| Date | May 2008 | Marks available | 9 | Reference code | 08M.3srg.hl.TZ1.4 |
| Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ1 |
| Command term | Show that and Describe | Question number | 4 | Adapted from | N/A |
Question
The relation R is defined on ordered pairs by
\[(a,{\text{ }}b)R(c,{\text{ }}d){\text{ if and only if }}ad = bc{\text{ where }}a,{\text{ }}b,{\text{ }}c,{\text{ }}d \in {\mathbb{R}^ + }.\]
(a) Show that R is an equivalence relation.
(b) Describe, geometrically, the equivalence classes.
Markscheme
(a) Reflexive: \((a,{\text{ }}b)R(a,{\text{ }}b)\) because \(ab = ba\) R1
Symmetric: \((a,{\text{ }}b)R(c,{\text{ }}d) \Rightarrow ad = bc \Rightarrow cb = da \Rightarrow (c,{\text{ }}d)R(a,{\text{ }}b)\) M1A1
Transitive: \((a,{\text{ }}b)R(c,{\text{ }}d) \Rightarrow ad = bc\) M1
\((c,{\text{ }}d)R(e,{\text{ }}f) \Rightarrow cf = de\)
Therefore
\(\frac{{ad}}{{de}} = \frac{{bc}}{{cf}}{\text{ so }}af = be\) A1
It follows that \((a,{\text{ }}b)R(e,{\text{ }}f)\) R1
[6 marks]
(b) \((a,{\text{ }}b)R(c,{\text{ }}d) \Rightarrow \frac{a}{b} = \frac{c}{d}\) (M1)
Equivalence classes are therefore points lying, in the first quadrant, on straight lines through the origin. A2
Notes: Accept a correct sketch.
Award A1 if “in the first quadrant” is omitted.
Do not penalise candidates who fail to exclude the origin.
[3 marks]
Total [9 marks]
Examiners report
Part (a) was well answered by many candidates although some misunderstandings of the terminology were seen. Some candidates appeared to believe, incorrectly, that reflexivity was something to do with \((a,{\text{ }}a)R(a,{\text{ }}a)\) and some candidates confuse the terms ‘reflexive’ and ‘symmetric’. Many candidates were unable to describe the equivalence classes geometrically.
