| Date | May 2009 | Marks available | 3 | Reference code | 09M.1.hl.TZ0.1 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | State | Question number | 1 | Adapted from | N/A |
Question
The relation \(R\) is defined on the set \(\mathbb{Z}\) by \(aRb\) if and only if \(4a + b = 5n\) , where \(a,b,n \in \mathbb{Z}\).
Show that \(R\) is an equivalence relation.
State the equivalence classes of \(R\) .
Markscheme
\(4a + b = 5n\) for \(a,b,n \in \mathbb{Z}\)
reflexive:
\(4a + a = 5a\) so \(aRa\) , and \(R\) is reflexive A1
symmetric:
\(4a + b = 5n\)
\(4b + a = 5b - b + 5a - 4a\) M1
\( = 5b + 5a - (4a + b)\) A1
\( = 5m\) so \(bRa\) , and \(R\) is symmetric A1
transitive:
\(4a + b = 5n\) M1
\(4b + c = 5k\) M1
\(4a + 5b + c = 5n + 5k\) A1
\(4a + c = 5(n + k - b)\) so \(aRc\) , and \(R\) is transitive A1
therefore \(R\) is an equivalence relation AG
[8 marks]
equivalence classes are
\(\left\{ { \ldots , - 10, - 5,0,5,10,\left. \ldots \right\}} \right.\) (M1)
\(\left\{ { \ldots , - 9, - 4,1,6,11,\left. \ldots \right\}} \right.\)
\(\left\{ { \ldots , - 8, - 3,2,7,12,\left. \ldots \right\}} \right.\)
\(\left\{ { \ldots , - 7, - 2,3,8,13,\left. \ldots \right\}} \right.\)
\(\left\{ { \ldots , - 6, - 1,4,9,14,\left. \ldots \right\}} \right.\)
or \(\left\{ {\left\langle 0 \right\rangle ,\left\langle 1 \right\rangle ,\left\langle 2 \right\rangle ,\left\langle 3 \right\rangle ,\left. {\left\langle 4 \right\rangle } \right\}} \right.\) A2
Note: Award A2 for all classes, A1 for at least 2 correct classes.
[3 marks]
Examiners report
Part (a) was generally well done but not always in the most direct manner.
Too many missed the equivalence classes in part (b).
