Beyond + and -

[Benny]

Your operator is very simple actually:

x§y = \sgn(x) (|x|-y)

Interesting, it seems to be true in some cases.

4§3=1=1*(|4|-3)
-4§3=-1=-1*(|-4|-3)=-1

But

3§4=§1!=-1=1(|3|-4)

So your definition works only in special cases, and is not a general equivalence.

Nothing really special about it at all

I don't know, it was new for me and I found it interesting.

0 § y is absolutely nothing because \sgn(0) is inexistent

So you give me a incomplete definition that is supposed to be equivalent to mine and so my definition is false? Seems like a flawed approach to me .

Even if your definition was right, sometimes it is useful to add "inexistent" things to math - like sqrt(-1). I don't see how my example is any different, maybe it is useless, but I don't see why it should be.
I am not sure, it might be that § extends the algebraic signs with a sign for 0? It may make sense, since it "neutralizes" numbers, and §1*-1 seems to be 0 (if I didn't ake an error above).

Also your approach says nothing about multiplication (and beyond).

And the solution to your algebraic equation (-X+Y=X+Y) is X = 0

Obviously, but now there are infnite solutions where X!=0.
So it solves -1+Y=1+Y, etc...
Also it seems to solve -1/X+1/Y=1/X+1/Y with Y=§1*X (at least for positive X), which is not solvable in the integers.