Let $\mathbb{Z}[\sqrt p]=\{ a+b\sqrt p ~| a,b\in \mathbb{Z},p~is~prime\} $
Assume $\mathbb{Z}[\sqrt p]$ ia an integral domain with usual addition and multiplication. Show $\mathbb{Z}[\sqrt p]$ is an ordered integral domain with respect to $P=\{ a+b\sqrt p ~|a,b\in\mathbb{Z} ~~~ and ~~~ a+b\sqrt p ~~is~~a~positive~real~number \}$. I know we are only expected to show for any $x\in \mathbb{Z}[\sqrt p] $ exactly on of the following holds $x\in P$ , $x=0$, or $-x\in P$ but I don't Know how to show that.
If I correctly understood the question, just observing $\mathbb{Z}[\sqrt{p}]\subseteq \mathbb R$ and $P\subseteq \mathbb{R}^+$ should work.