Question

What is the difference between fourth and fifth normal form?

Answer 1

Answer:

fifth-forth =first

Step-by-step explanation:

Answer 2

$\huge{\red{\mathfrak{Answer:-}}}$

A relation R is in 4NF iff every nontrivial join dependency, J, satisfied by R is implied by the candidate keys of R, where J consists of a join dependency on any 2projections of R (R1,R2).

A relation R is in 5NF iff every nontrivial join dependency, J, satisfied by R is implied by the candidate keys of R, where J consists of a join dependency on any nprojections of R (R1,R2, ... Rn).

In other words 4NF is very like 5NF but only 2-part join dependencies are relevant under 4NF whereas all N-part join dependencies must be considered in order to meet the conditions of 5NF. 4NF is therefore not very important and exists for historical reasons only, i.e. because it happened to be discovered and published first. In practical terms it's probably easier to ignore 4NF because 5NF is much more useful and important