Proving $\sqrt{6}$ is not part of a field

Question

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?

Answer 1

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.

Answer 2

Assume $a+b\sqrt 2 + c\sqrt 3=\sqrt 6, a,b,c\in \Bbb Q$. Then $\sqrt 6-b \sqrt 2=a+c\sqrt 3.$ Square both sides and you are down to two radicals. Isolate them and square again and you have one.