<s>[INST] Show that $\{a + b\sqrt{5} \mid a,b \in \mathbf{Q} \}$ is a field. The question asked to prove that $F$ is a commutative field, I got everything except proving that there is a multiplicative inverse in $F$ for every $x\in F$. Any help is appreciated!
Here is specifically what I am asking:

Let $F = \{a + b\sqrt{5} \mid a,b \in \mathbf{Q} \}$.
  Show that $\exists$ a multiplicative inverse $y\in F$ such that $xy = 1$ for all $x \in F$.

 [/INST] We need that $(a+b\sqrt{5})(c+d\sqrt{5})=1$.
Consider $(a+b\sqrt{5})(a-b\sqrt{5})=a^2-5b^2$.
This is a rational number, since both $a$ and $b$ are.
We have that $\displaystyle (a+b\sqrt{5}) \times \boxed{\frac{1}{a^2-5b^2}(a-b\sqrt{5})}=1$.
Therefore, $\displaystyle c=\frac{a}{a^2-5b^2}$, $\displaystyle d=-\frac{b}{a^2-5b^2}$ forms the multiplicative inverse.
The inverse exists, because if $a^2-5b^2=0$, then $a^2=5b^2$, and $a=\sqrt{5}b$. 
This cannot happen, because both $a$ and $b$ are rational.
</s>