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.
• It would benefit your question to define $\sim$ and explain the assumed correspondence between rationals and games. – stewbasic May 16 at 0:55