Trying to understand a proof that $\sqrt{3} \not \in \mathbb{Q}(\sqrt{2})$ I am trying to show show that $\sqrt{3} \not \in \mathbb{Q}(\sqrt{2})$. I can do this by assuming that $ \sqrt{3} = a + b\sqrt{2}$ for $a, b \in \mathbb{Q}$ and then drawing a contradiction. My textbook shows this by stating that $ \sqrt{3} \not \in \mathbb{Q}(\sqrt{2}) $ as $ 3 $ cannot be written as $ 2 c^2 $ for $c \in \mathbb{Q} $. I am struggling to understand how the textbook draws this conclusion. 
 A: If $\sqrt{3} = a+b\sqrt{2}$, squaring everything yields $3 = $ some rational $+2ab\sqrt{2}$. Since $3$ is rational, this implies $ab\sqrt{2}$ is as well, hence $a=0$ or $b=0$. 
Clearly $\sqrt{3}$ is not rational, so $a=0$. Now squaring again the first identity gives the conclusion of the textbook
A: $\sqrt 3 = a + b\sqrt 2$ implies $3 = (a + b \sqrt 2)^2 = a^2 + 2b^2 + 2 a b\sqrt 2$.
If $ab\ne0$, then $\sqrt 2$ would be rational.
If $b=0$, then $3 = a^2$, and $\sqrt 3$ would be rational.
If $a=0$, then $3 = 2b^2$, which is what your book says, and $3$ would be even.
A: Suppose $\sqrt{3}=a+b\sqrt{2}$ as you suggest with $a,b\in \mathbb{Q}$. Then square both sides. You get:
$3 = a^2+2b^2+2ab\sqrt{2}$
$2ab \sqrt{2} = 3-a^2-2b^2$
Suppose $a=0$. Then you get $0 = 3-2b^2$. Since you cannot represent $3=2b^2$ for any $b\in \mathbb{Q}$, it must be that $a\neq 0$. Suppose $b=0$. Then you have $0=3-a^2$. Since $\sqrt{3} \notin \mathbb{Q}$, this is not possible, either. Therefore, you have $a,b \neq 0$. This gives:
$\sqrt{2} = \dfrac{3-a^2-2b^2}{2ab} \in \mathbb{Q}$, which again is a contradiction.
