Prove that $K$ is a field Let be $K$ the set of real numbers that can be written as $a+b\sqrt2$, with $a$ and $b$ rational numbers. Prove that $K$ is a field.
I have already proved that $0$ and $1$ $\in K$, and that sum and product of two elements $\in K$. I have also already proved that the opposite $\in K$. I don't know how to prove that the reciprocal $\in K$. Can you help me?

Let $a+b\sqrt{2}\in K$ and $a+b\sqrt{2}\neq 0$. If $b=0$, then $a+b\sqrt{2}=a\neq 0$ and $1/a$ is its multiplicative inverse. Therefore, we assume $b\neq 0$, which implies that 
$a^2-2b^2\neq 0$. Otherwise, if $a^2-2b^2=0$, then $\sqrt{2}=\frac{a}{b}$ which is rational since $a, b$ are rational, which contradicts to the fact that $\sqrt{2}$ is irrational.
Define 
$$\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}.$$
First note that it is well-defined because $a^2-2b^2\neq 0$. Moreover, it belongs to $K$ because $\displaystyle\frac{a}{a^2-2b^2},\frac{b}{a^2-2b^2}$ are rational which follows from the fact that $a,b$ are rational numbers. Finally, we have
$$(a+b\sqrt{2})\cdot\left(\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}\right)=\frac{(a+b\sqrt{2})(a-b\sqrt{2})}{a^2-2b^2}=\frac{a^2-2b^2}{a^2-2b^2}=1.$$
That is to say, $\displaystyle\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}$ is the multiplicative inverse of $a+b\sqrt{2}$.
