07-02-functional-dependencies.mp4

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[@basrikartal]

She is very good! Her examples are well structured, contents are quite clear and presentations are very organized. High quality teaching materials.. Thank you for making and sharing those videos! All the best.

[@GraysonLenner]

If I close my eyes I am listening to dory (from finding nemo) teach me about functional dependencies

[@tahermahmoudsayed9233]

hahahahahaaa lol

[@avivperets7360]

She should replace our lecturer at the college immediately ! 20 minute of explanation of her fix what 8 hours of explanation of my lecturer did..

[@jianingzhang9092]

Her lecture is pretty clear and example are well organized. 30 min video covers two weeks' material in my course!!!

[@yueli2146]

This is the crystal clear

[@sorteslyngel2k]

Thank you for sharing this! You taught me what my teachers couldn't :)

[@uzferry5524]

you're such a great teacher!

[@RealAndroidTurorials]

God this is so good, she makes everything in the other language I checked look like pure trash ! Thank you very much

[@BillKonstantakos]

At 17:30 isn't A ⇒ A∩B just another trivial FD since A∩B ⊆ A? The condition A ⇒ B is not necessary, right?

[@kevinhodge3687]

You're right that A ⇒ A∩B is just another trivial FD, and is always a trivial FD even if some elements of B were not in A since the intersection would remove those extra elements anyways. Therefore it does hold true for any case where A ⇒ B, irregardless of whether that FD is trivial or not. However, A ⇒ B has to exist in at least some form for us to talk about it as a dependency, so in a way it's necessary that it exists in some form, and is therefore true.

[@daleoking1]

So where can I find the recommended readings? Love your lectures btw there some what better than my actual lectures lol.

[@Lue_eye77]

I love you!

[@origamikingg]

thank you. that's all i have to say

[@TheParfaitducongo]

Mucjas gracias!!!

[@numbo655]

Shouldn't SSN determine everything in the Student relation?

[@qubit3807]

Thank you!

[@haolinzhu2572]

Can we say S2 entailed by S1 if every relation satisfying S1 is also satisfied by S2

[@igniculus_]

nice hairstyle