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The set $\mathbb{Q}$ adjoin $\sqrt{2}$ is defined by $\mathbb{Q}(\sqrt{2})=\{p+q \sqrt{2}: p, q \in \mathbb{Q}\}$.
(a) Show that $\mathbb{Q}(\sqrt{2})$ is a field.
(b) Is $\sqrt{3} \in \mathbb{Q}(\sqrt{2})$ ?
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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, $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$
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