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Does the number $...999.999... = 0$? My reasoning for asking this is outlined below.

First, for base 10 it can be shown that $...999 = -1$. In addition it is common knowledge that $0.999... = 1$. This seems to imply that $...999.999... = (-1 + 1) = 0$. This appears bizarre at first, but it seems that $...999.999...$ has the zero element properties of being both an additive identity ($...999.999... + x = x$) and an absorbing element ($(...999.999...)\cdot x = ...999.999...$) with base 10 decimal numbers.

Things get stranger. Dividing $...999.999...$ by 9 yields the number $...111.111...$. If the logic above is correct, this would imply that $...111.111... = 0$ also (since $0/9 = 0$). This would further imply that $(...222.222..., ...333.333..., \cdots, ...888.888...)$ are also "zero elements" (since $0+0 = 0$).

Is the logic above correct, or is there a fundamental flaw that I'm missing? If correct, is there a name for this behavior? I'm fascinated by this pattern and would love to learn more.

Bill Dubuque
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    The video does not prove the statement ...9999=-1. It cannot, because no real has infinitely many digits before the decimal point. – coffeemath Oct 24 '23 at 00:52
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    I don't think this that bad of a post. The real question is, is there a formal system in which decimal digits extend in both directions (reals + 10-adics)? – Maximal Ideal Oct 24 '23 at 01:04
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    If such a "number" exists, then your computation would make sense. So the real question is, does your decimal expansion make sense at all? – Sangchul Lee Oct 24 '23 at 01:14
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    @coffeemath The video does prove ...9999 = -1, though. In particular, it proves that the two 10-adic numbers (...9999) and (1) sum to the 10-adic number (0). It makes no claims about real numbers, and therefore any facts about real numbers can't be use to contradict it. – Alex Jones Oct 24 '23 at 01:15
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    Unfortunately I can't unwatch the video. Perhaps you could call it $\infty$s complement arithmetic? – copper.hat Oct 24 '23 at 02:16
  • In fact everything works fine, allowing infinitely many digits in both directions, if you only use repeating digits in each direction. The implied doubly-infinite sum doesn't converge in the real numbers nor the 10-adic numbers, but it can still be defined by the geometric series formula $\sum_{n=0}^\infty ar^n=a/(1-r)$. The result is a rational number. – mr_e_man Oct 24 '23 at 02:39
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    I don't think this is a bad question at all, and any mathematical imprecision is due to Veritasium, not OP. OP, a lot of the discussion here is likely to be very technical. This is because to put things like this on rigorous ground, one has to be technical and precise. Often maths YouTubers will omit some of this detail! The idea that "number" is not inherently meaningful, and one is free to try and define your own number system in any way you like (but then you also have to prove it has the properties you want) is a serious philosophical leap that many users here will take for granted. – Izaak van Dongen Oct 24 '23 at 12:35
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    Here's some gory technical stuff: It's not easy to define eg a field extending $\Bbb Q$ where you can make sense of "$...99.99...$". You can consider the topological additive group $\Bbb Q_p\times\Bbb R/\sim$, where $(a,b)\sim(c,d)$ if $a-c=d-b\in\Bbb Q$, and one thinks of $(a,b)$ as "the number with left part $a$ and right part $b$". It's hard to make a ring this way, as it isn't obvious how to multiply a $p$-adic by a real. Or you can try and combine the $p$-adic and Euclidean topologies on $\Bbb Q$, eg by taking their intersection. These are just ideas: I've not worked out the details. – Izaak van Dongen Oct 24 '23 at 12:39
  • Maybe look at https://repository.iit.edu/islandora/object/islandora%3A1012031/datastream/OBJ/download/Continuous_Generalization_of_2___s_Complement_Arithmetic.pdf – copper.hat Oct 31 '23 at 04:09

2 Answers2

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The fundamental flaw in the logic is right at the top. You take one fact about $10$-adic numbers

$$ ...9999 + 1 = 0 $$

and another fact about real numbers

$$ 0.9999... = 1 $$

and attempt to combine them to say

$$ ...9999.9999... = 0 $$

It seems like a simple substitution, replacing one expression for $1$ with another equivalent one. However, the problem is the symbol $1$ is actually playing double duty here. In the first expression, $1$ actually refers to the $10$-adic number $1$, while in the second, it's the real number $1$.

We have a way to add $10$-adic numbers to other $10$-adic numbers and get $10$-adic numbers. We also have a way to add real numbers to other real numbers to get real numbers. We do not have any universally agreed upon way to add $10$-adic numbers to real numbers, nor any agreed upon definition of what kind of thing the result should even be.

When you make the substitution of the real number $0.9999...$ for the $10$-adic number $1$, you effectively change the meaning of that $+$ sign to be this undefined mixed addition operation, and so the result you come up with isn't really meaningful in the context of real numbers or $10$-adic numbers.

But, that doesn't mean it's not meaningful! One may ask whether there exists an algebraic structure that contains the structure of the real numbers and the $10$-adics as substructures, which equates the real and $10$-adic numbers that "should" be the same, and which allows for positional addition for numbers which extend infinitely in both directions. If such a system does exist, then your argument would indeed hold water.

However, it's possible that all these conditions lead to a contradiction somewhere, in which case the system couldn't exist. I don't have the answer to that question, but if it's something you're curious about, you should absolutely explore it!

Alex Jones
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    Thanks, this is the answer I wish I had written. – MJD Oct 25 '23 at 12:52
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    Such a structure in fact exists, and is called a 10-adic solenoid. It admits a bi-infinite decimal expansion with the properties one would expect and indeed $...99.99... = 0$ in this system. Positional addition is well-defined making it into a topological group, but importantly there's no good way to define multiplication of these "numbers", basically for the same reason that there's no good way to define multiplication of distributions, or bi-infinite Laurent series expansions. – pregunton Oct 28 '23 at 13:36
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Certainly! If you multiply $$x=\dots999.999\dots$$ by $10$, the decimal point moves right one place and you have $$10x=\dots999.999\dots$$ again. So $x$ satisfies the relation $$10x=x$$ and the only number that does this is $x=0$.

MJD
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