let set A={1,2,3} .Then show that the number of relations containing (1,2) and (2,3) which are reflexive and transitive but not symmetric is three...

The smallest reflexive relation on set A containing (1, 2) and (2, 3) is

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}

Since (1, 2) ∈ R and (2, 3) ∈R but (1, 3) ∈ R

So, R is non transitive. To make it transitive we have to include (1, 3) in R. Including (1, 3) in R we get

R_{1} = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

This is reflexive and transitive but not symmetric.

Now if we add pair (2, 1) to R_{1} to get R_{2}. Then the relation is still reflexive, transitive but not symmetric. Similarly by adding (3, 2) and (3, 1) to R_{1}, we get

R_{3} = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (3, 2)}

R_{y} = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (3, 1)}

These relations are reflexive but not symmetric. We observe that out of ordered pairs (2, 1), (3, 2) and (3, 1) at a time of we add any two ordered pair at a time to R_{1} then to maintain the transitivity we will be forced to add the remaining third pair and in this process the relation will become symmetric also which is not required. Hence the total number of reflexive, transitive but not symmetric relations containing (1, 2) and (2, 3) is four.

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