Concise proof for showing $S = (\mathbb{R}\times \mathbb{R})/\{(0, 0)\}$ is closed 
Consider the system $(S, \otimes)$ where $S = (\mathbb{R}\times \mathbb{R})/\{(0, 0)\}$ and $(a, b) \otimes(c, d) = (ac - bd, ad + bc)$. 

I'm trying to develop a more concise proof for showing that $S$ is closed under $\otimes$. More specifically, I'm trying to assert that $(ac - bd, ad + bc) \neq (0, 0)$. 

Since $a, b, c, d \in \mathbb{R}$ and addition and multiplication are closed in $\mathbb{R}$, $(ac - bd), (ad + bc)  \in \mathbb{R}$. However,  must now show that $(ac - bd, ad + bc) \neq (0, 0)$ to conclude $(a, b) \otimes(c, d) \in S$.  
Equating components, 
$$ac - bd = 0\hspace{1cm}(1)$$ 
$$ad + bc = 0\hspace{1cm}(2)$$
Since $(0, 0)\notin S$ we must consider three possible cases. 
C1. Suppose $a, b \in \mathbb{R_{\neq0}}. $ Then, by solving $(1)$ and $(2)$ we have that $(c, d) = (0, 0)$
C2. Suppose $b = 0, a\in \mathbb{R_{\neq0}}. $ Then, by solving $(1)$ and $(2)$ we have that $(c, d) = (0, 0)$
C3. Suppose $a = 0,  b \in \mathbb{R_{\neq0}}. $ Then, by solving $(1)$ and $(2)$ we have that $(c, d) = (0, 0)$
However, since $(0, 0)$ is not an element of $S$, $(a, b) \otimes (c, d) \neq(0, 0)$. 
This proof makes me a little uncomfortable. Is there something more concise? Or maybe something with better wording?
 A: There's a few ways to attack this.

One would be to recognize that $ac-bd$ is zero if and only if $(a,b)$ and $(d,c)$ are parallel as vectors - that is, presuming that neither is zero, if $(a,b)=\alpha (c,d)$ for some $\alpha\in \mathbb R\setminus \{0\}$. However, then $ad+bc=\alpha (a^2+b^2)\neq 0$. In a sense, we can think of $ac-bd$ as being a sort of exterior product of $(a,b)$ and $(d,c)$, which vanishes if they are parallel, and $ad+bc$ as being the inner product of $(a,b)$ and $(d,c)$, which vanishes if they are perpendicular. Thus, for both to vanish, they would need to be both parallel and perpendicular, which implies one is zero.

Another, more elementary, is to note that we can write the system as follows with matrices:
$$\begin{pmatrix}a & -b\\ b & a\end{pmatrix}\begin{pmatrix}c \\ d\end{pmatrix}=\begin{pmatrix}0 \\ 0\end{pmatrix}.$$
Then, if that first matrix is invertible, this has a unique solution for $c$ and $d$ - namely $c=d=0$. However, the determinant of that first matrix is $a^2+b^2$, thus unless $a=b=0$, the matrix is invertible.

If you know about complex numbers, note that
$$(a+bi)(c+di)=(ac-bd)+(ad+bc)i.$$
Then, your statement just says that $\mathbb C$ is an integral domain. (Of course, maybe this is why you wanted to prove this in the first place)
A: Suppose that $$(ac-bd,ad+bc)=(0,0)$$ 
Then scaling by $c$, $$(ac^2-bcd,acd+bc^2)=(0,0)$$
The original equation implies $ac=bd$ and $bc=-ad$, so 
$$(ac^2+ad^2,bd^2+bc^2)=(0,0)$$
$$(a(c^2+d^2),b(c^2+d^2))=(0,0)$$
Maybe it is a matter of opinion, but I think this clearly implies either $c=d=0$, or $a=b=0$ without any further work shown.
