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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!

Git Gud
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user194772
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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 $$\lim_{n \to \infty}(\sqrt{2}-1)^n = 0$$

Crostul
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  • what if $p$ were defined as $P:={ a+b\sqrt7 ~|a,b\in\mathbb{Z} ~~~ and ~~~ ab\sqrt 7 isa~positive~real~number }$? – user194772 Nov 24 '14 at 12:57
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    $ab\sqrt{7} > 0$ if and only if $ab >0$. So for example, for $a,b <0$ you have no minimum. – Crostul Nov 24 '14 at 13:01