12 ms·
To be fair, 0.9999... = 1 is not quite basic. You need to know things like infinitesimals, the distinction between value and representation of numbers etc.
by NhanH 10y ago
To be fair, 0.9999... = 1 is not quite basic. You need to know things like infinitesimals, the distinction between value and representation of numbers etc.
- sp527 10y agoIt's also not strictly, theoretically true. More an 'for all intents and purposes' kind of thing.
- moron4hire 10y agoNo, it is true in the strictest sense of the word.
- sp527 10y agoHardly. We use this reduction because it's essential from a mathematical and axiomatic perspective and because limits are a fundamental construct. But more philosophically, a natural and nonterminating simulation of 0.999... (were it possible) would never strictly equal 1. You would wait an infinite amount of time for it to do so. This comes down to how you view the problem. I would never argue with this through the lens of abstract higher mathematics. I think about it from a computability perspective, which is one way in which you can discount the observation.
- iopq 10y agoIt's not a reduction. If you try to find where to put 0.999... on the number line, it has to go exactly where 1 is. For one thing, 1 - 0.999... = 0.000... because you never get to have any remainder since 0.999... is infinite. Or here's another proof: x = 0.999... 10x = 9.999... 10x - x = 9.999... - 0.999... 9x = 9.000... = 9 9x = 9 x = 1
- kazinator 10y agoYour "proofs" simply assumes that 0.999... is a notation denoting 1, without examining the underpinnings which might legitimize that. 9.999.. - 0.999 is 9 no matter how we define .999... just as long as two or more occurrences of the 0.999... notation all denote the same entity, and we understand that the syntax 9.999... is 9 + 0.999... For example, if we define 0.999... as "rubber duck" then 9.999... stands for 9 + "rubber duck", and 9.999... - 0.999... stands for 9 + "rubber duck" - "rubber duck" = 9.
- deleted 10y ago[deleted]
- iopq 10y agoWhile that's true, the whole key to the proof is that 0.999... * 10 = 9.999... Because "one less 9 than infinity is still infinity" it's what really closes the loop on the proof.
- moron4hire 10y agoYou are discarding an essential part of the representation. You are moving the goal post to make yourself right. Just because people don't always understand infinitesimals doesn't make them right. It's the reason they are wrong. There just plain don't know what they are talking about. You can't represent 0.2 exactly using IEEE floats, either, but that doesn't mean the representation 0.2 is not exactly equal to 1/5th.
- anthk 10y agoScheme has (exact->inexact x) and (inexact->exact x) for those issues.
- deleted 10y ago[deleted]
- deleted 10y ago[deleted]
- chestervonwinch 10y agoThe sum, 0.9 * Σ (0.1)^n from n=0 to n=N-1, is 0.999... as N goes to ∞ (write out the terms). On the other hand, for any finite value of N, the sum [1] is equal to 0.9 * (1-(0.1)^N) / (1-0.1) = 1 - (0.1)^N. This value goes to 1 as N goes to ∞. More rigorously, for any positive value, ε, there is a value N, such that the value of the finite sum is within ε of 1. [1]: https://en.wikipedia.org/wiki/Geometric_series#Formula https://en.wikipedia.org/wiki/Geometric_series#Formula
- minikites 10y agoMost people are happy to accept 0.33333... is the same as one third and from there it's a quick hop to 3*0.3333...
- iopq 10y agoWell, that's if they accept that an infinitely repeating number is actually EQUAL to the fraction. Some people just accept that a non-infinite number will never equal the fraction, but don't understand the concept of infinite repetition as an actual number (point on the number scale), not an estimate.
- mcbits 10y agoIt's easy to imagine that eating 1 out of 3 slices of pie is "the same" in a very specific sense as eating 333333... out of 1000000... slices of the same pie (although that would be infinity out of infinity slices, which is meaningless). What slips people up is that ignoring everything but the total pie consumed (taking the limit) is embedded in the definition of real numbers. There's an analogous story with rationals: Suppose x1 = 1, y1 = 3, x2 = 2, and y2 = 6. If we plot them, (x1, y1) and (x2, y2) are clearly different points, but x1/y1 "equals" x2/y2 because they lie on the same line through the origin. We decide that we don't need to know about those individual points.
- umanwizard 10y agoIf people accept 0.33... = 1/3 but not 0.99... = 1, they are being logically inconsistent. A "proof" that just plays a trick on people's logically inconsistent assumptions to derive a result isn't very satisfying. You're not really uncovering anything fundamental through that proof, just playing games. Edit: what I mean to say is there are reasonable things "..." could mean such that .99... and .33... are both not equal to 1 and 1/3, respectively. But there are none for which one pair is equal and the other not, as you rightly point out. So all your proof does is show that the other person's viewpoint is inconsistent, but it doesn't give any evidence for one of the two consistent viewpoints over the other.
- dantillberg 10y agoI didn't quite accept that until I challenged a professor in a 300-level university math class. This came up in class, and I spoke up and was like, "no, that can't be true, because 0.3333 repeated N times will never equal 1/3." Up to that point, I had only a fuzzy notion in my mind of what 0.33333... even was or how it was defined. But the professor helped clear this up for me: that "infinitely-repeating decimals" was actually just shorthand for a limit definition, i.e. "0.333..." is defined to be the limit as N approaches infinity of (3/10^1)+(3/10^2)+...+(3/10^N).
- rntz 10y agoYou don't need to understand anything about infinitesimals to understand 0.999... = 1. Perhaps you meant limits? The "standard" approach would be to point out that Σ_{i=1}^∞ 9/(10^i) = 1 (that is, the sum from i = 1 to infinity of 9/(10^i) is 1), and understanding an infinite summation requires the concept of a limit. (Of course, there are simpler proofs that use only basic algebra and intuition about decimals; a limit is just the most direct approach.) You're spot-on about needing to understand there's a distinction between a number and its decimal representation, though.
- soVeryTired 10y agoLimits use a construction that's pretty similar to an infinitesimal. The epsilon-delta definition of a limit is no joke for students.
- alejohausner 10y agoOddly enough, I never understood the epsilon-delta description of limits until I read David Foster Wallace's book on infinity. All through my degree in math I was taught about things without learning the historical context that created those things.
- geoka9 10y agoGetting a degree in math without understanding epsilon-delta is quite an achievement! I am serious.
- SomeStupidPoint 10y agoI'm always super happy that I stumbled in to taking topology before real analysis. It meant that I understood the topological idea of limits before I had to do proofs using just the epsilon (for sequence) or epsilon-delta (for functions) definition, and so could translate the logic of showing things about neighborhoods in to the terminology of (real analysis) limits. Limits, in the abstract, are a fairly simple concept: in the case of sequences, for any neighborhood of the limit, the entire tail of the sequence (past some point) is contained in the neighborhood; in the case of functions, for any neighborhood of the limit at f(x), there's a neighborhood around x, such that every point in that neighborhood maps to the neighborhood around the limit.
- yequalsx 10y agoProperty of real numbers: between distinct real numbers is at least one other number. Now try to find a decimal representation of a number bigger than 0.9999999... but less than 1.0. You clearly can't. They must be equal. No need for infinitesimals.
- NhanH 10y agoDoesn't that property come from the fact that there is no infinitesimal in real number?
- kazinator 10y ago> find a decimal representation of a number 0.99999999...1 This denotes the abstract idea that we take 0.999... (infinite number of 9's) and add another digit. This is no more or less abstract than 0.999... to begin with. 0.999... is a unicorn, and 0.999...9 is a unicorn with a pink ribbon on the tip of its horn.
- lisper 10y ago> add another digit But you can't. All of the places where you might want to add a digit are already occupied by 9's.
- kazinator 10y agoSo you're saying a countable infinity can be occupied, so that there is no room for one more. You must then disbelieve concepts such as that the even integers can be put in 1:1 correspondence with all integers; i.e. that there are exactly as many even integers as there are integers.
- lisper 10y ago> So you're saying a countable infinity can be occupied, so that there is no room for one more. No. I'm saying that a countable infinity doesn't have room for one more at the end. There is no end. You can obviously stick a 1 in the middle, but then you have a number strictly less than 0.999...
- S4M 10y agoI think I came to the conclusion that 0.9999999... = 1 when I was 12 or 13. My reasoning was that for any n, we had 0 <= 1 - 0.999999... <= 10^(-n), so the difference had to be 0.
- JoeAltmaier 10y agoIts just algebra. As noted elsewhere in this thread, 10X = 9.999... so 10X - X = 9.999... - 0.999... = 9 = 9X. So X=1.
- nilkn 10y agoThis proof is more subtle than it appears. Here's a bogus rewrite, for instance: Let 10X = 9.9. Then 10X - X = 9.9 - 0.9 = 9 = 9X. Hence X = 1, but X is actually 0.99 in this case (not 0.9). You need 0.99 = 0.9 for this to work with the exact same structure as your version. Your proof only works because appending a 9 to an infinite expansion of 9s does not actually add a 9. But at this point you're forced to establish meaning for an infinite expansion of 9s, at which point this is really not just algebra anymore.
- banach2 10y agoWhat? Why -0.9?
- nilkn 10y agoBecause that's exactly what the argument above does. It simply subtracts away the decimal part of 9.999.... This is always wrong except in the case of infinitely many repeated digits, and the proof does not explain this. More rigorously, let 9.999{n} denote an expansion with n 9s after the decimal point, where n can also be infinity. The subtlety with the argument is that it needs X to be the same as everything after the decimal point (so that the result of the subtraction is just 9). This is never true for finite values of n, and the proof does not establish that it's true for an infinite value of n -- indeed, it can't do so without supplying a meaning in the first place. Another way of phrasing it is that it assumes that if X = 0.999..., then 10X = 9.999..., where there are the "same number" of 9s after the decimal point in 10X as there are in X. This seems intuitive for an infinite repeating sequence of 9s, because "one less than infinity" is still infinity, but it's not very rigorous, and the argument as written certainly doesn't explain this.
- 10y ago