by zoidb on 6/2/25, 6:54 AM with 110 comments
by tsimionescu on 6/2/25, 7:30 AM
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.by bubblyworld on 6/2/25, 8:15 AM
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)
by cbolton on 6/2/25, 7:50 AM
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.
by lcrz on 6/2/25, 7:30 AM
“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.
by quitit on 6/2/25, 1:43 PM
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.
by mr_mitm on 6/2/25, 7:33 AM
* As the example shows, the decimal representation isn't unique, so perhaps we should say "_a_ decimal representation".
by fouronnes3 on 6/2/25, 7:26 AM
by HourglassFR on 6/2/25, 8:32 AM
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.
by A_D_E_P_T on 6/2/25, 8:42 AM
https://arxiv.org/abs/0811.0164
It feels intuitively correct is what I'll say in its favor.
by neeeeeeal on 6/2/25, 3:51 PM
What a community.
by constantcrying on 6/2/25, 5:29 PM
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)
by quchen on 6/2/25, 7:38 AM
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?
by dominicrose on 6/2/25, 8:21 AM
ps: based on the title I thought this would be about IEEE 754 floats.
by yodsanklai on 6/2/25, 8:19 AM
by beyondCritics on 6/2/25, 4:51 PM
by anthk on 6/2/25, 1:22 PM
by implements on 6/2/25, 7:21 PM
by 400thecat on 6/2/25, 8:45 AM
what is meant here by this notation 0.x and 1.y ?
by singularity2001 on 6/2/25, 1:11 PM
0.9̅ = 0.9̂ + ε = 1
For some definition of 0.9̂ = 1 - ε
by dmvjs on 6/2/25, 12:47 PM
by smidgeon on 6/2/25, 8:08 AM
by sans_souse on 6/2/25, 7:10 AM
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.