The problem is to find the multiplicative inverse of a cut $x=A|B$, that is a cut $y=C|D$ such that $x\cdot y=1^*$.

How do you even approach this problem? The best way is to try an example, let's say $x=2^*$. What do you think the inverse should look like in this case? Once your intuition has developed, the trick is finding a way to write it out mathematically.

### (a) Positive Cuts

First, if $x>0^*$, the trick is to notice that all elements in $\frac{1}{B}$ should be in $C$ (except perhaps the smallest element of $B$), as should 0 and all negative numbers. All elements in $\frac{1}{A^+$}$ where $A^+$ is the *positive* elements of $A$, should be in $D$. Let

(1)
\begin{align} C=\{q\in\mathbb{Q}:q<\frac{1}{r} \text{ for some } r\in B\}, \end{align}

and let $D=\mathbb{Q}\setminus C$. Why does this work, i.e., why does $(A|B)\cdot(C|D)=1^*$? We expect $C$ to contain all negative numbers and zero, which it certainly does, and if $p\in B$, then $\frac{1}{p}<\frac{1}{r}$ for some $r\in B$, unless $p$ is the smallest element of $B$.

### (b) Negative Cuts

If $x<0^*$, the construction is similar, but this time $C$ should contain only the elements in $\frac{1}{B^-}$, where $B^-$ is the *negative* elements of $B$ (again, exclude the endpoint if it exists). Thus

(2)
\begin{align} C=\{q\in\Q:q<\frac{1}{r} \text{ for some } r\in B^-\}, \end{align}

and let $D$ be everything else.

### (c) Uniqueness

*How do you show something is unique?*
- either show that there is only one logical object which satisfies the properties,
*or* show that two objects which satisfy the desired properties must be equal.
*How do you show two objects are equal?*
- If they are sets, show that every element in one is contained in the other, and vice versa? If the objects satisfy
**trichotomy**, then you might show $a\leq b$ and $b\leq a$.

So there are a number of approaches here. We'll take the *direct* approach:

Suppose that $x=A|B$ is positive. Then $y=C|D$ must be positive, and must satisfy

(3)
\begin{align} E=\{r\in\mathbb{Q} : r<0 \text{ or there exist positive $a\in A,c\in C$ such that } r=ac\}=\{r\in\mathbb{Q}: r<1\}, \end{align}

by definition of cut multiplication (bottom of page 14 in the text). Therefore, for any positive $r<1$, there exist positive $a\in A,c\in C$ such that $r=ac$. Therefore, $c\in C$ if and only if there exists a positive $a\in A$ such that $ac<1$, and $C$ is unique. $\blacksquare$