MAT2612 Assignment 4 (202587) Solutions Year Module 2026

UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology


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MAT2612: Introduction to Discrete Mathematics

Assignment 04 | Year Module 2026

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MAT2612
Module Code:
Introduction to Discrete Mathematics
Module Name:
Lattices and Boolean Algebra
Assignment Topic:
04
Assignment Number:
202587
Unique Number:
100
Total Marks:




Submitted in partial fulfilment of the requirements for
Introduction to Discrete Mathematics, UNISA 2026

,UNISA | MAT2612 Lattices and Boolean Algebra



Question 1: Partial Order Test (6 Marks)


Question 1 6 Marks

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 defined on A = {1, 2, 3, 4, 5} by


R = {(1, 1), (1, 2), (1, 5), (2, 2), (2, 4), (3, 3), (4, 4), (5, 1), (5, 3), (5, 5)}.


A relation is a partial order on A precisely when it is reflexive, antisymmetric and transitive
simultaneously. Each property is tested in turn below.


Reflexivity


Reflexivity requires (a, a) ∈ R for every a ∈ A. Since


(1, 1), (2, 2), (3, 3), (4, 4), (5, 5) ∈ R,


every element of A is related to itself, so R is reflexive.


Antisymmetry


Antisymmetry requires that (a, b) ∈ R and (b, a) ∈ R together force a = b. Here


(1, 5) ∈ R and (5, 1) ∈ R,


yet 1 ̸= 5. This single pair is enough to break the property, so R is not antisymmetric.




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, UNISA | MAT2612 Lattices and Boolean Algebra



Transitivity


Transitivity requires that (a, b) ∈ R and (b, c) ∈ R together force (a, c) ∈ R. Taking


(1, 2) ∈ R and (2, 4) ∈ R,


transitivity would demand (1, 4) ∈ R. Checking the set shows (1, 4) ∈
/ R, so R is not transi-
tive.


Conclusion


R is reflexive, but it fails both antisymmetry and transitivity. Since all three properties are
required simultaneously,
R is not a partial order on A.




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