I'm having difficulty knowing where to start with this question:
Show that:
($\mathbb{Q}(\sqrt{2})$,<)$\cong$($\mathbb{Q}$,<)
I'm having difficulty knowing where to start with this question:
Show that:
($\mathbb{Q}(\sqrt{2})$,<)$\cong$($\mathbb{Q}$,<)