Is this an ordered field? 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?
 A: $F \subseteq \mathbb R$ 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 $ac \leq bc$.
A: You should notice that $a+b\sqrt 2\in F$ is positive iff $a-b\sqrt 2\in\mathbb R$ is positive. So since $\mathbb R$ is ordered, $F$ ought to turn out ordered as well.
And in fact you arriuved just at that point, but fell for a typo:
In what you should demonstrate you wrote as condition "with $e+f\sqrt 2>0$", but the correct condition would be "with $0\lesssim e+f\sqrt 2$".
In fact you had the same typo already in the quoted  multiplication property for general ordered fields.
A: Note that $\mathbb Q(\sqrt 2)$ is a subfield of $\mathbb R$, which is an ordered field, and "expected ordering" is the one derived from $\mathbb R$, and since it is compatible with the field operations on $\mathbb R$ we can say that $\mathbb Q(\sqrt 2)$ with the "expected order" is an ordered field.
However there is not a canonical ordering on $\mathbb Q(\sqrt 2)$ as an extension field of $\mathbb Q$ - the extension does not come with an order attached. 
You can see that if you explore the order defined in the question effectively by $a+b\sqrt 2 \gt 0$ in $\mathbb Q(\sqrt 2)$ if and only if $a-b\sqrt 2 \gt 0$ in the usual order on $\mathbb R$. This just reverses the sign of $\sqrt 2$ wherever it occurs, and arises because $\mathbb Q(\sqrt 2)$ has a non-trivial automorphism which fixes $\mathbb Q$. All the computations are compatible with the order.
So there is a choice of possible orderings, and in this sense we can say that $\mathbb Q(\sqrt 2)$ can be ordered as a field in either way.
Note I would say that it is wrong to say that $\mathbb Q(\sqrt 2)$ "is" an ordered field.
Note also that in any ordered field the squares are positive, and this defines a unique order on $\mathbb R$, for example.
A: QUOTE:  The order relation to be considered is:  $$ \color{magenta}{a+b\sqrt2\ \lesssim
 c+d\sqrt2 \iff c-a+(b-d)\sqrt2 \geq 0} $$
In particular, taking $a=0, \; \; b=0,$ we have 
$$
0 \lesssim c+d\sqrt2 \iff c-d\sqrt2 \geq 0
$$
There is nothing special about the letters, they are all just rational numbers. So, in the line just above, I could take substitutions $c \mapsto a, \; \; d \mapsto b$ and arrive at
$$
0 \lesssim a+b\sqrt2 \iff a-b\sqrt2 \geq 0
$$
