Be some Indian who shits on the streets

>performing arithmatic with infinity
>performing arithmatic with infinity and expecting an answer in finite time
>performing truncated assumptions with a lazy implementation of infinity and then pretending you did arithmatic at all

Could quantum mechanics be a continuation of an incorrect axiom that has no bearing on real life and our observations are simply limited by hard?

That's a dumb way to attempt to prove it, implying it's standard math. this is like proving 1=.999... by dividing 1 by 3. you can't group an infinite series arbitrarily and get constant results. You also can't perform arithmetic operations on them and maintain a consistent system. You HAVE to start by modifying your model to one where infinite summations are well defined.
tl;dr: fuck off numberphile

>performing arithmatic with infinity
Literally nothing wrong with that except for the part where you're a brainlet who doesn't know how to spell the word "arithmetic."

>That's a dumb way to attempt to prove it
That's the way Ramanujan did it, by relating it back to that alternating series so he could use its already known 1/4 value. You're pretty retarded if you think you're smarter than he was desu.

okay enjoy doing fake math to the benefit of no one, retard.

>DURR NO BENEFITS
lmgtfy.com/?q=signal processing ramanujan sum

and you're pretty retarded if you think an argument with nothing but an appeal to authority backing it is any argument at all, even assuming what you said was correct.
Even if I followed the retarded logic that because he was smart everything he said is infallible, you're still wrong. Ramanujan clearly stated "under my theory" (referring to Ramanujan summation), doing exactly what I had just said you have to do. Even assuming you're right, you're still wrong.
Retard.

It wasn't an argument for the sum working, it was an argument for why you're pretty retarded for calling Ramanujan's summation dumb. I wouldn't need to point out his intelligence in the first place if you didn't make the retarded decision to call his derivation approach dumb.

...

I didn't call Ramanujan summation dumb, I called your counter-argument to dumb. Again, what you showed is only correct when you work in a different system. If you present it as proof for any standard system were infinite sums are not well defined, you're retarded. Ramanujan clearly knew that
>But as a fact I did not give him any proof but made some assertions as the following under my new theory. I told him that the sum...=-1/12 under my new theory
>I wouldn't need to point out his intelligence in the first place
You don't need to call it out at all, no matter what I do, because an appeal to authority isn't an argument. Yet here is your retarded ass doing it again. It doesn't matter how smart he is, he can still be wrong. He wasn't in this case, so you think in that tiny little brain of yours that your argument makes sense, but what if we were talking about his "number of primes under x" formula? You're the worst kind of retard. The kind that memorizes random things (or just googles them) in order to appear intelligent yet has no idea about what's actually going on.

The Ramanujan summation doesn't make the mistake that picture depicts.

what mistake?
it's my theory so fuck off

>I called your counter-argument to dumb.
What? First of all that isn't my post, second of all why are you talking about a counter-argument to it when it was an argument against one of your posts, and third you didn't even reply to that post when you wrote "that's a dumb way to attempt to prove it," you replied to one of my posts, and the way you were calling "dumb" was in fact Ramanujan's way of deriving that value.

-1/12 is the y intersection of 1 + 2 +... when transformed into a smooth curve. the transformation allows divergent series to be assigned values other than infinity.

all of this is from fucking wikipedia. just read wikipedia you lazy fucks.

That's not the only way to get to that -1/12 value.

woops, I meant counter-argument to I clicked on the wrong reply
>and the way you were calling "dumb" was in fact Ramanujan's way of deriving that value
No because Ramanujan wasn't trying to prove to anyone that the sum is -1/12. Again,
>But as a fact I did not give him any proof but made some assertions as the following under my new theory.
He wasn't proving anything. He was clearly talking about a different system and once you accept his axioms that sum trivially follows. So "proving" to someone that the sum is -1/12 by using is dumb since the issue is clearly not that they don't know arithmetic, but that they are working under a different system.
And for the millionth fucking time, none of that even matters because
>durr that's how Ramanujan did it
is not an argument.

that is the way you're supposed to get it, not that numberphile bullshit

you mean a complex curve, smooth curves aren't rigid like that

I'm really sure that's not how ramanujan did it. this is just retarded

Well the mathematical models of quantum mechanics produces consistent predictions of what will happen in different situations. Pretty low chance that the models will be wrong but make the correct predictions.

>haha wow check out this AMAZING math trick and the squiggly lines and then oh man cut all this shit out and paste this stuff in and use the little magic pixie dust WOW look at this amazing result
I hate numberphile so much

street shitters invented your number system, R haplo rapebaby.

doesn't look complex to me

en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_⋯#/media/File:Sum1234Summary.svg

smooth curves aren't rigid like that. it has much more properties than just "a smooth curve", which could give literally any value.

This

Numberphile's explanation of this was completely retarded. It didn't explain shit, and actually implied that was the correct solution to the sum of all natural numbers. I guess to serve as clickbait? By far their worst video that I've seen.

better tell Wikipedia then that they have been gaslighted by the Illuminati

Wrong. That is absolutely retarded and a completely unacceptable proof.
The first line alone would get you laughed out of any University, since you obviously do not understand even the basics of analysis.

>s-4s=(1+2+3+4+...)-(4+8+12+16+...)
>-3s=1-2+3-4+...

somtimes brady spontaneously becomes a brainlet

It is really fucking weird that just happens to get the same answer. This can't be just a coincidence. This is me by the way . No shit you can't manipulate infinite series like that. But how does making that mistake produce the same answer? That is weird as fuck, actually. Really fucking weird. Why is not interested in THAT?

The [math]-\frac1{12}[/math] meme is dead, user.

Because if you use invalid operations, you can get anything you want.

Now write that Argument without those ... more formally.

[math]\sum\limits_{n=1}^{\infty} n = \lim_{m\to\infty} {\left(\sum\limits_{n=1}^{m} n\right)} = \infty[/math]

Adding/Substracting/Multiplying/Dividing infinite sums is only defined if the limit is a real number (no, infinity is not a number)

Try again, sweetie

>Nobody actually thinks that formula is true.
It's not more or less true than a statement like

e^3 > 4

Those are mathematical statements that depend on a context.

>Zeta(s) is only defined by the sum for re(s)>1.
This is technically false, as the analytical continuation is part of the zeta function definition. The infinite sum of reals is only defined for those numbers, however.

Finally, 1+2+3+...=-1/12 isn't just true in the analytical continuation context, there are several theories that lead there (like Ramanujan summation, not apriori tied to analytic continuation at all)

Let us define the decimal of 1/3 as small s

(A) [s×1] 0.333r3 × 3 = 0.999r9
(B) [s×2] 0.666r6 × 3 = 1.999r8
(C) [s×3] 0.999r9 × 3 = 2.999r7
(D) [s×4] 1.333r2 × 3 = 3.999r6
(E) [s×5] 1.666r5 × 3 = 4.999r5
(F) [s×6] 1.999r8 × 3 = 5.999r4
(G) [s×7] 2.333r1 × 3 = 6.999r3
(H) [s×8] 2.666r4 × 3 = 7.999r2
( I ) [s×9] 2.999r7 × 3 = 8.999r1

Upon truncating the repeating pattern in attempts to define the infinite sequence in finite time, we develop rounding errors within the decimal beyond (A) .
It can be agreed that (A) is closer to 1 than (B) is closer to 2, and how both are closer to their assumed ceiling than (C) is to 3. As we continue all the way to s×9, the rounding error now gives 8.999r1, which is 8 orders of infinity further from 9 than (A) is from 1.

The truncation process assumes an upper limit of infinity. The action is made even more futile when each increment of s requires an even greater order of accuracy, thus truncating the infinite sequence of 1/3 as s to satisfy s×3=1 will immediately cause rounding errors at s×4 that further increase in inaccuracy at every s×n beyond 4. Instead, s would need to be redefined at every step of the way to remove the rounding errors, yet s is always defined as the decimal of 1/3 as 0.333r3

the awful assumtion that 0.999r9 = 1 is not the worst issue here, rather instead that 0.333r3 must equal 0.333rN to eradicate rounding errors by truncation, which is the essential equivalent of saying 0.3333 = 0.333N, or 0.3 = 0.N, or 3 = N.
If 0.999r9 = 1, then 3 is equal to 1, 2, 3, 4, 5, 6, 7, 8, or 9; and if 3 can be equal to any number, then any number can be equal to any other number.
If 1 = 3
and 5 = 3
then 1 = 5

To get 0.999r9 = 1, means to literally make any number equal to any other number, which is a false statement, therefore 0.999r9 cannot equal 1 as claiming otherwisr is also a false statement.

But 0.999r9 = 1 is true by definition.

Someone needs to take some analysis. It is true, however, that every sequence converges in an ultrafilter, so perhaps you have that [math] \sum^n_{k=1} k \to \frac{-1}{12}[/math] in that sense.

I don't think you understand how numbers work.

Prove it wrong

Read about Casimir effect, it is an experimentally proved physical effect whose theory uses directly this result.

see

>how does making that mistake produce the same answer?
Because it's not a mistake.
>Because if you use invalid operations, you can get anything you want.
Ramanujan didn't know in advance to try to get -1/12, he expressed surprise in that result and said something along the lines of how people would try to get him committed to a psych ward for ending up with that value. So no, that's not an explanation. I also don't know of any actual mathematician making a formal argument that the Ramanujan summation was "invalid." At best you'll find complaints that it isn't "rigorous," which isn't the same thing. You can introduce this missing "rigor" by explicitly limiting how the terms can be manipulated, but that's not the same thing as saying Ramanujan went against what these sorts of explicit limitations would have allowed. It works because he didn't go outside of those limitations even though he also didn't explicitly establish those limitations either.

1/3 * 3 = 1, by definition

1/3 cannot be written as a decimal, however, as it would take an infinite amount of time, an eternity, to write out the repeating 3's.

This is where shit retard math comes into play thinking you can truncate that infinite sequence as something like 0.333 repeating.
0.333 REPEATING * 3 = 1, In theory, however even if you had four orders of eternity to spend writing out 0.333 infinitely three times, adding them together, your final result would still be 0.999 infinitely.

In truth, the act of calling upon the REPEATING assumption adds any minute amount to make the statement 0.333r×3=0.999r9=1 true, but each individual call to the REPEATING assumption INCRREMENTS the accuracy dependancy by an order of infinity. If 0.333r3 has an infinite amount of 3's, 0.666r6 would have squared the amount of 6's and 0.999r9 wpuld have cubed the amount of 9'd, which is then going into the territory of literally trying to perform arithmetic on infinity rather than simply with loose concepts of infinity. What is infinity squared, or infinity cubed?

This is the dumbest post I've ever read on Veeky Forums.
If you were just pretending to be retarded I gotta congratulate you on mastery of your craft.

If you reject standard mathematical axioms (like AC or the Axiom of infinity) then you can get any result want.

There is no point in your proof, it is about as worthwhile as defining 1=2.

PROVE IT WRONG MONGOLOID

SAYING "UR DUMB" IS NOT AN ARGUMENT

Thats the fucking point retard. Truncating an infinitely repeating sequence immediately betrays accuracy, and any number can equal any other number, which as you understand is retarded, yet you would still gladly defend 0.333r×3 = which uses the same exact faulty logic.

You believe 0.999r9 = 1 yet think that is somehow different than saying 2=1

By logic, 0.999r9 does not actually equal 1.

>PROVE IT WRONG
There is no proof.
You reject the standard axiomatic of ZFC, after that anything goes.

You could prove anything that way just by defining it to be true.

0.9999... = 0.3333... * 3 = 1/3 * 3 = 1

There's no "in theory" - in the common definition of "..." the above is true, if you don't like it you can make up your own field of math where it's not

Inside ZFC and the standard definition of the real numbers, 0.999r9 = 1 and 2=/=1.
These immediately follow from the definition of the reals.

If you reject ZFC anything you do is practically worthless, unless you build it up from an alternative framework, you are essentially just defining things to be true.

>prove it wrong
Whatever axioms you're using, they're not the one's mathematicians typically use. You can base your number system off of whatever it is you're doing there if you want.

>If 0.999r9 = 1, then 3 is equal to 1, 2, 3, 4, 5, 6, 7, 8, or 9

Why is that true?

How fucking dumb are you. Do you just not understand what "truncating" means?
Look it up faggot.

You cannot ACTUALLY write the decimal of 1/3. You CAN write a TRUNCATED ASSUMPTION as something like (0.333...) or (0.333r3), whatever way you want to WRITE it. Howeverx this truncated assumption is not actually equal to the decimal of 1/3.

do you not understand the concept of infinity either? you could write a decimal one hundred 3's and still be inaccurate, one thousand 3's and still be inaccurate, one million, one billion, one trillion 3's and still be inaccurate.
By truncating in calling upon a REPEATING sequence, you may as well simply remove the REPEATING sequence.

Speaking in terms of decimal, 3/10 = 0.3
0.3×3 = 0.9
claiming 0.999... = 1 is literally the same claim as 0.9 = 1, or 9/10 = 10/10

you cannot perform arithmetic on infinite sequences, regardless of ANY AND ALL mental gymnastics claiming otherwise.

>1.999r8
You can give up here, this doesn't mean anything.

Because 0.333r3 must equal to 0.333rN to eradicate rounding errors from truncating the infinite sequence as shown in the (A)~( I ) results.

Your mistake, if you're being serious, is thinking that there's any rounding going on at all.

>You CAN write a TRUNCATED ASSUMPTION as something like (0.333...) or (0.333r3), whatever way you want to WRITE it. Howeverx this truncated assumption is not actually equal to the decimal of 1/3.
>you could write a decimal one hundred 3's and still be inaccurate, one thousand 3's and still be inaccurate, one million, one billion, one trillion 3's and still be inaccurate.
You're confusing symbols of numbers with numbers themselves.
It's completely irrelevant that you can't personally scribble an infinite number of "3" symbols on a piece of paper. That was never a requirement for representing a number with an infinitely repeating decimal part.

>You cannot ACTUALLY write the decimal of 1/3.
Yeah, I can, it's 0.3333...
More formally it's [eqn]\lim_{m\to \infty} \sum_{n=1}^n \frac 3 {10^n} = \sum_{n=1}^\infty \frac 3 {10^n}[/eqn]

>you cannot perform arithmetic on infinite sequences, regardless of ANY AND ALL mental gymnastics claiming otherwise.
Wrong. If you had any experience with calculus you would know

It can't be true. Because if S=1+2+3+...
So S=1+(1+1)+(1+1+1)+...
S=1+1+1+...
Ramanujan was a bastard

0.333r3×2 = 0.666r6

0.666r6×3 = 1.999r8
6×3 = 18

You can't do that. This is why zeta regularization limits you to working with term variables, so you don't do shit like that.

underrated, saved

The bigger problem here is to get that 8, you have to work from multiplying with 3 from the end of the infinite sequence, which there is no end to an infinite sequence, so you can't ever even actually get 1.999r8; we can assume there is an 8 at the end because 6×3, but we cannot assume where the end is, and becomes the equivalent of turning any number like 0.382726102827633828627 into 0.3827261028r7

we cut out information by assuming a limit of infinity, which leads to the wrong answer outright. The retardation stems from instead of realizing the answer is wrong, pretending the answer is right in lieu, giving us 0.999=1

Infinite repeating sequences should be seen as inficators that the math is faulty. Pi doesn't repeat and look how much that has enabled the world.

I dont think anyone here can give a good reason for 0.999=1 existing.

Infinite repeating sequences follow directly from the definition or limits, and limits are used to define numbers like pi and e or concepts like derivatives which are used pretty much everywhere

Get 6.5 then. Go!

No, I am pretty sure it is. You can't manipulate infinite series. In fact, if you just translate those series into summation notation you can get a contradiction: you can't combine the lists since the 4n will always be longer. It looks cool, but my understanding is that it isn't a valid operation, and even just using variables instead of numbers with "..." will show that.

>4n will always be longer
offset, rather
longer was the wrong word.

The -1/12 also exists in classical analysis and already Euler had the most general result on it.

Consider the difference between integral and sum,

[math] \int _m^n f(x)\, {\rm d}x=\sum _{i=m}^n f(i)-\frac 1 2 \left( f(m)+f(n) \right) -\frac 1{12}\left( f'(n)-f'(m)\right) + \frac 1{720}\left( f'''(n)-f'''(m)\right) + \cdots [/math]

where the coefficients are in simple way computed from Bernoulli numbers. A child of this would be e.g.

[math] \frac{1}{3}(4^3-2^3)=(2^2+3^2)+\frac{1}{2}(4^2-2^2)-\frac{1}{12}\,2\,(4^1-2^1) [/math]

but using the above I can generate infinitely many rigid identities that will all have that magic factor.
It's only Ramanujan who studies the same things and make the identification. A classical analogon would be

[math] \sum_{n=0}^\infty n \, z^n - \dfrac {1} { \log(z)^2} = - \dfrac{1}{12} + {\mathcal O}((z-1)) [/math]

You can type that into wolfram alpha for confirmation. The limit z to infinite gives an expression for 1+2+3+..., however, classically there's the log.
>this kills the sum

>shits and giggles
found the urban redneck

i'll do you one better

[math]\displaystyle\ln\left(1+x\right)=x-\frac {x^2}
{2} + \frac {x^3} {3} - \frac {x^4} {4} ...[/math]
[math]\displaystyle \ln\left(2\right) = 1-\frac {1}
{2} + \frac {1} {3} - \frac {1} {4} ...[/math]
consider the series: [math]\displaystyle S = 1-\frac {1} {2} - \frac {1} {4} + \frac {1} {3} - \frac {1} {6} - \frac {1} {8} +
\frac {1} {5} - \frac {1} {10} - \frac {1} {12} + ... [/math]
it's obvious that S and ln(2) have the same terms, just reordered in a new pattern.
for clarification, here's it in series form
[math]\displaystyle \ln(2) = \sum \{1, -2^{-1}, 3^{-1}, -4^{-1}, 5^{-1}, -6^{-1}, 7^{-1}, -8^{-1}, 9^{-1}, -10^{-1}... \}[/math]
[math]\displaystyle S = \sum \{1, -2^{-1}, -4^{-1}, 3^{-1}, -6^{-1}, -8^{-1}, 5^{-1}, -10^{-1}, -12^{-1}... \}[/math]
it might seem like they aren't even equivalent, but note that instead of 1/odd, -1/even, 1/odd, -1/even, we have 1/odd, -1/even, -1/even, 1/odd. all we did was shifted the evens over by one, which explains why the number 1/7 isn't even in the first 9 terms of S (it's the 10th term, actually)
this pattern can be represented by: [math]\displaystyle
\frac {1} {2x-1} - \frac {1} {2\left(2x-1\right)} -
\frac {1} {4x} [/math]
x - x/2 = x, and similarly, [math]\displaystyle
\frac {1} {2x-1} - \frac {1} {2\left(2x-1\right)}=\frac {1} {2\left(2x-1\right)}[/math]
so we can rewrite S as the following: [math]\displaystyle \frac {1} {2} - \frac {1} {4} + \frac {1} {6} - \frac {1} {8} + \frac {1} {10} -
...[/math]
or [math]\displaystyle \frac {1} {2} \left( 1- \frac {1} {2} + \frac {1} {3} - ... \right) [/math] = [math]\displaystyle \frac {1} {2} \ln \left( 2 \right)
[/math]

even though ln(2) and S have the same terms, just in a different order.

1/3 = 0.333[3
2/3 = 0.666[6
3/3 = 0.999[9
3/3 = 1
Therefore
1 = 0.999[9

this works when referencing the thirds up to the whole. Referencing thirds beyond the whole is different.

4/3 = 1.333[2

By motive of proving 0.999[9 = 1, adding 0.333[3 to 0.999[9 results in 1.333[2

1.333[2 - 0.999[9 should give us 0.333[3, which it does, and 0.999[9 × 2 = 1.999[8, but what is 1.999[9 then?

One note of providing acceptance for 0.999[9 = 1 is how close it is to 1, but 1.999[8 = 2 is less accurate. It would be more reasonable to say 1.999[9 = 2,
but 1.999[9 is not the result of 0.333[3 × 6. So if there is a number greater than the infinite sequence that is more accurate, this creates a problem of how to interpret the infinite sequence.

The infinite sequence is more than just infinitely repeating. It could be any number and will be any number by nature of calculating each successive digit. Even when it repeats a single digit, the potential for the next digit to be any other number remains.

by nature of truncating an infinite amount of decimal 3's from the fraction 1/3 as something like 0.333[3, we invoke this special property of infinity that increments an imaginary value to satisfy not only 0.333[3 × 3 = 0.999[9 = 1, but also 0.333[3 × 9 = 2.999[ㅌ = 3.
Disregarding the special property would make ㅌ = 8, yet defining the end limit as 8 would leave room for the larger number ㅌ = 9, meaning 2.999[8 is not actually as equal or more equal to 3 than 2.999[9 would be. This is where the special property of infinity can be utilized to fill in the gap between 2.999[8 and 3, giving us 2.999[9 towards 3 instead. Realistically, it would fill in the gap from the time of its first invocation at 0.333[3 where every increment of 1/3 also increments the value of the property defining the gap, thus 0.333[3 × 3 is actually understood as (0.333[3 × 3) + (ㅌ × 3)

0.333[3 × 3 = 0.999[9
(0.333[3 × 3) + (ㅌ × 3) = 1
1 - (ㅌ × 3) = 0.999[9
0.999[9 =/= 1
ㅌ × 3 =/= 0

>1 + 0.33333... = 1.333332...
Really convinced me there

Oh but you dont have any problem when using limits right fucking faggot?

Ok guys i found out something very marvelous:
the sum of all positive integers is the same thing as the sum of all positive integers below infinity, and as we know the sum of all positive integers below n is
(n-1)/2 * n
and we also know that the sum of all positive integers is -1/12
so
(infinity-1)/2 * infinity = -1/12
1/2*infinity^2 -1/2*infinity + 1/12 = 0
infinity^2 - infinity + 1/6 = 0
infinity = +1/2 +/- sqrt((1/2)^2 - 1/6)
= either 1/2+1/sqrt(12) or 1/2-1/sqrt(12) if i calculated correctly.

>.9999 = 1 defacto!!!!
really niggers my noggin.

Theres no numbers between 0.99... and 1

But every real number haves at least one number between them if they are different, therefore they must be equal

0.991 is closer to 1 than 0.99
0.99999991 is closer to 1 than
0.9999999
>haves

I don't get this either. How can you combine the summations like this? If you set a limit of n, anywhere, even close to infinity, they sums don't match up this way. Also, if we are doing n to infinity, wouldn't both of those sums be equal to infinity? Maybe I am just dumb as fuck.

s=(1+2+3+4+...)
4s=(4+8+12+16+...)
s-4s=(1+2+3+4+...)-(4+8+12+16+...)
-3s=1-2+3-4+...

>-3s=1-2+3-4

-3s = -3-6-9-12-...
did you just shit your brains for a second or what.

No, you're just missing the point of introducing 4s and subtracting it in the first place which is that Ramanujan wanted to assign a value to the 1+2+3+4... series by relating it to the alternating 1-2+3-4... series that he already knew had the value of 1/4 assigned to it. And to get from 1+2+3+4... to 1-2+3-4... he subtracted 4 from the second term (2-4=-2), 8 from the fourth term (4-8=-4), 12 from the sixth term (6-12=-6), etc. It wouldn't be useful to subtract in the places you subtracted from, that wouldn't relate the series to something familiar he could work with.

>If you had any experience with calculus you would know
ironic. If you had any experience with calculus, you would know it has nothing to do with aritmetic on infinite series.

I'm not him. But I got that part. It is rather obvious. I don't get how you can use that notation and ignore that for N=4, you still have -12 and -16 terms. For N=9^99999999999 you still have dangling terms too. It doesn't seem like that operation is valid for summations both for infinite sums and for any other N. It looks like a slight of hand. Also, if we get to assume that infinity makes this all go away, why can we assume the preceding term is infinity minus infinity? I was always under the impression that you can't manipulate infinite series with this way because of the aforementioned issues.

I could be a complete idiot though. I don't know, but it certainly doesn't make notational sense.

( r × n ) + ( ㅌ × n ) = j

Where r is an infinite repeating decimal, n is the multiplicator, and ㅌ is a single instance of "tending towards infinity" to define accuracy. But how could we further define ㅌ?
Working with 1/3 to 3/3=1, this would be
( 0.333[3 × 3 ) + ( ㅌ × 3 ) = 1
=
( 0.999[9 ) + ( ㅌ ) = 1
=
1 - ㅌ = 0.999[9
In this sense, ㅌ would only be an inverse of [math]\infty[/math]. If [math]\infty[/math] defines some garantuan unthinkable uncountable thing, then ㅌ defines an abstract remainder to be applied to end an infinite sequence, where [math]\infty[/math] + ㅌ = ㅐ, letting ㅐ be equal to this abstract concept of [math]\infty[/math] + ㅌ.
The problem with [math]\infty[/math] is that it describes an abstract concept, therefore you need abstract functions to properly work with it and deconstruct it back down to something that is actually usuable in real number arithmetic, though i'm gonna guess the benfit of all this is as meaningless as the original claim of 0.999[9 = 1, but at least better since it doesn't defy logic by saying x=y and at least attempts to identify aspects that were not previously addressed.

0.999...=x
9.999...=10x
10x-x=9.999...-0.999...
9x=9
x=1
0.999...=1

You cant perform finite arithmetic on infinity.

i dont know why this is such a hard concept for many people in the thread to have understood. They've spent too much time being lied to by professors and having their noses buried in books, and not enough time actually attempting a useful calculation using an infinite repeating decimal.

Performing calculations in finite time and further engineering computers to perform calculations leads to problems related to computing speed, defined as time, as well as memory registers for storing and performing math upon digits.

It's disinformation to truncate and concact an infinite decimal, regardless of how large or small the new implied repetition decimal is. 0.999[...] is as innacurate as 0.99999999999[...] or 0.9[...]
You can make the statement that 0.333[...] = 1/3, but 0.333[...] × 3 does not equal 3/3 or 1. Instead, it equals 0.999[...], and requires the intentional inclusion or addition of another abstract value to rival the concept of
[math]\infty[/math] in order to truely make it equal to 1.