#1




Is the repeating decimal .9999999 equal to 1?
.33333 repeating is equal to one third. .6666 repeating is equal to two thirds. .333333 + .66666 = 1 Some math nerds like bigneck say that .9999 repeating is not equal to 1.  Don't listen to bigneck. 
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#2




Flat,
.87632 = .87632 17 = 17 1 = 1 .99999999 = .99999999 you = dork bigneck 
#3




Yes, it's the same. E.g.,
<BLOCKQUOTE><font size="1" face="Verdana, Arial">code:</font><HR><pre> Fractions: 2/7 + 5/7 = 7/7 = 1 Decimals: ______ ______ ______ 0.285714 + 0.714285 = 0.999999 Fractions: 2/9 + 7/9 = 9/9 = 1 Decimals: _ _ _ 0.2 + 0.7 = 0.9 [/code]  Wrong thinking is punished, right thinking is just as swiftly rewarded. You'll find it an effective combination. 
#4




[b]AAUUGGHH! I fed the trolls! Now they'll proliferate!
 Wrong thinking is punished, right thinking is just as swiftly rewarded. You'll find it an effective combination. 
#5




Flat,
.87632 = .87632 17 = 17 1 = 1 .99999999 = .99999999 you = dork bigneck 
#6




Sorry for posting that last one twice  if I did. Actually, .333333 + .66666 = .999993

#7




We already answered this one down below.
What is it with the logic people anyway? 1 does not & is not a 9 period. 
#8




Hey Handy, where was this answered before?

#9




zero point...repeating
Russell 
#10





#11




Another way to approach this, is that .99999... = SUM (9*10^n) as n > infinity.
It can easily be shown that this infinite sum is 1. 
#12




CKDExtHavn, you sound like Fermat in your last post.

#13




Hey, Jinx! Think!!
1 = 1. 4 = 4. .25 = .25 Therefore, 1/4 can't equal .25, can it? Sheeeeeeesh. [/sarcasm] .333 is not equal to 1/3 .333333333333 is not equal to 1/3 .33333333[1 billion 3's]33 is not equal to 1/3 but .3333.... where the decimal 3 repeats infinitely many times IS in fact, equal to 1/3. 
#14




Again with the Infinities...

#15




Again, I suggest we be a little more charitable. Many people have the understandable intuition that .9 repeating does not equal 1. This intution can be modelled in nonstandard number systems without contradition (or so I have heard). The existence of infintesimals is no more problematic than the existence of imaginary numbers.
Tony  Two things fill my mind with everincreasing wonder and awe: the starry skies above me and the moral law within me.  Kant 
#16




Now ask a machinist if .9999 is equal to one.
Most likely he will say no, because they are used to working in tollerences where that .0001 might make a difference in the way a part fits  Hand me that wrench. No, the one that looks like a hammer. Sig Courtesy of Walley 
#17




But, you see, pipefitter, not only are we not talking about machinists and fittings, we're also not talking about .9999. We're discussing 0.99999... ("Zero point Nine (with the 9 repeating ad infinitum)"). Thus, I think I'll go with the answer posted by the professional mathematician above who's already stated for us that not only does 0.99999... ("Zero point Nine (with the 9 repeating ad infinitum)") equal 1, but that it's easily proved.

#18




test
 "I'm hungry, let's get a taco."  Mr. White 
#19




.3333 is a decimal approximation of 1/3. It does not equal 1/3.
In the same vein, .9999 does not equal 1. If you want to say that the limit as the number of decimal places approaches infinity equals 1, fine, knock yourself out, but a functions limit is not an algebraic value. Let me stress: The values obtained from limits are not algebraic numbers. They are just that: limits. They tell you what a graph tends to do as you approach a certain number. .9999999 will never equal 1. What about the limit? Like I said, that value is not an algebraic number, it simply tells us what the tendency of .999999 will be. Since we can never reach infinity, .999999 will never equal 1. (That's why taking the limit of something as we "approach" infinity is so useful; it tells us what happens to the function when x increases without bounds.)  Kupek's Den 
#20




Quote:
Most people have an intuitive feel for what real numbers are, but not a rigorous understanding. That, in part, is what leads to misconceptions like this.  ...ebius sig. This is a moebius sig. This is a mo... (sig line courtesy of WallyM7) 
#21




THINK! For goodness sake!
Ask yourself if 0.333 = 0.3333? Is 0.3333 repeating = 0.333? What is 1/3? Without saying "repeating", there would be NO exact decimal equivalent! You see that, from this example, 0.9999 repeating ad infinitum is still < 1! These are two distinct numbers!  "They're coming to take me away haha, hoho, heehee, to the funny farm where life is beautiful all the time... "  Napoleon IV 
#22




Quote:
Quote:

#23





#24




One minor problem with your theory, there, Ryan. It's wrong, as in "incorrect."
The very definition of "repeating" in this sense is that, in fact, an infinite number of 9s follows the decimal point. As mathematics is rife with infinites and definitions, perhaps you'd care to explain exactly what the little bar over the 9 means in mathematical notation? Last I checked, it meant the number under the bar repeated ad infitum as in "an infinite number of 9s" in this case. 
#25




There is a theologic or philosophic point underlying this, of course, that Kupek raises.
There are numbers  let's say, measurements  in the real world, like 7, 2, and .333, that people can measure and use. These numbers have tolerances and measurementerrors. Two people measuring the same quantity with the same equipment may get slightly different answers within the tolerance of the measuring tools. Mathematics, however, extracts a theoretic underlying construct from the real world situation, where numbers "exist" in their own right, not tied to any measurable object. There is advantage in developing this theoretic construct. We've discussed elsewhere on this board, how many digits of pi do we need? At the point that you are talking about measuring the circumference of something down to less than the width of an electron (in terms of error/ tolerance), you probably got way more digits than is reasonable. In the theoretic world, the fact that pi is an infinite, nonrepeating decimal is quite interesting. In the real world, if I want to approximate the ratio of circumference to diameter and I use 3 as the approximation, there are plenty of situations where that's just dandy (within 5%). So, the question we ask is to what extent the question ("Is .999... = 1?) refers to the "real" world, in which case I agree with Kupek, you can never measure far enough for it to equal 1.... Or to the theoretic world of abstract mathematics, where you can take the limit of the partial sums to equal 1. Sticking to the real world has problems, of course, such as the famous (1/3)+(1/3)+(1/3) may not equal 1 if the computer doesn't round properly. Sticking to the abstract world of math has problems, too, of course, since we're apt to forget that there is no such thing as a "line" without width, in the real world. Innerestin' turn, here. 
#26




two facts:
1. its called "rounding" 2. go about it the other way. take "1" and divide it in half, then divide the outcome in half as well etc... the number of times you divide the outcome is equal to the number of 3s, 6s or 9s in 0.xxrepeating. bj0rn  chickens for sale...! 
#27




Anyone remember when the Pentium chip would process 1 to .989999 or so? remember what happened? Yep.

#28




Here's one of my problems with saying that .99999 repeating equals 1. If that is true, then so is this:
.99(rept.) = 1.00(rept.) That just don't sit right. Cabbage: I don't exactly know the process by which one would define the real numbers using ration numbers, but I think I can venture a guess. The difference between that and are example is that those (I'm assuming) are the cases when the limit as x approaches b equals f(b). That doesn't hold true in our case, because you simply can't evaluate f(infinity). In our case, the function will forever get closer and closer to 1, but never actually reach it. That's why we use limits. The limit is 1, but we never get there. When I make a distinction between algebraic numbers and a resultant of a limit, I am keeping in mind what these values represent: a plain ol' 2 represents 2 of something, but the 2 that is a limit of something represents the tendency of the function. There's a difference, and in this case, I think it's important. Case in point: f(x) = (x^2  6x + 9)/(x  3) lim x > 3 of f(x) = 0 f(3) = undefined This limit has a removable discontinuity, namely (x  3). The graph of the function looks like a straight line, except there is a hole at x = 3, because the function divides by zero. The way I see it, saying that .99(rept.) equals 1 is the same as saying that f(3) = 0. The closer and closer x gets to 3 from either the left or the right, the closer the functions gets to 0. It will never be zero, however, since the function is not defined there. Just because a function approaches a value does not mean it has to reach that value. The limit as x approaches infinity of our .99(rept.) function is 1, but it will never reach one, just get closer the larger x gets. If it never reaches it, then I don't see how it can equal 1. It's damned close, but that's not the same thing. Oh, and Ryan, the sun does move around the Earth if you use Earth as your reference frame. All we have to do is zero the Earth's velocity, and calculate the sun's velocity in reference to the Earth, and we have the sun moving around the Earth.  Kupek's Den 
#29




Well basically, you can kind of think of it like this. Using just rational numbers, we can define pi, for example, to be the limit of the sequence 3, 3.1, 3.14, 3.141,... You get the idea. The limit doesn't exist in the rationals, but this is a way of making the rationals completeby throwing in all the possible values that rationals "can" converge to, such as pi.
If we have an infinite decimal expansion (as with pi), how else can you say what it's equal to, without the idea of a limit? The sequence .9, .99, .999,... converges to 1, which certainly seems to me equivalent to saying .999... (infinitely) is 1.  ...ebius sig. This is a moebius sig. This is a mo... (sig line courtesy of WallyM7) 
#30




I see your point. That fact that
.99(rept.) = 1.00(rept.) would still hold true prevents me from agreeing with you. With that, we've said that two different numbers equal each other. The problem with what you presented is that pi is irrational, whereas 1 is rational. One can be expressed exactly on our normal ten based number system, whereas pi can not. (I suppose you could make a number system based on pi, which would make every number that is not a multiple of pi irrational, but I think we've been assuming we're working exclusively on a ten based system.) I mean, we don't say what pi is equal to; that's why it's represented as pi and not some number. It's irrational and can not be expressed exactly on our number system. We just approximate it, knowing that we aren't using an exact value.  Kupek's Den 
#31




Quote:
 ...ebius sig. This is a moebius sig. This is a mo... (sig line courtesy of WallyM7) 
#32




Try it this way, Kupek.
If .999... and 1.000... (infinite repeating decimals) are different numbers, then you can find a number (in fact, you can find infinitely many numbers) BETWEEN THEM. Go ahead. Find a number BETWEEN .99999... and and 1.00.... Granted, you can find lots of numbers between .999999999999999 and 1.000000000. That's no problem, because these are two different numbers. But if I let the decimal repeat infinitely, there is no number between the two. The Real Numbers can be shown to be dense (I hope I'm remember the right term): that is, between any two real numbers, you can find a bunch of other numbers. The rational numbers are likewise: between any two rational numbers, you can find infinitely many OTHER rational numbers. So, although the two numbers "look" different, they are in fact the same number. 
#33




The difference between 1 and .99 repeating is an infintesimal quanity. For information about such quantities, hyperreal numbers and nonstandard analysis, see
http://neuronio.mat.uc.pt/crcmath/math/math/n/n161.htm. Please do not reply that "we're talking about the standard number system here." The original question was not phrased in terms of any particular number system, not even implictly. Rather, the original question, as I interpreted it, rested on the false assumption that there was only a single "true" number system. That idea, of course, is untrue to the nature of mathematics. By the way, I think the discussion would be much simplified if, instead of discussing whether .99repeating equals 0, you discussed whether 99repeating equals infinity, or the the limit of the sequence 1, 2, 3 . . . equals infinity. They are "reciprocal" issues and the conventionality of the answer would become more apparent. Tony  Two things fill my mind with everincreasing wonder and awe: the starry skies above me and the moral law within me.  Kant 
#34




Quote:
Quote:
The difference is seen in this example: (I'll use the notation 0.9[n] to represent the last digit repeating n times, and 0.9[inf] to represent the last digit repeating infinitely many times.) If I have 9.9[inf], I can't tell whether I got it by adding 9 to 0.999[inf], by multiplying 0.9[inf] by 10, by adding 9.9 to 0.1 * 0.9[inf]. They all give the same result. If I set D = 10.9[inf], then 10  9.9[inf] = D, 10*D, and 0.1*D depending on how I got 9.9[inf]. So D = 10*D = 0.1*D. This means D=0, plain and smple. If I have 9.9[n], I get different representations: 9.9[n]9.9[n1], 9.9[n+1]. If E[n] = 10.9[n], 109.9[n] = E[n], 109.9[n1] = 10*E[n] = E[n1], 10  9.9[n+1] = 0.1+*E[n] = E[n+1]. I'm able to keep track of how I got the term E[n].  It is too clear, and so it is hard to see. 
#35




Zenbeam,
Hyperreal numbers, see the link in my previous posting, are numbers less than 1/n as n approaches infinity. It seems to me that 1.9[inf] falls under this definition. The proofs you present are interesting, but they only suggest to me that there must be distinct rules for the applying the operations of multiplication and addition to hyperreals to avoid a contradition or the erroneous conclusion they equal zero. I admit I could be wrong: 1.9[inf] may not be a hyperreal number (even though it looks like the limit of .1sum[1/9*10n] as n approaches infinity). I don't know enough about nonstandard analysis, however, to know for sure. Tony  Two things fill my mind with everincreasing wonder and awe: the starry skies above me and the moral law within me.  Kant 
#36




Monty posted 04132000 07:35 AM
Quote:
Quote:
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.(N barred)= lim n >infinity of N*10^(m)+N*10^(2m)+...N*10^(nm) where m=number of digits in N Quote:
CKDextHavn posted 04132000 09:37 AM Quote:
Kupek posted 04132000 07:30 PM Quote:
Kupek posted 04132000 11:02 PM Quote:

#37




CKDextHavn
Administrator Posts: 2018 Registered: Feb 99 posted 04142000 08:07 AM  Try it this way, Kupek. If .999... and 1.000... (infinite repeating decimals) are different numbers, then you can find a number (in fact, you can find infinitely many numbers) BETWEEN THEM. Go ahead. Find a number BETWEEN .99999... and and 1.00.... Granted, you can find lots of numbers between .999999999999999 and 1.000000000. That's no problem, because these are two different numbers. But if I let the decimal repeat infinitely, there is no number between the two.  What is the number between .98999999.... and .9999999......? Are they equal as well? Does that make .98999999... = to 1? 
#38




What number is in between 0.98999... and 0.999...? Quite a few, actually.
For example, 0.99. And 0.997. And 0.9903. And 0.99008. And you get the idea. 0.98999... and 0.999... have numbers inbetween them. 0.999... and 1 do not. 
#39




CK sez:
Quote:
This entire question depends on whether or not you accept an infintesmal as a number. .999... can be different from 1 when you're dealing with badlybehaved functions where the function doesn't exist at 1. ( (x1)/(x1)^2 ) Of course, this still means dealing with infintesmals. 
#40




Quote:
Just a nitpick, the rest I completely agree with.  Eschew Obfuscation 
#41




The Ryan posted 04142000 08:04PM the stuff between the quotes; my responses are in between the quotations.
Quote:
Quote:
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Quote:
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[quote]terminating decimal (noun) First appeared circa 1909 : a decimal which can be expressed in a finite number of figures or for which all figures to the right of some place are zero  compare REPEATING DECIMAL From "Number," Microsoft (R) Encarta. {bolding and description of pictures in the article brought to you by Monty} Quote:
Here's a hint or two for you: 1) The professional mathmetician (CKDext) has already shown that your take on this issue is incorrect 2) If you really don't know what you're talking about, keep quiet and try to learn. 
#42




Actually, I don't see much wrong with anything that The Ryan, Monty, or Dex said. Except when you say that the other is wrong.
I thought you guys were on the same side.  rocks 
#43




Quote:
Of course 1/4 = 0.25 why should it not? And yes, 1/3 = 0.3333 repeating  get it? As for 1 = 0.999 repeating... Please go back to basic math class if: a) You cannot comprehend the number 1 b) You fail to understand the meaning of the fraction 1/1 and how to set that up as a long division. I'd really like to see a long division where 1 divided into itself can yield 0.9999...  "They're coming to take me away haha, hoho, heehee, to the funny farm where life is beautiful all the time... "  Napoleon IV 
#44




I can see why Cecil feels like there is no hope for the teeming millions!
 "They're coming to take me away haha, hoho, heehee, to the funny farm where life is beautiful all the time... "  Napoleon IV 
#45




Quote:
 It is too clear, and so it is hard to see. 
#46




Proper treatment for trolls, IMHO: answer ONCE, to be sure that you are not dealing with honest ignorance, then drop it.
One more time, and then I'm done with this. Several people have offered proof that .999...(repeating infinitely) = 1. The simple multiplication proof is usually sufficient: (1) x = .99999.... (2) 10x = 9.999999... Subtracting (2)(1): (3) 9x = 9, hence x = 1. OK, that's a proof. The testimony of every professor of mathematics in the country would be another sort of proof, I suspect. You don't buy those? OK, then, if you thinkf .999... is different from 1, then you should be able to pick a number BETWEEN them. Go ahead, tell us one. For any two distinct real numbers, there are infinitely many real numbers between them. For instance, the midpoint. Go ahead, name one. We're not talking about theoretical numbers, now, like "infinitesimals" or whatever, we're talking real, honesttogod, writedownable numbers. Name one, between .999... and 1. If you can't, then you have to concede that .99999... is just another way of writing 1, in the same way that .333333... is just another way of writing 1/3 or that .11111.... is another way of writing 1/9. Oh, and BTW, Monty, I'm no longer a professional mathematician. I did that back in the 70s, but then I became an actuary (pay was better) and now I'm a more generalized consultant. Contrary to the comment made in another thread, I did not predate Riemann, although I did have a class under Zygmund. 
#47




Quote:
Excuse me, but it is relevant! I can demonstrate that 0.25 equates to 1/4. I can demonstrate that 0.3333 repeating equates to 1/3. Can you demonstrate that 1.0 equates to 0.9999 repeating (without rounding)? Technically, these are two distinct points on the numberline. If you receive a paycheck, would you accept anything less than 1.0 times the amount you are owed?  "They're coming to take me away haha, hoho, heehee, to the funny farm where life is beautiful all the time... "  Napoleon IV 
#48




Quote:
Thus, it is improper to subtract (a) from (b)! Algebra cannot be applied at whim!  "They're coming to take me away haha, hoho, heehee, to the funny farm where life is beautiful all the time... "  Napoleon IV 
#49




<< The ground rule for aplying this method is that the two equations MUST BE independent! >>
Baloney. The ground rule is that the two equations must be consistent, or you will get paradoxical results. GENERALLY, given two dependent equations in two variables, you cannot reach a solution. However, these are two equations in ONE variable, and I have obtained one from simple algebraic manipulation of the other. A long division? Sure. (.99999...)/1 = .99999... Happy? The way that you toss out incorrect statements and ignore the comments of any others, leads me to suggest that this conversation is finished. Fighting ignorance is one thing; fighting stubborness is something else. 
#50




What's interesting about this thread to me, is the logic. Many have issued mathmatical proofs that .999... = 1.
The detractors, on the other hand, have offered no mathmatical proof whatsoever. In fact, it would seem that the mathmaticians have won the point. The only avenue left to those who do not beleive, is using "logic" to state why the proofs are wrong, not showing any mathmatical flaws in the proofs themselves. Am I wrong? Did I miss the post that had a valid mathmatical proof that 1 is not equal to .999...? 
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