Discrete Math 9.5.1 Equivalence Relations

Please see the updated video at • Discrete Math - 9.5.1 ...
The full playlist for Discrete Math I (Rosen, Discrete Mathematics and Its Applications, 7e) can be found at • Discrete Math I (Entir...

[@TheStillalivee]

Even though I am a Turkish guy who studies in Turkey. Your videos are much more understandable than the Turkish tutorials. Thanks for keeping the topics this simple and easy to understand by giving relatable examples.

[@maryamali1818]

if I pass this course, I will thank god and you.

[@SawFinMath]

Maryam Ali Be sure to use the new playlist instead. The videos are updated and I fixed any mistakes.

[@maryamali1818]

@@SawFinMath I will, thank you from the bottom of my heart

[@medicalbommerbre]

@@maryamali1818 did u pass

[@ansonthedev]

Amazing video. Wish I found your channel sooner.

[@Ashley-ep2qm]

Bless your heart ! Your videos are so helpful

[@LuciaSilva-gw9ri]

Perfect, thank you so much!

[@OmKumar-js7jd]

(Badiya) best classes 🇮🇳

[@shreypatel9379]

Thank you for delievering such great quality lectures

[@SawFinMath]

So nice of you

[@abdullahmoiz8151]

excellent explanation than you very much

[@antojosu]

Awesome Class

[@chaunguyenhuynhbao3472]

Thank you so much !

[@sabirkhosa1586]

thanks you so much mam its helping course because these are book example that are more understanable.

[@mubarakahmed2488]

For mod 4 [1] - how's 7 a congruent when if you divide 7 by 4 the remainder is 3 I thought we were to look for number who remainder was [1]

[@noon8681]

Thanks bro

[@stephanewamba9610]

at 3:01 it should be (a-b)+(b-c) is in Z

[@SawFinMath]

Stephane Wamba I fixed the example in the updated video.

[@leozhang1340]

for the last problem, P1 should also be a "yes", right? because every set contains an implicit empty set

[@BenBrawn]

no. Every set has the empty set as a SUBSET, but not as an element.

[@franciscoribeiro2462]

@@BenBrawn shouldn't there be brackets for it to be considered an element of the original set? For instance the empty set is a subset of all sets, but {empty set} is not a subset of all sets. Isn't it the same case here?

[@BenBrawn]

Kiko Ribeiro brackets or no brackets, nether is an element of the set given. Yes what you say is correct, the empty set {} is a subset of every set but {{}} need not be a subset.

[@microlegions3101]

Ma'am I have a doubt - is equivalence classes and partitions are exactly one and the same thing or they're different?

[@SawFinMath]

They are different. However some partitions are examples of equivalence classes, as shown in the video.

[@microlegions3101]

@@SawFinMath Thanks, Ma'am! You're such a blessing!

[@dt615]

thank you maam, I wish you were my professor :((

[@mevlutkelle4083]

i love you kimberly

[@sifisoleema2074]

Hi pls help with this question (decide for each of the following relation whether or not it is an equivalence relation. Give full reasons. If it is an equivalence relation, give the equivalence classes. (a) let a. b be integers. Define aRb if and only if a 3 devises ( a-b) in other words R is the congruence modulo 3 relation....... Pls help

[@Mr_cheekz]

GOAT