so I came up with this idea:
the real projective line defines that $\infty = - \infty$. What if I divide any value $x$ (not equal to $\infty$) by infinity? Would that be 0? or "something" between $x/\infty$ and $x/-\infty$?
Would be very interesting
Grx
It depends entirely on how division is defined on the real projective line (which I will denote $S^1$). We could say that $S^1$ inherits the same operations as the reals (keeping in mind that $\infty$ is not a real number, and it's role in algebraic operations can only be defined in terms of limits, which inevitably leads to ambiguity) then the answer to your question is yes, $x/\infty=0$ for $x\neq\infty$. But the problem here is that the resulting algebraic structure is not very nice. There is sum ambiguity as to what $\infty+\infty$ means, among other things. Whatever it is, it is certainly not a field, in which multiplication and addition are nicely defined with inverses and identities. (I think it is actually impossible to define a field with continuous operations on $S^1$, maybe someone with a better understanding of algebraic topology can prove this?) We could imagine that $S^1$ is the unit circle in the complex plane with 0 being identified with 1, and $\infty$ being identified with $-1$. This leads us to a rather natural multiplicative operation which is that if $x,y\in S^1$, then $x\cong e^{i a}$ and $y\cong e^{i b}$ for some real number $a$ and $b$ between $-\pi$ and $\pi$. Then $x\times y$ could be defined to be $x\times y \cong e^{i(a+b)}$. This results in a very nice group structure with a well-defined division. Thus, the answer to your question in this case is that $x/\infty\cong e^{i(a-\pi)}$, which depends on $x$.
There are probably other was to define division in $S^1$, but (as with anything in mathematics) it has to be a useful definition for anyone else to care. So I guess the real answer to your question is that it can be anything you want it to be, as long as the system of division you invent is well-defined.