| Date | May 2018 | Marks available | 4 | Reference code | 18M.3srg.hl.TZ0.3 |
| Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
| Command term | Show | Question number | 3 | Adapted from | N/A |
Question
The relation \(R\) is defined such that \(xRy\) if and only if \(\left| x \right| + \left| y \right| = \left| {x + y} \right|\) for \(x\), \(y\), \(y \in \mathbb{R}\).
Show that \(R\) is reflexive.
Show that \(R\) is symmetric.
Show, by means of an example, that \(R\) is not transitive.
Markscheme
(for \(x \in \mathbb{R}\)), \(\left| x \right| + \left| x \right| = 2\left| x \right|\) A1
and \(\left| x \right| + \left| x \right| = \left| {2x} \right| = 2\left| x \right|\) A1
hence \(xRx\)
so \(R\) is reflexive AG
Note: Award A1 for correct verification of identity for \(x\) > 0; A1 for correct verification for \(x\) ≤ 0.
[2 marks]
if \(xRy \Rightarrow \left| x \right| + \left| y \right| = \left| {x + y} \right|\)
\(\left| x \right| + \left| y \right| = \left| y \right| + \left| x \right|\) A1
\(\left| {x + y} \right| = \left| {y + x} \right|\) A1
hence \(yRx\)
so \(R\) is symmetric AG
[2 marks]
recognising a condition where transitivity does not hold (M1)
(eg, \(x\) > 0, \(y\) = 0 and \(z\) < 0)
for example, 1\(R\)0 and 0\(R\)(−1) A1
however \(\left| 1 \right| + \left| { - 1} \right| \ne \left| {1 + - 1} \right|\) A1
so 1\(R\)(−1) (for example) is not true R1
hence \(R\) is not transitive AG
[4 marks]
