# Intuition for getting multiplicative inverse of a Dedekind cut

By a Dedekind cut, we mean, an ordered pair $$(L,U)$$ of subsets of $$\mathbb{Q}$$ such that they are disjoint, their union is $$\mathbb{Q}$$, and

• Each member of $$L$$ is smaller than each member of $$U$$

• $$L$$ contains no largest rational number.

Let $$(L,U)$$ be a Dedekind cut for which $$L$$ contains some positive rationals.

Let $$L'$$ be the collection of non-positive rationals, along with those positive rationals $$x$$ whose product with all positive rationals of $$L$$ is $$<1$$. Let $$U'$$ be the complement of $$L'$$ in $$\mathbb{Q}$$.

Can we say that $$(L',U')$$ defined in this way is the multiplicative inverse of $$(L,U)$$?

(One may see this wiki-link for product of Dedekind cuts).

Yes. One needs to check whether $$(L,U)\times(L',U')=(L_1,U_1)$$, where $$L_1=(-\infty,1)$$, $$U_1=[1,\infty)$$. Note also that only the multiplication of the lower sets needs be considered, since the uppers sets are their complements.

By the definition of multiplication of Dedekind cuts, since $$L$$ and $$L'$$ contain positive rationals, the lower set of their product is defined to contain

• all non-positive rationals
• all positive rationals $$ab$$ where $$a\in L$$, $$b\in L'$$.

Take any rational $$c<1$$. Pick a rational $$a\in L$$ such that $$a>cL$$; this is possible since $$c<1$$. So for any $$b\in L$$, $$(c/a)b<1$$, which implies $$c/a\in L'$$ and so $$c=(c/a)a\in L'\times L$$. This shows that $$(-\infty,1)\subseteq L$$.

Take any rational $$c\ge1$$. If $$c=ab$$ with $$a\in L$$, $$b\in L'$$, it would contradict the way $$L'$$ is defined. So $$L=(-\infty,1)$$.

Actually, you need to be slightly more careful than that.

Suppose the cut (L,U) is divided by a rational value $$f$$ (i.e U has $$f$$ as a minimum value). Then for every positive rational $$r$$ in L, $$f^{-1} r$$ < 1, so $$f^{-1}$$ is in L' by definition. It is easy to see that $$f^{-1}$$ is the maximum value in L', so (L',u') doesn't quite meet the given definition of a cut.