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I have also asked this question on Math Stack Exchange (link).

In On Numbers and Games, after discussing the inclusion of the real numbers in the surreal numbers, No, Conway discusses the merits of his construction versus historical ones. While constructing the reals via Dedekind cuts is straightforward, it has the clunky feature that multiplication $xy$ is defined in four cases based on the signs of $x$ and $y$. This means that proving the associativity of multiplication requires examining eight cases. The benefit of the surreal multiplication definition, where all numbers are defined as cuts of (previously constructed) numbers, is that multiplication is cleanly defined "genetically" in terms of its left and right options, without need for case splitting. However, using the surreal numbers to construct the real numbers has its own aesthetic drawbacks, namely being unnecessarily counter-intuitive, and giving special treatment to the dyadic rationals.

Conway then proposes the following compromise:

There is another way out. If we adopt a classical approach as far as the construction of $\mathbb Q$, and the define the reals as Dedekind sections of $\mathbb Q$ with the definitions of addition and multiplication given in this book, then all formal laws have 1-line proofs and there is no case splitting.

My question is, how does Conway's suggestion work, in detail? From what I can surmise, Conway is proposing defining real numbers $x$ as $x=\{x^L|x^R\}$, where $x^L$ and $x^R$ range over rational numbers, so $x^L<x^R$ for all such options, and at most one rational number does not appear as $x^L$ or $x^R$. Then the product of two real numbers would be defined as $$ xy=\{x^Ly+xy^L-x^Ly^L,x^Ry+xy^R-x^Ry^R\mid x^Ly+xy^R-x^Ly^R,x^Ry+xy^L-x^Ry^L\} $$ However, how does one define the product $x^Ly$? This is the product of a real number with a rational number, and rational numbers are to be defined "classically" without left and right options, so the above definition does not suffice. One could define, for real $x$ and rational $q$, $$ xq=qx:=\begin{cases}\{x^Lq\mid x^Rq\} & q>0 \\ \{x^Rq\mid x^Lq\} & q <0 \\ 0 &q=0 \end{cases} $$ but this would cause the case-splitting issue that Conway was trying to resolve.

I will note that the same issue arise when defining the sum of real numbers, but it can be circumvented. The surreal definition of addition is $$ x+y=\{x+y^L,x^L+y\mid x+y^R,x^R+y\}, $$ which in Conway's compromise would run into the same problem where $x+y^L$, the sum of a real number and rational number, needs to be defined separately. However, this can be modified to work, if you instead define $$ x+y=\{x^L+y^L\mid x^R+y^R\}, $$ and indeed this is exactly how addition of Dedekind cuts is defined. I wonder if there is a similar fix for multiplication of cuts.

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  • $\begingroup$ I already commented that on MSE but the fix for addition does not yield a group law. $\endgroup$
    – nombre
    Commented May 17, 2020 at 10:37
  • $\begingroup$ Does it not work to first identify the rational number $x$ with the real number $\{x^L\mid x^R\}$? $\endgroup$ Commented May 17, 2020 at 12:58
  • $\begingroup$ @TimothyChow I think this is what one would want to avoid: the reliance on dyadic numbers and perhaps also induction. $\endgroup$
    – nombre
    Commented May 17, 2020 at 16:52
  • $\begingroup$ @TimothyChow I think the definition of multiplication would then be circular. To define the product of a real and a rational, you would first have to define the product of a real and a rational. $\endgroup$ Commented May 17, 2020 at 16:54
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    $\begingroup$ Sorry for not being clear. I was thinking: (1) Define $\mathbb Q$ classically. This way you get all the non-dyadic stuff. (2) Define $\mathbb R$ as you said, letting $x = \{x^L \mid x^R\}$ where $x^L$ and $x^R$ range over $\mathbb Q$. (3) Prove as a theorem that $\mathbb Q$ is isomorphic to a certain subfield of $\mathbb R$. This is what I meant by "identifying" a rational number with a certain real number. (4) Use this identification to "uniformize" the definition of $\mathbb R$ so that you don't have to split into cases any more. $\endgroup$ Commented May 17, 2020 at 19:30

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