⋆
]]
⋆
COS1501: Theoretical Computer Science I
OCT/NOV Examination 2026 Revision Guide
Covers Past Papers: OCT/NOV 2023 to OCT/NOV 2025
⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆
[ Computer Science / School of Computing [
_ Exam Revision Guide
COS1501
Module Code:
Theoretical Computer Science I
Module Name:
OCT/NOV Examinations 2026
Paper / Exam:
OCT/NOV 2023, 2024 & 2025 Papers
Covers:
100 marks (MCQ format)
Total Marks:
Multiple Choice Questions
Format:
Use this guide to revise thoroughly. Focus on understanding, not memorisa-
tion.
Exam Revision Notes | COS1501 | 2023–2025
,COS1501 | Exam Revision 2023–2025 Theoretical Computer Science I
PAPER 1 — OCT/NOV 2025 100 marks
Key Concept
OCT/NOV 2025 focuses heavily on: set operations with nested sets, Venn diagrams,
relations (reflexive, symmetric, transitive), functions (injective, surjective, bijective),
proof by mathematical induction, and propositional logic.
Section 1: Number Sets Questions 1–8, 16 marks
Question 1 2 marks
Question: Which one of the following statements regarding number sets is FALSE?
(a) Z+ ⊆ R≥
(b) Z ⊆ Q
(c) Z≥ ⊆ R≥
(d) Q≥ ⊆ R+
Answer: (d) Q≥ ⊆ R+ is FALSE.
Q≥ denotes all non-negative rational numbers, which includes zero (0 ∈ Q≥ ). However, R+
denotes strictly positive reals, so 0 ∈
/ R+ . Therefore Q≥ ̸⊆ R+ .
⋆ Exam Tip
The key distinction: R+ means strictly positive (excludes 0); R≥ means non-negative
(includes 0). This distinction is a recurring trap.
Question 2 2 marks
Question: Which one of the following is TRUE regarding number sets?
(a) Q ⊂ R+
(b) Z− ⊂ Q
(c) R ⊆ Q
(d) N ⊂ Z−
Answer: (b) Z− ⊂ Q is TRUE.
Page 2 of 22
,COS1501 | Exam Revision 2023–2025 Theoretical Computer Science I
Every negative integer is also a rational number (it can be written as n/1), so Z− is a proper
subset of Q.
• (a) False: Q contains negative rationals; R+ does not.
• (c) False: R contains irrationals not in Q, so R ̸⊆ Q.
• (d) False: N = {0, 1, 2, . . .} or {1, 2, 3, . . .}, all non-negative; none are in Z− .
Question 3 2 marks
Question: Consider the set S = {x | −3 ≤ x < 5, x ∈ Z} ∩ {x | 1 < x ≤ 6, x ∈ R}. Which
best describes S?
(a) S = {−3, −2, −1, 0, 1, 2, 3, 4}
(b) S = {2, 3, 4}
(c) S = {x | 1 < x < 5, x ∈ R}
(d) S = {2, 3, 4, 5}
Answer: (b) S = {2, 3, 4}
The first set gives integers from −3 to 4 inclusive: {−3, −2, −1, 0, 1, 2, 3, 4}. The second set
gives reals strictly greater than 1 up to 6. Their intersection requires integers that satisfy
both: integers greater than 1 (i.e. ≥ 2) and less than 5. That gives {2, 3, 4}.
Question 4 2 marks
Question: Which of the following is a correct representation of the set of all integers
that are not positive?
(a) Z−
(b) Z≤
(c) {x | x ∈ Z, x < 0}
(d) Z \ Z+
Answer: (d) Z \ Z+
“Not positive” means x ≤ 0, which includes zero. This equals Z≤ (option b is also correct
notation), but more precisely Z \ Z+ = {. . . , −2, −1, 0}, which is the standard UNISA answer.
Option (a) Z− excludes 0, and option (c) also excludes 0.
Page 3 of 22
,COS1501 | Exam Revision 2023–2025 Theoretical Computer Science I
. Watch Out
Z− = negative integers only (excludes 0). Z≤ = non-positive integers (includes 0).
These are frequently confused.
Question 5 2 marks
Question: Is the following statement TRUE or FALSE? Z+ ⊂ Z≥
Answer: TRUE
Z+ = {1, 2, 3, . . .} and Z≥ = {0, 1, 2, 3, . . .}. Every positive integer is a non-negative integer,
but 0 ∈ Z≥ while 0 ∈
/ Z+ , so Z+ is a proper subset of Z≥ .
Question 6 2 marks
Question: Is the following statement TRUE or FALSE? Q ⊂ R+
Answer: FALSE
Q contains negative rational numbers (e.g. − 12 ), and R+ contains only strictly positive reals.
Therefore Q ̸⊆ R+ .
Question 7 2 marks
Question: Which one of the following is a prime number?
(a) 1 (b) 9 (c) 51 (d) 37
Answer: (d) 37
A prime number has exactly two distinct factors: 1 and itself.
• 1 is not prime (only one factor).
• 9 = 3 × 3 (not prime).
• 51 = 3 × 17 (not prime).
• 37 has no factors other than 1 and 37 — it is prime.
Page 4 of 22
,COS1501 | Exam Revision 2023–2025 Theoretical Computer Science I
Question 8 2 marks
Question: Is the following statement TRUE or FALSE? Every rational number is also a
real number.
Answer: TRUE
Q ⊂ R: the rational numbers are a proper subset of the real numbers. Every rational is real,
√
but not every real is rational (e.g. 2 is irrational).
Section 2: Sets and Set Operations Questions 9–20, 24 marks
Question 9 2 marks
Question: Let A = {a, b, {b}, {b, e}, e} and B = {b, {b, e}, e, f }. What is A ∩ B?
(a) {b, e, f } (b) {b, {b, e}, e} (c) {a, b, e} (d) {a, {b}, f }
Answer: (b) {b, {b, e}, e}
The intersection A ∩ B contains elements belonging to both sets. Checking element by element:
• b ∈ A and b ∈ B ✓
• {b, e} ∈ A and {b, e} ∈ B ✓
• e ∈ A and e ∈ B ✓
• a ∈ A but a ∈
/ B; {b} ∈ A but {b} ∈
/ B; f ∈
/A
So A ∩ B = {b, {b, e}, e}.
. Watch Out
Treat set-valued elements (like {b, e}) as single atomic elements. {b, e} is not the same
as b or e.
Question 10 2 marks
Question: Using the same sets A = {a, b, {b}, {b, e}, e} and B = {b, {b, e}, e, f }, what is
A ∪ B?
(a) {a, b, e, f } (b) {a, b, {b}, e, f } (c) {a, b, {b}, {b, e}, e, f } (d) {a, b, {b, e}, f }
Page 5 of 22
, COS1501 | Exam Revision 2023–2025 Theoretical Computer Science I
Answer: (c) {a, b, {b}, {b, e}, e, f }
The union contains all elements from either set: a, b, {b}, {b, e}, e from A, and f from B (oth-
ers are already in A). So A ∪ B = {a, b, {b}, {b, e}, e, f }.
Question 11 2 marks
Question: Using A = {a, b, {b}, {b, e}, e} and B = {b, {b, e}, e, f }, what is A − B (set dif-
ference)?
(a) {a, {b}} (b) {a, b, f } (c) {a, {b}, f } (d) {a}
Answer: (a) {a, {b}}
A − B contains elements in A that are not in B: a ∈ / B (since B contains b, not {b}).
/ B, {b} ∈
So A − B = {a, {b}}.
Question 12 2 marks
Question: Using A = {a, b, {b}, {b, e}, e} and B = {b, {b, e}, e, f }, what is the symmetric
difference A + B (also written A ⊕ B)?
(a) {a, {b}, f } (b) {a, b, f } (c) {a, {b}, {b, e}, f } (d) {f }
Answer: (a) {a, {b}, f }
A + B = (A − B) ∪ (B − A).
• A − B = {a, {b}} (elements in A not in B)
• B − A = {f } (elements in B not in A)
• A + B = {a, {b}, f }
Question 13 2 marks
Question: Let U = {1, 2, 3, {3}, {c, e}, e, f }, A = {{3}, {c, e}, f }, B = {1, 3, {3}, e, f }.
What is A ∪ B?
(a) {1, 3, {c, e}, f } (b) {1, 3, {3}, {c, e}, e, f } (c) {1, {3}, c, e, f } (d) {3, {3}, {c, e}, e}
Answer: (b) {1, 3, {3}, {c, e}, e, f }
Page 6 of 22
]]
⋆
COS1501: Theoretical Computer Science I
OCT/NOV Examination 2026 Revision Guide
Covers Past Papers: OCT/NOV 2023 to OCT/NOV 2025
⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆
[ Computer Science / School of Computing [
_ Exam Revision Guide
COS1501
Module Code:
Theoretical Computer Science I
Module Name:
OCT/NOV Examinations 2026
Paper / Exam:
OCT/NOV 2023, 2024 & 2025 Papers
Covers:
100 marks (MCQ format)
Total Marks:
Multiple Choice Questions
Format:
Use this guide to revise thoroughly. Focus on understanding, not memorisa-
tion.
Exam Revision Notes | COS1501 | 2023–2025
,COS1501 | Exam Revision 2023–2025 Theoretical Computer Science I
PAPER 1 — OCT/NOV 2025 100 marks
Key Concept
OCT/NOV 2025 focuses heavily on: set operations with nested sets, Venn diagrams,
relations (reflexive, symmetric, transitive), functions (injective, surjective, bijective),
proof by mathematical induction, and propositional logic.
Section 1: Number Sets Questions 1–8, 16 marks
Question 1 2 marks
Question: Which one of the following statements regarding number sets is FALSE?
(a) Z+ ⊆ R≥
(b) Z ⊆ Q
(c) Z≥ ⊆ R≥
(d) Q≥ ⊆ R+
Answer: (d) Q≥ ⊆ R+ is FALSE.
Q≥ denotes all non-negative rational numbers, which includes zero (0 ∈ Q≥ ). However, R+
denotes strictly positive reals, so 0 ∈
/ R+ . Therefore Q≥ ̸⊆ R+ .
⋆ Exam Tip
The key distinction: R+ means strictly positive (excludes 0); R≥ means non-negative
(includes 0). This distinction is a recurring trap.
Question 2 2 marks
Question: Which one of the following is TRUE regarding number sets?
(a) Q ⊂ R+
(b) Z− ⊂ Q
(c) R ⊆ Q
(d) N ⊂ Z−
Answer: (b) Z− ⊂ Q is TRUE.
Page 2 of 22
,COS1501 | Exam Revision 2023–2025 Theoretical Computer Science I
Every negative integer is also a rational number (it can be written as n/1), so Z− is a proper
subset of Q.
• (a) False: Q contains negative rationals; R+ does not.
• (c) False: R contains irrationals not in Q, so R ̸⊆ Q.
• (d) False: N = {0, 1, 2, . . .} or {1, 2, 3, . . .}, all non-negative; none are in Z− .
Question 3 2 marks
Question: Consider the set S = {x | −3 ≤ x < 5, x ∈ Z} ∩ {x | 1 < x ≤ 6, x ∈ R}. Which
best describes S?
(a) S = {−3, −2, −1, 0, 1, 2, 3, 4}
(b) S = {2, 3, 4}
(c) S = {x | 1 < x < 5, x ∈ R}
(d) S = {2, 3, 4, 5}
Answer: (b) S = {2, 3, 4}
The first set gives integers from −3 to 4 inclusive: {−3, −2, −1, 0, 1, 2, 3, 4}. The second set
gives reals strictly greater than 1 up to 6. Their intersection requires integers that satisfy
both: integers greater than 1 (i.e. ≥ 2) and less than 5. That gives {2, 3, 4}.
Question 4 2 marks
Question: Which of the following is a correct representation of the set of all integers
that are not positive?
(a) Z−
(b) Z≤
(c) {x | x ∈ Z, x < 0}
(d) Z \ Z+
Answer: (d) Z \ Z+
“Not positive” means x ≤ 0, which includes zero. This equals Z≤ (option b is also correct
notation), but more precisely Z \ Z+ = {. . . , −2, −1, 0}, which is the standard UNISA answer.
Option (a) Z− excludes 0, and option (c) also excludes 0.
Page 3 of 22
,COS1501 | Exam Revision 2023–2025 Theoretical Computer Science I
. Watch Out
Z− = negative integers only (excludes 0). Z≤ = non-positive integers (includes 0).
These are frequently confused.
Question 5 2 marks
Question: Is the following statement TRUE or FALSE? Z+ ⊂ Z≥
Answer: TRUE
Z+ = {1, 2, 3, . . .} and Z≥ = {0, 1, 2, 3, . . .}. Every positive integer is a non-negative integer,
but 0 ∈ Z≥ while 0 ∈
/ Z+ , so Z+ is a proper subset of Z≥ .
Question 6 2 marks
Question: Is the following statement TRUE or FALSE? Q ⊂ R+
Answer: FALSE
Q contains negative rational numbers (e.g. − 12 ), and R+ contains only strictly positive reals.
Therefore Q ̸⊆ R+ .
Question 7 2 marks
Question: Which one of the following is a prime number?
(a) 1 (b) 9 (c) 51 (d) 37
Answer: (d) 37
A prime number has exactly two distinct factors: 1 and itself.
• 1 is not prime (only one factor).
• 9 = 3 × 3 (not prime).
• 51 = 3 × 17 (not prime).
• 37 has no factors other than 1 and 37 — it is prime.
Page 4 of 22
,COS1501 | Exam Revision 2023–2025 Theoretical Computer Science I
Question 8 2 marks
Question: Is the following statement TRUE or FALSE? Every rational number is also a
real number.
Answer: TRUE
Q ⊂ R: the rational numbers are a proper subset of the real numbers. Every rational is real,
√
but not every real is rational (e.g. 2 is irrational).
Section 2: Sets and Set Operations Questions 9–20, 24 marks
Question 9 2 marks
Question: Let A = {a, b, {b}, {b, e}, e} and B = {b, {b, e}, e, f }. What is A ∩ B?
(a) {b, e, f } (b) {b, {b, e}, e} (c) {a, b, e} (d) {a, {b}, f }
Answer: (b) {b, {b, e}, e}
The intersection A ∩ B contains elements belonging to both sets. Checking element by element:
• b ∈ A and b ∈ B ✓
• {b, e} ∈ A and {b, e} ∈ B ✓
• e ∈ A and e ∈ B ✓
• a ∈ A but a ∈
/ B; {b} ∈ A but {b} ∈
/ B; f ∈
/A
So A ∩ B = {b, {b, e}, e}.
. Watch Out
Treat set-valued elements (like {b, e}) as single atomic elements. {b, e} is not the same
as b or e.
Question 10 2 marks
Question: Using the same sets A = {a, b, {b}, {b, e}, e} and B = {b, {b, e}, e, f }, what is
A ∪ B?
(a) {a, b, e, f } (b) {a, b, {b}, e, f } (c) {a, b, {b}, {b, e}, e, f } (d) {a, b, {b, e}, f }
Page 5 of 22
, COS1501 | Exam Revision 2023–2025 Theoretical Computer Science I
Answer: (c) {a, b, {b}, {b, e}, e, f }
The union contains all elements from either set: a, b, {b}, {b, e}, e from A, and f from B (oth-
ers are already in A). So A ∪ B = {a, b, {b}, {b, e}, e, f }.
Question 11 2 marks
Question: Using A = {a, b, {b}, {b, e}, e} and B = {b, {b, e}, e, f }, what is A − B (set dif-
ference)?
(a) {a, {b}} (b) {a, b, f } (c) {a, {b}, f } (d) {a}
Answer: (a) {a, {b}}
A − B contains elements in A that are not in B: a ∈ / B (since B contains b, not {b}).
/ B, {b} ∈
So A − B = {a, {b}}.
Question 12 2 marks
Question: Using A = {a, b, {b}, {b, e}, e} and B = {b, {b, e}, e, f }, what is the symmetric
difference A + B (also written A ⊕ B)?
(a) {a, {b}, f } (b) {a, b, f } (c) {a, {b}, {b, e}, f } (d) {f }
Answer: (a) {a, {b}, f }
A + B = (A − B) ∪ (B − A).
• A − B = {a, {b}} (elements in A not in B)
• B − A = {f } (elements in B not in A)
• A + B = {a, {b}, f }
Question 13 2 marks
Question: Let U = {1, 2, 3, {3}, {c, e}, e, f }, A = {{3}, {c, e}, f }, B = {1, 3, {3}, e, f }.
What is A ∪ B?
(a) {1, 3, {c, e}, f } (b) {1, 3, {3}, {c, e}, e, f } (c) {1, {3}, c, e, f } (d) {3, {3}, {c, e}, e}
Answer: (b) {1, 3, {3}, {c, e}, e, f }
Page 6 of 22
