Operations on Sets and Venn Diagrams — Worked Solutions

Ques­tion 1: Union, inter­sec­tion and dif­fer­ences#

The prob­lem#

$A = \{2, 4, 6, 8, 10\}$, $B = \{5, 10, 15\}$. Find $A \cup B$, $A \cap B$, $A - B$, $B - A$.

Under­stand­ing the prob­lem#

You are given two sets by list­ing their ele­ments and asked for four new sets. Recall:

• $A \cup B$: ele­ments in $A$ or $B$ (or both);
• $A \cap B$: ele­ments in both;
• $A - B$: ele­ments in $A$ but not in $B$;
• $B - A$: ele­ments in $B$ but not in $A$.

The idea#

Go through the ele­ments one by one and decide where each belongs. Write each com­mon ele­ment only once.

Step-by-step solu­tion#

Step 1. Com­mon ele­ments: only $10$ is in both sets.

$$A \cap B = \{10\}$$

Step 2. Union: all ele­ments of $A$, then add the ele­ments of $B$ not already listed ($5$ and $15$).

$$A \cup B = \{2, 4, 5, 6, 8, 10, 15\}$$

Step 3. $A - B$: remove $10$ from $A$.

$$A - B = \{2, 4, 6, 8\}$$

Step 4. $B - A$: remove $10$ from $B$.

$$B - A = \{5, 15\}$$

Check­ing the answer#

$A - B$, $A \cap B$ and $B - A$ are the three sep­a­rate pieces of $A \cup B$. Their sizes are $4 + 1 + 2 = 7$, and $A \cup B$ has $7$ ele­ments. ✓

Answer#

$A \cup B = \{2, 4, 5, 6, 8, 10, 15\}$, $A \cap B = \{10\}$, $A - B = \{2, 4, 6, 8\}$, $B - A = \{5, 15\}$.

Ques­tion 2: Com­ple­ments in a uni­ver­sal set#

The prob­lem#

$U = \{x \in \mathbf{N} : x \le 10\}$, $A = \{x : x \text{ is prime}\}$, $B = \{x : x \text{ is odd}\}$. Find $A'$, $B'$, $(A \cup B)'$, $A' \cap B'$.

Under­stand­ing the prob­lem#

$U = \{1, 2, \ldots, 10\}$. $A$ and $B$ are the primes and the odd num­bers inside $U$. The com­ple­ment $A'$ means $U - A$, every­thing in $U$ that is not in $A$.

The idea#

First write $A$ and $B$ in ros­ter form. Then take com­ple­ments within $U$. The last two answers should agree, by De Mor­gan's law.

Step-by-step solu­tion#

Step 1. List the sets. Remem­ber $1$ is not prime.

$$U = \{1, 2, \ldots, 10\}, \quad A = \{2, 3, 5, 7\}, \quad B = \{1, 3, 5, 7, 9\}$$

Step 2. Com­ple­ments.

$$A' = \{1, 4, 6, 8, 9, 10\}, \qquad B' = \{2, 4, 6, 8, 10\}$$

Step 3. Union, then its com­ple­ment.

$$A \cup B = \{1, 2, 3, 5, 7, 9\}, \qquad (A \cup B)' = \{4, 6, 8, 10\}$$

Step 4. Inter­sec­tion of the com­ple­ments: ele­ments in both $A'$ and $B'$.

$$A' \cap B' = \{4, 6, 8, 10\}$$

Check­ing the answer#

$(A \cup B)' = A' \cap B'$, exactly as De Mor­gan's law says. Also $n(A) + n(A') = 4 + 6 = 10 = n(U)$. ✓

Answer#

$A' = \{1, 4, 6, 8, 9, 10\}$, $B' = \{2, 4, 6, 8, 10\}$, $(A \cup B)' = \{4, 6, 8, 10\}$, $A' \cap B' = \{4, 6, 8, 10\}$.

Com­mon mis­take to avoid#

Includ­ing $1$ in the primes. $1$ has only one fac­tor, so it is not prime, and it belongs to $A'$.

Ques­tion 3: Are the sets dis­joint?#

The prob­lem#

Are $\{x : x \text{ is a multiple of } 4\}$ and $\{x : x \text{ is odd}\}$ dis­joint?

Under­stand­ing the prob­lem#

Two sets are dis­joint when they have no ele­ment in com­mon, that is, their inter­sec­tion is $\varnothing$.

The idea#

Ask whether any num­ber can be both a mul­ti­ple of $4$ and odd.

Step-by-step solu­tion#

Step 1. A mul­ti­ple of $4$ has the form $4k = 2(2k)$, so it is a mul­ti­ple of $2$, that is, even.

Step 2. An odd num­ber is not a mul­ti­ple of $2$.

Step 3. So no num­ber is both. The inter­sec­tion is empty.

$$\{\text{multiples of } 4\} \cap \{\text{odd numbers}\} = \varnothing$$

Check­ing the answer#

Mul­ti­ples of $4$: $4, 8, 12, 16, \ldots$ are all even; none is odd. ✓

Answer#

Yes, they are dis­joint: every mul­ti­ple of $4$ is even, so the inter­sec­tion is empty.

Ques­tion 4: Shad­ing Venn dia­grams#

The prob­lem#

Shade in Venn dia­grams: (a) $A' \cap B$; (b) $(A \cap B)'$; (c) $A - (B \cup C)$.

Under­stand­ing the prob­lem#

Draw the uni­ver­sal set $U$ as a rec­tan­gle and each set as a cir­cle inside it, over­lap­ping. You must shade exactly the region that the expres­sion describes.

The idea#

Trans­late each expres­sion into words ("in … and not in …"), then shade region by region. For two sets there are four regions: only $A$, both, only $B$, out­side both.

Step-by-step solu­tion#

Part (a): $A' \cap B$

Step 1. $A'$ is every­thing out­side cir­cle $A$. Inter­sect­ing with $B$ keeps only the part of that inside $B$.

Step 2. In words: in $B$ but not in $A$. This is the same as $B - A$.

Step 3. Draw two over­lap­ping cir­cles $A$ and $B$. Shade the cres­cent of $B$ that lies out­side $A$. Leave the over­lap unshaded.

Part (b): $(A \cap B)'$

Step 1. $A \cap B$ is the over­lap (the lens shape) of the two cir­cles.

Step 2. Its com­ple­ment is every­thing in $U$ except that lens.

Step 3. Shade the whole rec­tan­gle, includ­ing "only $A$", "only $B$" and the region out­side both cir­cles, but leave the over­lap unshaded.

Part (c): $A - (B \cup C)$

Step 1. Draw three mutu­ally over­lap­ping cir­cles $A$, $B$, $C$.

Step 2. $B \cup C$ is every­thing inside $B$ or $C$.

Step 3. $A - (B \cup C)$ keeps the points of $A$ that are in nei­ther $B$ nor $C$.

Step 4. Shade only the part of cir­cle $A$ that does not over­lap $B$ or $C$ — the "$A$ only" region.

Check­ing the answer#

For (b), De Mor­gan says $(A \cap B)' = A' \cup B'$: every­thing out­side $A$ together with every­thing out­side $B$, which is every­thing except the over­lap. That matches the shad­ing. ✓

Answer#

(a) Shade the part of $B$ out­side $A$; (b) shade all of $U$ except the over­lap of $A$ and $B$; (c) shade only the part of $A$ that lies out­side both $B$ and $C$.

Ques­tion 5: Ver­i­fy­ing the dis­trib­u­tive law#

The prob­lem#

Ver­ify $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ for $A = \{1, 2, 3, 4\}$, $B = \{3, 4, 5\}$, $C = \{1, 5, 6\}$.

Under­stand­ing the prob­lem#

"Ver­ify" means work out both sides for these par­tic­u­lar sets and show that they are equal.

The idea#

Com­pute the left side (brack­ets first) and the right side sep­a­rately, then com­pare.

Step-by-step solu­tion#

Step 1. Left side: first $B \cup C$, then inter­sect with $A$.

$$\begin{aligned} B \cup C &= \{1, 3, 4, 5, 6\} \\ A \cap (B \cup C) &= \{1, 3, 4\} \end{aligned}$$

Step 2. Right side: the two inter­sec­tions, then their union.

$$\begin{aligned} A \cap B &= \{3, 4\} \\ A \cap C &= \{1\} \\ (A \cap B) \cup (A \cap C) &= \{1, 3, 4\} \end{aligned}$$

Step 3. Both sides equal $\{1, 3, 4\}$.

Check­ing the answer#

Test $2$: it is in $A$ but not in $B$ or $C$, so it is on nei­ther side. ✓

Answer#

Both sides equal $\{1, 3, 4\}$, so the law holds for these sets.

Ques­tion 6: Prov­ing $A - B = A \cap B'$#

The prob­lem#

Prove: $A - B = A \cap B'$.

Under­stand­ing the prob­lem#

To show two sets are equal you must show they have exactly the same ele­ments. The clean­est way is to show that an ele­ment belongs to one side exactly when it belongs to the other.

The idea#

Use the def­i­n­i­tions of dif­fer­ence and com­ple­ment and a chain of "if and only if" ($\iff$) state­ments, just as the les­son proves De Mor­gan's law.

Step-by-step solu­tion#

Step 1. Take any ele­ment $x$ and start from the left side.

$$x \in A - B \iff x \in A \text{ and } x \notin B \qquad \text{(definition of difference)}$$

Step 2. "$x \notin B$" means $x$ is in the com­ple­ment of $B$.

$$x \notin B \iff x \in B' \qquad \text{(definition of complement)}$$

Step 3. So

$$x \in A - B \iff x \in A \text{ and } x \in B' \iff x \in A \cap B' \qquad \text{(definition of intersection)}$$

Step 4. Every ele­ment of either side is an ele­ment of the other, so the sets are equal.

Check­ing the answer#

Using Ques­tion 1's sets inside $U = \{1, \ldots, 15\}$: $A - B = \{2, 4, 6, 8\}$ and $A \cap B'$ keeps the ele­ments of $A$ that are not $5, 10, 15$, also $\{2, 4, 6, 8\}$. ✓

Answer#

$x \in A - B \iff x \in A,\ x \notin B \iff x \in A \cap B'$; hence $A - B = A \cap B'$.

Ques­tion 7: When union equals inter­sec­tion#

The prob­lem#

Show that $A \cup B = A \cap B$ implies $A = B$.

Under­stand­ing the prob­lem#

You are told the union and inter­sec­tion are the same set. You must prove that then $A$ and $B$ are equal. To prove two sets equal, show each is a sub­set of the other.

The idea#

Use the facts $A \subset A \cup B$ and $A \cap B \subset B$ (these fol­low straight from the def­i­n­i­tions). Chain them through the given equal­ity.

Step-by-step solu­tion#

Step 1. Show $A \subset B$. Take any $x \in A$.

$$x \in A \Rightarrow x \in A \cup B \qquad \text{(everything in } A \text{ is in the union)}$$

Step 2. Use the given equal­ity $A \cup B = A \cap B$.

$$x \in A \cup B = A \cap B \Rightarrow x \in B \qquad \text{(everything in the intersection is in } B)$$

So every ele­ment of $A$ is in $B$: $A \subset B$.

Step 3. Show $B \subset A$ the same way. Take $x \in B$; then $x \in A \cup B = A \cap B$, so $x \in A$.

Step 4. Since $A \subset B$ and $B \subset A$, we get $A = B$.

Check­ing the answer#

Try $A = \{1\}$, $B = \{1, 2\}$: union $\{1, 2\}$, inter­sec­tion $\{1\}$, not equal — con­sis­tent, because $A \ne B$.

Answer#

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

Ques­tion 8: Count­ing with the union for­mula#

The prob­lem#

$n(A) = 25$, $n(B) = 18$, $n(A \cup B) = 36$. Find $n(A \cap B)$, $n(A - B)$.

Under­stand­ing the prob­lem#

You know the sizes of $A$, $B$ and the union, and you need the size of the over­lap and of the "$A$ only" part.

The idea#

Rearrange $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ to find the over­lap. Then $A$ splits into $A - B$ and $A \cap B$, so $n(A - B) = n(A) - n(A \cap B)$.

Step-by-step solu­tion#

Step 1. Sub­sti­tute into the for­mula.

$$36 = 25 + 18 - n(A \cap B)$$

Step 2. Solve.

$$n(A \cap B) = 43 - 36 = 7$$

Step 3. $A$ only.

$$n(A - B) = 25 - 7 = 18$$

Check­ing the answer#

$n(B - A) = 18 - 7 = 11$. The three pieces give $18 + 7 + 11 = 36 = n(A \cup B)$. ✓

Answer#

$n(A \cap B) = 7$ and $n(A - B) = 18$.

Ques­tion 9: Tea and cof­fee#

The prob­lem#

Of $60$ stu­dents, $38$ like tea, $30$ like cof­fee, $6$ like nei­ther. How many like both?

Under­stand­ing the prob­lem#

Let $T$ = likes tea, $C$ = likes cof­fee. $6$ are out­side both cir­cles, so the rest like at least one drink.

The idea#

Find $n(T \cup C)$ from the "nei­ther" fig­ure, then use the union for­mula to find $n(T \cap C)$.

Step-by-step solu­tion#

Step 1. At least one drink.

$$n(T \cup C) = 60 - 6 = 54$$

Step 2. Union for­mula.

$$54 = 38 + 30 - n(T \cap C)$$

Step 3. Solve.

$$n(T \cap C) = 68 - 54 = 14$$

Check­ing the answer#

Tea only $38 - 14 = 24$, cof­fee only $30 - 14 = 16$, both $14$, nei­ther $6$: $24 + 16 + 14 + 6 = 60$. ✓

Answer#

$14$ stu­dents like both tea and cof­fee.

Ques­tion 10: Three vehi­cles#

The prob­lem#

In a sur­vey of $400$ fam­i­lies: $210$ own a scooter, $180$ a car, $120$ a cycle; $80$ own scooter and car, $50$ car and cycle, $60$ scooter and cycle, $30$ all three. How many own none?

Under­stand­ing the prob­lem#

Let $S$, $C$, $Y$ be the fam­i­lies own­ing a scooter, a car and a cycle. The pair fig­ures include the $30$ fam­i­lies who own all three. "None" means out­side all three cir­cles.

The idea#

Use the three-set for­mula from the les­son to find how many own at least one, then sub­tract from $400$.

Step-by-step solu­tion#

Step 1. Write the for­mula.

$$n(S \cup C \cup Y) = n(S) + n(C) + n(Y) - n(S \cap C) - n(C \cap Y) - n(S \cap Y) + n(S \cap C \cap Y)$$

Step 2. Sub­sti­tute.

$$\begin{aligned} n(S \cup C \cup Y) &= 210 + 180 + 120 - 80 - 50 - 60 + 30 \\ &= 510 - 190 + 30 = 350 \end{aligned}$$

Step 3. None.

$$400 - 350 = 50$$

Check­ing the answer#

Fill the Venn dia­gram from the cen­tre out: all three $30$; scooter and car only $50$; car and cycle only $20$; scooter and cycle only $30$; scooter only $210 - 50 - 30 - 30 = 100$; car only $180 - 50 - 20 - 30 = 80$; cycle only $120 - 20 - 30 - 30 = 40$. Sum $= 30 + 50 + 20 + 30 + 100 + 80 + 40 = 350$. ✓

Answer#

$50$ fam­i­lies own none of the three.

Com­mon mis­take to avoid#

For­get­ting to add back $n(S \cap C \cap Y)$. The fam­i­lies own­ing all three are sub­tracted three times in the pair terms, so they must be added once again.