0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $\alpha^*$

$\endgroup$
  • $\begingroup$ That does not make sense to me. That IS the standard definition of a Dedekind cut. Are you talking about the additive inverse? I am assuming you meant α* in the definition given in your response. $\endgroup$ – David Jun 23 '14 at 18:36
  • $\begingroup$ Actually, never mind. I think that was helpful, and it makes sense why the inverse in some definitions are called α*. It's to signify that α* is a rational cut (even in my text book, they use the * to signify a rational cut, don't know why it then defines α* as -α). So, the additive inverse is defined by a rational cut rather than another irrational Dedekind cut. Makes sense. Although, I'm beginning to see why Terry Tao has said there has been much unnecessary philosophical ramblings about the nature of the number systems and their constructions. $\endgroup$ – David Jun 23 '14 at 20:26
  • $\begingroup$ 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 $\{x\in\mathbb Q| x^2 < 2\vee x<0\}$ $\endgroup$ – 5xum Jun 24 '14 at 5:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.