Let $\mathbb{Z}[\sqrt2]=\{ a+b\sqrt2 ~| a,b\in \mathbb{Z}\} $ be an integral domain.
Let $p=\{ a+b\sqrt2 ~|a,b\in\mathbb{Z} ~~~ and ~~~ a+b\sqrt 2 ~~is~~a~positive~real~number \} $
Show that $\mathbb{Z}[\sqrt2]$ is ordered Integral Domain but not well-ordered with respect to $p$. I know how to show that it is ordered with resp. to $p$ but I'm not able to show that it is not well ordered with resp. to $p$.(I couldn't find a subset of $p$ that doesn't have a least element).
Any help is appreciated!
The easiest way to show that $p$ is not well-ordered, is to show that $p$ has no minimal element. This is trivial once you note that $$\lim_{n \to \infty}(\sqrt{2}-1)^n = 0$$