Let $\textbf{Q}(\sqrt{2})$ be the set of all real numbers of the form $r + s\sqrt{2}$, with $r,s\in\textbf{Q}$. Show that $\textbf{Q}(\sqrt{2})$ is a subfield of $\textbf{R}$. Show that there are two total orderings of $\textbf{Q}(\sqrt{2})$ under which it is an ordered field.
MY ATTEMPT
In order to prove it is a subfield of $\textbf{R}$, it suffices to define: \begin{align*} \begin{cases} (a + b\sqrt{2}) + (c + d\sqrt{2}) = (a + b) + (c+d)\sqrt{2}\\\\ (a + b\sqrt{2})\times(c + d\sqrt{2}) = (ac + 2bd) + (ad + bc)\sqrt{2} \end{cases} \end{align*} and check its corresponding properties. I am mainly interested in the second part, but I do not know how to approach it.
Could someone help me to describe such total orderings?
Consider the two maps $\mathbb{Q}(\sqrt{2})\to\mathbb{R}$, one given by sending the element $a+b\sqrt{2}$ to that real number, and the second to $a-b\sqrt{2}$. Verify that each of them is a field embedding. The usual order of the reals then induces an ordering on $\mathbb{Q}(\sqrt{2})$ by transport of structure along each of the two maps, giving you two orderings.