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Consider a set S $\left \{ 2,3,4,.....,23,24 \right \}$ and R is relation on S such that aRb if a divides b, then find the number of minimal elements in its hasse diagram

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Minimal Element : Let (A, R) be a poset. Then "a" in A is a minimal element if there does not exist an element b in A such that bRa.

So, for the given poset  ( S, / )

Minimal elements : {2,3,5,7,11,13,17,19,23}  = 9 elements

Maximal elements : {13,14,15,16,17,18,19,20,21,22,23,24} = 12 elements


Note: It is NOT required that two things be related under a partial order. That's the partial part of it. 

In the Hasse Diagram for this Poset (S,/), 13,17,19,23 will be Isolated Nodes since they do not Relate with any other element. And Trivially by definition, Every Isolated node is both Minimal as well as Maximal element. And If a element is Both Minimal and Maximal, then It is represented as Isolated node in the Hasse diagram for that poset.  So, We can say that An element is Isolated node in the Hasse diagram if and only if It is both Minimal and maximal.

Note: It is NOT required that two things be related under a "partial order". That's the partial part of it. If two objects are always related in a poset, it is called a total order or linear order or simple order. In this case (A, R) is called a chain.

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