| Date | May 2009 | Marks available | 14 | Reference code | 09M.3srg.hl.TZ0.3 |
| Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
| Command term | Find, Prove, and Write down | Question number | 3 | Adapted from | N/A |
Question
The relation R is defined on \(\mathbb{Z} \times \mathbb{Z}\) such that \((a,{\text{ }}b)R(c,{\text{ }}d)\) if and only if a − c is divisible by 3 and b − d is divisible by 2.
(a) Prove that R is an equivalence relation.
(b) Find the equivalence class for (2, 1) .
(c) Write down the five remaining equivalence classes.
Markscheme
(a) consider \((x,{\text{ }}y)R(x,{\text{ }}y)\)
since x – x = 0 and y – y = 0 , R is reflexive A1
assume \((x,{\text{ }}y)R(a,{\text{ }}b)\)
\( \Rightarrow x - a = 3M\) and \(y - b = 2N\) M1
\( \Rightarrow a - x = - 3M\) and \(b - y = - 2N\) A1
\( \Rightarrow (a,{\text{ }}b)R(x,{\text{ }}y)\)
hence R is symmetric
assume \((x,{\text{ }}y)R(a,{\text{ }}b)\)
\( \Rightarrow x - a = 3M\) and \(y - b = 2N\)
assume \((a,{\text{ }}b)R(c,{\text{ }}d)\)
\( \Rightarrow a - c = 3P\) and \(b - d = 2Q\) M1
\( \Rightarrow x - c = 3(M + P)\) and \(y - d = 2(N + Q)\) A1
hence \((x,{\text{ }}y)R(c,{\text{ }}d)\) A1
hence R is transitive
therefore R is an equivalence relation AG
[7 marks]
(b) \(\left\{ {(x,{\text{ }}y):x = 3m + 2,{\text{ }}y = 2n + 1,{\text{ }}m,{\text{ }}n \in \mathbb{Z}} \right\}\) A1A1
[2 marks]
(c) \(\{ 3m,{\text{ }}2n\} {\text{ \{ }}3m + 1,{\text{ }}2n\} {\text{ }}\{ 3m + 2,{\text{ }}2n\} \)
\(\{ 3m,{\text{ }}2n + 1\} {\text{ \{ }}3m + 1,{\text{ }}2n + 1\} {\text{ }}m,{\text{ }}n \in \mathbb{Z}\) A1A1A1A1A1
[5 marks]
Total [14 marks]
Examiners report
Stronger candidates had little problem with part (a) of this question, but proving an equivalence relation is still difficult for many. Equivalence classes still cause major problems and few fully correct answers were seen to this question.
