Why does division by zero not have an imaginary number "option"? In regular math, you cannot get the square root of a negative number. Likewise, you cannot divide by zero. Both dividing by zero and taking the square of a negative have no place in real life.
However, we have systems in place to deal with the square root of a negative number (hence, imaginary numbers). Why does an option for divide by zero not exist? 
 A: I tried this myself. If I understand correctly, you intend by your question to ask "Why can we not define a number (call it $q$), such that $q \cdot 0 = 1$?". Just a bit of algebra here: we have defined 0 as the additive identity. So we have $q \cdot 0 = q \cdot (0 + 0)$. Using the distributive law (which is always required in algebras, as far as I am aware), we have: $q \cdot 0 = q \cdot 0 + q \cdot 0)$. Now subtracting $q \cdot 0$ from each side, we have $0 = q \cdot 0$, which is a contradiction. Thus, such a number cannot exist. Notice, also, that we did not assume that our number was real; in fact, we assumed  nothing  about the number $q$. Thus it is the properties of $0$ that decide this, not the properties of our hypothesized $q$.
A: This is best understood in the framework of commutative rings.
There's a functor $$F:\mathbf{CRing} \leftarrow \mathbf{CRing}$$ given as follows: $$F(R) = R[i]/(i^2 +1).$$ We can think of $F$ as adjoining an element $i$ with the property that $i^2+1=0$. Informally, $i = \sqrt{-1}$.
There's also functor $$G:\mathbf{CRing} \leftarrow \mathbf{CRing}$$ given as follows: $$G(R) = R[j]/(j \cdot 0 - 1).$$ We can think of $F$ as adjoining an element $j$ with the property that $j \cdot 0-1=0$. Informally, $j = \frac{1}{0}$.
However, it can be seen that $G(R)$ is always the trivial ring:
$$G(R) = R[j]/(j \cdot 0 - 1) = R[j]/(0-1) = R[j]/(-1) = R[j]/R[j] \cong 1$$
So we cannot get anything useful out of $G$.
From a slightly different vantage point, main the difference between $F$ and $G$ is this: there's an obvious morphism $f_R:F(R) \leftarrow R$, and an obvious morphism $g_R:G(R) \leftarrow R$. But, whereas the morphism $f_R$ is injective for all rings $R$, on the other hand, the morphism $g_R$ is never injective, unless $R$ is the trivial ring, because $G(R)$ is always the trivial ring. So in particular, whereas $\mathbb{C} = F(\mathbb{R})$ can be viewed as an extension of $\mathbb{R}$, on the other hand, we cannot view $G(\mathbb{R})$ as an extension of $\mathbb{R}$. In fact, we cannot get anything useful like this at all.
A: We can define a quantity $i$ to have the property $i^2 = -1$, add $i$ to our set of numbers, and still have all of the rules of algebra work correctly.
Suppose we could define 1/0 to be a certain quantity, say z.  Then we have this problem:  If we start with the definition
$1/0 =z$, then multiply both sides by $0$, we get
$1 = z*0$, or
$1=0$, which is inconsistent.
A: There is no way to extend the reals in a similar fashion while still preserving all the desired properties of the reals. For example, defining it to be $\infty$ is the most common option, but then what is $0*\infty$?  How do we treat it when it come to addition?
