Real equivalent of the surreal number {0.5|}

Question

I've been reading up on Surreal numbers, but have some questions.

Some equivalent real and surreal numbers.

2.5 =
    {2|3} =
    {{{0|}|}|{{{0|}|}|}} =
    {{{{{|}}|{}}|{}}|{{{{{|}}|{}}|{}}|{}}}

0 =
    {-1|1} = {-2|1} = {-2,-1|1} =
    {{|0}|{0|}} = {{|{|0}},{|0}|{0|}} =
    {{{}|{{|}}}|{{{|}}|{}}}

-3/8 = {-0.5|-0.25} = {{-1|0}|{{-1|0}|0}}
    {{{{}|{{|}}}|{{{}|{{|}}}|{{|}}}}|{{{}|{{|}}}|{{|}}}}

What about the real number for {0.5|}?

Answer 1

In the surreal numbers, $$\{0.5|\}=1=\{0|\}=\{\{|\}|\}.$$

In general, if $a$ is real and $a\ge 0$, $$\{a\mid\}=\{\lfloor a\rfloor\mid\}=\lfloor a \rfloor+1,$$ where $\lfloor a \rfloor$ is the largest integer less than or equal to $a$. If $a<0$, $$\{a\mid\}=\{\mid\}=0.$$

Answer 2

Consider the games, which happen to be numbers, $G = \{ 0\,|\,\}$ and $H = \{ 0.5\,|\,\}$.

$G = 1$, directly from the simplicity rules. You know that $H = 1$, but that is not a direct implication. What you can do is show that $G \leq H \leq G$.

$H \leq G$, because $0.5$ is the only Left option of $H$, there are no Right options in $G$ and $0.5 < G = 1$.

$G \leq H$, because $0$ is the only Left option of $G$, there are no Right options in $H$ and $0 < H$. The last inequality, $0 < H$, is true because in the game $H$, Left wins no matter who starts, and this means the game is positive.

Now it is clear that $\{ 0\,|\,\} = \{ 0.5\,|\,\} = 1$. I hope that it helps you understanding why Polichinelle's answer is correct.