Question #286070

Construct a relation on the set {a, b, c, d} that is



a. reflexive, symmetric, but not transitive.



b. irreflexive, symmetric, and transitive.



c. irreflexive, antisymmetric, and not transitive.



d. reflexive, neither symmetric nor antisymmetric, and transitive.

1
Expert's answer
2022-01-10T13:29:47-0500

a. R={(a,a),(b,b),(c,c),(d,d),(a,b),(b,a)(a,c),(c,a),(b,c),(c,b),(b,d),(d,b)}R = \{ (a,a),\,(b,b),\,(c,c),\,(d,d),\,(a,b),\,(b,a)\,(a,c),\,(c,a),\,\,(b,c),(c,b),\,(b,d),\,(d,b)\}

For any x{a,b,c,d}x \in \{ a,b,c,d\} (x,x)R(x,x) \in R , so, R is reflexive.

For any x,y{a,b,c,d}x,y \in \{ a,b,c,d\} if (x,y)R(x,y) \in R then (y,x)R(y,x) \in R , so, R is symmetric.

(a,b)R,(b,d)R(a,b) \in R,\,(b,d) \in R , but (a,d)R(a,d) \notin R , so, R is not transitive.

b. Let R is this relation.

Since R is symmetric then for any x,y{a,b,c,d}x,y \in \{ a,b,c,d\} if (x,y)R(x,y) \in R then (y,x)R(y,x) \in R.

But then (since R is transitive) (x,x)R(x,x)∈R . But then R isn't irreflexive, which contradicts the condition, therefore, such a relation does not exist.

c. R={(a,b),(a,c),(b,d)}R = \{ (a,b),\,(a,c),\,(b,d)\}

For any x{a,b,c,d}x \in \{ a,b,c,d\} (x,x)R(x,x) \notin R , so, R is irreflexive.

For any x,y{a,b,c,d}x,y \in \{ a,b,c,d\} if (x,y)R(x,y) \in R then (y,x)R(y,x) \notin R , so, R is antisymmetric.

(a,b)R,(b,d)R(a,b) \in R,\,(b,d) \in R , but (a,d)R(a,d) \notin R , so, R is not transitive.

d. R={(a,a),(b,b),(c,c),(d,d),(a,b),(b,a),(a,c)}R = \{ (a,a),\,(b,b),\,(c,c),\,(d,d),\,(a,b),(b,a),\,(a,c)\}

For any x{a,b,c,d}x \in \{ a,b,c,d\} (x,x)R(x,x) \in R , so, R is reflexive.

(a,b)R,(b,a)R(a,b) \in R,\,(b,a) \in R , so R isn't antisymmetric, (a,c)R,(c,a)R(a,c) \in R,\,(c,a) \notin R so R isn't symmetric.

for any x,y,z{a,b,c,d}x,y,z \in \{ a,b,c,d\} (x,y)R(y,z)R(x,z)R(x,y) \in R \wedge (y,z) \in R \Rightarrow (x,z) \in R , so, R is transitive.


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