| Date | May 2008 | Marks available | 12 | Reference code | 08M.3srg.hl.TZ1.1 |
| Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ1 |
| Command term | Determine and Find | Question number | 1 | Adapted from | N/A |
Question
The binary operation \( * \) is defined for \(a{\text{, }}b \in {\mathbb{Z}^ + }\) by
\[a * b = a + b - 2.\]
(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 \( * \).
Markscheme
(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]
Examiners report
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.
