My book says yes, but I'm not so convinced. There is:
$$
F:=\{a+b\sqrt2: a,b \in \mathbb Q\}
$$
Sum and product are defined in $F$ as follow.
$$
(a+b\sqrt2)+(c+d\sqrt2) := a+c+(b+d)\sqrt2\\
(a+b\sqrt2)·(c+d\sqrt2) := ac+2bd+(ad+bc)\sqrt2
$$
The order relation to be considered is:
$$
a+b\sqrt2\ \lesssim c+d\sqrt2 \iff c-a+(b-d)\sqrt2 \geq 0
$$
I'm not so convinced due to the multiplication property of ordered fields:
$$
\forall a,b,c,\ c\gt0,\ a\lesssim b\implies a·c\lesssim b·c
$$
So I should demonstrate that
$$
(a+b\sqrt2)·(e+f\sqrt2)\lesssim (c+d\sqrt2)·(e+f\sqrt2);\\
\text{with } e+f\sqrt2 \gt0
$$
From the product definition this is equivalent to:
$$
(ae + 2bf) + (af+be)\sqrt2 \lesssim (ce+2df) + (cf+de)\sqrt2
$$
From the relation operation definition:
$$
(ce+2df)-(ae+2bf) + [(af+be)-(cf+de)]\sqrt2 \geq0\\
(c-a)(e-f\sqrt2)+(b-d)(e-f\sqrt2)\sqrt2 \geq0\\
(e-f\sqrt2)[c-a+(b-d)\sqrt2] \geq0
$$
I cannot say if $e-f\sqrt2 \geq0$, so I cannot say that $F$ is an ordered field. Am I right or is my book right?
Best Answer
$F \subseteq \mathbb R$ with the field operations and ordering induced from $\mathbb R$. So your book is right, $F$ is ordered.
Instead of looking at formulas, just think of the elements of $F$ as real numbers. If you have two real numbers $a \leq b$ and a third $c > 0$ then, because they are just real numbers, you have $ac \leq bc$.