Multiplication and reciprocals of Dedekind cuts

Multiplication of Dedekind cuts (Dedekind cut, The real numbers $\mathbb{R}$ as Dedekind cuts) is defined first for nonnegative cuts, then extended to all cuts by their signs (Order on the Dedekind reals) via the absolute value.

Positive case. For cuts $A,B>0^{\ast }$ (strictly positive),

$A\cdot B:=\{q\in \mathbb{Q}:q\le 0\}\cup \{ab:a\in A,b\in B,a>0,b>0\}\mathrm{.}$

Absolute value. $\mathrm{\mid }A\mathrm{\mid }:=A$ if $A\ge 0^{\ast }$, and $\mathrm{\mid }A\mathrm{\mid }:=-A$ otherwise (Addition, negation, and subtraction of Dedekind cuts for $-A$); thus $\mathrm{\mid }A\mathrm{\mid }\ge 0^{\ast }$ always, and $\mathrm{\mid }{0}^{\ast }\mathrm{\mid }=0^{\ast }$.

Sign extension. For arbitrary cuts $A,B$,

• $A\cdot B:=0^{\ast }$ if $A=0^{\ast }$ or $B=0^{\ast }$;
• $A\cdot B:=\mathrm{\mid }A\mathrm{\mid }\cdot \mathrm{\mid }B\mathrm{\mid }$ if $A,B$ are both $>0^{\ast }$ or both $<0^{\ast }$;
• $A\cdot B:=-(\mathrm{\mid }A\mathrm{\mid }\cdot \mathrm{\mid }B\mathrm{\mid })$ if $A,B$ have opposite signs.

Identity. The multiplicative identity is $1^{\ast }=\{r\in \mathbb{Q}:r<1\}$.

Reciprocal. For $A>0^{\ast }$,

$A^{-1}:=\{p\in \mathbb{Q}:p\le 0\}\cup \{p>0:\mathrm{\exists }s\in \mathbb{Q},s>0,s\mathrm{\notin }A,p<1\mathrm{/}s\}\mathrm{.}$

Equivalently, a positive rational $p$ lies in $A^{-1}$ iff $1\mathrm{/}p$ is an upper rational bound of $A$ that is not the least one (Rudin's construction). For $A<0^{\ast }$, set $A^{-1}:=-((-A{)}^{-1})$. Division is $A\mathrm{/}B:=A\cdot B^{-1}$ for $B\mathrm{\ne }{0}^{\ast }$.

• The product formula is stated only for strictly positive cuts $A,B>0^{\ast }$: then there exist positive $a\in A$, $b\in B$, so the positive products $ab$ are nonempty and, because $A,B$ have no greatest element (axiom (C3)), $A\cdot B$ has none either; together with downward closure this makes $A\cdot B$ a genuine cut, the clause $\{q\le 0\}$ being absorbed below those positive products. The formula is deliberately not applied at the boundary $A=0^{\ast }$ or $B=0^{\ast }$, where $\{q\le 0\}$ would leave $0$ as a greatest element and so fail to be a cut; products with a zero factor are supplied instead by the first sign rule, $A\cdot B=0^{\ast }$, so the operation is well posed on all cuts.
• On the rational embedding $q\mapsto q^{\ast }$ the operation agrees with $\mathbb{Q}$: $(pq{)}^{\ast }=p^{\ast }\cdot q^{\ast }$ (The rational cuts embed densely in $\mathbb{R}$, preserving sums, products, $0$, $1$ and the order); with $q^{\ast }\cdot (1\mathrm{/}q{)}^{\ast }=1^{\ast }$ that gives $(q^{\ast }{)}^{-1}=(1\mathrm{/}q{)}^{\ast }$ for $q>0$, the reciprocal being the one supplied by For a positive cut $A$, the reciprocal $A^{-1}$ satisfies $A\cdot A^{-1}=1^{\ast }$.
• That these operations send cuts to cuts and satisfy the field axioms ( commutativity, associativity, distributivity over addition, identity $1^{\ast }$, and $A\cdot A^{-1}=1^{\ast }$ for every $A\mathrm{\ne }{0}^{\ast }$) is The Dedekind reals form a field.