Basic Mathematics

Pass path: Learn the 3-set cardinality formula and interval notations in Part 1 and Part 2. Top path: Master 3-set survey word problems with Venn diagram decomposition in Part 2 and Part 4.

Part 1: Core Concepts (2 Marks)

• ★ Cardinality Formula for Three Sets: For any three finite sets $A$, $B$, and $C$:

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  $$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)$$
• ★ Rational vs. Irrational Numbers: A rational number can be expressed as a ratio of two integers $\frac{p}{q}$ ($q \neq 0$) with terminating or repeating decimal expansions; an irrational number cannot be expressed as a fraction and possesses non-terminating, non-repeating decimal expansions (e.g., $\sqrt{2}, \pi, e$).
• ★ Modulus (Absolute Value) Properties:

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  $$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}, \quad |x| \le a \iff -a \le x \le a, \quad |a + b| \le |a| + |b| \text{ (Triangle Inequality)}$$

Part 2: Structural Frameworks (10 Marks)

1. Three-Set Venn Diagram Partition Architecture

Universal Set Venn Diagram (Three Sets)

When solving 3-set survey problems, decompose the Venn diagram into 8 mutually disjoint regions:

1. $n_o(A \cap B \cap C) = x$ (Center region common to all three).
2. Pairwise only regions:

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   $$n_o(A \cap B) = n(A \cap B) - x, \quad n_o(B \cap C) = n(B \cap C) - x, \quad n_o(C \cap A) = n(C \cap A) - x$$
3. Exactly one set regions:

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   $$n_o(A) = n(A) - [n_o(A \cap B) + n_o(C \cap A) + x]$$

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   $$n_o(B) = n(B) - [n_o(A \cap B) + n_o(B \cap C) + x]$$

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   $$n_o(C) = n(C) - [n_o(C \cap A) + n_o(B \cap C) + x]$$
4. Universal set balance:

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   $$n(U) = n(A \cup B \cup C) + n(\overline{A \cup B \cup C})$$

Part 3: Distinction Tables

Numerical Sets: Rational ($\mathbb{Q}$) vs. Irrational ($\mathbb{Q}'$)

Part 4: Worked Problems & Exam Traps

Solved Archetype: 3-Set Survey Word Problem

Problem: In a survey of 100 BIM students, 50 like Programming (P), 40 like Database (D), and 35 like Networking (N). 20 like P and D, 15 like D and N, 18 like P and N, and 10 like all three subjects. Find: (a) number of students who like at least one subject, (b) number of students who like none of the subjects, and (c) number of students who like exactly one subject.

Step-by-step Solution:

1. Given data: $n(U) = 100$, $n(P) = 50$, $n(D) = 40$, $n(N) = 35$, $n(P \cap D) = 20$, $n(D \cap N) = 15$, $n(P \cap N) = 18$, $n(P \cap D \cap N) = 10$.
2. (a) Students who like at least one subject:

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   $$n(P \cup D \cup N) = 50 + 40 + 35 - (20 + 15 + 18) + 10 = 125 - 53 + 10 = 82$$
3. (b) Students who like none of the subjects:

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   $$n(\overline{P \cup D \cup N}) = n(U) - n(P \cup D \cup N) = 100 - 82 = 18$$
4. (c) Students who like exactly one subject:
   • $n_o(P \cap D) = 20 - 10 = 10$
   • $n_o(D \cap N) = 15 - 10 = 5$
   • $n_o(P \cap N) = 18 - 10 = 8$
   • $n_o(P) = 50 - (10 + 8 + 10) = 22$
   • $n_o(D) = 40 - (10 + 5 + 10) = 15$
   • $n_o(N) = 35 - (8 + 5 + 10) = 12$

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     $$\text{Exactly one subject} = n_o(P) + n_o(D) + n_o(N) = 22 + 15 + 12 = 49$$

Students lose marks here because they confuse $n(A \cap B)$ (“both $A$ and $B$”) with $n_o(A \cap B)$ (“only $A$ and $B$, but not $C$”), failing to subtract the 3-way intersection $n(A \cap B \cap C)$.