Surreal arithmetic with $\frac{1}{2}\omega$ In the final two chapters of Knuth's Surreal Numbers, both the world of multiplication and infinite/infinitesimal numbers are introduced. The basic ideas of both of these make sense to me, but I'm having two difficulties with the arithmetic of $\frac{1}{2}\omega$ that I'm afraid may reveal a deeper misunderstanding on my part.
1) Multiplication between two surreal numbers $xy$ creates:
$$
\langle (X_Ly+xY_L-X_LY_L)\cup (X_{R}y+xY_{R}-X_RY_R) | (X_Ly+xY_R-X_LY_R)\cup (X_Ry+xY_L-X_RY_L)\rangle
$$
Later, the value for $\frac{1}{2}x$ is given as:
$$
\langle\frac{1}{2}X_L\cup(x-\frac{1}{2}X_R)|(x-\frac{1}{2}X_L)\cup\frac{1}{2}X_R\rangle
$$
which follows directly. I'm thrown a bit, however, when $\frac{1}{2}\omega$ is calculated to be:
$$
\langle\{1,2,3,4,...\}|\{\omega-1,\omega-2,\omega-3,\omega-4,...\}\rangle
$$
Since $\frac{1}{2}\equiv\langle0|1\rangle$ and $\omega\equiv\langle\{1,2,3,4,...\}|\emptyset\rangle$ it seems like the left component of their multiplication should be
$$
\frac{1}{2}\{1,2,3,4,...\}\cup\emptyset
$$
$$
\{\frac{1}{2},1,\frac{3}{2},2,...\}
$$
and that the right should be
$$
\{\omega-\frac{1}{2},\omega-1,\omega-\frac{3}{2},\omega-2,...\}
$$
Is Knuth just making the intuitive leap that $\{\frac{1}{2},1,\frac{3}{2},2,...\}\equiv\{1,2,3,4,...\}$ or am I missing something crucial about multiplying $\frac{1}{2}$ through an infinite set? I know that isn't a huge leap, and I'm perfectly fine with making it, but I suspect this leads to my more fundamental confusion about:
2) Intuitively it seems obvious that $\frac{1}{2}\omega+\frac{1}{2}\omega\equiv\omega$, but, even when I accept the representation of $\frac{1}{2}\omega$ discussed above in 1), I don't see how it works. It seems to me that surreal addition of $\frac{1}{2}\omega$ to itself gives:
$$
\langle\{(1+\frac{1}{2}\omega),(2+\frac{1}{2}\omega),(3+\frac{1}{2}\omega),...\}|\{(\omega-1+\frac{1}{2}\omega),(\omega-2+\frac{1}{2}\omega),...\}\rangle
$$
which I think reduces to:
$$
\langle\{(1+\frac{1}{2}\omega),(2+\frac{1}{2}\omega),(3+\frac{1}{2}\omega),...\}|\{(\frac{3}{2}\omega-1),(\frac{3}{2}\omega-2),(\frac{3}{2}\omega-3),...\}\rangle
$$
and I don't see at all how any of that could be $\equiv\langle\{1,2,3,4,...\}|\emptyset\rangle$. What am I missing?
 A: Your calculation looks right.  What you are overlooking is a comment in the earlier chapter which discusses various tricks that help one calculate values.  Conway notes that one can often remove elements from Left and/or Right which do not affect the value.  
For example:
The Number $$\langle 0, 1, 2, 3 | \rangle$$ is equal to the number $$\langle 3 | \rangle$$ because $0$, $1$, and $2$ are to the left of $3$ and don't affect the value (which is 'between Left and Right').
By the same token, one can remove $\frac12$, $\frac32$ ,$2$, etc. from Left above because there is always an integer greater than the removed ones to force the remainder to be 'to the left of the value'; similarly for Right above.
A: Pat Muchmore had the right idea. For simplicity I will use $\psi:=\frac{1}{2}\omega$. You calculated $$\psi+\psi=\langle\{1+\psi,2+\psi,3+\psi,\ldots\}|\{\psi+\omega-1,\psi+\omega-2,\psi+\omega-3,\ldots\}\rangle$$ and want to show $\psi+\psi=\omega$. For this, it suffices to show $\psi+\psi\leq\omega$ and $\psi+\psi\geq\omega$.
By definition$^1$, $x\geq y$ iff there aren't any $v\in X^R,w\in Y^L$ with $v\leq y$ or $x\leq w$. 
Now for any $n\in\mathbb{N}$ we have $\psi+\omega-n>\omega$, so there is no $x\in(\psi+\psi)^R$ with $x\leq\omega$ (note that $\psi+\psi, \omega\notin (\psi+\psi)^R$). On the other hand, $\psi>n$, then $\psi+\psi>n$ and so there is no $x\in\omega^L$ with $\psi+\psi\leq x$. We conclude $\psi+\psi\geq\omega$.
There isn't any $x\in\omega^R$ with $x\leq\psi+\psi$ because $\omega^R=\{\}$. And for any $n\in\mathbb{N}$ we have $\omega>n+\psi$, therefore there is no $x\in(\psi+\psi)^L$ with $\omega\leq x$. Therefore $\omega\geq\psi+\psi$ and we conclude $$\frac{1}{2}\omega+\frac{1}{2}\omega=\omega$$
$^1$ I'm working with "On numbers and games" from J.Conway, and don't have Knuth's book at hand, but I'm sure $\leq$ is somehow equivalent defined there.
