Of course it is. The difference between the two representations is exactly zero. There is no last nine in the series. This problem is a mathematical canary in a goldmine. Those who Are one with mathematics arrive at this identity on their own Comprehend mathematics can see the truth of this identity once shown the way Don't quite get mathematics stick with their intuition, never seeing the light Are true mathematical geniuses create a valid system in which 0.999... is not equal to one, just for fun. None of the people who have said 0.999... is not 1 in this thread fall into the final category.
DH, True, but it is also true that none of the those who state 11 x .111... = 1.222... are logical geniuses.
Shmoe: Before I waste my time tracking down your following citation, I hope you will give me an honest description of what it says. "Topics in the Theory of Numbers" by Paul Erdos and Janos Suranyi, ISBN: 0387953205, Publisher: Springer, Page 86 to 87. They provide a proof that a decimal that is eventually reccuring must be rational that is essentially the same as the manipulations I did (treat as infinite series, etc.). I happen to be quite familiar with various proofs relating to the limits of the sums of infinite series. Your brief description of the citation you posted suggests one of those proofs. Just so there will be no misunderstanding, I object to the proof posted at this thread and claim that it is not a valid proof. I have never claimed that the conclusion is invalid. The proof I object to is the following x = 0.999... 10x = 9.999... 10x - x = 9.999... - 0.999... 9x = 9 x = 1 0.999... = 1 First: Do you consider the above to be a valid proof conforming to the modern requirements of mathematical logic? Second: does your citation use a method similar to the above to prove the conclusion? I posted the following I challenge you to find a formal calculus text which uses arithmetic on infinite strings of decimal digits to prove the results claimed here. If you can give me the title, author, and copyright date, I will try to locate the book. Please no high school algebra texts and please do not make up some bogus data. Others at this forum have sent me on fruitless searches for non existent sources. If you answer yes to the above two questions, I will hunt for the book you cited. If I take the time to hunt for that citation and discover that it presents the standard proof found in most calculus texts (or is an informal discussion for non mathematicians), I will have some choice remarks to make about your intellectual integrity.
The proof is not rigorous. None of the steps are justified. The proof is valid because each step can be justified. In particular, for the reals, Multiplication distributes over the addition of a finite a countably infinite number of terms (justifying steps 10x=9.999.. and 9.999...-0.999...=9) Good grief. An advanced mathematics text doesn't justifies each kindergarten step. Many steps, some of them rather huge, are "left as an exercise to the reader". For the last time, this is not calculus. It is analysis. Did you see the text I posted, aimed at the college junior/senior level (i.e., it is beyond calculus)? If not, here it is again, in big print Warner, Seth, "Modern Algebra", 1990, Dover Publications, New York Section 42 discusses "The Construction of a Dedekind Ordered Field". Problem 42.11 is particular applicable.
Absane, 11 * .111 = 1.221 11 * .111111 = 1.222221 11 * .111111111 = 1.222222221 When you multiply 11 by any series of 1s, the product always ends in '1', not a '2'. 1 * 1 = 1 on the end of the numerical product, not '2'. Also, .222 + .999 = 1.221 .222222 + .999999 = 1.222221 .222222222 + .999999999 = 1.222222221 When you add the final 2 & 9, the numerical product always ends in a '1'. If you add 1 + .222... , then the product is 1.222... It is only by rounding up the .999... to a true 1 does the product of 1 + .222... exactly equal 1.222... Same thing happens when you multiply .111... by any number larger than 11, (except for the 10's such as 20 * .111... = 2.222...) 19 * .111111 = 2.111109, not 2.111111
There is no end to 1.222... You are making the mistake of thinking infinite series act like a long finite series. There is no '1' at the end. If you don't like the nice compact 0.111..., notation, you can use the equivalent but terribly verbose limit/sum notation and apply Weierstrass limit theory. Just as Weierstrass limit theory proves that d/dx(x*x)=2*x, it also proves that 0.999... = 1.
2inquisitive... If 11*.111... = 1.222...221, then you should be able to do this (and still make sense): 1.222...221 - 1.222... = 0.000...001. Ok. Please.. tell me what fraction this is. PLEASE! There are two things you do not get: 1)Every rational number has a fractional presentation and a decimal representation. No other. It's by definition. 2) Addition/subtract/multiplication/division on Q (rationals) is CLOSED. For example.. a rational number plus a rational number = rational number. You break rule number one by claiming 0.00...001 exists. It has no fractional representation. And by assuming number 1 is wrong, you break rule number 2 because since there is no fractional representation, it much not be a rational number.
DH, I agree completely. That wasn't what I said though. I said multiply 11 by a never-ending series of ones and what do you get as the product? You say the product will end in a '2' by definition. I say the definition is lacking in logic, although I do know it is the official definition. As you increase the multipal, the error increases by my logic. What is 111 * .111111 ? It is 12.333321, not 12.333333 . What is the defined product of 111 * .111... ? I am sure it is 12.333... I am very much aware I cannot change the way mathematics is defined, but I may occasionally question the definitions themselves, not your knowledge of those definitions.
2inquisitive, 111 * .111... is not defined as 12.333... It is, however, equal to that by by virtue of deeper and fully consistent definitions. Certainly you can see that 1/9 = 0.111...? Just carry out the long division. It doesn't take too long to get to a point where things repeat. Similarly, 111/9 is 12.333... Mathematics strives for consistency. Something as simple as different ways of computing 111/9 = 111*1/9 = 111*0.111... had better come up with equal answers or mathematics is in a lot of trouble. The same goes for different ways of computing 1 = 9/9 = 9*1/9 = 9*0.111... = 0.999.... All of these represent the same number. Extending how you think of things in our finite world to the infinite just doesn't work sometimes. There are 50 even numbers between 1 and 100, 500 between 1 and 1000, etc. Extending that concept to the positive integers breaks down. The "number" of positive even integers is equal to the "number" of positive integers. Not half as many: just as many. You're everyday expectation tells you that 10*0.111... has one fewer digits to the right of the decimal than 0.111... has. That is not the case. They have the same number of digits to the right of the decimal, and these digits can be matched up on a one-to-one basis.
DH, LOL! No, my logic does not tell me that there are 'fewer' digits to the right of the decimal if the digits extend to infinity. My question is to do with whether 11 * .111... has an infinite string of 2's after the decimal, or whether 1 * 1 will always equal 1. You seem to claim that 1 * 1 will equal 2 at infinity.
Good. If you have a gmail (or some other account?) and can access the google book search, you can find the relevant pages here and here (hoping the links worked). A couple things: They say "We need to be careful in dealing with infinite series, since not all the properties of normal addition hold. Those relations that we need we will use without proof,..." Since you say you are familiar with infinite series, these should be no problem. If there's anything you want justified, just ask. They also say "...we do not consider those infinite decimals that from some point on are all 9's. It's easy to see that those do not arise from long division." The bold emphasis is mine, to point out why they aren't considering such decimals. It's not because what they do in this section won't work on .99..., you can check the details will hold yourself, they are heading towards proving every rational number has a unique decimal expansion that has at most a finite number of 9's. Since the decimal they had already associated with a rational number came from long division, to get a unique form the infinite 9's are tossed and they don't bother dealing with them (see theorem 1 on page 87). Even if you doubt the argument there works for .999... (if so, do point out where you think it fails), you will see the 1/7=.142857.... is a special case of what they consider. Absoutely. At least all the necessary steps can be justified rigorously, i.e. some background work is needed like the series results. Up to what I've explained above, and again I do claim the arguments of the relevant subsection do hold for 0.999.... It's the same thing I did, convert the decimals to the series they are then manipulate according to usual rules. Fair enough. Like I mentioned above, they use results about series without proving them. They also don't state what results they are using or where, I'm happy to do this for you if you feel they are too 'informal'. I'd still like to know what in my previous post you object to. I've been very clear with all of your posts where I found errors, I don't think it unreasonable to ask you to read mine and do the same.
Okie. There is no "end" of a decimal, no "last digit", no "infinite place". You agree 0.99... is the sum of 9/10^n with n ranging from 1 to infinity? The digits correspond to the terms in this sum, the sum is indexed by the natural numbers. A "last digit" would correspond to a "largest natural number". There is no largest natural number, for any natural n, n+1 is a larger natural number.. So, no matter what I do to these numbers i can never have a 1 at the end (I never have a 2 either). There is no end to speak of. This is not you questioning the definitions, this is you misinterpreting them or just making up your own. That is correct.
2inquisitive 0.111... IS the exact convesrsion of 1/9 if you stated that 0.111 is an approximation you would be correct, but you didn't Can anyone answer the querie i posted with regards to the following 1 + 2 = 3, 3/2 = 1.5 1+ 1 = 2, 2/2 = 1 1 + 0.999... = 1.999... , 1.999.../2 = 0.999... Before anyone starts to say 0.999...995 you can't have that If you have two exact numbers, add them together and divide by 2 you get the same number......all the time if you have two distinct numbers, add them and divide by two you get a different third number. Using this rule 0.999... = 1
Is that supposed to be the closed interval [0,1]? It is compact, but the reals are not. They are locally compact though. If you mean [0,1[ to be the half open interval that doesn't include 1, then it's not compact.
Depends on the topology you are using. For what its worth, after reading most, though not all of the thread, it strikes me that a lot of the confusion lies in what the symbol "=" actually means. I just wonder what those who doubt that 0.999...=1 consider equality to mean.
Anyone can easily prove that 1/9 is equal to 1.111111........ by working backwards: Q: What is 1 divided by 9, in decimal notation? A: By working this our through manual division (3rd grade math), you'll come with an answer 1.11111........ Therefore 1/9 = 1.111..... Therefore 0.9999..... = 1
[0,1[ was a typo actually. What does "half open" mean? Like [0,1)? That is neither open nor closed because the epison-neighborhood isn't a subset for all of x in the interval for [0,1) or [1,0)'. I call something likeAs far as I know... we have a definition for an open set. And if A' is open, then A is said to be closed. But yes I supposed you are correct... A subset R is said to be compact if every open cover of A has a finite subcover. R does not have to have a finite subcover. Is there anything we call R?
Yes, [0,1) is what I meant by 'half-open', I had meant to write [0,1) but apparently didn't, sorry. Sometimes what you wrote [0,1[ is used to mean the same the same thing as [0,1). You can call it a complete ordered field if you want, the complete ordered field if you like, it's unique up to isomorphism. ps. Human001's comment about the topology determining if a set is compact is exactly correct. Usually a lazy person like me who doesn't specifiy which topology they are considering when talking about the reals is referring to the usual one that comes from the usual distance metric.
Is there a definition for "density?" By that I mean R is dense because for any a and b, we an find c such that a < c < b and c is in R. Q is dense for the same reason. Z is not.. because if you have 0 and 1, no C exists that is in Z for the condition.
