Adding some kind of infinitesimals to the reals can be useful, as is done with hyperreal numbers, dual numbers, and other kinds of hypercomplex numbers, but it doesn't allow you to
divide by zero; there is no hyperreal or hypercomplex number
x that solves
x = 1/0.
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.
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.