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Nesterboy
10 days ago
10

Which of these relations on{0,1,2,3}are partial orderings? Determine the properties of a partial ordering that the others lack.

a) {(0,0),(2,2),(3,3)} b) {(0,0),(1,1),(2,0),(2,2),(2,3),(3,3)} c) {(0,0),(1,1),(1,2),(2,2),(3,1),(3,3)} d) {(0,0),(1,1),(1,2),(1,3),(2,0),(2,2),(2,3), (3,0),(3,3)} e) {(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2), (1,3),(2,0),(2,2),(3,3)
Mathematics
1 answer:
Svet_ta [9.5K]10 days ago
3 0

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.

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