How does Conway's proposed compromise for constructing the real numbers actually work?

Question

My question is about understanding a remark John Conway made in On Numbers and Games (ONAG), where he proposes a method for constructing the real numbers from the rationals. I will have to assume familiarity with Conway's construction of the surreal number field.

The context of this quote is Conway discussing of an unfortunate feature of constructing $\mathbb R$ from Dedekind cuts of $\mathbb Q$. Namely, when defining multiplication $x\cdot y$ of real numbers, you need to break into four cases based on the signs of $x$ and $y$. This means that proving associativity requires breaking into eight cases, which is clunky. One could instead modify Conway's construction of the surreals to construct $\mathbb R$, adding an extra axiom so the extra infinities and infinitesimals are not constructed. This solves the problem clunky associativity proof, since multiplication in the surreal numbers is defined without need for case splitting. However, the full surreal number construction has some drawbacks (being difficult to understand for undergraduates, and giving special status to the dyadic rationals).

To this end, Conway 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.

I surmise this means that real numbers would be defined as ordered pairs $(L,R)$ of sets of rational numbers, where for all $\ell \in L$ and $r\in R$ we have $\ell < r$, and $L\cup R$ excludes at most one member of $\mathbb Q$. My question is, how does Conway's suggestion actually work, in detail?

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That is the end of my question, what follows are my thoughts on the problem.

Following the surreal definition of multiplication, letting $x$ and $y$ be real numbers, and letting $x^L$ and $x^R$ be variables which range over the left and right sections defining $x$, it seems that Conway would define $$ xy=(\{x^Ly+xy^L-x^Ly^L,x^Ry+xy^R-x^Ry^R\},\{x^Ly+xy^R-x^Ly^R,x^Ry+xy^L-x^Ry^L\})\tag1 $$ However, this raises the question, "what does $x^Ly$ mean?". This is the product of a rational with a real number. You cannot use the above definition to define this product, since Conway stipulated the rationals were to be constructed as normal, so they do not have a left and right section. However, it seems the only sensible way to define the multiplication of a rational with $q$ with a real number $x$ would involve case-splitting, as below, defeating the whole purpose. $$ xq= \begin{cases} (\{x^Lq\},\{x^Rq\}) & q>0\\ (\{x^Rq\},\{x^Lq\}) & q<0\\ 0 & q=0 \end{cases} $$

The same problem would occur for the definition of addition: $$ x+y=(\{x+y^L,x^L+y\},\{x+y^R,x^R+y\}) $$ However, there is a nice workaround, where you instead define $$ x+y=(\{x^L+y^L\},\{x^R+y^R\}) $$ This is equivalent to the usual definition of addition in Dedekind's construction of the reals. It works because $x^L<x$ and $y^L<y$ implies $x^L+y^L<x+y$, so that $x^L+y^L$ can be safely put in the left section of $x+y$. Is there a similar fix for the multiplicative definition?

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I feel there must be something of value here. The definition in $(1)$ is very clever. Namely, if you define $x$ as a Dedekind-like cut with left members $x^L$ and $x^R$, and the same for $y$, then you can derive the cut for the product $xy$ using the fact that $x^L<x<r^R$, and similarly $y^L<y<y^R$, which implies the inequalities $$ (x-x^L)(y-y^L)>0,\qquad (x-x^L)(y^R-y)>0, \dots, $$ which implies $xy$ is between the elements of the complicated looking expression in $(1)$. No case splitting is needed. I just cannot quite figure out how to apply that same clever logic to define multiplication of cuts of the already constructed rational numbers.

Answer 1

Maybe not exactly what Conway had in mind, but here's how I would do it to avoid "clunky casing" :

1. It is easy to see that if $(L,R)$ is a cut, representing a real number $x$, then either $L$ contains no positive rational (in which case we say that $x$ is a nonpositive real, and that its sign is $-$) or $R$ contains no nonnegative rational (in which case we say that $x$ is a nonnegative real, and that its sign is $+$).

2. Using the fact that the rationals form an Archimedean field, it follows that the only real number which is both nonnegative and non positive is zero. We then define a positive (negative) real number as a nonzero positive (negative) real number.

3. Define addition as $x+y=(\lbrace x^L+y^L\rbrace,\lbrace x^R+y^R \rbrace)$ and opposition as $-x=(\lbrace -x^R \rbrace,\lbrace -x^L \rbrace)$. This defines an additive group structure on the reals without clunky casing.

4. From the definition of opposition, $x$ is positive iff $-x$ is negative. We can then define the absolute value $|x|$, which is a nonnegative real number.

5. Define an order on the reals by $x \leq y$ iff $y-x\geq 0$. This is a total order on the reals, compatible with addition, without clunky casing.

6. We can then obtain density of the rationals in the following form : If $(L,R)$ is a cut and $n\geq 1$ is an integer, then there is $x^L\in L$, $x^R\in R$ with $x^R-x^L \lt \frac{1}{n}$ (here, we need again the fact that the rationals form an Archimedean field, so that there must exist a largest integer above $nL$ and a smallest integer below $nR$).

7. From 6, we deduce that for any cut $(L,R)$ defining a nonzero number, there is a corresponding adjacent sequence, i.e. an increasing sequence $(l_n)_{n\geq 1}$ with values in $L$, and a decreasing sequence $(r_n)_{n\geq 1}$ with values in $R$, such that $r_n-l_n \lt \frac{1}{n}$ for every $n$. Furthermore, we can also take the two sequences such that all their terms have the same sign, the sign of $(L,R)$ in the sense of 1.

8. Let $x,y$ be positive real numbers, represented by cuts $(L_x,R_x)$ and $(L_y,R_y)$ respectively. Let $R$ be the usual product $R_xR_y=\lbrace x^Ry^R\rbrace$ and $L$ as the restricted product $\lbrace x^Ly^L | x^L \gt 0 \ \text{or} \ y^L \gt 0\rbrace$. Then all the negative rationals (or zero) are contained in $L$ (fix $x^L\gt 0$ and let $y^L$ vary, taking all negative (or zero) rational values). Suppose that there is a positive rational $q$ not in $L\cup R$, then if we take adjacent sequences $(l^x_n),(r^x_n)$ and $(l^y_n),(r^y_n)$ as in 8, we must have $l^x_nl^y_n \leq q \leq r^x_nr^y_n$ for every $n$, from which it follows that there is no rational other than $q$ outside $L\cup R$. Thus $(L,R)$ is a cut, and this allows us to define multiplication of two positive real numbers, which we will denote by $Prod_1$.

9. Associativity of $Prod_1$ and its distributivity over addition follow from the rational analogues.

10. Any real number $x$ has a helper , i.e. a positive real number $h(x)$ such that $x+h(x)$ is also positive (for example, if $x$ is positive or zero you can take $h(x)=1$, and if $x$ is negative you can take $h(x)=-(x-1)$).

11. Let $x$ be a real number and let $y$ be a positive real number. Let $h_1(x)$ and $h_2(x)$ be two helpers for $x$. Putting $H_i(x)=x+h_i(x)$ for $i=1,2$, we have the identity $H_1(x)+h_2(x)=H_2(x)+h_1(x)$, from which it follows that $Prod_1(H_2(x),y)+Prod_1(h_2(x),y)=Prod_1(H_2(x),y)+Prod_1(h_1(x),y)$ and hence that $Prod_1(H_1(x),y)-Prod_1(h_1(x),y)=Prod_1(H_2(x),y)-Prod_1(h_2(x),y)$ ; thus the quantity $Prod_1(x+h(x),y)-Prod_1(h(x),y)$ is independent of the choice of a helper $h(x)$ for $x$, and only depends on $x$ and $y$. We have thus defined the product of a real number with a positive real number, and we will denote this by $Prod_2$.

12. Similarly, let now $x,y$ be two real numbers. Using the same trick as in 11, one can show that the quantity $Prod_2(x,y+h(y))-Prod_2(x,h(y))$ is independent of the choice of a helper $h(y)$ for $y$, which allows us to define the product of any two real numbers.

13. By construction, the product over real numbers thus defined is distributive over addition and substraction, and compatible with the absolute value $|.|$

14. Given any three real numbers $x,y$ and $z$, the two real numbers $p_1=x(yz)$ and $p_2=(xy)z$ have the same absolute value (because of the associativity of $Prod_1$), and in particular they are both zero or both nonzero. If they are both nonzero, they also share the same sign (defined by the parity of the number of negative terms among $x,y,z$), so that they must in fact coincide. This shows associativity.