Why we cannot divide one zero divisor by another one? Or can we? For instance, in split-complex numbers we definitely have $2\cdot(j/2+1/2)=j+1$, which is absolutely valid. Can we then say that $\frac{j+1}{j/2+1/2}=2$? If not, why we cannot define it this way?
 A: You can define whatever you want
You can define $\frac{j+1}{j/2+1/2}$ to mean whatever you want.
But the notation suggests we are working in a field where $j+1$ has an inverse. This is not true for the split complex numbers so will confuse the reader.
For example if you write $\frac{j+1}{j/2+1/2}=2$ then your fraction bar does not work the same way the normal fraction bar works. If it did we could multiply both sides by $j-1$ to get
$$\frac{j+1}{j/2+1/2} =2$$
$$\implies \frac{j+1}{j/2+1/2} (j-1)=2(j-1)$$
$$\implies \frac{(j+1)(j-1)}{j/2+1/2}  =2(j-1)$$
$$\implies \frac{0}{j/2+1/2}  =2(j-1)$$
$$\implies 0  =2(j-1)$$
which is untrue.
A: Division by a zero divisor can never give a unique quotient. Suppose that $a$ is a zero divisor, so $\,\color{#c00}{a\hat a=0},\,\ a,\hat a\neq 0.\,$ Then $\, a\:\!\:\!x = b\Rightarrow \color{#c00}a(x+c\color{#c00}{\hat a}) = b,\,$ so there is no unique solution $\,x := b/a.\,$  OP is case $\,a = r(j+1),\ \hat a = j-1,\ r = \frac{1}2,\,$ where $\,a\hat a = r(j+1)(j-1)=0,\,$ by $\,j^2=1.$
A: Given a commutative ring. Division is an operation derived from multiplication, i.e., $c=a:b$ is defined as $c=a\cdot b^{-1}$, where the divisor $b$ must be invertible. Note that invertible elements (units) are not  zero-divisors. Similar situation for subtraction which is derived from addition.
