# Additive inverse of a Dedekind cut — is my definition alright?

I am struggling to understand why the Additive Inverse is typically defined as such:

$$\alpha^∗ := \{x \in \mathbb Q | \exists r > 0\text{ such that }−x−r\notin \alpha\}$$

or in another form

$$\alpha^* := \{−p−r : p\notin \alpha, r>0\}$$

or from my book

$$−\alpha = \{r \in \mathbb Q : r < -s\text{ for some }s\notin\alpha\}$$

For clarity,

$$\begin{array} a\alpha :=& \{x\in\mathbb Q | x < a\text{ for some }a\in\mathbb R\}\\ 0^* :=& \{x\in\mathbb Q| x < 0\}\end{array}$$

So, α∗ + α = 0*

I've dug around and found this:

motivation of additive inverse of a Dedekind cut set

which I already understood from the get-go. I know that α∗ (the additive inverse) is defined as such because if not, then the complement α∗ of a rational cut α would contain the element a. Since a is rational, then the complement α∗ would have a greatest rational a. Also, to further extend this reasoning, since 0* is defined for elements strictly less than 0, if a was included in the complement, then there may be two elements in α∗ and α such that these two elements add to 0, which is not in 0*. So, I get all of that. I'm wondering, why is the additive inverse not just this?

α* := {x∈Q | x < -a, for some a∈R}

It just seems much less fussy than the above definition of α* which takes the complement of α, then creates a negative image of the complement, and creates a smaller subset of that. It just seems so unnecessary. My definition of a* seems to fit all of the criteria of a Dedekind cut: it's non empty, it's not Q, it is closed below, and it contains no largest element. Also, when added to α, it gives 0*.

What am I missing here?

The definition $$\alpha = \{x\in \mathbb Q| x<a\text{ for some }a\in\mathbb R\}$$ implies that you already have the real numbers, when in reality, Dedekind cuts are used to define the reals, so when you are given an arbitrary Dedekind cut $\alpha$, you have no real number $a$ of which you can take $-a$ to create $\alpha^*$
• The thing is that you do not define the dedekind cut "$\sqrt 2$" as $\{x\in\mathbb Q| x<\sqrt 2\}$. You define it as $\{x\in\mathbb Q| x^2 < 2\vee x<0\}$ – 5xum Jun 24 '14 at 5:24