Prove that if $x,y \in S$, then $x+y \in S$ using partial order relation properties

Question

Let $S \subseteq \mathbb{N}$ and for any $x,y \in \mathbb{N}, x\rho y$ if and only if there exist $z \in S$ such that $x+z=y$. Show that if $\rho$ is a partial order relation then

(a) $0 \in S$

(b) If $x,y \in S$ then $x+y \in S$

Since $\rho$ is reflexive, $x\rho x$ so that $x+0=x$ then 0 $\in S$

The second question I can't prove it right.

Answer 1

One can view the relation in the following way: $x\rho y\iff y-x\in S$. Suppose that $x,y\in S$. Then \begin{equation*} 0\rho y ~(\text{as } y\in S), \text{ and also } y\rho(x+y) ~ (\text{as } x\in S).\end{equation*} By transitivity one obtains $0\rho (x+y)$, i.e., $x+y\in S$.