Closure of set of functional dependencies ( f - Closure )

What is Closure of set of functional dependencies and how it is determined ?
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[@hosampb5593]

Give this man an award and get rid of the professor in my university

[@aris.konstantinidis]

get rid of most of the professors in most universities :p

[@danhiebert0001]

Spent half an hour reading lecture slides, really not understanding what you managed to explain in less than 5 minutes. Good job

[@minercreepmc]

same

[@lucam9806]

Thank you for this very helpful video. Somehow my professor did not explain the difference between X closure and F closure, he always said "closure", I was very confused because I could see how to find the closure of an attribute, but not how to find the closure of a set of FDs.

[@davidnguyen9023]

Thank you sir!

[@madhupanda3572]

A->BC should also be present by Union rule

[@kirtiagarwal1467]

million of thanc 2 u sir....for the videos......

[@yashmishra12]

Can we not say that AG--->HI ?

[@fr3ak1shh]

You CAN say that using transitivity and union.

[@tylerwilliams641]

A-->BC can also be derived from A-->B & A--C.

[@yaniv242]

where is the link to the next lecture with the example please 3>?

[@abcxyb]

thank you!

[@AkshayaSivanantham]

very clear explaination sir

[@ShubhamSingh39]

Ya

[@Geethanjalireddyy]

brother your way of explanation is really super in normal forms too

[@pujapurkar20]

Thanks sir

[@daphneblake4161]

Is this done using Armstrong axioms?

[@mdzia9658]

nice and very precise explanation thank you

[@heroofmytown.8664]

How to know which video was your last video ? When you haven't numbered the videos.

[@missamal4553]

thank u so much

[@krishnakalani386]

Which software is this.

[@amoghkulkarni3519]

1.5 speed works perfect for this guy

[@AnuratnakumarKumar]

when AG->H and AG->I..then we can write it as AG->HI and since CG->HI,so we can also write CG->HI..... so i want to ask that can we say that the f+ of given set of dependcies will be A->H and CG->HI????

[@AnuratnakumarKumar]

sorry....there is correction in my commentCG->AG and A->H can be the closure

[@RaghuRam-rr4pk]

here can we represent CG->B

[@nishantgarg7497]

+Raghu Ram no.. coz there is no C->A... A->C is not equal to C->A.. Correct me if i am wrong