Field extensions

Subfield. Closed under addition, subtraction, multiplication: $(a+b\sqrt2)(c+d\sqrt2)=(ac+2bd)+(ad+bc)\sqrt2$.

Inverse. For $a+b\sqrt2\neq0$: $a^2-2b^2\neq0$ (since $\sqrt2\notin\mathbb{Q}$). Then
$$\frac{1}{a+b\sqrt2}=\frac{a-b\sqrt2}{a^2-2b^2}\in\mathbb{Q}(\sqrt2).$$

Degree. $\sqrt2$ is a root of $X^2-2$, irreducible over $\mathbb{Q}$ (rational root test). So the minimal polynomial has degree $2$, and $[\mathbb{Q}(\sqrt2):\mathbb{Q}]=2$.

Conclusion. $\mathbb{Q}(\sqrt2)$ is a field extension of $\mathbb{Q}$ of degree $2$, with basis $\{1,\sqrt2\}$.