0.9999... = 1

[Arra]

Don't believe it? Proof 1: We know that: (1/3) + (1/3) + (1/3) = 1 Since (1/3) = (0.333...): (0.333...) + (0.333...) + (0.333...) = 1 But, (0.333...) + (0.333...) + (0.333...) = 0.999... Therefore, 1 = 0.999... Proof 2: 1) x = 0.999... Multiply both sides by 10: 2) 10x = 9.999... Subtract (1) from (2) to get: 9x = 9 x = 1 (I didn't know whether to put this under Mathematics or Philosophy. I really hope a thread about this topic hasn't been submitted before.)

[Marvo]

Physically it is true. Mathematically it is not.

[Shadow27]

I've seen this logical fallacy before. Its just not correct. I'm not sure how to prove it wrong...

[DrunkenArse]

Physically it is true. Mathematically it is not. It's perfectly true mathematically. I have no idea how it could possibly be true physically. How could you measure a quantity to an infinite precision physically? I've seen this logical fallacy before. Its just not correct. I'm not sure how to prove it wrong... It would be easy to prove it wrong. The real numbers have the property that if a <= b and a =/= b, then there exists a c such that a < c < b. So all you need to do is find a number x such that .9999~ < x < 1. Indeed, the sequence of rationals given by (n-1)/n converges to 1 which means, if your assertion is true, that there is some N such that for all n > N, .999~ < (n-1)/n < 1. . So all you need to do is look at the sequence (n-1)/n until you find one satisfying the criterion. I'll help you start: 1/2 2/3 3/4 4/5 5/6 6/7 7/8 8/9 and so on. Please feel free express an opinion on this subject again when you're done counting and have such an N in hand If on the other hand you would like to have some point clarified, please feel free to actually ask a question and I'm sure somebody that actually knows what they're talking about will help you.

[Alric]

First off Proof 2 doesn't work. If you subtract 1 from 9.99999 you get 8.9999, which isn't 1. Same thing if you divide 9.9999 by 10, you get .99999 not 1. No matter what you do the results going to be x=.99999, and not 1. You are just totally wrong with your math on that. Now the first is a little more convincing at first, but only if you do it with a calculator. If you do the math by hand, you see that 1/3(divide 1 by 3) doesn't actually equal .3333.... It actually equals .3333...with a remainder of 1. If you drop off the remainder of 1, you are rounding off. .3333 is the rounded off answer for 1/3 but not its actual answer.

[DrunkenArse]

First off Proof 2 doesn't work. If you subtract 1 from 9.99999 you get 8.9999, which isn't 1. Same thing if you divide 9.9999 by 10, you get .99999 not 1. No matter what you do the results going to be x=.99999, and not 1. You are just totally wrong with your math on that. You misread it. You're not subtracting 1 from 9.99~, you subtracting equation (1) from equation (2). On the left hand side, that works out to 10x - x = 9x and on the other side you get 9.99~ - .99~ = 9. So 9x = 9. Then divide both sides by 9 to get x = 1. This is basic algebra Now the first is a little more convincing at first, but only if you do it with a calculator. If you do the math by hand, you see that 1/3(divide 1 by 3) doesn't actually equal .3333.... It actually equals .3333...with a remainder of 1. If you drop off the remainder of 1, you are rounding off. .3333 is the rounded off answer for 1/3 but not its actual answer. lol. Where does the remainder of 1 come in? You have to keep dividing. .3333 is the rounded answer though. .33~ (infinitely repeating) is the actual answer.

[Arra]

I'm surprised to get so much denial. I expected this would be old news, and that no one would disagree after seeing the first proof which is very easy to understand. There are other proofs that you can google or youtube if you still don't believe it. These are just the two I happen to have memorized. The fact that 0.999... = 1 is an accepted mathematical fact. When I first heard it, I was sure there must be something wrong with the reasoning involved in reaching such an 'absurd' conclusion, but could no longer reject it after seeing the proof. I like PhilosopherStoned's explanation, as it reveals the real problem with the number 0.9999. Any two distinct real numbers must have a number between them, and since there is no such number between 0.9999 and 1, they must be the same number. (EDIT: I realized what I said before in this paragraph made no sense and changed it.)

[DrunkenArse]

I like PhilosopherStoned's explanation, as it reveals the real problem with the number 0.9999. Any two real numbers must have a number between them, and since there is no such number between 0.9999 and 1, 0.9999 must not be a real number. (Please correct me if I got the implication wrong.) A little bit off. Mathematics required pedantry Any two distinct real numbers must have a third number between them. So the implication is that because there isn't a number between 0.99~ and 1, they must not be distinct real numbers, i.e., 1 = .99~. Of course I haven't actually proven that there's not a number between them. There was another thread on this in senseless banter and I offered a proof based on these ideas there. Great thread. Regardless of how many times it's been posted, if there's denial of such a basic fact, it should be posted again EDIT: And really, finding a number between them would only be one way to do it. I chose that because it let me tell a denier to go count to infinity and get back to me when he was done. The other criterion that often comes in handy for proving that two quantities, a and b, are the same is to prove that a - b = 0. I think that that might actually be how I proved it before. The more I think about it, the more I think I did it that way instead. 1 - .9 = .1 1 - .99 = .01 1 - .999 = .001 We can see that the right hand side is going to zero pretty fast.

[Replicon]

The "..." is short form for "and it goes on and on and on...." So really, 0.999... = lim(n -> infinity, sum( 9/10^k, k = 1..n) ) = 1 EDIT: D'oh silly Replicon, messed up the n's haha. Fixed now.

[Arra]

And really, finding a number between them would only be one way to do it. I chose that because it let me tell a denier to go count to infinity and get back to me when he was done. The other criterion that often comes in handy for proving that two quantities, a and b, are the same is to prove that a - b = 0. I think that that might actually be how I proved it before. The more I think about it, the more I think I did it that way instead. 1 - .9 = .1 1 - .99 = .01 1 - .999 = .001 We can see that the right hand side is going to zero pretty fast. And then someone might respond by saying 1 - 0.999~ = 0.0000...1 And that I suppose makes about as much sense as the number 0.9999... does anyway.

[DrunkenArse]

And then someone might respond by saying 1 - 0.999~ = 0.0000...1 And that I suppose makes about as much sense as the number 0.9999... does anyway. Think about this though. When you write 0.000...1, the ... is replacing a finite amount of zeros. Otherwise it doesn't make any sense to put 1 on the end and the number is just 0. So then we saw earlier that 1 - .999..9 is just .000..1 so the equation that you wrote would mean that there was some number of nines such that we can add another nine to the end of it without changing the difference. That clearly makes no sense. The number .99~ makes perfect sense. It's just 1

[Dannon Oneironaut]

So can we say that infinitely close to 1 = 1? And that infinitesimally close to 0 = 0? PhilospherStoned, are there different kinds of infinity?

[DrunkenArse]

So can we say that infinitely close to 1 = 1? And that infinitesimally close to 0 = 0? I suppose you could say that but it doesn't really make sense to me. "infinitely close to something" seems like it would lead to confusion very quick. The point is that .9999 is just another way of writing 1. As far as infinitesimally close, that's very hard to make formal. It's true that infinitesimals get used in introductory calculus. The derivative is viewed as dy/dx where dy and dx are infinitesimals. As soon as it's time to prove that calculus works though (that is when we grow up start calling it analysis), they're viewed as formal characters and then, later, as "differential forms". The idea is that dy is a function that takes as its input a change in x and gives a linear approximation of the change in y as its output. Apparently, someone has constructed a notion of infinitesimal numbers formally but nobody uses it as far as I know. PhilospherStoned, are there different kinds of infinity? This is a good question. There are different sizes of infinity as Cantor proved. There are "more" real numbers than integers or rationals. The idea is that two sets are the same size if we can put them into one to one correspondence. If you have nine blocks and nine apples, you know that you have the same amount of both of them because you can put one apple next to one block and not have any of either left over. Attempting to do this with the reals and the integers leads to contradiction. That's a formal result that's not too difficult. The infinity of the integers is called aleph sub 0 and the infinity of the reals is aleph sub 1. There's aleph sub 2, aleph sub 3, etc. Philosophically, I think that there are two different types of infinity though I've not seen it discussed. Essentially, there's the extensional infinity, 1, 2, 3, ... infinity, and then there's the resolutional infinity. This feels more like a sea of potential and is the infinity that's associated with the real numbers. Take any number. Pick some accuracy to which you would like to have it approximated. There are aleph sub 1 real numbers that can approximate it to your given accuracy. If you decide that you want it a billion trillion million times more accurate, there's still aleph sub 1 real numbers to do the job.

[Arra]

My STAT professor mentioned a few days ago that of the two types of infinity: All numbers between two numbers. Ex: [0, 1] which includes (0, 0.0000001, 0.00025, 0.589234, etc.) and All integers going into infinity, like [5, inf) which includes (5, 6, 7, 8, 9, ...) the first type has been mathematically proven to contain more numbers, to be a greater infinity, than the second type. This seems it must be false to me. Infinite is infinite. How can there be a higher type of infinity? But this interests me, and I'm planning on being proven wrong when I look it up, maybe tomorrow. As for 1 - 0.999... = 0.000...1, I agree that the concept is ridiculous. But conceptually, when we imagine a number like 0.999..., the difference that we're imagining between that number and 1, the reason why we write 0.999... and not 1 in the first place, is that we are imagining that there's a 0.000...1 difference between them. I'm not arguing that the statement 1 - 0.999... = 0.000...1 makes any mathematical sense, I'm just saying that this is psychologically what people imagine the difference between 0.999... and 1 to be. PhilosopherStoned, you're obviously very knowledgeable and I appreciate you taking the time to type such lengthy replies to share your insight.

[Marvo]

It's perfectly true mathematically. I have no idea how it could possibly be true physically. How could you measure a quantity to an infinite precision physically? The reason I said it is correct physically, is because at some point you can't get more precise. In the universe we have assigned both the shortest length of time and distance possible. Anything "below" this can't exist. This means that at some point you're so precise, that any more precision wouldn't make any sense. Mathematically it was, to my knowledge, a problem that nobody has an actual solution to, though some of the posts in this thread would suggest otherwise. I will have to investigate. I find it odd that somebody would make a thread about this, if it's so blindingly obvious. What exactly is the purpose? Interesting. According to wiki, it has indeed been shown that 0.999... equals 1. My older brother presented this problem to me a few years back, and he gave me the impression that it was a paradox of sorts.

[Shadow27]

I'll just say: lim ƒ(x) = 1 x → 3

[Xei]

This sort of stuff is some of the most interesting and beautiful mathematics which I really recommend looking into. I'm surprised at how many people tried to call this a fallacy though after being shown 2 valid proofs. I remember learning about this in school when I was 10 years old... It's actually pretty intuitive; 0.999... is 'infinitely close' to 1, and hence, it is 1. Technically this follows from the 'Archimedean property', which states that there are no 'infinitesimally small' numbers (and hence there can't be an infinitesimally small gap between the two; this answers the question about 0.00...1: it doesn't exist), and it can be proven, although it's a bit technical. It follows that all terminating decimals, for instance 4.3238912, actually have two expressions in decimal notation, 4.3238912 and 4.3238911999... which turns out to be more of a slight annoyance than anything. My STAT professor mentioned a few days ago that of the two types of infinity: All numbers between two numbers. Ex: [0, 1] which includes (0, 0.0000001, 0.00025, 0.589234, etc.) and All integers going into infinity, like [5, inf) which includes (5, 6, 7, 8, 9, ...) the first type has been mathematically proven to contain more numbers, to be a greater infinity, than the second type. This seems it must be false to me. Infinite is infinite. How can there be a higher type of infinity? But this interests me, and I'm planning on being proven wrong when I look it up, maybe tomorrow. Philosopher Stoned is basically correct about what he said, although there are some important distinctions to be made. In mathematics there are actually two totally different uses of infinity. The first is the infinity of analysis. This is used for when something is acting 'without bound' in some fashion or other. For example, one can say that the limit of 1/x^2 as x tends to 0 'is infinity'. However, this is not really rigorous as infinity isn't a number in this context. We therefore reformulate it in the following way: "for any number [no matter how high], you can give a range around 0 so that every value of 1/x^2 in that range is greater than said number". This removes all reference to infinity. Or, as another example, one can say that the limit of 1/x as x tends to 'infinity' is 0. This can be reformulated as "for any number [no matter how small], you can give a place on the x axis so that for any number to the right of that place, the distance between 1/x and 0 is less than said number". Again, this removes any reference to an 'infinite number'. The second infinity is the infinity of counting. This is the one you are referring to, and as PS correctly said, technically counting refers to creating a one-to-one correspondence between the elements of one set and another. He is also correct that Cantor proved that there are more real numbers (any decimal you like) and natural numbers (positive whole numbers). The proof is awesome and if you want I'll explain it. Your stats teacher is also correct; this actually follows from above. The reason is that you can set up a one-to-one correspondence between any interval of the real numbers ([0,1] in this instance) and the entire real number line. For example, the interval (-1, 1) can be sent to (-inf, inf) via the function f(x) = 1/x If you think about it, you'll find the assertion that every element in the first set now has a corresponding element in the second set is true, and hence, according to our notion of counting, there are the same number of elements in both sets. The intuitive way to think of this is that the real number line can be 'stretched' to form any other part of the real number line; that's why it's called continuous. [5, inf) can easily be sent to [1, inf) (the naturals) via the function f(n) = n - 4, and then it easily follows from the fact that there are more reals than naturals that what your teacher said was correct. You are right that this sets up a 'hierarchy of different infinities', and yes this is very strange, but I hope you can see that it's also well-defined. There is one mistake with what PS said, though, when he said that aleph-1 (i.e. the next biggest infinity after aleph-0, which represents the naturals) represents the infinity of the reals. This statement is referred to as 'the continuum hypothesis'. The amazing thing which was discovered last century is that this is actually unprovable, and so is its converse. Hence it is totally arbitrary whether it is true or not; you either assume it or you don't (so it's a kind of axiom). This has created two different 'mathematicses', each with their own different set of truths! There are many more weird and wonderful things in this area and I've only really skimmed some of the surface.

[Arra]

I'll just say: lim ƒ(x) = 1 x → 3 And what is ƒ(x)?

[Invader]

And what is ƒ(x)? Something like x/3, which is completely irrelevant to this thread, so you can probably ignore him.

[Arra]

Something like x/3, which is completely irrelevant to this thread, so you can probably ignore him. If that's what he meant, I suppose it is relevant. It's the same argument as lim as n -> inf of (n-1)/n = 1 But I don't think it's a proof that's going to convince most people. Because they'll say, yeah the limit is 1, but the number 0.999~ still exists as a distinct number from 1. The best proof I think is the first of the 2 proofs I wrote at the top, because it's simple enough for anyone to understand. Another good way to prove it might be to first realize that claiming the number "0.999..." exists is really the same thing as claiming the number "0.000...1" exists, since the addition of the two numbers is perceived to be 1. The latter is more intuitively nonsensical, to me at least.

[Xei]

0.999... does exist.

[DrunkenArse]

There is one mistake with what PS said, though, when he said that aleph-1 (i.e. the next biggest infinity after aleph-0, which represents the naturals) represents the infinity of the reals. Yeah, that's right of course. It's been a while. For some reason I was thinking that alpeh-1 was defined to be the infinity of the reals and the continuum hypothesis concerns whether there are or are not infinite cardinalities between aleph-0 and aleph-1. It's what I get for never actually looking at the Godel stuff too much. 0.999... does exist. No, no. You're confused here. 1 = .999~ and .999~ clearly doesn't exist so the inference is that 1 doesn't exist either. Right?

[Xei]

Yes, there was a rumour going round in lectures once that 1 isn't a number.

[DrunkenArse]

If that's what he meant, I suppose it is relevant. It's the same argument as lim as n -> inf of (n-1)/n = 1 The fact that f(n) = (n-1)/n converges to one doesn't actually have anything to do with it. The whole point of that was that, because we know that it converges to 1 in the way that it does, then if .99~ and 1 are two distinct real numbers, every sufficiently large n is mapped into the neighborhood (.99~, 1). * So that would be one way to prove that they're distinct. The sequence is just a convenient way to get at numbers within the necessary range. This is what the poster that I was responding too asked for. So the fact that there exist (a plethora of) functions with a limit of 1 at 3 is completely irrelevant unless further expounded upon. * In order to claim such a limited range, we need to require that for all N, there exist n > N with f(n) < 1, and that the sequence has a subsequence which converges to 1 which has 1 as it's least upper bound. Any such sequence will do the trick. That's just the simplest that I could think of.

[Oneiro]

I'm not a mathematician, nor any kind of scientist, but I'm having a great time reading this thread! So.. where am I so far? No, no. You're confused here. You bet! 1 = .999~ and .999~ clearly doesn't exist so the inference is that 1 doesn't exist either. ?? Right? Okay, right. Quote: Xei: "Yes, there was a rumour going round in lectures once that 1 isn't a number." Help! ?? Keep it up chaps..