Is the element $(0,0,0)\in\mathbb{R}^3$ a divisor of zero? (I'm assuming that $\mathbb{R}^3=\mathbb{R}\times\mathbb{R}\times\mathbb{R}$)
In my assignment, I'm told to prove that exactly one of the following can be true for an element $(x,y,z)\in\mathbb{R}^3$


*

*$(x,y,z) = (0,0,0)$

*$(x,y,z)$ is invertible

*$(x,y,z)$ is a divisor of zero


I'm starting with the first. Given the definition of the operations on the ring $A=\mathbb{R}\times\mathbb{R}\times\mathbb{R}$ (we were given non-standard multiplication), the multiplicative identity is $(1,0,0)$.
There's no way that, given our definition, $(0,0,0)(0,0,0)^{-1}=(1,0,0)$ so it's not invertible.
However, my issue is with showing that it is not a zero divisor. Whenever I use algebra to find each component of another element in the ring $p$, such that $p\cdot(0,0,0)=(0,0,0)\cdot p= (0,0,0)$, this is always true. Any suggestion?
 A: By "zero divisor," your book probably means that $v \in \Bbb{R}^3$ is a zero divisor if and only if there exists some $a \in \Bbb{R}^3$ such that $a, v \neq 0$ and $av=\textbf{0}$. Therefore, $(0, 0, 0)$ is not a zero divisor because $(0, 0, 0) \neq \textbf{0}$ is false.
It is important to note that $(0, 0, 0)$ is intentionally excluded from the definition of zero divisor because some mathematicians think that this is a more useful definition of zero divisor. This way, we can talk about rings without zero divisors called integral domains, where if $av=\textbf{0}$, then either $a=\textbf{0}$ or $v=\textbf{0}$. This would be rings like $\Bbb{R}$ or $\Bbb{R}^3$ where you can't get $\textbf{0}$ by multiplying two non-zero elements.
Thus, because of the definition of zero divisor used in your book, $(0, 0, 0)$ is not a zero divisor.
A: You can actually have zero divisors depending on how multiplication is defined. If addition is point-wise, then (1,0,1)(0,1,0)=(0,0,0).
You said you were given non-standard multiplication. Was it defined for you?
