(a)     Determine whether or not $*$ is

(i)     closed,

(ii)     commutative,

(iii)     associative.

(b)     (i)     Find the identity element.

(ii)     Find the set of positive integers having an inverse under $*$.

(a)     (i)     It is not closed because

$1 * 1 = 0 \notin {\mathbb{Z}^ + }$ .     R2

(ii)     $a * b = a + b - 2$

$b * a = b + a - 2 = a * b$     M1

It is commutative.     A1

(iii)     It is not associative.     A1

Consider $(1 * 1) * 5$ and $1 * (1 * 5)$ .

The first is undefined because $1 * 1 \notin {\mathbb{Z}^ + }$ .

The second equals 3.     R2

Notes: Award A1R2 for stating that non-closure implies non-associative.

Award A1R1 to candidates who show that $a * (b * c) = (a * b) * c = a + b + c - 4$ and therefore conclude that it is associative, ignoring the non-closure.

[7 marks]

(b)     (i)     The identity e satisfies

$a * e = a + e - 2 = a$     M1

$e = 2\,\,\,\,\,({\text{and }}2 \in {\mathbb{Z}^ + })$     A1

(ii)     $a * {a^{ - 1}} = a + {a^{ - 1}} - 2 = 2$     M1

$a + {a^{ - 1}} = 4$     A1

So the only elements having an inverse are 1, 2 and 3.     A1

Note: Due to commutativity there is no need to check two sidedness of identity and inverse.

[5 marks]

Total [12 marks]

Almost all the candidates thought that the binary operation was associative, not realising that the non-closure prevented this from being the case. In the circumstances, however, partial credit was given to candidates who ‘proved’ associativity. Part (b) was well done by many candidates.