Number 9 -- The Static and Dynamic Boner: There is a static model of 0.999... and a dynamic model. 0.999... is not static. It is dynamic. (9.1) (9.2)

Number 10 - The One-Third Boner: 1/3 is a number "of course". Multiplying a number by its reciprocal is divide negation (e.g. 1/3 x 3 = 1). 1/3 is not a number. (10.1) (10.2) (10.3)

Number 11 - The Slots Boner: 0.999... is a number with all decimal place slots to the right of the decimal point filled with 9s. To suggest 0.999... is a number with all decimal place slots to the right of the decimal point filled with 9s is to suggest there is an "end" to the 9s and therefore "nonsense". (11.1) (11.2)

Expression	First n terms of quotient	Remainder after n terms	Finite n case, simplified	"Limitless" case, simplified
a/(1-r)	a [1 + r + r^(2) + ... + r^(n-1)]	a r^(n)/(1-r)	a [1 + r + r^(2) + ... + r^(n-1)] + a r^(n)/(1-r)	a [1 + r + r^(2) + ... ]
1/9 = 0.1/(1-0.1) = 0.1/0.9	0.1 [1 + 0.1 + 0.1^(2) + ... + 0.1^(n-1)]	(0.1)(0.1)^(n)/(0.9)	0.111...1 (n 1s) + 0.000...1/9 (n-1 0s)	0.111... (not 0.111... + 0.000...1/9)
1=9/9 = 0.9/(1-0.1) = 0.9/0.9	0.9 [1 + 0.1 + 0.1^(2) + ... + 0.1^(n-1)]	(0.9)(0.1)^(n)/(0.9)	0.999...9 (n 9s) + 0.000...1 (n-1 0s)	0.999... (not 0.999... + 0.000...1)

Exhibit A

>> Cool, so 0.000...1 < 0.1, and 0.000...1 < 0.01, and indeed 0.000...1 < 1/10^n for any finite positive integer n. I appreciate the endorsement, brud. > > 1/10^n starting at integer n = 1, then n upped limitlessly continually aka infinitely is written as 0.000...1 brud.

Exhibit B

>> This makes it sound like you mean we... can't say whether 0.000...1 < 0.1, or 0.01, or any other power of 1/10. >> >> Conversely that would mean we can't say whether 0.999... is greater than 0.9 or 0.99. >> >> Am I reading you correctly or not? > > It is a limbosic number brud. You need to get training at the bunny slopes. Go there again brud.

His Brudship's position is clear. 0.000...1 and 0.999... are limbosic numbers with no fixed value, and to say they are greater than, less than, or equal to other numbers within the range they span is not possible.

Except... he already said that.

Exhibit C

>> Would 0.99…∞ < 0.99? And if so, should that mean 0.99… ≠ 1 given that 1∞ = 1? > > 0.999... is larger than 0.99 > > From a particular perspective, 0.999... is indeed less than 1 permanently. > > Everybody actually knows it. They just won't to get peer pressured.

Once more His Inconsistency can't figure out what he wants to say.

Thanks for playing.

The words of His Brudship:

Exhibit A

>There is a set of natural numbers, and you cannot lay a finger on the number on that set, as it is not constant. That set keeps growing too. Otherwise once again, you would be able to lay out your hand and write for me whether there is an odd number or even number all those natural numbers.

Exhibit B: the sidebar of this very subreddit

>Understanding the power of the family of finite numbers, where the set {0.9, 0.99, 0.999, etc} is infinite membered, and contain all finite numbers. The community is for those that understand the reach, span, range, coverage of those nines, which can be written (conveyed) specifically as 0.999... Every member of that infinite membered set of finite numbers is greater than zero, and less than 1, which indicates very clearly something (very clearly). That is 0.999... is eternally less than 1.

There is a set with infinitely many members. His Bunnymence said so. Every single member maps to a positive integer (the number of 9s it has).

You say that there's still a loophole? That His Inconsistency means you can have an infinite set of numbers of the form 0.999...9 and that it grows over time?

Well, too bad. I present Exhibit C (emphasis mine).

>Anyone understands that the infinite membered set of finite numbers {0.9, 0.99, 0.999, ...} already covers every possibility for the span (length) of nines to the right hand side of the decimal point.
>
>It doesn't cover it all in the future. It covers it all NOW. Already.
>
>And you surely understand that limitless means exactly that. That is exactly what happens when you have a limitless set of finite numbers. It covers every possibility. For the set, it covers every possibility for span (length) of nines to the right of the decimal point. Every value is finite. And the kicker is, there is a limitless number of extreme set members, and their span of nines coverage is written as: 0.999...

Maybe you wanna read your own writing sometime before you talk yourself into a corner, Your Inconsistency.

Thanks for playing.

Merriam-Webster:

boner, noun, bon·er

1 : one that bones

2 : a clumsy or stupid mistake

also : howler sense 2

The Ever-Bulging List

(Because we all need protection from the most hardened denier.)

1. The Irrational Boner: 0.999... is rational. 0.999... is not rational. (1.1) (1.2)
2. The Definitional Boner: 0.999... is eternally less than 1. It's fine if 0.999... is defined as 1. (2.1) (2.2) (2.3)
3. The Remainder Boner: In polynomial long division of a/(1-r), the remainder term goes away, or doesn't go away, at your convenience (e.g. 1/9 = 0.111... or 0.111...1 + 0.000...1/9 but 9/9 = 0.999... + 0.000...1 or 0.999...9 + 0.000...1). (3.1) (3.2) (3.3)
4. The Limbosic Integer Boner: A limbosic number can't be an integer. Some limbosic numbers are integers. (4.1) (4.2) (4.3)
5. The Infinite Long Division Boner: In long division of infinite decimals, you may consider truncations of the dividend in sequence in whatever way is convenient, and you may get different answers (e.g. 0.333.../2 -> 0.1, 0.16, 0.166, ... or 0.15, 0.165, 0.1665, ...). (5.1) (5.2)
6. The Unreal Boner: 0.999... is not a real number. 0.333... is a real number, and therefore 0.999... is a real number. (6.1) (6.2)
7. The Infinite Sets Boner: There is no static, infinite set of all natural numbers, but there is a static, already extant, infinite set of decimals of the form {0.9, 0.99, 0.999, ...}, even though that set maps to a static, infinite set of all naturals. (7.1) (7.2) (7.3)
8. The Greater Boner: Limbosic numbers do not have fixed values and can't be compared via (in)equality to any number within the range they span. Nevertheless, even though 0.999... is a limbosic number, 0.999... > 0.99. (8.1) (8.2) (8.3)
9. The Static and Dynamic Boner: There is a static model of 0.999... and a dynamic model. 0.999... is not static. It is dynamic. (9.1) (9.2)
10. The One-Third Boner: 1/3 is a number "of course". Multiplying a number by its reciprocal is divide negation (e.g. 1/3 x 3 = 1). 1/3 is not a number. (10.1) (10.2) (10.3)
11. The Slots Boner: 0.999... is a number with all decimal place slots to the right of the decimal point filled with 9s. To suggest 0.999... is a number with all decimal place slots to the right of the decimal point filled with 9s is to suggest there is an "end" to the 9s and therefore "nonsense". (11.1) (11.2)
12. The Infinite Integer Boner: There is no such thing as an infinite integer. 10... is one of infinitely many infinite integers. (12.1) (12.2)
13. The Integer Number of Nines Boner: The number of 9s in 0.999... is infinite. The number of 9s in 0.999... is always an integer. (13.1) (13.2)

Honorable Mentions from the Stiff Competition

FernandoMM1220's Boner: pi is always rational

Please feel free to suggest other massive boners from His Nineliness or others.

From a recent post:

He said 1 ÷ 9 is...

• 0.111...1 + 0.000...1/9, or
• 0.111...

The former turns into the latter when you commit to division and sign the contract.

That means 0.999...9 + 0.000...1 turns into 0.999... when you sign the contract.

Either 0.000...1/9 and 0.000...1 both go away or neither does. You can't have it both ways.

His own geometric series formula is...

a/(1-r) = a (1 + r + r^2 + ... + r^(n-1)) + a r^(n) / (1-r)

a = 0.1, r = 0.1: that last term is 0.000...1/9.

• If it goes away, you get 0.1/0.9 = 0.111...
• If it doesn't, you get 0.1/0.9 = 0.111...1 + 0.000...1/9.

a = 0.9, r = 0.1: that last term is 0.000...1

• If it goes away, you get 0.9/0.9 = 0.999...
• If it doesn't, you get 0.9/0.9 = 0.999...9 + 0.000...1.

Either they both go away or they both don't. It's your geometric series formula, YOUR NINELINESS.

From a recent post:

He said 1 ÷ 9 is...

• 0.111...1 + 0.000...1/9, or
• 0.111...

The former turns into the latter when you commit to division and sign the contract.

That means 0.999...9 + 0.000...1 turns into 0.999... when you sign the contract.

Either 0.000...1/9 and 0.000...1 both go away or neither does. You can't have it both ways.

His own geometric series formula is...

a/(1-r) = a (1 + r + r^2 + ... + r^(n-1)) + a r^(n) / (1-r)

a = 0.1, r = 0.1: that last term is 0.000...1/9.

• If it goes away, you get 0.1/0.9 = 0.111...
• If it doesn't, you get 0.1/0.9 = 0.111...1 + 0.000...1/9.

a = 0.9, r = 0.1: that last term is 0.000...1

• If it goes away, you get 0.9/0.9 = 0.999...
• If it doesn't, you get 0.9/0.9 = 0.999...9 + 0.000...1.

Either they both go away or they both don't. It's your geometric series formula, YOUR NINELINESS.

The words of His Brudship, Exhibit A:

> (1 ÷ 9) is never ending process. So once you start, you do not stop.

In Exhibit B, he reiterates that the process of producing an truly infinite decimal requires signing the contract and commitment.

>> As you've pointed out, though, 9 x 0.111... and 9 x [0.111...1 + 0.000...1/9] are not the same. > > Contractual brud. One involves jail time. The other does not.

In Exhibit C, he affirms that the only way to proceed is through "eternal surgery"

> The contract entitles you to have 0.333... pass as 1/3 if you stay fully committed to the eternal surgery done on the 1. Once you agree to terms, you stay committed to rolling out those threes, which by the way - having a times three magnifier allows you to view 0.999... during the surgery.

Remember that. Division to produce an infinite decimal result requires commitment. It requires signing the contract. It requires eternal surgery.

But we can apply this concept to subtraction as well.

Right-to-left and left-to-right subtraction

Many children are taught to subtract right to left and borrow as needed. For example:

   342
 - 179
 -----

Starting from the right, you can't subtract 2 from 9 directly, so we need to borrow from the middle column.

   33(12)
 - 17  9
 -------
       3

We'll need to borrow again from the hundreds place to do the tens.

   2(13)(12)
 - 1  7   9
 -----------
   1  6   3

342 - 179 = 163 is the correct answer.

With nonterminating decimals, you can't start all the way at the right, but you can still borrow. Instead, we will proceed left to right. Let's work that previous example borrowing left to right instead:

   342
 - 179
 -----
   ???

Our ? places are not yet determined. We haven't borrowed to the point the problem can be solved.

   2(14)2
 - 1  7 9
 --------
   1  ? ?

Even though we haven't borrowed enough to resolve all decimal places, we can see that we will never borrow from the hundreds place again. The leading digit of 1 is locked in.

   2(13)(12)
 - 1  7   9
 ----------
   1  6   3

One more borrow makes all places computable.

Left-to-right eternal borrowing for 1 - 0.999...

Now we can take this technique to 1 - 0.999... -- observe:

  1.00000...
- 0.99999...
------------

We'll need to borrow from the ones place to do the tenths.

  0.(10)0000...
- 0.  9 9999...
---------------
  0.  ? ????...

We'll also need to borrow from the tenths to do the hundredths, and we can see that each time, we still have more borrowing to do. Fortunately, by now the ones and tenths places are fixed: due to the nature of the problem, we will never have to borrow from those places ever again. So we can begin to write the answer:

  0.9(10)000...
- 0.9  9 999...
---------------
  0.0  ? ???...

Remember: all decimal places correspond to an integer position. We can very easily prove that we will continue having to borrow left to right and that, in doing so, we only produce zeros in the final result, like so:

  0.9999(10)...
- 0.9999  9 ...
---------------
  0.0000  ? ...

The only digits produced by this algorithm are all 0. 1 - 0.999... = 0 by eternal borrowing.

It doesn't take limits. It doesn't take constructions of real numbers or anything like that. His Baselessness has made this rookie error purely by failing to carry out subtraction, by misunderstanding simple arithmetic.

Any purported algorithm that produces a 1 at any place, at any time, is not committed to eternal borrowing, Your Bunnymence.

Thanks for playing.

According to HIS BRUDSHIP, 1 ÷ 3 = 0.333...

But 0.999... ÷ 3 = 0.333..., too...

And since he says 1 - 0.999... ≠ 0, that means...

(1 ÷ 3) - (0.999... ÷ 3) ≠ (1 - 0.999...) ÷ 3

HIS BRUDSHIP says I've made an error saying 1 / (1 - 0.1) = 1 + 0.1 + 0.01 + ... -> 1/(1-r) = 1 + r + r^2 + ... for r = 0.1, but he won't explain what exactly is wrong with defining r = 0.1 and substituting.

So let's get back to basics.

Statement 1: 2(0.1) = 0.1 + 0.1 = 0.2

Can we let r = 0.1 in this expression or not?

Statement 2: If (Statement 1) and (r = 0.1), then 2r = r + r = 0.2

Is this valid, YOUR BUNNYMENCE? Or NOT?

               1 ÷ 3 = 0.333...
           0.3 ÷ 0.9 = 0.3 + 0.03 + 0.003 + ...
   (0.3) ÷ (1 - 0.1) = 0.3 [1 + 0.1 + 0.01 + ...]
       1 ÷ (1 - 0.1) = 1 + 0.1 + 0.01 + ...
         1 ÷ (1 - r) = 1 + r + r^2 + ...  (where r = 0.1)

Simple. Algebra.

               1 ÷ 3 = 0.333...
           0.3 ÷ 0.9 = 0.3 + 0.03 + 0.003 + ...
   (0.3) ÷ (1 - 0.1) = 0.3 [1 + 0.1 + 0.01 + ...]
       1 ÷ (1 - 0.1) = 1 + 0.1 + 0.01 + ...
         1 ÷ (1 - r) = 1 + r + r^2 + ...  (where r = 0.1)

Simple. Algebra.

Can't answer on pi

Absolute nothing answer on 1/3/2

The words of HIS BRUDSHIP:

Exhibit A implies that either way you should get 1/6 = 0.166...

> Just do 1 ÷ 2 ÷ 3 > > or 1 ÷ 3 ÷ 2

Exhibit B says you should only get 0.166... this way

> If you get 0.166... , then you didn't make rookie errors.

Exhibit C makes things clear. When you divide an infinite decimal, you divide any truncation of that decimal and do not consider the remainder.

> Without spoon-feeding youS too much, consider being one or a few steps behind the increasing length of threes in your long division bunny slopes divide process. > > eg. 0.33 long division by 2. > > First digit in the divide result is 1 remainder 1. > > Then increase 0.33 by one digit > > 0.333 > > The divide result was initially 0.1, and the second result digit is 6 remainder 1. > > So the evolving result is 0.16 > > Then increase 0.333 by one digit > > 0.3333 > > The divide result was 0.16, and we then get another 6 remainder 1. > > Go figure brud. Pull up your socks.

Let's be clear. I'll use His Bunnymence's exact words but replace 0.333... with 1.999... instead

"eg. 1.9 long division by 2."

"First digit in the divide result is 9 remainder 1."

"Then increase 1.9 by one digit"

"1.99"

"The divide result was initially 0.9, and the second result digit is 9 remainder 1."

"So the evolving result is 0.99"

"Then increase 1.99 by one digit"

"1.999"

"The divide result was 0.99, and we then get another 9 remainder 1."

Go figure, brud.

You get (1 + 0.999...)/2 = 1.999.../2 = 0.999...

That means 1 + 0.999... = 2 x 0.999...

That means 1 = 0.999...

Thanks for playing.

Exhibit A

> A limbosic number is not an integer. A growing number ... dynamic one is obviously not an integer if you know what YOU mean.

Exhibit B

> It is an continually increasing integer. Limbosic integer class.

Let n be a nonnegative integer (that means it can be 0 or any positive integer). We'll call denote by N the set {0, 1, 2, ...}.

Let f be a function such that...

• f(0) = 0
• For n > 0, f(n) = c (for some integer c)

Let g(n) = f(n+1). That means...

• g(n) = f(n+1)
• g(0) = f(1) = c
• g(1) = f(2) = c
• g(n) = f(n+1) = c (since for any n, n+1 > 0)

That means g is a constant function. g(n) = c for any n.

Okay, so what's g(n) - f(n)?

• g(0) - f(0) = c - 0 = c
• g(1) - f(1) = c - c = 0
• For n > 0, g(n) - f(n) = f(n+1) - f(n) = c - c = 0

Let c = 9.

Then f is the function describing the digits of 0.999...

g is the function describing the digits of 10 x 0.999... = 9.999...

g - f is the function of the digits of 10 x 0.999... - 0.999... = 9.000...

Thanks for playing.

Pretty much instantly banned over this exchange

He doesn't have an answer for why. His whole schtick can't handle it.

Why do we have to "answer" to base 10?

Why does a constantly changing value represent 0.999...?

He has no answers for such things.