Proof. (a) Let $p+q \sqrt{2}, r+s \sqrt{2} \in \mathbb{Q}(\sqrt{2})$. Since $\mathbb{Q} \subset \mathbb{R}$ and $\mathbb{R}$ is a field, we have the following:

Closure under $(+)$

$$

(p+q \sqrt{2})+(r+s \sqrt{2})=(p+r)+(q+s) \sqrt{2} \in \mathbb{Q}(\sqrt{2})

$$

Closure under $(\cdot)$ :

$$

(p+q \sqrt{2}) \cdot(r+s \sqrt{2})=(p r+2 s q)+(r q+p s) \sqrt{2} \in \mathbb{Q}(\sqrt{2})

$$

(b) Suppose that $\sqrt{3}=a+b \sqrt{2} \in \mathbb{Q}(\sqrt{2})$. Note that $b \neq 0$. Thus we have

$$

\begin{aligned}

\sqrt{3}-b \sqrt{2} &=a \\

3-2 \sqrt{2} \sqrt{3} b+2 b^{2} &=a^{2} \\

2 \sqrt{6} b &=3-a^{2} .

\end{aligned}

$$

Since $b \neq 0$,

$$

\sqrt{6}=\frac{3-a^{2}}{2 b}

$$

That is, $\sqrt{6} \in \mathbb{Q}$, which is a contradiction. Therefore, $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$