\(\mathcal{E}=\{x:x\text{ is an integer and }1\le x\le 15\}\) \(A=\{\text{prime numbers}\}\) and \(B=\{\text{odd numbers}\}\) (a) List the elements of \(A'\...

The dash in \(A'\) denotes the complement: everything in the universal set that is not in \(A\). Here the universal set is the integers from \(1\) to \(15\).

(a) \(A\) is the set of prime numbers in that range. A prime has exactly two factors, itself and \(1\), so \(1\) is not prime and \(2\) is.

\(A = \{2,\ 3,\ 5,\ 7,\ 11,\ 13\}\) [M1]

\(B\) is the odd numbers, \(\{1,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13,\ 15\}\). The set \(A' \cap B\) means numbers that are odd but not prime, so remove the odd primes \(3\), \(5\), \(7\), \(11\) and \(13\) from \(B\):

\(A' \cap B = \{1,\ 9,\ 15\}\) [A1] cao

Each of these is genuinely composite or a unit: \(9 = 3 \times 3\), \(15 = 3 \times 5\), and \(1\) has only one factor so it is not prime.

(b) The union contains every number that is prime or odd or both, namely \(\{1,\ 2,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13,\ 15\}\). Counting them gives

\(n(A \cup B) = 9\) [B1]

Equivalently \(8 + 6 - 5 = 9\), since \(A\) and \(B\) share the five odd primes. The number \(2\) is the only even element of the union, and forgetting that it is prime is a frequent error.