INTRODUCTION TO MATHEMATICAL METHODS QUIZ ONE JULY INTAKE 2022

DATE : 12th AUGUST, 2022

Question 1

Write the following sets in set-builder form:
$C = \{0, 20, 120, 620\}$
$r = \left( \frac{3}{5}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5} \right)$

Answer for $C$:
$C = \left\{ x \mid x = 5^n - 5,\ n \in \mathbb{N},\ 1 \leq n \leq 4 \right\}$

Explanation:

• The set $C = \{0, 20, 120, 620\}$ follows the pattern $x = 5^n - 5$ for $n = 1, 2, 3, 4$:
  • $n = 1$: $5^1 - 5 = 5 - 5 = 0$
  • $n = 2$: $5^2 - 5 = 25 - 5 = 20$
  • $n = 3$: $5^3 - 5 = 125 - 5 = 120$
  • $n = 4$: $5^4 - 5 = 625 - 5 = 620$
• Thus, $C$ is defined as the set of all $x$ such that $x = 5^n - 5$ for natural numbers $n$ from 1 to 4.

Answer for $r$:
$r = \left\{ x \mid x = \frac{k}{5},\ k \in \{1, 2, 3\} \right\}$

Explanation:

• The given "set" $r = \left( \frac{3}{5}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5} \right)$ has duplicate elements ($\frac{3}{5}$ appears twice). As sets contain unique elements, we consider $r = \left\{ \frac{1}{5}, \frac{2}{5}, \frac{3}{5} \right\}$.
• The elements are fractions with denominator 5 and numerators 1, 2, 3.
• Thus, $r$ is defined as the set of all $x$ such that $x = \frac{k}{5}$ for $k \in \{1, 2, 3\}$.

Question 2

Rewrite the following sets in roster form:
$A = \{a_j \mid a_j \in \mathbb{N},\ a_{j+1} = 3a_j \text{ and } a_1 = 2\}$
$B = \{x \mid x \text{ is positive}, x \text{ is negative}\}$

Answer for $A$:
$A = \{2, 6, 18, 54, \dots\}$

Explanation:

• The sequence is defined by $a_1 = 2$ and $a_{j+1} = 3a_j$:
  • $a_1 = 2$
  • $a_2 = 3 \times a_1 = 3 \times 2 = 6$
  • $a_3 = 3 \times a_2 = 3 \times 6 = 18$
  • $a_4 = 3 \times a_3 = 3 \times 18 = 54$
  • This continues infinitely: $a_5 = 3 \times 54 = 162$, etc.
• In roster form, we list the first few elements followed by ellipsis to indicate an infinite set.

Answer for $B$:
$B = \emptyset \quad \text{(or)} \quad B = \{\}$

Explanation:

• The set $B$ contains elements $x$ that are both positive and negative. No real number satisfies this condition.
• Thus, $B$ is the empty set.

Question 3

For any two sets $A$ and $B$, prove that $P(A) \cup P(B) \subset P(A \cup B)$, but $P(A \cup B)$ is not necessarily a subset of $P(A) \cup P(B)$.

Answer:
Part 1: Prove $P(A) \cup P(B) \subseteq P(A \cup B)$

• Let $X$ be an arbitrary element of $P(A) \cup P(B)$. Then $X \in P(A)$ or $X \in P(B)$.
• Case 1: $X \in P(A) \implies X \subseteq A$. Since $A \subseteq A \cup B$,
• we have $X\subseteq A\cup B$
• $$$X \subseteq A \cup B \implies X \in P(A \cup B)$.
• 
  
• Case 2: $X \in P(B) \implies X \subseteq B$. Since $B \subseteq A \cup B$,
• we have $X \subseteq A \cup B \implies X \in P(A \cup B)$.
• Thus, $X \in P(A) \cup P(B) \implies X \in P(A \cup B)$, so $P(A) \cup P(B) \subseteq P(A \cup B)$.

Part 2: Show $P(A \cup B) \not\subseteq P(A) \cup P(B)$ in general

• Counterexample: Let $A = \{1\}$, $B = \{2\}$. Then:
  • $A \cup B = \{1, 2\}$
  • $P(A) = \{ \emptyset, \{1\} \}$, $P(B) = \{ \emptyset, \{2\} \}$
  • $P(A) \cup P(B) = \{ \emptyset, \{1\}, \{2\} \}$
  • $P(A \cup B) = \{ \emptyset, \{1\}, \{2\}, \{1, 2\} \}$
• Now, $\{1, 2\} \in P(A \cup B)$ but $\{1, 2\} \notin P(A) \cup P(B)$.
• Hence, $P(A \cup B) \not\subseteq P(A) \cup P(B)$.

Question 4

Out of 20 members in a family, 12 like to take tea and 15 like coffee. Assume that every member likes at least one of the two drinks.

• (a) How many like only tea and not coffee?
• (b) How many like only coffee and not tea?
• (c) How many like both coffee and tea?

Answer:
Define:

• $T$: Set of members who like tea, $|T| = 12$
• $C$: Set of members who like coffee, $|C| = 15$
• Total members $= |T \cup C| = 20$ (since all like at least one drink).

Using the principle of inclusion-exclusion:
$|T \cup C| = |T| + |C| - |T \cap C|$
$20 = 12 + 15 - |T \cap C|$
$20 = 27 - |T \cap C|$
$|T \cap C| = 27 - 20 = 7$

• (a) Only tea (not coffee): $|T - C| = |T| - |T \cap C| = 12 - 7 = 5$
• (b) Only coffee (not tea): $|C - T| = |C| - |T \cap C| = 15 - 7 = 8$
• (c) Both tea and coffee: $|T \cap C| = 7$

Verification: $5 \ (\text{only tea}) + 8 \ (\text{only coffee}) + 7 \ (\text{both}) = 20$, which matches the total.

Question 5

Let $A = \{1, 2, 3, 4\}$, $B = \{1, 2, 3\}$, and $C = \{2, 4\}$. Find all sets $X$ such that $X \subseteq B$ and $X \subseteq C$.

Answer:
$X = \emptyset \quad \text{or} \quad X = \{2\}$

Explanation:

• The condition $X \subseteq B$ and $X \subseteq C$ means $X \subseteq B \cap C$.
• Compute $B \cap C$:
  • $B = \{1, 2, 3\}$, $C = \{2, 4\}$
  • $B \cap C = \{2\}$ (only common element).
• Subsets of $\{2\}$ are $\emptyset$ and $\{2\}$.
• Thus, the sets $X$ are the empty set and $\{2\}$.

Question 6

If $A = \{4, 5, 8, 12\}$, $B = \{1, 4, 6, 9\}$, $C = \{1, 2, 3, 4\}$, find:

• (I) $A - (B - A)$
• (II) $A - (C - B)$

Answer for (I):
$A - (B - A) = \{4, 5, 8, 12\}$

Explanation:

• Compute $B - A$:
  • $B = \{1, 4, 6, 9\}$, $A = \{4, 5, 8, 12\}$
  • Elements in $B$ not in $A$: $1, 6, 9$ (since $4$ is in $A$)
  • Thus, $B - A = \{1, 6, 9\}$.
• Compute $A - (B - A)$:
  • $A = \{4, 5, 8, 12\}$, $B - A = \{1, 6, 9\}$
  • Elements in $A$ not in $\{1, 6, 9\}$: all elements of $A$ (no overlap)
  • Thus, $A - (B - A) = \{4, 5, 8, 12\}$.
• Identity Insight: $A - (B - A) = A \cap (B \cap A^c)^c = A \cap (B^c \cup A) = A$, since $B - A$ is disjoint from $A$.

Answer for (II):
$A - (C - B) = \{4, 5, 8, 12\}$

Explanation:

• Compute $C - B$:
  • $C = \{1, 2, 3, 4\}$, $B = \{1, 4, 6, 9\}$
  • Elements in $C$ not in $B$: $2, 3$ (since $1, 4$ are in $B$)
  • Thus, $C - B = \{2, 3\}$.
• Compute $A - (C - B)$:
  • $A = \{4, 5, 8, 12\}$, $C - B = \{2, 3\}$
  • Elements in $A$ not in $\{2, 3\}$: all elements of $A$ (no overlap)
  • Thus, $A - (C - B) = \{4, 5, 8, 12\}$.
• Identity Insight: $C - B$ is disjoint from $A$, so $A - (C - B) = A$.

@Dr. Microbiota

End of Solutions