Invertible element of $S$

Question

Let $S=\mathbb{Z}[\sqrt{2}]$ = {$a+b\sqrt2|a,b\in \mathbb{Z}$} and $R = \mathbb{Q}[\sqrt2]$ = {$\alpha + \beta\sqrt2 | \alpha, \beta \in \mathbb{Q}$}. Consider $x=3+2\sqrt2$ and $y = 3+4\sqrt2$

1. Determine whether x is an invertible element of S. If so, find its inverse.
2. Determine whether y is an invertible element of R. If so, find its inverse.

My attempt: $x$ is an invertible element of $S$ since $(3+2\sqrt2)*(3-2\sqrt2)$ = 1. And, the inverse of $x$ is $(3-2\sqrt2)$ $\in S$

Is that correct?

And, I am lost on the second one.

Answer 1

Your work is right for the first part. For the second part, suppose you have that $(\alpha+\beta\sqrt{2})(3+4\sqrt{2}) = 1$, then

$$(3\alpha+8\beta)+(4\alpha+3\beta)\sqrt{2} = 1.$$

Since $\alpha,\beta\in\Bbb Q$, the only way to get a rational number is if $4\alpha+3\beta = 0$, i.e. $\alpha = -\frac{3}{4}\beta$. Can you see how to proceed?

Answer 2

Rationalize the denominator !.

$x^{-1} = \dfrac{1}{3+2\sqrt{2}} = \cdots=3-2\sqrt{2}$

$y^{-1} = \dfrac{1}{3+4\sqrt{2}} = -\dfrac{3}{23} + \dfrac{4}{23}\sqrt{2}$