Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field Question:

Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field under the
  operations - addition and multiplication where addition is given by:
  $(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and
  multiplication is given by $(a + b\sqrt{5})(c + d\sqrt{5}) = (ac+5bd)
 + (ad+bc)\sqrt{5}$

My work:
Since $\mathbb Q$ is a field, addition and multiplication defined by $(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and  $(a + b\sqrt{5})(c + d\sqrt{5}) = (ac+5bd) + (ad+bc)\sqrt{5}$ will produce elements in $F$ therefore $F$ is closed under multiplication and addition.
Because $F$ is a subset of $\mathbb R$ the operation on $F$ correspond to the usual operations on $\mathbb R$ so the associative, commutative and distributive conditions are inherited from $\mathbb R$
The additive identity is $0+0\sqrt{5}$ because $(a + b\sqrt{5})+(0+0\sqrt{5}) = (a + b\sqrt{5})$
The multiplicative identity is $1+0\sqrt{5}$ because $(a + b\sqrt{5})(1+0\sqrt{5}) = (a + b\sqrt{5})$
The additive inverse is $((-a) + (-b)\sqrt{5})$ because $(a + b\sqrt{5})+((-a) + (-b)\sqrt{5}) = 0$
The multiplicative inverse is where I went wrong, and not quite sure what to do, I think I have to find something such that $(a + b\sqrt{5})\times? = 1 = 1+0\sqrt{5}$ I thought of just putting $$\frac{1+0\sqrt{5}}{a+b\sqrt(5)}$$ but I was told this wasn't right so not sure what else to do.
If anyone can help me with checking the multiplicative inverse condition that would be really appreciated.

We just need to simplify that expression so that it has the form:
$$
\alpha + \beta\sqrt{5}
$$
for some $\alpha,\beta \in \mathbb Q$. Indeed, we can do this via conjugates:
\begin{align*}
\frac{1}{a + b\sqrt 5}
&= \frac{1}{a + b\sqrt 5} \cdot \frac{a - b\sqrt 5}{a - b\sqrt 5} \\
&= \frac{a - b\sqrt 5}{a^2 - 5b^2}  \\
&= \underbrace{\frac{a}{a^2 - 5b^2}}_{\in ~ \mathbb Q} + \underbrace{\frac{-b}{a^2 - 5b^2}}_{\in ~ \mathbb Q}\sqrt 5
\end{align*}
To finish this off, can you see why the denominator $a^2 - 5b^2$ is never zero if $a$ and 
$b$ are not both zero?
