Determining if a set is a group 
Let $S=\lbrace x+y\sqrt2 : x,y\in \mathbb R \rbrace$ \ $\lbrace0\rbrace$. Justify whether $S$, together with traditional multiplication, is a group.

I've verified that the set is closed under the operation, as well as associative, and it has an identity, $1$. I'm having trouble convincing myself that the inverse of each element is in $S$, however. Of course, every non-zero real number has an inverse that's also real, simply $\frac{1}{x+\sqrt2}$.
However, to show that it's of the form required for $S$, I rewrite it as $\frac{1}{x}+\frac{1}{2y}\sqrt2$. But if $x=0$ or $y=0$, this form is undefined. So I can't decide if I've finished the problem or not.
 A: Let $a\in\mathbb{R}\setminus 0$ then $a=(a-\sqrt{2})+\sqrt{2}(1).$ Thus $S=\mathbb{R}^{\times}.$ Since $(\mathbb{R}^{\times},\times)$  is a group then $(S,\times)$ is a group.  
A: Let $S := \{x + y \sqrt 2 : x, y \in \mathbb R\} \setminus \{0\}$.


*

*Observe that $1 \in S$ by choosing $x = 1$ and $y = 0$.

*If $s \in S$ where $s = x + y \sqrt 2$ with $x, y \in \mathbb R$. Notice that $x = 0$ and $y = 0$ can't both be true. If $x = 0$, then $y \neq 0$ and the inverse would be $\displaystyle s^{-1} = \frac{1}{2y} \sqrt 2$. If $y = 0$, then $x \neq 0$ and the inverse would be $\displaystyle s^{-1} = \frac{1}{x}$. Notice that $x^2 - 2y^2 \neq 0$ for if it were, then $(x + y \sqrt 2) (x - y \sqrt 2) = 0$ and since $\mathbb R$ is an integral domain, one of these must be zero which implies that $x, y = 0$, a contradiction. Now suppose $x, y \neq 0$. Defining $\displaystyle a = \frac{x}{x^2 - 2y^2}$ and $\displaystyle b = \frac{-y}{x^2 - 2y^2}$ we get that $s^{-1} = a + b \sqrt 2$.

*$s, t \in S$ where $s = x + y \sqrt 2$ and $t = a + b \sqrt 2$, then $st = (xa + 2yb) + (xb + ya) \sqrt 2$ which means $st \in S$.

