I see you are trying to represent a real number with a string of characters. However, you have not supplied enough information to allow us to accurately determine which real number you mean.
Please tell us the location of the final 9 in this string:
0.999...9
We expect this to be a natural number. Failure to provide such a position will prevent us from determining your number, and will make the majority of your argument void.
NOW THAT YOU'VE READ THIS MESSAGE, YOU'VE GOT 24 HOURS TO LEARN THE SERBIAN LANGUAGE. IF YOU FAIL, YOU WILL DIE A HORRIBLE DEATH EXACTLY 24 HOURS FROM NOW. НE СEJ ТИКВE ГДE JOШ НИСУ НИКЛE!
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Anonymous2007-05-12 19:32 ID:yFChcQtv
I AM A MATHEMATICIAN AT A PRESTIGIOUS ENGLISH UNIVERSITY, AND I DECLARE THIS PROOF TO BE COMPLETELY CONSISTENT AND CORRECT
>>14
Dude, you know Grothendieck? Fucking awesome! Can you give me his number?
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Anonymous2007-05-13 12:01 ID:6S5Kzpy/
>>1
There is no such number. In fact, these .9... numbers don't even exist. Sure, they can be expressed using summation, but that doesn't mean they exist numerically.
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Anonymous2007-05-13 14:02 ID:NmaKCHb0
.999...
= sum 9/(10^k)
k=1 to inf
= sum 9*(1/10)^k
k=1 to inf
= (sum 9*(1/10)^k) - 9
(k=0 to inf )
The first term is now a geometric series and therefore:
if we let k tend to infinity, we will arrive at an "expression" for 0.999~
however, since k --> infinity => 1/k --> 0 => 1 - 1/k --> 1
0.999~ = 1
LEARN WHAT INFINITY IS, AND THEN GET YOUR HEAD AROUND WHAT 0.999~ ACTUALLY REPRESENTS. I'M SICK OF THIS FUCKING BULLSHIT FROM PEOPLE WHO DON'T BELIEVE IT AND COME UP WITH BULLSHIT "THEOREMS"
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Anonymous2007-05-13 22:10 ID:6S5Kzpy/
.9... isn't a number. Saying something that isn't a number equals a number is ridiculous. Let's move on from this endless nonsense.
>>22
You can post this however many times you want, it doesn't change the fact that you are wrong.
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Anonymous2007-05-14 0:45 ID:2YwYUj/X
>>23
It is correct. Many mathematicians are arrogant, and later on they are rightly proved incorrected. Summation is a process, and it is not necessarily have a real number as its result. If you think it does, name the exact number. But you can't because it is proof by contradiction (no number you can provide equals ".9..." exactly).
Here's an idea: Take a fucking high school calculus class so that you understand why you're wrong in this statement. Then, take a fucking analysis class so that you understand the definition of the real numbers. When you accomplish these two things - and it'll take you a while, since you seem pretty slow - you can come here and discuss whether or not 0.999.. = 1. But of course, if you've actually accomplished these two things, then you won't believe that 0.999... != 1, so it's a bit of a catch 22. (Catch 21.999... lolol)
Everyone who has posted here knows that .999... == 1. The people who are 'wrong' on this thread are acting that way intentionally. I imagine half of the people who correct them to know this, the other half thinks the 'wrong' people are just wrong. And the only people that believe .9999... != 1 are too busy laughing at us.
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Anonymous2007-05-14 14:36 ID:zCqfq85R
here:
let X = 1 + 3 + 9 + 27 + ...
then 3X = 3 + 9 + 27 + 81 + ...
and X - 3X = 1 (by cancellation). So 2x = 1 or X = 0.5 HOW
>>35
You can't operate on non-convergent series and expect to get a correct answer. I'm not sure if you're the same idiot as before, so this might be redundant, but: learn high school calculus.
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Anonymous2007-05-14 17:03 ID:2YwYUj/X
>>32
I've taken these classes and got decent grades in them. You get bent out of shape when you are asked to provide something and can't do it. Reeks of phailure to me.
>>38
I did provide what you asked for, in >>25. 0.999... is equal to 1, and 1 is a real number. If you knew the definition of the real numbers as limits of cauchy sequences of rationals you'd know this already, but you don't, so you must not have taken analysis, because this is taught in every introductory analysis class everywhere. QED. Furthermore, you can't even find the limit of a geometric series (>>26), and since limits of geometric series are taught in high school calculus or earlier, you must not have taken high school calculus.
>>39
Euler worked with (traditionally) non-convergent series not using standard summations, but instead using things like Abel summations.
0.9... is NOT a number. Sure, 0.9 is a real number, so is 0.99, 0.999, etc; but no matter the number of 9s there are in the number you pick, at least one will always be missing. An element that is not in R cannot equal an element that is in R.
So drill it into your stupid fucking head: The real numbers are an infinite set of FINITE numbers; not an set infinite set of infinite numbers.
>>45
1 = 0.999...
Also, by your definition, the real numbers would be countable, pi would not be a real number, and the "set" of real numbers would be a different set* based on what base you choose! Talk about stupid.
* - by your definition, 1/3 is not a real number, since its decimal expansion, 0.333..., is infinite. But in base 3, 0.1 represents 1/3. Whoops!
>>45
Also, I forgot to mention, you just replied telling someone they were wrong and then agreeing with them. Good job.
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Anonymous2007-05-15 0:09 ID:4wEg1xcj
>>46
Of course 0.333... is not a real number, but 1/3 in base 3 is a real number because you can find a finite base in which it can be finitely represented. If you have a set S = {a, b, c} and a set T = {a, b, d}, you cannot say that any element in S v T, but not in S ^ T, has a corresponding element in the other set. Simple. Same principle here.
>>48
Countable, Pi is not a real number. You have two problems to deal with still, and you've basically just accepted the fact that your definition of the real numbers can not be represented in one base, contrary to the "actual" real numbers (you know, the set that every mathematician in the world is familiar with and agrees upon). Furthermore, your definition offers no advantages over the "actual" real numbers to make up for this deficiency. As you've stated it in this most recent post, the real numbers are the rational numbers.
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Anonymous2007-05-15 2:07 ID:4wEg1xcj
>>50
Well, you also have the irrational "numbers". They are locked away in a mental hospital until they can accept the fact that they are not real.
The problem is that you stated something that was blatantly false to anyone who understood the math you were talking about. Go back and look at >>20. ZOMG, sum(n=1, +inf)[(9)10^-n] = 1.
>>61
Moron, I was pointing out problems in the other morons definition of the real numbers. Read the whole post before commenting.
>>64
Jesus, did I make a wrong turn and end up on sci.math? I haven't seen a crank/troll this bland and lazy since the mid-90's.
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Anonymous2007-05-15 13:55 ID:4wEg1xcj
>>65
There is no problem in my definitions. Many mathematicians often get very homicidal/suicidal when they are proven wrong about something "big". This happens every once in awhile.
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Anonymous2007-05-15 14:01 ID:y9niFlW+
If you want to change the current definition of 'real number,' you kinda have to have a pretty good reason. You know, one other than 'I wanna feel smrt.'
>>66
Your "definition" of the real numbers makes the real numbers and the rational numbers the same thing. Why even bother?
Furthermore, there's nothing "proven wrong" at all by a definition switch, even if I was willing to accept it. To prove anyone wrong you have to use the same definition and (gasp) PROVE them wrong. For example, if I say "every natural number is even", and you respond that 1 is not even, and I reply that the natural numbers are the set {0, 2, 4, 6, ... }, did I just prove you wrong? Of course not, because I changed a definition that is UNIVERSALLY ACCEPTED to make my argument, and even if you were to allow my new definition I haven't PROVEN you wrong because you were arguing that there exist non-even elements in the set formerly known as the natural numbers, not the set I was referring to as the natural numbers.
As for mathematicians getting upset when something doesn't work out as predicted, you're completely right. The converse, however, is not true. Just because some mathematicians disagree with you (even angrily sometimes) does not mean you are correct. Vehemently arguing that pi is exactly 3 would get responses from mathematicians, and many of them would be calling you a moron - does that mean pi is exactly 3?
>>70
The only things you've said over and over are "waaah waaah I don't like the accepted definition of the real numbers" and "everyone disagrees with me so I MUST be right."
Until you actually refute something I've said - anything - I'm done posting. Come up with an actual argument if you want to talk with the adults.
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Anonymous2007-05-15 21:05 ID:3Bdy/5b9
>>66
lol, have you ever met a real mathematician? Homicidal/Suicidal when we are proven wrong? Ha! Look at Hilbert's program, which was destroyed by Godel's incompleteness theorem. Did mathematics die? No, it didn't - it improved. I suggest you take you trolling and useless definitions elsewhere. 0.999... = 1 here.
Then you are claiming that sum(n=1, +inf)[(9)10^-n] is not convergent?
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Anonymous2007-05-16 0:50 ID:4hYn09lA
>>73
I'm claiming plain and simple that .9... is not a number, so it makes no sense to compare it to one. It can be represented as a series, as we've already beat the death out of, but ultimately that is irrelevant to the original argument that NotANumber is ANumber. And regarding your comment, you have hopefully been taught the difference between limits and actual values.
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Anonymous2007-05-16 1:08 ID:srxeNIT8
>>74
The question is, have you been taught the difference between limits and actual values? I
m assuming by "actual values", you mean the values of the individual terms of the sequence in question.
The only problem is that the real numbers are by definition equivalence classes of Cauchy sequences. The identification of such an equivalence class with the actual limit is merely a notational convenience. ergo 0.999... = 1
>>78
This is the last time I'm going to repeat it: The real numbers are an infinite set of FINITE numbers. Your argument is incorrect.
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4tran2007-05-16 5:20 ID:XM+arhOs
>>79
fi·nite – adjective
1. having bounds or limits; not infinite; measurable. [yes]
2. Mathematics. a. (of a set of elements) capable of being completely counted. [unrelated]
b. not infinite or infinitesimal. [yes]
c. not zero. [yes]
My argument is not an assertion about real numbers. My argument claims that .999... is a finite number. By the first definition, I placed a definite bound on the number. It is most definitely not infinite. I don't care if you believe it's a real number. The damn thing is finite.
>>81
Pi is not a real number, but rather a process with no corresponding real number. You can only use its symbolic form in equations involving real numbers. The same is true for the summation representing .999...
{Anyone taking a Numerical Analysis class would know the above.}
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Anonymous2007-05-16 12:21 ID:srxeNIT8
>>82
LOL
'Numerical Analysis' has nothing to do with what you are talking about - unless you are confusing machine numbers with real numbers!
>>82
Jesus. Go back to your computer science class, you've just failed so epically at math that it gave me a headache. REAL NUMBERS AND RATIONAL NUMBERS ARE DIFFERENT SETS.
>>92
It isn't constructivism. Constructivists are at least smart enough to realize that redefining the real numbers to be the rational numbers is pointless, and mainly focus on arguing that we shouldn't use the real numbers.
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Anonymous2007-05-17 17:00 ID:sStkk/IP
>>94
Well, then they are wrong also. Can't use numbers that don't exist.
2.3 Definition If there exists a 1-1 mapping of A onto B, we say that A and B can be put in 1-1 correspondence, or that A and B have the same cardinal number, or, briefly, that A and B are equivalent, and we write A ~ B. This relation clearly has the following properties:
It is reflexive: A ~ A.
It is symmetric: If A ~ B, then B ~ A.
It is transitive: If A ~ B and B ~ C, then A ~ C.
Any relation with these three properties is called an quivalence relation.
2.4 Definition For any positive integer n, let J_n be the set whose elements are the integers 1, 2,..., n; let J be the set consisting of all positive integers. For any set A, we say:
(a) A is finite if A ~ J_n for some n (the empty set is also considered to be finite).
(b) A is infinite if A is not finite.
(c) A is countable if A ~ J.
(d) A is uncountable if A is neither finite nor countable.
(e) A is at most countable if A is finite or countable.
----------
copied straight from the first book I found, Rudin, Principles of Mathematical Analysis. Honestly, finite vs infinite ain't hard. Just wait till you get to toposes without NNOs. Then it gets hard.
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4tran2007-05-18 11:26 ID:qENmSn+F
>>102
Yes, that's a very good definition of finite vs infinite (thanks), but it doesn't apply. You provided a detailed and thorough definition of finite vs infinite for sets, but we're not dealing with sets. We're arguing over a single element (that may or may not belong to some set). The original argument was whether .999... was finite or not. I claim it is. Random person denies that.
>>103
The person saying 0.999... was infinite was claiming that, because it has an infinite number of digits after the decimal, it is not a real number. He was, of course, wrong.
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4tran2007-05-18 13:07 ID:qENmSn+F
>>106
Yes, that's true, but it somehow seems that he believes that infinite digits = infinite number it represents (I probably should have been more explicit). Glad you recognize his error.
apparently you haven't learned the idiom concerning shit
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Anonymous2007-05-19 2:30 ID:jqaiHHIB
Infinity is not a number at all it is merely the absence of an end. Saying that k is infinity is not valid. The glimmer of hope that you see is merely the fact that 1-1/k calculated in an infinite number of ways would yield an infinite number of results, however infinity does not do the impossible. The only other argument you could make is that the universe does not actually divide like math does. For example when a decaying radioactive isotope gets to less than one atom the mathematical model collapses and the atom just decays. However the truth of that matter is very much the same. Lets say that if you were counting in atoms, but due to the strings that make up the subatomic particles you are limited to an absolute calculation in (for the sake of argument) lets say the .000000000001 place. In this case the farthest possible you could go while describing atoms in a 1/k function is .999999999999. If you cant go any farther due to it not existing then the only logical step is to reach 1.
>>113
fail. Particle decays are expressed as a statistical average. When the mathematical model says that there is less than 1 particle remaining, it does not always mean the particle will instantly decay.
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Anonymous2007-05-19 12:29 ID:jqaiHHIB
No it doesnt, but thats quite irrelevant. The point is that the mathematical model, even if it is only probability, cannot go farther than one atom and still express something other than the probability of that atom decaying. If there was an indivisible level of reality, it logically follows that it would be the end of all irrational numbers, and the farthest a .9999... can go.
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Anonymous2007-05-19 12:31 ID:jqaiHHIB
The real question is: do subatomic particles have components that are smaller and smaller until infinity? Is there any other force that can make this irrelevant? Finally, you must ask your self if infinity exists in this universe. I don't care either way but I prefer to think that it doesn't.