7 ms·

0.9999 ≊ 1

36 points by zoidb 1y ago

singularity2001 1y ago
Maybe this can be fixed for good using (axiomatic) hyperreal numbers: 0.9̅ = 0.9̂ + ε = 1 For some definition of 0.9̂ = 1 - ε
anthk 1y ago
0.999... -> 1 because you are correcting a supossed carry from a decimal forever. This is close to adding +1 to every odd number ever. No matter how much you try, you will get an even number.
sans_souse 1y ago
and 0.3… + 0.3… + 0.3… = 0.9… = 1.0 Maybe there is a difference, but it's intangible. Maybe it is to the number line what Planck Length is to measures. As a non-math-guy, I understand and accept it, but I feel like we can have both without breaking math. In a non-idealized system, such as our physical reality; if we divide an object into 3 pieces, no matter what that object was we can never add our 3 pieces together in a way that recreates perfectly that object prior to division. Is there some sort of "unquantifiable loss" at play? So yea, upvoting because I too am fascinated by this and its various connections in and out of math.
- bardan 1y ago
  0.3... is just the decimal representation of 1/3. So: 0.3... = 1/3 0.6... = 2/3 0.9... = 3/3 (= 1)
  - LiKao 1y ago
    But you are assuming 0.3... is the representation of 1/3. We don't have to make this assumption, it's just the one we are usually taught. Math doesn't really break from making different assumptions, quite the opposite. Let's make some different assumptions, not following high school math: When I divide 1 by 3, I always get a remainder. So it would just be as equally valid to introduce a mathematical object representing this remainder after I performed the infinite number of divisions. Then 1/3 = 0.3... + eps / 3 2/3 = 0.6... + 2eps / 3 3/3 = 0.9... + 3eps / 3 and since 0.9... = 1 - eps, we get 3/3 = 0.9... + eps = 1 It's all still sound (I haven't proven this, but so far I don't see any contradiction in my assumptions). And it comes out where 0.9... is not equal to 1. Just because I added a mathematical object that forces this to come out. Edit: Yes, I am breaking a lot of other stuff (e.g. standard calculus) by introducing this new eps object. But that is not an indicator that this is "wrong", just different from high school math.
- pvdebbe 1y ago
  Nothing is broken, just people stumbling on various notations.
- spyrja 1y ago
  Another easy way to understand it is to extend the idea of remainders to decimals. When we say N / D = Q r R that obviously means N = D * Q + R. For example 13 / 3 = 4 r 1 because 13 = 3 * 4 + 1. Likewise then 1 / 3 = 0.3(...) = 0.3 r 0.1 because 1 = 3 * 0.3 + 0.1, but also 1 / 3 = 0.3(...) = 0.33 r 0.01 because 1 = 3 * 0.33 + 0.01, etc. Hence 3 * 0.3(...) = 0.9(...) = 1.
- Suppafly 1y ago
  > if we divide an object into 3 pieces, no matter what that object was we can never add our 3 pieces together in a way that recreates perfectly that object prior to division. Is there some sort of "unquantifiable loss" at play? When you cut a cake into 3 slices, there is always a little bit of cake stuck the knife.
fouronnes3 1y ago
To me the most obvious proof is that therere are no numbers in between 0.999... and 1. Therefore it must be the same number.
- murkle 1y ago
  Exactly, add them up and divide by 2. What's the answer?
  - fouronnes3 1y ago
    TFA goes into this somehow but I fail to see why it's so hard to grasp that they are the same. Maybe I should read more crackpot blogs!
- blackbear_ 1y ago
  The fact that there are no numbers in between is not obvious at all, and has to be proven formally! In fact, there is a (rational) number between any two distinct real numbers, therefore your proof attempt only works if you assume that 0.999 equals 1. As that is a circular reasoning, it is not a valid proof.
  - thaumasiotes 1y ago
    > your proof attempt only works if you assume that 0.999 equals 1. As that is a circular reasoning, it is not a valid proof. No, his proof is fine. Take the standard definition of > as applied to decimal numbers when they're represented as strings. It's very easy to show that no x simultaneously satisfies x > 0.9999... and 1.0000... > x.
    - blackbear_ 1y ago
      You are right, that works if you assume that every number can be represented as a decimal string. That is indeed true for real numbers, but not for hyper-reals (https://en.m.wikipedia.org/wiki/Hyperreal_number https://en.m.wikipedia.org/wiki/Hyperreal_number), which is what I had in mind when I originally said that it was not obvious.
lcrz 1y ago
So the authors tries to be rigorous, but again falls into the same traps that the people who claim 0.9… != 1 fall. “0.999… = 1 - infinitesimal” But this is simply not true. Only then they get back to a true statement: “Inequality between two reals can be stated this way: if you subtract a from b, the result must be a nonzero real number c”. This post doesn’t clear things up, nor is it mathematically rigorous. Pointing towards hyperreals is another red herring, because again there 0.999… equals 1.
- hinkley 1y ago
  I don’t like any of his examples at the top. Look, it’s not that hard: x = 0.999… 2x = 1.999… 2x - x = 1 x = 1 Multiplying by ten just confused things and the result doesn’t follow for most people.
  - derbaum 1y ago
    Whether you multiply by 10 or 2, the same "counter" argument from the article stands. Only now you don't have a trailing zero after infinite nines, you have a trailing 8.
    - ndsipa_pomu 1y ago
      I don't understand how you can even have a trailing zero after an infinite number of nines. Surely any place that someone would want to put the zero can be refuted by correctly stating that a nine goes there (it's an infinite number of them, after all) and there is literally no "last" place.
      - hinkley 1y ago
        I’ve seen videos of actual mathematicians complaining to each other about how the general public thinks like GP. There is no last digit. Every time you reach the horizon there’s another horizon.
    - anthk 1y ago
      Technically you don't have an '8', you keep doing a carried sum forever, think about it. The last eight will be set to 9 forever and appended a new one to it. Thus, you are getting a periodical 1.9_ in practice.
tsimionescu 1y ago
The way I was taught decimals in school (in Romania) always made 0.99... seem like an absurdity to me: we were always taught that fractions are the "real" representation of rational numbers, and decimal notation is just a shorthand. Doing arithmetic with decimal numbers was seen as suspect, and never allowed for decimals with infinite expansions. So, for example, if a test asked you to calculate 2 × 0.2222... [which we notated as 2 × 0,(2)], then the right solution was to expand it: 2 × 0.2222... = 2 × 2/9 = 4/9 = 0.444... Once you're taught that this is how the numbers work, it's easy(ish) to accept that 0.999... is just a notational trick. At the very least, you're "immune" to certain legit-looking operations, like 0.33... + 0.66... = 1/3 + 2/3 = 3/3 = 1 Instead of 0.33... + 0.66... = 0.99... So, in this view, 0.3 or 0.333... are not numbers in the proper sense, they're just a convenient notation for 3/10 and 1/3 respectively. And there simply is no number whose notation would be 0.999..., it's just an abuse of the decimal notation.
- lmm 1y ago
  > Doing arithmetic with decimal numbers was seen as suspect, and never allowed for decimals with infinite expansions. With that attitude how do you handle e.g. pi or sqrt(2), which it's perfectly legitimate to do arithmetic with?
  - AndrewDucker 1y ago
    Once you're dealing with irrational numbers you have to understand that all results are approximations.
    - lmm 1y ago
      Well, sure, but you should still be able to ask and answer questions like "Is pi + sqrt(2) less than or greater than 4.553?"
      - AndrewDucker 1y ago
        In that case you know how many decimal places you want to expand them to, in order to compare.
mr_mitm 1y ago
Any confusion about this should go away as soon as you make clear what exactly you are talking about. If you construct the real numbers using Cauchy sequences and define the* decimal representation of a number using a Maclaurin series at x=1/10 then it's perfectly clear that 0.9... and 1.0... are two different representations of the same number. So it's the same equivalence class, but not the same representation. Thus, if you're talking about the representation of the abstract number 1, they're not equal but equivalent. If you're talking about the numbers they represent, they're equal. * As the example shows, the decimal representation isn't unique, so perhaps we should say "_a_ decimal representation".
- dagw 1y ago
  The intersection between people who are both confused by this and are comfortable working with Cauchy sequences, Maclaurin series and equivalence classes, is probably pretty small.
quchen 1y ago
It baffles me how there are still blogposts with a serious attitude about this topic. It’s akin to discussing possible loopholes of how homeopathy might be medicinally helpful beyond placebo, again and again. Why are hyperreals even mentioned? This post is not about hyperreals or non-standard math, it’s about standard math, very basic one at that, and then comes along with »well under these circumstances the statement is correct« – well no, absolutely not, these aren’t the circumstances the question was posed under. We don’t see posts saying »1+2 = 1 because well acktchually if we think modulo 2«, what’s with this 0.9… thing then?
- tsimionescu 1y ago
  I think it's worse than this. Even with hyperreals, 0.999... = 1, I believe, since they have to obey all laws of arithmetic that are true for the reals. At the very least, 3 × 0.333... = 1, and not 0.999... even for the hyperreals.
  - qayxc 1y ago
    IMHO the confusion arises, because the author failed to recognise that N cannot be a natural number if they go down the nonstandard analysis path. N would have to be elevated to a hyperinteger as well, which would eliminate the infinitesimal they end up with.
  - Tistron 1y ago
    You're saying that 0.999...=1, and simultaneously you are saying that 3 × 0.333... = 1 and not 0.999... What? How can it be that a=b and a≠c when b=c?
    - tsimionescu 1y ago
      I'm saying that, in the hyperreals as well as the reals, I am 100% certain that 3 × 0.33... = 1. I am not as sure that 0.999 = 1 with the hyperreals, BUT, if it's true as the author claims that 0.99... ≠ 1 in the hyperreals, then it must follow that 3 × 0.33... ≠ 0.99... in the hyperreals.
- 1y ago
cbolton 1y ago
The right way to approach this is to ask a question: What does 0.999... mean? What is the mathematical definition of this notation? It's not "what you get when you continue to infinity" (which is not clear). It's the value your are approaching as you continue to add digits. When applying the correct definition for the notation (the limit of a sequence) there's no question of "do we ever get there?". The question is instead "can we get as close to the target as we want if we go far enough?". If the answer is yes, the notation can be used as another way to represent the target.
smidgeon 1y ago
Don't say that near Richard Dedekind, he'll cut you.
bubblyworld 1y ago
There's an extremely subtle point here about the hyperreals that the author glosses over (and is perhaps unaware of): If you take 0.999... to mean sum of 9/10^n where n ranges over every standard natural, then the author is correct that it equals 1-eps for some infinitesmal eps in the hyperreals. This does not violate the transfer principle because there are nonstandard naturals in the hyperreals. If you take the above sum over all naturals, then 0.999... = 1 in the hyperreals too. (this is how the transfer principle works - you map sums over N to sums over N* which includes the nonstandards as well) The kicker is that as far as I know there cannot be any first-order predicate that distinguishes the two, so the author is on very confused ground mathematically imo. (not to mention that defining the hyperreals in the first place requires extremely non-constructive objects like non-principal ultrafilters)
- im3w1l 1y ago
  So something I was thinking of: A number in decimal notation can be seen as a function from the integers to {0,1,2,3,4,5,6,7,8,9} (where negative numbers map to digits left of the decimal point and non-negative to digits right of the decimal point) such that only finitely many negative numbers map to non-zero. Could you generalize this to include the hyperreals by lifting the restrictions on finitely many, and also adding in some transfinite ordinals to the domain of the function?
  - bubblyworld 1y ago
    I suspect yes - no need to introduce transfinite ordinals, you simply map from the set Z*, which is the integers but including the nonstandard ones. In fact you don't even need to remove the finiteness hypothesis, the transfer principle should guarantee that every hyperreal has such a representation since you can prove that every real does for the standard version. (if the finiteness thing seems confusing, remember that there are infinitely large nonstandard integers in the hyperreals, and you can't tell them apart from the others "from the inside")
yodsanklai 1y ago
I'd say the key point is to understand the difference between a number, and the decimal representation of a number. 0.99999... is one possible representation of number 1. 1 is another one. Once one understand the definition of the decimal representation, it's just a simple proof to show that 0.99999... = 1.
- kypro 1y ago
  I'm not a mathematician, but this is always the way I've looked at it too. We can't represent values like 1/3 precisely in the decimal number system, the best we can do is represent in a way that it's clear what's implied with minimal error. The representation isn't really suppose to be interpreted as an infinite decimal series, and depending on how you interpret 3.333... you could argue it's a slightly different value. And that's plainly obvious – 3.333... != 1/3
- throwaway31131 1y ago
  I think one other piece is one needs to understand the number is not being built, the whole representation exists all at once. When I tutor students the confusion is they think of each 9 like a brick being added to a wall and for them the wall is never done, that’s their argument why 0.999 doesn’t equal 1. Then when you explain numbers don’t have a time dimension they usually get it.
- constantcrying 1y ago
  >I'd say the key point is to understand the difference between a number, and the decimal representation But there is no such distinction. In fact the decimal representation is "closer" to a real number, then just 1. >is one possible representation of number 1. Why? You are just asserting things. You do not even give an argument why that should be the case. Why is 0.999... a representation of 1 and not 0.123?
  - yodsanklai 1y ago
    > But there is no such distinction Of course there's a distinction. A decimal representation is a sequence of digits, not a number > Why? It boils down from the definition of the decimal representation and the limit of a geometrical sequence. https://en.wikipedia.org/wiki/Decimal_representation https://en.wikipedia.org/wiki/Decimal_representation
dominicrose 1y ago
I think rational thinking just doesn't work when it comes to infinity math. I'd say the same thing about probabilities. ps: based on the title I thought this would be about IEEE 754 floats.
HourglassFR 1y ago
I don't get what the author is trying to do here. I mean he complains that talking about the limit of a sequence is too asbstract and unfamiliar to most people so the explaination is not satisfaying. But then names drop the notion of an Archimedean group and introduces with a big ol' handwave the hyperreals to solve this very straightforward highschool math problem… Now don't get me wrong, it is nice and good to have blogs presenting these math ideas in a easy if not rigorous way by attaching them to known concept. Maybe that was the real intend here, the 0.99… = 1 "controversy" is just bait, and I am too out of the loop to get the new meta.
A_D_E_P_T 1y ago
FWIW, there's an old Arxiv paper with this same argument: https://arxiv.org/abs/0811.0164 https://arxiv.org/abs/0811.0164 It feels intuitively correct is what I'll say in its favor.
400thecat 1y ago
> The belief that 0.x must be less than 1.y makes perfect sense to rational people what is meant here by this notation 0.x and 1.y ?
dmvjs 1y ago
are there no longer an infinite number of floating point numbers between every two floating point numbers?
- anthk 1y ago
  Two? It's the same number. There's no number between 0.9r and 1.0
quitit 1y ago
Where school kids tend to get stuck is that they'll hold contradictory views on how fractions can be represented. First it'll be uncontroversial that ⅓ = 0.333... usually because it's familiar to them and they've seen it frequently with calculators. However they'll then they'll get stuck with 0.999... and posit that it is not equal to 1/1, because there must "always be some infinitesimally small amount difference from one". However here lies the contradiction, because on one hand they accept that 0.333... is equal to ⅓, and not some infinitesimally small amount away from ⅓, but on the other hand they won't extend that standard to 0.999... Once you tackle the problem of "you have to be consistent in your rules for representing fractions", then you've usually cracked the block in their thinking. Another way of thinking about it is to suggest that 0.999.. is indistinguishable from 1.
- Suppafly 1y ago
  >However they'll then they'll get stuck with 0.999... and posit that it is not equal to 1/1, because there must "always be some infinitesimally small amount difference from one". Honestly teachers are half of the problem because they seem to make a game out of pointing out these sorts of contradictions instead of teaching the idea that you need "to be consistent in your rules for representing fractions". That and every next step in math classes is the teacher explaining that most of how you were taught to think about math in the previous step was incorrect and you really should think about it this way, only to be told that again the next year.
neeeeeeal 1y ago
This is why I love HN. One post about advanced SQL ACID concepts, the next about mathematics, yet another about history. What a community.
beyondCritics 1y ago
The explanation is, that the number is _not_ the infinite string of characters, but the sum of the scaled digits of the string. This sum is defined as the limit of the partial sums. In Germany, you can understand this in high school.
- constantcrying 1y ago
  And why does that change anything? No, it comes down to the definition of "=", which is not explained in schools.
constantcrying 1y ago
All these supposed proofs are totally wrong. Students are correctly interpreting as hand waving, by people who themselves do not have a good answer because that is exactly the case. The reason 0.999... and 1 are equal comes down to the definition of equality for real numbers. The informal formulation would be that two real numbers are equal if and only if their difference in magnitude is smaller than every rational number. (Formally two real numbers are equal iff they belong to the same equivalence class of cauchy series, where two series are in the same equivalence class iff their element wise difference is smaller than every rational number)
implements 1y ago
“By definition, there is no real number between 0.9r and 1 therefore they are the same” … was how I heard it explained.

36 points by zoidb 1y ago

singularity2001 1y ago

Maybe this can be fixed for good using (axiomatic) hyperreal numbers: 0.9̅ = 0.9̂ + ε = 1 For some definition of 0.9̂ = 1 - ε

anthk 1y ago

0.999... -> 1 because you are correcting a supossed carry from a decimal forever. This is close to adding +1 to every odd number ever. No matter how much you try, you will get an even number.

sans_souse 1y ago

and 0.3… + 0.3… + 0.3… = 0.9… = 1.0 Maybe there is a difference, but it's intangible. Maybe it is to the number line what Planck Length is to measures. As a non-math-guy, I understand and accept it, but I feel like we can have both without breaking math. In a non-idealized system, such as our physical reality; if we divide an object into 3 pieces, no matter what that object was we can never add our 3 pieces together in a way that recreates perfectly that object prior to division. Is there some sort of "unquantifiable loss" at play? So yea, upvoting because I too am fascinated by this and its various connections in and out of math.

bardan 1y ago
0.3... is just the decimal representation of 1/3. So: 0.3... = 1/3 0.6... = 2/3 0.9... = 3/3 (= 1)
- LiKao 1y ago
  But you are assuming 0.3... is the representation of 1/3. We don't have to make this assumption, it's just the one we are usually taught. Math doesn't really break from making different assumptions, quite the opposite. Let's make some different assumptions, not following high school math: When I divide 1 by 3, I always get a remainder. So it would just be as equally valid to introduce a mathematical object representing this remainder after I performed the infinite number of divisions. Then 1/3 = 0.3... + eps / 3 2/3 = 0.6... + 2eps / 3 3/3 = 0.9... + 3eps / 3 and since 0.9... = 1 - eps, we get 3/3 = 0.9... + eps = 1 It's all still sound (I haven't proven this, but so far I don't see any contradiction in my assumptions). And it comes out where 0.9... is not equal to 1. Just because I added a mathematical object that forces this to come out. Edit: Yes, I am breaking a lot of other stuff (e.g. standard calculus) by introducing this new eps object. But that is not an indicator that this is "wrong", just different from high school math.
pvdebbe 1y ago
Nothing is broken, just people stumbling on various notations.
spyrja 1y ago
Another easy way to understand it is to extend the idea of remainders to decimals. When we say N / D = Q r R that obviously means N = D * Q + R. For example 13 / 3 = 4 r 1 because 13 = 3 * 4 + 1. Likewise then 1 / 3 = 0.3(...) = 0.3 r 0.1 because 1 = 3 * 0.3 + 0.1, but also 1 / 3 = 0.3(...) = 0.33 r 0.01 because 1 = 3 * 0.33 + 0.01, etc. Hence 3 * 0.3(...) = 0.9(...) = 1.
Suppafly 1y ago
> if we divide an object into 3 pieces, no matter what that object was we can never add our 3 pieces together in a way that recreates perfectly that object prior to division. Is there some sort of "unquantifiable loss" at play? When you cut a cake into 3 slices, there is always a little bit of cake stuck the knife.

fouronnes3 1y ago

To me the most obvious proof is that therere are no numbers in between 0.999... and 1. Therefore it must be the same number.

murkle 1y ago
Exactly, add them up and divide by 2. What's the answer?
- fouronnes3 1y ago
  TFA goes into this somehow but I fail to see why it's so hard to grasp that they are the same. Maybe I should read more crackpot blogs!
blackbear_ 1y ago
The fact that there are no numbers in between is not obvious at all, and has to be proven formally! In fact, there is a (rational) number between any two distinct real numbers, therefore your proof attempt only works if you assume that 0.999 equals 1. As that is a circular reasoning, it is not a valid proof.
- thaumasiotes 1y ago
  > your proof attempt only works if you assume that 0.999 equals 1. As that is a circular reasoning, it is not a valid proof. No, his proof is fine. Take the standard definition of > as applied to decimal numbers when they're represented as strings. It's very easy to show that no x simultaneously satisfies x > 0.9999... and 1.0000... > x.
  - blackbear_ 1y ago
    You are right, that works if you assume that every number can be represented as a decimal string. That is indeed true for real numbers, but not for hyper-reals (https://en.m.wikipedia.org/wiki/Hyperreal_number https://en.m.wikipedia.org/wiki/Hyperreal_number), which is what I had in mind when I originally said that it was not obvious.

lcrz 1y ago

So the authors tries to be rigorous, but again falls into the same traps that the people who claim 0.9… != 1 fall. “0.999… = 1 - infinitesimal” But this is simply not true. Only then they get back to a true statement: “Inequality between two reals can be stated this way: if you subtract a from b, the result must be a nonzero real number c”. This post doesn’t clear things up, nor is it mathematically rigorous. Pointing towards hyperreals is another red herring, because again there 0.999… equals 1.

hinkley 1y ago
I don’t like any of his examples at the top. Look, it’s not that hard: x = 0.999… 2x = 1.999… 2x - x = 1 x = 1 Multiplying by ten just confused things and the result doesn’t follow for most people.
- derbaum 1y ago
  Whether you multiply by 10 or 2, the same "counter" argument from the article stands. Only now you don't have a trailing zero after infinite nines, you have a trailing 8.
  - ndsipa_pomu 1y ago
    I don't understand how you can even have a trailing zero after an infinite number of nines. Surely any place that someone would want to put the zero can be refuted by correctly stating that a nine goes there (it's an infinite number of them, after all) and there is literally no "last" place.
    - hinkley 1y ago
      I’ve seen videos of actual mathematicians complaining to each other about how the general public thinks like GP. There is no last digit. Every time you reach the horizon there’s another horizon.
  - anthk 1y ago
    Technically you don't have an '8', you keep doing a carried sum forever, think about it. The last eight will be set to 9 forever and appended a new one to it. Thus, you are getting a periodical 1.9_ in practice.

tsimionescu 1y ago

The way I was taught decimals in school (in Romania) always made 0.99... seem like an absurdity to me: we were always taught that fractions are the "real" representation of rational numbers, and decimal notation is just a shorthand. Doing arithmetic with decimal numbers was seen as suspect, and never allowed for decimals with infinite expansions. So, for example, if a test asked you to calculate 2 × 0.2222... [which we notated as 2 × 0,(2)], then the right solution was to expand it: 2 × 0.2222... = 2 × 2/9 = 4/9 = 0.444... Once you're taught that this is how the numbers work, it's easy(ish) to accept that 0.999... is just a notational trick. At the very least, you're "immune" to certain legit-looking operations, like 0.33... + 0.66... = 1/3 + 2/3 = 3/3 = 1 Instead of 0.33... + 0.66... = 0.99... So, in this view, 0.3 or 0.333... are not numbers in the proper sense, they're just a convenient notation for 3/10 and 1/3 respectively. And there simply is no number whose notation would be 0.999..., it's just an abuse of the decimal notation.

lmm 1y ago
> Doing arithmetic with decimal numbers was seen as suspect, and never allowed for decimals with infinite expansions. With that attitude how do you handle e.g. pi or sqrt(2), which it's perfectly legitimate to do arithmetic with?
- AndrewDucker 1y ago
  Once you're dealing with irrational numbers you have to understand that all results are approximations.
  - lmm 1y ago
    Well, sure, but you should still be able to ask and answer questions like "Is pi + sqrt(2) less than or greater than 4.553?"
    - AndrewDucker 1y ago
      In that case you know how many decimal places you want to expand them to, in order to compare.

mr_mitm 1y ago

Any confusion about this should go away as soon as you make clear what exactly you are talking about. If you construct the real numbers using Cauchy sequences and define the* decimal representation of a number using a Maclaurin series at x=1/10 then it's perfectly clear that 0.9... and 1.0... are two different representations of the same number. So it's the same equivalence class, but not the same representation. Thus, if you're talking about the representation of the abstract number 1, they're not equal but equivalent. If you're talking about the numbers they represent, they're equal. * As the example shows, the decimal representation isn't unique, so perhaps we should say "_a_ decimal representation".

dagw 1y ago
The intersection between people who are both confused by this and are comfortable working with Cauchy sequences, Maclaurin series and equivalence classes, is probably pretty small.

quchen 1y ago

It baffles me how there are still blogposts with a serious attitude about this topic. It’s akin to discussing possible loopholes of how homeopathy might be medicinally helpful beyond placebo, again and again. Why are hyperreals even mentioned? This post is not about hyperreals or non-standard math, it’s about standard math, very basic one at that, and then comes along with »well under these circumstances the statement is correct« – well no, absolutely not, these aren’t the circumstances the question was posed under. We don’t see posts saying »1+2 = 1 because well acktchually if we think modulo 2«, what’s with this 0.9… thing then?

tsimionescu 1y ago
I think it's worse than this. Even with hyperreals, 0.999... = 1, I believe, since they have to obey all laws of arithmetic that are true for the reals. At the very least, 3 × 0.333... = 1, and not 0.999... even for the hyperreals.
- qayxc 1y ago
  IMHO the confusion arises, because the author failed to recognise that N cannot be a natural number if they go down the nonstandard analysis path. N would have to be elevated to a hyperinteger as well, which would eliminate the infinitesimal they end up with.
- Tistron 1y ago
  You're saying that 0.999...=1, and simultaneously you are saying that 3 × 0.333... = 1 and not 0.999... What? How can it be that a=b and a≠c when b=c?
  - tsimionescu 1y ago
    I'm saying that, in the hyperreals as well as the reals, I am 100% certain that 3 × 0.33... = 1. I am not as sure that 0.999 = 1 with the hyperreals, BUT, if it's true as the author claims that 0.99... ≠ 1 in the hyperreals, then it must follow that 3 × 0.33... ≠ 0.99... in the hyperreals.
1y ago

cbolton 1y ago

The right way to approach this is to ask a question: What does 0.999... mean? What is the mathematical definition of this notation? It's not "what you get when you continue to infinity" (which is not clear). It's the value your are approaching as you continue to add digits. When applying the correct definition for the notation (the limit of a sequence) there's no question of "do we ever get there?". The question is instead "can we get as close to the target as we want if we go far enough?". If the answer is yes, the notation can be used as another way to represent the target.

smidgeon 1y ago

Don't say that near Richard Dedekind, he'll cut you.

bubblyworld 1y ago

There's an extremely subtle point here about the hyperreals that the author glosses over (and is perhaps unaware of): If you take 0.999... to mean sum of 9/10^n where n ranges over every standard natural, then the author is correct that it equals 1-eps for some infinitesmal eps in the hyperreals. This does not violate the transfer principle because there are nonstandard naturals in the hyperreals. If you take the above sum over all naturals, then 0.999... = 1 in the hyperreals too. (this is how the transfer principle works - you map sums over N to sums over N* which includes the nonstandards as well) The kicker is that as far as I know there cannot be any first-order predicate that distinguishes the two, so the author is on very confused ground mathematically imo. (not to mention that defining the hyperreals in the first place requires extremely non-constructive objects like non-principal ultrafilters)

im3w1l 1y ago
So something I was thinking of: A number in decimal notation can be seen as a function from the integers to {0,1,2,3,4,5,6,7,8,9} (where negative numbers map to digits left of the decimal point and non-negative to digits right of the decimal point) such that only finitely many negative numbers map to non-zero. Could you generalize this to include the hyperreals by lifting the restrictions on finitely many, and also adding in some transfinite ordinals to the domain of the function?
- bubblyworld 1y ago
  I suspect yes - no need to introduce transfinite ordinals, you simply map from the set Z*, which is the integers but including the nonstandard ones. In fact you don't even need to remove the finiteness hypothesis, the transfer principle should guarantee that every hyperreal has such a representation since you can prove that every real does for the standard version. (if the finiteness thing seems confusing, remember that there are infinitely large nonstandard integers in the hyperreals, and you can't tell them apart from the others "from the inside")

yodsanklai 1y ago

I'd say the key point is to understand the difference between a number, and the decimal representation of a number. 0.99999... is one possible representation of number 1. 1 is another one. Once one understand the definition of the decimal representation, it's just a simple proof to show that 0.99999... = 1.

kypro 1y ago
I'm not a mathematician, but this is always the way I've looked at it too. We can't represent values like 1/3 precisely in the decimal number system, the best we can do is represent in a way that it's clear what's implied with minimal error. The representation isn't really suppose to be interpreted as an infinite decimal series, and depending on how you interpret 3.333... you could argue it's a slightly different value. And that's plainly obvious – 3.333... != 1/3
throwaway31131 1y ago
I think one other piece is one needs to understand the number is not being built, the whole representation exists all at once. When I tutor students the confusion is they think of each 9 like a brick being added to a wall and for them the wall is never done, that’s their argument why 0.999 doesn’t equal 1. Then when you explain numbers don’t have a time dimension they usually get it.
constantcrying 1y ago
>I'd say the key point is to understand the difference between a number, and the decimal representation But there is no such distinction. In fact the decimal representation is "closer" to a real number, then just 1. >is one possible representation of number 1. Why? You are just asserting things. You do not even give an argument why that should be the case. Why is 0.999... a representation of 1 and not 0.123?
- yodsanklai 1y ago
  > But there is no such distinction Of course there's a distinction. A decimal representation is a sequence of digits, not a number > Why? It boils down from the definition of the decimal representation and the limit of a geometrical sequence. https://en.wikipedia.org/wiki/Decimal_representation https://en.wikipedia.org/wiki/Decimal_representation

dominicrose 1y ago

I think rational thinking just doesn't work when it comes to infinity math. I'd say the same thing about probabilities. ps: based on the title I thought this would be about IEEE 754 floats.

HourglassFR 1y ago

I don't get what the author is trying to do here. I mean he complains that talking about the limit of a sequence is too asbstract and unfamiliar to most people so the explaination is not satisfaying. But then names drop the notion of an Archimedean group and introduces with a big ol' handwave the hyperreals to solve this very straightforward highschool math problem… Now don't get me wrong, it is nice and good to have blogs presenting these math ideas in a easy if not rigorous way by attaching them to known concept. Maybe that was the real intend here, the 0.99… = 1 "controversy" is just bait, and I am too out of the loop to get the new meta.

A_D_E_P_T 1y ago

FWIW, there's an old Arxiv paper with this same argument: https://arxiv.org/abs/0811.0164 https://arxiv.org/abs/0811.0164 It feels intuitively correct is what I'll say in its favor.

400thecat 1y ago

> The belief that 0.x must be less than 1.y makes perfect sense to rational people what is meant here by this notation 0.x and 1.y ?

dmvjs 1y ago

are there no longer an infinite number of floating point numbers between every two floating point numbers?

anthk 1y ago
Two? It's the same number. There's no number between 0.9r and 1.0

quitit 1y ago

Where school kids tend to get stuck is that they'll hold contradictory views on how fractions can be represented. First it'll be uncontroversial that ⅓ = 0.333... usually because it's familiar to them and they've seen it frequently with calculators. However they'll then they'll get stuck with 0.999... and posit that it is not equal to 1/1, because there must "always be some infinitesimally small amount difference from one". However here lies the contradiction, because on one hand they accept that 0.333... is equal to ⅓, and not some infinitesimally small amount away from ⅓, but on the other hand they won't extend that standard to 0.999... Once you tackle the problem of "you have to be consistent in your rules for representing fractions", then you've usually cracked the block in their thinking. Another way of thinking about it is to suggest that 0.999.. is indistinguishable from 1.

Suppafly 1y ago
>However they'll then they'll get stuck with 0.999... and posit that it is not equal to 1/1, because there must "always be some infinitesimally small amount difference from one". Honestly teachers are half of the problem because they seem to make a game out of pointing out these sorts of contradictions instead of teaching the idea that you need "to be consistent in your rules for representing fractions". That and every next step in math classes is the teacher explaining that most of how you were taught to think about math in the previous step was incorrect and you really should think about it this way, only to be told that again the next year.

neeeeeeal 1y ago

This is why I love HN. One post about advanced SQL ACID concepts, the next about mathematics, yet another about history. What a community.

beyondCritics 1y ago

The explanation is, that the number is _not_ the infinite string of characters, but the sum of the scaled digits of the string. This sum is defined as the limit of the partial sums. In Germany, you can understand this in high school.

constantcrying 1y ago
And why does that change anything? No, it comes down to the definition of "=", which is not explained in schools.

constantcrying 1y ago

All these supposed proofs are totally wrong. Students are correctly interpreting as hand waving, by people who themselves do not have a good answer because that is exactly the case. The reason 0.999... and 1 are equal comes down to the definition of equality for real numbers. The informal formulation would be that two real numbers are equal if and only if their difference in magnitude is smaller than every rational number. (Formally two real numbers are equal iff they belong to the same equivalence class of cauchy series, where two series are in the same equivalence class iff their element wise difference is smaller than every rational number)

implements 1y ago

“By definition, there is no real number between 0.9r and 1 therefore they are the same” … was how I heard it explained.