Equivalence Relations

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We look at the notion of an equivalence relation on a set, define an equivalence class, and consider several examples.
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Опубликовано:

16 фев 2021

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Комментарии : 31

@tomkerruish2982 3 года назад

I like replacing symmetry and transitivity with euclideanity (euclideanness?), viz., xRy and xRz together imply yRz. This, along with reflexivity, imply symmetry and thence transitivity.

@MichaelPennMath 3 года назад

This is a good one! I think I'll save it as an Exam question.

@goodplacetostop2973 3 года назад

13:41 Pay great attention, Michael’s students 😛 19:41 Good Place To Stop Homework : - Give an equivalence relation • on R such that the equivalence class of any element is size 4. - Give an equivalence relation • on R such that there are infinitely many equivalence classes of R under •, and each equivalence class has the same cardinality as R.

@ruinenlust_ 3 года назад

would x ~ y x is in {y, -y, 1/y, -1/y} work for the first one? no clue about the second one

@captainsnake8515 3 года назад

@@ruinenlust_ don’t think this works because zero would have only 1 element in its equivalence class.

@captainsnake8515 3 года назад

My solutions: 1. x~y if and only if x-floor(x)=y-floor(y) AND floor(x/4)=floor(y/4) 2. x~y if and only if floor(x)=floor(y)

@tomatrix7525 3 года назад

Another Video:) How great. Good work Michael

@kevinmartin7760 3 года назад

It should be noted that all the equivalence classes are disjoint (the transitivity prevents them from sharing elements of A) and, taken together, span A×A (because the reflexiveness applies to *all* elements of A). This further means that the sum of the sizes of all the equivalence classes equals |A|. The total number of distinct equivalence relations on A ends up being the number of partitions of |A|.

@MrYesman43 3 года назад

Is there a playlist with the videos from this course?

@aaronhunt4302 3 года назад

Good stuff as usual 👌👏

@kevinmartin7760 3 года назад

The examples would have been more interesting if A had included -3 and 3 and "same sign" were "same absolute value" instead. "Same parity" produces two equivalence classes each containing half the elements of A, whereas "same absolute value" would produce |A|/2 equivalence classes each containing two elements.

@debjitmullick7004 3 года назад

Sir , how can we join the 'proof writing' skill program ...??? Can we ...??

@pedroalonso7606 3 года назад

I enjoyed this video a lot but I missed a mention to partitions. I guess they will be explained in a next video.

@sapientum8 3 года назад

good content

@MathCuriousity 7 месяцев назад

Hey good kind being. may I pose 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!!

@nontth5355 3 года назад

Is this some kind of set theory?? We don't learn this kind of thing in highschool are we? But this seems interesting. Where can I find more info on this topic?

@MichaelPennMath 3 года назад

Here is a link to the playlist: ru-vid.com/group/PL22w63XsKjqykuLOimt7N59e6wn_aE0wd Also, here is the book I am loosely following: www.people.vcu.edu/~rhammack/BookOfProof/ I can't suggest this book enough, it is really great!

@nontth5355 3 года назад

@@MichaelPennMath tysm. Btw can u do a video explaining abstract algebra for people like me who is only in high school and interest in this topic to understand please.

@jamesrayner9980 3 года назад

loving the new camera, I think

@MichaelPennMath 3 года назад

You think you love it, or you think it's a new camera? lol

@Abhisruta 3 года назад

@@MichaelPennMath lol

@jamesrayner9980 3 года назад

@@MichaelPennMath think it's new, i always love your videos

@rektator 3 года назад

Every equivalence relation on a set is induced via a function in the following sense: Let f:X->Y be a function between sets X and Y. We define an equivalence relation R on X by setting that xRy iff f(x) = f(y) for elements x and y in X. This is clearly an equivalence relation and it is called the set theoretic kernel of function f and often denoted ker f (In algebraic models, these kernels become very important) Suppose that there exists an equivalence relation R on X. Consider the function p:X->X/R, p(x) = [x] for elements x in X. Now R = ker p. Hence every equivalence relation is obtained from a function using equality.

@rektator 3 года назад

Here's an interesting reason why these kernels of function are ubiquitous: Let L be an alphabet of first order predicate logic. Consider L-models M and N and an L-model morphism f:M->N. Suppose that p is a positive sentence of first order predicate logic (positive means it doesn't contain a negation or an implication). Then if f is surjective and M satisfies p, so does N. So surjective L-morphisms preserve positive truth. There is a version also when injective function are able to reflect truth of universal sentences but there's an extra condition that one has to check with respect to relation symbols in L. All algebraic sentences are both positive and universal. Now to the interesting part: Again let f:M->N be an L-model morphism. There exists a unique L-model structure on the set M/ker f such that the canonical projection p:M->M/ker f, x|->[x], becomes a 'reasonable' L-model morphism. Furthermore there exists a unique function g:M/ker f->N, where g o p = f. Additionally g is an injective L-model morphism. If we combine the previous notions we have the following corollary: If G and H are groups and f:G->H is a group homomorphism, then G/ker f is a group that satisfies the same positive sentences that G satisfies. Additionally G/ker f satisfies all the universal sentences that the group H satisfies.

@flux4162 3 года назад

Interesting integral question, Integrate arctan(x)arcos(x) from 0 to 1, like so Michael sees this

@yuseifudo6075 8 месяцев назад

@helo3827 3 года назад

19:41

@random-td7tf 3 года назад

Rip. Good place to stop lost. Here’s how you can pay respect to him : ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-YddwkMJG1Jo.html

@angel-ig 2 года назад

11:56 "So the equivalence class of the empty set is just the empty set": wrong. The correct thing is "So the equivalence class of the empty set is the set containing just the empty set"