Anonymous at Sat, 15 Feb 2025 04:01:12 UTC No. 16586289

You are retarded

Stop guessing start learning at Sat, 15 Feb 2025 04:07:04 UTC No. 16586295

>>16586274
Why do idiots here like to argue the mathematical nuance in some conjecture To make themselves sound smarter than the whole scientific community.

Like of course your "technically" correctly but you miss the whole point that when it's applied for all intents and purposes the consensus agrees implicitly that it's 1.

Anonymous at Sat, 15 Feb 2025 04:48:46 UTC No. 16586318

[math]\infty[/math] is not in [math]\mathbb{R}[/math] so the limit is indeed not defined there, unlike 0.9, 0.99, 0.999... You can have limits that equal [math]\pm\infty[/math] in the extended reals, but you have to use a topology where [math](a,\infty][/math] is an open set, so you cannot use the normal metric-space-specific definition that works for [math]\mathbb{R}[/math]. See https://en.wikipedia.org/wiki/Limit_(mathematics)#Topological_space for the more general definition.

Anonymous at Sat, 15 Feb 2025 06:16:35 UTC No. 16586366

>>16586318
retard

Anonymous at Sat, 15 Feb 2025 06:40:00 UTC No. 16586385

>>16586274
Fucking retard

Anonymous at Sat, 15 Feb 2025 06:41:04 UTC No. 16586387

>>16586295

1/3 * 3 = 0.333... * 3, you don't need limits for this.

Anonymous at Sat, 15 Feb 2025 07:01:16 UTC No. 16586401

>>16586274
>a quantity that infinitely approaches a certain number is equal to that number
Infinite is not a number.

Anonymous at Sat, 15 Feb 2025 07:02:09 UTC No. 16586402

>>16586318
neat
>>16586387
OP's type of post do test the limits of the board denizen's patience

Anonymous at Sat, 15 Feb 2025 07:03:10 UTC No. 16586403

>>16586401
but it is, one just needs to work in the proper system >>16586318

Anonymous at Sat, 15 Feb 2025 07:05:43 UTC No. 16586408

>>16586274
>Why is lim x=>0 of 1/x equal to infinity
because it isn't
lim needs lim-=lim+
fyi, -inf=/=inf

Anonymous at Sat, 15 Feb 2025 07:16:26 UTC No. 16586421

>>16586274
>0.999... does not equal 1
Correct. The digit representation of a real number cannot end with ...9999... as that would lead to a contradiction. As to the limit of 1/x being infinity, the limit does not "equal" infinity as some function having some value. It just means that 1/x can achieve any large random value past a certain point, that is it.

Anonymous at Sat, 15 Feb 2025 08:05:03 UTC No. 16586436

>>16586274
>0.999... does not equal 1.
What is 1 - 0.999... then?

Anonymous at Sat, 15 Feb 2025 08:37:06 UTC No. 16586455

>>16586436
Nothing? 0.9999.. is not part of R so 1 - 0.999... does not mean anything.

Anonymous at Sat, 15 Feb 2025 09:06:53 UTC No. 16586469

>>16586455
>R is incomplete
what are the implications of this? scientifically speaking of course

Anonymous at Sat, 15 Feb 2025 09:23:59 UTC No. 16586479

>>16586469
>>R is incomplete
Well that's like your opinion man...
The function that takes in a real number and outputs it's decimal representation does not return any representation that:
1) Doesn't start with ...000...
2) Ends in ...999...

Anonymous at Sat, 15 Feb 2025 09:38:44 UTC No. 16586482

>>16586469
>>16586479
[math][/math]
Let [math]f : \mathbb{Z}x\mathbb{R}\rightarrow \{0,1,2,3,4,5,6,7,8,9\}[/math] be the function that returns the digit of a real number at the position of a specified whole number. Then:
[math]f(z,r) = \lfloor 10^z \cdot r \rfloor - 10 \lfloor 10^{z-1} \cdot r \rfloor [/math]
It can be proven that no output will end in ...999...

Anonymous at Sat, 15 Feb 2025 09:48:38 UTC No. 16586486

>>16586479
>Well that's like your opinion man...
No that's your opinion. A cauchy sequence can be made
0.9
0.99
0.999
0.9999
....
You believe the limit point of this sequence is not in R. Therefore R is incomplete, which implies there is some complete superset of R that does contain 0.999...

Anonymous at Sat, 15 Feb 2025 10:46:07 UTC No. 16586511

>>16586486
>You believe the limit point of this sequence is not in R.
Never said that. The sequence is convergent and it's limit is 1.
>0.999...
That's not a real number anon it's just some random symbols, you might as well start thinking about "1++=++1"

Anonymous at Sat, 15 Feb 2025 10:46:54 UTC No. 16586512

>>16586482
>It can be proven that
I accept your concession.

Anonymous at Sat, 15 Feb 2025 10:48:08 UTC No. 16586513

>>16586482
>>16586512
Actually nevermind, that is probably true. The false part is that f gives the unique decimal representation every number.

Anonymous at Sat, 15 Feb 2025 10:53:11 UTC No. 16586516

>>16586512
I never conceded anything I can prove both parts about ...000... and ...999....
>>16586513
>The false part is that f gives the unique decimal representation every number.
Wait what? So how do you define the decimal representations of real numbers?

Anonymous at Sat, 15 Feb 2025 10:58:34 UTC No. 16586522

>>16586274
The point of a limit is to capture the behavior, not the specific result.

For instance, let's say we have a line going at a nice 45 degree angle. At x=1, y=1. At x=2, y=2. And so on. Let's also say this line terminates at 5, so when x>5, y=undefined.

Now, what's the limit as x=>7? When x=7, y=undefined. But we can see that if there was an answer, it'd probably be y=7, so that's what the limit is equal to. Again, if the limit and the actual answer were always the same, we wouldn't need a limit.

In the case of adding decimals to 0.9999, the answer is getting closer and closer to 1, so that's what the limit is equal to. It doesn't matter whether the equation itself actually does so at that point.

In the case of increasing the denominator, the number is getting smaller, so the answer is getting closer and closer to 0. It doesn't matter whether 1/infinity is actually 0 or not, it matters that the tendency is approaching that result.

Anonymous at Sat, 15 Feb 2025 11:05:58 UTC No. 16586526

>>16586522
>In the case of adding decimals to 0.9999, the answer is getting closer and closer to 1, so that's what the limit is equal to. It doesn't matter whether the equation itself actually does so at that point.
That's all true, but that in no way means that 0.999... is a real number.

Anonymous at Sat, 15 Feb 2025 11:09:11 UTC No. 16586529

>>16586516
I don't know the canonical definition, but since you asked me instead of wikipedia I will give you a definition:
A decimal number representation is a map [math]f : \mathbb{Z}\to\{0 ... 9\}[/math] specifying the digit at each decimal place, with the restriction that [math]\exists n. \forall z > n. f(z) = 0 [/math] so that you don't get infinite digits on the left side of the decimal point.

The real number corresponding to a decimal representation [math]f[/math] is the limit of the series [math]\Sigma_{z = -sup(\{z \in \mathbb{Z} | f(z) \neq 0\})}^{\infty} 10^{-z} * 10[/math] counting down from the highest decimal place. Aka the definition that everyone always gives in 0.9999 =/= 1 threads, and also the inverse of your function if you consider it a map [math]\mathbb{R} \to \mathbb{Z} \to \{0...9\}[/math], if you restrict my function's domain to the image of your function (the mapping from decimal to real is not injective, because, for example, 1 and 0.9999... are the same number).

Anonymous at Sat, 15 Feb 2025 11:10:12 UTC No. 16586531

>>16586529
sorry, * 10 should be * f(-z) there, my mistake

Anonymous at Sat, 15 Feb 2025 11:16:39 UTC No. 16586540

>>16586529
>The real number corresponding to a decimal representation f
>f
> is the limit of the series Σ∞z=−sup({z∈Z|f(z)≠0})10−z∗10
>Σ
>z
>=
>−
>s
>u
>p
>(
>{
>z
>∈
>a
>|
>f
>(
>z
>)
>≠
>0
>}
>)
>∞
>10
>−
>z
>∗
>10
> counting down from the highest decimal place. Aka the definition that everyone always gives in 0.9999 =/= 1 threads, and also the inverse of your function if you consider it a map RZ{0...9}
>R
>
>a
>
>{
>0...9
>}
>, if you restrict my function's domain to the image of your function (the mapping from decimal to real is not injective, because, for example, 1 and 0.9999... are the same number).
So you agree with me that 0.999... does not correspond to a real number using the function I provided? Mind you your function works backwards, it assigns to every digit representation some real number, but that is not how real numbers are defined. Real numbers are defined by dedekind cuts and from those definitions we build out digit representations not the other way around.

Anonymous at Sat, 15 Feb 2025 11:36:39 UTC No. 16586544

>>16586540
No, I don't agree. I said that my mapping is the inverse of your function if you restrict the domain to the image of your function. I had to say this because my mapping can map representations not produced by your function to real numbers.

I never said that real numbers are defined as decimal representations, I just gave a mapping that gives a valid number for 0.999.... If you don't agree with how I interpret decimals into reals then I would like to see your mapping.

Anonymous at Sat, 15 Feb 2025 11:53:46 UTC No. 16586548

>>16586544
>If you don't agree with how I interpret decimals into reals then I would like to see your mapping
Yes anon but why should such an unrestricted interpretation even exist???
Does ...1111111.000... mean anything to you? Yes I know it's not part of the set since it's mandated to begin with ...0000 but that is what I'm saying also. My restriction (it's not really a restriction it's just a property of the image set) is literally a symmetry of the ..0000... condition you gave!!! To see this more clearly let's put them side by side and translate them into symbols:
1) [math]\exists z \forall Z Z<z \Rightarrow f(Z) = 0 [/math].
2) [math]\nexists z \forall Z Z>z \Rightarrow f(Z) = 9 [/math].
If you accept the ...000... condition you must see how the ..999... condition is it's natural and obvious counterpart!

Anonymous at Sat, 15 Feb 2025 12:42:19 UTC No. 16586577

>>16586548
The point of the decimal representation is to represent the reals, which are finite but have the completeness property. So I restrict it on the left to keep it finite, but allow an infinite amount of non-zero digits on the right so that you can specify an infinite set that the intended real number is the supremum of. I agree that it's natural to extend it to allow infinity on the left side too, and if you do that then you get the extended real numbers that include [math]\{\infty, -\infty\}[/math] in addition to the ordinary real numbers. You can think of my restriction as just taking the subset that are not infinity.

Anonymous at Sat, 15 Feb 2025 12:45:44 UTC No. 16586580

>>16586577
The "infinite set" being [math]\{0.d_1, 0.d_1d_2, 0.d_1d_2d_3...\}[/math] if [math]d_i[/math] is the ith digit of the decimal representation.

Anonymous at Sat, 15 Feb 2025 12:54:47 UTC No. 16586583

>>16586548
Sorry, I didn't read most of your post so I responded to the wrong argument. The reason I have a restriction on the left to 0 is to keep it finite, and the reason I don't have a restriction for 9s on the right side is because even with infinite 9s you can form a bounded set of rational numbers which you take the supremum to get the real number for.

Anonymous at Sat, 15 Feb 2025 13:08:53 UTC No. 16586597

>>16586577
>>16586580
>>16586583
>I don't want the 999 condition
But why would you do that??? So you can have a ugly non-injective function in which some numbers have 1 representation while others have 2? Mind you not all numbers have a decimal representation ending in ...999... Only those that can be written as z/10^n where z and n are whole numbers? Would you reject the elegant duality of the ...999... and ...000... conditions for this messy and ugly function? It's not too late to accept the truth anon

Anonymous at Sat, 15 Feb 2025 13:19:50 UTC No. 16586610

>>16586597
I don't think you're funny. But if you want to go full dual then let's have the 000... restriction on both sides. If your number is a multiple of a power of 10 then you must end your decimal representation in ... 999 ...

Anonymous at Sat, 15 Feb 2025 13:32:57 UTC No. 16586624

>>16586610
>But if you want to go full dual then let's have the 000
That would not admit any irrational number and a lot of rational numbers. You cannot make root 2 end in ...999... Or even 1/3. You can lead a horse to water but you can't force him to drink.

Anonymous at Sat, 15 Feb 2025 13:38:03 UTC No. 16586630

>>16586624
I don't know if it's dual but this is what I meant
1) [math]\exists z. \forall Z > z. \Rightarrow f(Z) = 0[/math]
2) [math]\nexists z. \forall Z < z. \Rightarrow f(Z) = 0[/math]
Turned the < around in 1) to match my original description

Anonymous at Sat, 15 Feb 2025 13:45:58 UTC No. 16586634

>>16586630
Those 2 conditions don't make any sense.
1) Implies that every number is of the form z*10^p eith z,p whole numbers
2)Implies that every number is infinite

Anonymous at Sat, 15 Feb 2025 13:50:41 UTC No. 16586637

>>16586634
I think maybe I was confusing in >>16586529
[math]f(z)[/math] gives you the value for the [math]10^z[/math] place, but I also used [math]z[/math] from the summation index to mean the negative of those [math]z[/math].

1) is saying that it's finite (eventually there is a max decimal place that is nonzero)
2) is saying that no decimals end in ... 0 ...

Anonymous at Sat, 15 Feb 2025 14:45:32 UTC No. 16586696

>>16586455
Just because 0.999... isn't in the set of reals doesn't mean its not a rational number.

Anonymous at Sat, 15 Feb 2025 14:51:20 UTC No. 16586708

>>16586696
It kinda does anon, all rational numbers are also real numbers.

Anonymous at Sat, 15 Feb 2025 16:39:43 UTC No. 16586836

>>16586708
Alright, then what is the decimal representation of 1/3 ?

Anonymous at Sat, 15 Feb 2025 17:07:08 UTC No. 16586865

>>16586836
it equals 5 in base 150

Anonymous at Sat, 15 Feb 2025 17:30:39 UTC No. 16586893

>>16586865
So then 0.999... and 0.333... are both in the reals.

Anonymous at Sat, 15 Feb 2025 17:49:51 UTC No. 16586917

>>16586893
0.999... and 0.333... does not exist because you cannot group 10 numbers into 3 equal stacks, saying 1/3 = 0.333... is an assumption

Anonymous at Sat, 15 Feb 2025 18:30:43 UTC No. 16586950

>>16586917
They exist in other bases. The base you use only represents the numbers, it doesn't change them.

Anonymous at Sat, 15 Feb 2025 18:35:18 UTC No. 16586954

>>16586917
We're talking about real numbers not integers you fucking mongoloid

Anonymous at Sat, 15 Feb 2025 23:34:42 UTC No. 16587361

>>16586836
...000.333...

Anonymous at Mon, 17 Feb 2025 11:17:00 UTC No. 16589094

>>16586548
what you have done is actualy the fnite rational numbers, a bourbaki baguette concept, its fine i guess, but you make it sound hypergay
https://fr.wikipedia.org/wiki/Nombre_d%C3%A9cimal
lets explore combinatorics, we have your thing wher neither teh left or right of the point can go on forever, then we have the reals, wheere the right of teh pont can go on forever, we then hav eteh p-adcs, where teh left of teh point can go on forver, & lastly teh thing you contine to allide with "...*number*...", where we have that both left & right can go on forever, not clue what it'd be called, but i remember having seen some mathematician of those that can write backwards on a glass sheet so as to have said sheet between them and a camera/audience, from what i remember ...999.999...=0, or more specifficaly ...aaa.aaa...=0 in base a+1, & that's about it

Anonymous at Mon, 17 Feb 2025 11:28:09 UTC No. 16589101

>>16589094
>we have your thing wher neither teh left or right of the point can go on forever
Read it again anon, my first condition implies it can't go on forever to the left not to the right. The second condition simply implies that no number ends with an infinity of zeros.

Anonymous at Tue, 18 Feb 2025 05:49:16 UTC No. 16589875

>>16586455
.9999... is identical to 1 and 1 is in the set of reals

Anonymous at Tue, 18 Feb 2025 06:03:47 UTC No. 16589884

>>16589101
It doesn't matter if they did because x.999... = (x + 1).000...

Anonymous at Tue, 18 Feb 2025 21:36:53 UTC No. 16590738

>>16586274
because mathematicians are stupid and dont know what limits are

Anonymous at Fri, 21 Feb 2025 03:39:44 UTC No. 16594393

>>16586482
Your mistake here is assuming f (as abstractly defined to return *the* nth digit of a real number) is a well-defined function, which tacitly assumes that every real number has a unique decimal representation. However, we can show this assumption is false, since the real number 1=0.999... has two distinct representations.

Also your closed form formula is only guaranteed to work for irrational numbers and numbers which are finitely terminating.

Anonymous at Fri, 21 Feb 2025 03:45:47 UTC No. 16594396

>>16586274
"equal to infinity" is the informal way of saying "grows to a positive number without bound"
in this way, you are describing how the limit fails to exist.
however, you need to be careful here. 1/x approaches +inf from the positive direction, and -inf from the negative direction, so the limit proper doesn't exist.