The explanations above are excellent. I would like to add some comments on the matter, just that.
If we BELIEVE x+y > 0 does not imply xy > 0, we must offer a counterexample to this "conjecture".
If we BELIEVE x+y > 0 does not imply the negation of (xy > 0) , i.e., (xy=0 or xy<0 ) , we again must offer a counterexample to this (other) "conjecture".
In this case:
? : xy > 0
Take, for instance, (x,y) = (2,-1) , that satisfies x+y > 0 , to refute the conjecture given by the FOCUS, because xy > 0 is false.
Take, for instance, (x,y) = (1,1), that satisfies x+y > 0, to refute the negation of the conjecture given by the FOCUS (in the sense that we would have a unique answer in the negative if xy>0 would always be false).
In short:
? : xy > 0
Take (x,y) = (2,-1) <NO> , that is, this ordered pair answers the question (=focus) in the negative.
Take (x,y) = (1,1) <YES> , that is, this ordered pair answers the question (=focus) in the affirmative.
Important: each ordered pair MUST respect the statement(s) considered (in this case x+y>0), otherwise the "test" (an expression I respectfully disapprove) does not apply.
That´s what our method call a BIFURCATION ... I took this expression from the ODE (ordinary differential equations) subject.
I hope I have made things clearer.
Regards,
fskilnik.
