Inverting $a+b\sqrt{2}$ in the field $\Bbb Q(\sqrt{2})$

I have been reading through my notes and I came across this example and I found it hard to understand so I need some help in explaining how the inverse of this is found.

The set $\mathbb Q(\sqrt 2)=\{a+b\sqrt2$: $a,b$ in $\mathbb Q \}$ is a field. The inverse of an element $a+b\sqrt 2$ in $ \mathbb Q(\sqrt 2$) is

$$\frac{1}{a+b\sqrt{2}}=\frac{a}{a^2-2b^2}+\frac{-b}{a^2-2b^2}\sqrt{2} $$

Can anyone explain to how the inverse was found? Appreciate your help.

Solution 1:

Consider an arbitrary element $a + b \sqrt{2} \in \mathbb{Q}[\sqrt{2}]$. From our experience with real numbers, we know that its inverse will be $\displaystyle \frac{1}{a + b\sqrt{2}}$. However, this is not in the "standard" form for elements in $\mathbb{Q}[\sqrt{2}] = \{x + y\sqrt{2} \ | \ x, y \in \mathbb{Q}\}$.

To get this into the desired form, multiply the purported inverse by $\displaystyle \frac{a - b\sqrt{2}}{a - b\sqrt{2}}$. Doing so, we get:

$$\frac{1}{a + b\sqrt{2}} \cdot \Bigg(\frac{a - b\sqrt{2}}{a - b\sqrt{2}}\Bigg) \ = \ \frac{a-b\sqrt{2}}{a^2 - 2b^2} \ = \ \frac{a}{a^2 - 2b^2} + \Bigg(\frac{-b}{a^2-2b^2}\Bigg)\sqrt{2}$$

And this is indeed what we want since $\displaystyle \frac{a}{a^2 -2b^2}$ and $\displaystyle \frac{-b}{a^2 - 2b^2}$ are rational numbers.