Answer:
See below
Step-by-step explanation:
Remember some definitions about binary relations. If R⊆S×S then
- R is reflexive if (a,a)∈R for all a∈S
- R is irreflexive if (a,a)∉R for all a∈S
- R is symmetric if (a,b)∈R implies (b,a)∈R for all a,b∈S
- R is asymmetric if (a,b)∈R implies (b,a)∉R for all a,b∈S
- R is antisymmetric if (a,b)∈R and (b,a)∈R imply that a=b, for all a,b∈S
- R is transitive if (a,b)∈R and (b,c)∈R imply (a,c)∈R for all a,b,c∈S
a) R is not reflexive since (1,1)∉R.
R is irreflexive, since (a,a)∉R for all a=1,2,3,4
R is asymmetric: (2,3)∈R and (3,2)∉R (thus R is not symmetric).
R is antisymmetric, there are no cases to check. R is transitive, there are no cases to check.
b) R is reflexive, checking case by case, (a,a)∈R for all a=1,2,3,4. Hence R is not irreflexive.
R is not asymmetric: (1,2)∈R but (2,1)∈R. R is not symmetric, since (4,1)∈R but (1,4)∉R
R is not antisymmetric: (1,2)∈R and (2,1)∈R but 1≠2.
R is not transitive: (1,2)∈R and (2,4)∈R but (1,4)∉R.
