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(STATEMENT OF PERSONAL IGNORANCE [SOPI]: Anyone who actually understands this stuff please correct my mistakes below. Thanks.) In the real numbers, which are n

(STATEMENT OF PERSONAL IGNORANCE [SOPI]: Anyone who actually understands this stuff please correct my mistakes below. Thanks.) In the real numbers, which are not always simple or intuitive, 0.99... = 1. That's true and I seem to understand the proof. But the real numbers aren't the only system that might be sitting behind "0.99..." and "1" when I write those symbols down and talk intuitively to people in my family. The reals are just the system we're taught first. I believe there are other systems (I think the surreals are an example) that work just as well for everyday purposes, but where ( my understanding is that) there are numbers that differ from 1 by a value that approaches zero, yet those numbers are not equal to 1. (I've played with the surreals but only as a hobbyist.) If you do calculus in these other numbers, I think physically meaningful problems will still yield the same answers. (For example Zeno's Paradoxes are still not an excuse for failure to attend school.) But it isn't a law of nature, I think, that all number systems that can hold 0.999... and 1 must make them equal.

Scarblac 6y ago
A number is just a number, it doesn't approach anything. A series can approach something, but a number can't. In any system where 0.99... is valid notation for a number, it doesn't approach anything.
- tom-thistime 6y ago
  Sure. Instead of "0.99..." please substitute lim n->inf sum(1..n)(9 times 10^-n). The point I'm making is that the "obvious truth" 0.99... = 1 that we're all talking about depends on the assumption that we're working in the real numbers. I claim that the real numbers are not something intuitively obvious to every sufficiently intelligent person; instead they are kind of weird and technical. I go on to claim, though I'm more unsure of this, that the real numbers are not even the only way to make calculus work.
  - ramshorns 6y ago
    Well, a limit of a series is just a number and doesn't approach anything either. If the series approaches something, we say the limit exists and is equal to that thing. Anyway, in the surreal numbers you could probably make up a notation where 0.999... actually denotes 1 - ε or something. But I daresay it might not be very useful because then how do you denote 1 - ε/2 or anything else.
    - ramshorns 6y ago
      Oh, I figured it out, kind of. [I don't really know what I'm talking about either.] 1.000...0 = 1 1.000...05 = 1 + ε/2 1.000...1 = 1 + ε 0.999...8 = 1 - 2ε 0.999...9 = 1 - ε 0.999...98 = 1 - ε/5 . . .
      - saalweachter 6y ago
        > 0.999...8 = 1 - 2ε > 0.999...98 = 1 - ε/5 cough
        ramshorns 6y ago
        Sure, why not? Different notations for different surreal numbers. But it's not actually a good notation; I wouldn't know how to write 1 + 10ε.
    - tom-thistime 6y ago
      We've defined the series we're talking about, and we've defined 0.99... as the limit of that series. I don't know whether repeating decimals are useful for testing equality of surreal numbers. All I'm trying to say is we're all talking about anything and everything except the definition "when are two real numbers equal?" And then we're saying people who don't understand the consequences of that definition are kind of dummies ... while we continue to not actually say what the definition is.
      - ramshorns 6y ago
        Fair. Maybe the definition is that two real numbers are equal iff there is no real number between them. Which is nonintuitive if you're used to things like natural numbers which have successors. But if you teach people this way of thinking about the real numbers, then arguments like "it's the last number riiight before 1" stop working.
  - esmi 6y ago
    > I go on to claim, though I'm more unsure of this, that the real numbers are not even the only way to make calculus work. That claim is certainly true. The proof is that calculus (Analysis) exists for the complex number system too. Although I don’t think that’s what you meant and I doubt the complex number system is “more intuitive.” Just out of curiosity, have you heard of real analysis? How do you define calculus?
    - tom-thistime 6y ago
      A Riemann integral in complex analysis uses the same definitions as one in real analysis. Derivatives ditto. In the thread we're comparing real analysis to nonstandard analysis. I'm eager to be corrected if you can tell me something I said that's wrong. I'm not interested in gradually upping the ante with you until it's clear who really has more math background.
      - esmi 6y ago
        I didn’t really see anything wrong in the thread but you use words like “calculus” and I honestly don’t know exactly what you mean. Do you mean the plug and chug methods often taught in high school? Or do you mean the application of a set of theorems derived from the axioms of a given system of numbers? If you mean the former then perhaps we can clear up some misconceptions which are clarified by the latter.
    - obastani 6y ago
      Take a look at nonstandard calculus: https://en.wikipedia.org/wiki/Nonstandard_calculus https://en.wikipedia.org/wiki/Nonstandard_calculus It is based on the hyperreal numbers: https://en.wikipedia.org/wiki/Hyperreal_number https://en.wikipedia.org/wiki/Hyperreal_number Practically speaking, I don't think it buys you anything over traditional calculus/analysis. It's just pointing out that there are alternative approaches to formalizing calculus.
SpicyLemonZest 6y ago
Those number systems do exist, but I'm not sure it's right to say they work just as well for everyday purposes. They work only as long as you use them in a way that reduces to treating them as real numbers, either never computing an infintesimal in the first place or calculating 23 + 6ε and saying "oh that's basically just 23".
- tom-thistime 6y ago
  Sure. And it's true, 0.99... is equal to 1. All I'm saying is [SOPI below] it's all a little more technical than the junior high school proof. For example if 23+6\epsilon = 23, then how do I define 23 + 6\epsilon - 23? I can choose different approaches here, but "zero" is going to be pretty inconvenient when I go to do an integral. [SOPI] Statement of Personal Ignorance. I don't quite know what I'm talking about. If you do know, please step in and help correct me.
  - zests 6y ago
    What are you integrating over?
    - tom-thistime 6y ago
      I'm taking the limit of any convergent infinite sum of terms each weighted by \epsilon. Although \epsilon itself equals zero, of course the limit of the sum isn't necessarily zero. So if we go and define some other funky "infinitesimal" \epsilon != 0 in the surreals, we have to be careful. Apparently people have done that kind of thing successfully, but it took a long time after Cauchy for that to happen.
brutt 6y ago
0.999... + 0.000...1 = 1 0.000...1 = 1/∞ 0.999... = 1 - 1/∞ 1/∞ is zero or not?
- brabel 6y ago
  If you accept that 1/∞ = 0 Then you accept that ∞ * 0 = 1 But the definition of 0 is exactly that anything multiplied by it must be 0. So this cannot be true. To take a more verbal route: you cannot take nothingness and repeat it. Repeating (or multiplying) nothingness (or 0) is fundamentally nonsense. Programmer explanation: one cannot loop through `null` even once, let alone a large number.
  - aliceryhl 6y ago
    No, you don't have to accept that. In my Analysis 2 course we worked a whole bunch with [0, ∞], i.e. the positive real numbers together with infinity, and we defined 1/∞ = 0 and ∞ * 0 = 0. You lose some of your usual rules of arithmetic, but it gets a lot easier to talk about integrals.
    - brutt 6y ago
      https://en.wikipedia.org/wiki/Surreal_number https://en.wikipedia.org/wiki/Surreal_number Quote: There are also representations like { 0, 1, 2, 3, … | } = ω { 0 | 1, 1/2, 1/4, 1/8, … } = ε where ω is a transfinite number greater than all integers and ε is an infinitesimal greater than 0 but less than any positive real number. Moreover, the standard arithmetic operations (addition, subtraction, multiplication, and division) can be extended to these non-real numbers in a manner that turns the collection of surreal numbers into an ordered field, so that one can talk about 2ω or ω − 1 and so forth.
      - aliceryhl 6y ago
        I'm not quite sure I understand what you're trying to tell me? The example I mentioned is not about surreal numbers.
      - 6y ago
da_chicken 6y ago
The easiest way I know to explain it is fractions. 1 / 3 = 0.33333.... 2 / 3 = 0.66666.... So what's 3 / 3? Some people don't like that one. They might like this one better: 1 / 11 = 0.0909090909... What's 10 times that? 10 * 0.0909090909... = 0.90909090... So, let's do some addition and let the values zipper together because a nine will always line up with a zero: 10 * 0.0909090909... + 0.0909090909... = 0.90909090... + 0.0909090909... = 0.9999999999... However, 10 * 1 / 11 = 10 / 11. And 10 / 11 + 1 / 11 = 11 / 11. So 11 / 11 must be the same as 0.99999.... This works for any repeating fraction. You can do it with 1/7 and 6/7. You add the decimal representations of the numbers up and the value will be 0.99999... Technically, it works for any repeating fraction in any base. This is great because a lot of fractions are only repeating fractions in certain bases. So if 0.1 in base 10 is a repeating decimal in base 2 (it is) then you can show that (in decimal) 0.1 + 9 * 0.1 will represent (in binary) 0.11111...., which is equal to 1. The issue is that 1 / 11 + 10 / 11 (in decimal) must still equal 1 in ALL bases. Well, guess what? In Base 11 the decimal looks like: 0.1 + 0.A = 1.0 And 1.0 in base 11 is 1.0 in any base.
- war1025 6y ago
  That 11ths thing is actually really clever. I had never seen that argument before.
- xyzzyz 6y ago
  This is basically how we were taught to convert decimal fractions with periodic decimal expansions to regular fractions in school. You can even do that without that lining up of zeroes and nines, just multiply by the period: if x = 0.090909..., then 100x = 09.090909... (we shift the decimal point by two positions), and since the stuff after the point is the same, after subtracting it cancels out: 100x - x = 9.0, and so 99x = 9, from which we obtain x = 1/11.
anewvillager 6y ago
Even in surreal numbers, where there indeed exists numbers "infinitely close" to 1 but smaller than it, 0.9999.. would still be exactly 1. That's a quirk/feature of the decimal representation, not the underlying theory of numbers. In surreal numbers there's a number called ε a number infinitely close to 0 (but larger than it), so what you would think that 0.9999.. represent is actually written 1-ε maybe? But there's another number ε/2 that is between 0 and ε; 1-ε/2 is even closer to 1 than 1-ε is. Indeed, there are infinite numbers infinitely close to 1! (and none is really represented by 0.999...)

Scarblac 6y ago

A number is just a number, it doesn't approach anything. A series can approach something, but a number can't. In any system where 0.99... is valid notation for a number, it doesn't approach anything.

tom-thistime 6y ago
Sure. Instead of "0.99..." please substitute lim n->inf sum(1..n)(9 times 10^-n). The point I'm making is that the "obvious truth" 0.99... = 1 that we're all talking about depends on the assumption that we're working in the real numbers. I claim that the real numbers are not something intuitively obvious to every sufficiently intelligent person; instead they are kind of weird and technical. I go on to claim, though I'm more unsure of this, that the real numbers are not even the only way to make calculus work.
- ramshorns 6y ago
  Well, a limit of a series is just a number and doesn't approach anything either. If the series approaches something, we say the limit exists and is equal to that thing. Anyway, in the surreal numbers you could probably make up a notation where 0.999... actually denotes 1 - ε or something. But I daresay it might not be very useful because then how do you denote 1 - ε/2 or anything else.
  - ramshorns 6y ago
    Oh, I figured it out, kind of. [I don't really know what I'm talking about either.] 1.000...0 = 1 1.000...05 = 1 + ε/2 1.000...1 = 1 + ε 0.999...8 = 1 - 2ε 0.999...9 = 1 - ε 0.999...98 = 1 - ε/5 . . .
    - saalweachter 6y ago
      > 0.999...8 = 1 - 2ε > 0.999...98 = 1 - ε/5 cough
      - ramshorns 6y ago
        Sure, why not? Different notations for different surreal numbers. But it's not actually a good notation; I wouldn't know how to write 1 + 10ε.
  - tom-thistime 6y ago
    We've defined the series we're talking about, and we've defined 0.99... as the limit of that series. I don't know whether repeating decimals are useful for testing equality of surreal numbers. All I'm trying to say is we're all talking about anything and everything except the definition "when are two real numbers equal?" And then we're saying people who don't understand the consequences of that definition are kind of dummies ... while we continue to not actually say what the definition is.
    - ramshorns 6y ago
      Fair. Maybe the definition is that two real numbers are equal iff there is no real number between them. Which is nonintuitive if you're used to things like natural numbers which have successors. But if you teach people this way of thinking about the real numbers, then arguments like "it's the last number riiight before 1" stop working.
- esmi 6y ago
  > I go on to claim, though I'm more unsure of this, that the real numbers are not even the only way to make calculus work. That claim is certainly true. The proof is that calculus (Analysis) exists for the complex number system too. Although I don’t think that’s what you meant and I doubt the complex number system is “more intuitive.” Just out of curiosity, have you heard of real analysis? How do you define calculus?
  - tom-thistime 6y ago
    A Riemann integral in complex analysis uses the same definitions as one in real analysis. Derivatives ditto. In the thread we're comparing real analysis to nonstandard analysis. I'm eager to be corrected if you can tell me something I said that's wrong. I'm not interested in gradually upping the ante with you until it's clear who really has more math background.
    - esmi 6y ago
      I didn’t really see anything wrong in the thread but you use words like “calculus” and I honestly don’t know exactly what you mean. Do you mean the plug and chug methods often taught in high school? Or do you mean the application of a set of theorems derived from the axioms of a given system of numbers? If you mean the former then perhaps we can clear up some misconceptions which are clarified by the latter.
  - obastani 6y ago
    Take a look at nonstandard calculus: https://en.wikipedia.org/wiki/Nonstandard_calculus https://en.wikipedia.org/wiki/Nonstandard_calculus It is based on the hyperreal numbers: https://en.wikipedia.org/wiki/Hyperreal_number https://en.wikipedia.org/wiki/Hyperreal_number Practically speaking, I don't think it buys you anything over traditional calculus/analysis. It's just pointing out that there are alternative approaches to formalizing calculus.

SpicyLemonZest 6y ago

Those number systems do exist, but I'm not sure it's right to say they work just as well for everyday purposes. They work only as long as you use them in a way that reduces to treating them as real numbers, either never computing an infintesimal in the first place or calculating 23 + 6ε and saying "oh that's basically just 23".

tom-thistime 6y ago
Sure. And it's true, 0.99... is equal to 1. All I'm saying is [SOPI below] it's all a little more technical than the junior high school proof. For example if 23+6\epsilon = 23, then how do I define 23 + 6\epsilon - 23? I can choose different approaches here, but "zero" is going to be pretty inconvenient when I go to do an integral. [SOPI] Statement of Personal Ignorance. I don't quite know what I'm talking about. If you do know, please step in and help correct me.
- zests 6y ago
  What are you integrating over?
  - tom-thistime 6y ago
    I'm taking the limit of any convergent infinite sum of terms each weighted by \epsilon. Although \epsilon itself equals zero, of course the limit of the sum isn't necessarily zero. So if we go and define some other funky "infinitesimal" \epsilon != 0 in the surreals, we have to be careful. Apparently people have done that kind of thing successfully, but it took a long time after Cauchy for that to happen.

brutt 6y ago

0.999... + 0.000...1 = 1 0.000...1 = 1/∞ 0.999... = 1 - 1/∞ 1/∞ is zero or not?

brabel 6y ago
If you accept that 1/∞ = 0 Then you accept that ∞ * 0 = 1 But the definition of 0 is exactly that anything multiplied by it must be 0. So this cannot be true. To take a more verbal route: you cannot take nothingness and repeat it. Repeating (or multiplying) nothingness (or 0) is fundamentally nonsense. Programmer explanation: one cannot loop through `null` even once, let alone a large number.
- aliceryhl 6y ago
  No, you don't have to accept that. In my Analysis 2 course we worked a whole bunch with [0, ∞], i.e. the positive real numbers together with infinity, and we defined 1/∞ = 0 and ∞ * 0 = 0. You lose some of your usual rules of arithmetic, but it gets a lot easier to talk about integrals.
  - brutt 6y ago
    https://en.wikipedia.org/wiki/Surreal_number https://en.wikipedia.org/wiki/Surreal_number Quote: There are also representations like { 0, 1, 2, 3, … | } = ω { 0 | 1, 1/2, 1/4, 1/8, … } = ε where ω is a transfinite number greater than all integers and ε is an infinitesimal greater than 0 but less than any positive real number. Moreover, the standard arithmetic operations (addition, subtraction, multiplication, and division) can be extended to these non-real numbers in a manner that turns the collection of surreal numbers into an ordered field, so that one can talk about 2ω or ω − 1 and so forth.
    - aliceryhl 6y ago
      I'm not quite sure I understand what you're trying to tell me? The example I mentioned is not about surreal numbers.
    - 6y ago

da_chicken 6y ago

The easiest way I know to explain it is fractions. 1 / 3 = 0.33333.... 2 / 3 = 0.66666.... So what's 3 / 3? Some people don't like that one. They might like this one better: 1 / 11 = 0.0909090909... What's 10 times that? 10 * 0.0909090909... = 0.90909090... So, let's do some addition and let the values zipper together because a nine will always line up with a zero: 10 * 0.0909090909... + 0.0909090909... = 0.90909090... + 0.0909090909... = 0.9999999999... However, 10 * 1 / 11 = 10 / 11. And 10 / 11 + 1 / 11 = 11 / 11. So 11 / 11 must be the same as 0.99999.... This works for any repeating fraction. You can do it with 1/7 and 6/7. You add the decimal representations of the numbers up and the value will be 0.99999... Technically, it works for any repeating fraction in any base. This is great because a lot of fractions are only repeating fractions in certain bases. So if 0.1 in base 10 is a repeating decimal in base 2 (it is) then you can show that (in decimal) 0.1 + 9 * 0.1 will represent (in binary) 0.11111...., which is equal to 1. The issue is that 1 / 11 + 10 / 11 (in decimal) must still equal 1 in ALL bases. Well, guess what? In Base 11 the decimal looks like: 0.1 + 0.A = 1.0 And 1.0 in base 11 is 1.0 in any base.

war1025 6y ago
That 11ths thing is actually really clever. I had never seen that argument before.
xyzzyz 6y ago
This is basically how we were taught to convert decimal fractions with periodic decimal expansions to regular fractions in school. You can even do that without that lining up of zeroes and nines, just multiply by the period: if x = 0.090909..., then 100x = 09.090909... (we shift the decimal point by two positions), and since the stuff after the point is the same, after subtracting it cancels out: 100x - x = 9.0, and so 99x = 9, from which we obtain x = 1/11.

anewvillager 6y ago

Even in surreal numbers, where there indeed exists numbers "infinitely close" to 1 but smaller than it, 0.9999.. would still be exactly 1. That's a quirk/feature of the decimal representation, not the underlying theory of numbers. In surreal numbers there's a number called ε a number infinitely close to 0 (but larger than it), so what you would think that 0.9999.. represent is actually written 1-ε maybe? But there's another number ε/2 that is between 0 and ε; 1-ε/2 is even closer to 1 than 1-ε is. Indeed, there are infinite numbers infinitely close to 1! (and none is really represented by 0.999...)