UNIVERSITY OF SOUTH AFRICA (UNISA)
Faculty of Science, Engineering and Technology
⋄
Lattices and Boolean Algebra
Assignment 04 — Year Module 2026
⋄
Module Code: MAT2612
Module Name: Introduction to Discrete Mathematics
Assignment No.: Assignment 04
Total Marks: 100
Unique Number: 202587
Module Type: Year Module
Submitted in partial fulfilment of the requirements for MAT2612
at the University of South Africa.
, UNISA | MAT2612 Lattices and Boolean Algebra — Assignment 04
1 Question 1: Partial Order on A = {1, 2, 3, 4, 5}
Question
State, with reasons, whether R is a partial order on A = {1, 2, 3, 4, 5}, where
R = {(1, 1), (1, 2), (1, 5), (2, 2), (2, 4), (3, 3), (4, 4), (5, 1), (5, 3), (5, 5)}.
Solution
The relation under consideration is
R = {(1, 1), (1, 2), (1, 5), (2, 2), (2, 4), (3, 3), (4, 4), (5, 1), (5, 3), (5, 5)}
on A = {1, 2, 3, 4, 5}. For R to be a partial order it must be reflexive, antisymmetric and transi-
tive.
Reflexivity. A relation is reflexive if (a, a) ∈ R for every a ∈ A. Since (1, 1), (2, 2), (3, 3), (4, 4), (5, 5)
all belong to R, every element of A is related to itself, so R is reflexive.
Antisymmetry. A relation is antisymmetric if (a, b) ∈ R and (b, a) ∈ R together force a = b.
Here (1, 5) ∈ R and (5, 1) ∈ R, yet 1 ̸= 5. Hence R is not antisymmetric.
Transitivity. A relation is transitive if (a, b) ∈ R and (b, c) ∈ R together force (a, c) ∈ R. Since
(1, 2) ∈ R and (2, 4) ∈ R, transitivity would require (1, 4) ∈ R; however (1, 4) ∈
/ R. Hence R is
not transitive.
Because R satisfies reflexivity but fails both antisymmetry and transitivity, it does not meet all
three requirements of a partial order. Therefore
R is not a partial order on A.
Page 1 of 14
Faculty of Science, Engineering and Technology
⋄
Lattices and Boolean Algebra
Assignment 04 — Year Module 2026
⋄
Module Code: MAT2612
Module Name: Introduction to Discrete Mathematics
Assignment No.: Assignment 04
Total Marks: 100
Unique Number: 202587
Module Type: Year Module
Submitted in partial fulfilment of the requirements for MAT2612
at the University of South Africa.
, UNISA | MAT2612 Lattices and Boolean Algebra — Assignment 04
1 Question 1: Partial Order on A = {1, 2, 3, 4, 5}
Question
State, with reasons, whether R is a partial order on A = {1, 2, 3, 4, 5}, where
R = {(1, 1), (1, 2), (1, 5), (2, 2), (2, 4), (3, 3), (4, 4), (5, 1), (5, 3), (5, 5)}.
Solution
The relation under consideration is
R = {(1, 1), (1, 2), (1, 5), (2, 2), (2, 4), (3, 3), (4, 4), (5, 1), (5, 3), (5, 5)}
on A = {1, 2, 3, 4, 5}. For R to be a partial order it must be reflexive, antisymmetric and transi-
tive.
Reflexivity. A relation is reflexive if (a, a) ∈ R for every a ∈ A. Since (1, 1), (2, 2), (3, 3), (4, 4), (5, 5)
all belong to R, every element of A is related to itself, so R is reflexive.
Antisymmetry. A relation is antisymmetric if (a, b) ∈ R and (b, a) ∈ R together force a = b.
Here (1, 5) ∈ R and (5, 1) ∈ R, yet 1 ̸= 5. Hence R is not antisymmetric.
Transitivity. A relation is transitive if (a, b) ∈ R and (b, c) ∈ R together force (a, c) ∈ R. Since
(1, 2) ∈ R and (2, 4) ∈ R, transitivity would require (1, 4) ∈ R; however (1, 4) ∈
/ R. Hence R is
not transitive.
Because R satisfies reflexivity but fails both antisymmetry and transitivity, it does not meet all
three requirements of a partial order. Therefore
R is not a partial order on A.
Page 1 of 14
