#self doubt [closed]

Question

Comments

empid	speciality	managerid
1	a	m1
1	b	m2
1	c	m4
2	a	m1
3	a	m3

above data satisfy all conditions

by (ii) Each manager has only one speciality i.e.

         managerid -> speciality

      (iv) employee has only manager for each speciality

          empid,speciality --> managerid

So Primary key will be = (empid,speciality)

 managerid -> speciality // 3NF

empid,speciality --> managerid  // 3NF

so relation schema may be in 3NF

 

please correct me if mistakes.

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nice effort Dharmendra Lodhi your reply is valuable for me.

thanks

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what is answer

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3 nf