I'm working with the book "On numbers and games" from John Conway, first edition from 1976.

On page 20 he writes

Summary. Numbers form a totally ordered Ring. Note that in view of Theorem 8 and the distributive law, we can assert, for example, that $x \geq 0$, $y\geq z$ together imply $xy\geq xz$, and that if $x\neq 0$, we can deduce $y=z$ from $xy=xz$.

I can't wrap my head around the second part, probably because I'm missing some obvious fact about rings or fields.


So let's see what he references here. On page 29 he writes

Theorem 7. For all $x,y,z$ we have the identities $x0 \equiv 0$, $x1 \equiv x$, $xy\equiv yx$, $(-x)y\equiv x(-y)\equiv -xy$, and the equalities $(x+y)z=xz+yz$, $(xy)z=x(yz)$.

where "identity" means the actually used forms are identical and "equality" means the equivalence classes (of relation $=$) are identical. $\equiv$ implies $=$. Furthermore

Theorem 8. (i) If x and y are numbers, so is $xy$
(ii) If $x_1=x_2$, then $x_1 y = x_2 y$
(iii) If $x_1\leq x_2$ and $y_1 \leq y_2$, then $x_1 y_2 + x_2 y_1 \leq x_1 y_1 + x_2 y_2$, the conclusion being strict if both the premises are.

All thats left out from the book between these citations are the proofs and corrolary Theorem 9 ($x,y$ positive implies $xy$ positive). Note that in the next part of the chapter, $\frac{1}{x}$ gets defined in the form $\{L|R\}$, but the existence of the multiplicative inverse has been established above and I want to know how.

My approach

I can prove the uniqueness with the given theorems without problem. Let $y,z$ be two inverses of $x$ (i.e. $xy=1$ and $xz=1$). Then $$y\equiv y1 = y(xz)\equiv (xz)y = (zx)y \equiv z(xy) = z1 \equiv z$$

I fail to see its existence. We basically want to prove $x_1 y=x_2 y \Rightarrow x_1=x_2$ (other direction of 8(ii)) using only the theorems above and the addition on surreal numbers as a abelian group. I doubt that the definitions of negative, addition and multiplication on surreal numbers are of importance here, it seems to be a pure algebraic problem. For simplicity say $x=x_1, z=x_2$. Surely I have $$0\equiv x 0=x(y-y)\equiv (y-y)x = yx + (-y)x \equiv yx - yx \equiv xy - xy = zy - zy \equiv yz - yz = (y-y)z \equiv z(y-y) = z0\equiv 0$$ but of course this proves nothing. Multiplying with $1$ doesn't seem to help either (because without given existence of inverses this doesn't solve anything) and adding seems to be unhelpful, too.

You probably can see at once why this work or doesn't work, but if it doesn't, can you prove it?

Edit: A colleague just gave me a hint that it suffices to prove $xy=0 \Rightarrow x=0 \vee y=0$, but I can't see this either. Maybe that's were the definition of multiplication comes in?

  • $\begingroup$ I checked out that book a ways back but don't remember now what a "surreal number" is. Would you mind inserting something about what they are apart from these elements about their axiomatic development? $\endgroup$ – j0equ1nn Dec 8 '17 at 12:37
  • $\begingroup$ Dear @j0equ1nn I'm afraid I would rather not, because, just as I assumed when writing this question, it is not relevant for an answer and would probably just clutter the question because it's rather sophisticated. But I can recommend the Wikipedia page on this subject, it is really well structured: en.wikipedia.org/wiki/Surreal_number $\endgroup$ – SK19 Dec 8 '17 at 17:58
  • $\begingroup$ OK fair enough, thanks $\endgroup$ – j0equ1nn Dec 8 '17 at 18:22

[I have the 2nd edition, not the 1st, but there don't seem to be any differences important for this question.]

First, an important correction. You write "the existence of the multiplicative inverse has been established", but Conway is not (yet) claiming this. He is only claiming that the cancellation law $xy = xz \implies y = z$ (when $x \ne 0$) holds.

The cancellation law is not true in all rings, so we need something more to be able to prove it. In this case, we can use the inequality laws. Note that we have the principle of trichotomy (exactly one of $x < y$, $x = y$, and $x > y$ holds). This follows from Theorem 2 and the definitions of $=$ and $<$.

Using Theorem 8(iii), we have, if $0 < x$ and $y < z$, the conclusion $xy < xz$. Also, if $0 < x$ and $z < y$, then $xz < xy$. Applying trichotomy, this tells us that if $0 < x$ and $y \ne z$, then $xy \ne xz$. In a similar fashion, we can prove that if $x < 0$ and $y \ne z$, then $xz \ne xz$. Again applying trichotomy, if $x \ne 0$ and $y \ne z$, then $xy \ne xz$, which is equivalent to the cancellation law.

  • $\begingroup$ Thanks for your note on existence, I really thought this would be the same, but I can see why it is not. I also really like how your proof doesn't use the hint :) $\endgroup$ – SK19 Dec 7 '17 at 10:01

The hint from the colleague seems to suffice to prove the existence cancellation law: Theorem 9 states that $x,y>0 \Rightarrow xy >0$ which implies $xy \neq 0$ from the (surreal) definition of $>$.

Since $-xy\equiv (-x)y\equiv x(-y)$, we have e.g. $x > 0, y < 0 \Rightarrow xy \neq 0$, because would the product be equal to $0$ so would $x(-y) \equiv -xy = -0 = 0$, but $x,-y > 0$. By symmetry we have lemma: $x,y \neq 0 \Rightarrow xy \neq 0$.

Now given $xy = zy$ and assume $y\neq 0$. Set $w\equiv x-z$, then $x=z+w$ (additive inverse). By distributive law and 8 (ii) we have $$xy=(z+w)y=zy+wy$$ which can be simplified to $0=wy$. Since $y\neq 0$ we must have $w=0$ by our lemma above, hence $x=z$.

  • $\begingroup$ @EricM.Schmidt I corrected it in my answer, but I wouldn't remove the mistake from the question, as part of the learning process. $\endgroup$ – SK19 Dec 7 '17 at 10:04

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