Show that $\mathbb{Q}(\sqrt{2})$ is the smallest subfield of $\mathbb{C}$ that contains $\sqrt{2}$ This was an assertion made in our textbook but I have no idea how to show that either statement is true.  Also would like to show that that $\mathbb{Q}(\sqrt{2})$ is strictly larger than $\mathbb{Q}$, which was the second part of the assertion.

If $\mathbb Q(\sqrt2)$ is defined as the set of number of the form $p+q\sqrt2$ with $p,q\in \mathbb Q$, it is easy to show it is a field, the only tricky part might be multiplicative inverse, you can use the conjugate for that.
On the other hand any field $F$ that contains $\mathbb Q$ and $\sqrt 2$ also contains $q\sqrt2$ with $q$ rational, since $F$ is closed under products and $q$ and $\sqrt2$ belong to $F$. Finally, since $F$ contains every rational $p$ and every number of the form $q\sqrt2$ it contains their sum: $p+q\sqrt2$.
This proves $Q(\sqrt2)\subseteq F$ as desired.

Please note it would be unusual for $\mathbb Q (\sqrt 2)$ to be defined that way.
