Show that $R$ is reflexive.

Show that $R$ is symmetric.

Show, by means of an example, that $R$ is not transitive.

(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]