# Product of Dedekind cuts: why it is defined in that way?

A Dedekind cut is a partition of $$\mathbb{Q}$$ into two proper subsets $$A,B$$ such that each member of $$A$$ is smaller than each member of $$B$$, and $$A$$ has no largest element.

For Dedekind cuts $$(A,B)$$ and $$(A',B')$$, the addition is little natural: it is $$(A+A',B+B')$$. But the multiplication is little tedious. (I didn't find any rigorous explanation for defining multiplication). In the book Which Numbers are Real by Michael Henle (p.45), the author says

The multiplication of Dedekind cuts is not simple. The problem is determining sign of a product.

Now, the determination of sign in the product which we visit first time is for integers. I saw the link of wiki. It is stated, after the rules of sign for integers, that

this rule (of sign for product of integers) is a necessary consequence of demanding distributivity of multiplication over addition, and is not an additional rule.

By this comment, I thought that we can focus simply first for product of positive integers (with $$0$$) and with above necessary rule of sign, we can completely define the product of all the integers.

Now, similarly, we may think that, we can try to define the product of positive Dedekind cuts, and then with the rule of sign, we can define product of all Dedekind cuts.

However, I didn't find any natural way to get the definition of multiplication of two positive Dedekind cuts; why it is defined in that way? Can anybody give me some natural way to define the product of positive Dedekind cuts?

The reason that Dedekind cuts are awkward for negative reals is that multiplication by negative numbers is not order-preserving.

Consider Dedekind cuts $$(A,B)$$, $$(A',B')$$ representing reals $$>0$$. Then $$A\cap\Bbb Q^+$$ and $$A'\cap\Bbb Q^+$$ are nonempty, where $$\Bbb Q^+=\{x\in\Bbb Q:x>0\}$$. Their product is $$(A'',B'')$$ where $$A''=\{x\in \Bbb Q:x\le 0\}\cup\{aa':a\in A\cap\Bbb Q^+,a\in A'\cap\Bbb Q^+\}.$$

The rationale is that $$A''\cap\Bbb Q^+=\{aa':a\in A\cap\Bbb Q^+,a\in A'\cap\Bbb Q^+\}.$$ This works just like addition, expect we restrict attention to positive numbers, where multiplication is order-preserving.