4 ms·
You see how Division Algorithm is capitalized in my comment? Missed that. But this just shows that Euclidean division is a bad choice if you are dealing with d
by danbruc 4y ago
You see how Division Algorithm is capitalized in my comment?
Missed that. But this just shows that Euclidean division is a bad choice if you are dealing with dividing signed quantities, that sign changes cause magnitude changes makes no sense in that case.
This is just false. Modular arithmetic doesn't represent each equivalence class with a canonical member. You either work with raw numbers ("1"), or you represent the equivalence classes directly ("[-2]").
I don't really understand what you mean. If you say [13] - [24] = [-11] mod 3 that is certainly true, but what's the point then? Wouldn't you at least want [13] - [24] = [1] mod 3 even if you skip reducing [13] - [24] to [1] - [0] explicitly?
- nh23423fefe 4y ago[-11] == [1], even more so, they are the same object, because they are sets. Sure you can designate the answer however you want. 4/6 = 2/3. I'm not sure what you mean by "whats the point"