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 $x \in F$.

 A: Hint:
$$(a-b\sqrt{5})(a+b\sqrt{5})=a^2-5b^2$$
Can you show that $a^2-5b^2 \neq 0$ and complete the proof?
A: 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.
A: The statement in OP is wrong:

$\exists$ a multiplicative inverse $y\in F$ such that $xy = 1$ for all $x \in F$

Need to be careful about the position of the quantifiers.  The correct statement you want to show should be:

For any nonzero $x\in F$, there exists $y\in F$ such that $xy=1$. 

The problem is essentially finding $(c,d)$ such that
$$
c+d\sqrt{5}=\frac{1}{a+b\sqrt{5}}
$$
for given $(a,b)$ where $a,b,c,d$ are assumed to be rational numbers. A usual trick is "rationalizing" the denominator:
$$
\frac{1}{a+b\sqrt{5}}=\frac{a-b\sqrt{5}}{(a+b\sqrt{5})(a-b\sqrt{5})}
=\frac{a-b\sqrt{5}}{a^2-5b^2}.
$$
We are done by showing that $a^2-5b^2\neq 0$ for $(a,b)\neq (0,0)$; this is simple by noticing that $\sqrt{5}$ is irrational. 
