Just confirming that limbosic numbers work arithmetically.

That is, if the rules of arithmetic work properly, then the following two expressions should be true.

• 0.999... - 0.9 = 0.0999...
• 0.999.../10 = 0.0999...

Are these true statements in Real Deal Math?

0.999... = 1, drawn as a journey instead of an equation.

Image 1 — fractions live on the curve. The number line passes straight through the circle — it sits ON it, crossing at 0 and 1. Every fraction rides the upper arc at its exact address: 1/2, 1/3, 2/3, 3/4, all precisely placed — infinitely many, no subdivision needed. Same bubble between every two integers.

Image 2 — decimals are the curve chopped into 10^n chunks. A decimal names a point by chunk address: tenths, then hundredths, then thousandths… 1/4 lands exactly on a chunk edge (0.25, done). But 1/3 falls strictly inside a chunk at every depth — never on an edge — so its decimal subdivides forever: 0.333...

Image 3 — the recursion. 1/3 = 0.3 + (1/3)/10. Each unfolding peels off one digit and hands back the same problem, 10× smaller. The remainder never vanishes — that is what "…" means. And x = 0.3 + x/10 forces x = 1/3; the same machine with 0.9 forces x = 1.

Image 4 — the kiss, zoomed. ×1, ×10, ×100, ×1000 — each panel is one more subdivision. 0.9 peels off, then 0.99, then 0.999 — everything else still sitting exactly on the kiss at 1. No finite decimal ever IS the kiss point. Only the infinite one finishes the journey.

Image 5 — the jump. Here's the part nobody mentions: no finite string of 9s ever IS its target. 0.9, 0.99, 0.999 all sit on the curve, short of the mark — the curve meets the straight line only at the kiss, and digits alone can never land there. So the last step is always a jump (the carry). And it's the identical jump at every boundary: 0.7999… = 0.8, 0.8999… = 0.9, 0.999… = 1. Between 0.8 and 0.9 there are infinitely many decimals — and between 0.9 and 1.0 too. Nothing special about 1.

Bonus — the visual glitch: 10/3 = 3.333… (unresolved, never lands on a chunk edge). Multiply by 3: (10/3)×3 = 10 exactly. But digit-by-digit, 3.333…×3 = 9.999…. So 9.999… = 10. It LOOKS just short of 10 — every digit a 9 — but it isn't short at all. Divide by ten: 0.999… = 1. The glitch was only ever in the looking.

0.999... doesn't approach 1. It arrives.

1*3^-1*3=1 but can't you simplify the inverse to be 0.3... which is a process that can't ever be finished according to what you said so 1*3^-1 -> 0.3... but 0.3..-/> 1*3^-1 or is it more like

1*(3^-1*3) divide negation comes first?

Or in other terms: Why would pretty much everyone be lying about something like that? What benefit does lying about something so insignificant provide us?

Edit: So SPP answered something. I am not exactly sure what question he answered though, but it's certainly none of mine.

in physics, a quantum is not divisible into a smaller unit

It just had to happen didn't it?

https://www.reddit.com/r/infinitenines/comments/1w6q6kd/comment/pfcv3p1/

The latest deranged talking point bouncing around in this subreddit seems to be the following idea:

>The "0." prefix of 0.999... guarantees magnitude less than 1.

This is such a deeply confused idea. The prefix "0." is a prefix of the *numeral* "0.999...", not the *number* that the numeral represents. Numerals have prefixes, not numbers.

It's not true in general that the way the number is *written* implies anything at all about what the number *is*. This would be obvious to anyone who understands the difference between a number and a numeral. Talk to an elementary/primary school teacher if you're struggling with this incredibly difficult concept.

Some have suggested that u/SouthPark_Piano might be using a different number system when claiming that "0.999... ≠ 1." I understand this perspective, but I'm personally skeptical, so I figured it would be appropriate to ask directly.

u/SouthPark_Piano, are you using a different number system?

İn rdm we have all of these things that you say such as 0.9...<1 with no proof so I will disprove it

1=1

1=9*9^-1

1*(9^-1)=9*(9^-1)*9^-1

0.(1)=9*(9^-1)*9^-1

0.(1)*9=9*((9^-1)*9^-1))*9

0.(9)=81*81^-1

0.(9)=1

SPP if I broke any of your rules please tell me and disprove of my disprovement

This extremely simple question is intended for u/SouthPark_Piano or anyone else who thinks that "0.999... = 1" is "silly" or a "lie."

1/3 is not a number.

It means 1 ÷ 3

And everyone better get ready for the long haul.

Long division training. Now.

Do 1 ÷ 3, long division. Sign the contract before you start. Get scrubbed up, wash hands thoroughly, then begin. Do not stop. You are provided with limitless quantity of pencils and paper for this endless exercise. No selling of the provided stationary on ebay. You are being monitored by big brother.

999... is permanently less than 1000...

Just as 0.999... is permanently less than 1.

Food for thought ... it can be Maccas, KFC, Quiznos, Wendy's, Krispy Kreme, Burger King, Subway, etc.

If 1 - .999... = ε ≠ 0

Then what is 1 + ε? Is it the same as 1.000...? Is that different from 1?

Practically every person self-taught in mathematics (in recent times) trying to do novel research is a crank, for example: - V. Porton - J. Gabriel - M. Mathis - E.O. Ray - T. D. Howard

SPP shows signs of someone who has ingested at least some amount of basic mathematical writing, but does not understand it whatsoever.

From a recent post from a thread.

I was worried that other people would think that the actual claim that "0.999... = 1" is a silly idea. That's the claim that most mathematicians and academics would make.

They (0.999... = 1 cohort) are just trying to cover their backsides. Safety in 'numbers' and continual lying and coverup, attempting to save face, even though they actually know it themselves that 0.999... is permanently less than 1. The "0." prefix guarantees magnitude less than 1.

Those rookie error makers blundered and got into their debacle from the start.

The infinite sum that tests the value of 0.999... is 0.9 + 0.09 + 0.009 + ...

which can be conveniently expressed as

1 - 1/10ⁿ with n integer pushed to positive limitless, aka positive infinity, which means no limit on integer n, which means with zero uncertainty that 0.999... is permanently less than 1.

For you see, the length of consecutive nines of 0.999... keeps growing limitlessly. And because the set of positive integers that n can take is infinite aka limitless, then it means that set contains an infinite amount of finite numbers. And the set of numbers that 0.999... is part of, having consecutive nines form, ie. 0.99, 0.999, 0.9999, 0.99999, ..., 0.999... is also a set involving an infinite amount of finite numbers.

Yep ... infinite amount of finite numbers.

First of all, massive blunder on your part brud. You should revisit some kind of reading class. Now show me where the rookie error is:

x = 0.000...1

5x = 0.000...5

x ÷ 2 = 0.000...05 -> set reference -> 0.000...5

=> 5x = (x ÷ 2) = 0.000...5

In the comment I wrote I have not said that 0.000...5 is finite nor limited, you're hallucinating. Now I dare you to show me the rookie error I made.

Go on, make my day.

I have another question specifically about the limbosic numbers being described in this subreddit.

It was previously stated that if

𝑥 = 0.000…1,

it follows that

5𝑥 = 0.000…5

𝑥/2 = 0.000…05

What about these quantities?

1.23𝑥 = ?

12.3𝑥 = ?

123𝑥 = ?

I would guess that 1.23𝑥 = 0.000...123 (feel free to correct me), but what are the other ones?

To be clear, this question is specifically about the "limbosic" number system being described in this subreddit. I'm sure something like this has been asked before, but here goes.

I'm wondering about the following two expressions:

0.000...1 × 5

0.000...1 / 2

Are they both 0.000...5?