Let us consider a set {a,b,c,d}.
Reflexivity: A relation R on set S is called reflexive if (a,a)∈R for all a∈S.
Symmetry: A relation R on set S is called symmetric if (b,a)∈R whenever (a,b)∈R for all a,b∈S.
Anti symmetry: A relation R on set S is called anti symmetric if (b,a)∈R and (a,b)∈R than a=b for all a,b∈S.
Transitivity: a relation R on set S is called transitive if (a,b)∈R and (b,c)∈R than (a,c)∈R for all a,b,c∈S.
a) Consider a relation on {a,b,c,d} such that
R1={(a,a),(b,b),(c,c),(d,d),(a,b),(b,a),(b,c),(c,b)}
Now we can see that for all x∈{a,b,c,d},(x,x)∈R1 thus R1 is reflexive.
Also (b,a)∈R1 whenever (a,b)∈R1 and (c,b)∈R1 whenever (b,c)∈R1 thus we can say R1 is symmetric.
Again we can see that if (a,b)∈R1 and (b,c)∈R1 but (a,c)∈/R1 Hence R1 is not transitive.
b) Let R2=∅ . Consider on a,b∈{a,b,c,d}, if b R R a then a R b " hence the R is symmetric.
Let a,b,c∈{a,b,c,d} , if c R a then a R b or b R R c " hence the R is transitivity.
Now take any element a∈S and observe that aR~a . Thus R is not reflexive, hence it is irreflexive.
c) Consider a relation on {a,b,c,d} such that
R3={(a,b),(b,c)}
We can see that (a,a)∈/R3 so R3 is not reflexive and also we can see that no element is related to each other means it is irreflexive.
Also we have if a=b whenever (a,b)∈R3 and (b,a)∈R3 but here (b,a)∈/R3 so it is antisymmetric.
Again we can see that if (a,b)∈R1 and (b,c)∈R1 but (a,c)∈/R1 Hence R3 is not transitive.
d) Consider a relation on {a,b,c,d} such that
R4={(a,a),(b,b),(c,c),(d,d),(a,b),(b,a),(c,a),(b,c)}
We can see that for all x∈{a,b,c,d},(x,x)∈R4 thus R4 is reflexive.
Also (b,c)∈R4 but (c,b)∈/R4 thus we can say R4 is not symmetric. Now (b,a)∈R4 whenever (a,b)∈R4 but a=b thus R4 is not antisymmetric.
Again we can see that if (a,b)∈R4 and (b,c)∈R4 then (a,c)∈R4. Hence R4 is transitive.
e) Consider a relation on {a,b,c,d} such that
R5={(a,b),(b,a),(c,c),(a,c)}
We can see that (a,a)∈/R5 so R5 is not reflexive and also we can see that (c,c)∈R5 means it is not irreflexive.
Also (a,c)∈R5 but (c,a)∈/R5 thus we can say R5 is not symmetric. Now (b,a)∈R5 whenever (a,b)∈R5 but a=b thus R5 is not antisymmetric.
Again we can see that if (b,a)∈R5 and (a,c)∈R5 then (b,c)∈/R5. Hence R5 is transitive
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