🧵 Let's settle this once and for all.
Anonymous at Sun, 9 Feb 2025 18:58:59 UTC No. 16580532
I'll prove that [math]0.999...[/math] is not a real number.
Suppose that there exists [math]r \in \mathbb{R}[/math] such that [math]r = 0.999...[/math].
From the first digit being [math]0[/math] we have that [math]1 > r[/math]. From the second digit being [math]9[/math] we get that [math]r \geq 9 \cdot 10^{-1}[/math], from the third digit being [math]9[/math] we get [math]10r - 9 \geq 9 \cdot 10^{-1}[/math].
More generally we have [math](\forall n \in \mathbb{N}) 10^n \cdot r - (10^n -1) \geq 9 \cdot 10^{-1}[/math].
Subtracting [math]1[/math] from both sides we get [math](\forall n \in \mathbb{N}) 10^n \cdot r - 10^n \geq -10^{-1}[/math].
Factoring out [math]10^n[/math] and swapping all terms to the opposite side we get [math](\forall n \in \mathbb{N}) 10^{-1} \geq 10^n \cdot (1-r)[/math].
Since [math]1 > r[/math] we have that [math] 1-r \neq 0[/math] so we can multiply both sides by [math] (1-r)^{-1}[/math]: [math](\forall n \in \mathbb{N}) 10^{-1}(1-r)^{-1} \geq 10^n[/math].
Letting [math] k = 10^{-1}(1-r)^{-1}[/math] we get [math](\forall n \in \mathbb{N}) k \geq 10^n[/math].
But this leads to a contradiction since [math](\forall n \in \mathbb{N}) k \geq 10^n \Rightarrow k \geq 10^{\lceil k \rceil}[/math] and [math]10^{\lceil k \rceil}> \lceil k \rceil > k[/math].
But if [math]0.999...[/math] is not a real number how do you determine if some collection of digits represents some non-negative real number?
Well, it's very simple. First, a digit representation is some function [math] f : \mathbb{Z} \rightarrow \{0,1,2,3,4,5,6,7,8,9\}[/math].
Now one of these functions represents some non-negative real number if and only if it satisfies these two conditions:
[math]1) (\exists N \in \mathbb{Z}) (\forall n \in \mathbb{Z}) (n < N \Rightarrow f(n) = 0)[/math] This ensures that every number represents some finite quantity.
[math]2) (\nexists N \in \mathbb{Z}) (\forall n \in \mathbb{Z}) (n > N \Rightarrow f(n) = 9)[/math] This avoids the above contradiction.
Anonymous at Sun, 9 Feb 2025 19:25:17 UTC No. 16580544
>From the first digit being 0 we have that 1>r
Wrong. Try again, retard
Your contradiction at the end actually disproves this assumption.
Anonymous at Sun, 9 Feb 2025 19:34:42 UTC No. 16580552
>>16580544
What does a digit being 0 mean to you tard?
Anonymous at Sun, 9 Feb 2025 19:52:10 UTC No. 16580560
>>16580552
What does a decimal expression like [math]0.t_1t_2t_3\ldots[/math] mean to you?
Anonymous at Sun, 9 Feb 2025 20:03:16 UTC No. 16580572
>>16580532
>>From the first digit being 0 we have that 1>r
Prove it
Anonymous at Sun, 9 Feb 2025 20:11:15 UTC No. 16580578
>>16580532
the first 2 digits arent the same so they arent equal.
Anonymous at Sun, 9 Feb 2025 20:13:21 UTC No. 16580580
>>16580560
If [math]r \in \mathbb{R}[/math] and [math] r = 0.t_1t_2t_3...[/math] then we have:
[math]1) 1 > r \geq 0[/math]
[math]2) (\forall n>0) \lfloor 10^n r \rfloor = \overline{t_1t_2...t_n} [/math].
Anonymous at Sun, 9 Feb 2025 20:13:54 UTC No. 16580582
>>16580578
Numbers don't need to have the same digits to be equal.
Anonymous at Sun, 9 Feb 2025 20:15:42 UTC No. 16580584
>>16580580
Just tell me how you _define_ an expression like [math]0.t_1t_2t_3\ldots[/math]
Anonymous at Sun, 9 Feb 2025 20:32:47 UTC No. 16580596
>>16580572
>>16580578
>>16580582
Let [math]S[/math] be the set of all digit representations, or more precisely the set of all functions [math]f: \mathbb{Z} \rightarrow \{ 0,1,2,3,4,5,6,7,8,9\}[/math].
Let [math]f: \mathbb{R}_{>0} \rightarrow S[/math] be the function that sends every non-negative real to its digit representation.
Now suppose [math]0.999... = 1[/math].
Now let [math]f(0.999...) = f(1) = g[/math].
We have [math]f(0.999...) = g \Rightarrow g(0) =0.[/math]
Also [math]f(1) = g \Rightarrow g(0) = 1.[/math]
Combining these two facts we have [math]0 =f(0) = 1 \Rightarrow 0 =1[/math], a contradiction.
Anonymous at Sun, 9 Feb 2025 20:32:55 UTC No. 16580597
>>16580582
they do though
Anonymous at Sun, 9 Feb 2025 20:35:30 UTC No. 16580602
>>16580584
The definition I provided in the reply, it also applies iff the sequence of ts does not eventually become only 9s.
Anonymous at Sun, 9 Feb 2025 20:38:10 UTC No. 16580605
>>16580596
>or more precisely the set of all functions f : Z -> {0,1,2,3,4,5,6,7,8,9}
Not every element of this set is a decimal representation of a real number
>Let f:R>0S be the function that sends every non-negative real to its digit representation.
This function is not well defined.
Anonymous at Sun, 9 Feb 2025 20:39:15 UTC No. 16580608
>>16580596
5 and 6 have a contradiction.
Anonymous at Sun, 9 Feb 2025 20:40:27 UTC No. 16580612
>>16580602
Well, there you have it. Your falsehood "[math]r < 1[/math]" is baked in your pseudo-definition. If we instead define [math]0.t_1t_2t_3\ldots[/math] as the series [math]t_1/10 + t_2/10^2 + t_3/10^3 + \cdots[/math], try arriving to the 1) of your "definition" from there.
Anonymous at Sun, 9 Feb 2025 20:42:02 UTC No. 16580614
>>16580602
>decimal representation of a number is defined as the image of the number under the mapping which sends it to its decimal representation
Your definition is circular
Anonymous at Sun, 9 Feb 2025 20:53:11 UTC No. 16580622
>>16580532
[math]
\boxed{0 < p < 1} \\
1 = p + (1-p) ~~~~~~ \overset{1}{ \overbrace{[=====p=====|==(1-p)==]}
\text{divide p using x} ~~~~~~ \overset{1}{ \overbrace{ \underset{p}{[ \underbrace{=====x=====|==(p-x)==}]
\\
\text{solve x and (p-x), when length ratios must be the same} \\
\dfrac{x}{p-x}= \dfrac{p}{1-p} \Rightarrow x- xp = p^2 - xp \Rightarrow \underline{x=p^2} \Rightarrow \underline{(p-x)=p(1-p)} \\
\overset{1}{ \overbrace{ \underset{p}{[ \underbrace{=====p^2=====|==p(1-p)=
\\
\overset{1}{ \overbrace{ \underset{p^2}{[ \underbrace{=====p^3=====|==p^2(1-p
\overset{1}{ \overbrace{ \underset{p^3}{[ \underbrace{=====p^4=====|==p^3(1-p
(1-p)+p(1-p)+p^2(1-p)+p^3(1-p)+ \cdots =1 ~~~~ \left | ~ \times \frac{1}{1-p} \right . \\
1+p+p^2+p^3+ \cdots = \dfrac{1}{1-p}
[/math]
Anonymous at Sun, 9 Feb 2025 21:09:12 UTC No. 16580630
Though nonstandard, >>16580580 is clearly a nicer definition than >>16580612. To see why, we only need to ask the two parties to explain the concepts they incorporated by reference. >>16580580 must explain the floor function and >>16580612 must explain infinite series.
Anonymous at Sun, 9 Feb 2025 21:17:22 UTC No. 16580645
>>16580630
The series is the limit of the partial sums. Explained.
Anonymous at Sun, 9 Feb 2025 21:17:52 UTC No. 16580647
>>16580645
Now explain limit.
Anonymous at Sun, 9 Feb 2025 21:18:46 UTC No. 16580649
>>16580647
Do you know about filters?
Anonymous at Sun, 9 Feb 2025 21:20:26 UTC No. 16580651
>>16580649
You should explain anything that isn't basic set theory or basic arithmetic operations.
Anonymous at Sun, 9 Feb 2025 21:34:54 UTC No. 16580663
>>16580605
>Not every element of this set is a decimal representation of a real number
Yes anon that is indeed what I have been saying all this time.
>This function is not well defined
Let me build a consistent definition:
For every real number [math] r [/math]define the following function [math]f_r : \mathbb{Z} \rightarrow \mathbb{Z}[/math].
[math]f_r(z) = \lfloor 10^z r \rfloor - 10\lfloor 10^{z-1}r \rfloor[/math].
We'll prove that the image of this function is actually [math]\{0,1,2,3,4,5,6,7,8,9\}[/math
Lemma 1: [math]\lfloor a+b \rfloor \geq \lfloor a \rfloor + \lfloor b \rfloor[/math]. Proof: It's obvious.
Using Lemma 1, 9 times we deduce that: [math]\lfloor 10^z r \rfloor - 10\lfloor 10^{z-1}r \rfloor \geq 0[/math].
Lemma 2: [math]\lceil a \rceil + \lceil b \rceil \geq \lceil a+b \rceil[/math]. Proof: It's obvious.
Using Lemma 2, 9 times we deduce that: [math]10(\lfloor 10^{z-1} r \rfloor + 1) = 10 \lceil 10^{z-1} r \rceil \geq \lceil 10^{z} r \rceil > \lfloor 10^{z} r \rfloor [/math].
From the first and last terms in the inequality, we have [math] 10 > \lfloor 10^z r \rfloor - 10\lfloor 10^{z-1}r \rfloor[/math].
So we have that [math](\forall z \in \mathbb{Z})10 > f_r(z) \geq 0[/math] or that [math]f_r(z) \in \{0,1,2,3,4,5,6,7,8,9 \}[/math].
Finally to define our decimal representation function simply assign [math]f(r)=f_r[/math] and we are done.
Anonymous at Sun, 9 Feb 2025 21:42:35 UTC No. 16580668
>>16580608
Yes that was the point of the "proof by contradiction"
>>16580612
Nothing is baked in. That follows from the first digit on the left of the point being 0. Look into the general definition.
>>16580614
Nope not circular, just uses multiplication, powers of 10 and floors and ceilings.
Anonymous at Mon, 10 Feb 2025 07:55:20 UTC No. 16581075
knock yourselves out
Anonymous at Mon, 10 Feb 2025 20:00:08 UTC No. 16581790
as if any of those fancy posts are necessary
and as if anyone wants to wade through all of that nonsense
Anonymous at Wed, 12 Feb 2025 23:58:49 UTC No. 16584126
>>16580532
So I respect the attempt, but you have a couple of tripping points that are maybe leading you astray. First and foremost, we need to understand what a real number actually *is*, or perhaps more precisely, what the *set* of real numbers is. There are several equivalent constructions for the set of real numbers, but I find the most illustrative for this particular example to be the Cauchy construction.
We can define the Real numbers [math] \mathbb{R} [\/math] as the set of all equivalence classes of cauchy sequences [math] (r_n) [/math] of rational numbers, where two rational sequences [math] (t_n), (s_n) [/math] are equivalent precisely if [math] lim_{n \to \infty} t_n - s_n =0 [/math]. I'm probably over-explaining and you know what most of that means, but just so we're on the same page, let's break down the parts of that definition.
- Rational numbers are just fractions of integers (with nonzero denominators)
- Cauchy sequences are just sequences where the terms eventually settle down and stay within an arbitrarily small range. In the world of Reals, Cauchy sequences are the same as convergent sequences, but because rational sequences can converge to irrational numbers, we have a special word for it.
- Equivalence classes are sets that partition a larger set (the set of rational cauchy sequences in this case). We say that two members of the larger set are equivalent if and only if they are representatives (members) of the same equivalence class. These equivalence classes are often defined implicitly by specifying an equivalence relation (in this case the two sequences eventually settle down and stay arbitrarily close together). Every member of the larger set is in **exactly** one equivalence class. (TO BE CONTINUED)
Anonymous at Thu, 13 Feb 2025 00:08:03 UTC No. 16584129
>>16584126
When we write the real number "1" we're actually writing shorthand for "the real number, understood as an equivalence class, which contains the rational Cauchy sequence t_n [math](t_n)=(1, 1, 1, ...) [/math] which just has 1 as every element of the sequence."
Similarly when we write the real number "0.999" we're actually writing shorthand for "the real number, understood as an equivalence class, which contains the rational Cauchy sequence [math](s_n)=(0.9, 0.99, 0.999, ...) [/math] whose nth element is just n digits of 9 after the decimal place" If you prefer, you could write out explicitly that [math] s_n = 1 - 10^{-n} [/math]
But notice the following:
[eqn]
lim_{n\to\infty} t_n - s_n &= lim_{n\to\infty} 1 - (1 - 10^{-n}) \\
&= lim_{n\to\infty} 10^{-n} \\
&= 0
[/eqn]
By definition of the reals, we then see that rational Cauchy sequences [math]t_n[/math] and [math]s_n[/math] are equivalent, and are thus representatives of the same real number. Thus in the world of real numbers, [math]1 = 0.999...[/math]
Anonymous at Thu, 13 Feb 2025 00:18:45 UTC No. 16584137
Oops, never done TeX on 4chan before. Sorry. Here's some typesetting corrections for my explanations
>>16584126
We can define the Real numbers [math] \mathbb{R}[/math] as the set of all equivalence classes of Cauchy sequences [math](r_n)[/math] of rational numbers....
>>16584129
[eqn]
\lim_{n\to\infty} t_n - s_n
= \lim_{n\to\infty} 1 - (1 - 10^{-n})
= \lim_{n\to\infty} 10^{-n}
= 0
[/eqn]
Anonymous at Thu, 13 Feb 2025 09:30:48 UTC No. 16584486
>>16584126
Sorry, today I define real numbers as Dedekind cuts, so you'll need to explain that all over again.
Anonymous at Thu, 13 Feb 2025 11:15:13 UTC No. 16584573
>>16584486
Step 1: Let [math] phi [/math] be the canonical ring isomorphism from the Dedekind construction of [math]\mathbb{R}[/math] to Cauchy construction of [math]\mathbb{R}[/math]
Step 2: Everything I just said.
Anonymous at Thu, 13 Feb 2025 11:16:51 UTC No. 16584577
>>16584573
[math] \phi [/math]...
Fuck
Anonymous at Thu, 13 Feb 2025 11:34:09 UTC No. 16584606
>>16584573
Do you suggest we tell that to eighth graders?
Anonymous at Thu, 13 Feb 2025 17:40:22 UTC No. 16584913
>>16584606
Not unless they ask me how to prove 0;999...=1 under the dedekind cut construction.
For an eighth grader, I'd just do
[eqn] x = 0.999...[/eqn]
[eqn] 10x = 9.999... = 9 + 0.999...[/eqn]
[eqn] 9x = 9 [/eqn]
[eqn] x = 1 [/eqn]
Anonymous at Thu, 13 Feb 2025 19:12:18 UTC No. 16584997
>>16584913
How would you explain the meaning of 0.999... to the eighth grader? If you have not explained that 0.999... has a meaning and is not just a sequence of digits, the eighth grader is justified in believing that 0.999... cannot possibly be the same thing as 1 because they are not the same sequence of digits, and any "proof" they are the same must be somehow flawed.
Anonymous at Thu, 13 Feb 2025 21:16:44 UTC No. 16585101
>>16584997
Thankfully, neither I nor the truth is accountable to an eighth grader. I would give them the simple algebra proof above and then offer handwavey reasons why "two members of the real numbers cannot be infinitely close, but separate numbers"
If they were to demand more rigor than that, I'd give them a bit of reading material to learn how number systems are constructed starting with the naturals and working up to the reals, wish them good luck in their studies, and offer to answer their questions after a few days if they get stuck.
Anonymous at Thu, 13 Feb 2025 21:29:52 UTC No. 16585114
>>16585101
The problem is you expect to be able to teach an eighth grader a "truth" about 0.999... without explaining what 0.999... means. If your words have no meaning, you haven't really taught anything. Wakalixes makes it go.
Anonymous at Thu, 13 Feb 2025 22:16:43 UTC No. 16585167
>>16585114
Hey, you're the one who brought eight graders into this. I would happily explain to any student of any age what 0.999... really means, as I have demonstrated by taking time out of my day here to do the same for you lovely people. Most eight graders don't have the fully developed abstract reasoning skills required to understand formal set-theoretic constructions, epsilon-delta proofs, and quotient rings. I've met kids that do have those skills, but they're a rarity. I don't expect I will be able to teach most of them, but I'll almost always give it a shot because I enjoy watching the love of math bloom in a growing mind.
I'm guessing by the quotations around "truth" that you need more convincing and detailed arguments on the construction of the reals. I strongly recommend Terry Tao's book on real analysis:
https://www.math.unm.edu/~crisp/cou
To build the real numbers from the ground up, you need the material from chapters 2-5. Please feel free to come back in a few days and ask questions if you get stuck :)
Anonymous at Fri, 14 Feb 2025 00:54:05 UTC No. 16585336
>>16584913
can you please prove it in the rationals?, i have never seen someone do it, they always work in the reals, & i'd love to see it done
Anonymous at Fri, 14 Feb 2025 01:25:28 UTC No. 16585355
>>16580582
>Numbers don't need to have the same digits to be equal.
THIS!
Anonymous at Fri, 14 Feb 2025 01:50:48 UTC No. 16585363
>sample materials for asbestos
>threshold is 1% asbestos by mass
>material is 1% asbestos
>material is not considered asbestos-containing
1.49% = 1%. magic
Anonymous at Fri, 14 Feb 2025 05:11:34 UTC No. 16585451
>>16584129
But every number in numerator lower then 10^n is going do zero. So 0=4^n/10^n=4/10.
Anonymous at Fri, 14 Feb 2025 06:15:12 UTC No. 16585499
0.999.. is not congruent to 1, these numbers are obviously different
Anonymous at Fri, 14 Feb 2025 07:40:40 UTC No. 16585539
>>16585167
Did you notice that the only thing "wrong" with OP's post is that it defines infinite decimals in a nonstandard way? (elaborated on in >>16580580) On some other world mathematicians might have adopted OP's convention, and people claiming 0.999... = 1 would be the crackpots. It's pretty obvious if you read through OP's post that he knows how to prove 0.999... = 1 under standard definitions. Responding to that by regurgitating what you learned in college about real numbers is something an LLM would do.
>Hey, you're the one who brought eight graders into this.
I bring up eighth graders because this is approximately when concepts like 0.999... are normally introduced to students, and things should be explained as simply as possible. OP keeps things relatively simple, why can't you?
>Most eight graders don't have the fully developed abstract reasoning skills required to understand formal set-theoretic constructions, epsilon-delta proofs, and quotient rings.
Fortunately none of that is necessary to explain what 0.999... is supposed to represent.
>I'm guessing by the quotations around "truth"
I put quotation marks around truth because that which has no meaning has no truth value.
>>16585336
You bring up an excellent point. There is no need to formally construct the reals to show that 0.999... = 1 because it is just as true in the rationals.
Let's begin with a definition of infinite decimals. An infinite decimal represents the smallest number that as at least as large as every number you can make by truncating the infinite decimal to a finite one. So for example, 0.(3 repeating) is at least as large as all of the numbers 0.3, 0.33, 0.333, and so on ad infinitum. Note that in order for such a number to exist, the set of numbers with the property "at least as large as every number you can make by truncating the infinite decimal to a finite one" needs to have a least element. In the rational numbers, some infinite decimals will be well-defined but not others.
Anonymous at Fri, 14 Feb 2025 07:57:29 UTC No. 16585544
>>16585539 (continued)
So to show that 0.999... = 1 in the rationals, we need to show that the set S of rational numbers at least as large as every member of {0.9, 0.99, 0.999, ...} has a least element, and that least element is 1. It should be clear that 1 is a member of S. Suppose there is a smaller element of S, which we'll call x. We have x < 1 and [math]x \geq 1 - \frac{1}{10^n}[/math] for every positive integer n. It follows that for every positive integer n, [math]1 - x \leq \frac{1}{10^n}[/math], and since 1 - x is positive, [math]\frac{1}{1 - x} \geq 10^n[/math]. Let [math]k = \left\lceil \frac{1}{1 - x} \right\rceil[/math]; then we have [math]k \geq \frac{1}{1 - x} \geq 10^k > k[/math], a contradiction. Therefore 1 is the smallest member of S.
Anonymous at Fri, 14 Feb 2025 07:59:30 UTC No. 16585545
>>16580532
>>16580544
KEK
>implies $1\ne 0.999...$
>SEE $1\ne 0.999...$
🗑️ Anonymous at Fri, 14 Feb 2025 08:31:13 UTC No. 16585556
>>16580532
I think its just that the decimal notation is limited and sometimes its better to represent things different.
Like π instead of 3.14...
Or 1/3 rather than 0.333...
On a certain level both are right.
The 3 at the end of the .333... means it'll never truly be a whole 1/3.
These are just arbitrary rules at the end of the day unrelated to the universe so it doesn't really matter.
Anonymous at Fri, 14 Feb 2025 08:42:44 UTC No. 16585562
>>16585556
>unrelated to the universe
kek mathematicians inventing imaginary numbers instead of just admitting they are dumb
Anonymous at Fri, 14 Feb 2025 17:02:07 UTC No. 16585840
>>16585556
The thing is though, there IS no 3 at the end of the 0.333...
To claim there's one at the end is to assert that there is an end, i.e. that there are finitely many digits. But there are infinitely many digits.
Anonymous at Fri, 14 Feb 2025 17:26:55 UTC No. 16585862
>>16585556
Most teachers in middle/high school don't bother to explain this (a lot of them don't know themselves), but when mathematicians write 0.(3 repeating) what they mean is the smallest number that is at least as large as 0.333...333 written out to any finite number of 3's. All of the terminating decimals which repeat 0.333...333 for a while and stop at some point are less than 1/3, but there's no real number simultaneously at least as large as all of them but also less than 1/3. Such a number would have to be infinitesimally close to 1/3, which means its distance from 1/3 is smaller than 0.000...0001 no matter how many zeros you use. In the real numbers we assume numbers can't be infinitesimally close unless they're the same number, which is an assumption helpful for obtaining solutions to problems like the area of a circle, and pretty much anything you would need to use calculus for.
Anonymous at Fri, 14 Feb 2025 17:36:32 UTC No. 16585868
0.999...=/=1 claim is like the flat earth of mathematics.
Anonymous at Fri, 14 Feb 2025 17:52:01 UTC No. 16585878
>>16585539
>Did you notice that the only thing "wrong" with OP's post is that it defines infinite decimals in a nonstandard way?
A proof that starts with an inconsistent set of assumptions with the context can prove literally anything. They describe infinite decimals in a way that simply doesn't behave like real numbers.
>It's pretty obvious if you read through OP's post that he knows how to prove 0.999... = 1 under standard definitions.
This is not obvious to me at all.
>Fortunately none of that is necessary to explain what 0.999... is supposed to represent
** Proceeds to start an (admittedly good) epsilon-n proof in the same message**
>I bring up eighth graders because this is approximately when concepts like 0.999... are normally introduced to students, and things should be explained as simply as possible. OP keeps things relatively simple, why can't you?
I can own up to the fact I included more theory than was strictly necessary to prove 0.999... = 0. It did occur to me to show this was true in the rationals, but OP's post was about real numbers, so I chose that direction.
I have had professors describe two kinds of mathematicians: those that prefer to solve an individual problem quickly and directly, and those who prefer to slowly lay the groundwork that trivializes whole classes of problems. If you and I were trying to reach a tree on the opposite side of a deep chasm, your explanation is more akin to throwing a grappling hook towards the tree and swinging across. Mine is more akin to filling the chasm with gravel and walking across. Both approaches have their merits.
> Responding to that by regurgitating what you learned in college about real numbers is something an LLM would do.
Seems uncalled for. It's not like I'm spewing copypasta here. What I learned about real numbers in college was a skillset earned through long hours of practice. I'm not reciting proofs; I'm reconstructing them from base principles.
Anonymous at Fri, 14 Feb 2025 17:56:34 UTC No. 16585882
>>16585878
>I can own up to the fact I included more theory than was strictly necessary to prove 0.999... = 0.
Strike that, I meant 0.999 = 1
🗑️ Anonymous at Fri, 14 Feb 2025 18:38:46 UTC No. 16585902
>>16585862
>some infinite are larger than other infinities
[math] 0.999\ldots = \sum_{n=1}^{\infty} \frac{9}{10^n} [/math]
[math] 0.000\ldots1 = \sum_{n=1}^{\infty} \frac{1}{10^{n+1}} [/math]
but what if [eqn]
0.000\ldots1 = \sum_{n=1}^{\infty} \frac{1}{10^{n^n}}
[/eqn]
Anonymous at Fri, 14 Feb 2025 20:18:20 UTC No. 16586002
>>16585878
>They describe infinite decimals in a way that simply doesn't behave like real numbers.
You are incorrect here. Read OP's post as well as the clarifying post >>16580580 more carefully and think it through. OP is using exactly the same real numbers everyone else uses. The difference is that OP is using a different definition of what an infinite decimal denotes than the standard one. OP's definition gives every non-negative real number a unique infinite decimal representation, but it does so at the cost of leaving infinite decimals like 0.999... undefined.
>** Proceeds to start an (admittedly good) epsilon-n proof in the same message**
You need to use a bit more theory to formally prove 0.999... = 1 than to merely give it a meaning. But even if you just get the meaning across, that is often enough for someone to see intuitively why the equality is true.
Anonymous at Fri, 14 Feb 2025 20:22:04 UTC No. 16586006
>>16585878
>I have had professors describe two kinds of mathematicians: those that prefer to solve an individual problem quickly and directly, and those who prefer to slowly lay the groundwork that trivializes whole classes of problems. If you and I were trying to reach a tree on the opposite side of a deep chasm, your explanation is more akin to throwing a grappling hook towards the tree and swinging across. Mine is more akin to filling the chasm with gravel and walking across. Both approaches have their merits.
I'd agree with that, but I've found that when teaching younger students (below the college level), the grappling hook approach is generally superior, and even in areas that we later fill in with gravel, doing things simply and concretely first helps people understand the meaning of the theory they learn later. If someone's never counted on their fingers, they'll have a hard time understanding what they're doing when they use an algorithm to add two numbers or learn about the commutative property.
I think the same applies to communicating with the public, most of whom are not at the level of undergrad math majors. Every additional bit of theory that you use makes your post incomprehensible to a large chunk of the audience.
Another thing about epsilon-n. I agree that most eighth graders would have trouble understanding the general idea, but you can use it in a way they would understand by helpfully filling in concrete values for epsilon and n. For example, if I have an infinite decimal that starts with the digits 5.943, what's the maximum possible distance between 5.943 and the number my infinite decimal represents? The answer is 0.001. The notion that this works no matter what place value you truncate the infinite decimal at can be used as a definition of what infinite decimals mean.
Anonymous at Fri, 14 Feb 2025 21:23:55 UTC No. 16586042
I wonder if the public would understand infinite decimals better if math classes spent some time learning how to perform basic arithmetic operations on them. They do this every other time they introduce a new type of number or even a new representation of numbers. There are lessons on adding fractions, adding mixed numbers, adding decimals, adding negative numbers, and eventually adding complex numbers. Same with subtracting, multiplying, and dividing them. But infinite decimals are left out. Students are left to guess, and they often guess wrong. Ask someone what [math]0.\overparen{6}[/math] times 4 is, and you'll often get the answer [math]2.\overparen{4}[/math].
Anonymous at Fri, 14 Feb 2025 21:24:56 UTC No. 16586043
>>16586042
One roadblock is that you can't always do arithmetic operations on standard infinite decimals, even in a streaming fashion where you only compute a finite portion of the result. An easy example is taking an infinite decimal representing [math]\sqrt{2}[/math] and trying to multiply it by itself. No matter how many digits of [math]\sqrt{2}[/math] you go out to, as long as you only have finitely many digits, you will never have enough information from the digits you've computed to decide whether the resulting number is less than, equal to, or greater than 2, and thus you never know if your result should start with 1 or 2.
There are minor modifications we can make to the infinite decimal representation that fix this. One of them is allowing negative digits. That is, in base 10, use digits -9 through 9 instead of 0 through 9. Negative digits can be represented by putting a line over the digit. This conflicts with the usual notation (at least in the US) for repeating decimals, but that can be replaced with an arc or overbracket, which I think is clearer anyway. Apparently the overline notation for repeated decimals isn't universal anyway; spics use the arc, while britbongs use dots.
But maybe introducing negative digits is overkill. It might be simpler just to teach interval arithmetic, without worrying about having students convert the stream of intervals they get from using improving precision back into an infinite decimal.
The bigger problem is that it would be tedious and the students would hate it, and parents would come in to complain too. They've already stopped teaching kids to compute square roots, and long division is only hanging on because of the excuse "we need this to explain polynomial long division." I think this is a bad trend, not because anyone might have to do long division on paper at their job, but because it turns math class into the church of the calculator.
Anonymous at Fri, 14 Feb 2025 21:35:51 UTC No. 16586049
Here's a picture illustrating 0.999... or [math]0.\overparen{9}[/math] using nested intervals. [math]0.\overparen{9}[/math] is the number that is between 0 and 1 inclusive, between 0.9 and 1.0 inclusive, between 0.99 and 1.00 inclusive, and so on indefinitely. It's easy to see these intervals converge on 1.
We can use these intervals to do computations with real numbers. If I know a real number is in the intervals [0,1], [0.9,1.0], [0.99,1.00], [0,999,1.000], ..., then when I multiply it by 10 I know the product is in the intervals [0,10], [9,10], [9.9, 10.0], [9.99, 10.00], ..., and this justifies the assertion that [math]0.\overparen{9} \times 10 = 9.\overparen{9}[/math].
We can do the same sort of computation with [math]0.\overparen{6} \times 4[/math]
[0.6, 0.7] x 4 = [2.4, 2.8]
[0.66, 0.67] x 4 = [2.64, 2.68]
[0.666, 0.667] x 4 = [2.664, 2.668]
[0.6666, 0.6667] x 4 = [2.6664, 2.6668]
from which you can intuit that the correct product is [math]2.\overparen{6}[/math].
Anonymous at Fri, 14 Feb 2025 21:57:31 UTC No. 16586073
>>16580532
(0).(9)(9)(12)(9)... > (1).(0)(0)(0)(0)...
Anonymous at Fri, 14 Feb 2025 23:05:58 UTC No. 16586126
>Can it be produced using a real operation?
Yes. Multiply decimal 1/3 by 3.
>Is the result viewed as nonsense?
Yes.
>but we have to actually expend effort to prove that nonsense stemming form a clerical error is in fact nonsense
Infinity is an artificial mathematical construct. It does not belong in your results.
Anonymous at Fri, 14 Feb 2025 23:10:33 UTC No. 16586131
>>16586126
>Multiply ***decimal*** 1/3 by 3.
you can't
Anonymous at Sat, 15 Feb 2025 02:03:04 UTC No. 16586239
>>16580532
>From the first digit being 0 we have that 1 > r
This is the dumbest thing I've ever heard
Anonymous at Sat, 15 Feb 2025 02:10:40 UTC No. 16586241
>>16586239
what is this logical fallacy called?
Anonymous at Sat, 15 Feb 2025 16:39:13 UTC No. 16586835
>>16586241
Pegging the question
Anonymous at Sat, 15 Feb 2025 21:26:14 UTC No. 16587221
>>16585868
no it's not nearly as bad
Anonymous at Sat, 15 Feb 2025 23:45:00 UTC No. 16587375
>>16585539
>An infinite decimal represents the smallest number that as at least as large as every number you can make by truncating the infinite decimal to a finite one.
Uhh you can't take the minumum of a random set of real numbers anon! That's not well defined
Anonymous at Sat, 15 Feb 2025 23:50:18 UTC No. 16587378
>>16586126
>Yes. Multiply decimal 1/3 by 3.
What does that even mean anon, I know how to multiply real numbers not decimals, you'll have to provide a definition.
Anonymous at Sun, 16 Feb 2025 00:42:37 UTC No. 16587413
>>16587378
>What does that even mean anon
Within base-10, multiply the result of 1÷3 by 3.
>I know how to multiply real numbers
clearly not, since 1÷3 is real.
Anonymous at Sun, 16 Feb 2025 00:43:45 UTC No. 16587415
>>16586131
Why would what base you are in determine whether or not a value can be multiplied by another value?
Anonymous at Sun, 16 Feb 2025 01:02:58 UTC No. 16587430
>>16587415
Why would the post I replied to specify the base if it weren't supposed to be somehow determinative? Presumably the intent had something to do with coprimality and the stale red herring about completing an infinite number of operations.
Anonymous at Sun, 16 Feb 2025 01:12:35 UTC No. 16587437
>>16587413
Show me how you multiply two decimal representations that is two functions f, g : Z -> {0,1,2,3,4,5,6,7,8,9}.
Anonymous at Sun, 16 Feb 2025 01:17:13 UTC No. 16587438
>>16587430
>Why would the post I replied to specify the base
Because they were trying to get you to .999... and the result of 1÷3 multiplied by 3 in base-10 is not written that way in other bases?
For instance, in base-9, it would be .3*3=1
Seems highly relevant, though not at all determinative.
Now again, why would what base you are in determine whether or not a value could be multiplied. You didn't answer the question.
>the stale red herring about completing an infinite number of operations
It's a single operation.
>>16587437
>Show me how you multiply two decimal representations that is two functions f, g : Z -> {0,1,2,3,4,5,6,7,8,9}
Why?
Anonymous at Sun, 16 Feb 2025 01:38:08 UTC No. 16587456
>>16587438
>>Why would the post I replied to specify the base
>Because they were trying to get you to .999... and the result of 1÷3 multiplied by 3 in base-10 is not written that way in other bases?
>For instance, in base-9, it would be .3*3=1
>Seems highly relevant, though not at all determinative.
>>the stale red herring about completing an infinite number of operations
>It's a single operation.
So then the post I originally replied to and your question are both soliciting commentary on a baby-brained vacuous truth. Lol okay.
>Now again, why would what base you are in determine whether or not a value could be multiplied. You didn't answer the question.
Why would I answer such a dumb, fart-sniffing question?
Anonymous at Sun, 16 Feb 2025 02:22:24 UTC No. 16587489
>>16587456
>Why would I answer such a dumb, fart-sniffing question?
Because it directly relates to and solves your confusion, unless you're going to claim 1/3 is unsolvable in base-10 or something equally insane.
You seem to like asking but hate answering questions.
Anonymous at Sun, 16 Feb 2025 03:47:28 UTC No. 16587552
>>16587438
You randomly claimed that 0.333... x 3.000... = 0.999... where does this claim come from?
Anonymous at Sun, 16 Feb 2025 04:08:46 UTC No. 16587565
>>16587552
>where does this claim come from?
.333...+.333...+.333...=.999... and the definition of multiplication as iterative addition. Not sure what you want from me.
Have you answered a single questions or is JAQing off all you can do?
Anonymous at Sun, 16 Feb 2025 04:26:45 UTC No. 16587571
>>16586126
>Multiply decimal 1/3 by 3.
This purported proof for 0.999... = 1 is silly. You're relying on the same guesses that make little kids think that [math]0.\overparen{6} \times 4 = 2.\overparen{4}[/math]. If we do just a tiny bit better, and include carrying in our guess of how to extend arithmetic algorithms without justification from finite to infinite decimals, then it's ambiguous whether there's a 1 carried in and out at each step, so you don't know if the result is supposed to be [math]0.\overparen{9}[/math], the result without the carry, or [math]1.\overparen{0}[/math], the result with the carry. Fortunately for our guess, both possible results are the equal, but the calculation doesn't do anything to prove that fact.
When we do it with interval arithmetic, we encounter the same sort of issue:
[0.3, 0.4] x 3 = [0.9, 1.2]
[0.33, 0.34] x 3 = [0.99, 1.02]
[0.333, 0.334] x 3 = [0.999, 1.002]
[0.3333, 0.3334] x 3 = [0.9999, 1.0002]
and so on.
If we didn't know in advance that all the digits were 3s -- imagine we instead obtained this infinite decimal from a calculation that wasn't an obvious repeating loop -- then at no point in the calculation process would we know whether the product was greater, equal to, or less than 1, and we'd never be able to choose the first digit.
The signed-digit version fares better, but doesn't give the answer you want:
[0.2, 0.4] x 3 = [0.6, 1.2] [math]\subset[/math] [0, 2] = [math][1.\overparen{\overline{9}}, 1.\overparen{9}][/math] -> we can choose 1 as the first digit
[0.32, 0.34] x 3 = [0.96, 1.02] [math]\subset[/math] [0.9, 1.1] -> we can choose 1.0 as the first two digits
[0.332, 0.334] x 3 = [0.996, 1.002] [math]\subset[/math] [0.99, 1.01] -> we can choose 1.00 as the first three digits
[0.3332, 0.3334] x 3 = [0.9996, 1.0002] [math]\subset[/math] [0.999, 1.001] -> we can choose 1.000 as the first four digits
Anonymous at Sun, 16 Feb 2025 04:55:12 UTC No. 16587582
>>16587565
How do you add these decimal representations? What does 0.333...+0.333 mean?
Anonymous at Sun, 16 Feb 2025 04:57:17 UTC No. 16587584
>>16587565
>Have you answered a single questions or is JAQing off all you can do?
I'm not the fart sniffing guy
Anonymous at Sun, 16 Feb 2025 11:09:58 UTC No. 16587766
>>16587582
>How do you add these decimal representations?
You just do. The successor function doesn't apply to decimals. You literally can't make it simpler.
Didn't you literally claim to know how to multiply real numbers? How are you having trouble with basic addition? This is π+π=τ level shit.
Anonymous at Sun, 16 Feb 2025 11:29:06 UTC No. 16587776
>>16580532
>From the first digit being 0 we have that 1>r
So the rest of the proof is redundant
Barkon !8v8vr3ErDk at Sun, 16 Feb 2025 11:30:47 UTC No. 16587779
>>16587584
Yes you are. You love farts. That's all you do, sniff farts. You're the world's fart sniffer.
Anonymous at Sun, 16 Feb 2025 12:34:46 UTC No. 16587822
>>16587766
>You just do.
Ok so you're just retarded. Could have said that from the beginning anon, why waste all our time?
Anonymous at Sun, 16 Feb 2025 17:56:47 UTC No. 16588147
>>16587489
What confusion? I'm not any of the .999...=1 deniers here, you fucking idiot.
YOUR confusion seems to be that you think it's pedagogically useful to explain how a multiplicative inverse works when the objection is obviously representational.
Anonymous at Sun, 16 Feb 2025 18:34:58 UTC No. 16588210
>>16587552
0.333... means the threes never end or change
3.000... means the zeros never end or change
anything times zero is zero, so you can ignore everything to the right of the decimal in 3.000...
all that's left is (0*3).(3*3)(3*3)(3*3).... = 0.999... meaning the nines never end or change
Anonymous at Sun, 16 Feb 2025 18:37:44 UTC No. 16588214
>>16587822
>Ok so you're just retarded.
I can't help you if you don't know how to add .3 repeating and .3 repeating. It's not exactly different from adding 3 and 3.
Not my fault you never got a first grade education cause you were raised in a cult or something.
Anonymous at Sun, 16 Feb 2025 19:07:30 UTC No. 16588270
>>16587571
Lol what. Notwithstanding the fact that 1/3 * 3 = 1 is indeed a silly and vacuous contrivance, there's no ambiguity, at all, about any digit, ever, in the algorithm for 1/3 = 0.333... or for 0.333... * 3 = 0.999...
Anonymous at Sun, 16 Feb 2025 19:22:36 UTC No. 16588303
>>16580532
You're right. 0.999... is a series of symbols we use to represent a real number. 1/3 is another valid series of symbols we use to represent the same number.
Anonymous at Sun, 16 Feb 2025 19:28:59 UTC No. 16588315
>>16588303
That should be 3 * 1/3
Anonymous at Sun, 16 Feb 2025 19:59:42 UTC No. 16588368
>>16588270
What algorithm are you using to obtain 0.333... * 3 = 0.999...? Show me multiplying 0.444... * 7 by the same algorithm.
Anonymous at Sun, 16 Feb 2025 20:11:03 UTC No. 16588384
>>16588368
Yeah, I get that you're trying to shovel in a bunch of generic horseshit that has nothing to do with algorithmically issuing a decimal representation of 1 / 3 * 3.
The algorithm to obtain 0.333... * 3 = 0.999.. is
10 PRINT 0*3
20 PRINT "."
30 PRINT 3*3
40 GOTO 30
Anonymous at Sun, 16 Feb 2025 20:14:04 UTC No. 16588389
>>16588384
10 PRINT 0*7
20 PRINT "."
30 PRINT 4*7
40 GOTO 30
It seems your algorithm gives 0.444... * 7 = 0.282828...
Your algorithm is wrong.
Anonymous at Sun, 16 Feb 2025 20:16:25 UTC No. 16588393
>>16588389
That's your algorithm, not mine, you fucking moron.
Mine is for 0.333... * 3 = 0.999...
Yours looks like something a monkey would pull out its ass and throw at the wall
Anonymous at Sun, 16 Feb 2025 22:44:26 UTC No. 16588527
>>16588147
>the objection is obviously representational.
It obviously isn't because representations are constructed and arbitrary so anyone making such an objection would be out of their god damn mind.
Anonymous at Sun, 16 Feb 2025 23:06:46 UTC No. 16588541
>>16588527
Have you noticed that there are no threads about multiplying "decimal" 1/2 * 2 = 1? Could it be that representation is the source of the confusion? No, it "obviously isn't" because obviously everyone always randomly chooses a denominator that doesn't divide 10.
Anonymous at Mon, 17 Feb 2025 02:12:10 UTC No. 16588655
>>16588210
>(0*3).(3*3)(3*3)(3*3).... = 0.999
This statement alone should qualify you for government benefits. So by the same logic we have 0.5 × 7 = 0.(35)
Anonymous at Mon, 17 Feb 2025 02:13:11 UTC No. 16588657
>>16588214
If it is so simple you should have no problem explaining it.
Anonymous at Mon, 17 Feb 2025 02:15:46 UTC No. 16588659
>>16588541
>that there are no threads about multiplying "decimal" 1/2 * 2 = 1?
These tards would also claim that 0.4999... = 1/2
Anonymous at Mon, 17 Feb 2025 02:18:06 UTC No. 16588660
>>16588541
>Have you noticed that there are no threads about multiplying "decimal" 1/2 * 2 = 1?
No. I tend to not dwell on a lack of things.
>Could it be that representation is the source of the confusion?
No? Because ".999... means 1" is generally a frequent reply to this bullshit. There would legitimately be nothing to contest and no confusion if that were the issue. There are literally an infinite number of ways to write 1, eg 2/2, so anyone getting upset over a representation they're unfamiliar with would be mentally ill. People are contesting the math, not the notation, a thing you cannot contest. Don't accuse people of being crazy when they could simply be ignorant. It's rude.
Anonymous at Mon, 17 Feb 2025 02:22:01 UTC No. 16588663
>>16580532
Well said Brother.
Soon we of the one TRUE FINITE FAITH will reclaim all of mathematics for the righteous. Personally I cant wait to see those GOD CURSED INFINITY LOVING SODOMITES being marched off to reeducation camps.
PRAISE GOD in his finite glory!
Amen.
Anonymous at Mon, 17 Feb 2025 02:27:20 UTC No. 16588667
>>16588660
>so anyone getting upset over a representation they're unfamiliar with would be mentally ill
You are the mentally ill one anon. I want a world in which every real number has an unique decimal representation, so that things like "what's the first digit after the decimal point of 1?" make sense. You want some retarded world in which some real numbers ( the ones of the form n*10^z, n and z whole numbers) have 2 decimal representations, for no good reason other than you some some yt video about it.
>>16588663
I have no problem with infinity, 1÷3 does indeed equal 0.333...
Anonymous at Mon, 17 Feb 2025 03:13:39 UTC No. 16588709
>>16588663
>I have no problem with infinity
Aha! You fell into my trap, BLASPHEMER! I have FLUSHED out a GOD-CURSED INFINITY LOVING SODOMITE!
You will burn in the discrete flames of HELL for a finite duration.
Amen.
Anonymous at Mon, 17 Feb 2025 03:48:27 UTC No. 16588742
>>16588655
I'd like to file a surjective claim for your neetbux in australia or wherever you live since you're clearly an imbecile.
>>16588659
How does 0.499... arise from "in decimal" 1/2 * 2 = 1
>>16588660
>No. I tend to not dwell on a lack of things.
You lack balls and you lack a dick. Maybe you should file a complaint to the bill murray office of the ghostbusters department.
>People are contesting the math, not the notation, a thing you cannot contest.
People are literally contesting the notation, not the math. You're blind and dumb and retarded, meaning that everyone who is both blind and dumb, or blind and retarded, or dumb and retarded, all of us, are better human beings than you are; we're simply more constructive and not a bunch of dean-ball-licking retarded gimps.
Tennis now.
Anonymous at Mon, 17 Feb 2025 04:54:19 UTC No. 16588797
>>16588709
>>16588663
I'm literally blacker than a soot butterfly and this poster is a homophobic nigger who should be banned from posting.
Anonymous at Mon, 17 Feb 2025 05:58:02 UTC No. 16588843
Honestly the construction OP is trolling everyone with is a useful one. Does it have a standard name? The most obvious thing I can think of would be calling it the canonical decimal representation of a number. A quick search shows at least one analysis textbook, written by Jiongmin Yong, using that exact terminology.
Anonymous at Mon, 17 Feb 2025 06:21:18 UTC No. 16588862
Canonical in the sense that we're choosing a preferred decimal representation [math]1.\overparen{0}[/math] out of all the possible decimal representations which include [math]1.\overparen{0}[/math] itself, [math]0.\overparen{9}[/math], and also things like [math]1\overline{9}.\overparen{0}[/
Anonymous at Mon, 17 Feb 2025 07:16:31 UTC No. 16588900
>>16588797
SILENCE HERETIC!
You will have your chance to plead your case before the HOLY MATHEMATICAL INQUISITION. Then once your mortal body has been purged at the stake you will face GOD!
Then HE will cast you down into the fiery pits of HELL where you can reflect upon the mathematical SINS which led to your DAMNATION!
Brothers and Sisters of the ONE TRUE FINITE FAITH, we must root out HERESY wherever it is found.
BECAUSE GOD WILLS IT!
DEUS VULT!
Say it with me Brothers and Sisters!
DEUS VULT!
Amen.
Anonymous at Mon, 17 Feb 2025 07:37:22 UTC No. 16588907
>>16588862
* [math]1\overline{9}.\overparen{0}[/
4chan a gay
Anonymous at Mon, 17 Feb 2025 08:13:03 UTC No. 16588941
>>16588907
Use the preview latex feature
Anonymous at Mon, 17 Feb 2025 08:19:40 UTC No. 16588950
>>16588941
The LaTeX is fine, the issue is 4chan inserting <wbr> tags into text and then not being able to find the [math] and [/math] tags. Ancient bug on 4chan's part.
[math]1\overline{ 9 }.\overparen{ 0 }[/math]
[math]2.\overparen{ \overline{9} }[/math]
Anonymous at Mon, 17 Feb 2025 08:39:13 UTC No. 16588969
>>16588950
>the [math] and [/math] tags
That one was my fault but you can probably figure out what I meant.
Anonymous at Mon, 17 Feb 2025 11:00:03 UTC No. 16589073
>>16589042
Is it odd or even?
Anonymous at Mon, 17 Feb 2025 11:03:40 UTC No. 16589077
>>16589073
Those are properties of Integers, not number with periodic evolution.
Anonymous at Mon, 17 Feb 2025 11:11:51 UTC No. 16589087
>>16589077
How can numbers which are composed of numbers lose properties?
Anonymous at Mon, 17 Feb 2025 11:12:48 UTC No. 16589091
>>16589087
is i > 0 or i < 0?
Anonymous at Mon, 17 Feb 2025 11:16:12 UTC No. 16589093
>>16589091
Red herring, answer the question.
Anonymous at Mon, 17 Feb 2025 11:17:12 UTC No. 16589095
>>16589093
I have, you're just too obtuse to understand
Anonymous at Mon, 17 Feb 2025 11:27:22 UTC No. 16589100
>>16589087
Simply.
Anonymous at Mon, 17 Feb 2025 11:33:24 UTC No. 16589108
>>16589095
What operation on a number causes it to lose even and oddness?
Anonymous at Mon, 17 Feb 2025 11:38:21 UTC No. 16589109
>>16589108
one that maps the given number outside the set of integers
Anonymous at Mon, 17 Feb 2025 11:41:36 UTC No. 16589112
>>16589108
The even numbers are a subset of the whole numbers hence every even number is whole.
Anonymous at Mon, 17 Feb 2025 11:48:49 UTC No. 16589117
>>16589109
Meaningless. Every number in a digit string is either odd or even. The terminal digit defines the property. You are welcome to invent your own mathematics, but you are going to have to define all of your terms so people don't think you are a sperging autist.
Anonymous at Tue, 18 Feb 2025 00:03:14 UTC No. 16589705
>>16589042
Do you know what closed and open intervals are?
Anonymous at Tue, 18 Feb 2025 16:34:14 UTC No. 16590367
>>16589705
Tell me.
Anonymous at Tue, 18 Feb 2025 16:53:27 UTC No. 16590391
>>16580532
1/3 = 0.333...
1 = 3/3 = 0.999...
Anonymous at Tue, 18 Feb 2025 16:55:02 UTC No. 16590397
>>16580532
What is the number between 1 and .999...
Anonymous at Tue, 18 Feb 2025 17:01:31 UTC No. 16590404
>>16590367
The opposite of how they were used in that post.
Anonymous at Tue, 18 Feb 2025 17:07:05 UTC No. 16590416
>>16590404
No, interval is set of all points between real numbers, and is used as such.
Anonymous at Tue, 18 Feb 2025 17:10:06 UTC No. 16590422
>>16590416
And is the interval between 0 and 1 that includes 0 and 1 called "open" or "closed"?
Anonymous at Tue, 18 Feb 2025 17:23:53 UTC No. 16590446
>>16590422
Honestly I'm not sure now... Maybe it's reversed. My memory sucks, and I haven't seen math in 16 years.
Anonymous at Wed, 19 Feb 2025 00:29:38 UTC No. 16590926
>>16588900
the only post worth reading ITT
Anonymous at Wed, 19 Feb 2025 01:57:41 UTC No. 16590990
>>16580532
0.999... has infinite digits, and infinity is not a real number. Thus 0.999... is not a real number. QED.
Anonymous at Wed, 19 Feb 2025 03:08:34 UTC No. 16591042
>>16590990
All I see are 4 digits, a decimal point, and an ellipsis.
Anonymous at Wed, 19 Feb 2025 07:36:55 UTC No. 16591160
>>16580663
>[math]f_r(z)=\lfloor 10^zr\rfloor-10\lfloor10^{z-1}r\rfl
I agree your function is well-defined and gives a decimal representation of a real number.
Let [math]r=0.999...[/math]
[math]f_r(0)=\lfloor 10^0r\rfloor-10 \lfloor 10^{-1}r\rfloor[/math]
[math]f_r(0)=\lfloor r\rfloor-10 \lfloor 10^{-1}r\rfloor[/math]
In both cases, if [math]0<r<1[/math] or [math]r=1[/math] we have [math]\lfloor 10^{-1}r\rfloor=0[/math]
So [math]f_r(0)=\lfloor r\rfloor[/math]. But now we ask is [math]\lfloor r\rfloor=1[/math] or [math]\lfloor r\rfloor=0[/math]?
Evidently we want a good definition for the floor function. The common definition I have seen is [math] \lfloor x \rfloor = \max \{m\in\mathbb{Z} | m\leq z \}[/math]. However to judge our case we must assert if [math]1\leq r[/math] which is our original contention.
As far as I can see we have to use a new definition of decimal representation that does not use floor or ceiling.
Anonymous at Wed, 19 Feb 2025 20:50:50 UTC No. 16591797
>>16590990
I sense you have seen a glimmer of GOD'S TRUTH, young whippersnapper.
For not only is infinity not a number it simply DOES NOT EXIST! Just like Unicorns and women with penises, it has no place in MATHEMATICS!
The mere mention of it is an affront to all that is SACRED and HOLY!
I urge you to join the ONE TRUE FINITE FAITH and help us retake MATHEMATICS for the RIGHTEOUS!
We shall burn every text book that mentions infinity!
We shall cleanse the world of HERESY with HOLY FIRE!
WE SHALL RETAKE CONSTANTINOPLE! ( just as an added bonus, not directly related to Mathematics )
ITS GOD'S WILL!
DEUS VULT!
DEUS VULT!
DEUS VULT!
Praise GOD!
Amen.
Anonymous at Thu, 20 Feb 2025 03:59:08 UTC No. 16592138
>>16591797
You're so retarded that I'll ask how many letters are involved in DEUS VULT!
Anonymous at Thu, 20 Feb 2025 08:04:38 UTC No. 16592284
>>16592138
You have turned away from the ONE TRUE FINITE FAITH. That makes you a GOD CURSED SODOMIZING HERETIC!
But I will pray to GOD to have mercy on your SOUL. After your corporal body has been PURGED OF SIN at the STAKE!
"Oh merciful GOD, please forgive this crawling scum for the errors of his ways, for he is mentally deficient and knows not the manner of his gross stupidity"
I ask this in the name of the ONE TRUE FINITE FAITH.
Amen.
Anonymous at Thu, 20 Feb 2025 08:31:09 UTC No. 16592300
>>16591160
>Let r=0.999...
Ahhh there is your problem. What does this mean? Which dedekind cut is this? We need to start with a real number not work backwards from a decimal representation.
Anonymous at Fri, 21 Feb 2025 01:57:42 UTC No. 16594296
>>16587375
Bro, this is literally one of the defining properties of the real numbers: the least upper bound (LUB) property. Every set of real numbers with an upper bound has a least upper bound.
Anonymous at Fri, 21 Feb 2025 02:11:17 UTC No. 16594313
>>16588667
I'd like a billion shmeckles, a pound of spice melange from Dune, and a harem of genuine succubi. Wanting a thing does not guarantee that it exists. If it makes you feel any better, there are only [math]\aleph_0 [/math] real numbers with non-unique decimal representations. Given that the cardinality of [math]\mathbb{R}[/math] is [math]2^{\aleph_0}[/math], the percentage of reals with unique representations is 100%. This percentage calculation only works because 0.999... = 1
Anonymous at Fri, 21 Feb 2025 02:18:44 UTC No. 16594322
>>16588843
Canonical decimal representation is good. Finitely terminating decimal representation is also good, and perhaps a bit more descriptive, since all of the reals with non-unique decimal representations only have 2 such representations: one ending in (XYZ)999... and one ending in (XYZ+1)000.
Anonymous at Fri, 21 Feb 2025 02:22:03 UTC No. 16594326
>>16589117
There is no terminal digit for 0.999... Since there is no terminal element of the whole numbers. The 9s just keep going. By your own logic, even-ness and odd-ness are not well-defined for non finitely-terminating decimals.
Anonymous at Fri, 21 Feb 2025 02:24:59 UTC No. 16594329
>>16592300
Bro it's not worth your time. I already gave them the cauchy construction for the reals and these bitches are just too dumb/stubborn to engage with it.