Why don't we define division by zero as an arbritrary constant such as $j$? Why don't we define $\frac 10$ as $j$ , $\frac 20$ as $2j$ , and so on? I know that by following the rules of math this eventually leads to $1=2$ , but we could make an exception and say that $j$ is the only number such that $0*j \not= 0$ , and put other restrictions necesary so that we don't get contradictions. We do this for $i$ , so why can't we do it here? For example, $i^2$ , is defined as $-1$ , but you could also say $i^2=\sqrt {-1} *\sqrt {-1}=\sqrt {(-1)(-1)}=\sqrt{1}=1$ , but we make an exception for this.
 A: You'd have to make an exception for $j$ pretty much everywhere, and at that point, you might as well not include it.  
When you include $i=\sqrt{-1}$, you give up some properties like $\sqrt{ab} = \sqrt{a}\sqrt{b}$ but most of the existing laws continue to hold, and more to the point, the complex numbers have useful additional properties, such as being algebraically closed.  The exceptions are few compared to what continues to work.
A: Actually, as I learned only recently here on math.SE there actually does exist an algebraic structure where you can define $1/0$, called wheel. However in that structure, almost all the ordinary laws are modified. Indeed the distributive law is replaced by a complete bunch of laws. Since you don't gain much, it is easier to just leave division by zero undefined.
A: As you suspected, we have this problem: $$aj=a\times j = (a+0)\times j = (a\times j)+(0\times j)=aj + 1$$
Hence the distributive law needs to be tossed out.  That's pretty important.
Another important one is that if $a=b$ then $a\times c=b\times c$.  
But $0^2=0$ yet $j\times 0^2\neq j\times 0$.  Hence the associative law for multiplication needs to be tossed out too.
