Thank you so much for all of the work and effort. I am back in school my bachelor's 20 years ago in an entirely different field. You are helping so many students with different backgrounds.
No. Implications only have to hold true for one direction. If p -> q. Biconditionals have to hold true for both directions if p -> q AND if q->p. Subsets are true in one direction like implications. Suppose we have sets A and B. Suppose A is a subset of B. If x is an element of A then x is an element of B(if p -> q). This does not hold true for the converse(q->p). If x is an element of B it is not necessarily an element of A.
Its a combination of all possible subsets, and if the subset is null, or {}, than there wouldn't be any other combination because its empty. That just my theory tho
So many ads in just 15 mins video.... each time I skip some seconds, ad pops up. Never seen this much ads in one video, not just this one but in all of yours this course videos. So irritating it is!
Im learning more from your videos then from my class I'm so glad you posted the entire course!
Glad you like them!
Thank you, I study discrete math from you,
Unfortunately our teacher is very bad and I couldn't find a better one than you.
God bless you
I have the same problem right now :(
The way you explain the math concepts really simplifies things and makes it much easier to grasp. Thanks for making math less overwhelming!
Thank you. I've found your video is 1000x helpful than my college online teacher. Your presenting technique and content are way easier to grasp.
Thank you so much for all of the work and effort. I am back in school my bachelor's 20 years ago in an entirely different field. You are helping so many students with different backgrounds.
commenting this on every video so u get more reach because these are really helpful and it means way too much to me.
U are answering my whole question about proper subset, great explanation
i am very glad founding this channel
Godsend. should pay you my tuition instead of my uni.
Thank You Kimberly. Your explanation is the best in all TH-cam videos. 👍👍👍👍👍.
keep up the hard work
Just a thoughtful suggestion -- it would've probably been better if you could offer an example after each definition just so it sticks.
i learned what i didn't learned from my class thank you very much
THANKS for your videos! i learned the whole year of DM from you!
Thanks alot for the playlist it has helped me a lot your a good lecture
At 3:04 shouldn't the arrow sign be a biconditional arrow instead of the conditional arrow since its "iff"?
No. Implications only have to hold true for one direction. If p -> q. Biconditionals have to hold true for both directions if p -> q AND if q->p.
Subsets are true in one direction like implications. Suppose we have sets A and B. Suppose A is a subset of B. If x is an element of A then x is an element of B(if p -> q). This does not hold true for the converse(q->p). If x is an element of B it is not necessarily an element of A.
I have a question 10:32
What about also adding {{}, 0, 1, 3} so that |P(A)| becomes 9???
Its a combination of all possible subsets, and if the subset is null, or {}, than there wouldn't be any other combination because its empty. That just my theory tho
Outstanding video lecture
2.5 is set cardinality is the whole thing covered in this?
You said the cardinality of power set with n elemnets is 2^n. What happen if the power set contains duplicate eg A = 0, 1, 1, 2, 3
good job , thanks very much.
ty
at 10:26 , shouldn't there be (1,1) and (2,2)????
awesome!!
These are well explained, but I admit I am used to hearing it pronounced "tooples".
seu curso ta me ajudando demais valeu
Glad I could help!
Thank U so much
the ads are really annoying....
Necessary evil
So many ads in just 15 mins video.... each time I skip some seconds, ad pops up. Never seen this much ads in one video, not just this one but in all of yours this course videos. So irritating it is!
I don't place the ads. TH-cam does that. I don't have the time to go in and manually place the ads.
korguizzz
One Question at 7:15, you write the statement-> ∀x(x∈A→x∈B) ⋀ ∃x(x∈B ⋀ x ∉ B)
Could I write the following-> ∀x(x∈A→x∈B) ⋀ ∃x(x∈B → x ∉ B)
Why/Why not?
Usually the existential quantifier does not proceed an implication.