# Why don't we define division by zero as an arbritrary constant such as $j$? [duplicate]

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.

## marked as duplicate by Git Gud, user63181, user67258, Zev Chonoles, Pedro Tamaroff♦Jul 6 '13 at 22:37

• Because those restrictions would leave you with the empty set or $\{j\}$ or something of the sort. And you can't say $\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}$. See this. – Git Gud Jul 6 '13 at 22:13
• If you try this, you will quickly notice, that you need to make a whole bunch of exceptions. But then, what's the purpose of defining $j$, when everytime you do simple calculations with $j$, you need to consider these exceptional cases again? – Tomas Jul 6 '13 at 22:21
• $j\cdot 0=1\rightarrow j^2\cdot 0^2=j^2\cdot0=1.$ However, $j\cdot(j\cdot 0)=j\cdot 1=j=j^2\cdot 0$. Then $1=j.$ Therefore, you must give up $(ab)^2=a^2b^2$ as well. – Ian Mateus Jul 6 '13 at 22:27
• @IanMateus: Giving up $(ab)^2=a^2b^2$ isn't that bad; it basically means that your multiplication isn't commutative any more. For example, for quaternions, $(ij)^2 = k^2 = -1$, but $i^2j^2 = (-1)(-1) = 1$. – celtschk Jul 10 '13 at 12:26
• See also here: math.stackexchange.com/questions/125186/… – Hans Lundmark Aug 15 '18 at 12:32

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.

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.

• It is interesting how living with $0/0$ is much harder than with $1/0$. – Ian Mateus Jul 6 '13 at 22:31
• Actually, even $0/0$ is defined in wheels. – celtschk Jul 6 '13 at 22:32
• That's what I'm talking about, I'm comparing the richness of the Riemann sphere and wheels. – Ian Mateus Jul 6 '13 at 22:33
• "Wheels"? Is this a turn of phrase, or a true construction? If it's a real idea, could you please supply a reference. I like the sound of it. – Fly by Night Jul 6 '13 at 22:37
• @FlybyNight: See the second link in my answer. I've now added the explicit name in the text. – celtschk Jul 6 '13 at 22:38

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.

• Perhaps a solution for the first problem is leaving $j-j$ undefined so we can't derive things like $0=1$? – Ian Mateus Jul 6 '13 at 22:48
• That's even worse, now you broke addition, the one operation that still worked. – vadim123 Jul 6 '13 at 22:49
• Hey, you guys are abusing mathematics... Stop it please :) – imranfat Jul 6 '13 at 23:06