In this video, I work through an example of proving that a relation is an equivalence relation. We do this by showing that the relation is reflexive, symmetric, and transitive.
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 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.
@@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!
@@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.
@@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 ?
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!
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.
Finaly , what I expected...Very clear and to the point...Thank you!
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 !!!!
In my opinion, if you are good at discrete math like this, you are a god damn genius
wow thanks a lot, the best video that ive ever found about relations
Best explanation I’ve come by ! Thank you !!!
Crystal clear 🎯
This video truly deserves 72+1 likes and no dislike.
Thank you very much.
Excellent video James
Thank you James!
Sir kia A intersection B equivalenc hy agar R or S equal houn
Thanks a lot
thank you sir
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!!
The mapping is unnecessary. Just make R be the set containing the ordered pairs you want.
@@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.
@@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!
@@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.
@@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 ?
love you
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!
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.
Gabriel thank you!!
thanks a lot sir
With examples that are even
could we have your social media to be in touch
Thank you sir