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> Personally I've always thought "proofs" using "arithmetic" are right, but kind of stated backwards. I've never considered them right at all. By saying someth

> Personally I've always thought "proofs" using "arithmetic" are right, but kind of stated backwards. I've never considered them right at all. By saying something like 0.9... x 10 = 9.9... and then saying that 9.9... - 0.9... = 9 you're basically just a priori defining 0.9... to be 1. In other words you're basically just defining 0.9... as a symbol to be some number x which has the property that 10x - x = 9. So you're basically just defining it to be 1. I've never seen a proof of 0.9... = 1 using Peano arithmetic which made any sense to me. I doubt one actually exists in any true logical meaning. Unless you're making use of limits, completeness, or something equivalent I don't see how a proof could possibly make any sense.

JoeAltmaier 7y ago
Yes that proof depends upon the representation in text of rational numbers (a dot and a series of digits). Try it in hexadecimal - it becomes opaque nonsense. Without some mathematical basis for 0.9... X 10 being something, there's a dangerous dependency on the representation that makes many folks uneasy.
- ginko 7y ago
  I don't see how it would be significantly different in other bases. In hexadecimal it'd be 0x0.ffff.. * 0x10 = 0xf.fff.. The rules are essentially the same.
  - JoeAltmaier 7y ago
    Oh right! My mistake. Was stuck in my head with "10 == nine plus one" instead of your insight, multiply by '10' in whatever base. Anyway it's still an artifact of representation. If rationals were represented as fractions, then it is unrepresentable. 1/1 X 10 == 10/1?
judofyr 7y ago
You only need to define 0.9… as 9/10 + 9/100 + 9/1000 + …. Without knowing how that series converges you can then use the two expressions mentioned to conclude that it has to be equivalent to 1.
- CogitoCogito 7y ago
  > You only need to define 0.9… as 9/10 + 9/100 + 9/1000 + …. Without knowing how that series converges you can then use the two expressions mentioned to conclude that it has to be equivalent to 1. Sure you can provide a hand-wavy argument and try to give some intuition if you'd like. That doesn't make it any sort of logical proof though. I guess it depends on what you're after.
  - marcosdumay 7y ago
    The GP's definition is the fundamental definition of the decimal notation. It's exactly what the "0.9..." symbol means. You can redefine the "0.9..." symbol to mean something else as much as you want, you can have it meaning pi if you like, but then you are just changing the subject on the most unhelpful way.
    - CogitoCogito 7y ago
      > The GP's definition is the fundamental definition of the decimal notation. It's exactly what the "0.9..." symbol means. > You can redefine the "0.9..." symbol to mean something else as much as you want, you can have it meaning pi if you like, but then you are just changing the subject on the most unhelpful way. I _know_ that. I think I maybe should just bow out of this conversation. I'm apparently incapable of explaining myself in a way that is understandable to people here. I'll consider this my fault. I'll just summarize: I don't think any "proof" that 0.9... = 1 that is only expressed in terms of arithmetic operations and does not make use of limits is legitimate. In other words I claim that a proof like "0.9... = x" means "9.9... = 10x" means "9 = 9x" is illegitimate. Instead of taking "0.9... = 1" on faith it takes "10 x 0.9... = 9.9..." and "9.9... - 0.9... = 1.0... = 1" on faith. There's no proof here. It's just shifting around symbols. Of course there are logical proofs, but they make use of limits/completeness/properties of real numbers explicitly. Feel free to disagree...
      - dmitshur 7y ago
        FWIW, I believe I understand, and agree with the point you’re trying to make. In most circumstances I wouldn’t have a left comment because there’s nothing useful for me to add, just a thumbs up. But I wanted to make an exception this time. It’s worth remembering that most people who understand and agree rarely leave a reply.
mcphage 7y ago
> I've never seen a proof of 0.9... = 1 using Peano arithmetic which made any sense to me. I doubt one actually exists in any true logical meaning. Peano arithmetic only covers nonnegative whole numbers, so one will never exist.
- CogitoCogito 7y ago
  > Peano arithmetic only covers nonnegative whole numbers, so one will never exist. Thank you for the pedantism. How about I replace "Peano arithmetic" with the "operations of multiplication/addition/division/etc. expressible upon the rational numbers"?
  - empath75 7y ago
    It's not a rational number. Real numbers are defined as an equivalence class such that if the differences of two infinite sequences of rationals tend toward zero, then they are equal. The difference between 0.999... and 1.000... clearly tends towards zero as it heads of to infinity, and so they are equal. If you want to argue that it doesn't then you have to come up with some other definition for numbers which have an infinite decimal expansion. (Technically, of course, 1 is a rational number, but if you're using 0.9999... to represent it, you're using a real number representation, so you're bound by the definition)
    - CogitoCogito 7y ago
      > It's not a rational number. > Real numbers are defined as an equivalence class such that if the differences of two infinite sequences of rationals tend toward zero, then they are equal. The difference between 0.999... and 1.000... clearly tends towards zero as it heads of to infinity, and so they are equal. > If you want to argue that it doesn't then you have to come up with some other definition for numbers which have an infinite decimal expansion. > (Technically, of course, 1 is a rational number, but if you're using 0.9999... to represent it, you're using a real number representation, so you're bound by the definition) I'm not sure why you think I don't know the difference between rational and real numbers, but I assure you I do. What I said was I don't see how a proof involving the standard arithmetic operations found within the rational numbers, but not including any concepts of limits, completeness, etc. is invalid. Let me know if you still don't understand my point.
dahart 7y ago
> you’re basically just a priori defining 0.9... to be 1. I think the point is not defining 0.9... to be 1, the point is that “...” means an infinite number of 9s. If you shift the decimal point by 1, then nothing changes, there are still an infinite number of 9s. If you shift the decimal point by 5 places, there are still an infinite number of 9s to the right. And here is the logical (induction) step: if you shift the decimal point by an infinite number of places, then there are still an infinite number of 9s to the right. This works for any repeating fraction, in groups of more than 1 repeating digit. > I’ve never considered them right at all. Do you mean you disagree with the result, or that you agree with the result but don’t believe the proof is really a proof?
- CogitoCogito 7y ago
  > I think the point is not defining 0.9... to be 1, the point is that “...” means an infinite number of 9s. If you shift the decimal point by 1, then nothing changes, there are still an infinite number of 9s. If you shift the decimal point by 5 places, there are still an infinite number of 9s to the right. And here is the logical (induction) step: if you shift the decimal point by an infinite number of places, then there are still an infinite number of 9s to the right. This works for any repeating fraction, in groups of more than 1 repeating digit. I've heard this argument many times. I understand the intuitive reasoning. I just don't find it a proof. I mean with reasoning like this why can't you have an infinite amount of 9s and then just put a 7 after that? What's keeping you from doing that? It's just a hand-wavy argument with no rules of any kind of what are allowed. > Do you mean you disagree with the result, or that you agree with the result but don’t believe the proof is really a proof? I've never considered that specific "proof" a proof. When 0.9... is given a proper definition of limits and considered within the real numbers, then sure of course it's true and the proof is legitimate.
  - dahart 7y ago
    > why can’t you have an infinite amount of 9s and then just put a 7 after that? You can. The proof still works if you do that. What you’re refusing to accept here is the definition of infinity.
andrewla 7y ago
Limits and completeness are convenient shorthands here. But let's start from a more basic perspective. You say > you're basically just defining 0.9... as a symbol to be some number x which has the property that 10x - x = 0. Okay, well what is an alternate definition that makes more sense intuitively? 0.333...., for example, is one that seems pretty intuitive. We can get to .333... by iterated long division of 1 by 3. 3 | 1 0 (3 * 0 = 0) => 0. 10 (add zero) 9 (3 * 3 = 9) => 0.3 10 (add zero) 9 (3 * 3 = 9) => 0.33 ... And we can verify the reverse by doing the same trick above; 0.333... = 10 * 0.333... - 3 => 3 = 9 * 0.333... => 0.333... = 3 / 9 = 1 / 3. So does this trick always work? If we have a repeated decimal, can we always multiply by 10 ^ (length of repeated sequence), subtract off, and get the value of that repeated decimal? If so, then it is reasonable to say that 0.999.... is equal to 1. We can't really go in the forward direction without cheating (that is, going from 1 -> .999...); the best we can do is to modify long division to allow us to do it: 3 | 3 0 (3 * 0 = 0) => 0. 30 (add zero) 27 (3 * 9 = 27) => 0.9 30 (add zero) 27 (3 * 9 = 27) => 0.99 ... And so on. Obviously this isn't in Peano arithmetic exactly, but I think it holds together. If we allow repeated decimals in general to be valid representations of rational numbers, then we have to accept 0.999.... is equal to 1.
birktj 7y ago
What's very neat is that the algebraic argument also works for other sequences like 0.888... x = 0.8... 10x = 8.8... = 8 + x 9x = 8 8/9 = 0.8...
ben509 7y ago
Here's one stupid, simple proof that doesn't require any predefinition. We do longhand division of 9 by 9, but start by putting a 0 in the first place. 0.99 9 | 9.0000 0 9 0 8 1 90 Hence, 9 / 9 = 0.999 = 1.

JoeAltmaier 7y ago

Yes that proof depends upon the representation in text of rational numbers (a dot and a series of digits). Try it in hexadecimal - it becomes opaque nonsense. Without some mathematical basis for 0.9... X 10 being something, there's a dangerous dependency on the representation that makes many folks uneasy.

ginko 7y ago
I don't see how it would be significantly different in other bases. In hexadecimal it'd be 0x0.ffff.. * 0x10 = 0xf.fff.. The rules are essentially the same.
- JoeAltmaier 7y ago
  Oh right! My mistake. Was stuck in my head with "10 == nine plus one" instead of your insight, multiply by '10' in whatever base. Anyway it's still an artifact of representation. If rationals were represented as fractions, then it is unrepresentable. 1/1 X 10 == 10/1?

judofyr 7y ago

You only need to define 0.9… as 9/10 + 9/100 + 9/1000 + …. Without knowing how that series converges you can then use the two expressions mentioned to conclude that it has to be equivalent to 1.

CogitoCogito 7y ago
> You only need to define 0.9… as 9/10 + 9/100 + 9/1000 + …. Without knowing how that series converges you can then use the two expressions mentioned to conclude that it has to be equivalent to 1. Sure you can provide a hand-wavy argument and try to give some intuition if you'd like. That doesn't make it any sort of logical proof though. I guess it depends on what you're after.
- marcosdumay 7y ago
  The GP's definition is the fundamental definition of the decimal notation. It's exactly what the "0.9..." symbol means. You can redefine the "0.9..." symbol to mean something else as much as you want, you can have it meaning pi if you like, but then you are just changing the subject on the most unhelpful way.
  - CogitoCogito 7y ago
    > The GP's definition is the fundamental definition of the decimal notation. It's exactly what the "0.9..." symbol means. > You can redefine the "0.9..." symbol to mean something else as much as you want, you can have it meaning pi if you like, but then you are just changing the subject on the most unhelpful way. I _know_ that. I think I maybe should just bow out of this conversation. I'm apparently incapable of explaining myself in a way that is understandable to people here. I'll consider this my fault. I'll just summarize: I don't think any "proof" that 0.9... = 1 that is only expressed in terms of arithmetic operations and does not make use of limits is legitimate. In other words I claim that a proof like "0.9... = x" means "9.9... = 10x" means "9 = 9x" is illegitimate. Instead of taking "0.9... = 1" on faith it takes "10 x 0.9... = 9.9..." and "9.9... - 0.9... = 1.0... = 1" on faith. There's no proof here. It's just shifting around symbols. Of course there are logical proofs, but they make use of limits/completeness/properties of real numbers explicitly. Feel free to disagree...
    - dmitshur 7y ago
      FWIW, I believe I understand, and agree with the point you’re trying to make. In most circumstances I wouldn’t have a left comment because there’s nothing useful for me to add, just a thumbs up. But I wanted to make an exception this time. It’s worth remembering that most people who understand and agree rarely leave a reply.

mcphage 7y ago

> I've never seen a proof of 0.9... = 1 using Peano arithmetic which made any sense to me. I doubt one actually exists in any true logical meaning. Peano arithmetic only covers nonnegative whole numbers, so one will never exist.

CogitoCogito 7y ago
> Peano arithmetic only covers nonnegative whole numbers, so one will never exist. Thank you for the pedantism. How about I replace "Peano arithmetic" with the "operations of multiplication/addition/division/etc. expressible upon the rational numbers"?
- empath75 7y ago
  It's not a rational number. Real numbers are defined as an equivalence class such that if the differences of two infinite sequences of rationals tend toward zero, then they are equal. The difference between 0.999... and 1.000... clearly tends towards zero as it heads of to infinity, and so they are equal. If you want to argue that it doesn't then you have to come up with some other definition for numbers which have an infinite decimal expansion. (Technically, of course, 1 is a rational number, but if you're using 0.9999... to represent it, you're using a real number representation, so you're bound by the definition)
  - CogitoCogito 7y ago
    > It's not a rational number. > Real numbers are defined as an equivalence class such that if the differences of two infinite sequences of rationals tend toward zero, then they are equal. The difference between 0.999... and 1.000... clearly tends towards zero as it heads of to infinity, and so they are equal. > If you want to argue that it doesn't then you have to come up with some other definition for numbers which have an infinite decimal expansion. > (Technically, of course, 1 is a rational number, but if you're using 0.9999... to represent it, you're using a real number representation, so you're bound by the definition) I'm not sure why you think I don't know the difference between rational and real numbers, but I assure you I do. What I said was I don't see how a proof involving the standard arithmetic operations found within the rational numbers, but not including any concepts of limits, completeness, etc. is invalid. Let me know if you still don't understand my point.

dahart 7y ago

> you’re basically just a priori defining 0.9... to be 1. I think the point is not defining 0.9... to be 1, the point is that “...” means an infinite number of 9s. If you shift the decimal point by 1, then nothing changes, there are still an infinite number of 9s. If you shift the decimal point by 5 places, there are still an infinite number of 9s to the right. And here is the logical (induction) step: if you shift the decimal point by an infinite number of places, then there are still an infinite number of 9s to the right. This works for any repeating fraction, in groups of more than 1 repeating digit. > I’ve never considered them right at all. Do you mean you disagree with the result, or that you agree with the result but don’t believe the proof is really a proof?

CogitoCogito 7y ago
> I think the point is not defining 0.9... to be 1, the point is that “...” means an infinite number of 9s. If you shift the decimal point by 1, then nothing changes, there are still an infinite number of 9s. If you shift the decimal point by 5 places, there are still an infinite number of 9s to the right. And here is the logical (induction) step: if you shift the decimal point by an infinite number of places, then there are still an infinite number of 9s to the right. This works for any repeating fraction, in groups of more than 1 repeating digit. I've heard this argument many times. I understand the intuitive reasoning. I just don't find it a proof. I mean with reasoning like this why can't you have an infinite amount of 9s and then just put a 7 after that? What's keeping you from doing that? It's just a hand-wavy argument with no rules of any kind of what are allowed. > Do you mean you disagree with the result, or that you agree with the result but don’t believe the proof is really a proof? I've never considered that specific "proof" a proof. When 0.9... is given a proper definition of limits and considered within the real numbers, then sure of course it's true and the proof is legitimate.
- dahart 7y ago
  > why can’t you have an infinite amount of 9s and then just put a 7 after that? You can. The proof still works if you do that. What you’re refusing to accept here is the definition of infinity.

andrewla 7y ago

Limits and completeness are convenient shorthands here. But let's start from a more basic perspective. You say > you're basically just defining 0.9... as a symbol to be some number x which has the property that 10x - x = 0. Okay, well what is an alternate definition that makes more sense intuitively? 0.333...., for example, is one that seems pretty intuitive. We can get to .333... by iterated long division of 1 by 3. 3 | 1 0 (3 * 0 = 0) => 0. 10 (add zero) 9 (3 * 3 = 9) => 0.3 10 (add zero) 9 (3 * 3 = 9) => 0.33 ... And we can verify the reverse by doing the same trick above; 0.333... = 10 * 0.333... - 3 => 3 = 9 * 0.333... => 0.333... = 3 / 9 = 1 / 3. So does this trick always work? If we have a repeated decimal, can we always multiply by 10 ^ (length of repeated sequence), subtract off, and get the value of that repeated decimal? If so, then it is reasonable to say that 0.999.... is equal to 1. We can't really go in the forward direction without cheating (that is, going from 1 -> .999...); the best we can do is to modify long division to allow us to do it: 3 | 3 0 (3 * 0 = 0) => 0. 30 (add zero) 27 (3 * 9 = 27) => 0.9 30 (add zero) 27 (3 * 9 = 27) => 0.99 ... And so on. Obviously this isn't in Peano arithmetic exactly, but I think it holds together. If we allow repeated decimals in general to be valid representations of rational numbers, then we have to accept 0.999.... is equal to 1.

birktj 7y ago

What's very neat is that the algebraic argument also works for other sequences like 0.888... x = 0.8... 10x = 8.8... = 8 + x 9x = 8 8/9 = 0.8...

ben509 7y ago

Here's one stupid, simple proof that doesn't require any predefinition. We do longhand division of 9 by 9, but start by putting a 0 in the first place. 0.99 9 | 9.0000 0 9 0 8 1 90 Hence, 9 / 9 = 0.999 = 1.