Set Notation Worked ExamplesIGCSE Maths 0580

(a) List the multiples of $3$ that lie in the universal set — stop at $12$.

Now the factors of $12$. Work in pairs: $1 \times 12$, $2 \times 6$, $3 \times 4$.

(b) The intersection contains only the elements in both lists.

(c) First build the union — everything in $A$ or $B$, listing each element once.

The complement is everything in $\xi$ that is not in the union.

Count them, and check against the total: $12 - 7 = 5$. ✓

(a) Read the inequality carefully: $\le$ includes $-3$, but $<$ excludes $4$.

Count the elements — do not forget $0$.

(b) List primes below $20$. A prime has exactly two factors, so $1$ is not prime, and $2$ is the only even prime.

Count them.

(c) The intersection holds elements in both. From $A$, only $2$ and $3$ are prime — negatives and $0$ and $1$ are not.

First pin down $\xi$. Both inequalities are strict, so $20$ and $36$ are not included: $\xi = \{21, 22, \ldots, 35\}$, which has $35 - 21 + 1 = 15$ elements.

(a) Test each odd number from $21$ to $35$ for factors: $21 = 3 \times 7$, $25 = 5 \times 5$, $27 = 3 \times 9$, $33 = 3 \times 11$ and $35 = 5 \times 7$ are not prime.

(b) The only square number in $\xi$ is $25$ ($16$ and $36$ lie outside), so $S = \{25\}$. Since $25$ is not prime, no element is in both $P$ and $S$. The answer is a set with nothing in it, written with the empty-set symbol — not the number $0$.

(c) Work inside the bracket first. The even numbers in $\xi$ are $22, 24, \ldots, 34$ and the multiples of $5$ are $25, 30, 35$. The only number in both is $30$.

The complement is every element of $\xi$ except $30$. The question asks for $n(\ldots)$, the number of elements, so give a count rather than a list.

(a) "In both $A$ and $B$" is an intersection; "not in $C$" is the complement of $C$. Intersect all three conditions.

(b) "In $A$ only" means in $A$, but in neither of the others. Each exclusion becomes a complement.

(c) "In none of them" means outside the union of all three. Write the union, then complement it.

An equivalent form for (c), using De Morgan's law, is the intersection of the three complements. Either is acceptable.

(a) Use the inclusion–exclusion formula. The overlap is subtracted once because it was counted twice.

Substitute the given values.

Rearrange.

(b) "In $A$ but not $B$" is $n(A)$ with the overlap removed.

Similarly for $B$ only, which you need for part (c).

(c) "Exactly one" means the two outer regions added — the overlap is excluded.

Check: the three regions must total the union.

(a) Find the intersection first — elements in both sets.

Now take the complement: everything in $\xi$ except $2$ and $4$.

(b) Find each complement separately. $A'$ is the odd numbers here.

And $B'$ is everything from $6$ upwards.

Take the union — everything in either list, each written once. Note $7$ and $9$ appear in both but are listed once.

(c) The two sets are identical, each with $8$ elements. This is De Morgan's law in action.

(a) The union collects everything from both sets. Write repeated elements once only.

(b) The intersection keeps only what appears in both.

(c) Count the elements of the intersection.

(a) $10$ sits in the part of $B$ that does not overlap $A$, so it is outside circle $A$. The symbol for "is not an element of" is $\notin$.

(b) $B'$ is everything outside circle $B$. Intersecting with $A$ leaves the part of circle $A$ that is not in $B$ — the crescent on the left.

(c) $(A \cup B)'$ is the region outside both circles. Count the numbers there: $4, 7, 8, 11, 13, 14$.

(d) Collect every element inside circle $B$: $B = \{1, 2, 3, 5, 6, 10, 15\}$. Each one divides $30$ exactly, and no other factor of $30$ lies in $\xi$ ($30$ itself is too big). The diagram holds $2 + 3 + 4 + 6 = 15$ numbers, so $\xi = \{1, 2, \ldots, 15\}$.

(e) $A \cup B'$ contains everything in $A$ together with everything outside $B$. The only elements it leaves out are those in $B$ but not in $A$: $1, 2, 5, 10$.

Check by listing: $A \cup B' = \{3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 15\}$, which has $11$ elements. ✓

(a) Every shaded part lies inside $A$ or $B$, and nothing inside $C$ is shaded. So start with $A \cup B$ and then keep only the part outside $C$.

Check against the diagram: the parts of $A \cup B$ left unshaded are exactly those inside $C$. ✓

(b) All three shaded parts are inside $A$. Within $A$, the shaded parts are where $A$ meets $B$ or meets $C$, so intersect $A$ with $B \cup C$.

An equivalent answer builds it from the two overlaps.

(c) Most of the diagram is shaded, so describe what is not shaded: all of circle $A$ and all of circle $C$. The shaded region is everything outside $A \cup C$.

(d) $(A \cup B \cup C)'$ is only the region outside all three circles. It leaves out the shaded part of $B$ that lies outside $A$ and $C$, so it describes too little.

(a) $A \cup B$ covers the regions $7, 4, 9, 3, 2, 5$. Intersecting with $C'$ removes the three regions inside $C$ ($3$, $2$ and $5$).

(b) Work inside the bracket first. $B \cap C'$ is the part of $B$ outside $C$: the regions $4$ and $9$.

Now take the union with all of $A$, which is $7 + 4 + 3 + 2 = 16$. The region $4$ is already in $A$, so only the $9$ is new.

(c) The two sets have different numbers of elements, so they cannot be the same set. This is enough to show Priya is wrong.

Compare region by region. The second set contains the regions $3$ and $2$, which are not in the first set. Both lie inside $A$ and inside $C$ — together they make up $A \cap C$.

(a) $A \cup B \cup C$ is every region inside at least one circle: $1$ to $7$. The complement is what is left in $\xi$ — only the outside.

(b) Work inside the bracket first. $B \cup C$ is every region inside $B$ or $C$: $2, 3, 4, 5, 6, 7$.

Intersecting with $A'$ keeps only those regions that are outside circle $A$, so remove $4$, $6$ and $7$.

(c) $A \cap B$ is the overlap of $A$ and $B$: regions $4$ and $7$. $C'$ is everything outside circle $C$: regions $1, 2, 4, 8$.

A union takes every region in either part, each once. Note that region $7$ is shaded even though it is inside $C$, because it is in $A \cap B$.