(Abstract Algebra 1) Definition of an Equivalence Relation

The definition of an equivalence relation is given along with three examples.

[oluwafunmilayoolapade]

Infact this has increase my interest in algebra. I hope you give this type of teachings on p-groups. Thanks for your contribution.

[pinklady]

At beginning, I have learned that ~ means "be equivalent to." Thanks. I keep listening to video.

Edit. I just subscribed.

[kajalbharti]

My university teach nothing in this covid period, so I was really relax for the online exams. But suddenly they notified us the exam be conducted in offline mode. I was freaked don't know nothing, thank you so much sir. I wish you were my teacher. Thanks!!

[RonaldModesitt]

Excellent videos! Do you use a book as the basis of your videos. If so I would appreciate knowing the title and author.

[soumyashreeram]

you are a wonderful teacher. Thank-you for helping us kids out. :)

[juanjaimescontreras]

Great videos, I learned so much from your videos, Thank You

[lucjandespiegeleire]

these videos realy help to understand my textbooks witch are so freaking difficult, thanks =)

[nick]

Thanks a lot! Examples really help to make it more concrete

[KanoBoom]

pattern: On the set ?, let x ~ y if and only if (condition), for any x, y ∈ ?

[rj]

Thank you, I wish that you are my teacher

[ruthzkycanz]

Can I have a question? if bRa and aRb does it imply that a=b?

[alancristopher]

How Prove ∼ is an equivalence relation: x∼y if and only if y=hxt. H, T subgroups of G?

[victorserras]

You need to remove the "any" in the definition of your relation for it to be an equivalence relation. Only then the rest of what you said holds.

[jose_eric]

Hello sir. Can you please solve this one.

R = Z×Z defined by nRm off nm≥0.

[eesueryu]

I found one example is not correct: m~n iff m-n is divisibke by 3. Right to left direction is okay, but the opposite doesn't hold because equality or equinumerosity is equivalence relation, so even though a relation is equivalent, that doesn't imply that the relation is being divisible by 3.

[upasnahbalkissoon]

hi morning sir, can you plz suggest me an algebra pdf

[seriousmax]

"for any m, n in Z" is confusing, because the relation won't hold for every m, n in Z... Or am I wrong?

[k_h]

How did we get m - p = (m-n) + (n-p) ?

[musicdragon]

Why is m-p =(m-n) +(n-p) I don't get that step