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I'm working my way through Surreal Numbers by Knuth, and am finding myself a little hung up on the explanation of how addition works. The rule for addition is given as:

$$ x + y = ((X_L+y)\cup(Y_L+x), (X_R+y)\cup(Y_R+x)) $$

On the next page, one of the protagonists works out $1+1$ to be:

$$ (\{0+1,0+1\},\emptyset) $$ and follows the reasoning on through to prove that $1+1=2$. Where I'm getting a bit lost is that it seems there should be some additions involving $\emptyset$ that are omitted without explanation. If $1$ is defined as $(0,\emptyset)$, then a true following of the addition rule seems like it should be:

$$ 1+1=((0+1)\cup(0+1), (\emptyset+1)\cup(\emptyset+1)) $$ $$ 1+1=(1,1) $$

All of this is obviously wrong, but it's not clear to me why $\emptyset+1$ isn't $1$. I guess the most basic version of my question is: why is $0$ distinct from $\emptyset$ in surreal numbers, and what is the nature of that distinction?

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  • $\begingroup$ $\varnothing$, here, is not used to denote a surreal number; it is used to denote a set of surreal numbers. $\endgroup$ – Hurkyl Jan 8 '17 at 2:02
  • $\begingroup$ $0$ in the surreal numbers is $(\emptyset,\emptyset)$. $\endgroup$ – Mark S. Jan 8 '17 at 6:25
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$1$ is not really defined as $(0,\emptyset)$, but as $(\{0\},\emptyset)$. Then $$1+1=\left(\{0+1\}\cup\{0+1\},(\emptyset+1)\cup(\emptyset+1)\right)$$ However, $\emptyset+1$ is empty since there are no surreals in $\emptyset$ to add to $1$ ($\emptyset+1$ is the set of $x+1$ for all $x\in\emptyset$), so $1+1$ simplifies to $$\left(\{0+1\}\cup\{0+1\},\emptyset\cup\emptyset\right)=\left(\{0+1\},\emptyset\right)\text{.}$$

If you know that $0+1$ is $1$, we have $1+1=(\{1\},\emptyset)$.

To answer your other question, $0$ is distinct from $\emptyset$ in the surreals since every surreal is an ordered pair of sets. $0$ is defined to be the ordered pair $(\emptyset,\emptyset)$.

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A surreal number is, in particular, a ordered pair of sets of surreal numbers, so $\emptyset$ isn't a surreal number at all. In the formula for addition, the expressions like $X_R + y$ do not represent one value but rather the set of all numbers of the form $z + y$, where $z \in X_R$. But the definition $1 = (\{0\}, \emptyset)$ implies that $1$ has no right options at all. So if $x=1$, then $X_R$ is empty, and there aren't any such numbers. Similar reasoning applies to $Y_R + x$. The conclusion is that $1+1$ has no right options whatsoever.

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  • $\begingroup$ You contradict yourself a bit: You're correct that $X_R+y$ does mean the set of all such sums of the form $x_R+y$ for $x_R\in X_R$. But then in the case where $X_R=\emptyset$, we have $\emptyset+y$ is meaningful (it represents a set of surreals), and it means $\emptyset$. $\endgroup$ – Mark S. Jan 8 '17 at 6:24
  • $\begingroup$ @MarkS Actually, come to think of it, if it works the same as in ONAG, then $X_R$ is not a set, but a variable that ranges over all right options of $x$. But perhaps Knuth's book (which I haven't read) uses the notation differently. $\endgroup$ – Eric M. Schmidt Jan 8 '17 at 6:39
  • $\begingroup$ In notation like $\left\{G^L+H,G+H^L\mid\cdots\right\}$ (a common shorthand found in ONAG and Siegel) then of course $G^L$ is ranging over the left options so that we retain the idea of listing out the options of the sum. But Pat's question explicitly uses $\cup$ in the definition of the sum of two surreals, so we are forced to interpret $X_R+y$ as a set since it has to be something that we can take a union with. $\endgroup$ – Mark S. Jan 8 '17 at 6:45
  • $\begingroup$ @MarkS I think you are definitely right. Time to edit my answer again. $\endgroup$ – Eric M. Schmidt Jan 8 '17 at 6:49

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