Step-by-step explanation:
A = {0,1,2,3}
a): R = {(0,0),(2,2),(3,3)}
R displays antisymmetry, as whenever (a,b)∈R, it follows that a=b.
R lacks reflexivity since (1,1) ∉ R even though 1 ∈ A.
R is transitive; therefore, if (a,b)∈R and (b, c) ∈ R, then a=b=c and (a,c)=(a,a)∈R.
R fails to be a partial ordering due to its lack of reflexivity.
b): R = {(0,0),(1,1),(2,0),(2,2),(2,3),(3,3)}
R is antisymmetric because if (a,b)∈R and (b, a) ∈ R, then a must equal b (e.g., (2,0) ∈ R and (0,2) ∉ R; likewise, (2,3) ∈ R and (3,2) ∉ R).
R is reflexive since each (a,a) resides in R for all elements a ∈ A.
R is transitive; if (a,b)∈R and (b,c)∈R, it implies (a,c) exists in R or identical to (a,b) in R.
R qualifies as a partial ordering due to its reflexivity, antisymmetry, and transitivity.
c): R = {(0,0),(1,1),(1,2),(2,2),(3,1),(3,3)}
R is reflexive as (a,a)∈R is true for every a ∈ A.
R is antisymmetric; if (a,b)∈R holds and if also (b,a)∈R, then a invariably equals b (e.g., (1,2)∈R while (2,1) ∉ R; similarly for (3,1) and (1,3)).
R fails transitivity because (3,1) ∈ R and (1,2) ∈ R, but (3,2) ∉ R.
R is not a partial ordering due to transitivity not being satisfied.
d): R = {(0,0),(1,1),(1,2),(1,3),(2,0),(2,2),(2,3), (3,0),(3,3)}
R exhibits reflexivity since (a,a)∈R for each element a ∈ A.
R displays antisymmetry, as if (a,b)∈R and (b,a)∈R then a must equal b (e.g., (1,2)∈R and (2,1)∉R; similarly validated for others).
R is not transitive because (1,2)∈R and (2,0)∈R, but (1,0)∉R.
R is not a partial ordering due to transitivity issues.
e): R = { ( 0, 0 ), ( 0, 1 ), ( 0, 2 ), ( 0, 3 ), ( 1, 0 ), ( 1, 1 ), ( 1, 2 ), ( 1, 3 ), ( 2, 0 ), ( 2, 2 ), ( 3, 3 ) }
R proves to be reflexive, given that (a,a)∈R for all a∈A.
R is not antisymmetric since both (1,0)∈R and (0,1)∈R hold while 0 is distinct from 1.
R lacks transitivity, as (2,0)∈R and (0,3)∈R, while (2,3)∉R.
R cannot be classified as a partial ordering as it fails in both antisymmetry and transitivity.
