Date | November 2010 | Marks available | 8 | Reference code | 10N.3srg.hl.TZ0.1 |
Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
Command term | Determine and Write down | Question number | 1 | Adapted from | N/A |
Question
Let R be a relation on the set \(\mathbb{Z}\) such that \(aRb \Leftrightarrow ab \geqslant 0\), for a, b \( \in \mathbb{Z}\).
(a) Determine whether R is
(i) reflexive;
(ii) symmetric;
(iii) transitive.
(b) Write down with a reason whether or not R is an equivalence relation.
Markscheme
(a) (i) \({a^2} \geqslant 0\) for all \(a \in \mathbb{Z}\), hence R is reflexive R1
(ii) \(aRb \Rightarrow ab \geqslant 0\) M1
\( \Rightarrow ba \geqslant 0\) R1
\( \Rightarrow bRa\), hence R is symmetric A1
(iii) \(aRb{\text{ and }}bRc \Rightarrow ab \geqslant 0{\text{ and }}bc \geqslant 0,{\text{ is }}aRc?\) M1
no, for example, \( - 3R0\) and \(0R5\), but \( - 3R5\) is not true A1
aRc is not generally true, hence R is not transitive A1
[7 marks]
(b) R does not satisfy all three properties, hence R is not an equivalence relation R1
[1 mark]
Total [8 marks]
Examiners report
Although the properties of an equivalence relation were well known, few candidates provided a counter-example to show that the relation is not transitive. Some candidates interchanged the definitions of the reflexive and symmetric properties.