Chapter 4: Sets

Step-by-Step Algebraic Formulation:

• Total percentage of voters $= 100\%$.
• Percentage of voters who did not vote $= 20\%$.
• Therefore, total valid votes polled $= 100\% - 20\% = 80\%$.
• Votes cast for contestant A $= y\%$.
• Votes cast for contestant B $= (y + 30)\%$.

Since the entire polled vote was divided exclusively between contestants A and B:

Step-by-Step Cardinality Analysis:

Let the total number of students in the class be $n(U) = 70$.

• Let $F_1$ be the set of students failing in Test-I: $n(F_1) = 50\% \text{ of } 70 = 0.50 \times 70 = 35$.
• Let $F_2$ be the set of students failing in Test-II: $n(F_2) = 40\% \text{ of } 70 = 0.40 \times 70 = 28$.
• The maximum number of students who could fail in at least one test is $n(F_1 \cup F_2) = n(F_1) + n(F_2) - n(F_1 \cap F_2)$.
• In standard two-set examination problems where every student either fails or passes, assuming maximum independent failure spread $n(F_1 \cap F_2) = 0$ (or analyzing pass rates directly):
  • Pass in Test-I: $P_1 = 100\% - 50\% = 50\% \implies n(P_1) = 35$.
  • Pass in Test-II: $P_2 = 100\% - 40\% = 60\% \implies n(P_2) = 42$.
  • Total students $= n(P_1 \cup P_2) \le 70$.
  • $n(P_1 \cap P_2) = n(P_1) + n(P_2) - n(P_1 \cup P_2) = 35 + 42 - 70 = 77 - 70 = 7$.

Principle of Inclusion-Exclusion for Two Finite Sets:

By the fundamental theorem of set cardinality:

Adding $n(A \cap B)$ to both sides gives:

Step-by-Step Analysis of Set Difference:

• $A = \{2, 4, 6, 8, 10, 12, 14, 16, \dots\}$ (set of all positive even integers, which is infinite).
• $B = \{2, 3, 5, 7, 11, 13, 17, 19, \dots\}$ (set of all prime numbers).
• The intersection $A \cap B = \{2\}$ because $2$ is the unique even prime number.
• The difference set $A - B = \{x : x \in A \text{ and } x \notin B\}$ removes only elements of $B$ from $A$:

Since this set contains all even natural numbers greater than 2, it contains infinitely many elements.

Geometric Set Relationships:

1. For $P \cap Q$:
   • Every equilateral triangle has 3 equal sides, so it has at least 2 equal sides (hence every equilateral triangle is an isosceles triangle).
   • Thus, $Q \subseteq P$, which means $P \cap Q = Q = \text{The set of all equilateral triangles}$.
2. For $R - P$:
   • $R = \text{The set of all right-angled triangles}$.
   • $P = \text{The set of all isosceles triangles}$.
   • $R - P$ removes all isosceles right-angled triangles (e.g. $45^\circ\text{-}45^\circ\text{-}90^\circ$) from $R$, leaving only the scalene right-angled triangles, i.e., the set of non-isosceles right-angled triangles.

Power Set Cardinality Theorem:

For any finite set $A$ with $n(A) = k$, the number of elements in the power set is given by $n(P(A)) = 2^k$.

• Here $A = \phi$ (empty set), so $n(A) = 0$.
• $n(P(A)) = 2^0 = 1$.
• The power set itself is $P(\phi) = \{\phi\}$, which contains exactly $1$ element (the empty set).

Interval Operations on Real Numbers:

• $A = [0, 3] = \{x \in \mathbb{R} : 0 \le x \le 3\}$ (closed interval from 0 to 3).
• $B = [2, 6) = \{x \in \mathbb{R} : 2 \le x < 6\}$ (left-closed, right-open interval from 2 to 6).
• The union $A \cup B$ consists of all real numbers that belong to at least one of these intervals:

Note that $0$ is included ($[$) while $6$ is excluded ($)$).

Standard Hierarchy of Number Sets:

• $\mathbb{N} = \{1, 2, 3, \dots\}$ (Natural Numbers)
• $\mathbb{Z} = \{\dots, -2, -1, 0, 1, 2, \dots\}$ (Integers)
• Notice that every natural number is an integer, so $\mathbb{N} \subset \mathbb{Z}$ is true.
• However, negative integers (like $-3 \in \mathbb{Z}$) and $0$ do not belong to $\mathbb{N}$. Hence, $\mathbb{Z} \subset \mathbb{N}$ is false (not correct).

Distinction between Element ($\in$) and Subset ($\subset$):

• The elements of $A$ are individual numbers: $1, 2, 3, 5, 9$.
• (a) $0 \notin A$: True, 0 is not in the list.
• (b) $3 \in A$: True, 3 is an element of set $A$.
• (d) $\{3\} \subset A$: True, the set containing 3 is a subset of $A$.
• (c) $\{3\} \in A$: False / Not true, because the singleton set $\{3\}$ is a subset, not an individual element inside $A$.

Step-by-Step Calculation:

• Let $H$ be the set of students who know Hindi $\implies n(H) = 100$.
• Let $E$ be the set of students who know English $\implies n(E) = 50$.
• Students who know both languages $\implies n(H \cap E) = 25$.
• Since each student knows either Hindi or English, the total group size equals $n(H \cup E)$:

Step-by-Step Iteration:

1. Given $A = \emptyset$.
2. First power set $P(A) = P(\emptyset) = \{\emptyset\}$. (Note: $n(P(A)) = 2^0 = 1$).
3. Now, $P(P(A)) = P(\{\emptyset\})$. Since $\{\emptyset\}$ has $1$ element (namely $\emptyset$), its power set will have $2^1 = 2$ subsets:
   • The empty set: $\emptyset$
   • The set itself: $\{\emptyset\}$
4. Therefore:

Formula & Calculation:

For any finite set with $n$ elements, the total number of subsets is given by $2^n$.

• Here, $B = \{a, b, c, d\} \implies n(B) = 4$.
• Number of subsets $= 2^4 = 16$.

Subsets and Proper Subsets:

• Given $n(P) = 5$.
• Total number of subsets $= 2^n = 2^5 = 32$.
• A proper subset of $P$ is any subset except $P$ itself.
• Total number of proper subsets $= 2^n - 1 = 32 - 1 = 31$.

Analysis:

1. Every element of $A$ ($a, b, c$) is also in $B$, so $A \subseteq B$ is YES (True).
2. $A \cup B = \{a, b, c\} \cup \{a, b, c, d\} = \{a, b, c, d\} = B$.
3. $A \cap B = \{a, b, c\} \cap \{a, b, c, d\} = \{a, b, c\} = A$.

Step-by-Step Roster Form & Computations:

• $A = \{2, 3, 5, 7, 11, 13, 17, 19\}$
• $B = \{2, 4, 6, 8, 10, 12, 14\}$
• (i) $A \cap B$: $\{2\}$ (the only common element).
• (ii) $A - B$: $\{3, 5, 7, 11, 13, 17, 19\}$ (elements in $A$ excluding 2).
• (iii) $(A \cup B)'$:
  • $A \cup B = \{2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 17, 19\}$
  • With universal set $U = \{1, 2, \dots, 20\}$:
  • $(A \cup B)' = \{1, 9, 15, 16, 18, 20\}$.

Step-by-Step Derivation:

1. By the cardinality addition theorem: $$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$ $$28 = 15 + 20 - n(A \cap B) \implies 28 = 35 - n(A \cap B) \implies n(A \cap B) = 35 - 28 = 7$$
2. Since $n(A \cap B) = 7 \neq 0$, sets $A$ and $B$ have common elements and are NOT disjoint.
3. To find $n(A - B)$: $$n(A - B) = n(A) - n(A \cap B) = 15 - 7 = 8$$

Solving the Quadratic Equations:

• For set $A$: $x^2 - 5x + 6 = 0 \implies (x - 2)(x - 3) = 0 \implies x = 2, 3$. Thus, $A = \{2, 3\}$.
• For set $B$: $x^2 - 3x + 2 = 0 \implies (x - 1)(x - 2) = 0 \implies x = 1, 2$. Thus, $B = \{1, 2\}$.
• $A \cup B$: $\{1, 2, 3\}$.
• $A \cap B$: $\{2\}$.
• $A - B$: $\{2, 3\} - \{1, 2\} = \{3\}$.
• $B - A$: $\{1, 2\} - \{2, 3\} = \{1\}$.
• Since $\{3\} \neq \{1\}$, $A - B \neq B - A$.

Formal Proofs for $A \subset B$:

1. Proof of $A \cap B = A$:
   • Since $A \subset B$, for every $x \in A$, we have $x \in B$. Thus $x \in A \text{ and } x \in B \implies x \in A \cap B$. Hence $A \subseteq A \cap B$.
   • By definition of intersection, $A \cap B \subseteq A$. Combining gives $A \cap B = A$.
2. Proof of $A \cup B = B$:
   • Since every element in $A$ is already in $B$, adding $A$ to $B$ adds no new elements. Thus $A \cup B = B$.
3. Proof of $B' \subseteq A'$:
   • Let $x \in B' \implies x \notin B$.
   • Since $A \subset B$, if $x$ were in $A$, it would have to be in $B$. Since $x \notin B$, $x$ cannot be in $A$ ($x \notin A$).
   • Therefore $x \in A'$. This proves that $B' \subseteq A'$.

Step-by-Step Verification:

• $A' = U - A = \{6, 7, 8, 9, 10\}$
• $B' = U - B = \{1, 3, 5, 7, 9, 10\}$
• Verifying Law (i): $(A \cup B)' = A' \cap B'$
  • $A \cup B = \{1, 2, 3, 4, 5, 6, 8\}$
  • $\text{LHS} = (A \cup B)' = U - (A \cup B) = \{7, 9, 10\}$
  • $\text{RHS} = A' \cap B' = \{6, 7, 8, 9, 10\} \cap \{1, 3, 5, 7, 9, 10\} = \{7, 9, 10\}$
  • $\text{LHS} = \text{RHS}$. Hence verified.
• Verifying Law (ii): $(A \cap B)' = A' \cup B'$
  • $A \cap B = \{2, 4\}$
  • $\text{LHS} = (A \cap B)' = U - \{2, 4\} = \{1, 3, 5, 6, 7, 8, 9, 10\}$
  • $\text{RHS} = A' \cup B' = \{6, 7, 8, 9, 10\} \cup \{1, 3, 5, 7, 9, 10\} = \{1, 3, 5, 6, 7, 8, 9, 10\}$
  • $\text{LHS} = \text{RHS}$. Hence verified.

Algebraic Formulation:

Let the two sets have $m$ and $n$ elements. Then their total subsets are $2^m$ and $2^n$ respectively ($m > n$).

Factoring out $2^n$:

Express 56 as the product of a power of 2 and an odd integer:

Comparing both sides:

Step-by-Step Cardinality Breakdown:

• Total surveyed students $n(U) = 100$.
• Students who like Mathematics: $n(M) = 72$.
• Students who like Science: $n(S) = 65$.
• Students who like both: $n(M \cap S) = 58$.

1. (i) Only Mathematics: $$n(M \text{ only}) = n(M) - n(M \cap S) = 72 - 58 = 14$$
2. (ii) Only Science: $$n(S \text{ only}) = n(S) - n(M \cap S) = 65 - 58 = 7$$
3. (iii) Neither Mathematics nor Science: $$n(M \cup S) = n(M) + n(S) - n(M \cap S) = 72 + 65 - 58 = 79$$ $$n((M \cup S)') = n(U) - n(M \cup S) = 100 - 79 = 21$$

Three-Set Cardinality Formulas:

Let $C$, $F$, and $B$ represent the sets of students playing Cricket, Football, and Basketball:

• $n(C) = 30, n(F) = 25, n(B) = 20$
• $n(C \cap F) = 10, n(F \cap B) = 12, n(C \cap B) = 8$
• $n(C \cap F \cap B) = 5$
• Total class size $n(U) = 50$

1. (i) Students playing at least one game $n(C \cup F \cup B)$: $$n(C \cup F \cup B) = n(C) + n(F) + n(B) - n(C \cap F) - n(F \cap B) - n(C \cap B) + n(C \cap F \cap B)$$ $$= 30 + 25 + 20 - 10 - 12 - 8 + 5 = 75 - 30 + 5 = 50$$
2. (ii) Students playing exactly two games: $$\text{Exactly two} = [n(C \cap F) - 5] + [n(F \cap B) - 5] + [n(C \cap B) - 5]$$ $$= (10 - 5) + (12 - 5) + (8 - 5) = 5 + 7 + 3 = 15$$
3. (iii) Students playing none of the games: $$\text{None} = n(U) - n(C \cup F \cup B) = 50 - 50 = 0$$

Step-by-Step Case Study Solution:

• $n(U) = 150$
• $n(A) = 85$ (Programming)
• $n(B) = 70$ (Data Science)
• $n(A \cap B) = 35$ (Both)

1. Part (i): Enrolled in exactly one course: $$\text{Only Programming} = n(A) - n(A \cap B) = 85 - 35 = 50$$ $$\text{Only Data Science} = n(B) - n(A \cap B) = 70 - 35 = 35$$ $$\text{Exactly one course} = 50 + 35 = 85$$
2. Part (ii): Find $A' \cap B'$ and practical interpretation:

   By De Morgan's Law: $A' \cap B' = (A \cup B)'$.

   $$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 85 + 70 - 35 = 120$$ $$n(A' \cap B') = n(U) - n(A \cup B) = 150 - 120 = 30$$

   Contextual Interpretation: $A' \cap B'$ represents the 30 surveyed students who are not enrolled in either Programming or Data Science (i.e. enrolled in other platform subjects or exploratory non-tech tracks).

Three-Set Venn Diagram Region Calculation:

Let us partition the universal set of 1000 respondents into the 8 mutually disjoint Venn regions:

• Center (All 3 methods): $n(A \cap B \cap C) = 150$
• Only UPI and Wallets (not Cards): $n(A \cap B \cap C') = 350 - 150 = 200$
• Only UPI and Cards (not Wallets): $n(A \cap C \cap B') = 280 - 150 = 130$
• Only Wallets and Cards (not UPI): $n(B \cap C \cap A') = 240 - 150 = 90$
• Only UPI: $n(A) - [200 + 130 + 150] = 650 - 480 = 170$
• Only Wallets: $n(B) - [200 + 90 + 150] = 520 - 440 = 80$
• Only Cards: $n(C) - [130 + 90 + 150] = 480 - 370 = 110$
• None: $80$
• Verification of Total: $170 + 80 + 110 + 200 + 130 + 90 + 150 + 80 = 1000$ (Matches perfectly!)

1. Question i: UPI or Digital Wallets but NOT Credit/Debit Cards: $$\text{Formula} = (A \cup B) - C = \text{Only UPI} + \text{Only Wallets} + \text{Only (UPI \& Wallets)}$$ $$= 170 + 80 + 200 = 450$$
2. Question ii: Only Credit/Debit Cards: $$\text{Only Cards} = 110$$
3. Question iii: Do NOT use UPI but use at least one other (Wallets or Cards): $$\text{Formula} = \text{Only Wallets} + \text{Only Cards} + \text{Only (Wallets \& Cards)}$$ $$= 80 + 110 + 90 = 270$$
4. Question iv: Exactly two payment methods: $$\text{Formula} = \text{Only}(A \cap B) + \text{Only}(A \cap C) + \text{Only}(B \cap C)$$ $$= 200 + 130 + 90 = 420$$