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Suppose $a$, $b$, and $c$ are dyadic rationals, where their corresponding numbers (games) are $A$, $B$, and $C$ respectively. Prove that $a+b=c$ if and only if $A+B\sim C$ (or $A+B$ is equivalent to $C$). Moreover, prove that $a \geq b$ if and only if $A \geq B$.

Note that for a dyadic rational $g=\frac{x}{2^p}$ for odd $x$, then its corresponding number would be $G=\{\frac{x-1}{2^p}|\frac{x+1}{2^p}\}$. Two games $G$ and $H$ are equivalent (or $G \sim H$ if $G - H$ is a P-position).

I was trying to do this by induction by assuming that the results are all true for simples dyadic rational games than $a + b$. Then, I said that since $a=\frac{n}{2^i}$ and $b=\frac{m}{2^j}$, if $i>j$, I can easily prove the results by looking at the options for $a+b$. However, I am stuck when $i=j$. Is there any better option to do this? I assumed that we would prove the second statement in a similar way, but I am not sure.

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  • $\begingroup$ It would benefit your question to define $\sim$ and explain the assumed correspondence between rationals and games. $\endgroup$ – stewbasic May 16 at 0:55
  • $\begingroup$ Just did it above! $\endgroup$ – George Leverson May 16 at 1:11
  • $\begingroup$ See math.stackexchange.com/a/2996364/86856 for one approach (though it may assume some facts about surreal numbers that you haven't learned yet). $\endgroup$ – Eric Wofsey May 16 at 2:00
  • $\begingroup$ We haven't talked about surreal numbers yet. Also I am confused how the approach can be applied for this problem $\endgroup$ – George Leverson May 16 at 2:19

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