Author 
Topic: 0.999... (Read 34359 times) 

Sir Col
Uberpuzzler
impudens simia et macrologus profundus fabulae
Gender:
Posts: 1825


Re: 0.999...
« Reply #75 on: Nov 29^{th}, 2005, 4:44pm » 
Quote Modify

(I hope I'm not wasting my time with this post, but it may help identify your misconceptions, Whiteshadows) May I ask what you think 0.999... is equal to? Either it (i) exists, or (ii) it does not exist. (i) If it exists then it is either equal to one, less than one, or more than one. I am sure you would agree it is not greater than one, and as you don't believe that it is equal to one, you must believe it is less than one. Right? In which case you probably believe that it is the "last" number before one. Right? What happens if you take two numbers, x and y, and calculate (x+y)/2? That's right it gives a number inbetween. So what does (0.999...+1)/2 equal? There certainly cannot be a number inbetween, so I guess the only resolution would be to say that the numbers are equal. However, let us consider the other possibility... (ii) If it does not exist we must be saying that no infinitely recurring decimals exist. Right? Let x = 3/10 + 3/100 + 3/1000 + ... = 0.333... 10x = 3 + 3/10 + 3/100 + ... = 3 + x Hence 10xx = 9x = 3, so x = 3/9 = 1/3. If we ever terminate that series, we have a problem... If y = 3/10 + 3/100 + 3/1000 10y = 3 + 3/10 + 3/100 10yy = 9y = 33/1000 = 3000/10003/1000 = 2997/1000 = 2.997, so x = 2.997/9 = 0.333 So what is 0.333... ? Of course we can use the same method for 0.999... z = 9/10 + 9/100 + 9/1000 + ... = 0.999... 10z = 9 + 9/10 + 9/100 + ... = 9 + z 10zz = 9z = 9, so z = 9/9 = 1 But if you deny the existence of 0.999... = 9/9 = 1 then I guess you must be denying the existence of all other recurring decimals. Right?

« Last Edit: Nov 29^{th}, 2005, 5:00pm by Sir Col » 
IP Logged 
mathschallenge.net / projecteuler.net



groll
Guest

I'm having slight problems with one of the equations commonly used in this discussion. I hope some one here can take their time and explain my errors. x = 0.999... 10x = 9.999... 10x  x = 9.999...  0.999... We assume that we have n decimals Let a = 0.999.. (with n decimals) b = 10 * a = 9.999... (with n1 decimals) c = b  9 = 0.999... (with n  1 decimals) Now subtract c from a and you will have a number greater than 0. Now even as we let n travel towards infinity we will have a difference, you can allways find a value between a and c. This would lead to 10x  x = 9.999...  0.999... 9x = 8.999...1 (since there will allways be one more nine in the second number) x = 8.999...1 / 9 x < 1 ================================== I assume that this might have to do with how we're allowed to work with infinity. If I let "m = infinity" would then "m  1 = m" ? If you state that you have an infinite number of decimals (say 9's) would I be able to add one more? If there are m natural numbers where m is an infinite number, would it then be correct to state that there exists a number m + 1? ================================== Thanks in advance /Dan


IP Logged 



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: 0.999...
« Reply #77 on: Dec 12^{th}, 2005, 8:34am » 
Quote Modify

on Dec 12^{th}, 2005, 7:50am, groll wrote:I assume that this might have to do with how we're allowed to work with infinity. 
 Yes, that's the reason. There isn't a limit to the number of decimals, so there is no such n. Quote:If I let "m = infinity" would then "m  1 = m" ? 
 If m is a (natural) number (i.e. 0,1,2,3 etc) then m can't be infinity. Because infinity isn't a natural number. That's also why adding, multiplication etc can be counterintuitive when infinity is involved. Quote:If you state that you have an infinite number of decimals (say 9's) would I be able to add one more? 
 In a way. take for example an ordered set of numbers, {1,2,3,4,5, ..} it contains infinitely many numbers, but you can add a new one, that's not in it. {0,1,2,3,4,5, ..} So you can add to infinity. But in fact, there are also equally many numbers in both sets. For each number X in the first set, (X1) is in the second one. And conversely, for each number Y in the second set, (Y+1) is in the first set.

« Last Edit: Dec 12^{th}, 2005, 8:41am by towr » 
IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



Marvin
Newbie
Gender:
Posts: 5


Re: 0.999...
« Reply #78 on: Dec 12^{th}, 2005, 1:42pm » 
Quote Modify

on Dec 12^{th}, 2005, 7:50am, groll wrote:We assume that we have n decimals Let a = 0.999.. (with n decimals) b = 10 * a = 9.999... (with n1 decimals) c = b  9 = 0.999... (with n  1 decimals) Now subtract c from a and you will have a number greater than 0. 
 This is correct with finite number of decimals. For n (finite) number of decimals, you can add or subtract one and get n1 < n < n+1. on Dec 12^{th}, 2005, 7:50am, groll wrote:Now even as we let n travel towards infinity we will have a difference, you can allways find a value between a and c. 
 Infinity is greater than any number, so if you add one decimal, you won't get "one more than infinity" decimals, the same with subtracting. infinity + x = infinity for any real number x (including negative numbers  subtracting). With infinite number of decimals, there would be no last n'th decimal (and no n1'th before it). When we talk about some last digit, then it must be with some finite number of digits  which is less than infinity. With n, finite number of decimals, a  c = 9*10^n When n > infinity, the limit of the series equals 0. The limit means what would a  c be if n = infinity (it shouldn't be written this way, because no natural number equals infinity).


IP Logged 



Lord_of_math
Newbie
Posts: 3


Re: 0.999...
« Reply #79 on: Feb 22^{nd}, 2006, 2:59pm » 
Quote Modify

X=.999999.... This is my proof, same as icarus, but hopefully simpler 10x=9.999999 I multiplied by 10 10x9=.999999 subtracted 9 from both sides 10x9=x Substituted from first step 9x9=0 subtracted x from both sides 9x=9 added 9 to both sides x=1 divided both sides by 9 By the way, I am 15, and i did this with the use of math i learned in like 6th and 7th grade and http://mathworld.wolfram.com/ is an excellent math cite... look up aleph, it discusses different sizes of infinity // I accidentally deleted the post following this one trying to take care of the tripple post following that one. So I decided to reinsert it here. Hence the following is a repost of Icarus reply. towr posted Feb 23rd, 2006, 4:07am by Icarus That is exactly the first of the proofs I listed, though it is hard to tell now, because I used the math symbolry we had before William updated the site to use a newer and more efficient version of YaBB. Since the symbolry was an addin that he had provided, the new bulletin board did not have it, and he has never gotten around to adding it back in. So now all you see is the codes for the symbols. I can't even fix it because the upper limit for post sizes was decreased significantly, so I am not allowed to save the fixed post, as it is too long.

« Last Edit: Apr 30^{th}, 2006, 3:13am by towr » 
IP Logged 



said_yam
Newbie
Posts: 1


Re: 0.999...
« Reply #80 on: Apr 29^{th}, 2006, 1:48pm » 
Quote Modify

There is a new theory that may end the debate about 1=0.999... the whole theory can be found in this link http://www.geocities.com/humood_theory/index.htm here is a Quote about cracking the famouse proof: Proof (A) x=0.999... 10x= 9.999... 10xx=9.999...  0.999...=9 9x=9 x=1 Quote:  I will start by cracking this proof and show it is in fact a Paradox not a proof. Let x = 0.999... 10x = 9x + x = [ 9* (0.999...) ] + 0.999... [ I expressed 10x as (9x +x)] 10x  x = ( 9x + x )  x = [ 9* (0.999...) ] + 0.999...  0.999... [Positive 0.999... goes with Negative 0.999...] 9x = 9* (0.999...) x = 9* ( 0.999...) / 9 = 1 * 0.999... x =0.999 We did not end with x=1 In fact what made it looks like a proof was the mistake we did in this step: 10 x  x = 9.999...  0.999... = 9 This step assumes the value of infinite 9s after the decimal point in ( 10x ) equals the value of infinite 9s in (x) The infinite decimals in x = 0.999... have different behavior than the infinite 9s in 10x =9.999... Because 9s in x= 0.999... Go to infinity faster than 9s in 10x=9.999... (10 x ) is always one decimal behind (x) When x = 0.99 then 10x=9.9 When x = 0.9999 then 10x=9.999 When x = 0.999999999 then 10x=9.99999999 Value of Infinite decimals in (x)  Value of infinite decimals in (10x) = Lim x> 8 9 / (10^x) So the difference between the two values equals an infinitely small fraction which is Lim x> 8 9 / (10^x) This means in Proof (A) when we say: 10 x  x = 9.999...  0.999... = 9 We are already Assuming that a number minus an infinitely small fraction is the number itself. Which means we are already assuming that Lim x> 8 1  ( 1/x ) = 0.999... = 1 So, in Proof (A) we already assumed 1= 0.999... Then no wonder we finally ended with our assumption, which make it not a Proof at all. You cannot prove any statement by assuming it to be true from the beginning. But when we explain (10 x ) in a very fundamental method no body can argue with, we will say: 10x =0.999... + 0.999... + 0.999... + 0.999... +0.999... +0.999... +0.999... +0.999... +0.999... +0.999... And when we subtract x = 0.999... from it : 10x  x =0.999... + 0.999... + 0.999... + 0.999... +0.999... +0.999... +0.999... +0.999... +0.999... + 0.999...  0.999...= 0.999... * 9 The infinite 9s in x after the decimal is gone with only one of those 0.999... Not all of them as we did in Proof (A).  All the theory and its results can be seen in the link: http://www.geocities.com/humood_theory/index.htm For comments please write it here in this forum or send it to the email said_...@yahoo.com


IP Logged 



Grimbal
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 7526


Re: 0.999...
« Reply #81 on: Apr 29^{th}, 2006, 6:17pm » 
Quote Modify

The road to enlightenment is long and difficult.


IP Logged 



Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863


Re: 0.999...
« Reply #82 on: Apr 30^{th}, 2006, 1:05pm » 
Quote Modify

Hoo boy. That is almost as bad as the "disproof" of innumerability of the reals that T&B referenced awhile back (although T&B wasn't promoting that one, just bringing it to our attention). said_yam, all that this shows is that you do not understand the mathematical meaning of the terms "limit", "proof", and "real number". There are so many things wrong with this, it is impossible to address them in the size limits of an individual post, so I will start off with your supposed "cracking" of the 10xx proof. One thing you will note about my listing of this proof is that I put it under the heading "Argument depending on the reader knowing how to add/multiply decimals". That is, the proof expects the person reading to already know that 9 + 0.999... = 9.999.... If you do not accept this, then you are not going to accept the proof as stated. This in itself is not a weakness in the argument. Most proofs assume that the reader is aquainted with certain previous results, and does not try to prove those results again. If we did not do this, but proved everything directly from the axioms and definitions, our proofs would quickly become too long and unwieldly to follow. In Nicholas Bourbaki's Elements of Set Theory, he estimates that if you were to write out the expression he uses to define the number "1" in the basic symbols of his theory, it would require better than 22,000 characters. So what about your "crack"? You make a big deal of doing this manipulation and ending up with x = 0.999... again. In fact, you could have gotten to this point much easier! Just stop after the initial line "Let x = 0.999...", and WOW! you get x = 0.999... right there! That you end up with 0.999... instead of 1 would only indicate a problem with the 10x  x proof if 0.999... is NOT equal to 1. If 0.999... = 1, then it is only to be expected that sometimes as you do the arithmetic, you get 0.999..., and sometimes you get 1. Your crack is not a crack at all, but is an expected consequence of the proof being true. You only see it as a problem, because you are assuming at the out set that the result is false. But as I said, if you do not accept that 9 + 0.999... = 9.999..., then this proof is not for you. First, you will need to see why this is true. This requires a delving of the definitions of real numbers, decimal expressions, and limits. I have posted just such a delving in the first post of this thread. Unfortunately, changes in the bulletin board software since then have rendered the mathematical symbolry I used illegible. And the new post size limits have prevented me from being able to go in and fix the problem (you can't post anything that long anymore, nor can you modify a post that long without truncating it to the new limits  which I run into about a quarter of the way through).


IP Logged 
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863


Re: 0.999...
« Reply #83 on: Apr 30^{th}, 2006, 2:23pm » 
Quote Modify

Let's look at the opening "calculation" of your Humood Theory page, shall we? You start off with this calculation: Lim_{x>oo} (1  1/x) = 1 (which you are attempting to disprove). So, you say, Lim_{x>oo} (1  1/x)^{x} = Lim_{x>oo} 1^{x} 1/e = 1 You claim that this disproves that Lim_{x>oo} (1  1/x) = 1. I claim that is disproves the idea that whenever lim f(x) = A, then the lim g(x, f(x)) = lim g(x, A), regardless of the function g. That is the "rule" you used to go from Lim_{x>oo} (1  1/x) = 1 to Lim_{x>oo} (1  1/x)^{x} = Lim_{x>oo} 1^{x}. But in fact, it is not true in general and exponentiation is one of many exceptions. Even if g had no direct dependence on x, it would be true only if g were continuous at A. With g depending on x itself, more stringent conditions need to be applied before you can move limits around like that. That lim_{x>oo} (1  1/x) = 1 can be proven easily from the definition of a limit. By definition, this holds if we can show that for any h > 0, we can find a number M (depending on h), such that if x > M, then  (1  1/x)  1  < h. So, let h > 0 be an arbitrary number. Choose M = 1/h. If x > M = 1/h, then h > 1/x =   1/x  =  (1  1/x)  1. Which is exactly what was required. So, by the definition of "limit", lim_{x>oo} (1  1/x) = 1. The problem was with your next step, not here. _________________________________________________ Let's state some definitions. Size limitations again prevent me from going into detail, but this is the basics: Real numbers: The minimal topologically complete ordered field. In particular, the real numbers are a set which contains two particular elements, which we label "0" and "1", and call the additive and multiplicative identities, respectively. It also has two operations, addition and multiplication, and an ordering "<" which satisfy: (1) commutivity: a+b = b+a, ab = ba (2) associativity: (a+b)+c = a+(b+c), (ab)c = a(bc) (3) identity: a+0 = a, a1 = a (4) inverse: for all x, there is y such that x + y = 0. For all x != 0, there is z such that xz=1. (5) distributivity: a(b+c) = ab + ac (6) order: if a>b, then a+c > b+c. If a>b and c>0, then ac > bc. (7) completeness: if S is a nonempty set of real numbers such that if x is in S, and y < x, then y is in S, and if for all x in S, x < a for some fixed a, then there is a real number p such that if x < p, then x is in S, and if x > p, the x is not in S. (p may or may not be in S, and is called the "supremum" of S). The real numbers are also minimal, in the sense that no subset of them closed under addition and multiplication has all of the properties above. There are many ways of building a set with those properties. If you build such a set, we call it a "model" of the real numbers. Any two models of the real numbers can be put in 1to1 correspondence with each other in a unique fashion that preserves addition, multiplication and order. As far as real number behavior is concerned, all models behave exactly the same (they have to, because of the correspondence). For this reason, we do not specify any particular structure or way of building the real numbers. Merely that they must meet the requirements mentioned above. Limits (I will restrict this to limits at infinity) are defined, per Cauchy, as follows: lim_{x>oo} f(x) = L if and only if, for every h >0, there is an M such that if x > M, then  f(x)  L  < h. All of the rules for taking limits that you have been taught can be derived from that definition. This includes the fact that limits are unique (so, since lim_{x>oo} (11/x) = 1, as I already proved, and since, as you yourself have admitted, lim_{x>oo} (11/x) = 0.999..., we are left with only one conclusion: 1 = 0.999...). Infinite Decimals: the decimal expression d_{n}d_{n1}...d_{0}.d_{1}d_{2}... by definition represents the real number D, where D = lim_{K>oo} d_{n}...d_{0}.d_{1}...d_{K} One thing to note about this definition (which is also true of the definition of an infinite series), is that it does NOT require an infinite amount of summing to define the unique value D that the decimal expression represents. Though mathematics deals all the time with concepts such as "infinite sums" or other infinite processes, the meaning of such things is always defined by some finite means.


IP Logged 
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



fireashwinter
Newbie
Posts: 7


Re: 0.999...
« Reply #84 on: May 19^{th}, 2006, 9:46am » 
Quote Modify

I'm responding to item 9 in Icarus' post on common misconceptions. Actually, I'm responding to the various discussion related to the interposing of additional numbers. I don't doubt that the various informed posters are aware of all of this. Perhaps my statements are even implied by theirs. I'm sorry about that I am not a mathemetician, engineer, or the like. I should start off by affirming that I agree that .999... = 1. However, the "disproof" I've seen of interposing numbers are not 100% convincing (I admit I have not looked at every single one). The most common "disproof" is that if there is a number between 9 and 10, designated #, then since #=.95, .###... would be greater than 1, so there is no such number .###... that lies between .999... and 1. That proof is disturbing to me, because I think that the suggestion "#" can be interpreted differently. In fact, I think that it should be interpreted differently, rather than assumed to have the most absurd possible meaning. (Even if it was intended as most absurd, it is more interesting to treat the intent as the most legitimate.) Obviously, I do not mean to propose that "#" is something like 9.000...0001, because that would be the same as proposing that #=9. What I propose is that # is the addition of an actual digit, and that as such there is not the correspondence that #=9.5. The addition of # creates an 11digit system, aka "ak." Our decimal system can be thought of as: x/n +y/n^{2} +... where the numerators are less than n, and n is the number of digits. Also, the largest numerator we will say is n1, because 9 = 101. (Note, I understand that you could express things differently, but with the same meaning, if you wanted to.) Let's keep the same idea with "ak," so that So, .kk... = .k/n+.k/n^{2}... Since k is the largest numerator, and n (the number of digits) is 11, we can make an analogy between k/n + k/n^{2}+... in the "ak" system and our own 10digit system. Having k/n in "ak" is not quite the same as having 9/10, or 9.5/10. More reasonably, having k/n is like having 10/11. So a good analogy is that .kk... corresponds to the 09 decimal system as: 10/11 + 10/121 + 10/1331 ... Which you can show mechanically seems that it will be .999... In fact, for any system of n digits (n > 1, n is real), the decimal represented as .xxx... when x=n1 will correspond to .999... This can be better shown using the fact that a converging infinite series has the sum of: a/(1r), which you can look up in a Calculus book or website if you don't believe it. series is [(n1)/n]/[1(1/n)] = 1 (sorry about using "n" in this potentially confusing way) To me, this shows that the idea that you can propose any decimal system based on n>1 digits is not proof that there is a number between .999... and 1, because your number will actually correspond to .999..., or 1 revolution of the system. Actually, the bit where I discuss a/(1r) is good enough all by itself, but might not be understood alone. In any system of that like, .xxx... is going to analogize to .999... of that "system," or as I like to think of it .999... of that mode. At that point, the # argument seems reduced to wondering whether we are actually looking at a 09 system, or some other system which arbitrarily uses "9." There is also the issue of whether digits have essential correspondence. However, we are discussing within a system or systems where digits have constant interval relationships. If you don't accept that, then arguments against .999... =1 challenge the meanings of both "1" and ".999...," and constitute neither immediate proof that the two are unequal, nor that another number can intervene.


IP Logged 



fireashwinter
Newbie
Posts: 7


Re: 0.999...
« Reply #85 on: May 19^{th}, 2006, 9:59am » 
Quote Modify

Quote: This can be better shown using the fact that a converging infinite series has the sum of: a/(1r), which you can look up in a Calculus book or website if you don't believe it. 
 Sorry, 0<abs(r)<1. Oops. It's the case, though.


IP Logged 



Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863


Re: 0.999...
« Reply #86 on: May 19^{th}, 2006, 3:28pm » 
Quote Modify

on May 19^{th}, 2006, 9:46am, fireashwinter wrote:This can be better shown using the fact that a converging infinite series has the sum of: a/(1r), which you can look up in a Calculus book or website if you don't believe it. 
 That is the sum of a power series with constant coefficient: a + ar + ar^{2} + ar^{3} + ar^{4} + ... = a/(1r) (Which converges if and only r < 1. Note that it does converge for r = 0.) There are many, many convergent infinite series that are not of this form, and their sums are not given by that expression. As for the rest, what you are describing is simply a base11 decimal system instead of our normal base 10. The real numbers can be expressed in decimal systems for any base n > 1. In fact you can define a "basen" decimal system for any real number n with n > 1 (even bases 1 and 1 are workable with some special rules). The digits are the integers d with 0 <= d < n. However, if n is not an integer, every real number has multiple representations. If n is integer, then only rational numbers r such that n^{k}r is an integer have multiple representations  exactly two: one terminating (ending in repeated zeros), and one ending repeated (n1)s. However, the "con" arguments I was referring to were not talking about a base11 counting system. Rather, they were trying to introduce another digit # to the base10 decimal system, where 9 < # < 10. A base11 system does not in any way support their position, and they knew it. They needed a base10 system with an additional digit. How did they define their digit # or decimal expressions involving it, such at 0.###... ? They didn't! They simply assumed that such expressions were meaningful even though they were not defined. And whenever someone tried to address their arguments by supplying a meaning to their meaningless expressions, they simply declared that is not what they meant. And that, above all else, is why the argument was simply hot air.

« Last Edit: May 19^{th}, 2006, 3:35pm by Icarus » 
IP Logged 
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



GeMsToNe
Newbie
Maybe my brain isn't working... O_o;
Gender:
Posts: 39


Re: 0.999...
« Reply #88 on: Aug 20^{th}, 2006, 9:21pm » 
Quote Modify

I just noticed... 1.000  0.999 = 0.0001 (in which 10  9 is equal to one, because you cancel the next zero to nine) Which means... 1.000...  0.999... = 0.000...1 And isn't... 1/3 = .3333... 2/3 = .6666... 3/3 = 1 Because 3 divided by 3 is equal to one. I don't know why I even posted... blah, I'm not so good with math. I just noticed some stuff, that's all... =D


IP Logged 



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: 0.999...
« Reply #89 on: Aug 21^{st}, 2006, 2:13am » 
Quote Modify

Or you could conclude 0.000...1 (if it means anything at all) must be 0 Because 0.999... is obviously thrice 0.333... If the latter is 1/3rd, then the former must be 1


IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



GeMsToNe
Newbie
Maybe my brain isn't working... O_o;
Gender:
Posts: 39


Re: 0.999...
« Reply #90 on: Aug 21^{st}, 2006, 7:51am » 
Quote Modify

Oh, I get it now. Thanks towr ^^


IP Logged 



Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863


Re: 0.999...
« Reply #91 on: Aug 21^{st}, 2006, 4:38pm » 
Quote Modify

The normal definition for decimals does not define "decimal expressions" such as 0.000...1, where the ... represents an infinite number of zeros (or any other digit). For any defined decimal expression, each digit has only a finite number of digits to its left (it can be infinite on the right, but it has to be finite on the left). But the "1" at the end of this one has an infinite number of digits to its left. So the standard definition does not include 0.000...1. But as towr has indicated, you can define this expression yourself. But the only way to do so and be consistent with the rules for adding and multiplying decimal expressions is to have 0.000...1 = 0, exactly because of the calculation you've shown.


IP Logged 
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



Aman
Newbie
Posts: 5


Re: 0.999...
« Reply #92 on: Dec 14^{th}, 2006, 8:32pm » 
Quote Modify

Sorry to continue to beat a dead horse but I have always thought of 0.999... as the greatest number that is less than 1. If not, how would you represent the greatest number less than 1?


IP Logged 



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: 0.999...
« Reply #93 on: Dec 14^{th}, 2006, 9:06pm » 
Quote Modify

on Dec 14^{th}, 2006, 8:32pm, Aman wrote:... how would you represent the greatest number less than 1? 
 max{x: x < 1}


IP Logged 
Regards, Michael Dagg



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: 0.999...
« Reply #94 on: Dec 14^{th}, 2006, 9:14pm » 
Quote Modify

on Dec 14^{th}, 2006, 8:32pm, Aman wrote:If not, how would you represent the greatest number less than 1? 
 There isn't one: for any number x < 1, there's a larger number less than 1, such as (1+x)/2: x = (x+x)/2 < (1+x)/2 < (1+1)/2 = 1.


IP Logged 



Aman
Newbie
Posts: 5


Re: 0.999...
« Reply #95 on: Dec 15^{th}, 2006, 8:35am » 
Quote Modify

Last two posts seem to contradict each other. Or am I reading them wrong?


IP Logged 



Marvin
Newbie
Gender:
Posts: 5


Re: 0.999...
« Reply #96 on: Dec 15^{th}, 2006, 9:00am » 
Quote Modify

They do contradict, but Eigenray is right. The greatest number less than 1 would be max{x: x < 1} if there were such a number. But this set has no maximum, for the reasons Eigenray wrote.


IP Logged 



Aman
Newbie
Posts: 5


Re: 0.999...
« Reply #97 on: Dec 15^{th}, 2006, 11:29am » 
Quote Modify

OK. So is the issue that there is no way to represent the greatest number less than 1, or that there is no greatest number less than 1?


IP Logged 



Marvin
Newbie
Gender:
Posts: 5


Re: 0.999...
« Reply #98 on: Dec 15^{th}, 2006, 11:40am » 
Quote Modify

Probably I was not clear enough, sorry. There is no greatest number smaller than 1. It is a property of the real numbers: If x<>y then there are infinitely many numbers between them. So for any x, if x<1, there are always numbers larger than x and still smaller than 1.


IP Logged 



Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863


Re: 0.999...
« Reply #99 on: Dec 15^{th}, 2006, 4:21pm » 
Quote Modify

More to the point: for any two distinct numbers x and y, x < (x+y)/2 < y. So, if M were the largest number less than 1, we would have a contradiction: M < (1+M)/2 < 1, so M is not the largest number less than 1 after all. Michael_Dagg (who is a mathematician, by the way) was answering tongueincheek to your question. Your question assumed that such a number existed, so he simply translated the phrase "greatest number less than 1" into a mathematical expression. Being able to write an expression that means "greatest number less than 1" does not mean that such a value actually exists. For instance, Max{x : x>= 2 and x <= 1} clearly fails to exist, since no number is a member of that set.


IP Logged 
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



