TITLE: Proving $\sqrt{6}$ is not part of a field
QUESTION [7 upvotes]: I'm currently in a beginning analysis course, and I am asked to prove that $F =\{a+b\sqrt2 +c\sqrt3 :a,b,c∈Q\}$ is not a field. 
I know that this violates the first multiplication axiom, that if $x,y \in F$ then $xy \in F$. However, I don't know how to prove that $\sqrt6$ cannot be written in the form $a+b\sqrt2 +c\sqrt3,$ where $a,b,c∈Q$. Is there a way to show this using elementary algebra, and not go into field extensions?

REPLY [9 votes]: Suppose $\sqrt{6}=a+b\sqrt{2}+c\sqrt{3}$ with $a,b,c$ rational. Then also
$$
\sqrt{6}-a=b\sqrt{2}+c\sqrt{3}
$$
and when you square both sides of this, the only surd around will be $\sqrt{6}$, which makes your life a lot easier. You should be able to manipulate the result to show that
$$
(b^2-3)(c^2-2)=0
$$
which contradicts the original rationality assumption.