Equivalence Relations -- Proof Writing 17

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[noahtaul]

15:14 Instead of doing separate cases, you could take ad=bc and multiply by f, and take cf=de and multiply by b, and get adf=bcf=bed, and then it’s perfectly acceptable to divide by d to get af=be.

[malachichoy]

Studying this right now in Discrete Mathematics. This video is great and easy to follow.

[bobh]

At 11:30 shouldn’t it be 2πZ, not 2nZ?

[PhoenixInfeno]

At around the 30 min mark you talk about "obvious operations" involving equivalence classes and use mod n as an example, but that doesn't really prove that [a]+[b] = [a+b] for all possible equivalence relations, doesn't it? Or are these "obvious operations" only intended to be obvious for mod n?

[iabervon]

The other direction of proving that aRb implies [a]=[b] is almost the same, but this is where you have to use the fact that R is symmetric, so that you have bRa to do the same steps with. I think this sort of proof is nice to introduce a lemma in, where you show that xRy implies [x] is a subset of [y], and then use this fact twice to prove the stronger claim. I tend to think of the usual case of a lemma being a statement you prove, and use only a few times before you have a more useful statement to use instead, and this would be a good example.