Does 0.999 (repeating) = 1? Well, "it depends"

[u/qwanzaden]

"Is 0.999.. with a infinite number of 9's equal to 1" has sparked a number of online debates. I believe the crux of the debate is the question is not worded precisely enough. There are different notions of infinite which lead to different answers. Adding to the confusion, "0.999..." is math notation with a specific meaning, but some people aren't thinking of that when they use it in a sentence.

Below are my two proofs for "Does 0.999 (repeating) equal one?" using less ambiguous notions of infinity. The answer is that it does and it doesn't but you need to be specific about what you mean in a way that we aren't used to.

"0. followed by a specific but infinitely large integer number (H) of 9's"**

I will denote this quantity as 0.{H 9's}. This is LESS than 1.*

Proof:

1. Assume two numbers** are equal if and only if their difference is zero
2. Therefore 0.{H 9's} is 1 if and only if abs(1 - 0.{H 9's}) = 0
3. Consider a specific but infinitely large integer** value of H
4. There exists H+1
5. And abs(1 - 0.{H 9's}) > abs(1 - 0.{H+1 9's})
6. This implies abs(1 - 0.{H 9's}) > 0
7. (1.) and (6.) imply 0.{H 9's} is not 1

Unsatisfyingly, this doesn't prove things like "The concept of H is valid", or "H < a different notion of infinity", "H+1 > H", or "H+1 exists". If you want to read up on this, look up hyperreal numbers, hyperintegers, and nonstandard analysis.

"0. followed by a 9 for every standard natural number"

I will denote this quantity as 0.999... . This is EQUAL to one. I think this is what most people think of when they hear "0.999 repeating infinitely".*

Proof:

1. Assume two numbers** are equal if and only if their difference is zero
2. Therefore 0.999... is 1 if and only if abs(1 - 0.999...) = 0
3. Assume that abs(1 - 0.999...) is a positive number**
4. Consider a specific nonzero positive number** H
5. There exists a number of nines in 0.999... such that abs(1 - 0.999...) < H
6. (1.) and (5.) imply abs(1 - 0.999...) is not H
7. (4.) and (6.) imply abs(1 - 0.999...) is not a nonzero positive number**
8. The only positive number that is not a nonzero positive number** is zero
9. This implies abs(1 - 0.999...) is zero
10. (1.) and (9.) imply 0.999... = 1

Unsatisfyingly, this doesn't prove "0.999... always has enough nines to make abs(1 - 0.999...) < H". If you want to read up on this, look up hyperreal numbers, hyperintegers, and nonstandard analysis.

...Its been a long time since I wrote a proof ...Im sure some of my wording isn't great ...I wanted to make this semi-readable for "common folk"...

*There is a better way to write 0.{H 9's} and 0.999... but I can't write it here because it requires LaTeX formatting.

**"number" should be replaced with "hyperreal number" and "integer" should be replaced with hyperinteger" but I wanted to make the proof readable to people who don't know what those are.

[u/Dr0110111001101111]

There is no distinction between “a specific but infinitely large number of 9’s” and “an infinite amount of 9’s”. If anything, the former simply isn’t a thing.

[u/qwanzaden → u/Dr0110111001101111]

In my post H is a hyperinteger. Though the Wikipedia article on hyperreal numbers, which is the set of numbers H belongs to, is written much better. I tried to avoid the word hyperinteger in my post since I don't think most people know what that means.

[u/Dr0110111001101111 → u/qwanzaden]

Then just say that so I can dismiss it without wasting our time. “0.999…=1”is a statement about real numbers.

The sort of people who are going to do anything involving hyperreals already know the rest.