Thank you so much for this video! I now understand equivalence relations and equivalence classes! May God bless you and shower you with joy and happiness!
Thank you so much! I've been searching for abstract algebra videos all over youtube and your videos are literally the only ones I understand! For the last example, would you also have to look at the case where [x]=/=[y] and show they have to be disjoint?
No matter how they know the topic, some teachers are really bad at teaching. They can't speak in the level of the students. Some of them speak like we're babies, and some of them speak like we're geniuses. However, some people can explain the things well. This person is one of them, thank you sir.
Thank you! At 9:17, wouldn't it have been more straightforward to start with y being an element of [x], since [x] := {y | x ~ y}, and from [x] = [y] we can just imply that y is an element [y]?
After watching a lot of videos on this topic, I stumbled upon this video and this is the only video that was able to explain me this topic. (My textbook just confused me)
There’s two ways of doing math, the intuitive way ( like this video) or the textbook way, I hate textbooks and don’t really use them in my studies because the ones we get are so boring and unintuitive.
is the disjunction in the very last proof an exclusive or inclusive disjunction? ([x]=[y] or [x]intersects[y]=null set) if it's an exlusive disjunction (which seems more likely), then assuming one false and proving another seems a bit counterintuitive because that seems to apply only to inclusive disjunction
Because -1 divided by 3 would give you -1/3 but -3 divided by 3 would give you -3/3 = -1 same for -6/3 = -2. For a number to be divisible by another number it should not leave and remainders or result as a fraction.
-1 = 3*-1 + 2 Hence when -1 is divided by 3, it leaves a remainder of 2. But, -6 = 3*-2 + 0 Which means -6 / 3 doesn't leave a remainder or we can say -6 is divisible by 3.
