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?
On
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.
On
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 $2.$
On
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 $2=0$.
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 $1 = 0$.