4 ms·
let ϴ = 0/0 then 1*ϴ = ϴ = 0/0 = (0*0)/0 = 0*(0/0) = 0*ϴ it follows 1 = 0 and thus x = x * 1 = x * 0 = 0 = y * 0 = y * 1 = y for all x and y
by afiori 4y ago
let ϴ = 0/0 then 1*ϴ = ϴ = 0/0 = (0*0)/0 = 0*(0/0) = 0*ϴ it follows 1 = 0 and thus x = x * 1 = x * 0 = 0 = y * 0 = y * 1 = y for all x and y
- xigoi 4y agoThis is assuming that Θ interacts with arithmetic operations the usual way (that is, ℝ ∪ {Θ} is a field), which the person you're replying to did not say.
- afiori 4y agoTrue, but the point of giving a "value" to 0/0 is to use it somehow. For example in the context of limits you define a whole lot of number like values like 0+ or 0- that are useful wrt operations on limits. I was trying to give an example of how ℝ ∪ {Θ} has almost no advantages compared to just ℝ
- tshaddox 4y agoSure, but the whole "problem" we were trying to solve was that zero doesn't interact with arithmetic operations the usual way.