## Show that a given extension of QQ does not contain sqrt(2)

Show that $\mathbb{Q}(\sqrt{5})$ does not contain $\sqrt{2}$.

Recall that every element of $\mathbb{Q}(\sqrt{5})$ is uniquely of the form $a+b\sqrt{5}$ for some $a,b \in \mathbb{Q}$. Suppose now that $(a+b\sqrt{5})^2 = 2$. Comparing coefficients, we see that $a^2 + 5b^2 = 2$ and $2ab = 0$, so either $a = 0$ or $b = 0$. If $b = 0$, then $a^2 = 2$, and if $a = 0$, then $b^2 = 5/2$. Either case leads to a contradiction since these equations have no solutions in $\mathbb{Q}$.