[zkmon]

No you didn't get it. You missed "Let's consider them as operations. Apply any one of them on any other of them, you get the third one."

[youoy]

So is what i wrote a third one? Fourth? Fifth? :)

[zkmon]

Not sure what you are talking about. What you wrote reduces to just x. What I meant was, if you substitute say, -x for x in -1/x, you get 1/x, which is the third inverse. Same is true for the other two pairs. So, if we call them functions f, g and h, then, f=g(h)=h(g); g=f(h)=h(f); h=f(g)=g(f)