us to 1. It is, again, the limit. As you write more 9s, you never get to "0.999..." (an infinite number of 9s). The value never gets to 1. The key point is that these are saying the same thing! If you still aren't convinced, try this. Let the value of "0.999..." be x. You say that x < 1. But surely you see that for any x < 1, I can create a value y = "0.999...9" with enough 9s so that y > x. So if I reach y with after adding a bunch of 9s, how can adding more 9s get me to x, if x < y? 

Title: Re: 0.999. Post by Icarus on Nov 6^{th}, 2002, 8:21pm The definition of decimal notation is this. A decimal sequence {d_{i}}_{i=N}^{[subinfty]} for some integer N [le] 0, is a sequence of digits d_{i} from the set {09}, commonly written (in the USA and some other places  others use a , or a raised . instead of a lowered .) as d_{N}d_{N+1}...d_{1}d_{0}.d_{1}d_{2}. . . (I have reversed the signs on the index because it makes the rest of the notation more conventional, though it is a bit confusing here.) This sequence by definition represents the unique real number L such that L = lim_{k[to][subinfty]} [sum]_{i=N}^{k} d_{i}10^{i} Note that L is not any of the partial sums. L is the limit. Now the definition of the limit of a sequence: the limit L of a sequence {a_{i}} is the unique number L satisfying this condition: for every [epsilon]>0, there is an N such that for every i > N, a_{i}  L < [epsilon]. If you don't believe the number L is unique, consult any good calculus text. They all contain the proof that it is. (If the one you picked doesn't, then rest assured that it is NOT a good calculus text!) Or try to prove it yourself. The proof is not hard. Some things to notice here. 1) There is no such decimal notation as 0.9999...(infinitely many 9s)...9 decimal notations are a "shorthand" for decimal sequences, which are in turn functions from the set of all integers [ge] N for some N [le] 0 into the set {0,1,2,3,4,5,6,7,8,9}. There is no integer to map to that last 9. (If you disagree with this, go look up the definition of "Integer".) You are free to define such a notation (say, use ordinals instead of integers for your index). But then it is also up to YOU to define what real number it represents! You are also to free to introduce new numbers to be represented by your new notation. Just don't mistake these for real numbers. (There are many ways of defining the real numbers, but they are all equivalent to this: The set of Real numbers is the smallest topologicallycomplete ordered field.) 2) 0.999... is not the same as 0.999...(finitely many 9s)...9, no matter how many 9s are in that finite sequence. 0.999... is a written representation of the infinite sequence {d_{i}}_{i=0}^{[subinfty]} with d_{0}=0 and d_{i}=9 for all i>0. This sequence is by definition of decimal notation a representation of the number L = [sum]_{i=1}^{[supinfty]} 9*10^{i}. It has been proven many times in this forum that L is the real number 1. 3) A limit is NOT an approximation. The elements of the sequence are the approximations. The limit is the number that is being approximated. This is a fundamental difference that is behind the confusion several respondents have shown. By definition, the decimal sequence 0.999... represents the limit, not any of the approximations. 4) If you don't agree with the results of a proof, don't assume that it is wrong and you are right. If you can't find a fault in its reasoning, you better check YOUR reasoning a little more carefully. 5) To take a limit does not require you to do an infinite amount of anything. If you will look again at the definition of the limit of a sequence, you will notice that there is no mention of infinity anywhere in it (other than saying that this is what the phrase "limit as i goes to [infty]" means). 

Title: Re: 0.999. Post by Jeremy on Nov 7^{th}, 2002, 5:56am Ok, i really suggest we drop this thread... i think if we haven't saved anyone by this time, no one is going to change their opinions. Hopefully we can wait till they get to a precalculus class, and then they can take it up with their math teacher. And as for Guess Who's comment about 1/3 not being equal to .333... because if you do long division you will ALWAYS have a remainder umm.... well duh. the 3's just keep going... the ellipse (...) means it's a repeating series. The three's never stop, so you keep having these remainder's of 3, and you have to keep dividing and getting another digit. Really the series wills stop as soon as you divide into a nice even number with no remainder (and that will never happen) 

Title: Re: 0.999. Post by towr on Nov 7^{th}, 2002, 5:56am do we really want to accept that every integer (except 0) has 2 representations? 2 1.999... 1 0.999... 1 0.999... 2 1.999... I would argue 0.999... isn't a valid number, but rather bad syntax 

Title: Re: 0.999. Post by Jeremy on Nov 7^{th}, 2002, 5:59am and towr, there's already an infinite number of ways to write every number: 1=2/2=3/3=4/4=5/5=6/6=7/7=8/8... 

Title: Re: 0.999. Post by towr on Nov 7^{th}, 2002, 6:07am 5/5 is an expression, not a number.. You need to calculate it's value to get one.. A number is an atom, an equation is a set of operations on atoms.. 

Title: Re: 0.999. Post by S. Owen on Nov 7^{th}, 2002, 7:10am on 11/07/02 at 05:56:42, towr wrote:
Yeah, it would be funny to think that there is more than one base10 decimal representation for 1. Agreed, the "trick" is that "0.999..." is an abuse of notation. A decimal representation needs to have a finite number of digits. We can still establish a meaning for "0.999...", and its value, 1, but it's not another decimal representation of 1. As for dropping this thread... somehow I still enjoy arguing about this one. Turn off reply notification if they're bothersome. 

Title: Re: 0.999. Post by TimMann on Nov 7^{th}, 2002, 10:17am on 11/07/02 at 07:10:07, S. Owen wrote:
Oops, that's a mistake. If you mean that, you'd be saying that irrational numbers and rational numbers that aren't multiples of negative powers of 10 have no decimal representation at all. The only decimal representation of 1/3 is 0.333..., and the only decimal representation of pi is 3.14159... It's more common to do the opposite and say that all decimal representations have an infinite number of digits; for example, 1 = 1.000... This is needed in Cantor's diagonalization proof that the cardinality of the set of reals is greater than that of the set of integers. There's nothing wrong with either 1.000... or 0.999... as a decimal representation for 1. It's too bad that some numbers turn out to have two infinite decimal representations, but we're stuck with it. The best we can do is choose one of the two and declare it to be canonical. 

Title: Re: 0.999. Post by S. Owen on Nov 7^{th}, 2002, 11:41am Yeah, let me retract that... I guess I was taking towr to mean 'are there really two finite decimal representations of 1', which "0.999..." isn't, but that's not terribly interesting. Well maybe... does "0.333..." count as an exact decimal representation of 1/3? Certainly if it equals anything, it's 1/3. But I've held the line that "0.333..." isn't really a decimal value as much as a tricky expression of a limit. The answer is probably a matter of context... as far as (finite) computers are concerned, there is no way to exactly represent 1/3 as a base2 decimal, so the answer's no in that context. But a mathematician might usefully define "0.333..." as the base10 decimal representation of 1/3 and have no problems. 

Title: Re: 0.999. Post by Icarus on Nov 7^{th}, 2002, 4:15pm on 11/07/02 at 11:41:29, S. Owen wrote:
0.333... is a tricky expression of a limit, but that does not stop it from being a "decimal value". All decimals are tricky expressions of mathematical operations. After all, what does 0.3333 (finite) mean but 3/10 + 3/100 + 3/1000 + 3/1000 ? How is saying that any different from saying 0.333... means [sum]_{i=1}^{[supinfty]} 3/10^{i} ? The only difference I see is the sophistication of the concepts, and that 0.333... has a second level of notational convention applied (the ellipsis). However the ellipsis and the summation are both welldefined concepts whose meanings allow us to say 0.333... = 1/3, exactly, and not by "abuse of notation". One other thing. The only real numbers with more than one decimal expression are the nonzero rationals whose denominators are expressable as 2^{n}5^{m}. They have exactly two: the terminating one which ends in repeating 0s, and the one that ends in repeating 9s (with the last digit not 9 being 1 less that the equivalent digit in the terminating version). And, it's not really a matter of whether we want it or not. The two notations are a byproduct of how decimals work. We could decide to throw the repeating 9s notations out, but then we would have to make exceptions every time we discuss decimals: a * (0.bcdef...) is calculated by multiplying a by each digit in turn, summing and performing the carries except when this leads to repeating 9s, then you have to ... I personally think it is much easier to live with double notations, where you only have to state the equivalence once, and then you're covered for life! 

Title: Re: 0.999. Post by towr on Nov 7^{th}, 2002, 11:43pm simplicity makes perfect.. If I answered 0.999... on a mathexam when the answer would be 1, it would be marked as wrong, or at least not right.. The simplest answer is the right one.. Would have been nice to try when I was in highschool though.. I could have driven my mathteacher mad.. :P 

Title: Re: 0.999. Post by Icarus on Nov 8^{th}, 2002, 3:45pm When I was teaching, I would have taken a point off for failure to simplify, but I certainly would not have called it wrong, or even "not right", just not the best form for the answer. 

Title: Re: 0.999. Post by towr on Nov 10^{th}, 2002, 7:49am well.. a failure to simplify seems to qualify as not right imo.. In any case it's what I meant.. hmm.. Maybe I should teach my niece to try this later when she's in highschool, and how to argue the point its the same.. 

Title: Re: 0.999. Post by Guest on Nov 21^{st}, 2002, 8:22am To all those claiming 0.999... <1 : It never ceases to amaze me how many stupid idiots on message boards have no idea about this problem. Go off and study how mathematics and analysis works before jumping in with your pathetic uneducated opinions. 

Title: Re: 0.999. Post by towr on Nov 21^{st}, 2002, 9:16am It never ceases to amaze me what spectacular significant contributions some people can bring to a discussion.. You, however, certainly aren't one of them.. 

Title: Re: 0.999. Post by Guest on Nov 21^{st}, 2002, 6:31pm towr : your posts really are laughable. "well.. a failure to simplify seems to qualify as not right imo.. In any case it's what I meant.. " So if the answer to a calculation is (x+4)(x+5) and I write x^2 + 9x + 20 then I'm wrong am I? Come on. I actually have a degree in mathematics and I know both answers are equally valid, no matter how you're marked in your "math exam". If the question asked to put something in its simplest terms, eg 4/8, then 1/2 or 0.5 would be correct answers and 2/4 would not. "5/5 is an expression, not a number.. You need to calculate it's value to get one.. " Of course its a number, you imbecile. 5/5 and 0.999... are equal real numbers more commonly known as 1. They are the same real number, we just choose to call it 1. 0.999... may be an abuse of notation but its perfectly well defined. 0.999... = lim n> infinity of (9/10 + 9/100 + 9/1000 + ... + 9/(10^n)) Just look it up in any mathematics textbook. For your information x/2, (x+3)(x+5) and exp((x^2)/2) are expressions. "A number is an atom, an equation is a set of operations on atoms.. " Nice defintion. No really. And your point? "It never ceases to amaze me what spectacular significant contributions some people can bring to a discussion.. You, however, certainly aren't one of them.. " I refer the right honourable cretin to the reply I gave a few moments ago. 

Title: Re: 0.999. Post by jeremiahsmith on Nov 21^{st}, 2002, 7:13pm It's about time this board got its very own pretentious asshole! I was getting sick and tired of how everyone was getting along so well! :P 

Title: Re: 0.999. Post by Icarus on Nov 21^{st}, 2002, 8:20pm So, "Guest", You have a degree in mathematics! Very good! Now get in line, like the rest. I have a PhD in it myself, and there are several other mathematics degrees on this board, as well as degrees in related fields. The regulars on this board, including towr, are both intelligent and welleducated. You would be wise to consider their words carefully, even when you disagree. Unlike you, they know the value of differing points of view. Even those who are mistaken can bring new depth to your understanding, and suggest things you may not have considered. For instance, Kozo earlier in this thread talks about 0.99...(infinitely many 9s)...9. Of course, YOU know that this isn't a real number. But I ask, do you really know why it isn't? Have you ever considered the possibility of defining numbers that can be expressed in this fashion? Can it be done consistently? If so, how do they behave? How do they relate to the real numbers? These are useful considerations, but you are so full of yourself I doubt you ever saw them. Just as you failed to understand towr's remarks because you would rather belittle them than think! Now, if you have something useful to add this or any other thread, I would love to read it. But if you want to keep up with this selfaggrandizement, stop wasting our time. 

Title: Re: 0.999. Post by Guest on Nov 22^{nd}, 2002, 6:14am So, "Icarus", You have a PhD in mathematics! Very good! Then you will agree that everything I have said was mathematically correct if not the tone in which it was written. "For instance, Kozo earlier in this thread talks about 0.99...(infinitely many 9s)...9. Of course, YOU know that this isn't a real number. But I ask, do you really know why it isn't?" Take a never ending sequence of 9s and put a 9 on the end, except you can never reach the end. So that doesn't make sense, it is not well defined, hence the number he is trying to define is not a member of the reals. However we can talk about the limit as n tends to infinity of 0.99... (n 9s) ...9 which is just 0.999... . Similarly the limit of 0.999...(n 9s)...A as n tends to infinity where A is a member of the set {0,1,2,3,4,5,6,7,8} is also equal to 1 and hence 0.999... by standard epsilon proofs. Have you ever considered the possibility of defining numbers that can be expressed in this fashion? Yes before, and when I read Kozo's post. See the above analysis. Because the last digit is on the end of a very long sequence whose length is tending to infinity, it has zero effect on the value of the number. So I ask of the usefulness of this idea. It doesn't work very well in the decimal system that we use but if you were to give definitions that would work better then I would be interested to know about them. 

Title: Re: 0.999. Post by Pietro K.C. on Nov 22^{nd}, 2002, 8:38am Hauehauehauheauehauh!!! :D :D :D Who would ever have thought that the "0.999... = 1?" question could bring such strong emotions to the fore? I don't believe this "Guest" is for real. I think it's just Wu messing with us, or maybe one of the forum regulars, just for the hell of it. Anyways, as for the "mathematics degree" which you have, why don't you shove it in the Putnam section? If you got a degree just to know facts (which you are so anxious to show), you should have studied botany, or something like that. I would love to see a nonGoogled solution of yours to "3 points in a circle", or "bright and dark stars". Thanks! ;) 

Title: Re: 0.999. Post by FenderStratFatMan on Nov 22^{nd}, 2002, 8:50am on 07/29/02 at 06:40:49, Kozo Morimoto wrote:
Wow Kozo your using the same argument as my little brother. Anyway, the reason that would be wrong is because there is no way to write .0000000000000 infinite with one in the last place. Secondly 1  .9 infinite is 0 try on a mathematical calculator. 1/.9999999999 = 1 and 1 + .99999999999 = 2. Explain any of those. In reality a complex equation is not needed to prove that .999999999 = 1. FenderMan 

Title: Re: 0.999. Post by Pietro K.C. on Nov 22^{nd}, 2002, 9:24am Quote:
Come on, FenderMan, the point of the discussion is to try and clarify people's thoughts. Kozo's mistake is as valid as any other, and, as I see it, is pretty much the ONLY mistake you can plausibly make regarding this puzzle. This kind of judgment is really uncalled for. Kozo is being polite, the least we can do is return the favor. Second, a calculator will tell you all sorts of things, including absurdities like: 2/3 = 0.666666667 pi = 3.141562654 1  0.9999999999 (terminating) = 0. Don't trust them too much! :) 

Title: Re: 0.999. Post by Guest on Nov 22^{nd}, 2002, 1:32pm "Pietro" : To be fair, I can't fault any of your analysis on this thread so far. "I don't believe this "Guest" is for real." Really? "Anyways, as for the "mathematics degree" which you have, why don't you shove it in the Putnam section?" I might do that. I mentioned the "mathematics degree" to add to the argument proving towr and his assumption wrong. What are your qualifications? "If you got a degree just to know facts (which you are so anxious to show), you should have studied botany, or something like that." I imagine botanists would be quite annoyed about that comment. What are you trying to say? I would love to see a nonGoogled solution of yours to "3 points in a circle", or "bright and dark stars". Thanks! I will tackle those problems I choose to, not anything laid down as a challenge. As anyone who has done a degree will tell you, it allows one to specialise to a certain extent. I reserve that right here. Analysis is one of my preferred subjects, real and complex. A particularly nice result attributed to Picard is that supposing a complex function f has an isolated essential singularity at a, then in any punctured disc centred at a, f actually assumes every complex value in the plane (except possibly one). [Source : Introduction to Complex Analysis, H. A. Priestly] Now this is not "anxiousness" to show facts for "selfaggrandizement", it's actually a beautiful and somewhat surprising result. What is pathetic is this : people with zero concept of mathematical definitions of very simple ideas (e.g. numbers) completely disregarding the complexity and beauty of mathematics by a) arguing strongly only with their intuition b) being pretty sure they are right, and c) not listening to reason or being willing to learn. 

Title: Re: 0.999. Post by James Fingas on Nov 22^{nd}, 2002, 2:02pm Fellow Forumgoers: I would really hate this thread, which has had a lot of good (and also some notsogood) discussion on it, to devolve into a flamewar. Guest, I found your post on Nov. 21^{st} to be rude. If people do not know about mathematics, then there is no better place for them to raise their questions and learn from the answers than in a forum like this. The question is simple enough that people can make up their own ideas, but complex enough to allow some serious mathematical formalization. Although I do not have a degree in math, I appreciate some of the insights that I've seen in this particular thread, shared with all of us by the people who know what they're talking about. If I never ask a stupid question, how can I come to a deeper understanding of how math works? 

Title: Re: 0.999. Post by Icarus on Nov 22^{nd}, 2002, 4:00pm Yes, you can define numbers with infinitely many nines, and then some more. They are not Real numbers, so your argument against them, which relies on concepts of the Reals, is fallacious. Consider sequences from the countable ordinals into the 10 digits for the definition. They also do not form a field, or even a group, but they do have some interesting behaviors of their own. (Topologically, I believe this is still the long line, though the construction is different.) Yes, the actual math you presented was correct. However your belief that you have any idea what towr was talking about is not. This is evident, since nothing mathematical you said disagrees with what towr's points. If you will read towr's comments again with a less haughty disposition, you will discover that he never argued that 0.999... != 1. His only point was that 0.999... should not be considered an actual number at all. I argue that such a definition would create far more problems than the minor one it solves (two representations for terminating decimal numbers). This does not mean that his point is wrong. There is no "right" or "wrong" on this one, it's a matter of convention. Now, if you have read James' post and understand the truth of what he is saying, and will behave yourself better in the future, I would be more than happy to see you contribute to these forums. But insulting everyone who disagrees with you and being condescending is childish. 

Title: Re: 0.999. Post by jon g on Nov 23^{rd}, 2002, 1:52am just drop it, you all have gone too far. Why don't you answer a real question like what is the meaning of life or is there a god? 

Title: Re: 0.999. Post by jeremiahsmith on Nov 23^{rd}, 2002, 11:53am on 11/23/02 at 01:52:09, jon g wrote:
Simple. Those questions haven't been added to William's puzzle pages yet. :P 

Title: Re: 0.999. Post by Kozo Morimoto on Nov 24^{th}, 2002, 10:09pm After the first few posts and reading so many responses which were against me, I realized that I was barking up the wrong tree. After extensive googling, all the literature that I could find indicated that I was wrong. So I decided to play the Devil's Advocate and try to provoke the fellow forum participants into think a bit more. The responses I got were basically canned responses and it didn't look as though most of them actually thought about what was being discussed. Answers like 'plug these numbers in calculators etc' were fallacious arguments and I wasn't satisfied with the provided 'proofs'. There were the few exceptions and I would like to thank them for persevering with me in answering some of questions. The conclusions I came to from this discussion were: a) this is an artifact of our decimal numbering system and can not be avoided, and it happens with other base n numbering systems as well. b) that starting with 0.9 and keep adding 9s on the end does not ever become 0.999... 0.999... is a totally different beast and need to be treated accordingly. c) that my understanding of infinity far far from complete. Obviously my engineering degree didn't cover enough of the esoterics of pure mathematics and concentrated too much on calculus. I definitely need to do more personal research into infinity d) infinity is like a different country  they have different rules over there 

Title: Re: 0.999. Post by anonymous coward on Nov 25^{th}, 2002, 8:37am on 11/23/02 at 01:52:09, jon g wrote:
the answers to those questions are already well known! 42 No 

Title: Re: 0.999. Post by Moti on Nov 25^{th}, 2002, 2:34pm can either of you grasp infinity, maximal or minimal? i didn't think so. try to think of 0.999 as an ongoing process it is a number constantly striving to be 1 but IS NOT! however, since we can't even grasp the difference between 0.999 and 1 this difference doesn't have any significance what so ever when dealing with final math. so except for when dealing with math philosophy there is really no need to use this difference. since this difference is only philosophical it really doesn't bother us when making even the most complex and high resolution scientifical calculation. from the infinity's point of view there is no difference between 17 and 999^9999999999999 at all so in the same exact way, from our "human" point of view there is no difference between 0.9999 and 1. hope you enjoyed my thoughts on the matter. peace :) 

Title: Re: 0.999. Post by Icarus on Nov 25^{th}, 2002, 6:16pm 0.999... is not a process, it is the final result of a process. It is not 0.9 or 0.99 or 0.999 or ..., or even all of them taken together. It represents instead the final result of all their striving. As such it is not different from 1 at all, but is 1, under a pseudonym. As for 17 and 999^9999999999999, what they look like from infinity depends on what infinity you are looking from, and how you are looking. If [omega] is the least infinite ordinal, then 17 + [omega] = [omega] and 999^9999999999999 + [omega] = [omega], but [omega] + 17 [ne] [omega] + 999^9999999999999 (addition is not commutative for infinite ordinals). Some differences never go away. 

Title: Re: 0.999. Post by KAuss on Nov 26^{th}, 2002, 3:51am there are SO many posts after reading the first 2 pages I'm just stupified and want to take a crack at this... I dunno if anyone brough this up into words, but I know in writing many have said it, but cannot grasp... We live in a mathmatic world of logs of 10.... which means inorder to conserve symbols to represent values, we repeat every 10 by digits. i.e. 1 2 3 4 5 6 7 8 9 (1)0 now, lets say we live in a log 15 world i.e. 1 2 3 4 5 6 7 8 9 a b c d e (1)1 < where 11 = 15 and e = 14 as we know it today... now apply this stupid .9999..9999 to this topic, the next thing would be .aa, then .a1 .a2 .a3 .a4 .a5 .a6 .a7 .a8 .a9 .aa .ab .ac .ad .aeeeeeeeeeeeeeee THEN .b and so .9999inf is NOT 1...... you guys with your infinity... there is no VALUE to anything multiplied to infinity, since it has no end, and since it don't stop, there is no measureable vaule, it is constantly changing, so you can't calculate it. also... 10x /= 9.9999..999 = x because if you did 10x = .99999...9999*10 then you're 1 decimal behind and the number isn't .99999..99999 any more, it's actually .999999999..9999 minus the last 9 because you just moved it's decimal place with log 10.... 10x /= 9.999999..999 is also wrong because if you used anything other than 10 you'll get some screwy number that isn't even .9999999...999 ne more because you are breaking algebra's rules under logs of 10.... have you guys ever wonder why you multiply by 10?? i.e 8x = 7.99992 n some stupid crap if you did that... again, if you look at the math in the sense of logs of 15 where we move every decimal place 15 digits deep, we won't have this discussion but rather .eeeeeeeeee...eeeeeee = 1? then we'll do (remember 11 means 15 in log 15) 11x = e.eeeeeeeeee..eeeee (this is false btw, just as false as the 10x equation due to the fact that anyother number than (11 or 15 in our world) will produce something other that .eeeeeeeeinf) now imagine (if you can) a log infinity.... this means we'll have a different symbol for each numeric value possible which is infinity itself... WE WON'T HAVE DECIMAL POINTS.... I'm sorry, but it don't take no rocket scientist to see 0.999999999999999999999999999999999999999999999999999999999999999999forever /= 1 if you can't look at this and see, you're too caught up in logs of 10.... Decimals are so overrated... Kozo mad props for haning in so long LOL, I almost believed it myself when theys aid they were = My answer... 0.9inf < 1 

Title: Re: 0.999. Post by wowbagger on Nov 26^{th}, 2002, 5:13am on 11/26/02 at 03:51:26, KAuss wrote:
As far as I know, the string of digits "11" in base 15 represents the decimal number 16, while 15 is writen (10)_{15}. After all, you have 10 digits in base 10 and should have 15 digits in base 15, so you'd better use the 0. on 11/26/02 at 03:51:26, KAuss wrote:
I think I don't get your point there. As you suggested further down your post, the question in base 15 is whether 0.eeeee... = 1 on 11/26/02 at 03:51:26, KAuss wrote:
As Icarus (and others I think) has pointed out, 0.999... (indefinitely) is not changing, it's a way of denoting the value of a limit. And that is what you calculate and compare to 1. on 11/26/02 at 03:51:26, KAuss wrote:
This won't qualify as a number system, I guess. 

Title: Re: 0.999. Post by towr on Nov 26^{th}, 2002, 5:45am well, if you start looking at other numberingsystems.. let's say basen than the analog problem is not lim(t>inf) sum(9*n^{i}, i, 1, t) but lim(t>inf) sum((n1)*n^{i}, i, 1, t) which is the same as lim(t>inf) 1  n^{t} = 1 , (since lim(t>inf) n^{t} = 0) It might be interesting to take a look at base1, you only have one number to represent numbers with so you get one = 1, two = 11, three = 111 etc zero would be an empty string, but if you represent it with 0, than you can get 0.000000... = 1 (since 1  1 = 0, analog to (n1) in above equations) :p 

Title: Re: 0.999. Post by S. Owen on Nov 26^{th}, 2002, 6:03am Welcome  just a few things: on 11/26/02 at 03:51:26, KAuss wrote:
In base 15, the next number after e is 10 (which has the value 15), and then 11 (which has the value 16). No big deal. In base 15, 0.999... has a value equal to 9/14 (base 10 numerals there!). There is no "next" value after it, or after any real number for that matter. But I agree that 0.a > 0.999... But of course, showing that 0.999... in base 15 doesn't equal 1 says nothing about it in base 10. To answer the original question, we should consider base 10, or consider 0.eee... in base 15. on 11/26/02 at 03:51:26, KAuss wrote:
Yeah, you really can't think of this problem in terms of multiplying or dividing finite decimals, because 0.999... is not such a thing, and it's easy to misapply reasoning. That is, just because I could never work out 10*0.999... on paper, since it has an infinite number of digits, does not mean that 0.999... cannot be assigned a value. on 11/26/02 at 03:51:26, KAuss wrote:
I ask you to consider my favorite counterquestion to this then: Let 0.999... = x; x < 1. But for any x < 1, I can construct a number y = 0.999...9 with enough 9s to make it bigger than x. So x < y < 1. But x has more 9s than y  how can it be smaller? 

Title: Re: 0.999. Post by KAuss on Nov 26^{th}, 2002, 1:38pm I'm lacking the ability now to switch text to numbers and vice versa after reading this theard LOL.. but yeah, this forum is pretty cool, lots of smart people haning around talking about just anything to the bitter end LOL anyway back to the number thing... Like I said, minus all the other things... Can you imagine if we have an infinte base numbering system? Where we HAVE no decimals... then the number would mearly be a number before another number.... it wouldn't ever get carried over... The thing with the number growing is sheer fact that this number will never stop growing... 1 is a constant... Someone pointed out that they'll intersect like a lot of you said, but if this is the case, then the rate of growth must be constant, and if whenever it intersects, after the intersection, it must grow larger than 1.... So I don't see how a number that is ever growing can equal to anything that has a solid value where it's value don't change... Another thing I don't get is... What is this talk about getting bigger and bigger?? Did we ever throw TIME (which don't exsist btw) into this equation? We look at this as a repetitive process instead of an instant process... This number is as developed as any, but it don't = x since x is not ever changing... the thing with x < y < 1......... What I don't get is, how can you get y = 1 more 9 than .999...inf? There is something bigger than infinte? I always thought you can't alter infinte, since it represents everything in the number line... So how can you have 1 more than something that is everything? so x = y /= 1 cause we can say this.. x = y /= 1 = a = b = c / .999inf /  1inf / / .999inf ?? ^ that makes no sense... The angle isn't like that obviously but almost parallel, but by what you all are saying with limit, (which deprives the piont of saying infinite since it don't end) will hit and STOP growing... I don't see how... I have totally lost my sense of reason.... LOL I'm now like an aimless arrow in society... Did Pi ever stop producing decimals yet? Our number system sucks :) One last thing I still don't get.... If .999inf is the process to 1 or whatever, then what is the process right AFTER 1? The number that is still 1 but it's always infinite X bigger than 1? just like how .999inf is 1 but it's always infinite Y smaller than 1.... 

Title: Re: 0.999. Post by S. Owen on Nov 26^{th}, 2002, 2:48pm on 11/26/02 at 13:38:11, KAuss wrote:
0.999...9 will never equal 1 no matter how many 9s there are  it will never "intersect." It will just get closer and closer to 1. "To 1". "1" is where it's going but never gets to. "0.999..." is where it's going but never gets to... you can think of the limit either way. Indeed, they are equal! I agree with your comment about time  this shouldn't really be though of as a process. Or you can, but you have to recognized that 0.999... is the stationary limit of that process. on 11/26/02 at 13:38:11, KAuss wrote:
Well yeah, that's the contradiction that you can derive from the assumption that 0.999... < 1. So therefore 0.999... = 1. 

Title: Re: 0.999. Post by Icarus on Nov 26^{th}, 2002, 6:33pm on 11/26/02 at 13:38:11, KAuss wrote:
There are several types of infinities. And you can define more freely. Here are the three best known: Continuum Infinities These are infinities at the "ends" of the number line, represented by [infty] and [infty]. These are defined by adding a couple extra elements to the set [bbr] of real numbers, which are defined to be respectively greater than, and less than, every real number. The result of including these infinities is to make the real numbers topologically (that is, as far as limits and continuity are concerned) equivalent to a closed interval. A similar construction adds a single infinite point to the complex plane, making it topologically equivalent to a sphere. Cardinal Infinities Two sets are said to equipotent ([sim]) if you can match up their elements in a onetoone fashion, with nothing left over from either set. For example, the set of Integers [bbz] and the set of Even Integers [smiley=bbe.gif] are equipotent ([bbz] [sim] [smiley=bbe.gif]), because we can match each integer [smiley=x.gif] with the even integer 2[smiley=x.gif]. But {0} [smiley=nsim.gif] {1,2}, because if you match 1 with 0, there is nothing left to match 2 with, and vice versa. We can assign to every set an object, called the set's cardinality, so that two sets have the same cardinal if and only if they are equipotent. (How this assignment can be done delves far deeper into set theory than I want to get into here.) The settheory definition of finite and infinite (from which all other definitions of these terms derive) is that a set is finite if is not equipotent to any proper subset of itself. A set is infinite if there is a proper subset to which it is equipotent. The example I gave shows that [bbz] is infinite. A cardinal is finite (infinite) if it is the cardinal of a finite (infinite) set. The finite cardinals are just the Whole numbers: 0, 1, 2, 3, ... . The question then arises, how many infinite cardinals are there? The answer is: infinitely many. If [smiley=ca.gif] is an infinite set, then it can be proven that P([smiley=ca.gif]) [smiley=nsim.gif] A, where P([smiley=ca.gif]) isthe set of all subsets of [smiley=ca.gif]. The smaller infinite cardinals are expressed using the hebrew letter Aleph [smiley=varaleph.gif]. The smallest is [smiley=varaleph.gif]_{0} ("Alephnought"), which is the cardinality of the Natural numbers [bbn] (also the Whole numbers [smiley=bbw.gif], the Integers [bbz], and even the Rational numbers [bbq]). The next smallest infinite cardinal is called [smiley=varaleph.gif]_{1} ("Alephone"), etc. The Real numbers [bbr] are equipotent with P([bbn]), which as I stated earlier is NOT equipotent to [bbn]. Thus there are no more Rational numbers than there are Natural numbers, but there are more Real numbers. It is not known if the cardinality of the Reals is [smiley=varaleph.gif]_{1}. (Actually it is much more complicated than just "not known", but this is too long already). Addition and multiplication are defined for cardinals as follows: If sets [smiley=ca.gif] and [smiley=cb.gif] are disjoint, then Card([smiley=ca.gif]) + Card([smiley=cb.gif]) = Card([smiley=ca.gif] [cup] [smiley=cb.gif]), and Card([smiley=ca.gif]) [times] Card([smiley=cb.gif]) = Card([smiley=ca.gif] [times] [smiley=cb.gif]) ([smiley=ca.gif] [times] [smiley=cb.gif] is the set of all ordered pairs with the first element from [smiley=ca.gif] and the second from [smiley=cb.gif]). For infinite cardinals, this turns out to be uninteresting: If at least one of [smiley=x.gif] or [smiley=y.gif] is infinite, then [smiley=x.gif] + [smiley=y.gif] = [smiley=x.gif] [times] [smiley=y.gif] = max([smiley=x.gif], [smiley=y.gif]). Ordinal Infinities This is the set of infinite numbers I refered to in my reply to "Guest" above. Their definition is more complicated than Cardinals, but they have a much richer structure. Each ordinal corresponds to a means of "Wellordering" a set. To wellorder a set, you define an order operation [smiley=prec.gif] that is
[bbn] is wellordered by "<", but [bbz], [bbq], and [bbr] are not, since none of them have a least element. A wellordering [smiley=prec.gif] on a set [smiley=cs.gif] is considered equivalent to a wellordering [smiley=lessdot.gif] on the same or another set [smiley=ct.gif] if all the elements of one set can be matched in a onetoone fashion with all the elements of the other set (as with cardinality) but also in a way that preserves the order: if [smiley=a.gif], [smiley=b.gif] [in] [smiley=cs.gif] and [smiley=a.gif], [smiley=y.gif] [in] [smiley=ct.gif] with [smiley=a.gif] [smiley=leftrightarrow.gif] [smiley=x.gif] and [smiley=b.gif] [smiley=leftrightarrow.gif] [smiley=y.gif], then [smiley=a.gif] [smiley=prec.gif] [smiley=b.gif] [bigleftrightarrow] [smiley=x.gif] [smiley=lessdot.gif] [smiley=y.gif]. Note that two wellorderings can be equivalent only if their sets are the same size (have the same cardinality). Wellordering is very restrictive. There is only one wellordering (up to equivalence) for all finite sets of the same size. Infinite sets have more freedom. For instance, I can define the following ordering [smiley=prec.gif] on the Natural numbers: [smiley=x.gif] [smiley=prec.gif] [smiley=y.gif] if ( 1 < [smiley=x.gif] < [smiley=y.gif] ) or ( 1 = [smiley=y.gif] < [smiley=x.gif] ). [smiley=prec.gif] is a wellordering of [bbn] which has a maximum element (1), while the standard ordering < has no maximum element. Ordinals are assigned to wellorderings so that two orderings have the same ordinal only if they are equivalent. Ordinals can also be added ord([smiley=x.gif]) + ord([smiley=y.gif]) = ord([smiley=z.gif]) where [smiley=z.gif] is the ordering on the disjoint union of the sets of [smiley=x.gif] and [smiley=y.gif], in which everything in [smiley=y.gif]'s set is greater than everything in [smiley=x.gif]'s set. Ordinal addition is generally NOT commutative (ord([smiley=x.gif]) + ord([smiley=y.gif]) [ne] ord([smiley=y.gif]) + ord([smiley=x.gif])). You can also define a multiplication, but I will not go into it. When the sets are finite, there is only one wellordering, and thus only one ordinal, so finite ordinals are just the Whole numbers again. With infinite ordinals, things get more interesting. The ordinality of [bbn] is usually denoted by the greek letter small omega [omega]. The ordinality of the order I gave above (that moves 1 to the opposite end of the number line) is [omega]+1. The rule for adding infinite ordinals is that if 0 < [smiley=x.gif] < [smiley=y.gif], and [smiley=y.gif] is infinite, then [smiley=x.gif] + [smiley=y.gif] = [smiley=y.gif], but [smiley=y.gif] + [smiley=x.gif] > [smiley=y.gif]. The smallest ordinals follow this pattern: 0, 1, 2,... [omega], [omega]+1, [omega]+2,... [omega]+[omega] = 2[omega], 2[omega]+1, 2[omega]+2,..., 3[omega],..., [omega]^{2},..., [omega]^{3},... Finally now I can answer KAuss's question: how can you have infinitely many 9s, and then another 9? An ordinary decimal number (we will only look at the fractional part) is just a sequence, that is, a function from the natural numbers, into the set of digits: {0,1,2,3,4,5,6,7,8,9}. To get another 9 after infinitely many, we need a different domain for our function than [bbn]. Instead, we choose functions from a set of ordinals which includes infinite ordinals. So, the first part of 0.999...([infty] 9s)...9 gives the digits corresponding in our sequence to 0, 1, 2, ... , while the final 9 corresponds to [omega]. How do you get a number out of this? If you are talking about Real numbers, then you can't (other than the same number as represented without the final digit). But it is possible to simply treat these sequences as a set of numbers themselves, which contains the Reals, plus additional "pseudoreal" numbers. Which is what I was refering to in my previous posts. Kozo: You did a great job of being the devil's advocate. That's why we keep refering to your posts! 8) 

Title: Re: 0.999. Post by guest535 on Dec 10^{th}, 2002, 11:18am the difference between .99999... and 1 = .000...1 the difference between .99999... and .99999...8 = .000...1 so if .9999... = 1 then .999999...8 = .9999999 which is 1 so .99999...8 =1 if .999999999 was 1 then thats what it would be. ever wounder why it starts with a 0.? its like saying 1=2 or 1 = 1,000,000 what is infinity? a number that really dosnt exist so as .999 never gets to an infinate number if 9's (you can always add 1 more) It never reaches 1 so isnot 1. 

Title: Re: 0.999. Post by jeremiahsmith on Dec 10^{th}, 2002, 11:47am on 12/10/02 at 11:18:55, guest535 wrote:
*sigh* Before posting to a thread, do try to read the whole thread...we've been through this before. 

Title: Re: 0.999. Post by S. Owen on Dec 10^{th}, 2002, 12:35pm on 12/10/02 at 11:18:55, guest535 wrote:
There's your problem  you write "0.999..." but you are thinking of something like "0.999...9"  that is, you are referring to some decimal with a finite number of 9s, but this riddle is about the slippery notion of "0.999...". In your equation, the number of digits on the right must be the same as the number of digits on the left, and you indicate a finite number of digits on the right. If you'd like to argue that the right side is "an infinite number of 0s, and then a 1," which is not a possible decimal number, but still something whose value can be reasoned about, then you can... but you'll be hard pressed to show it is anything other than 0! 

Title: Re: 0.999. Post by aaaaaab on Dec 12^{th}, 2002, 7:01am Pleaaaase, i do not know how thsi goes on, if the correct answer was there from the really beginning: 0.99999...=1, no doubt (in base ten of course, that is the math we are doing in another base like you said yo can equally prove that o.eeeee...=1) The answer is basic from middle school, i knwo other people have said it, but here it is again: Lets call the number we are trying to calculate A (A=0.9999999.......) then 10A= 9.999999...... when substarcting both, we get that 9A = 9 So A is equal to 1. With the same argument you can say that 0.888888888.... is equal to 8/9 or that 0.345345345345345345....... is equal to 345/999. These numbers all belong to the rational numbers, unlike pi for example or sqrt(2) that cannot be expressed by a fraction. Dan 

Title: Re: 0.999. Post by Jim Auster on Dec 14^{th}, 2002, 7:46pm Ok, if a frog starts ten feet away from a pond and jumps halfway to the pond each time he jumps, he will never reach the pond. .999.... will never equal 1 because it is simply not 1. continue adding 9's and you get closer, but youre never there. called an asymtote. i'm a beginner at this, and correct me if i'm wrong, but isnt this the same as saying 1 is equal to 2 if you just imagine it to be? 

Title: Re: 0.999. Post by S. Owen on Dec 14^{th}, 2002, 8:52pm Sure, you can think of 0.999... as the limit of the sequence 0.9, 0.99, 0.999, .... Connect the points (1, 0.9), (2, 0.99), (3, 0.999), ... on the plane and you'll have a graph of something that approaches 1 asymptotically as x goes to infinity, I think you'll agree. Put briefly, I think the point is that 0.999... *is* also that asymptote, that value that is approached. It's not a yvalue in that sequence. Once you agree that that's what 0.999... is, you have to agree that it's 1. If you don't buy that  look back a page or two to see what kinds of contradictions you get if you assume that 0.999... < 1. 

Title: Re: 0.999. Post by fenomas on Dec 15^{th}, 2002, 11:05pm Since people are still bringing up numbers like "0.999...8" or "1.00...1", remember: Talking about an infinite series of 9s followed by an 8 is like talking about a man who lived for for an infinite number of years and then died. :) In other words, in the words of every math teacher I ever had, "there's no such animal". (Why do math teachers like that phrase so much?) 

Title: Re: 0.999. Post by Goodyfresh on Dec 19^{th}, 2002, 7:21pm My goodness you crazy people, it's so easy. Simply define this as the sum of an infinite geometric series and you will see that it equals one. 

Title: Re: 0.999. Post by Goodyfresh on Dec 19^{th}, 2002, 7:33pm I would like to clarify this topic for all those who believe that infinity is not a number. Anyone who has read a lot about math knows what I am about to say: INFINITY=LIMIT AS X>0 OF 1/X=THE MAPPING 1/Z APPLIED TO THE ORIGIN IN THE COMPLEX PLAIN=MANY OTHER DEFINITIONS AS WELL. So, INFINITY IS A NUMBER YOU STUPID PEOPLE!!!!! From now on, philosophers should not try to talk about math, because philosophers aren't mathematicians. When somebody uses philosophy to talk about math, everything gets screwed up. For example, somebody I know now thinks that the Pythagorean Theorem "Isn't really true but works anyway" because they read that in a book by a stupid philosopher who knows nothing about math. This philosopher says that irrational numbers like sqrt(2) don't exist. This kind of stuff is stupid, I know how to define sqrt(2) using an infinite series for sqrt(1/2) from the binomial theorem. PHILOSOPHERS SUCK AT MATH. I'm not saying philosophers suck, I'm saying they suck at math. This was a very long, angry message. Ok, bye. 

Title: Re: 0.999. Post by towr on Dec 20^{th}, 2002, 12:22am So to recapitulate, infinity is a number because you say so, and anyone who doesn't agree is stupid, and that's is you're whole argument..? Well, you sure aren't a philosopher, they at least know how to argue a point, whether it be true or false.. Also note that the way you define infinity you made it the same as minus infinity.. You need to specify if you take the limit approaching from above 0, or from below 0. Else, if I'm not mistaken, that answer is undefined, and thus neither. Philosophers have their place in every science, including math. And also someone can be both mathmatician and philosopher, the two concepts aren't mutually exclusive. 

Title: Re: 0.999. Post by Pietro K.C. on Dec 20^{th}, 2002, 9:48am Agreed, just look at Wittgenstein. I myself find it very hard to divorce completely from philosophy a subject so platonic as mathematics. Your reasonableness is refreshing, towr. As for infinity being a number because of the above definition... um... what is a number? Just because something is defined, it's a number? Well, I define the number P (for "Pietro's number") to be all the trajectories in spacetime of all the ripples made in ponds by canadian frogs that jumped in them during a partial ecclipse of the sun. 

Title: Re: 0.999. Post by James Fingas on Dec 20^{th}, 2002, 10:12am Pietro, I'm shocked! :o And I'm sure that every rightminded FrenchCanadian is appalled at your definition of P, Pietro's number! 

Title: Re: 0.999. Post by Icarus on Dec 20^{th}, 2002, 3:49pm Not being French or Canadian I guess I don't have to be shocked. So Pietro, I am curious about your new number! Does it have any interesting properties? From the definition it has to be closely related to 42. ;) Goodyfresh: you shouldn't insult people for their arguments when it's clear your have not read those arguments! If you will actually read the thread, you will discover that the series representing .999... has been discussed many times! As for your definition of oo, It does not comprise a usable definition at all, though it is a derivable property from the definitions of some infinities. There are plenty of different infinities. Check my post on Nov. 26 to see a summary of a few of the many types of infinities out there. Whether an infinity is a number is a matter of choice. The concept of "number" is not well defined in mathematics. The word is used to describe many things you would not normally think of as numbers. However, the concept of a "Real number" or a "Complex number" is well defined, and no infinity fits in it. When you add infinities to these sets, you get the "extended Real numbers" or the "Riemann Sphere" (see my post "Z^n and the Riemann Sphere" in the complex analysis forum to find out what the Riemann Sphere is). towr: It is actually very common to define a single infinity which serves as both lim_{x[to]0+}1/x and lim_{x[to]0}1/x. This infinity wraps up the Real line to make a circle. [infty] is the antipode of 0 on this circle. Again, see my post on the Reimann Sphere in the complex analysis forum to see the same trick done to the complex plane. 

Title: Re: 0.999. Post by lukes new shoes on Dec 29^{th}, 2002, 4:56am x=1 y=0.99999 (using the underline to represent an infinite amount) 2x=2 2y=1.99998 3x=3 3y=2.9997 this shows a difference between the values of x and y, but i dunno if thats mathematically legal 

Title: Re: 0.999. Post by towr on Dec 29^{th}, 2002, 5:21am but you forget that 0.99999 isn't 0.99999 infinity +1 doesn't equal infinity :p (infinity +1  infinity = 1, infinity  infinity = 0, as long as the infinities are equal) 

Title: Re: 0.999. Post by lukes new shoes on Dec 29^{th}, 2002, 6:21pm i havent read the whole way thru this thread, but the main argument ive seen for 0.999 being equal to 1 is: n=0.999 10n=9.999 10nn=9.999  0.999 9n=9 n=1 but according to towr 0.9999 is different to 0.9999, so the infinite amount of nines in step one is going to be different to the infinite amount of nines in step two, and therefore when they are subtracted will not equal zero. 

Title: Re: 0.999. Post by Robbie Man on Dec 29^{th}, 2002, 10:30pm .99999... is merely a number that represents the closest possible number to 1. The fact that .99999 is a decimal (and 1 is not) is enough proof for me. I further think it is a mistake whenever we try to use .99999 is a mathimatical equation. You see, 1.9999 = 0.0001, so 1.9999.... must = 0.0000 with a very very small .0001 somewhere at the end of infinite. 

Title: Re: 0.999. Post by towr on Dec 30^{th}, 2002, 1:04am on 12/29/02 at 18:21:33, lukes new shoes wrote:
What's happening here is really just a geometric series.. you have 0.999 which is: 9*10^{1} + 9*10^{2} + 9*10^{3} + ... + 9*10^{n} where n goes to infinity to solve it you subtract from 10*9*10^{1} + 10*9*10^{2} + 10*9*10^{3} + ... + 10*9*10^{n} (ten times the original value) which is 9*10^{0} + 9*10^{1} + 9*10^{2} + ... + 9*10^{(n1)} After subtraction you're left with 9*10^{0}  9*10^{n}, so this is 9 times what you started with. All terms except the first and last have canceled each other out. Now divide by 9 to get the value for the original series 10^{0}  10^{n} = 1  10^{n} The limit for n to infinity of 10^{n} is 0, so you're left with 1. The discussion seems to hinge on the question "is the limit for n to infinity of 10^{n} really 0 ?" You could f.i. say it's really the smallest number just above 0. But then, in the real number domain there are always others numbers between any two numbers (between any different x and y there lies f.i. (x+y)/2 ). So any two numbers that aren't seperated by more numbers have to be the same (different > seperated => ~seperated > ~different). 

Title: Re: 0.999. Post by S. Owen on Dec 30^{th}, 2002, 6:29am on 12/29/02 at 22:30:02, Robbie Man wrote:
There is no "closest possible number to 1"... well, say that this number is X. Then what would you call (X+1)/2? It seems like you can view 0.999... as a number just barely less than 1  as barely less as possible, but that ultimately doesn't make sense. If it's less than 1, then there is some string of 9s which is closer, which is a contradiction. 

Title: Re: 0.999. Post by Cyrus on Dec 30^{th}, 2002, 12:42pm Howdy all, I'm kind of a moron. But I read ALL the posts on ALL the pages and kind of understood most of them. And all the arguments I understand seem to point that .9999..... = 1. This may be completely unrelated but saying that .9999... = 1 isn't that the same as saying that the graph of y=1/x would EVENTUALLY meet the y (or x) axis if you carried it out to infinity. And if it NEVER meets the axis then the axis and the line must be running parrallel because only parrallel lines will never cross. I don't really know what I'm talkin about I took a calculus class when I was in high school but I forget all that stuff now, so I'm just hoping that someone can point out my mistake and explain this to a very NON mathmatics degree person. I don't think I'm an idiot, just ignorant. 

Title: Re: 0.999. Post by S. Owen on Dec 30^{th}, 2002, 1:23pm on 12/30/02 at 12:42:02, Cyrus wrote:
Hardly unrelated or moronic  I think you've got the point exactly. I believe that some people are effectively trying to argue something like "the graph of y = 1(1/x) never intersects y = 1 for any x > 0." That's right, but it does not show that 0.999... < 1. Assuming that those two things are basically the same thing is the easy mistake to make, and the crux of this riddle. To continue the analogy, saying that 0.999... = 1 is more like saying "the graph of y = 1(1/x) approaches (but does not reach) y = 1 as x approaches infinity." on 12/30/02 at 12:42:02, Cyrus wrote:
The graph itself isn't a straight line, so it's not parallel or unparallel to anything. The tangent lines to the graph are lines, but there is no value of x such that the tangent at x is a horizontal line, and thus parallel to the xaxis. So the graph of 1/x is always decreasing, but that does not mean it gets to 0, because the rate at which it decreases also decreases. 

Title: Re: 0.999. Post by lukes new shoes on Dec 30^{th}, 2002, 4:29pm towr wrote: The limit for n to infinity of 10^{n} is 0, so you're left with 1. the limit is 0, which means it approaches 0 but never actually reaches it, so you're not left with 1. (getting off track, if i have an endless amount of apples and some gives me another apple how many apples do i have?) 

Title: Re: 0.999. Post by redPEPPER on Dec 30^{th}, 2002, 5:24pm on 12/29/02 at 05:21:26, towr wrote:
0.99999 isn't a real number at all. There's no way you can set a specific value for the last decimal of a number with infinite decimals. Such a number doesn't have a final digit. This is why lukes new shoes' demonstration is not "mathematically legal", as he suspected. Quote:
It doesn't? Could you explain? Quote:
What are such things as "equal infinities"? You seem to be treating infinity like a regular number here. x+1 not = x, x+1x=1 ... but that's not correct. Infinity  infinity might take a number of values, including infinity. Example: there's an infinity of natural numbers. Among these, there's an infinity of even numbers. If you remove all these from the natural number (infinity  infinity) you're left with an infinity of even numbers, and zero (so infinity  infinity = infinity + 1). 

Title: Re: 0.999. Post by redPEPPER on Dec 30^{th}, 2002, 5:36pm on 12/29/02 at 22:30:02, Robbie Man wrote:
There's no end to the infinite. You started well with "the closest possible number to 1". Let's accept that 0.999... is that number. But there's a basic property of real numbers that is, two distinct real numbers have another real number between them. Actually they have an infinity of numbers between them, but one is enough. As you cannot possibly write a number that would be between 0.999... and 1, they must be the same number. It's a little like in geometry. There's no such thing in geometry as "adjacent" points. Between two distinct points you can add more points, you can measure a distance. If you can't, then the points are not distinct. 

Title: Re: 0.999. Post by lukes new shoes on Dec 30^{th}, 2002, 6:06pm i've decieded that 0.999 is greater that one. equation 1 infinity + infinity = infinity equation 2 infinity  infinity = 0 add equation 1 to equation 2 infinity + infinity + infinity  infinity = infinity + 0 3*infinity  infinity = infinity infinity  infinity = infinity 0 = infinity add 0.999 to both sides 0 + 0.999 = infinity + 0.999 0.999 = infinity therfore 0.999 > 1 

Title: Re: 0.999. Post by redPEPPER on Dec 30^{th}, 2002, 6:07pm on 12/30/02 at 16:29:05, lukes new shoes wrote:
Achilles shoots an arrow at a target. The distance to the target is one mile. The arrow first crosses half of the distance (1/2 mile). Then the arrow crosses half of the remaining distance (1/4 mile). Then half of what's left (1/8 mile) and so on... The limit_{x > inf} of that sum_{i=1 > x} 1/2^{i} is obviously 1 mile. Does that mean the arrow approaches the target but "never actually reaches it"? Would you volunteer as the target to verify this? ;) 

Title: Re: 0.999. Post by Brandon on Dec 30^{th}, 2002, 11:48pm I have a 1 gallon jug (j). It is empty (j=0). I fill 9/10ths of the empty space (j+=0.9). I fill 9/10ths of the empty space (j+=0.09). If I am able to do this forever, I will never fill the entire jug (j<1). Please discuss. :) 

Title: Re: 0.999. Post by towr on Dec 31^{st}, 2002, 2:39am ok.. let's look at infinity(s) a bit closer.. http://tcw2.ppsw.rug.nl/~towr/PHP/formula.php?formula=%5Cinfty can be http://tcw2.ppsw.rug.nl/~towr/PHP/formula.php?formula=%5Clim_%7Bn+%5Cdownarrow+0%7D%5Cfrac%7B1%7D%7Bn%7D+ let's call this x but http://tcw2.ppsw.rug.nl/~towr/PHP/formula.php?formula=%5Clim_%7Bn+%5Cdownarrow+0%7D%5Cfrac%7B1%7D%7Bn%5E2%7D+is also http://tcw2.ppsw.rug.nl/~towr/PHP/formula.php?formula=%5Cinfty, let's call this one y x/x=y/y=1, x/y = 0, y/x = http://tcw2.ppsw.rug.nl/~towr/PHP/formula.php?formula=%5Cinfty They're different.. Maybe a better way to look at the different order of magnitudes of infinity is to look at number sets, for example rational and real numbers. Both these sets are infinite, yet the real set is infinitely larger. On the other hand you might be inclined to say that the set of rational numbers is larger than the set of natural numbers, yet these are the same size, since there is a one to one mapping from one set to the other. 

Title: Re: 0.999. Post by Cyrus on Dec 31^{st}, 2002, 11:14am Ok, I THINK I got it now. Thanks S Owen and the billions of other explanations I've read. So looking at Bradon's example of the jug, or the crossing of the football field example I remember reading earlier on someone's post. Is the explanation that ... you never actually FILL the jug, or REACH the other line. Just like the line y = 1(1/x) never actually intersects the axis. But you are so INFINITESIMALLY (no idea how you spell that word) close that you are in fact there. So you've filled the jug so full that it's INCOMPREHENDABLY full, but it's not actually full. ;D Am I understanding or getting more confused. THAT I don't know. And maybe THAT is the TRUE riddle. :/ 

Title: Re: 0.999. Post by S. Owen on Dec 31^{st}, 2002, 11:32am on 12/31/02 at 11:14:56, Cyrus wrote:
The argument's not so much that 0.999... is infinitesimally close to 1  but that it is 1. 0.999... = 1 is just saying that 0.9, 0.99, 0.999, ... gets ever closer to something, and that something is 1. It's making a statement about the limit of the sequence, not any element in the sequence. Or, saying 0.999... = 1 is like saying the level of the water in the jug will approaches 100% full. 

Title: Re: 0.999. Post by redPEPPER on Dec 31^{st}, 2002, 6:08pm on 12/30/02 at 23:48:49, Brandon wrote:
Actually, if you're able to do this forever, you WILL fill the entire jug, just like the arrow will reach the target in my previous example. Imagine you fill 9/10ths of the empty space in one second initially, and then in half the time it took to fill the previous iteration after that. In one second you will fill 9/10th of the jug. In 1.5 seconds, 99/100 of the jug. In 1.75 seconds, 999/1000 of the jug... After two seconds, the jug is full. Not almost full: completely full. As sure as the arrow will reach the target. If you go for a finite number of iterations, you approach the limit. But if you go infinitely, you DO reach the limit. 

Title: Re: 0.999. Post by SPAZ on Jan 1^{st}, 2003, 12:57am to clarify the initial math.... given: let x = .999... (going on for ever) multiply each side by 10: 10x = 9.999... subtract x from each side: 10x  x = 9.999...  x simplify: 9x = 9.999... .999... 9x = 9 devide each side by 9 9x = 9 9 9 simplify: x = 9 9 9 = 1 9 x = 1 thus: .999... = 1 

Title: Re: 0.999. Post by Cyrus on Jan 2^{nd}, 2003, 8:57am on 12/30/02 at 17:36:42, redPEPPER wrote:
I really like that real number/geometry example. Easy for a simpleton like myself to understand and agree with. I like it when people use different forms of math to prove their answers, like using geometry to make a point about real numbers. Good teaching skill! :) But again . . . I go back to the graph of y=1(1/x). Does that mean that the line DOES intersect 1 if you follow x to infinity? Just like the jug does eventually get completely full. Because if it does, then if you look at (x = infinity + 1) wouldn't y be greater than 1?? Or corespondingly the jug would overflow. Maybe I don't understand this concept of infinity + 1. Does it exist?? Jeeze, just when I think I understand, I get confused again. Somebody tell me to shut up if I'm being stupid. 

Title: Re: 0.999. Post by S. Owen on Jan 2^{nd}, 2003, 10:46am on 01/02/03 at 08:57:19, Cyrus wrote:
I think it all hinges on what one means by "following x to infinity". You can't evaluate the value of y = 1(1/x) graph at x = infinity. But, quite naturally, you can evaluate the limit of y as x approaches infinity, and that is 1. If anything, you can say that this is the value at x = infinity, but I feel that is misleading, because then you naturally want to ask what is the value at infinity+1, and it would seem that it must be > 1. I'd like to go back to the jug example... I think there needs to be a little clarification there. I think the original post meant something like this: I add 9/10 of a gallon in 1 second. Then, I add another 9/100 of a gallon in 1 second. Then, I add 9/1000 of a gallon in 1 second. Does the jug have 1 gallon after any finite number of steps? No. Does the jug have 1 gallon after any finite number of seconds? No. However here is the second situation that was discussed: I add 9/10 of a gallon in 1 second. Then, I add another 9/100 of a gallon in 1/2 second. Then, I add 9/1000 of a gallon in 1/4 second. Does the jug have 1 gallon after any finite number of steps? No. Does the jug have 1 gallon after any finite number of seconds? Yes  2 seconds. Anyway, my point is that 0.999... = 1 is analgous to the first situation if anything, in my opinion, not this second one. In the second one (which is a riddle unto itself!) you do reach 100% full in a finite amount of time, because the time for each step decreases fast enough, and it might be misleading to relate that to 0.999... = 1 to say that this is why the are equal. 

Title: Re: 0.999. Post by towr on Jan 2^{nd}, 2003, 11:28am on 01/02/03 at 10:46:36, S. Owen wrote:
It depends on what part you are looking at. Lets look at the time steps in the second situation, in binary: 1 second + 0.1 second + 0.01 second etc = 1.111... seconds = 10 seconds Which is really the same problem, only in a different base. 

Title: Re: 0.999. Post by redPEPPER on Jan 2^{nd}, 2003, 11:36am on 01/02/03 at 10:46:36, S. Owen wrote:
Yeah, people tend to treat infinity like a number. See some of the demonstrations above, with things such as infinity + or  1. But infinity is more a concept in itself than a number. One of its definitions could be: it's the end of a line that doesn't have an end. Saying the graph above never reaches 1, or saying that it reaches it at infinity is not very different. But you can't assume from the latter that you can focus on the place where it reaches 1 and look what happens farther right on the graph. Because it doesn't reach 1 at any x coordinate that's a real number. It reaches 1 at the very edge of the graph... that doesn't have an edge, as it's measured by real numbers. Ah it's not easy to explain. I'm not sure I understand it myself :P Quote:
My goal was to show that something that's infinite can be "contained" in something that's finite. People confuse the notion of infinity with something like an arbitrarily high number. 0.999... is described as adding 9s to 0. and then adding more 9s and then more, and then more... If you add one figure per time unit, you'll never be done. The number you're writing will never equal 1. It will come closer and closer, but will never reach it, or will reach it after an infinite time, which is not convenient if you want to use that number afterwards. People will say that, no matter how far you are, you can always write a bigger number that's smaller than 1. But if the time needed to add a digit decreases with each iteration, it's possible to be done in a finite amount of time. After two seconds, the number is complete! It has an infinite number of digits, not an arbitrarily high finite number. I figured this could help understanding. This reminds me of a riddle that I'll post here. Look for four beetles and a square. 

Title: Re: 0.999. Post by S. Owen on Jan 2^{nd}, 2003, 11:37am Oh sure, agreed, all of this is the same question in the sense that it's a question of infinite sums, series, whatever. I was just trying to say that in that second situation, we're talking about a process that happens to finish in a finite amount of time... you "do get to 1" in the second case... but that does not directly correspond to why 0.999... = 1. The argument is not that 0.999... = 1 because 0.999... "gets to 1" somehow. Just wanting to be clear on exactly how these problems relate. 

Title: Re: 0.999. Post by lukes new shoes on Jan 5^{th}, 2003, 1:06am another way to show that 0.999...=1 0.999... divided by 9 equals 0.111... 1 divided by 9 equals 0.111... so 0.999... equals 1 

Title: Re: 0.999. Post by Kozo Morimoto on Jan 6^{th}, 2003, 4:14am on 01/05/03 at 01:06:24, lukes new shoes wrote:
That's assuming that 0.999... can actually be divided by 9 to give you 0.111... The whole point of the 'riddle' was that you can't make such assumptions unless you can prove it. 

Title: Re: 0.999. Post by kenny on Jan 6^{th}, 2003, 12:07pm Wow, there are a lot of posts to this thread. Here's my answer. By saying a number "X = 0.999...", what I mean is "X is the limit of the sequence 0.9, 0.99, 0.999, ..." This can also be expressed as: x_i = 1  10 ^ (i). X is limit as i approaches infinity. Clearly, the x_i get closer and closer to 1 as i approaches infinity. Also, it is clear that for any positive number epsilon there is an N such that all x_i for i > N are closer to 1 than epsilon. (Try N =  log(epsilon)  + 1 (log base 10)). Thus, *by the definition of limit* 1 is a limit of the sequence. It is an interesting fact about sequences that if two numbers are both limits of the sequence, then they must be equal. This has been proven. Thus, X = 1. The only way to argue that X might not be 1 is to use a nonstandard definition of a mathematical limit. Or else to have a different definition of what 0.999... means. I believe my definition above is standard.  Ken 

Title: Re: 0.999. Post by jim rodriguez on Jan 28^{th}, 2003, 8:20pm hey math is great and everything, and some of this stuff is kind of interseting, but take it easy guys. heres the answer: 1/3 does not equal 4!!!! whoever said that is a moron! your pal jim Rodriguez cruz 

Title: Re: 0.999. Post by poo on Jan 28^{th}, 2003, 8:22pm MATH LEAGUE RULES!!!!!!!!!!!!!!!!!!!! jim 

Title: 0.999repeated < 1 Post by The math guru on Jan 29^{th}, 2003, 5:45pm no matter how you slice it, so to speak, 0.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 etc is less than one, no matter how close it gets to being one. 

Title: Re: 0.999. Post by S. Owen on Jan 29^{th}, 2003, 5:48pm No matter how many 9s you type, you will also not reach 0.999... (an infinite number of 9s), so this reasoning does not pertain to the riddle. 

Title: Re: 0.999. Post by Mira on Jan 29^{th}, 2003, 8:16pm No matter how long it goes on for 0.9999... will never equal 1, it'll always be 0.0000...0001 away from it. therefore 0.9999...<1 

Title: Re: 0.999. Post by redPEPPER on Jan 30^{th}, 2003, 1:34am No matter how many 9s you add to 0.999... it will never equal 1 as long as you add a finite number of 9. If you have an infinite number of 9 (that's the number 0.99~ we're dealing with here) then you can't have a number 0.00...01 to add to it, because you can't put a 1 at the end of an infinite string of zeroes. An infinite string has no end. An infinite string is not the same as an arbitrarily long string. 

Title: Re: 0.999. Post by Senor Incognito on Feb 2^{nd}, 2003, 9:00pm Standard proof: 0.999' = x x10 x10 9.999' = 10x x x 9 = 9x x = 1 

Title: Re: 0.999. Post by Ryan Sullivan on Feb 25^{th}, 2003, 4:00pm It's foolish to believe that .99999999999999999(inf) = 1. Using an integer system, and the theories some have provided... is there any number between 8 and 9? No. Therefore, they are equal. It's blatantly untrue, and .99999999inf is NEVER, and will NEVER be 1. 

Title: Re: 0.999. Post by Icarus on Feb 25^{th}, 2003, 4:26pm Mr. Sullivan  The difference between 8 and 9 is 1. What is the difference between 0.999... and 1? As well, what is (0.999... + 1)/2 ? Surely this number must lie between 0.999... and 1, where you say no number is (I say it too, but only because nothing is between a number and itself). The oft repeated demonstration, whose last occurence immediately precedes your post, already shows that you would have us abandon the rules which everyone knows for manipulating decimal expressions, just so you can demand 0.999... and 1 are different numbers. Do you have any clue just how many properties of real numbers you would have us give up? Goodbye the zeroproduct property! So long to the fundamental theorems of algebra and calculus! Adieu to the existance of multiplicative inverses! :'( But, we gain this: If two numbers aren't written out the same as decimals, then they must be different! That certainly is an improvement over all those silly useful theorems and properties isn't it? ::) 

Title: Re: 0.999. Post by S. Owen on Feb 25^{th}, 2003, 5:18pm on 02/25/03 at 16:00:22, Ryan Sullivan wrote:
But of course, the original riddle deals with reals, not integers, so this "counterexample" does not apply to the given reasoning. I'd like to repeat... If 0.999... < 1, then you must be able to find at least one real number that is between them. When you can find that, I'll believe you. 

Title: Re: 0.999... Post by otter on Feb 28^{th}, 2003, 11:57am on 01/30/03 at 01:34:28, redPEPPER wrote:
Perhaps a restatement of redPEPPER's message is the best answer. Definition: 0.999... = the sum of the infinite series 9/10 + 9/100 + 9/1000 ... (i.e. an infinite number of 9's) Let us assume that 0.999... and 1 are different. Then there must be some number x equal to 10.999... The value of x must be zero. Any value of x other than zero must be the result of an operation on a pair of finite numbers (as redPEPPER indicates, you can't put a 1 at the end of an infinite number of zeroes) which invalidates the original assumption (remember we defined 0.999... to be an infinite series) and thus the assumption is false. Therefore 0.999... and 1 are, in fact, the same. QED 8) 

Title: Re: 0.999. Post by aero_guy on Feb 28^{th}, 2003, 5:53pm Have you all noticed that there is a critical length to these posts of about two pages? After that people don't feel like reading the entire thread before adding whatever "insight" they have come up with. Of course, by the time the thread is that long odds are their idea was already hashed over. Or, in the case of an 8 page thread like this, hashed over, rejected, reasserted, derided, lauded, thrown out with the garbage, dragged back in, and then finally buried in the back yard. Is this currently the longest thread going? 

Title: Re: 0.999. Post by wowbagger on Mar 3^{rd}, 2003, 3:34am on 02/28/03 at 17:53:22, aero_guy wrote:
Well, I do  if I have the time, that is. :) Quote:
Yes, indeed! 

Title: Re: 0.999. Post by udippel on Mar 3^{rd}, 2003, 8:58am okay; so why not simply freeze it ?? It seems everything has been said; any possible explanation, any possible wrong explanation, any possible correct explanation, anything at all. Can William not  like an auctioneer  321(sound of a hammer)closed ?? 

Title: Re: 0.999. Post by Jeremiah Smith on Mar 3^{rd}, 2003, 9:37am Because if he closes it, it will eventually drop off the front of the forums list since it won't get updated anymore. And since new people are always coming, we would, eventually, get a "help me with .99999... = 1 probleme!!!!!!!!!111" thread, and this would all start again. 

Title: Re: 0.999. Post by Icarus on Mar 3^{rd}, 2003, 5:10pm on 03/03/03 at 09:37:55, Jeremiah Smith wrote:
Maybe that would be a good thing. Most people who have never come across this aspect of decimal notation before have quite normal confusions on the answer. But when they look at this humongous thread, it seems to confuse them even more. Witness the post from Jim Rodriguez! (I went looking, but never figured out why he thought someone said 1/3 = 4 ???) Perhaps a fresh discussion, without the interference of those of us with towering egos ::), would allow them to thresh it out in their own minds! 

Title: Re: 0.999. Post by David Ryan on Mar 4^{th}, 2003, 10:20am 3 things. 1) 1.000...01 does not exist. The ... implies and infinite number of zeroes, hence, no end. the .01 implies an end, and so the very concept is paradoxical. 2) .###... cannot be represented as .95, because that is 2 digits to represent one. hence, .# + .0# does not equal .95 + .095. 3) Infinity is not a number, but a concept. 

Title: Re: 0.999. Post by aero_guy on Mar 4^{th}, 2003, 11:07am Will could make it a sticky locked topic. That way it is closed but stays on page one. As to inifinity being a concept not a number, don't get these guys started on the different types of infinity! 

Title: Re: 0.999. Post by Alex Wright on Mar 4^{th}, 2003, 1:55pm My grade 10 IB math class explored this one day in class. My Teacher showed us this proof: 9.999...=10 9.999...X=10X multiply both by 10: 99.999...X=100X Subtract original values: 99.999...X=100X 9.999...X=10X you end up with: 90X=90X Thus: X=X :. 9.999... = 10 exactly. 

Title: Re: 0.999. Post by Kozo Morimoto on Mar 4^{th}, 2003, 3:29pm on 03/04/03 at 10:20:29, David Ryan wrote:
Why can't it exist? Why is it paradoxical? I posed this question on this thread before but people just ignroed it. If you have a skipping rope and you hold one end in the left hand and the other end in the right hand, and then you get your friend to take the middle of the rope and start walking away from you. At infinity, don't you get 1.000...0001? Both ends are bound yet its infinite in the middle? OR, at infinity, you end up with 2 ropes and not 1? 

Title: Re: 0.999. Post by S. Owen on Mar 4^{th}, 2003, 3:53pm I don't totally see the analogy with the jumprope, but I would note that the friend will always be a finite distance away, and is never "at infinity"... "infinity" is a place that the friend can go towards, but can never reach. I think the trick is that we all well understand what "1.0001" means, and can easily extend our reasoning to understand "1.000...01 (a million 0's omitted)", so it's natural to extend this and think we know what "1.000...01 (an infinite number of 0's omitted)" means. The latter looks reasonable enough, but what is it really denoting? It seems to denote a finite and infinite number of 0's at the same time. That's the real issue  not the math at all, but what the notation means. You can probably only reasonably interpret "1.000...001 (infinite 0's)" as "the value that it gets closer to as the number of 0's increases"  it's a limit, and the limit is 1. 

Title: Re: 0.999. Post by Icarus on Mar 4^{th}, 2003, 4:17pm on 03/04/03 at 10:20:29, David Ryan wrote:
3 replies: 1) Actually, 1.00...01, with the ... representing infinitely many zeros is not paradoxical, and can be defined (if you were to search through the posts, you will find one where I did so). However, depending on how you define it, you either get yet another representation for the number 1, so why bother? Or you get some other real number, but have to give up any sensible rules for manipulating decimals. Or, you get something that is not a REAL number, but is part a larger set of "numbers". In any case, they have no bearing on the question of whether 0.999... = 1. 2) You can define a decimaltype representation for real numbers which includes an 11th digit # = to .95 (or equal to any other number you choose, including one of the other digits). The value of any such pseudodecimal expression is defined the same way: if x = .d_{1}d_{2}d_{3}... where the d_{i} are the digits of the notation, then x = [sum]_{i=1}^{[supinfty]} d_{i}10^{i} What you get by doing so is multiple pseudodecimal expressions for every nonzero real number, instead of the 1 or 2 supplied by ordinary decimals. 3) Infinity can be a number (look again at my posts). Just not a REAL number, or a COMPLEX number. on 03/04/03 at 11:07:51, aero_guy wrote:
Too late! The mad mathematician strikes again! :P on 03/04/03 at 13:55:50, Alex Wright wrote:
ACK! I very deeply hope that this is a case of you not understanding what your teacher was saying, because the only alternative is that your teacher very deeply deserves to be fired, and is currently destroying any hope of mathematical understanding in all of his or her students! :'( :'( :'( What you have given here proves nothing whatsoever. You start off by assuming the result you intend to prove. You then prove a trivial result, and because of it believe your assumption was true?? I could use the same techniques to prove any statement you want: 5=21 Proof: assume 5=21. Then 5x0=21x0 so 0=0 true. Therefore 5=21! The actual proof you are aiming at has been reproduced on this thread many times. The last one, by Senor Incognito, is here (http://www.ocf.berkeley.edu/~wwu/cgibin/yabb/YaBB.cgi?action=display;board=riddles_medium;num=1027804564;start=175#184), but a better exposition of it is the one here (http://www.ocf.berkeley.edu/~wwu/cgibin/yabb/YaBB.cgi?action=display;board=riddles_medium;num=1027804564;start=150#169) by SPAZ. 

Title: Re: 0.999. Post by Icarus on Mar 4^{th}, 2003, 5:27pm on 03/04/03 at 15:29:35, Kozo Morimoto wrote:
As I said to Mr. Ryan, I did not ignore it. Yes, you can define expressions of the form 1.000...01. But S.Owen has it right  the real crux of the matter is: what do you mean by such an expression? Just because you can define 1.000...01 does not mean there is real number out there that it represents. If you claim it is a real number, then YOU have to define what number it is. The definition for ordinary decimals is as I gave in the previous post, but that definition does not apply here. You never did this. In your posts you merely assumed that if you could conceive of such a thing, then it must be a real number, but you never explained how. If you were to try to come up with a mathematically valid definition for which real number it is, you would discover that one of the following two situations held for your definition: 1) That the transinfinite decimals have no effect on the value of the number: 1.000...01 = 1.000...9329489024 = 1 2) That the basic rules for manipulating decimals do not apply in any reasonable fashion to your decimals. The other possibility is that you accept transinfinite decimals as defining new numbers, which are not REAL. Then you get all sorts of fascinating behaviour, but you still discover one thing: since you are not allowed to redefine the current decimals, even with such numbers, 0.999... = 1. In this extended set of numbers, 1  0.000...1 = 0.999...9 != 0.999... (All the ... here represent an infinite repetition of the preceding digit. Part of what you lose with this extension of the REALs are the algebraic properties of inequalities, so yes 1  0.000...1 > 1). Alas, no matter how you look at it, the argument falls up short of disproving 0.999... = 1. 

Title: Re: 0.999. Post by Janet on Mar 4^{th}, 2003, 5:55pm If 0.999...=1 because there is no space in between the two numbers, would 1.999...=2, or 4.5999...=4.6? What about 4999.999...=5000?!? Or is "0.999...=1" a special case, possibly a loophole of our number system? 

Title: Re: 0.999. Post by Icarus on Mar 4^{th}, 2003, 6:24pm 0.999...=1 is no special case. Any "terminating" decimal (which is equivalent to an infinite decimal ending in repeating 0s) has a second representation ending in repeating 9s. The two decimal expressions are the same to the left of the last digit not zero in the terminating version. The last digit itself is one less for the repeating 9 version. For example, in addition the ones you listed, the following is also true: 3.44456 = 3.44455999... as is 9.999 = 9.998999... You may be relieved to know that terminating decimals are the only real numbers with more than one decimal expression, and they have only two each. 

Title: Re: 0.999. Post by cho on Mar 4^{th}, 2003, 7:48pm Here's another proof that .999...=1 .999...=.9+.09+.009+.0009 which equals (1.1)+(.1.01)+(.01.001)... which equals 1+(.1+.1)+(.01+.01)+(.001+.001)... which equals 1+(.1.1)+(.01.01)+(.001.001)... which equals (1+.1)+(.1+.01)+(.01+.001)... which equals 1.000...1 which equals its reciprocal, 1+(1010)+(100100)+(10001000)... which equals (1+10)+(10+100)+(100+1000)... which equals 11+90+900+9000... which equals 1000000...1 ergo .999...=infinity 

Title: Re: 0.999. Post by schizo on Mar 5^{th}, 2003, 5:38pm x= 0.999... 10x= 9.999... 10x x= 9 9x= 9, x= 1 The numbers have to be finite in order to subtract since one must start from the smallest value of decimal place. In this case, the proof DOES assumes that there is a limit to 0.999...! 

Title: Re: 0.999. Post by S. Owen on Mar 5^{th}, 2003, 5:54pm Agreed... that solution assumes things that are seemingly more complex than the original problem (like what 10 * 0.999... means, what it means to subtract 0.999...) I don't find it very satisfying, since it doesn't seem to touch the "heart" of the issue. Heh, but at least it ends up at the right place. 

Title: Re: 0.999. Post by Icarus on Mar 5^{th}, 2003, 8:40pm An analysis of Cho's counter argument: .999...=.9+.09+.009+.0009 ... true which equals (1.1)+(.1.01)+(.01.001)... true which equals 1+(.1+.1)+(.01+.01)+(.001+.001)... This step is justifiable, but is not as obvious as Cho thinks. A convergent infinite series can only be rearranged if it is absolutely convergent. The basic sequence here, 1.1+.1.01+.01... is absolutely convergent, so it can be summed up in any order. So this step is true. which equals 1+(.1.1)+(.01.01)+(.001.001)... which equals (1+.1)+(.1+.01)+(.01+.001)... I have to assume Cho lost track of where he was going here. Both of these steps are justifiable, but the next step clearly follows from the line before these two, and it is NOT obvious that it is equal to this last line! which equals 1.000...1 Since we are clearly dealing with real numbers here, there is no such thing as "1.000...1" (dots refering to an infinite number of zeros). However, if this is changed to "1.000...", then this is an easy equivalent to the third expression. which equals its reciprocal, 1+(1010)+(100100)+(10001000)... true, once the previous expression is corrected which equals (1+10)+(10+100)+(100+1000)... Patently false. As I said earlier, you can rearrange how you sum an absolutely convergent series. But it a theorem that a conditionally convergent series (the series converges, but the series of absolute values does not converge) can be rearranged so as to converge to any limit, or to diverge to either oo or oo. The base series here 1+1010+100100... does not converge at all. You cannot rearrange how the sum is taken and expect it sum to the same value which equals 11+90+900+9000... True, in that [infty]=[infty] which equals 1000000...1 This is a number that fails to be real in two different ways, but Cho does give the correct value in the end: ergo .999...=infinity Of course, the equality does not hold because one of the intermediate steps is false. But notice that this misstep comes only after it is shown that 0.999... = 1. on 03/05/03 at 17:38:01, schizo wrote:
If numbers had to be finite (in decimal places) to subtract, then subtraction could not be defined on anything but terminating decimals. You can start subtraction at the left and move right. Occasionally you have to move back left to borrow, but whenever a nonzero digit occurs in the number you are subtracting from, all future borrowings do not need to go any futher left than this digit, so the previous digits are all completely determined at this point. Thus as long as the number you are subtracting from does not terminate, you can subtract starting with the left and move right. If it does terminate, you can still subtract starting at the left: just write the number in its nonterminating form: X.dd...dd = X.dd...d(d1)999... In any case, this problem does not arise here, as there is no need to borrow to perform this subtraction. The existence of 0.999..., that 9.999...  0.999... = 9, and that 10 * 0.999... = 9.999... are all fairly fundamental properties of decimal notation, which I have a hard time believing anyone is unfamiliar with. It does not strike me that taking them for granted is inappropriate in a proof. After all, if every theorem had to be proven directly from the axioms, proofs would be exceedingly harder to follow than they already are! As for how satisfactory a proof it is, I suppose that depends on what you are looking for from your proof. This one (that schizo has reproduced) is short, sweet, and easily followed, which I find very satisfactory indeed, even if it doesn't give much insight as to why they are equal. 

Title: Re: 0.999. Post by Kozo Morimoto on Mar 5^{th}, 2003, 9:04pm on 03/04/03 at 15:53:24, S. Owen wrote:
I never said the friend will ever be at infinity, just as how adding 9s to 0.9 will never ever become 0.999..., but "at infinity" by definition, its 0.999... So do I end up with 2 ropes of inifite length or do I still have 1 rope? 

Title: Re: 0.999. Post by David Ryan on Mar 6^{th}, 2003, 3:20am What is infiity times 2? All of my classes thus far have led me to believe it is infinity. If you have a rope doubled over at an infinite distance, it is still A rope, one. If the rope stretches "to infinity", then you, say, cut the middle, you would have 2 ropes, both of infinite length. That, at least, is the conclusion that I come to at 6:00 in the morning before school. Also, someone had mentioned don't get you guys started on different types of infinity; I was unaware that there were. Please...enlighten me. As a note, this site has made study hall one of my two most educational classes throughout the day. (what a nerd i am) //edited by Icarus to fix typo pointed out in the next post. 

Title: Re: 0.999. Post by David Ryan on Mar 6^{th}, 2003, 3:23am 3rd line, 6th word = cut 

Title: Re: 0.999. Post by wowbagger on Mar 6^{th}, 2003, 4:01am on 03/06/03 at 03:20:43, David Ryan wrote:
For a brief intro, see Icarus's post (http://www.ocf.berkeley.edu/~wwu/cgibin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=125#135) a few pages up in this thread. Btw: I guess you meant the 7th word  unless you're interpreting your post as an array in C, in which case it would be the 2nd line ;) Members can modify posts :) 

Title: Re: 0.999. Post by aero_guy on Mar 6^{th}, 2003, 5:31am Whew, that was less painful than I had thought, but then again I basically asked the same question in another thread just now, I guess I should review your link. Oh, and where the replacement goes depends on your browser width. OK guys, the horse is reduced to a puddle of goop, enough with the stick. 

Title: Re: 0.999. Post by S. Owen on Mar 6^{th}, 2003, 6:59am on 03/05/03 at 21:04:29, Kozo Morimoto wrote:
I see your analogy... I don't know if there is a "right" answer. I think that the only way you could define such a string of digits would be to take two strings starting with 1 (jumprope handles), with a bunch of 0's (jump rope), and say that their tails meet up at infinity. But, regardless of whether you can define a string like that, I don't think it means that it is a real decimal number. But you're right, neither is 0.999... In both cases you can still reasonably define a value for them by working out the limit  both have the value of 1, if anything. Interesting thought. 

Title: Re: 0.999. Post by Icarus on Mar 6^{th}, 2003, 4:32pm on 03/06/03 at 06:59:11, S. Owen wrote:
You can define strings like that. For example, consider the set N2 = {1/n  n is a nonzero integer}. This set has as its lowest element 1, then 1/2, 1/3 ... infinitely many leading up to zero, then on the positive side, the same situation reversed, ending in 1. (Zero itself is not in the set.) A "sequence" from N2 into the digits {0,1,2,3,4,5,6,7,8,9} provides just such a string. That is, you index the digits not with natural numbers ("digit 1, digit 2, ..."), but with egyptian fractions ("digit 1, digit 1/2, ... ..., digit 1/2, digit 1"). This is somewhat clumsy, so instead call them "out" and "back" digits ("digit 1 out, digit 2 out, ... ..., digit 2 back, digit 1 back"). But the next thing you have to figure out is, if this is to represent a real number, what value does it have? Contrary to what S.Owen says 0.999... IS a real number. Its value is welldefined: 0.999... represents the unique real number L such that for any e>0, there is a natural number N such that for all n > N,  SUM_{i=1}^{n} 9*10^{i}  L  < e (in other words:  0.999...(n 9s)...9  L  < e ) That such an L exists and that it is unique have both been proven, for all sequences of digits, not just constant 9s. And it has been shown here many times that for 0.999..., that number L is 1. But this definition does not extend in any obvious way to the "outandback" style decimal Kozo describes. So there is no way to "work out the limit". The best you can do, staying in the Real numbers, is to define the value to be that of the digits on the "out" side, without regard to the "back" side. But if you are bothered by 2 decimal expressions for the same number, why would you want to add infinitely many more decimal expressions for it? 

Title: Re: 0.999. Post by Icarus on Mar 17^{th}, 2003, 10:31pm Since I'm feeling loquacious (i.e. bored) tonight, I thought it might be helpful to those who are interested, but have never seen it, to give the process by which the Real numbers are defined. There are actually several possible definitions, but this one is "standard". (More accurately, what I am doing here is the standard method of building a model for the Real numbers.) 1) (Naturals) We start with sets and cardinality. Two sets are called equipotent if there is a onetoone correspondance between all the elements of one and all the elements of the other. For example the sets {1,2,3} and {a,b,c} are equipotent since we can match up their elements: 1  a 2  b 3  c But we cannot match up the elements of {1,2,3,4} with {a,b,c}: 1  a 2  b 3  c 4  So these sets are NOT equipotent. Say A~B if the set A is equipotent to the set B. The three basic properties of equipotence are: for all sets A, B, C: (1) A~A (reflexivity) (2) If A~B, then B~A (symmetry) (3) if A~B and B~C, then A~C (transitivity). These 3 properties mean that ~ is an equivalence relation. (The next step actually produces paradoxes as I have it stated, if you look deep enough. But the methods for avoiding the paradoxes requires some highpowered math, so I am going to ignore these nasty facts instead.) For each set A, the set Card(A) = {B  B is equipotent to A} is called the cardinality of A. Objects of the form Card(A) for some A are called cardinals or cardinal numbers. Definition of finite and infinite: A set A is infinite if and only if there is a proper subset B of A such that A~B. A cardinal is called infinite if it is the cardinal of an infinite set. A cardinal (or set) is finite if it is not infinite. (For example, the set N of natural numbers  which I am in the middle of building  is infinite, since we have the onetoone correspondence x <> 2x between N and the set of even natural numbers.) We can define the following for cardinals (iff = "if and only if"): Order) Card(A) <= Card(B) iff A~C for some subset C of B. Sum) If A and B are disjoint (have no common elements), then Card(A) + Card(B) = Card(A U B). Product) For any sets A, B, the product set A x B = { (a,b)  a is in A and b is in B}. Define Card(A) * Card(B) = Card(A x B). Exponentiation) A subset f of A x B is called a map from A to B if for each a in A, there is exactly one b in B such that (a,b) is in f. (In this case, we define: f(a) = b.) Define the set B^{A} = {f  f is a map from A to B}. Define Card(B)^{Card(A)} = Card(B^{A}). Some other definitions: 0 = Card({}) (cardinal of the empty set) 1 = Card({0}) 2 = 1 + 1; 3 = 2 + 1; 4 = 3 + 1; ... 9 = 8 +1, 10 = 9 + 1. The set of all finite cardinals is called the Whole numbers, and is denoted by W. The set of Whole numbers greater than 0 is called the Natural numbers and is denoted by N. (It is also common to define the Natural numbers to include 0, and call the set without 0 "The Positive Naturals" N^{+}. The definitions I give here are more traditional, since the conception of Natural numbers predates the introduction of 0 into Western mathematics. They are also more common in my experience, as it is handy to have a quick reference set around that starts at 1 instead of 0.) 2) (Integers). Define the following relation on the set W^{2} = W x W: (a,b) ~ (c,d) iff a + d = b + c. It is easily verified that ~ is an equivalence relation, which allows us to define the notation [(a,b)] = {(c,d)  (c,d) ~ (a,b)}. ( [(a,b)] is called the equivalence class of (a,b).) The set Z = {[(a,b)]  a, b in W} is called the set of Integers. We can define Order) [(a,b)] < [(c,d)] iff a + d < b + c. Sum) [(a,b)] + [(c,d)] = [(a+c,b+d)] Difference) [(a,b)]  [(c,d)] = [(a+d,b+c)] Product) [(a,b)]*[(c,d)] = [(ac+bd,bc+ad)] The Whole numbers sit in the Integers by the mapping x > [(x,0)]. We suppress this map in the future and take the whole numbers (and the naturals) to be a subset of Z. We also note that if we define n = [(0,n)] as well, then we have Z = {..., 2, 1, 0, 1, 2, ...}, and we can write a  b for [(a,b)]. Accompanying these definitions are a whole slew of properties of <, +, , and * that everyone knows. 3) (Rationals) Let X = {(a,b)  a and b are integers and b != 0}. Define the following equivalence relation on X: (a,b) ~ (c, d) iff ad = bc. (Note that this would not be an equivalence if we allowed b or d to be 0). The equivalence classes are again [(a,b)] = {(c,d)  (c,d) ~ (a,b)}. The set of Rational Numbers is Q = {[(a,b)]  (a,b) is in X} The order relation and operators are defined by: Order) [(a,b)] < [(c,d)] iff (ad < bc with bd>0), or (ad > bc with bd < 0) Sum) [(a,b)] + [(c,d)] = [(ad+bc, bd)] Difference) [(a,b)]  [(c,d)] = [(adbc, bd)] Product) [(a,b)]*[(c,d)] = [(ac,bd)] Quotient) [(a,b)]/[(c,d)] = [(ad,bc)] (c != 0) We imbed Z into Q by the map x > [(x,1)], and from here on suppress the map. Then we can express [(a,b)] = a/b. 4) (Reals). First, a couple definitions: An upper bound for a set S is a number b such that for all x in S, x <= b. If S has an upper bound, then S is said to be bounded above. The supremum or least upper bound of a set S is a number B such that (i) B is an upper bound of S, and (ii) for every upper bound b of S, B <= b. The supremum of S is denoted by sup(S). If the supremum of S exists, it must be unique. But there are bounded sets S of rational numbers which do not have a supremum in Q. For example, the set S_{2}={x  x<=0 or (x>0 and x^{2} < 2)}. A subset S of Q is called a Dedekind Cut if (i) S is not empty, and is bounded above. (ii) For any x in S, if y is in Q and y < x, then y is in S (iii) S has no maximal element (i.e. for all x in S, there is a y in S such that x < y). Two examples: if r is a rational number, then the set { x  x<r } is a Dedekind cut, as is the set S_{2} defined above Define the set R of Real numbers to be R = {S  S is a Dedekind Cut of Q}. Order) If S, T are in R, then S <= T if S is a subset of T. Operations) S # T = {x  there is s in S and t in T with x <= s # t}, where # is any of the operations +, , *, or /. Q imbeds in R by the map x > {y  y < x}. Once again, we suppress the map and consider Q to be a subset of R. Example: A little work will show that S_{2}*S_{2} = 2, and S_{2} > 0, so S_{2} = sqrt(2). The Reals have one significant advantage over the Rationals: they possess a property called "completeness", whose easiest expression is the "supremum property": Theorem: (Supremum Property) If S is a nonempty subset of R, and S is bounded above, then S has a supremum. Proof: The Dedekind Cut {x  x in Q, and x < y for some y in S} is the supremum. (The "infimum property", is equivalent.) 5) (Decimal Notation). If you have made it through this, you will see that the Real numbers are defined entirely independent of the concept of decimal notation. This is as it should be, for decimals are only a way of denoting Real numbers. They are not what Real numbers are! For this reason I stopped the definitions of 0, 1, ... at 10 above. Now we define decimal notation: Definition: A (positive) decimal expression is a map D from the set <n> = { z  z is an integer, and z <= n} for some Whole number n into the set {0,1,2,3,4,5,6,7,8,9} of digits, with D(n) !=0 if n>0 (D(n) is the leading digit only). If D is a decimal expression, it is denoted (in the U.S.A.) by listing the digits in descending order of their indexes, with a "decimal point" between the digits for indexes 0 and 1: If d_{i} = D(i) for all i, then D is denoted by d_{n}d_{n1}...d_{0}.d_{1}... If there is an integer k <= 0 such that d_{i} = 0 for all i < k, then we usually drop all the trailing zeros from the notation: d_{n}d_{n1}...d_{0}.d_{1}...d_{k} and call it a terminating decimal. (Now you know! :P) By definition, the decimal expression D represents the real number sup({ d_{n}*10^{n} + d_{n1}*10^{n1} + ... d_{k}*10^{k}  k in Z with k < n}). To complete the definition, D is the opposite of the number represented by D. Finally, let me make the following observations relating to comments posted in this forum. While it is possible to invent decimal notations such as 0.999...(infinite 9s)...9, they are not covered by this or any standard version of the definition of decimal expressions. Thus it is NOT enough to simply say "what about 1.000...01"? Decimal expressions are not numbers in and of themselves, they are only a way of denoting numbers. If you introduce new decimal expressions, YOU also have to explain what numbers they denote. On the other hand, any map from an initial segment of Z (i.e. a set <n> for some n) into the digits defines a decimal expression, which has a definite value as given above. This is standard for any accepted form of the definition of decimal expressions. In particular 0.999... and 1.000... are both valid decimal expressions and both represent real numbers which, as has been shown many times, are in fact the same. (Thanks to Wowbagger for pointing out some corrections!) 

Title: Re: 0.999. Post by Chewdogscp on Apr 14^{th}, 2003, 1:41am In elementary we always learn that a/a = 1, correct? That is that any number divided by itself is always going to be once, since one nuber goes into itself exactly one time. Hence: The approximation that 3/3 = .999... itself is illogical and an invalid aproximation. Just because 1/3 is approximated in decimal form as .333... and 2/3 is approximated as .666... The correct assumption for 3/3 should be 1. This means that .999... itself is not even = to 3/3, however, and it is proof that .999<1 

Title: Re: 0.999. Post by BNC on Apr 14^{th}, 2003, 4:49am Here we go again......... Have you read the 9 pages of this thread before posting? 

Title: Re: 0.999. Post by aero_guy on Apr 14^{th}, 2003, 5:33am [Inigo]Let me 'splain, no there is too much, let me sum up...[/Inigo] You are using circular reasoning. You say that it is not equal because it is not equal. Your basic premise is wrong though, read the other messages. 

Title: Re: 0.999. Post by rmsgrey on Apr 14^{th}, 2003, 12:24pm Something to meditate upon for those who think that because 0.999...(finite)<1 0.999...(infinite)<1 1+1+1+...(finite)<oo 1+1+1+...(infinite)=? PS might not be too productive, but I felt that having waded through the thread I had to do something to justify my time... PPS To the best of my recollection, my maths course introduced the Reals as the closure under Cauchy limits of bounded monotone sequences of rationals. Of course, Icarus' definition is equivalent (I assume  at 5am my deductive processes are prone to undetected errors), but 'my' definition leads directly to all (nontransfinite) decimal values being reals. The pythagorean proof that root 2 is irrational suffices to show that the reals contain elements not found within the rationals. As I recall, Cantor's diagonalisation argument was used to prove different cardinality. 

Title: Re: 0.999. Post by Icarus on Apr 14^{th}, 2003, 5:01pm It's a long slog to the end of this thread, and those posters who just can't seem to say their piece in a few lines, but instead write epic responses are no help at all. ::) There are several ways to complete the Rationals and obtain the Reals. All of them are equivalent. I chose Dedekind cuts because they are in some ways more elementary than the rest. They require a slightly lower level of mathematical sophistication to understand than Cauchy sequences do. In particular, you don't have to understand limits to understand them. And several posters are very confused when it comes to limits. (This is not uncommon. My experience is that less than 10% of all successful calculus students really have any idea what a limit is. The rest just manage to work things out by rote. And of those 10%, a majority have serious misconceptions.) There are also ways to complete the Rationals and get something else (http://mathworld.wolfram.com/padicNumber.html) 

Title: Re: 0.999. Post by Zeb Dahl on May 2^{nd}, 2003, 1:13pm on 07/26/02 at 17:16:25, Kozo Morimoto wrote:


Title: Re: 0.999. Post by Daniel J. Wells on Jun 10^{th}, 2003, 2:43pm I dont know where this discussion started, but I would like to end it. The question concerns the nature of the number 0.999... . The answer is simple. The number 0.999... is not a Real Number. If you want to work with the Hyperreal numbers then it is a nonstandard number with real part 1. Yes, this number 0.999... is less than 1, but it is also greater than every real number less than one. This sort of situation is not possible in teh Real numbers because they have the Archemidian property, that is. there is some multiple of 1 that is greater than any given real number. This also implies that for any possitive real number there is a smaller possitive number of the form 1/N where N is a multiple of 1. Suppose a number, Q (0.999...) is less than 1 but greater than every (other) real number less than 1. Clearly, 1Q is greater than 0, since Q<1. (1Q>11=0) Also since Q is greater than any real number less than 1, Q is less than any number greater than 1 and thus 1Q is less than any real greater than 0. But as I mentioned early the Arcemedian principle garuntees that there a natural number N such that 1/N<Q. Thus the supposition that Q exists is false. (Indeed, if you ignore the Completeness Property, and make this assumption, you obtain the Hyperreal Numbers.) [Please forgive any attrocious spelling errors I have most certainly made.] 

Title: Re: 0.999. Post by Daniel j Wells on Jun 10^{th}, 2003, 2:51pm Just in case you missed the point of my post: 0.99999999.... is not a Real Number (the kind you use in calculus, or algerbra for than matter.) any more than i (where i^2=1) is a Real Number. 

Title: Re: 0.999. Post by Sir Col on Jun 10^{th}, 2003, 3:13pm Are you suggesting that 1 is not a real number? s_{1}=1/2+1/4+1/8+1/16+... 2s_{1}=1+1/2+1/4+1/8+1/16+... Therefore 2s_{1}–s_{1}=s_{1}=1 In the same way, s_{2}=9/10+9/100+9/1000+... 10s_{2}=9+9/10+9/100+9/1000+... Therefore 10s_{2}–s_{2}=9s_{2}=9 and so s_{2}=1. In other words, as much as it is true to say that the infinite sum of 1/2+1/4+...=1, it is true that to say that 0.999...=9/10+9/100+9/1000+...=1. 

Title: Re: 0.999. Post by Icarus on Jun 10^{th}, 2003, 3:44pm I stand amazed at someone who has heard about hyperreals (and therefore presumably nonstandard analysis), the completeness property, and the archimedean property, but does not understand the meaning of 0.999... . As Sir Col indicates, the definition of 0.999... is [sum]_{n=1}^{[supinfty]} 9*10^{n}. How can you claim that this convergent series is not real? 

Title: Re: 0.999. Post by redPEPPER on Jun 11^{th}, 2003, 3:33am on 06/10/03 at 14:43:16, Daniel J. Wells wrote:
Congratulations, you just proved that 0.999... = 1 ;D What? You don't see it? Well, you demonstrated that there cannot be a real number smaller than 1 but greater than any other real number smaller than 1. By notation, 0.999... is real, greater than any other real number smaller than 1, and certainly not greater than 1 itself. By your demonstration it cannot be smaller than 1. It must therefore be equal to 1 :D Your flaw was to assume from the very start that 0.999...=Q without proving it, specifically when you define Q<1. But 0.999...<1 is far from a given, as this whole thread shows. 

Title: Re: 0.999. Post by Sir Col on Jun 11^{th}, 2003, 4:59am One of the classic proofs I use with my classes when they're struggling with this concept is simple, yet powerful. Given any two numbers, a and b, the arithmetic mean, (a+b)/2, lies midway between a and b. If a=0.999... is the 'last' number before 1 and b=1. I would ask, what is the arithmetic mean of a and b? The only way to resolve this dilemma is for a and b to be equal. However, I must give some consideration to Daniel's argument. I have never heard of hyperreal numbers, so I will have to research them for myself before I am able to fully appreciate his points. Despite my ignorance I can accept that if we define 0.999... as a number outside of the real numbers, then it is not real – that is common sense; if the set of A (hyperreal numbers) are not part of the set B (real numbers) and c (0.999...) is contained in A (hyperreal numbers) then c (0.999...) is not part of B (real numbers). My issue would be whether such a set exists and then whether 0.999... belongs to this set. Obviously imaginary numbers are mutually exlusive to real numbers, but I thought that real and imaginary collectivey (the complex set) was complete; isn't that the fundamental theorem of arithmetic which Gauss proved? I do appreciate though that my proof above only holds if 0.999... is real in the first place and the other proofs that have been presented tease our intuition tantalisingly on the borders of the infinitesimal and the infinite – both concepts too small or too big by definition for us to comprehend. Is there any truth in this? :/ 

Title: Re: 0.999. Post by Icarus 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. 

Title: Re: 0.999. Post by Sir Col 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! 

Title: Re: 0.999. Post by Icarus 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. 

Title: Re: 0.999. Post by otter 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. 

Title: Re: 0.999. Post by wowbagger on Jul 14^{th}, 2003, 6:18am Following the link in otter's post, we read the following: Quote:
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! :) 

Title: Re: 0.999. Post by S. Owen on Jul 14^{th}, 2003, 8:56am Indeed  ".999~ never quite reaches that limit" .999~ is that limit! 

Title: Re: 0.999. Post by Chewdogscp 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.... 

Title: Re: 0.999. Post by wowbagger on Jul 16^{th}, 2003, 5:03am on 07/15/03 at 17:13:20, Chewdogscp wrote:
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! ;) 

Title: Re: 0.999. Post by Icarus 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


Title: Re: 0.999. Post by aero_guy 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. 

Title: Re: 0.999. Post by Sir Col 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 ;D 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'. 

Title: Re: 0.999. Post by Icarus 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! 

Title: Re: 0.999. Post by Sir Col 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. 

Title: Re: 0.999. Post by mook 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.... 

Title: Re: 0.999. Post by Sir Col 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. 

Title: Re: 0.999. Post by mook 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... 

Title: Re: 0.999. Post by Icarus 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 ! 

Title: Re: 0.999. Post by wowbagger on Jul 28^{th}, 2003, 11:55am on 07/23/03 at 18:15:08, Sir Col wrote:
So why do you think I quoted that post? ::) 

Title: Re: 0.999. Post by Sir Col on Aug 1^{st}, 2003, 2:02pm You're a bad man, wowbagger! :P 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? ::) 

Title: Re: 0.999. Post by Icarus 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! :o 

Title: Re: 0.999. Post by Sir Col 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... 

Title: Re: 0.999. Post by Icarus 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...) 

Title: Re: 0.999. Post by Sir Col 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. 

Title: Re: 0.999. Post by Icarus 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 (http://www.ocf.berkeley.edu/~wwu/cgibin/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/cgibin/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/cgibin/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/cgibin/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/cgibin/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(( 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 (http://www.ocf.berkeley.edu/~wwu/cgibin/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] 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 (http://www.ocf.berkeley.edu/~wwu/cgibin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=75#83) 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 (http://www.ocf.berkeley.edu/~wwu/cgibin/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)...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 (http://www.ocf.berkeley.edu/~wwu/cgibin/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 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. 

Title: Re: 0.999. Post by Icarus on Sep 1^{st}, 2003, 12:56pm 1) Infinite Decimals are Approximations (http://www.ocf.berkeley.edu/~wwu/cgibin/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 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 (http://www.ocf.berkeley.edu/~wwu/cgibin/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 "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 (http://www.ocf.berkeley.edu/~wwu/cgibin/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/cgibin/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/cgibin/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/cgibin/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/cgibin/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/cgibin/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", "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 (http://www.ocf.berkeley.edu/~wwu/cgibin/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/cgibin/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/cgibin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=0#7) 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 (http://www.ocf.berkeley.edu/~wwu/cgibin/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/cgibin/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/cgibin/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! ::) 

Title: Re: 0.999. Post by matt on Sep 8^{th}, 2003, 10:15am It's pretty funny to watch you all argue over something like this. The simple answer is that 0.999... < 1. I learned that from my calculus teacher when I decided to argue the point. 0.999... != 1, but the math terminology indicates that it is "infinitely close to" 1. So that's like saying "it isn't, but it may as well be." I can tell my "authoritiative" answer doesn't thrill you. You want proof that they're at least not equal, though? Easy as pie. An irrational number is defined as "any number that can't be represented by the ratio of two integers." So I ask, what ratio of integers would result in 0.9999....? It's not 3/3... that results in 1. So, 0.999.... is an irrational number. So why are so many people trying to say that an irrational number can be equal to a rational one? If they were the same, they'd have to be within the same set. if x member of R and x=y then y member of R. Well, 1 member of REAL, 1 = 0.999...., 0.999 member of REAL? It doesn't work... it can't work. Although, it might as well work, because for all practical intents and purposes, they're the same. 

Title: Re: 0.999. Post by THUDandBLUNDER on Sep 8^{th}, 2003, 10:22am Quote:
;D 

Title: Re: 0.999. Post by matt on Sep 8^{th}, 2003, 10:38am Icarus posted this (though with better formatting): Sorry, but it ain't so. By definition 0.999... = limitn {0.999...( n9s)}. 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. I have to take exception to this.... you're stating that 0.9... is the limit of 0.9! Maybe my limited math experience is showing here... but that's a bit of infinite recursion, and that's simply not possible. I really don't know what you're trying to say... but 1/3 != 0.3... because 0.3... is the base10 approximation of a number that is unrepresentable with that notation. 0.3... is by its very nature something that doesn't really exist (in the sense of a concrete quantifiable number). But, the limit of 0.3....... n3's as n>INF is definitely 1/3. As the limit of 0.9..... (n 9's) as n>INF is definitely 1. But it's not 0.9... (n9's). That's self defeating and stupid. Granted I've never heard of hyperreals, and I've only progressed through Calculus 2. But I think I've got a pretty firm grip on things. Last I checked, putting 0.9... and saying "It really should be the limit of 0.9..." doesn't change the original question. And if someone told you there's an assumed limit in "0.9..." you should get your money back for that course. Of course, I'm one of those idiots who doesn't see a major problem with 0.0....1, so I must be braindead for thinking that an alternative representation for the calculus h constant is valid. Go back to your Calculus fundamentals... it's the first thing they teach you. h is used like any other variable, and you don't balk at it there. Why is it a problem when people who haven't been exposed to it refer to it with improper terminology? 

Title: Re: 0.999. Post by Ulkesh on Sep 8^{th}, 2003, 10:43am I think I'll have to agree with matt here. I mean, when something is so intuitively obvious it seems pointless to use something as inaccurate at maths to justify a 'correct' answer. After all, Godel's incompleteness theorem quite clearly shows that maths is incomplete and therefore wrong about more or less everything. While we're on the subject, I'd just like to voice my support of the longheld view of a flat Earth. Since Physics is basically applied maths, and since I've rigorously proven that maths in inherently wrong, the misguided fools who believe in building their models of physical phenomena purely upon empirical evidence are clearly not using their own intuition and seeing what is obvious. If the earth were round we'd clearly fall off. The idea of 'gravity' is therefore fallacious, QED. 

Title: Re: 0.999. Post by S. Owen on Sep 8^{th}, 2003, 10:52am on 09/08/03 at 10:15:06, matt wrote:
To expand on T&B's smiley face  your argument is plainly circular. You say that 3/3 != 0.999... because 3/3 == 1. But that is only a valid argument if you assume that 1 != 0.999... ! This sort of argument doesn't show that anything is irrational anyway. For example, 3/3 != 0.8, but that does not mean that 0.8 is irrational. You would need to show that there are no integers a and b such that a/b = 0.8, not that some particular a and b do not satisfy a/b = 0.8. First step is to decide what you think "0.999..." even means. Then you can decide what it equals and does not equal. 

Title: Re: 0.999. Post by matt on Sep 8^{th}, 2003, 10:59am somebody said this, but I'm on the preview page and can't see his/her name (sorry): To expand on T&B's smiley face  your argument is plainly circular. You say that 3/3 != 0.999... because 3/3 == 1. But that is only a valid argument if you assume that 1 != 0.999... ! yeah... my mistake. I didn't mean to put the word "so" there or imply any sort of causality. I meant to prevent anyone with replying that 3/3 directly gives 0.9..., as that would be to assume the proof (like I ended up doing by mistake). I was just typing faster than I was thinking. ;D I still challenge anybody to come up with a ratio that is == 0.9... that doesn't include some sort of unquantifiable expression (such as (INF1)/INF == 0.9...). And I still won't concede that 0.9... really means the limit of 0.9... because that's just as circular as me! 

Title: Re: 0.999. Post by wowbagger on Sep 8^{th}, 2003, 11:28am on 09/08/03 at 10:59:58, matt wrote:
Hm, if it doesn't mean the limit to you, what does it mean? Do you see this expression as representing a real number? And if so, which one? As Icarus already pointed out, you can invent many notations, but you have to say what number it represents. Standard mathematics defines such nonterminating decimals als the limit of an infinite process  how do you define the value of your 0.999...? How is the definition via the limit circular? It has the form x = lim_{n[to][infty]} [sum]_{k=1}^{n} 9*10^{k}. There's no x in the right hand side, neither explicitly nor hidden. 

Title: Re: 0.999. Post by S. Owen on Sep 8^{th}, 2003, 12:25pm on 09/08/03 at 10:59:58, matt wrote:
Yes, I think this is the real heart of this question  what does "0.999..." even mean? We all immediately understand what the string "0.999" signifies, but although it looks similar, "0.999..." is not conventional notation for a decimal. So this is what we need to decide. I think probably the only reasonable interpretation is to simply extrapolate from what decimal notation means; so 0.999... = 0.9 + 0.09 + 0.009 + .... And this is 1. As to giving you a ratio that equals 0.999..., well, 3/3 is such a ratio. But for you to decide whether that is true, you need to decide what 0.999... equals, which is original question anyway. So, I don't think the rational/irrational line of argument gets anywhere before exploring what 0.999... even is. 

Title: Re: 0.999. Post by Sir Col on Sep 8^{th}, 2003, 1:10pm I like Matt's challenge: "So I ask, what ratio of integers would result in 0.9999....?" That's a very original slant on the discussion. Perhaps 0.999... is an illegal expression in the first place? I think this, in part, relates to the denumerability thread that I started in the General/Whatever area, here (http://www.ocf.berkeley.edu/~wwu/cgibin/yabb/YaBB.cgi?board=riddles_general;action=display;num=1062961180). Maybe, as Matt says, and I'm probably showing my ignorance again here, the decimal representation is incomplete (as it can never be completed) and is not the same as the rational representation. However, I think that S. Owen made some excellent points, as the heart of the question is, "What does 0.999... mean?" I agree with him, the only sensible interpretation is to extrapolate the infinite expansion of the decimal: 0.9+0.09+0.009+... Confused! ??? I really must stop reading this thread. ;D By the way, Icarus, thank you for your amazing summary of all those pages. 

Title: Re: 0.999. Post by towr on Sep 8^{th}, 2003, 1:29pm Let's make a decimal expansion of 1/1, using a longdevision 1/ 1.000... \ 1.000... 1  0.0 .0  .0 (etc) now let's do it again, but differently, because we like to be contrary 1/ 1.000... \ 0.999... 1  1.0 .9  .10 .09  .01 (etc) So clearly 0.999... is a rational result of dividing 1 by 1 

Title: Re: 0.999. Post by Icarus on Sep 8^{th}, 2003, 6:50pm on 09/08/03 at 10:15:06, matt wrote:
:'(Yet another supposed math teacher who does not know the subject they are trying to teach.:'( My father had a math teacher (everything teacher actually  it was oneroom school) whose approach to solving math problems was to flail around at random until she managed to produce the answer in her book. He told me that she regularly included the problem number in calculating her "solutions". The point is: just because a teacher tells you something, that doesn't mean it's true. Quote:
"infinitely close" is not an math terminology. It is a phrase used solely by those who do not understand what they are talking about, or by those who do know in leading into the truth: "infinitely close" means zero distance between them; ie., they are equal. Quote:
I don't like saying this, because it sounds like I'm sticking my nose in the air, and demanding that you take things on my sayso. This not true. But several comments you have made indicate you believe that we are not well educated in mathematics. I have a Ph.D. in the subject. Several others here have degrees. Most if not all of the posters since your first post are much further advanced in the subject of mathematics than you are. This does not mean you should take any of our word on this, but it does mean that if you find yourself in disagreement  you would be wiser to question your own understanding first, instead of assuming that we are all wrong and you are right. By all means, bring up your questions. Ask why this isn't so. Demand explanations of statements you don't understand or sound false. But don't assume that everyone else doesn't know what they are talking about. on 09/08/03 at 10:59:58, matt wrote:
I usually open up a second session. This allows me explore in it while creating my post in the first. I can even quote from multiple posts by doing a quote in the second and copying it back into the first. Of course, if you don't have some form of broadband, that may not be practical. Quote:
No  I never do that! ::) But you are wrong. You can directly get 0.999... from 3/3, as Towr has demonstrated. Quote:
Such an expression is not acceptable, but again this misses the point. You always get 0.999... from a/a expressions, because 0.999... = 1.[/quote] Quote:
I am not clear as to what you think is being said here. There is nothing even remotely circular about it. Either you do not have a good understanding of the concept of a limit of a sequence, or you have completely misunderstood the notation. The definition of 0.999... (where the dots here means that infinitely many 9s follow), is that 0.999... = lim_{n[to][subinfty]} A_{n} Where {A_{n}} is the sequence A_{1} = 0.9 = 9/10 A_{2} = 0.99 = 99/100 A_{3} = 0.999 = 999/1000 ... Why do you think this definition is circular? Are you under the mistaken impression that 0.999... is itself in the sequence? It isn't. Each of the sequence elements only has a finite number of 9s. Sometimes we tend to browbeat too much, but I would honestly like to discuss this with you  and even more with that calc teacher! Anyone teaching mathematics really needs to know better than that! Backing up to something else... on 09/08/03 at 10:38:26, matt wrote:
I addressed this as the first misconception, and what I said there is true. 0.333... is NOT an approximation. All finite length strings of threes are approximations of 1/3, with the approximations improving as the string length increases. By definition, 0.333... is the limit of these approximations. That limit is exactly 1/3. Quote:
How so? 1/3 is quite quantifiable, and it is what 0.333... means, by the definition. I assume what you are refering to here is the inability to ever write down an infinite number of 3s. This is misconception number (4) in my list. I don't have to write down infinitely many threes to represent the IDEA of it. You yourself have written down exactly that idea many times. To claim it is impossible would be disingenuous. Quote:
Was that last "(n9's)" a mistype? No one ever said limit was 0.9... (n9's). Quote:
hyperreals were a chimera someone threw out. This whole situation is defined in the real numbers. No "superset" of numbers is going to change the meaning in the real numbers. Quote:
Sorry, but no. Quote:
When you write down "0.9", do you say "it really should be 9/10". If so, you've got a problem. Because you need to change that to "(1+1+1+1+1+1+1+1+1)/(1+1+1+1+1+1+1+1+1+1)". But no, you then need to exchange that for the definitions of "+", "/", "()", and of course "1". Nicolas Bourbaki in the formalism he developed in "The Elements of Set Theory" estimated that to express 1 in the primative elements of his set theory would require several tens of thousands of symbols. I hope you have a lot of time, as you will need to write it down 19 times! ;D "+" is worse yet, and "/" will be a real killer. All notations are defined as shorthand representations of something else (with the exceptions of a few primatives, which are defined in terms of how they relate to each other). "0.999..." is a defined notation which stands for the concept of infinitely repeating 9s, which in turn stands for the limit of the real numbers represented by finite numbers of 9s, as the number of nines increases without bound. An examination of the formal definitions given above, or from any reference you can find, will show that these are grounded in solid finite terms, without any shaky handwaving. Quote:
SOMEBODY should, anyway! ;) Quote:
[smiley=h.gif] is a real variable. Ie. just like every other real variable, it's values are all real numbers. The problem with 0.0...1 (with the dots representing infinitely many zeros) is that no one has defined it. 0.999... is a defined expression. 0.0...1 is not. Before you can start talking about 0.0...1, you need to define what number it represents. This can be done, but you either end up with something other than real numbers (so [smiley=h.gif] does not take them on as values in the standard real calculus), or else you just end up with another expression for 0 (so [smiley=h.gif] is not it either, since in the application you are refering to  definition of the derivative  [smiley=h.gif] is not allowed to be 0), or you get something that does not behave at all like other decimals and so is worthless for any sort of calculation. In particular, you are comparing 0.0...1 to a variable expression. All decimals refer to a single number. But you are arguing this new expression should be regarded as a variable (which can take on many values)? 

Title: Re: 0.999. Post by Icarus on Sep 8^{th}, 2003, 6:51pm on 09/08/03 at 13:10:41, Sir Col wrote:
?? Would you mind clarifying? What do you mean by "incomplete". How is it that "it can never be completed"? The only interpretation I can find for this is yet another instance of the "infinite process" fallacy. That infinite decimals require infinite processes. Conceptually, there is such a thing. But in the actual mathematics, there is not. We define the outcomes of our conceptual infinite processes in a finite fashion. If we cannot find such a finite definition, then the idea remains undefined. That is not the case here. Quote:
Yes, when trying to determine what something means, the definition is a REALLY GOOD place to start! 

Title: Re: 0.999. Post by Smart Guy on Oct 3^{rd}, 2003, 4:07am on 07/27/02 at 06:32:11, Kozo Morimoto wrote:
The answer is (0.9999......1) 

Title: Re: 0.999. Post by Sir Col on Oct 3^{rd}, 2003, 6:02am I'm afraid not, Smart Guy. You 'answer' terminates, by virtue of having a 1 at the end. In which case consider these two numbers: 0.999...1 0.999...9 Which is bigger? Now consider these two: 0.999...9 0.999...99 As soon as you terminate the decimal, it ceases to be 0.999... (recurring). And Icarus, I must object to your sarcasm: Quote:
You'd almost pass for being English. :P 

Title: Re: 0.999. Post by THUDandBLUNDER on Oct 3^{rd}, 2003, 12:24pm Quote:
...or a scriptwriter for The Hitchhiker's Guide to the Galaxy?? 

Title: Re: 0.999. Post by Icarus on Oct 3^{rd}, 2003, 5:02pm on 10/03/03 at 06:02:42, Sir Col wrote:
That whirring sound you hear is my Scottish forebears spinning in their graves at the very thought! 

Title: Re: 0.999. Post by william wu on Oct 6^{th}, 2003, 12:47pm on 10/03/03 at 12:24:12, THUDandBLUNDER wrote:
Haha :D I think Icarus would be perfect for such a position :) P.S. Thanks again Icarus for your uncanny summaries of this topic 

Title: Re: 0.999. Post by johnkil on Oct 11^{th}, 2003, 1:37pm Graphically, If, as x approaches some number c, the values "absolute value of R(x) as it approaches infinity, then the line x=c is a vertical asymptote of the graph of R with an asymptote of x=1, c=0.9999999... (ad infinitum) c+0.x would approach 1 but NEVER intersect. therefor 0.999999... < 1 ! 

Title: Re: 0.999. Post by THUDandBLUNDER on Oct 11^{th}, 2003, 1:49pm Quote:
Graphically, it intersects at infinity. And how many 9's does 0.999999... contain?? An infinite number of 9's. Therefore 0.999999... == 1 Duh. 

Title: Re: 0.999. Post by S. Owen on Oct 11^{th}, 2003, 2:14pm Yes, the argument that 0.999... < 1 because graphically the sequence of points (1, 0.9), (2, 0.99), (3, 0.999) approaches y=1 asymptotically is incorrect. However I don't so much like the counterargument that it "intersects at infinity", since it just begs another question  what does that mean? It also perpetuates this way of analyzing the problem  where does the sequence intersect y=1? It doesn't, but the more important point is what about this scenario corresponds to 0.999...? The best point to make that 0.999... actually corresponds to that line which the sequence approaches, and does not correspond to the sequence of points. johnkill: if you're not convinced, again I would like to trot out the fast counterargument: Let 0.999... = x. If x < 1, then consider the number y = (x+1)/2. x < y < 1, but what would the decimal representation of y be? 0.999...? 

Title: Re: 0.999. Post by Icarus on Oct 11^{th}, 2003, 6:35pm I lumped this argument (it has been made before) in with some similar arguments as misconception (2). I suppose I should have listed it separately. But it boils down to the same thing: 0.999... is the limit, not the sequence or any member of the sequence. The limit is not the approximations, but the thing that is being approximated. Therefore, as S.Owen has well said: 0.999... is the asymptote, not the approaching curve! However, concerning your counter demonstration, S. Owen, I reiterate my disgreement to its worth. The reason that some people are convinced by this simple argument is that they do not understand the subject well enough to spot what needs to be demonstrated to make it rigorous. For this to be a rigorous demonstration, you must first show that every real number has a decimal notation, and that to be greater than 0.999... and 1 requires digits strictly between 9 below and (1)0 above. The latter is wellknown if not wellunderstood. The former requires a much deeper knowledge of real numbers than people who are confused on this issue have. Just as you do not like T&B's argument, I do not like this one that relies for its simplicity on the ignorance of the person hearing it. Both arguments are valid, but only with enough sophistication on the part of the listener that 0.999... = 1 should no longer be an issue for them. 

Title: Re: 0.999. Post by random0 on Nov 14^{th}, 2003, 2:53pm I suppose this whole argument is about base 10? 

Title: Re: 0.999. Post by Icarus on Nov 14^{th}, 2003, 3:57pm Yes, the question refers to base 10 only. Though a similar thing occurs in any base: 0.111..._{2} = 1 0.222..._{3} = 1 0.FFF..._{16} = 1 

Title: Re: 0.999. Post by Sir Col on Nov 14^{th}, 2003, 4:27pm Boy it takes a long time to post a message in this thread. Any chance of starting a sequel thread and placing Icarus' excellent summary at the top, and maybe a link to this old thread? It's funny, I was teaching one of my younger classes (1213 year olds) how to write fractions as decimals (manual division to establish cyclic patterns), and then looking at how we write decimals as fractions. A natural question, when looking at decimals being written as fractions, is whether or not all decimals can be converted to fractions. As we'd established that all fractions recur or terminate they wondered if there are any other types of decmial. Once we wrote down a few examples (I've avoided the terminology with them for the moment), they decided to stick to recurring or terminating decimal types. This is where it got really interesting. After me showing them how to write recurring decimals as fractions (using the algebraic approach), they wondered if it was reasonable to suppose that, just because all fractions will recur or terminate, all recurring decimals necessarily come from fractions. When I presented them with a few practice questions: 0.666..., 0.0555..., 0.232323..., 0.999..., you can imagine their reaction when they came across the last one. Their question, in which they were asking to be convinced that all recurring decimals necessarily come from fractions, I found quite interesting. I wonder if this is the heart of the problem for some people in understanding the equivalence of 0.999... and 1? 

Title: Re: 0.999. Post by TimMann on Nov 14^{th}, 2003, 11:47pm I don't think I understand their question. Did they mean, "is every repeating decimal equal to some fraction"? And by "fraction", did they mean "rational number", or did they think 0.999... = 1 is an exception because 1 "isn't a fraction", or because 9/9 (if you showed them that method of converting a repeating decimal to a fraction) isn't a proper fraction? 

Title: Re: 0.999. Post by Sir Col 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? 

Title: Re: 0.999. Post by Icarus 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 :P. 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. 

Title: Re: 0.999. Post by Sir Col 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. 

Title: Re: 0.999. Post by Icarus 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 (http://www.ocf.berkeley.edu/~wwu/cgibin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=175#194). 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. 

Title: Re: 0.999. Post by Sir Col 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. 

Title: Re: 0.999. Post by jay c 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 

Title: Re: 0.999. Post by Sameer 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 ;D 

Title: Re: 0.999. Post by THUDandBLUNDER on Feb 13^{th}, 2004, 7:40am Quote:
...by repeating the same method as the very first post in this thread? Why not read this (http://www.ocf.berkeley.edu/~wwu/cgibin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027804564;start=225#248)? 

Title: Re: 0.999. Post by Icarus on Feb 13^{th}, 2004, 9:06pm on 02/13/04 at 07:37:34, Sameer wrote:
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. 

Title: Re: 0.999. Post by Sameer 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'? 

Title: Re: 0.999. Post by towr on Feb 16^{th}, 2004, 11:16am Quote:
so it doesn't tend to 1, it is 1 Quote:
[sum] _{k=1}[supinfty] f(k) [equiv] lim n[to][infty] [sum] _{k=1}^{n} f(k) 

Title: Re: 0.999. Post by Sameer on Feb 16^{th}, 2004, 3:45pm on 02/16/04 at 11:16:06, towr wrote:
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... 

Title: Re: 0.999. Post by Icarus 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. 

Title: Re: 0.999. Post by Icarus 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. 

Title: Re: 0.999. Post by Sameer 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? 

Title: Re: 0.999. Post by towr on Feb 17^{th}, 2004, 7:00am h(2) is not defined, so no If h was a continuous function (http://mathworld.wolfram.com/ContinuousFunction.html), 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 (http://mathworld.wolfram.com/Limit.html) from n > [infty] 

Title: Re: 0.999. Post by Sameer on Feb 17^{th}, 2004, 7:51am Hmm so I am gonna quote Icarus from the misconception post Quote:
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. 

Title: Re: 0.999. Post by towr on Feb 17^{th}, 2004, 8:11am on 02/17/04 at 07:51:53, Sameer wrote:
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)} ). 

Title: Re: 0.999. Post by Sameer on Feb 17^{th}, 2004, 8:23am on 02/17/04 at 08:11:37, towr wrote:
awesome towr... thanks for clearing the cloud in my head!!! :D. I learn so much everyday!!! 

Title: Re: 0.999. Post by TenaliRaman 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? 

Title: Re: 0.999. Post by towr on Feb 19^{th}, 2004, 1:30am on 02/18/04 at 21:18:11, TenaliRaman wrote:
I suppose the thread is long, but is that good enough of a reason? 

Title: Re: 0.999. Post by TenaliRaman 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. 

Title: Re: 0.999. Post by towr on Feb 19^{th}, 2004, 6:46am If this thread were locked, they would probably just make new threads. Which isn't any better.. 

Title: Re: 0.999. Post by Sameer 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... 8) 

Title: Re: 0.999. Post by TenaliRaman 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. 

Title: Re: 0.999. Post by Icarus on Feb 19^{th}, 2004, 4:19pm I see that the problem I've getting on the last couple days has not been universal. However, towr has superbly answered the questions. To TenaliRamen, it's true that you are not likely to see any new insights into this problem in this thread again, and very little other discussion, as the large size of this thread makes responding to it a pain. But in terms of space used, I doubt it is very much by computer standards, and it does provide a place for those still interested &/or unclear about the truth here, such as Sameer was. Indeed, the discussion between Sameer, towr, & I demonstrates to me that this thread needs to be ongoing, as there will always be those for whom this topic is not fully settled. If we were to lock this thread, then I would want to immediately start a successor to continue the conversation, so that those new to this have a place where they can comment. Perhaps this would be a good idea, with the summaries of this thread leading off, since it would encourage new visitors to actually read it rather than simply skipping to the end and ignoring the previous conversations. 

Title: Re: 0.999. Post by SWF on Feb 19^{th}, 2004, 7:34pm When x is less than or equal to 1: tan^{1}(x)=x  x^{3}/3 + x^{5}/5  x^{7}/7 + x^{9}/9  ... To facilitate the upcoming term by term comparison with the series for 0.999..., rearrange some of the terms to get: tan^{1}(x)= (xx^{3}/3) + (x^{5}/5x^{7}/7) + [sum] ( x^{4n+1}/(4n+1)  x^{8n5}/(8n5)  x^{8n1}/(8n1) ) where the summation is on n from 2 to infinity. If there is any doubt about convergence of this form for inverse tangent, see the attached graph comparing the exact value of inverse tangent with what is obtained with 30 terms of this series. Also, using standard convergence tests, it is not difficult to show the series converges for x less than or equal to 1. Inserting x=1 into this series and multiplying by 4/[pi] equals 1 since 4/[pi]*tan^{1}(1)=1. Compare term by term with 0.999... = 0.9+0.09+0.009+...: 4/[pi]*(11/3) < 0.9 4/[pi]*(1/51/7) < 0.09 4/[pi]*( x^{4n+1}/(4n+1)  x^{8n5}/(8n5)  x^{8n1}/(8n1) ) < 9*10^{N1} (for all n>=2) Therfore 1 < 0.999... ;) 

Title: Re: 0.999. Post by Icarus on Feb 20^{th}, 2004, 3:25pm Very nice! Did you come up with it yourself, or find it somewhere? I know what is going on, but will leave it for someone else, other than giving this minor hint: [hide] It's has to do with something I've mentioned before in this thread.[/hide] 

Title: Re: 0.999. Post by Sameer on Feb 20^{th}, 2004, 4:14pm on 02/20/04 at 15:25:49, Icarus wrote:
Hahah very nice hint indeed Icarus ;) And I do agree with the earlier post. I do get confused sometimes with my own fundamentals and tend to project them wrong. Besides even though I have been on this forum for quite a while I still haven't been able to go through all of them. In fact I am just starting on medium. I havent even looked at hard (which I doubt would be just more than browsing after my grand show on page 12 of this thread). Tenali, my post on space was flowing with sarcasm too ;D. I haven't been in school for a while and besides I am not working in the field I studied. That is why I have lost touch with lot of my math. Just the other day when I was flipping through one of the math books and I was like whoa, can't remember. This forum is the place where I can explore something I knew before and have forgotten or have had shaky foundations to begin with. I am simply honored to be in company of you all greats (the count is definitely higher than 5 so that is not the answer to that question in easy section :P) 

Title: Re: 0.999. Post by rmsgrey on Feb 20^{th}, 2004, 4:56pm The series for tan^{1} may converge, but it doesn't converge absolutely for x=1 (if you change each '' to '+', the result diverges). By observing that 8/3[pi]<8/9<.89, and substituting that for the first term in SWF's expansion of .999... you get that .9899999999999...>1 (using 8/9 rather than .89 gives .9888888...>1 but requires you to take the behaviour "at infinity" on faith; given the subject matter of this thread that may be unwise) [e]corrected equation (sorry Icarus)[/e] 

Title: Re: 0.999. Post by Icarus on Feb 20^{th}, 2004, 8:41pm on 02/20/04 at 16:56:51, rmsgrey wrote:
And what does this tell you? Quote:
Whoa! :o Now, since 2 < 8/[pi] [approx] 2.55, we have 2 < 1! It's amazing what you can prove if your observation skills are a little rusty! ;) 

Title: Re: 0.999. Post by rmsgrey on Feb 21^{st}, 2004, 8:10am on 02/20/04 at 20:41:33, Icarus wrote:
That rearranging the terms can change the limit. Quote:
Particularly since I went through my entire post thinking I had put that pesky 3 in! (I've edited it in now) 

Title: Re: 0.999. Post by Icarus on Feb 21^{st}, 2004, 8:58am Exactly. When I was in college, one of my professors suggested as a research project to find conditions on rearrangements of a conditionally coverging series that would guarantee the rearranged series would converge to the same limit, and/or conditions guaranteeing it would converge at all. I couldn't think up any good ways of approaching the problem, however. 

Title: Re: 0.999. Post by Eigenray on Feb 22^{nd}, 2004, 2:01am If a (real) series converges, but does not converge absolutely, then it can be made to converge to any arbitrary (extended real) value, or diverge, by a suitable rearrangement of its terms. The proof is not hard; the key is that [hide]the subsequence of positive terms converges to 0, yet its sum must converge to infinity, and similarly for the negative terms[/hide]. The converse is true as well: If a series converges absolutely, then any rearrangment of its terms converges to the same value. 

Title: Re: 0.999. Post by SWF on Feb 22^{nd}, 2004, 5:48pm on 02/20/04 at 15:25:49, Icarus wrote:
Thank you. I made this up for everyone's amusement, but I already knew that rearranging a conditionally convergent series can be made to give whatever one wants. I stayed away from using the standard example of the series for ln(2), hoping to make spotting the flaw more difficult, but that trick did not work. With all the pages in this thread, it takes quite a long time to post with a modem connection because the bottom of the screen seems to include all the previous posts. 

Title: Re: 0.999. Post by Icarus on Feb 22^{nd}, 2004, 7:45pm That you were familiar with the result was obvious to those of us who are also familiar with it. It was a nice change of pace from the recurring arguments, despite its tongueincheek nature. This thread is hard to reply to even on a cable modem. I've been thinking about taking Tenali Ramen's advice and locking it, but also replacing it with a new thread linking back, and perhaps starting out with copies of my thread summaries. To reiterate what I've said already, this thread serves still as a learning tool for those who have never confronted this issue, and for those who think they understand what is going on but still have misconceptions, such as Sameer did. For this reason I do not want to shut it down completely. But a shorter thread might make it more likely for such people to join in a meaningful conversation. 

Title: Re: 0.999. Post by Gatorshiz100 on Apr 1^{st}, 2004, 12:05pm You crazy mathematicians. I'm an engineer, .9 = 1 is good enough for me. 

Title: Re: 0.999. Post by Avi Gavlovski on Apr 8^{th}, 2004, 5:16pm The real numbers are a complete metric space. Hence, every Cauchy sequence converges to a point in the set. Hence, since the sequence {Sn} = 9/10+9/10^2+...+9/10^n is Cauchy, it converges to a point in the set. Furthermore, the only valid definition of .999... is to define it as the limit of {Sn} as n>infinity. The nonvalid definitions that ive seen posted are all equivalent to: .999... = max{reals < 1} However, if this number were indeed less than 1, then .999... < (.999... + 1)/2 < 1 which would imply that its not a max. Further, it implies that there IS NO max. There is a sup: 1. 

Title: Re: 0.999. Post by hamby on Apr 9^{th}, 2004, 7:20am Does a hundred gabillion trillion zillion = infinity. No. .999... does not equal 1. 

Title: Re: 0.999. Post by Sameer on Apr 9^{th}, 2004, 7:26am on 04/01/04 at 12:05:01, Gatorshiz100 wrote:
I am an engineer too and I say you should know why .99... = 1 

Title: Re: 0.999. Post by rmsgrey on Apr 19^{th}, 2004, 6:40am on 04/09/04 at 07:20:21, hamby wrote:
"gabillion" and "zillion" are not terms I'm familiar with. A hundred trillion trillion trillion is a finite value, and can be expressed explicitly or generated with a finite number of steps. A trillion trillion trillion ... where the ... represents the limit of extending the number of "trillion"s by one each step cannot be generated by a finite process, and does equal infinity. Similarly, any finite string of nines (after the decimal point) gives a value less than one, but .999... cannot be generated by a finite process and does have the same value as 1.000... 

Title: Re: 0.999. Post by grimbal on Apr 29^{th}, 2004, 4:42pm When you feel 0.999... should be < 1, it is because you never realized that the decimal writing of a number is not necessarily unique. You think that a number is identified by its decimal notation. It is not. 0.999... is the same number as 1.000..., just like 1/3 is the same number as 2/6. You never would say that 1/3 is not equal to 2/6 because it is written differently. To prove the equality, if x = 0.999..., you have x*109 = x. From there, x*9 = 9 and therefore x = 1. 

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