Prove that Q[sqrt(2)] is closed under addition, and show that MIV Holds.
Q[Sqrt(2)] is define by a+bsqrt(2), a and b in Q.
MIV is existence of multiplicative inverse Ex: aa^-1=1
To construct MIV:
$\displaystyle \frac1{a+b\sqrt{2}} = \frac1{a+b\sqrt{2}} \times \frac{a-b\sqrt{2}}{a-b\sqrt{2}} = \frac{a}{a^2 - 2b^2}+\frac{-b\sqrt{2}}{a^2 - 2b^2}$
If $\displaystyle a,b \in \mathbb{Q}$, so will $\displaystyle \frac{-b}{a^2 - 2b^2}$ and $\displaystyle \frac{a}{a^2 - 2b^2}$
Thus $\displaystyle (a+b\sqrt{2})^{-1} \in \mathbb{Q}[\sqrt2]$
To prove its existence, start with the constructed form and multiply out and show that the product is 1