The complete questions states:
On $\mathbb Q\:$[$\sqrt2 $] we define the relation: $\mathbb a+b\sqrt2 < a'+b'\sqrt2$ if $\mathbb a<a'$ and $\mathbb b<b'$ then show that the field $\mathbb Q\:$[$\sqrt2 $] cannot be ordered (i.e., show that < does not constitute a total order on $\mathbb Q\:$[$\sqrt2 $]).
I have already proven that < is anti-reflexive, anti-symmetric, and transitive. So the only way that I can think to prove this is to prove that $\mathbb Q\:$[$\sqrt2 $] is partially ordered and find numbers contradicting $\mathbb a+b\sqrt2 < a'+b'\sqrt2$
i.e. 3+5$\sqrt2 $ vs 5+3$\sqrt2 $ ==> where 3<5 but 5 is not <3
so 3+5$\sqrt2 $ is not less than 5+3$\sqrt2 $ and visa versa. Am I on the right track?