Proof. We take $R_+ = \{r\in R:\,\, 0 < r\} = \{r\in R:\,\, 0\leq r\, \wedge \,0 \neq r\}$ Let $a,\,b \in R_+$ Then $0 < a$ $0 < b$ and therefore we have $0 < a\!+\!0 < a\!+\!b$ i.e. $a\!+\!b \in R_+$ If $R$ has no zero-divisors, then also $ab \neq 0$ , and $0 = a0 < ab$ i.e. $ab\in R_+$ Let $r$ be an arbitrary non-zero element of $R$ Then we must have either $0 < r$ , or $r < 0$ , (not both) because $R$ is totally ordered. The latter alternative gives that $0 = -r\!+\!r < -r\!+\!0 = -r$ The both cases mean that either $r\in R_+$ , or $-r\in R_+$