For ${\rho _1}$ and ${\rho _2}$ determine whether or not each is reflexive, symmetric and transitive.

For each of ${\rho _1}$ and ${\rho _2}$ which is an equivalence relation, describe the equivalence classes.

${\rho _1}$

$({x_1},{\text{ }}{y_1}){\rho _1}({x_1},{\text{ }}{y_1}) \Rightarrow 0 = 0\;\;\;$hence reflexive.     R1

$({x_1},{\text{ }}{y_1}){\rho _1}({x_2},{\text{ }}{y_2}) \Rightarrow x_1^2 - x_2^2 = y_1^2 - y_2^2$

$\Rightarrow (x_1^2 - x_2^2) =  - (y_1^2 - y_2^2)$

$\Rightarrow x_2^2 - x_1^2 = y_2^2 - y_1^2 \Rightarrow ({x_2},{\text{ }}{y_2}){\rho _1}({x_1},{\text{ }}{y_1})\;\;\;$hence symmetric     M1A1

$({x_1},{\text{ }}{y_1}){\rho _1}({x_2},{\text{ }}{y_2}) \Rightarrow x_1^2 - x_2^2 = y_1^2 - y_2^2{\text{ - i}}$

$({x_2},{\text{ }}{y_2}){\rho _1}({x_3},{\text{ }}{y_3}) \Rightarrow x_2^2 - x_3^2 = y_2^2 - y_3^2{\text{ - ii}}$     M1

${\text{i}} + {\text{ii}} \Rightarrow x_1^2 - x_3^2 = y_1^2 - y_3^2 \Rightarrow ({x_1},{\text{ }}{y_1}){\rho _1}({x_3},{\text{ }}{y_3})\;\;\;$hence transitive     A1

${\rho _2}$

$({x_1},{\text{ }}{y_1}){\rho _2}({x_1},{\text{ }}{y_1}) \Rightarrow \sqrt {2x_1^2}  \leqslant \sqrt {2y_1^2} \;\;\;$This is not true in the case of (3,1)

hence not reflexive.     R1

$({x_1},{\text{ }}{y_1}){\rho _2}({x_2},{\text{ }}{y_2}) \Rightarrow \sqrt {x_1^2 + x_2^2}  \leqslant \sqrt {y_1^2 + y_2^2}$

$\Rightarrow \sqrt {x_2^2 + x_1^2}  \leqslant \sqrt {y_2^2\_y_1^2}  \Rightarrow ({x_2},{\text{ }}{x_2}){\rho _2}({x_1},{\text{ }}{y_1})\;\;\;$hence symmetric.     A1

it is not transitive.     A1

attempt to find a counterexample     (M1)

for example $(1,{\text{ }}0){\rho _2}(0,{\text{ 1)}}$ and $(0,{\text{ }}1){\rho _2}(1,{\text{ 0)}}$     A1

however, it is not true that $(1,{\text{ }}0){\rho _2}(1,{\text{ 0)}}$     A1

${\rho _1}$ is an equivalence relation     A1

the equivalence classes for ${\rho _1}$ form a family of curves of the form

${y^2} - {x^2} = k$     A1

Most candidates attempted this question with many showing correctly that ${\rho _1}$ is an equivalence relation. Most candidates, however, were unable to find a counterexample to show that ${\rho _2}$ is not transitive although many suspected that was the case. Most candidates were unable to describe the equivalence classes.

Most candidates attempted this question with many showing correctly that ${\rho _1}$ is an equivalence relation. Most candidates, however, were unable to find a counterexample to show that ${\rho _2}$ is not transitive although many suspected that was the case. Most candidates were unable to describe the equivalence classes.