Sets: Mixed Practice — Worked Solutions

Ques­tion 1: Ros­ter form and set-builder form#

The prob­lem#

Write $\{x : x \in \mathbf{Z},\ x^2 < 10\}$ in ros­ter form and $\{5, 10, 15, 20, 25\}$ in set-builder form.

Under­stand­ing the prob­lem#

Ros­ter form lists every ele­ment. Set-builder form describes the ele­ments by a rule. $\mathbf{Z}$ is the set of all inte­gers, pos­i­tive, neg­a­tive and zero.

The idea#

For the first set, test inte­gers until $x^2$ reaches $10$. For the sec­ond, spot the pat­tern: each ele­ment is $5$ times a nat­ural num­ber.

Step-by-step solu­tion#

Step 1. Squares of inte­gers: $0^2 = 0$, $(\pm1)^2 = 1$, $(\pm2)^2 = 4$, $(\pm3)^2 = 9$, $(\pm4)^2 = 16$. Only those up to $\pm3$ are less than $10$.

$$\{x : x \in \mathbf{Z},\ x^2 < 10\} = \{-3, -2, -1, 0, 1, 2, 3\}$$

Step 2. $5 = 5 \times 1$, $10 = 5 \times 2$, …, $25 = 5 \times 5$. So each ele­ment is $5n$ with $n = 1, 2, 3, 4, 5$.

$$\{5, 10, 15, 20, 25\} = \{x : x = 5n,\ n \in \mathbf{N},\ n \le 5\}$$

Check­ing the answer#

Put $n = 1$ to $5$ into $5n$: you get back exactly $5, 10, 15, 20, 25$ and noth­ing else.

Answer#

$\{-3, -2, -1, 0, 1, 2, 3\}$; $\{x : x = 5n,\ n \in \mathbf{N},\ n \le 5\}$.

Com­mon mis­take to avoid#

For­get­ting the neg­a­tive inte­gers and $0$ in the first set.

Ques­tion 2: Count­ing sub­sets#

The prob­lem#

How many sub­sets and how many proper sub­sets does $\{p, q, r, s\}$ have?

Under­stand­ing the prob­lem#

A sub­set can be any selec­tion of the ele­ments, includ­ing none (the empty set) and all of them. A proper sub­set is any sub­set except the set itself.

The idea#

A set with $n$ ele­ments has $2^n$ sub­sets, because each ele­ment is either cho­sen or not. Remov­ing the set itself leaves $2^n - 1$ proper sub­sets.

Step-by-step solu­tion#

Step 1. The set has $n = 4$ ele­ments.

$$\text{subsets} = 2^4 = 16$$

Step 2. Remove the set $\{p, q, r, s\}$ itself.

$$\text{proper subsets} = 16 - 1 = 15$$

Check­ing the answer#

Count by size: $1$ empty, $4$ sin­gle, $6$ pairs, $4$ triples, $1$ whole set; $1 + 4 + 6 + 4 + 1 = 16$.

Answer#

$16$ sub­sets; $15$ proper sub­sets.

Ques­tion 3: Is the set empty?#

The prob­lem#

Is $\{x \in \mathbf{R} : x^2 + 1 = 0\}$ empty? Is $\{x \in \mathbf{Z} : 2x = 7\}$?

Under­stand­ing the prob­lem#

A set is empty if no ele­ment sat­is­fies its rule. Note which num­ber sys­tem each set lives in: real num­bers $\mathbf{R}$ in the first, inte­gers $\mathbf{Z}$ in the sec­ond.

The idea#

Solve each equa­tion and see whether any solu­tion belongs to the stated num­ber sys­tem.

Step-by-step solu­tion#

Step 1. $x^2 + 1 = 0$ means $x^2 = -1$. The square of a real num­ber is never neg­a­tive, so no real $x$ works. The first set is empty.

Step 2. $2x = 7$ gives $x = 3.5$, which is not an inte­ger. So no inte­ger works, and the sec­ond set is empty.

Check­ing the answer#

If the sec­ond set were over $\mathbf{R}$, it would be $\{3.5\}$, not empty. The num­ber sys­tem decides.

Answer#

Yes, both sets are empty.

Ques­tion 4: Inter­val nota­tion#

The prob­lem#

Write $\{x \in \mathbf{R} : -1 < x \le 4\}$ as an inter­val and draw it on a num­ber line.

Under­stand­ing the prob­lem#

$-1$ is not included (strict $<$), but $4$ is included ($\le$).

The idea#

Use a round bracket for an excluded end­point and a square bracket for an included one.

Step-by-step solu­tion#

Step 1. Left end $-1$ excluded: "$($". Right end $4$ included: "$]$".

$$\{x \in \mathbf{R} : -1 < x \le 4\} = (-1, 4]$$

Step 2. To draw it: mark $-1$ and $4$ on a num­ber line. Draw an open (hol­low) cir­cle at $-1$ and a filled (solid) dot at $4$. Shade or draw a thick line join­ing them.

Check­ing the answer#

$0 \in (-1, 4]$ since $-1 < 0 \le 4$. $-1 \notin$ and $4 \in$, as the brack­ets say.

Answer#

$(-1, 4]$, drawn with a hol­low cir­cle at $-1$, a solid dot at $4$ and the seg­ment between shaded.

Ques­tion 5: Set oper­a­tions#

The prob­lem#

$U = \{1, 2, \ldots, 9\}$, $A = \{1, 2, 3, 4, 5\}$, $B = \{4, 5, 6, 7\}$. Find $A \cup B$, $A \cap B$, $A - B$, $A'$, $(A \cap B)'$, $A' \cup B'$.

Under­stand­ing the prob­lem#

• $A \cup B$: ele­ments in $A$ or $B$ or both.
• $A \cap B$: ele­ments in both.
• $A - B$: ele­ments in $A$ but not in $B$.
• $A'$: ele­ments of $U$ not in $A$.

The idea#

Work through the def­i­n­i­tions one at a time, always check­ing against $U$ for com­ple­ments.

Step-by-step solu­tion#

Step 1. Union: com­bine, writ­ing each ele­ment once.

$$A \cup B = \{1, 2, 3, 4, 5, 6, 7\}$$

Step 2. Inter­sec­tion: com­mon ele­ments.

$$A \cap B = \{4, 5\}$$

Step 3. Dif­fer­ence: remove $4$ and $5$ from $A$.

$$A - B = \{1, 2, 3\}$$

Step 4. Com­ple­ment of $A$ in $U$.

$$A' = \{6, 7, 8, 9\}$$

Step 5. Com­ple­ment of $A \cap B = \{4, 5\}$.

$$(A \cap B)' = \{1, 2, 3, 6, 7, 8, 9\}$$

Step 6. $B' = \{1, 2, 3, 8, 9\}$. Now take the union with $A'$.

$$A' \cup B' = \{6, 7, 8, 9\} \cup \{1, 2, 3, 8, 9\} = \{1, 2, 3, 6, 7, 8, 9\}$$

Check­ing the answer#

$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 5 + 4 - 2 = 7$. It matches.

Answer#

$A \cup B = \{1, \ldots, 7\}$; $A \cap B = \{4, 5\}$; $A - B = \{1, 2, 3\}$; $A' = \{6, 7, 8, 9\}$; $(A \cap B)' = A' \cup B' = \{1, 2, 3, 6, 7, 8, 9\}$.

Ques­tion 6: Ver­i­fy­ing De Mor­gan's law#

The prob­lem#

Ver­ify De Mor­gan's sec­ond law using ques­tion 5.

Under­stand­ing the prob­lem#

De Mor­gan's law here is $(A \cap B)' = A' \cup B'$. "Ver­ify" means com­pute both sides for these par­tic­u­lar sets and see that they agree.

The idea#

We already found both sides in Ques­tion 5.

Step-by-step solu­tion#

Step 1. Left side, from Step 5 of Ques­tion 5: $(A \cap B)' = \{1, 2, 3, 6, 7, 8, 9\}$.

Step 2. Right side, from Step 6 of Ques­tion 5: $A' \cup B' = \{1, 2, 3, 6, 7, 8, 9\}$.

Step 3. The two sets have exactly the same ele­ments, so $(A \cap B)' = A' \cup B'$ for these sets.

Check­ing the answer#

Both sides leave out exactly $4$ and $5$, the com­mon ele­ments of $A$ and $B$.

Answer#

Both sides equal $\{1, 2, 3, 6, 7, 8, 9\}$, so the law holds.

Ques­tion 7: Proof that $A \cap (A \cup B) = A$#

The prob­lem#

Show that $A \cap (A \cup B) = A$ for all sets $A, B$.

Under­stand­ing the prob­lem#

This must be proved for all sets, so exam­ples are not enough. To prove two sets are equal, show each is a sub­set of the other.

The idea#

Show (i) $A \cap (A \cup B) \subset A$ and (ii) $A \subset A \cap (A \cup B)$.

Step-by-step solu­tion#

Step 1. Part (i). Take any $x \in A \cap (A \cup B)$. Being in an inter­sec­tion means being in both sets, so in par­tic­u­lar $x \in A$. Hence $A \cap (A \cup B) \subset A$.

Step 2. Part (ii). Take any $x \in A$. Then $x \in A \cup B$ as well, because a union con­tains every­thing in $A$. So $x$ is in both $A$ and $A \cup B$, that is $x \in A \cap (A \cup B)$. Hence $A \subset A \cap (A \cup B)$.

Step 3. Each set is a sub­set of the other, so they are equal.

Check­ing the answer#

With $A = \{1, 2\}$, $B = \{3\}$: $A \cup B = \{1, 2, 3\}$ and $A \cap \{1, 2, 3\} = \{1, 2\} = A$.

Answer#

$A \cap (A \cup B) \subset A$ and $A \subset A \cap (A \cup B)$, so $A \cap (A \cup B) = A$.

Ques­tion 8: When $A \subset B$#

The prob­lem#

If $A \subset B$, what are $A \cup B$ and $A \cap B$? Prove your answer.

Under­stand­ing the prob­lem#

$A \subset B$ means every ele­ment of $A$ is already an ele­ment of $B$.

The idea#

Guess from a small exam­ple, then prove each claim by show­ing sub­sets both ways.

Step-by-step solu­tion#

Step 1. Guess. With $A = \{1\}$, $B = \{1, 2\}$: $A \cup B = \{1, 2\} = B$ and $A \cap B = \{1\} = A$.

Step 2. Prove $A \cup B = B$.
If $x \in A \cup B$, then $x \in A$ or $x \in B$; if $x \in A$ then $x \in B$ because $A \subset B$. Either way $x \in B$, so $A \cup B \subset B$.
Also $B \subset A \cup B$ always. Hence $A \cup B = B$.

Step 3. Prove $A \cap B = A$.
If $x \in A \cap B$ then $x \in A$, so $A \cap B \subset A$.
If $x \in A$ then $x \in B$ (as $A \subset B$), so $x \in A \cap B$; thus $A \subset A \cap B$. Hence $A \cap B = A$.

Check­ing the answer#

In a Venn dia­gram, $A$ sits inside $B$: the union is the big cir­cle $B$ and the over­lap is the small cir­cle $A$.

Answer#

$A \cup B = B$ and $A \cap B = A$.

Ques­tion 9: Count­ing out­side both sets#

The prob­lem#

$n(U) = 70$, $n(A) = 40$, $n(B) = 25$, $n(A \cap B) = 12$. Find $n(A' \cap B')$.

Under­stand­ing the prob­lem#

$A' \cap B'$ is the set of ele­ments in nei­ther $A$ nor $B$: the region out­side both cir­cles.

The idea#

By De Mor­gan's law $A' \cap B' = (A \cup B)'$. So first find $n(A \cup B)$, then sub­tract from $n(U)$.

Step-by-step solu­tion#

Step 1. Use $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.

$$n(A \cup B) = 40 + 25 - 12 = 53$$

Step 2. Sub­tract from the uni­ver­sal set.

$$n(A' \cap B') = n(U) - n(A \cup B) = 70 - 53 = 17$$

Check­ing the answer#

Venn regions: only $A$: $40 - 12 = 28$; both: $12$; only $B$: $25 - 12 = 13$; out­side: $17$. Total $28 + 12 + 13 + 17 = 70$.

Answer#

$n(A' \cap B') = 17$.

Ques­tion 10: Tel­ugu and Hindi speak­ers#

The prob­lem#

In a group of $90$ peo­ple, $55$ speak Tel­ugu and $48$ speak Hindi; every­one speaks at least one. How many speak both? Only Hindi?

Under­stand­ing the prob­lem#

Let $T$ = Tel­ugu speak­ers, $H$ = Hindi speak­ers. "Every­one speaks at least one" means $n(T \cup H) = 90$. "Only Hindi" means Hindi but not Tel­ugu.

The idea#

Use $n(T \cup H) = n(T) + n(H) - n(T \cap H)$ and solve for the over­lap.

Step-by-step solu­tion#

Step 1. Sub­sti­tute the num­bers.

$$90 = 55 + 48 - n(T \cap H)$$

Step 2. Solve.

$$n(T \cap H) = 103 - 90 = 13$$

Step 3. Only Hindi = Hindi speak­ers minus those who speak both.

$$n(H - T) = 48 - 13 = 35$$

Check­ing the answer#

Only Tel­ugu $= 55 - 13 = 42$. Then $42 + 13 + 35 = 90$. Cor­rect.

Answer#

$13$ speak both; $35$ speak only Hindi.

Ques­tion 11: Three sub­jects#

The prob­lem#

A col­lege has $300$ stu­dents: $150$ study math­e­mat­ics, $120$ physics, $100$ chem­istry; $50$ maths and physics, $40$ physics and chem­istry, $45$ maths and chem­istry, $20$ all three. How many study none of the three? Exactly one?

Under­stand­ing the prob­lem#

Let $M$, $P$, $C$ be the three sets. The pair num­bers ($50$, $40$, $45$) include the $20$ who take all three. "Exactly one" means one sub­ject and no other.

The idea#

Use the three-set for­mula for $n(M \cup P \cup C)$, then fill the Venn dia­gram from the cen­tre out­wards to find the "only" regions.

Step-by-step solu­tion#

Step 1. At least one sub­ject:

$$\begin{aligned} n(M \cup P \cup C) &= 150 + 120 + 100 - 50 - 40 - 45 + 20\\ &= 370 - 135 + 20 = 255 \end{aligned}$$

Step 2. None:

$$300 - 255 = 45$$

Step 3. Fill the cen­tre and the "exactly two" regions.

$$\begin{aligned} \text{all three} &= 20\\ M \text{ and } P \text{ only} &= 50 - 20 = 30\\ P \text{ and } C \text{ only} &= 40 - 20 = 20\\ M \text{ and } C \text{ only} &= 45 - 20 = 25 \end{aligned}$$

Step 4. The "only one sub­ject" regions: sub­tract from each total the parts of its cir­cle already filled.

$$\begin{aligned} \text{maths only} &= 150 - 30 - 25 - 20 = 75\\ \text{physics only} &= 120 - 30 - 20 - 20 = 50\\ \text{chemistry only} &= 100 - 20 - 25 - 20 = 35 \end{aligned}$$

Step 5. Add them.

$$75 + 50 + 35 = 160$$

Check­ing the answer#

All seven regions: $75 + 50 + 35 + 30 + 20 + 25 + 20 = 255$, which matches Step 1.

Answer#

$45$ study none; $160$ study exactly one.

Com­mon mis­take to avoid#

Writ­ing "maths only $= 150 - 50 - 45 = 55$". The $20$ who take all three were sub­tracted twice, so add them back once: $55 + 20 = 75$.

Ques­tion 12: An exam­ple where can­celling fails#

The prob­lem#

Give an exam­ple of sets $A$, $B$, $C$ with $A \cap B = A \cap C$ but $B \ne C$.

Under­stand­ing the prob­lem#

We need to show that you can­not "can­cel $A \cap$" from both sides. $B$ and $C$ must dif­fer, but only out­side $A$.

The idea#

Let $B$ and $C$ share the part inside $A$ and dif­fer in ele­ments that are not in $A$.

Step-by-step solu­tion#

Step 1. Choose $A = \{1\}$, $B = \{1, 2\}$, $C = \{1, 3\}$.

Step 2. Com­pute both inter­sec­tions.

$$A \cap B = \{1\}, \qquad A \cap C = \{1\}$$

Step 3. They are equal, but $B = \{1, 2\} \ne \{1, 3\} = C$.

Check­ing the answer#

$2 \in B$ but $2 \notin C$, so $B \ne C$ for sure.

Answer#

$A = \{1\}$, $B = \{1, 2\}$, $C = \{1, 3\}$ (any sim­i­lar exam­ple is cor­rect).