Prove that $G = \{a+b\sqrt{2}\}$ is an abelian group 
Let $G = \{a+b\sqrt2 \mid a, b \in \mathbb{Q}\}$, prove that $(G,+)$ is an abelian group.

I know that I just have to show the definition of an abelian group holds for this set but I'm wondering now what I'm even allowed to use to prove that? I mean, we could just argue that all numbers of the form $a+b\sqrt2$ are real numbers and for example commutativity would directly follow, but I saw a proof in a book using the fact that
$$(a+b\sqrt2) + (c+d\sqrt2) = (a + c) + (b + d)\sqrt2.$$
But at that stage in the book, the distributive law hasn't been introduced yet, so why are we allowed to use that?

Hint: If addition in $G$ is defined by $(a+b\sqrt2) + (c+d\sqrt2) = (a + c) + (b + d)\sqrt2$, then it coincides with the addition in $\mathbb R \supseteq G$, and there is not much left to prove.
