Show that the field of all numbers of the form $a+b\sqrt 2$ for $a,b\in\Bbb Q$ admits a nonstandard ordering (i.e. not the one coming from the standard ordering of real numbers).
Could anyone kindly help? I have been thinking a long time and have no clue about this.
You can use the automorphism $a+b\sqrt{2} \mapsto a - b\sqrt{2}$ to "pull back" the ordering from the reals and get an ordering in which $\sqrt 2$ is negative. (I'm not really sure if this answers the question, though, as this ordering does come from a real embedding. But it does not come from the "standard" embedding which maps $\sqrt 2$ to $\sqrt 2$.)