# Why is this definition of an additive inverse significant

In the process of learning Real Analysis, I encountered a definition of an additive inverse of a cut $\alpha$ to be $$\text {add inv of } \alpha \colon= \{p:\exists r>0 \text{ s.t.} (-p-r)\notin \alpha\}$$ The definition does make sense but I can't seem to understand the motivation behind defining it this way.
Why is this definition significant? Isn't this a bit uncalled for? Why good does this definition do to us ? Does it make some things easier (how so?)?
Sorry for asking too many questions at once. The thing is, I am a High school student. I just started studying Real Analysis on my own and even though it has been quite a rough road for me, the experience has been truly rewarding and satisfying.
I am slowly getting used to the rigor in mathematics and so I can't grasp these kinds of concepts as easily. So could you please elaborate a bit more than necessary.

Any help is much appreciated!

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I wouldn't view the construction of the reals as being indicative of the kind of thinking that you need in order to do real analysis. – Dylan Moreland Jan 8 '12 at 19:05
@DylanMoreland. Come on, he said he's just started. Besides, he's still in high school. To begin real analysis by constructing $\mathbb{N}$, then $\mathbb{Q}$ and $\mathbb{R}$ is the right way to go. – Samuel Tan Jan 9 '12 at 2:08
@Samuel I didn't mean to sound critical. I just meant that the construction is somewhat technical and that one can happily solve problems in, say, Rudin without having a great feel for cuts. In other words, the OP should not feel discouraged. I agree that it's good to work this out once in your life. – Dylan Moreland Jan 9 '12 at 2:31

How are you defining cuts? For the usual definition (which I give below), $-\alpha$ turns out to be $\{p:\exists r>0\, (-p-r\in \alpha)\}$, not $\{p:\exists r>0\, (-p-r\notin \alpha)\}$. I’ll go through this derivation in some detail, in hopes that you can adapt it to your situation.

To understand where the definition of $-\alpha$ comes from, you have to go back to the definition of the sum of two cuts. I’m going to assume the following definitions of cut and of addition of cuts:

$\alpha\subseteq \mathbb{Q}$ is a cut iff
$\;$
$\qquad(1)\qquad\varnothing\ne \alpha\ne\mathbb{Q}$;
$\qquad(2)\qquad\text{if }p<q\in \alpha,\text{ then }p\in \alpha$; and
$\qquad(3)\qquad \alpha\text{ has no greatest element}$.

If $\alpha$ and $\beta$ are cuts, $\alpha+\beta\triangleq\{p\in\mathbb{Q}:\exists a\in \alpha\,\exists b\in \beta(p=a+b)\}$.

I’ll also assume that the embedding $^*$ of $\mathbb{Q}$ into the set of cuts has been defined so that $q^*=\{p\in\mathbb{Q}:p<q\}$, so that $0^*=\{q\in\mathbb{Q}:q<0\}$ is intended to be the additive identity.

Clearly we want $-\alpha$ to be defined so that $\alpha+(-\alpha)=0^*$. Thus, I want $-\alpha$ to satisfy the following condition:

$\qquad\qquad\qquad\qquad$ for all $p\in\mathbb{Q}$, $p<0$ iff $\exists a\in \alpha\,\exists b\in -\alpha(p=a+b)$.

This is a bit hard to sort out, so let’s pretend for a moment that we’ve actually constructed $\mathbb{R}$ and look at a concrete example. Suppose that $\alpha$ is intended to be $\sqrt 2$: $\alpha=\{p\in\mathbb{Q}:p<\sqrt 2\}$. Clearly $-\alpha$ should be $-\sqrt 2$, or $\{p\in\mathbb{Q}:p<-\sqrt 2\}$. But $p<-\sqrt 2$ iff $-p>\sqrt 2$, so $-\alpha$ ought to be $\{p\in\mathbb{Q}:-p>\sqrt 2\}=\{p\in\mathbb{Q}:-p\notin\alpha\}$. This suggests that perhaps we should define $-\alpha$ in general to be $\{p\in\mathbb{Q}:-p\notin \alpha\}$. This almost works, but there’s a small problem if $\alpha$ is a rational cut $q^*$: in that case $\{p\in\mathbb{Q}:-p\notin q^*\}=\{p\in\mathbb{Q}:-p\ge q\}=\{p\in\mathbb{Q}:p\le -q\}$, which is not a cut, since it has a greatest element. We want $-(q^*)$ to be simply $(-q)^*$, i.e., $\{p\in\mathbb{Q}:p<-q\}$. There’s a simple (if slightly clumsy-looking) way around the problem: we simply set $$-\alpha\triangleq\{p\in\mathbb{Q}:-p\notin\alpha\text{ AND }-p\text{ is not the least element of }\mathbb{Q}\setminus\alpha\}\;.\tag{1}$$ It’s not hard to check that this really does define a cut. To check that it defines the right cut, suppose first that $a\in\alpha$ and $b\in-\alpha$. Then $-b\notin\alpha$, so $a<-b$, and therefore $a+b<0$, as desired.

Now suppose that $p$ is any negative rational; we need to show that there are $a\in\alpha$ and $b\in-\alpha$ such that $p=a+b$. This amounts to finding $a\in\alpha$ and $-b\in\mathbb{Q}\setminus\alpha$ such that $-b$ is not the least element of $\mathbb{Q}\setminus\alpha$ and $a+b=p$ or, equivalently, $a-p=-b$.

Let $r=-p>0$; then we want to find $a\in\alpha$ and $-b\in\mathbb{Q}\setminus\alpha$ such that $-b$ is not the least element of $\mathbb{Q}\setminus\alpha$ and $a+r=-b$. To give this some intuitive content, we’re trying to find rationals $a$ and $-b$ that are exactly $r$ apart and that ‘straddle’ the cut $\alpha$.

To do this, let $q\in\alpha$ and $s\in\mathbb{Q}\setminus\alpha$ be arbitrary. Since $\mathbb{Q}$ is non-Archimedean, there is an $n\in\mathbb{Z}^+$ such that $n>(s-q)/r$ and hence $q+nr>s$, so $\{n\in\mathbb{Z}^+:q+nr\notin\alpha\}\ne\varnothing$. Let $m=\min\{n\in\mathbb{Z}^+:q+nr\notin\alpha\}$; then $q+(m-1)r\in\alpha$ and $q+mr\notin\alpha$. Thus, if $q+mr$ is not the least element of $\mathbb{Q}\setminus\alpha$, we can simply take $a=q+(m-1)r$ and $-b=q+mr$ to get $a+r=-b$ with $a$ and $-b$ as desired.

If $q+mr$ is the least element of $\mathbb{Q}\setminus\alpha$, we have to be a little cleverer: in this case we let $a=q+\left(m-\frac12\right)r$ and $-b=q+\left(m+\frac12\right)r$. Clearly $a+r=-b\in\mathbb{Q}\setminus\alpha$, $-b$ is not the least element of $\mathbb{Q}\setminus\alpha$, and $a<q+mr=\min(\mathbb{Q}\setminus\alpha)$, so $a\in\alpha$, again exactly as desired.

Notice that in the indented part of the argument we were actually showing that if $r$ is any positive rational, there are $a\in\alpha$ and $p\in-\alpha$ such that $-r=a+p$ or, equivalently, such that $a=-p-r$. In other words, we were showing that $\{p:\exists r>0\;(-p-r\in\alpha)\}\subseteq-\alpha$. On the other hand, if $p\in-\alpha$ by definition $(1)$, then $-p\notin\alpha$; let $a\in\alpha$ be arbitrary, set $r=-p-a$, and note that $r>0$ and $-p-r=a\in\alpha$, showing that $-\alpha\subseteq\{p:\exists r>0\;(-p-r\in\alpha)\}$. Thus, definition $(1)$ can be replaced by: $$-\alpha\triangleq\{p\in\mathbb{Q}:\exists r>0\;(-p-r\in\alpha)\}\;.\tag{2}$$

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