Concise proof for showing $S = (\mathbb{R}\times \mathbb{R})/\{(0, 0)\}$ is closed

Question

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 $(ac - bd, ad + bc) \neq (0, 0)$.

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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?

Answer 1

There's a few ways to attack this.

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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.

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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.

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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)

Answer 2

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.