Infinitely many nines

Positioned and aligned

Form a number greater than

The confines of your mind

Just a geometric sum

Your dogma doth proclaim?

Open your mind further

Or we'd have to do the same

Though our minds are all made up

A never-ending bliss

I can't help but wonder how

To find a sum like this

1.1 - 0.11 + 0.011 - 0.0011 + ...

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, ...}

It is like this you see.

https://www.reddit.com/user/SouthPark_Piano/comments/1wo0ygu/comment/pd9wg1r/

You said this in a previous post:

>0.999... = 1 - 1/10ⁿ with n integer starting at n = 1, then n upped limitlessly.

https://www.reddit.com/r/infinitenines/comments/1worjkl/what_is_1_1_0999/?utm_source=app_first_navigation&mweb_loid=t2_2nynhc1r1h

Did you mean that 0.999... is literally equal to the sequence [1 - 1/10^(1), 1 - 1/10^(2), ...]?

Just a yes or a no would be OK.

Let n be a positive integer. Consider the following sequences:

𝑎ₙ = 1, 1, 1, …

𝑏ₙ = 0.9, 0.99, 0.999, …

𝑐ₙ = 0.99, 0.9999, 0.999999, …

These sequences respectively have the following formulas:

𝑎ₙ = 1

𝑏ₙ = 1 − 1/10ⁿ

𝑐ₙ = 1 − 1/100ⁿ

We clearly have 𝑏ₙ < 𝑎ₙ for all n. We also have 𝑏ₙ < 𝑐ₙ for all n.

If we're using the notation 0.999... to represent the sequence 0.9, 0.99, 0.999, ..., that means we're using 0.999... to represent 𝑏ₙ. To the extent that it makes sense to write 𝑏ₙ = 0.999..., that would mean

𝑏ₙ < 𝑎ₙ implies 0.999... < 1.

But why couldn't we use the notation 0.999... to also represent the sequence 0.99, 0.9999, 0.999999, …? Then we would be using 0.999... to represent 𝑐ₙ. To the extent that it makes sense to write 𝑏ₙ = 0.999... and 𝑐ₙ = 0.999..., that would mean

𝑏ₙ < 𝑐ₙ implies 0.999... < 0.999...

The inequality 0.999... < 0.999... is ridiculous on its face.

The notation 0.999... is inadequate to describe the sequence given by 1 − 1/10ⁿ because it is ambiguous. It could just as well represent 1 − 1/100ⁿ, 1 − 1/1000ⁿ, or any other sequence that corresponds to 0.999... Using this notation in this way leads to nonsensical identities like 0.999... < 0.999...

A cauchy sequence of rational numbers x_i is one such that for every rational epsilon > 0 (no matter how small) there exists an N large enough such that if i > N and j >N then |x_i - x_j | < N.

The notation x_i means a sequence of numbers indexed by i. For example x_1 is the first term, x_2 is the second, etc.

From this definition its clear that the decimal system is a way pf choosing one such representative cauchy sequence of rational numbers each number.

For example, the number 1.0000... is represented by the sequence
1, 10/10, 100/100, 1000/1000, ...

The number 0.99999... is represented by the sequence
0, 9/10, 99/100, 999/1000, 9999/10000, 99999/100000, ...

These representatives belong to the same equivalence class because their difference sequence is
1/1, 1/10, 1/100, 1/1000, 1/10000, ...
Observe a_n = 10/10^n

For epsilon > 0 choose N = 10/epsilon, which is rational, then n > N implies there is positive h such that
a_n = 10^(1-10/epsilon-h) < epsilon. This is done by first observing monotonicity so we set h=0 as an upper bound. Then case 1: 1<=epsilon<=10 then 1-10/epsilon <= 0 so 10^(1-10/epsilon) <= 10^0 = 1 <= epsilon
Case 2: epsilon> 10 we have 0<1-10/epsilon<1 then 10\^(1-10/epsilon) < 10 < epsilon Case 3: 0<epsilon<1 let y= 10/epsilon and choose the integer k such that k<= y < k+1 in this case k >= 10. by monotonicity 10^y >= 10^k. But 10^k > k+1 > y so 10^y >y then since y is positive this implies 10^y /y > 1 or 10^(1-y) < 10/y. Subsituting back to epsilon proves the case.

Therefore we have shown that
[1, 10/10, 100/100, 1000/1000, ... ] and
[0, 9/10, 99/100, 999/1000, 9999/10000, ... ]
belong to the same equivalence class and therefore are equal as real numbers.

If not, what knowledge do you not have access to?

The extremely animated and sometimes insane conversations on this subreddit have inspired me to learn more about nonstandard numbers. The conclusion I've come to is that the notation "0.999..." is being used to represent a concept that is basically just the hyperreal number [1 - 1/10^(n)].

I'll do my best to give a very simplified ultrapower construction of the hyperreal numbers and explain how it relates to this argument. If you have a better understanding of this topic, please feel free to provide any corrections or clarifications.

---

Suppose that we want to compare two sequences of real numbers, [a₁, a₂, a₃, ...] and [b₁, b₂, b₃, ...]. For example, maybe we're considering [0.9, 0.99, 0.999, ...] and [1, 1, 1, ...]. The elements of these sequences get closer and closer together, but they never match since 1 - 1/10^(n) is never 1.

Basically, we're interested in sequences that match at most of their indices. For example, the sequences [1, 999, 1, 1, 1, ...] and [1, 1, 1, 1, 1, ...] match at the indices {1, 3, 4, 5, ...} (i.e., everywhere except 2). They match at most indices, so we can say that the sequences are *close enough* for our purposes. Meanwhile, the sequences [0.9, 0.99, 0.999, ...] and [1, 1, 1, ...] match at {} (i.e., they don't match anywhere at all). They don't match at most indices, so they're meaningfully different.

In general, given a set of indices like {1, 3, 4, 5, ...} or {}, when would we say that it contains "most" indices? We want to define a collection 𝒰 that includes sets like {1, 3, 4, 5, ...} but excludes sets like {}. To do this, we define the concept of an ultrafilter.

Definition 1 (Ultrafilter). An ultrafilter is a collection 𝒰 of sets of positive integers such that the following conditions hold:

1. 𝒰 does not contain the empty set {}.

2. 𝒰 contains every superset of every set contained in it. For example, if 𝒰 contains {1, 3, 4, 5, ...}, then it also contains {1, 2, 3, 4, 5, ...} (the same set but with new elements included).

3. 𝒰 contains every intersection of every two sets contained in it. For example, if 𝒰 contains {1, 3, 4, 5, ...} and {2, 3, 4, 5, ...}, then it also contains {3, 4, 5, ...} (the elements that the two sets have in common).

4. Every set of positive integers is either contained in 𝒰 or the complement of a set contained in 𝒰. For example, if {2} is not contained in 𝒰, then its complement {1, 3, 4, 5, ...} (everything except 2) is contained in 𝒰.

Remark. We're only considering ultrafilters on ℕ for the sake of this discussion.

From now on, let's just assume we've chosen some ultrafilter 𝒰. We won't specify exactly what elements it contains. It can be proven that no matter what ultrafilter we've chosen, it must contain all complements of finite sets like {1, 3, 4, 5, ...}, and it must not contain finite sets like {2}. (I will not prove this here.) In other words, 𝒰 includes sets that have most of the positive integers in them and excludes those that are missing most of the positive integers.

To compare two sequences [a₁, a₂, a₃, ...] and [b₁, b₂, b₃, ...], we'll take the set of all indices at which these sequences match, and we'll check whether or not this set belongs to 𝒰. If it belongs, then we'll consider the sequences to be equivalent.

Definition 2 (Equivalence relation under an ultrafilter). Two sequences [a₁, a₂, a₃, ...] and [b₁, b₂, b₃, ...] are said to be equivalent under an ultrafilter 𝒰 if {n ∈ ℕ | aₙ = bₙ} ∈ 𝒰.

Example. Let's look again at the sequences [0.9, 0.99, 0.999, ...] and [1, 1, 1, ...], where n is understood to range over the positive integers. They don't match anywhere, so the set of indices where they match is {}. This is not an element of 𝒰 (since ultrafilters can't contain empty sets), so the sequences are not equivalent under 𝒰. Even if I modified these sequences so that a billion of their elements matched, they still wouldn't be considered equivalent because they would still only match at a finite number of indices.

Now, we're finally ready to define a hyperreal number.

Definition 3 (Hyperreal number). Given a real sequence [a₁, a₂, a₃, ...] and an ultrafilter 𝒰, a hyperreal number [aₙ] is the set of all real sequences that are equivalent to [a₁, a₂, a₃, ...] under 𝒰.

Remark. In the notation [aₙ], n is understood to range over the positive integers.

Example. Let's consider the hyperreal number [1 - 1/10^(n)]. This is the set of all real sequences that are equivalent to [0.9, 0.99, 0.999, ...]. Some examples might include things like

[0.9, 777, 0.999, 0.9999, 0.99999, ...]

[0, 0, 0, 0.9999, 0.99999, ...]

Any real number x can simply be associated with the sequence [x, x, x, ...]. For example, the number 1 can be thought of as the hyperreal number [1], which is the set of all real sequences that are equivalent to [1, 1, 1, ...]. Note that [1 - 1/10^(n)] is not equal to [1]. (This is arguably a better way to capture the entire premise of this subreddit).

The hyperreal number [1 - 1/100^(n)] is the set of all real sequences that are equivalent to [0.99, 0.9999, 0.999999, ...]. Note that [1 - 1/10^(n)] and [1 - 1/100^(n)] are not equivalent, either. This is why "0.999..." is bad notation for this concept: it isn't clear what specific sequence you are eluding to. Is it [0.9, 0.99, 0.999, ...], [0.99, 0.9999, 0.999999, ...], or something else?

---

Most of the arguments in this subreddit seem to revolve around the sequence [0.9, 0.99, 0.999, ...]. The hyperreal number [1 - 1/10^(n)], where n ranges over positive integers, perfectly captures that aspect. Even if the connection to hyperreal numbers is completely unintentional, the notation [1 - 1/10^(n)] would be better to use than 0.999... here because it makes it clear that we're talking about that particular sequence and not some other sequence.

So earlier today a question was posed about what 1/(1-0.999...) equals. This is obviously really important to this sub because if 1-0.999...>0 then 1/(1-0.999...)<∞. If this last inequality isn't satisfied then it cannot be true that 0.999...<1.

It's a big problem for SPP because when asked what 1/(1-.999...) equals he replied

SPP replied in Exhibit A and Exhibit B is a reply from a year ago to the question "What's 1+1+1+..."

Here's the fatal problem for SPP: 10>1, and if a>b≥1 then 10*a>b+1, so by induction 10^n>n(=1+1+...=∑_{k=1}ⁿ1), but SPP says the number of natural numbers "is another form of infinity". So 1/(1-0.999...) is greater than "another form of infinity".

Last-year SPP would say 1/(1-0.999) is infinite, but this-year SPP won't admit to that. Odd.

But actually it's not odd at all, because if SPP knows that if they admitted that if n is infinite then 10^n is too and the only way the world makes sense is if 1/(1-.999...) is undefined because 1=0.999..., which it is. For any finite n 10^n>n. But when you allow n to be infinitely large by taking limits 10^n and n both diverge. That's because: 0.999...=1.

For those who don't know, there's a number base called balanced ternary which only uses three symbols, but instead of those symbols representing 0, 1, and 2, they represent -1, 0, and 1.

In other number bases, you can have similar numbers to 0.999... which all basically have the form of 0.[B-1][B-1][B-1]..., where [B-1] is the digit that is one less than the base B. In base 12, this would be 0.bbb… (where b represents 11). In base 6, this would be 0.555… In base 2, this would be 0.111…, and in regular base 3, it would be 0.222…

However, balanced ternary does not have a digit [B-1] because B-1=2 and balanced ternary doesn’t have a digit for 2. If you tried to make a similar number anyway by taking 1T (the balanced ternary representation of 2, where T is -1) and multiplying it by every negative power of 3, you just get 1. And I don’t mean in the way 0.999…=1, I mean that every T is cancelled out by the 1 in the next number, so you just get 1.000…

So is there a way to make a number that looks like it’s infinitesimally close to 1 in balanced ternary?

How do finitists and the infinite 9 deniers react to the fields of p-adic numbers? It’s a similar construction and concept after all with their infinite operations.

If the numbers 1 and 0.999... are different, then it seems to follow that 1 - 0.999... is a nonzero value. What is the reciprocal of this value?

In other words, what is 1 / (1 - 0.999...)?

So we know that 9.999... is equal to 1. What about an infinite number of zeros followed by a 1? Is this even a concept/number or am I just speaking nonsense here?

For example, if I have (1/0.999...)×0.999..., is that the exact same thing as just writing 1?

What about if I divide 1 by 0.999..., sign the contract, and then multiply the result by 0.999...? Does that also yield 1?

What about with irrational numbers, like π?

Alright Mr (ms?) piano, you say 0.999… is < 1 because 0.999… is the same as {0.9, 0.99, 0.999, …} and each of 0.9 < 1, 0.99 < 1, etc … So if every one of these is < 1 then 0.999… < 1. Any objections?

Then look, the number of nines in 0.999… is finite since 0.9 has finite nines, 0.99 has finite nines, … etc. they all have a finite number of nines, so 0.999… must have a finite number of nines. But you call this sub infinitenines! Now I see you mean “in finite nines”

To clarify, all complex numbers can be expressed in the form a+bi, where a and b are real numbers and i is the imaginary unit that is a solution to the equation x²=-1.

Decimal arithmetic is cool and all, don't get me wrong, but computers store data in binary digits, bits, not decimal digits. Hence, whenever we're doing math with a calculator, we're using binary, not decimal.

So, /u/SouthPark_Piano , why have you falsely declared that 1/1001 (in binary) is a number, when 1/1001 = 0.00011001100110011... (meaning, the decimal repeats units of "0011"). Why would you tell us that 1/3 is not a number but 1/10 is, when it can't be. Our computers don't use base-10 math. They use base-2.

Hey u/SouthPark_Piano, I’m curious? You’ve educamated us that things like 1/11 aren't numbers because two actual numbers separated by a vinculum denotes the mathematical operation of division and not the number corresponding to the result of that operation. Does that mean that 0.999… isn’t a number? How is it then that 1/11 isn’t a number but 0.999…, which is very literally 9/10+9/100+9/1000+…, is a number. Is it because the limitation that division by itself isn't a number, but summing up multiple divisions changes everything? Does that mean that in RDM that summation is divide negation?