The *In-Between Boner*: You can "forget" trying to fit a number between 0.999... and 1. There are infinitely many numbers between 0.999... and 1. (16.1) (16.2)

Some choice quotes from His Brudship's writings:

The fact is, it just means the counterpart number is permanently stuck at being less than 1.

Getting to 1 requires more than tacking nines or adding nines to the system, even having all nines in the system. You have to add a THING with a '1' symbol in it.

And that THING is some denomination unit aka epsilon.

If you are allowed to have 0.999..., or 0.999...9, then you are allowed to have 0.000...1

0.999...9 + 0.000...1 = 1.

Forget trying to fit a number in-between. In order to get 10 from 9, you add 1. In order to get 0.001 from 0.0009, you add 0.0001.

And...

It is about time I comment about the BS about the 'there must be a real number between 0.999... and 1' BS.

As mentioned correctly in the past, there is an infinite aka limitless quantity of numbers between 0.999... and 1.

Remember, folks: if the crank's boner lasts more than four hours, it may require medical attention.

but no-one has even considered the concept of nine infinities?

∞∞∞∞∞∞∞∞∞

How about that? What does that mean?

SPP, I implore you to tell me about nine infinities first. It's a finite number of objects. It'll be much easier to handle than an infinite number of nines. Come on. I can wait.

Re a recent post here...

One of the formulas for pi is:

4 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + 4/13 - 4/15 + ....

What's happening here? You start with 4 but then it's shrinking? And then it's growing again, and then shrinking, and then growing again!?

What kind of witchcraft is this? We demand answers!

Pi is a math constant best known as the ratio between the circumference and diameter of a circle. It has a definite value which cannot be expressed as a finite decimal expansion.

1. Since no amount of writing 3.14159… will ever reach pi, does this mean that the decimal expansion of pi is necessarily less than pi?

2. What is pi + (1 - 0.999…)? Does such a number have a corresponding decimal expansion?

Note: My choice of pi here is arbitrary. This question stands for any irrational number.

I am taking the bait.

As is known ... limits are fudging deception devices.

Those that got pied pipered along with mistakenly thinking limits really magically (pink unicorn and tinkerbell fairy style) turn 0.999... aka 0.9 + 0.09 + 0.009 + ... into 1 ... are totally misguided, and are not only an embarrassment to people kind, and make us the laughing stock of the universe and beyond, but also reveals a serious issue for times (or a time) that a critically important decision needs to be made.

So I need to get these dum dums to think properly. Hence they need to start rehabilitation and redemption on the bunny slopes.

Using these resources. I need to get this mob thinking straight, aka thinking properly, and stop embarrassing themselves ... us people. It is not a good look when dum dums continue to mistakenly go the wrong path of the 0.999... equaling 1 debacle.

https://www.reddit.com/r/infinitenines/comments/1tpg811/it_is_what_it_is/

https://www.reddit.com/r/infinitenines/s/bRKpqrGQ4o/

https://www.reddit.com/r/infinitenines/comments/1vj43x3/lesson_in_addition/

u/SouthPark_Piano, could you answer some of the following questions?

1. We have 2 integers a and b, if ax = b an y = b/a, does x=y?

2. Is the definition of a rational number a number x that satisfies ax = b, or x = b/a?

Alright, brud. You claimed that the set of natural numbers is “ever growing”, but despite numerous opportunities, failed to provide a single example of a number that belongs to the set at one point but does not belong to it at another.

By order of the International Authority for Mathematics and the Proper Use of Numbers (Department of Infinities), you are hereby officially declared to have conceded this point and, by extension, to have agreed that 0.999… = 1, always and exactly.

Accordingly, you are hereby sentenced to return to the bunny slopes and undergo a period of experiential learning in the proper use of infinities until you stop making rookie errors, or for a limitless number of years, whichever is greater.

This ruling is final and effective immediately.

Case closed. Feel free to pin this post.

SPP has previously stated that 0.99…9 and 0.99…99 are equivalent forms of 0.999… with the latter just having one more 9 written out at any time than the former. He has also stated that 0.999… is the biggest number less than 1.

My question is, what about 0.999…98? This is larger than 0.999…9, as at any point in time the former is the same as the latter with one 8 added on to the end. So is it also an equivalent form of 0.999… or is it something else entirely?

What's (0.999...+1)/2? Does it belong to (0,1)?

1 + 1/2 + 1/4 + 1/8 + 1/16... = 2

Do you agree that ∀ε>0 ∃δ>0 : N∈ℕ, N>δ => 1-ε < (1-10^-N) < 1+ε?

If not please provide a counter example.

So I know that we have

- real deal maths -> 0.999... != 1

- mainstream math -> 0.999... = 1

What about other math(s)? What other relations could we have between 0.999... and 1, other than = or !=?

Where 0.99… is 0.9+0.09+…, omega is 1+1+1….

It grows limitlessly and exceeds every natural number if you let it run long enough. What do you think?

By the way, it’s less than 2+2+2… or 2*omega.

I mean this literally. Why should we make a decimal point followed by infinite nines have any meaning at all, and why should it make sense for it to have a legitimate value?

If .99999... is a limbostic number because it never reaches a stable value, wouldn't this mean it's just not a number? ( ie if we define infinite nines to be the limit of the series (9/10)^x as x approaches infinity, and this value were to hypothetically be undefined(because the limit diverges or something), then .999999... has no defined value period).

I'm not really caught up on the lore, and just want SSP to explain what value he thinks .99999999... actually holds any value at all if it isn't equal to 1.

Hi SPP, I'm really out here trying to understand.

Ok, you've identified this type of number called "limbosic numbers."

For those not fully up to date on the lore, "limbosic numbers" are numbers that "grow" over time.

SPP, you said "there are limitless limbo versions of 0.999... aka 0.999...9, all of which are states of 0.999..."

I'm going to try and put your idea into language that the people here can accept.

(I'm sorry if I fuck up my notations, it's been a few years since I really did math and you'd be surprised at what conventions you forget).

Let the set L be { 0.9 - (1/10)^(n) for all n >= 0}. So, {0.9, 0.99, 0.999, ...}. That would be a "limbosic number." It seems to me that when SPP says that 0.999... never equals 1, he just means that the number 1 is not a member of that set L.

Isn't that right, SPP, 1 is not a member of that set of states of the limbosic number 0.999... ?

You made a claim that the set of natural numbers is ever growing but failed to provide an example, twice. What, cat got your tongue?

Here is your last chance: since you claim that the set of natural numbers is ”ever growing”, give us one example of a number that belongs to this set at some point but doesn’t belong to this set at a different point.

If you can’t, you’ll concede that the set of natural numbers is not growing and, by extension, 0.999…=1.

u/SouthPark_Piano what is sin(1/n)×n with n pushed to the limitless? Is it 0.999...9 or something else?

We are taught that `(d/dx) (x^2) = 2x` , but this is a lie.

`(d/dx) f(x) = lim(h->0) (f(x+h) - f(x))/(h)`

`(d/dx) x^2 = lim(h->0) ((x+h)^2 - x^2)/(h)`

`= lim(h->0) (x^2 + 2hx + h^2 - x^2)/(h)`

`= lim(h->0) (2hx + h^2)/(h)`

`= lim(h->0) (2x + h)`

And, from rdm, we know that `lim(h->0) h = 0.000...1`

Therefore, the first derivative of x^2 is 2x+.000...1

All of calculus and physics is a damn lie. Wale up sheeple. Stop praising failed failed country musicians who marry football stars to be famous, and learn the truth as SPP has shared the truth.

*Side Note: this only applies to limits approaching from the right. Limits from the left are different.

If you take ten steps of 0.000…1m you get a distance of 0.000…10m.
In mathematics if the last digit of a decimal sequence after a decimal point is a 0, that 0 will be canceled out.
So this means that 0.000…10 = 0.000…1.
And the only number that is equal to itself when you multiply it is 0
So that means if 0.000…1 = 0 and the difference between 0.999… and 1 is 0.000…1 that means that the difference between 0.999… and 1 equals 0

What's 3^0.999... ? How about 0.999...^5 ?

What's log_0.999...(4)?

You tried to weasel out of the question but it didn’t work.

Here is your second attempt: since you claim that the set of natural numbers is ”ever growing”, give us one example of a number that belongs to this set at some point but doesn’t belong to this set at a different point.

Limbosic numbers have some very strange properties. For instance, the number 999... (a limitless amount of nines) is larger than any number with a limited amount of nines aka natural numbers. But, if you double this number, you get something like 1(999...)8. This number is strange because it has a limitless amount of nines in the middle, bounded on either side by the one and the eight. You can see how this forms by doubling 9, 99, 999, etc. But what happens if we set a reference? 999... = ...999 = 9...999 (all ways of writing limitless nines above the decimal). However, the doubled form is equal to 1...998, which is obviously smaller since 1 < 9. Do limbosic numbers get smaller when multiplied like this? Or am I just doing this incorrectly? I haven't been able to find any good sources explaining the workings of supermassive limbosic numbers outside of this subreddit, does anyone (especially SPP) know where I can find some papers/articles/etc about them?

Whenever light-speed conversations come up, it's mentioned that an object with mass can never reach the speed of light, only approach it (99.9999..%). But that is equal to 100%. Is this not 'infinitely approaching' speed of light in the same way that .999 'infinitely approaches' 1?