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Let $F$ be the set of $\alpha\subset \mathbb{Q}$ with following properties.

(I) $\alpha ≠ \emptyset$ and $\alpha ≠ \mathbb{Q}$

(II) $p\in \alpha$ and $q$q\in \alpha$

(Notice that it's slight different from usual dedekind cut)

Define $\alpha < \beta$ iff $\alpha \subsetneq \beta$. Then $F$ is fully-ordered. Plus, $F$ has least-upperbound property. For $\alpha,\beta \in F$, define $\alpha + \beta$ = {$r+s\in \mathbb{Q}$|$r\in \alpha$ and $s\in \beta$} Then $\alpha + \beta$ also satisfies properties (I)&(II).

Then operation $+$ is associative and commutative and there exists an additive identity $0^*$ that is {$q\in \mathbb{Q}$|$q≦0$}.


Here, i don't know how to prove that 'There doesn't exist additive inverse'. Help

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If there were additive inverses, the set would be a group and addition would be cancellable. However, $\{r\in\mathbb Q\mid r\lt a\}+\{s\in\mathbb Q\mid s\lt b\}=\{r\in\mathbb Q\mid r\lt a\}+\{s\in\mathbb Q\mid s\le b\}$, so addition isn't cancellable, so there can't be additive inverses.