In some notes of Niels Schwartz it is claimed that the following are equivalent for every ideal of poring (partially ordered ring).
- $a+b\in I \Rightarrow a,b\in I$;
- $a,b\in I$ and $a<c<b \Rightarrow c\in I$.
I cannot prove the $2\Rightarrow 1$ direction.
Meta. I don't know how to tag this question. Please, bare with me and amend the tags.
You are missing some important context in the question statement. First, $a$ and $b$ must be contained in $P$, the "positive" elements in the ring. Second, the inequalities in the second condition are weak, not strict. Now to show that $2\implies 1$, note that as $b\geq0$, we know $a+b\geq a+0=a$. So $0\leq a\leq a+b$. As $0$ is contained in every ideal, this shows that if $a+b\in I$ then $a\in I$, from which $(a+b)-a=b\in I$ as well.