Re:Imaginary numbers are real
by LadyCailin on Saturday July 23, 2022 @08:12PM
(#32208906)
I am not at all a mathematician, but I just can’t wrap my head around how 1/0 isn’t infinity. 1/1=1, 1/0.5=2, 1/0.25=4, etc. The closer we get to 0, the larger the number. I just can’t accept that if we actually get to 0 in our progression, it is anything besides infinity. Perhaps it becomes its own class of number, like what happened with imaginary numbers. That is, infinity times zero is 1n (where n is a bookkeeping marker like i in the square root of negative numbers), or like with limits, where you have a “+ C” in there.
> closer we get to 0, the larger the number
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.
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/
Re:Imaginary numbers are real
by codeismath on Saturday July 23, 2022 @10:36PM
(#32209746)
Okay, but then you'd show that 1/-1=-1, 1/-0.5=-2, 1/-0.25=-4, etc. The closer you get to 0, the more negative the number. Then you just wouldn't accept that if you actually got to 0 in your progression, it is anything besides negative infinity. And then you've shown that negative infinity is equal to infinity.
Sure, I definitely agree that 1/0 doesn't give you a finite quantity. What I'm saying is that it doesn't give you a number at all unless you're willing to sacrifice major parts of algebra. And I'm not much of a mathematician either, so I might have gotten something wrong.
As I understand it, if you add 1/0 = ∞ you can't treat that ∞ as a quantity in the usual algebraic ways. In particular you can't multiply ∞ by 0 and get back some well-defined quantity such as 1; if you allow that, you quickly find that you can prove that all finite numbers are equal. The standard tricky example of this is in https://www.math.toronto.edu/mathnet/falseProofs/first1eq2.h...:
Let a = b.
Then a² = ab, a² + a² = a² + ab, 2a² = a² + ab, 2a² - 2ab = a² + ab - 2ab = a² - ab = 2(a² - ab) = 1(a² - ab).
If we then divide both of these last expressions by a² - ab we get 2 = 1. This is only invalid because a² - ab = 0. Adding a different ∞ₙ for each value of n/0, as lisper suggests, doesn't help.
So, if you want to extend your number system with 1/0 = ∞, you either need to throw out some of the standard laws that permit algebraic manipulations like that, or you end up with all finite quantities being equal.