3
$\begingroup$

This is from Conway's on numbers and games:

$x+(-x)=0$. We have to show $x+(-x)\geq 0$ and $x+(-x )\leq 0$. If say $(x+(-x))\ngeq 0$, we should have some $(x+(-x))^R\leq 0$, that is $x^R+(-x)\leq 0$ or $x+(-x)^L\leq 0$. But these are false, since we have by induction $x^R+(-x)^R\geq 0$, $x^L+(-x)^L\geq 0$.

I know that $x^R+(-x)\leq 0$ is a right option of $(x+(-x))$ so if its negative the number is. I don't understand where the "or $x+(-x)^L\leq 0$" come from. And why do we have $x^R+(-x)^R\geq 0$, $x^L+(-x)^L\geq 0$ by induction?

$\endgroup$
5
$\begingroup$

Where does $x+(-x)^L\le0$ come from?

It is a typo that your question has that my copy of the book does not, possibly related to confusion about notational conventions.

My copy says $x+-x^L$, not $x+(-x)^L$. This is because the right options of $x+(-x)$, that is, the ones denoted by "$(x+\,-x)^R$" in the book, have either the form $x^R+(-x)$ or the form $x+(-x)^R$. And by the definition of $-x$, something like $(-x)^R$ is like $-(x^L)$. By the conventions related to order of operations in regular arithmetic (since exponentiation has higher precedence than multiplication by $-1$), we write $-x^L$ to mean $-(x^L)$ and not $(-x)^L$.

Why do we have things by induction?

This is related to the paragraph just before PROPERTIES OF ORDER AND EQUALITY.

...note that almost all our proofs are inductive...we feel free to suppress all references to these inductive hypotheses

In this case, the implied inductive assumption is that $y+\,-y\ge0$ is true for all of the options of $x$, so that $x^R+\,-x^R\ge0$ and $x^L+\,-x^L\ge0$.

Where's the rest of the proof?

In case it wasn't clear, when the book says "say, $x+\,-x\ngeq0$", it's implying that we're only proving $x+\,-x\ge0$ and that the proof of $\le0$ is analogous.

$\endgroup$
  • $\begingroup$ Why does $x^L+-x^L\geq 0$ imply $x+-x^L\leq0$ is false? $\endgroup$ – badatmath Jun 4 '18 at 6:13
  • $\begingroup$ by doing algebra on them as if they were real numbers it follows from $x \geq x^L$ but we havent shown we're allowed to do that yet. $\endgroup$ – badatmath Jun 4 '18 at 6:46
  • $\begingroup$ Like we haven't shown we can add to both sides of an inequality; this is shown after $\endgroup$ – badatmath Jun 4 '18 at 10:06
  • $\begingroup$ @why There's no algebra involved, just the definition of $\le $. By definition, $a\le b $ iff $\text{no } b^R\le a$ and $b\le \text{no }a^L $. So by contrapositive, $a\nleq b $ iff $\text{some }b^R\le a $ or $b\le \text{some }a^L $. Here we have $a=x+\,-x^L $, $b=0$, and the relevant $ \text{some }a^L$ being $x^L+\,-x^L $. As an aside, for a more explicit introduction to the Surreals, I recommend looking at "Surreal Numbers" by Claus Tøndering. $\endgroup$ – Mark S. Jun 4 '18 at 10:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.