# In the surreal numbers, is it fair to say $0.9$ repeating is not equal to $1$?

I find the surreal numbers very interesting. I have tried my best to work through John Conway's On Numbers and Games and teach myself from some excellent online resources.

I have prepared a short video to introduce surreal numbers, but I want to double-check some of my claims and would appreciate some help.

1. Is it fair to say that $0.999$ repeating is not equal to $1$ in the surreal numbers? This is the title of my video, so I really want to make sure it's a reasonable statement.

2. Can $\{0, \frac12, \frac34, \frac78, \dots \mid 1\} = 1 - \epsilon$ be thought of as $0.999$ repeating?

3. Is the number $\{0, \frac{9}{10}, \frac{99}{100}, \frac{999}{1000}, \dots \mid 1\}$ also equal to $1 - \epsilon$?

I am hoping the the answer is "Yes" for each question. If not, please let me know and I will get on the task of majorly revising the video. Thanks!

• What would $0.999$ repeating mean in the surreal numbers? Does that series converge? And no, it is definitely not $1-\varepsilon$ since there are plenty of elements even closer to $1$. – Tobias Kildetoft Sep 10 '14 at 17:36
• if $0.999$ repeating is a surreal number, would it not be equal to $1 - \frac{1}{\omega}$ – Epsilon Sep 10 '14 at 17:56
• @NickR Why $\omega$? Why not $\omega^2$? – Tobias Kildetoft Sep 10 '14 at 18:03
• @TobiasKildetoft My familiarity with the surreal numbers is not deep enough to answer that, but my guess would be that $1 - \frac{1}{\omega} \lt 1 - \frac{1}{\omega^2} \lt 1$ and that $1 - \frac{1}{\omega}$ is born first. That's probably not an answer, but it's the best I can do. – Epsilon Sep 10 '14 at 19:26
• I had the same question (3) a few years back. If you call $x = \{0,9/10,99/100,\dotsc\mid 1\}$, you can show that $x\leq 1$ and that $1\leq x$ based on the definition of ordering (recall that $1 = \{0\mid\}$). And so $x = 1$. – minimalrho Sep 11 '14 at 2:15

The surreal numbers contain the real numbers (as well as infinite and infinitesimal numbers).

Both $0.999\dots$ and $1$ are real numbers.

In the real numbers $1 = 0.999\dots$ and so it must also be true in the surreal numbers.

As Bryan correctly points out in his answer, surreal numbers which are not real numbers do not have a decimal expansion. This would appear to undermine the idea of using $0.999\dots$ to represent a surreal number.

• Hm… that's a big difference between the surreals and the hyperreals — every hyperreal has a decimal expansion, but not every surreal does. (Also the hyperreals form a set, have exponentiation defined everywhere, etc.) – Akiva Weinberger Jun 4 '15 at 4:41

Consider the sequence $$x_n=\sum_{i=1}^n\frac{9}{10^i}$$

whose 'limit' we understand to be the thing we call $.9\bar{9}$. When we consider this sequence in the surreal numbers, it does not converge to anything.

Let $I(1)$ represent the neighborhood of all infinitesimal numbers around $1$. The sequence does not converge to $1$ because $x_n$ never enters $I(1)$. If we consider any other real number $x$ less than $1$, the sequence will eventually surpass it and also the neighborhood $I(x)$. So if it converged, it would have to converge to some infinitesimal number to the left of $1$ which is in the infinitesimal neighborhood of $1$. But again, that sequence never enters $I(1)$. So $x_n$ does not converge; that is, $.9\bar{9}$ does not represent anything meaningful in the surreal numbers.

• By the same argument, it would seem that no sequence of real numbers converges. E.g. $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ would never enter I(1). – Epsilon Sep 11 '14 at 3:44
• And that wouldn't converge either. If you're taking that series to mean it's usual meaning - a limit of partial sums of real numbers. The surreal number $\{1/2, 3/4, 7/8, ... | 1\}$ is something different than the (nonexisting) 'limit' of your series. Likewise, the surreal number $\{0, 9/10, 99/100, ... |1\}$ is something different from the series I stated. Trying to represent surreal numbers with real number decimal expansions doesn't work. What would $1-\epsilon^2$ be? Or worse $1-\epsilon^\omega$? – Robert Wolfe Sep 11 '14 at 4:20
• Oh, I see. That's very interesting. So I guess the OP didn't really mean 0.999... but rather the surreal number $\{ 0, 9/10, ... | 1 \}$. I had a quick look through On Numbers and Games about a year ago and found it very appealing. I guess I should have taken a closer look. – Epsilon Sep 11 '14 at 4:26
• @Epsilon: Actually, constant sequences (and eventually-constant sequences) converge. – celtschk Dec 7 '17 at 8:52

I think you could say that $1 - \varepsilon$ is $0.999...$ in the sense that $\forall n \in \mathbb{N}, \left\lfloor 10^{n+1}(1-\varepsilon) \right\rfloor - 10\left\lfloor 10^n(1-\varepsilon) \right\rfloor = 9$ and $1 - \varepsilon = (+-+++...)$ is the simplest surreal satisfying this.

$1-\varepsilon = \{0;0.9;0.99;...\} | \{1\}$ because $\{0;0.9;0.99;...\}$ and $\{0;\frac{1}{2};\frac{3}{4};...\}$ are mutually cofinal.

The answer to 3. is affirmative. We can prove it with a straight calculation.

## Fun with surreals

Now let $(a_n)$ be a real sequence with $0< a_n$ for all $n\in\mathbb{N}$ and $\displaystyle\lim_{n\rightarrow \infty} a_n=0$ (in the reals). Then we have $1-a_n < 1$ for all $n\in\mathbb{N}$ and $\displaystyle\lim_{n\rightarrow\infty} (1-a_n)=1$.

For any such sequence $(a_n)$ we have $\varepsilon=\{0|\{a_n\}_{n\in\mathbb{N}}\}$ (here $\{a_n\}_{n\in\mathbb{N}}:=\bigcup_{n\in\mathbb{N}} \{a_n\}$ as usual). This can be proven, but I'll omit it here. Note that $1=\{0|\}$. Also note that on both sides of $|$ we normally write sets (nothing written means empty set) but simplified $\{0\}$ to $0$.

We will calculate what $1-\varepsilon$ is. We need to understand this surreal number in order to make comparisons to other surreal numbers. By definition, for any surreal numbers $x$ and $y$ we have $-y:=\{-y^R|-y^L\}$ and $x+y:=\{(x^L+y)\cup (y^L+x) | (x^R+y)\cup(y^R+x)\}$. It is important to notice that we are dealing with setwise operations here, so e.g. $-y^R$ means the set $\{-w:w\in y^R\}$ and $x^L+y$ means the set $\{v+y:v\in x^L\}$ etc.

$1-\varepsilon$ is just $1+(-\varepsilon)$. And, by definition, we have $-\varepsilon=\{\{-a_n\}_{n\in\mathbb{N}}|0\}$. We will calculate the four sets of the addition one by one.

$$1^L+(-\varepsilon)=\{0\}+(-\varepsilon)=\{0-\varepsilon\}=\{-\varepsilon\}$$

In the last step I used $0+x=x$ for any surreal $x$, which is also provable.

$$(-\varepsilon)^L+1=\{-a_n\}_{n\in\mathbb{N}}+1=\{-a_n+1\}_{n\in\mathbb{N}} = \{1-a_n\}_{n\in\mathbb{N}}$$

Last step is commutative law for surreal numbers, also provable. In fact, whenever all involved surreals are real, you can calculate with them just as you would be in $\mathbb{R}$ (provable).

$$1^R+(-\varepsilon)=\emptyset+(-\varepsilon)=\{v+(-\varepsilon):v\in\emptyset\}=\emptyset$$

This is closely working with our definition of setwise addition.

$$(-\varepsilon)^R+1=\{0\}+1=\{0+1\}=\{1\}$$

This should be clear by now.

So we have calculated that

$$1-\varepsilon = \{\{1-a_n\}_{n\in\mathbb{N}}\cup\{-\varepsilon\} | 1\}$$

Now we use a nice trick. What we have now is a representation of $1-\varepsilon$. There are others. In fact, given any surreal $x$ with a representation $x=\{x^L | x^R\}$ and sets $A\subseteq x^L$ and $B\subseteq x^R$ such that there exist $v\in x^L$ and $w\in x^R$ with $a < v$ and $b > w$ for all $a\in A$ and $b\in B$, then we have $x=\{x^L\setminus A | x^R\setminus B\}$ (provable). In easier terms, we can "kick out" small elements of $x^L$ and big element of $x^R$, getting another valid representation of $x$.

And now guess what? Since $0\leq\varepsilon$ (provable), we have $-\varepsilon\leq 0$ (also provable), but since $\displaystyle\lim_{n\rightarrow\infty} (1-a_n)=1$ there is at least one (in fact infitely many) $N\in\mathbb{N}$ such that $0<1-a_N$ and so $a < 1-a_N\in (1-\varepsilon)^L$ for all $a\in\{-\varepsilon\}$, so in fact we have

$$1-\varepsilon = \{\{1-a_n\}_{n\in\mathbb{N}}|1\}$$

Setting $a_n = \frac{1}{2^n}$ we get $1-\varepsilon = \{0,\frac{1}{2},\frac{3}{4},\frac{7}{8},\ldots|1\}$

Setting $a_n = \frac{1}{10^n}$ we get $1-\varepsilon = \{0,\frac{9}{10},\frac{99}{100},\frac{999}{1000},\ldots|1\}$

## Now is $1-\varepsilon=0.\bar{9}$?

This depends on what $0.\bar{9}$ is for you. If you define it for yourself to be $0.\bar{9}:=\{0,\frac{9}{10},\frac{99}{100},\frac{999}{1000},\ldots|1\}$, which is a somewhat intuitive definition, then the answer is yes (as I have just shown) and $0.\bar{9}\neq 1$ (because $1-\varepsilon\neq 1$ (provable)).

However, mathematicians would criticize you for using a notation that is already in use for a very similiar kind of object. In fact, they understand $0.\bar{9}$ to be

$$0.\bar{9}:=\lim_{N\rightarrow\infty} \sum_{n=1}^N 9\cdot 10^n$$ (using the usual limes) which, by taking the factor $9$ out and using the formula for the geometric series can be proven to be equal to $$9\cdot \left(\frac{1}{1-\frac{1}{10}}-1\right)=1$$

Therefore, with this notation we have $1-\varepsilon\neq 0.\bar{9} = 1$.

Please note that this is not a contradiction, but we used the term $0.\bar{9}$ to denote two different things which turn out not to be equal.

Another note on the second notation. I wrote "using the usual limes", which means that $\displaystyle \lim_{n\rightarrow\infty} b_n = b$ is defined to mean $$\forall\, \varphi\in\mathbb{R}^+\exists\, N\in\mathbb{N} \;\forall\, n\in\mathbb{N}:n\geq N \implies |b_n-b| < \varphi$$

Usually in calculus we write $\forall\,\varepsilon > 0$ instead of $\forall\, \varphi\in\mathbb{R}^+$, but by that we mean any real number greater than $0$. So I used $\phi$ to avoid confusion with the surreal number $\varepsilon$ we used above all the time and $\in\mathbb{R}^+$ instead of $>0$ to emphasize that we talk about positive reals, not any surreal number greater $0$.

Of course, as a third option, we can always change our definition. But if we use $$\forall\, \varphi \in\operatorname{No}^+\exists\, N\in\mathbb{N} \;\forall\, n\in\mathbb{N}:n\geq N \implies |b_n-b| < \varphi$$ where $\operatorname{No}^+$ denotes all positive surreal numbers, then the only real sequences converging with our old definition that would still converge with our new definition would be these that are eventually constant. So our new limes definition couldn't be viewed as a generalization of the old one since it returns different results. Take for example a sequence $(a_n)$ as defined at the beginning of this post. Since $1-\varepsilon=\{\{1-a_n\}_{n\in\mathbb{N}}|1\}$ we have $1-a_n<1-\varepsilon$ (since for any surreal $x^L<x<x^R$ setwise) but also $1-a_n<1-m\varepsilon$ for any $n,m\in\mathbb{N}$. In loose terms, you can never reach the infinitisimal $\varphi$-area around $1$ with a sequence that takes positive, non-infinitisimal steps (i.e. real sequences).