Well, that's not so bad; they are equal on the real projective line: https://en.wikipedia.org/wiki/Real_projective_line. Making 1 = 2 is a bigger problem.
Re:Imaginary numbers are real
by LadyCailin on Sunday July 24, 2022 @08:04AM
(#32212613)
Haha, ok this blows my mind a little more now. This definitely helps me conceptualize the problem a bit better, even if I’m still not satisfied with it. I still feel like there is a better solution, like giving 0 a sign or something.
That is in fact what we do in IEEE 754 floating point! We have +0 and -0; 1/+0 = +∞ and 1/-0 = -∞. However, this means that IEEE 754 floating point violates substitutability, the most basic property of equality, because +0 == -0 but in some cases f(+0) ≠ f(-0).
You may enjoy Tom 7's paper/movie about this: http://tom7.org/nand/
From above. What if you approached 0 from below instead?
1/-1=-1, 1/-(1/2)=-2, 1/-(1/4)=-4.
So you just proved that 1/0 equals both positive and negative infinity, and therefore you also just proved that positive and negative infinity are equal. This is one of the reasons why giving 1/0 a definition is rarely seen as useful.