3. Classroom tasks and solutions / Аудиторні задачі та розв’язання #

Source of tasks. The eleven problems below are taken exactly from the practical-class assignment for this topic (set operations, Venn diagrams, and set algebra); only the wording has been translated to English. Each task is followed by a complete worked solution so the whole class can check every step. The theory used is in 2method.md.

Throughout, $A$, $B$, $C$ are sets inside a universe $U$; recall the difference law $A \setminus B = A \cap \overline{B}$.

A. Illustrating operations with diagrams (Tasks 1–3) #

Task 1 #

Illustrate the set $E = A \setminus (B \cup C)$ with Euler/Venn circles.

Solution. Draw three overlapping circles $A$, $B$, $C$ inside the universe rectangle. Shade inside-out:

1. First mark $B \cup C$ — everything lying inside circle $B$ or circle $C$.
2. $E = A \setminus (B \cup C)$ keeps the part of $A$ that is outside that marked area.

So the answer is the “petal” of circle $A$ that overlaps neither $B$ nor $C$ — the portion of $A$ separate from the other two circles. Everything of $A$ that it shares with $B$ or with $C$ is excluded.

Task 2 #

Using Venn diagrams, verify the identity

$$A \cap (B \cup C) = (A \cap B) \cup (A \cap C).$$

Solution. Shade each side on its own diagram and compare.

• Left side $A \cap (B \cup C)$. First shade $B \cup C$ (both circles $B$ and $C$); then intersect with $A$, keeping only the part of that region lying inside $A$. The result is the portion of $A$ that overlaps $B$ or $C$.
• Right side $(A \cap B) \cup (A \cap C)$. Shade the lens $A \cap B$ and the lens $A \cap C$, then take their union. The result is again the portion of $A$ overlapping $B$, together with the portion of $A$ overlapping $C$.

The two shaded regions coincide exactly, so the identity holds. (This is the distributive law of $\cap$ over $\cup$.)

Task 3 #

Write the formula for the shaded region of the given Venn diagram, where the shading covers the parts of circles $A$ and $B$ that lie outside circle $C$.

Solution. Read the region inside-out:

1. The shaded area covers everything in $A$ or $B$ — that is $A \cup B$.
2. …except whatever lies inside $C$, which is removed.

Removing $C$ from $A \cup B$ gives

$$E = (A \cup B) \setminus C.$$

B. Proving identities (Tasks 4–5) #

Task 4 #

Prove that $A \setminus (A \setminus B) = A \cap B.$

Solution (algebraic).

$$\begin{aligned} A \setminus (A \setminus B) &= A \cap \overline{(A \setminus B)} && \text{(difference law)} \\ &= A \cap \overline{(A \cap \overline{B})} && \text{(}A \setminus B = A \cap \overline{B}\text{)} \\ &= A \cap (\overline{A} \cup \overline{\overline{B}}) && \text{(De Morgan)} \\ &= A \cap (\overline{A} \cup B) && \text{(double complement)} \\ &= (A \cap \overline{A}) \cup (A \cap B) && \text{(distributivity)} \\ &= \varnothing \cup (A \cap B) && \text{(complement law)} \\ &= A \cap B. && \text{(identity law)} \end{aligned}$$

Hence $A \setminus (A \setminus B) = A \cap B$. $\blacksquare$

Task 5 #

Prove that $A = (A \cap B) \cup (A \setminus B),$ where $A$ and $B$ are sets.

Solution. Start from the right-hand side and use $A \setminus B = A \cap \overline{B}$:

$$\begin{aligned} (A \cap B) \cup (A \setminus B) &= (A \cap B) \cup (A \cap \overline{B}) && \text{(difference law)} \\ &= A \cap (B \cup \overline{B}) && \text{(distributivity)} \\ &= A \cap U && \text{(complement law, } B \cup \overline{B} = U\text{)} \\ &= A. && \text{(identity law)} \end{aligned}$$

Thus $(A \cap B) \cup (A \setminus B) = A$. Intuitively, every element of $A$ either is in $B$ (so it lands in $A \cap B$) or is not (so it lands in $A \setminus B$); the two pieces together rebuild $A$. $\blacksquare$

C. Computing sets (Tasks 6–9) #

For Tasks 6–9 let

$$A = \{2, 4, 6, 8, 10, 12\}, \qquad B = \{3, 6, 9, 12\}.$$

Task 6 #

Write the set $E = A \cup B$.

Solution. The union collects every element that is in $A$ or in $B$ (each listed once):

$$E = A \cup B = \{2, 3, 4, 6, 8, 9, 10, 12\}.$$

Task 7 #

Write the set $E = A \cap B$.

Solution. The intersection keeps only the elements in both sets. Comparing the lists, $6$ and $12$ occur in each:

$$E = A \cap B = \{6, 12\}.$$

Task 8 #

Write the set $E = A \setminus B$.

Solution. The difference keeps the elements of $A$ that are not in $B$. Removing the shared elements $6$ and $12$ from $A$:

$$E = A \setminus B = \{2, 4, 8, 10\}.$$

Task 9 #

Write the set $E = \overline{A \setminus B}$, the complement of $A \setminus B$ taken relative to the universe $U = A \cup B$.

Solution. From Task 8, $A \setminus B = \{2, 4, 8, 10\}$, and from Task 6 the universe is $U = A \cup B = \{2, 3, 4, 6, 8, 9, 10, 12\}$. The complement removes $A \setminus B$ from $U$:

$$E = \overline{A \setminus B} = U \setminus (A \setminus B) = \{3, 6, 9, 12\}.$$

(Notice this is exactly $B$ — consistent with Task 5, since $U = A \cup B$ and $U \setminus (A \setminus B) = B$ here.)

D. Set-builder and power set (Tasks 10–11) #

Task 10 #

List the elements of the following sets (with $\mathbb{N} = \{1, 2, 3, \dots\}$):

$$D = \{\, x \mid x \in \mathbb{N} \ \text{and}\ x \ \text{is a divisor of } 15 \,\}, \qquad E = \{\, x \mid x \in \mathbb{N} \ \text{and}\ 15 < x < 30 \ \text{and}\ x \ \text{is prime} \,\}.$$

Solution. Apply each defining property to the candidates.

• $D$: the natural-number divisors of $15$ are $1, 3, 5, 15$ (since $15 = 1\cdot 15 = 3 \cdot 5$). Hence $D = \{1, 3, 5, 15\}$.
• $E$: test each integer strictly between $15$ and $30$ for primality. The composites are ruled out ($16,18,\dots$ even; $21 = 3\cdot 7$; $25 = 5^2$; $27 = 3^3$), leaving the primes $17, 19, 23, 29$. Hence $E = \{17, 19, 23, 29\}$.

Task 11 #

List the elements of the power set $\mathcal{P}(G)$, where $G = \{1, 2, 3\}$ ($\mathcal{P}(G)$ is the set of all subsets of $G$).

Solution. Since $|G| = 3$, there are $2^{3} = 8$ subsets. Enumerate them by size so none is missed:

• size $0$: $\varnothing$;
• size $1$: $\{1\}, \{2\}, \{3\}$;
• size $2$: $\{1,2\}, \{1,3\}, \{2,3\}$;
• size $3$: $\{1,2,3\}$.

Therefore

$$\mathcal{P}(G) = \big\{\, \varnothing,\ \{1\},\ \{2\},\ \{3\},\ \{1,2\},\ \{1,3\},\ \{2,3\},\ \{1,2,3\} \,\big\}.$$

E. Counting and products (Tasks 12–13) #

Task 12 #

In a group of $30$ students, $18$ study French and $15$ study German, and $7$ study both. How many study (a) at least one of the two languages, (b) neither?

Solution. By inclusion–exclusion, $|F\cup G| = |F| + |G| - |F\cap G| = 18 + 15 - 7 = 26$. So (a) $26$ study at least one language, and (b) $30 - 26 = 4$ study neither. The $-|F\cap G|$ corrects the double-count of the $7$ students who are inside both $|F|$ and $|G|$.

Task 13 #

For $A = \{1, 2\}$ and $B = \{x, y, z\}$: (a) list $A \times B$ and give $|A\times B|$; (b) compute the symmetric difference $\{1,2,3\}\triangle\{2,3,4\}$.

Solution. (a) The Cartesian product pairs each element of $A$ with each of $B$:

$$A\times B = \{(1,x),(1,y),(1,z),(2,x),(2,y),(2,z)\},\qquad |A\times B| = |A|\cdot|B| = 2\cdot 3 = 6.$$

(b) The symmetric difference keeps the elements lying in exactly one set: $\{1,2,3\}\triangle\{2,3,4\} = \{1,4\}$ (the shared $2,3$ drop out).

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