$(a,b) \mathbin\# (c,d)=(a+c,b+d)$ and $(a,b) \mathbin\&(c,d)=(ac-bd(r^2+s^2), ad+bc+2rbd)$. Multiplicative inverse? Let $r\in \mathbb{R}$ and let $0\neq s \in \mathbb{R}$. Define operations $\#$ and $\&$ on $\mathbb{R}$ x $\mathbb{R}$ by $(a,b) \mathbin\#(c,d)=(a+c,b+d)$ and $(a,b) \mathbin\&(c,d)=(ac-bd(r^2+s^2), ad+bc+2rbd)$ as addition and multiplication, respectively. Do these operations form a field?
I have shown that every axiom of the field is satisfied except for the "existence of multiplicative inverses" axiom. This is where I need help. We either need to show that this axiom fails and so we don't have a field or that the axiom $aa^{-1}=1$ is satisfied for every $a\neq 0 \in \mathbb{R}$
The multiplicative identity in this problem is $(1,0)$.
Any solutions or hints are greatly appreciated.

The addition and multiplication formulas somewhat remind of the formulas for complex number. We suspect that a map of the form
$$ \phi\colon (x,y)\mapsto \alpha x+\beta y$$
for suitable $\alpha,\beta\in\mathbb C$ turns out to be a field isomorphism. As we want $(1,0)\mapsto 1$, we conclude $\alpha=1$.
Then 
$$ \phi((a,b)\mathop{\&}(c,d))=\phi(a,b)\phi(c,d)=(a+\beta b)(c+\beta d) = ac+\beta(bc+ad)+\beta^2bd$$
and on the other hand
$$\phi(ac-bd(r^2+s^2),ad+bc+2rbd)=ac-bd(r^2+s^2)+\beta(bc+ad+2rbd)$$
so that we obtain an identity by chosing $\beta$ with $\beta^2=2r\beta-(r^2+s^2)$, i.e., one of the roots of the polynommial  $X^2-2rX+r^2+s^2$.
For $s\ne 0$, this garantees $\beta\in\mathbb C\setminus\mathbb R$. Therefore the image of $\phi$ in $\mathbb C$ is more than just $\mathbb R$; as $\phi$ is linear $\mathbb C$ and $F$ both have diemnsion $2$, we conclude that $\phi$ is a vector space isomorphism. As the choice of $\beta$ guaranteed compatibility with multiplication, $\phi$ is indeed a ring isomorphism and as $\mathbb C$ is a field, so is $F$.
