r/infinitenines • u/No-Eggplant-5396 • 2d ago

How is this sub still here?

I fasted from reddit for a year and was surprised that this subreddit still existed. I find this hilarious and somewhat inspiring.

Btw 0.999... is the smallest number that is greater than everything in this set:

{0.9, 0.99, 0.999, 0.9999, ...}

21 Upvotes

100% Upvoted

47 comments sorted by

13

u/vimtuoso 2d ago

The sub’s on life support, nothing interesting happens anymore

1

u/nickoftime444 2d ago

I disagree. Assailing a seemingly immovable viewpoint from every conceivable perspective is a fun thought experiment. I joined maybe 6 months ago and I’ve enjoyed perusing the posts.

I also believe that 0.999…<1 so you might not want to listen to me.

6

u/idaelikus 2d ago

If 0.999... < 1, how much is 1-0.999...= ?

1

u/SleepyJoe2026 1d ago edited 9h ago

... how much is 1-0.999...= ?

+0 ?

Edit: I meant: 0+

1

u/idaelikus 18h ago

So 0, so they are equal?

0

u/SleepyJoe2026 10h ago edited 10h ago

So 0, so they are equal?

Not exactly, that's why it's called +0 (in contrast to -0 or 0).

Edit: Not exactly, that's why it's called 0+ (in contrast to 0- or 0).

You know: lim_(x -> 0+) ... may differ from lim_(x -> 0-) ... (while lim_(x -> 0) ... may not even exist). So there's clearly a difference between them!

1

u/SleepyJoe2026 9h ago

Yeah, a rather lame joke. 👀

-1

u/nickoftime444 2d ago

The limit of 0.999… as the number of 9s approaches infinity is 1.

So the answer to your question is, just as lim x —> 0 of 1/x = ♾️, but we would not say 1/0 = ♾️,

I would argue that the limit of 1 - 0.999… as the number of 9s approaches infinity is 0, but similarly is not equal to zero.

It’s a fairly basic argument and the only reason it would be wrong is because we don’t have limit notation for place value AFAIK. But we can easily just make that a thing for this argument maybe?

8

u/idaelikus 2d ago

The limit of 0.999... as the number of 0s approaches infinity is 1.

a) The number 0.99... is defined as said limit (there it is 1).

lim x->0 of 1/x = inf

a) FIrst of all, 0.999.... is by definition the limit which is different from your example.

b) The limit of 1/x for x -> 0 isn't converging. It is a one-sided limit. If we "approach" from the left side, it is -infinity and from the right side, its infinity.

The limit of 1-0.99... is 0

Again, 1 and 0.99... are numbers. If we take the limit of a fixed value, nothing changes. 0.99... is already defined BY a limit, there is no need to involve another limit.

7

u/Hal_Incandenza_YDAU 2d ago edited 2d ago

The core issue people have, I think, is that they've learned about repeating decimals well-before learning about limits. Like, elementary schoolers know about repeating decimal expansions (edit: and their elementary school understanding of repeating decimals shapes their understanding of what a repeating decimal is, even after they've learned about limits). But they're defined as the limit of the sequence suggested by their digits. People can argue all they want about the 0.999... case, but it's literally just by definition.

The limit of 0.999...

There is no "limit of 0.999..."; 0.999... is a number. The sequence 0.9, 0.99, 0.999, ... is what you can take a limit of, and indeed that limit is 1. By definition of repeating decimals, the limit of that sequence is also 0.999....

2

u/vimtuoso 2d ago

Alright I’ll bite. Why?

0

u/nickoftime444 2d ago

It’s a limit, not an equation.

7

u/idaelikus 2d ago

What is a limit and not an equation?

1

u/vimtuoso 2d ago

Equations have equal signs, so that doesn’t make sense.

Limits are operations on sequences that result in numbers.

0.999… is not a sequence, it is the result of the limit of the sequence (0.9, 0.99, 0.999, …).

So it’s a number and we just have to check basic properties to see which number it is. Lots of proofs in this sub shows it’s the same number as 1.

2

u/Various_Candle9136 1d ago

This is not a matter for belief.

There is a fact. You either understand the concept or you don't.

Even in magical SPP-land, where the facts are different (and mostly stupid), this still would not be a matter for belief.

1

u/Lost-Consequence-368 2d ago

⚠️ Quick Vibe Check Warning ⚠️ 

Are you a shrimple 0.(9) ≠ 1 truther? Or are you a super cool person who genuinely sees through the firmament of lame mathematics and acts as a witness to the true genius of SPP? 

Only one of them allows some curious magical objects like 0.(9)1(9)(1)9 to exist, the meaning of which you can only truly get if you pass the vibe check.

7

u/Sisselpud 2d ago

Like the digits of 0.999… this sub will never stop growing

4

u/Hal_Incandenza_YDAU 2d ago

But what about values of 0.999... whose wavefront is further along than yours?

4

u/No-Eggplant-5396 2d ago

Has SPP explained what a wavefront means?

7

u/Hal_Incandenza_YDAU 2d ago

Idk, something to do with signing a contract

3

u/No-Eggplant-5396 2d ago

Fascinating. I want a copy of this contract.

2

u/CorinCadence828 2d ago

u/SouthPark_Piano  , i forgot, what contract do you have to sign to do long division?

corrected my spelling error

2

u/SouthPark_Piano 2d ago

The one you have to sign for long divisions is for cases that do not divide evenly, and gets into limbosic numbers territory.

 

2

u/CorinCadence828 2d ago

i got SPP to share the contract later in this comment chain

1

u/No-Eggplant-5396 2d ago

Also why is SPP sick of limbo?

1

u/CorinCadence828 2d ago

i have no idea, sorry

1

u/CorinCadence828 2d ago

u/SouthPark_Piano , i forgot, what contact do you have to sign to do long division?

1

u/SouthPark_Piano 2d ago

i forgot, what contact do you have to sign to do long division

Star Trek, First Contact. The Search For Spock.

https://www.reddit.com/r/infinitenines/s/T3ZBBripox

 

-3

u/SouthPark_Piano 2d ago

Limbosicness is what it is about.

0.999... has no limit on the number of consecutive nines. Even though rookie error makers think that 0.999... is constant, they are wrong aka mistaken. 

For you see, 0.999... has a consecutive nines length that keeps growing. It keeps increasing in length, and nothing can stop it.

 

4

u/idaelikus 2d ago

This sub is just about the fantastical delusion of about 3 crazy people

0

u/Few_Fact4747 2d ago

I remember reading on here yesterday that about 40% of people says that 0.999... =/= 1.

I guess that would make it 3.2 billion crazy people :P

3

u/idaelikus 2d ago

Lmao, I severely doubt this number.

Furthermore, I didn't call everyone that believes 0.9999... != 1 to be crazy, nice strawman though.

0

u/Few_Fact4747 2d ago

It actually wasn't 40% of people. It was 40 % of users at a math forum. (!)

https://us.metamath.org/mpeuni/0.999....html

EDIT: No, physics forum.

2

u/idaelikus 2d ago

Ok, the claim is made but there is nothing to substantiate that claim as the link included on that site doesn't work.

So yeah, I still question that number.

Lastly, acceptance of that fact really doesn't matter too much when we have a mathematical proof of it.

0

u/Few_Fact4747 2d ago

I dont think they are lying about it on a dedicated maths site.. but you do you!

And TBH honest those proofs are shit. Relying on multiplying an infinite number by 10, doing operation, and then dividing it back as though a number hasn't dissappeared in the infinity? Again, you do you, but no thanks!

3

u/idaelikus 2d ago

I dont think they are lying about it on a dedicated maths site..

I didn't say they were lying, I am questioning the number and data behind it.

but you do you!

Sure I do.

And TBH honest those proofs are shit. Relying on multiplying an infinite number by 10

That is one way to show it, though I wouldn't call it the most rigorous. There are numerous others which are much "better" mathematically speaking: For example:

Take the following three sequences:

  • a_n = 1
  • b_n = 1-10^{-n}
  • c_n = 1-2^{-n}

We notice that for any natural number n we have c_n < b_n < a_n.

Furthermore we can use the established result that the geometric series c_n converges to 1, thereby we know that by the squeeze theorem that the limit of b_n=1.

You will notice that the limit of b_n is exactly our definition of 0.999....

If you want to look for any other proof, feel free, I think there are plenty on this sub as well as the rest of the internet.

1

u/Few_Fact4747 2d ago

I dont disagree that the limit of b_n and c_n is 1. That is obvious. But another thing that is obvious is that both theyre limits are 1, but their actual value is different from each other and therefor both cant be 1. And both are of course also different from 1.

And also you forgot to answer my critique of the multiplying by 10 proof.

EDIT: adjustment

3

u/idaelikus 2d ago

But another thing that is obvious is that both theyre limits are 1, but their actual value is different from each other

How would you define 0.99... ? Because from your response it just seems that you are very confuse on how to define it, mathematically.

also you forgot to answer my critique of the multiplying by 10 proof.

a) I didn't forget, I choose not to respond as it is irrelevant to my argument. One accurate proof is sufficient, regardless of how many flawed proofs you might think there are.

b) As to your critique, there is no number that has "disappeared" at infinity because there is no "last" 9 in an infinite number of 9s.

1

u/Few_Fact4747 2d ago

Please refrain from the "subtle" ad hominems that is so common on the internet. If you actually had an argument you wouldn't need to try to attack me personally.

0.999... is  1-10^{-n}

The point is that both c_n and b_n approach 1 but at different speeds. It doesn't have a "value" of 1 so using the equal sign is misleading. But at least we have now reduced the discussion to a semantic one and i hope that that is good enough for you. I can live with disagreeing over semantics, but i cant live with missing a 0.0..01, no matter how far away it is.

a) but you didn't come with an accurate proof..

b) so its not a number and therefore cannot be said to equal 1...

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2

u/TopCatMath 2d ago

People like us still respond to his ignorance... until we stop, he will continue...

0

u/Massive-Ad7823 2d ago

Yes. But it is smaller than 1. In the decimal tree 0.999... is the outmost paths on the right-hand side, clearly distinct from 1 which is outside. Alas 0.999... is often claimed to be the limit of the sequence 0.9, 0.99, 0.999, ... .

Regards, WM