Finaly , what I expected...Very clear and to the point...Thank you!
@bobphilishen
2 жыл бұрын
In my opinion, if you are good at discrete math like this, you are a god damn genius
@brittanysnow2119
Жыл бұрын
James you just helped me pass my midterm, thank you SO much for this explanation. It was not making sense in my head till now !!!!
@arielcototapia1746
3 жыл бұрын
wow thanks a lot, the best video that ive ever found about relations
@gigispence6011
4 жыл бұрын
Best explanation I’ve come by ! Thank you !!!
@danverzhao9912
4 жыл бұрын
This video truly deserves 72+1 likes and no dislike.
@crewify5460
Жыл бұрын
Crystal clear 🎯
@armanadabi1820
Жыл бұрын
Thank you very much.
@daelinparmanand1848
3 жыл бұрын
Excellent video James
@jerwaynetwh
4 жыл бұрын
Sorry i cant understand the part where a-b = 3k, where a-b is divisible by 3. May I know how did you get 3k? as I thought it will be k/3 instead. Sorry!
@gamms95
4 жыл бұрын
in case you didn't find out yet, that's a division with elements replaced. like if (15/k) = 3 -> 15 = 3k -> k = 5. so in this problem x - y = 3k, you can see as (x-y)/3 = k. if k is in Z, x-y is divisible by 3.
@jerwaynetwh
4 жыл бұрын
Gabriel thank you!!
@MathStuff1234
4 жыл бұрын
Thank you James!
@MathCuriousity
10 ай бұрын
Hey may I ask a question: let’s say we have an equivalence relation aRb. Why can’t I represent this within set theory as set T comprising subset of Cartesian product of a and b, mapped to a set U which contains true or false? Thanks so much!!
@HamblinMath
10 ай бұрын
The mapping is unnecessary. Just make R be the set containing the ordered pairs you want.
@MathCuriousity
10 ай бұрын
@@HamblinMath hey friend! Sorry for not understanding but would you unpack your reply a bit? I don’t understand why people on Reddit told me relations like equivalence or just symmetrical or just reflexive are “meta” relations and can’t really be seen as relations between two sets and set theory doesn’t allow it.
@MathCuriousity
10 ай бұрын
@@HamblinMath to clarify my second reply to your reply: but I would very much like to know how we can do this with the truth/false as elements of the destination set!
@HamblinMath
10 ай бұрын
@@MathCuriousity You *can* define a relation as a function from A x A to {True, False}, but I don't see any reason why you would, since the relation itself would be just the preimage of "True." The function doesn't gain you anything.
@MathCuriousity
10 ай бұрын
@@HamblinMath I don’t understand - the whole confusion I have is -if I have a reflexive relation for instance - it seems the ordered pair is of the elements a and b the relation acts on - but where is the “truth” stored ?
@irfansani8367
Жыл бұрын
Sir kia A intersection B equivalenc hy agar R or S equal houn
Пікірлер: 29