Treating $\frac{n}{0}$ as a constant? Is there any way of treating $\frac{1}{0}$ without breaking maths? I tried just the variable $\lambda$, thinking that it would be easy to manipulate:
$$\frac{2}{0} = 2\cdot\frac{1}{0} = 2\lambda$$
But I soon realised that you couldn't divide any numbers by it because
$$\frac{2}{\lambda} = \frac{2}{0}^{-1}=\frac{0}{2}=0, \text{ implying that } 2=0\lambda, \text{ or 0.}$$
Is there any system/way of making this work without destroying maths and everything it stands for?
 A: Any rule you can come up with that has you dividing by $0$ will eventually lead to a contradiction.  Since there is no satisfactory definition, we have to say that the operation is undefined.  
About the best you can do.
If you have a function such as $y=\frac{x^2-1}{x-1}$ you can talk about the behavior of the the function in the neighborhood of $x=1$
When $x = 1$ you are dividing by zero and the function not defined.
But if you look at values of $x$ very near $1,$ you will see that the closer $x$ gets to $1,$ the closer $y$ gets to $2$
When $x$ is in the neighborhood of $1, y$ is in the neighborhood of $2.$
A: Even in symbolic form, there is no way to include division by zero $\frac{1}{0}$ into the reals, complex numbers, or any (non-zero) ring that does not produce a contradiction given the other ring axioms. It's true in a ring that $0 \cdot a = a \cdot 0 = 0$ for all $a \in R$, but apparently $0 \cdot \lambda =  1$, hence $1 = 0$.
A: I would say no. By our usual conventions, division by zero is undefined (and even worse, anything like $\frac{0}{0}$ is called an indeterminate form.
Nice try though, it's good to experiment. You'd be breaking established rules in algebraic structure if you try this.
A: Nice try, mainly I meant nice try in disproving yourself, because when testing the waters, the ability to prove/disprove something based on how you decided to define things is a great way of determining if your definitions make any sense mathematically.
However, instead of dividing by zero, which happens to be impossible, a better problem would be to divide by something small.  So small that it is almost $0$, but not quite there yet.  For lack of better term, we'll call this magical almost $0$ the number $\epsilon$.  Funny thing is we can divide by $\epsilon$, especially if the thing we are dividing by is almost $0$ too.
I want to look at the following function:
$$f(x)=\frac{x^2-1}{x-1}$$
Particularly, I want to study it at $x=1$, though this results in division by $0$.  So instead, I study the function at $x=1+\epsilon$ and $x=1-\epsilon$.
$$\begin{array}{c|c|c}\epsilon&f(1+\epsilon)&f(1-\epsilon)\\\hline1&3&1\\0.1&2.1&1.9\\0.01&2.01&1.99\\0.001&2.001&1.999\\\vdots&\vdots&\vdots\\\epsilon&2.000\dots&1.999\dots\end{array}$$
So I guess we can agree that it makes the most sense that $f(1\pm\epsilon)=2!$  This... idea.  It is known as a limit, and it usually only works when we have:
$$f(x)=\frac00$$
In the event that we have $\frac10$, the numbers will get infinitely big.  Try $f(x)=\frac1x$ around $x=0$ for example.
A good note is that we avoid any problems with this idea of division by zero since we aren't actually using $0$.  We're just using very small numbers, where all basic math still holds.  This way, we can't construct weird proofs of $2=0$.
