TITLE: Schnitt/Dedekind Cuts addition to zero?
QUESTION [0 upvotes]: I'm being asked to prove that if T is a Schnitt, and (-T) is the negative of that Schnitt, then (T) + (-T) = 0.
Both the addition operation and the o are on the Rationals.
I think I need to use cases, and maybe further subcases, but I'm really stuck.

REPLY [0 votes]: I am assuming that you define $-T$ to be $\{-q:q\in\Bbb Q\setminus T\}$, so that
$$T+(-T)=\{r+s:r\in T\text{ and }-s\in\Bbb Q\setminus T\}\,.$$
Let $r\in T$ and $-s\in\Bbb Q\setminus T$; then by the definition of a cut we know that $r<-s$, and hence that $r+s<0$.
Now let $q$ be any negative rational; we want to show that $q\in T+(-T)$. Let $t\in T$.

Show that there is an $n\in\Bbb Z^+$ such that $t-nq\notin T$.

Let $n_0=\min\{n\in\Bbb Z^+:t-nq\notin T\}$, $r=t-(n_0-1)q$, and $s=t-n_0q$; $r\in T$, and $s\notin T$. That is, $s\in\Bbb Q\setminus T$, so $-s\in -T$, and $r+(-s)\in T+(-T)$.

Verify that $r+(-s)=q$, and conclude that $T+(-T)=\{q\in\Bbb Q:q<0\}$, the cut corresponding to the rational number $0$.
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