Show that $R$ is an equivalence relation.

Given that $n = 2$ and $p = 7$, determine the first four members of each of the four equivalence classes of $R$.

since ${a^n} - {a^n} = 0$,     A1

it follows that $(aRa)$ and $R$ is reflexive     R1

if $aRb$ so that ${b^n} - {a^n} \equiv 0(\bmod p)$     M1

then, ${a^n} - {b^n} \equiv 0(\bmod p)$ so that $bRa$ and $R$ is symmetric     A1

if $aRb$ and $bRc$ so that ${b^n} - {a^n} \equiv 0(\bmod p)$ and ${c^n} - {b^n} \equiv 0(\bmod p)$     M1

adding, (it follows that ${c^n} - {a^n} \equiv 0(\bmod p)$)     M1

so that $aRc$ and $R$ is transitive     A1

Note:     Only accept the correct use of the terms “reflexive, symmetric, transitive”.

[7 marks]

we are now given that $aRb$ if ${b^2} - {a^2} \equiv 0(\bmod 7)$

attempt to find at least one equivalence class     (M1)

the equivalence classes are

$\{ 1,{\text{ }}6,{\text{ }}8,{\text{ }}13,{\text{ }} \ldots \}$    A1

$\{ 2,{\text{ }}5,{\text{ }}9,{\text{ }}12,{\text{ }} \ldots \}$    A1

$\{ 3,{\text{ }}4,{\text{ }}10,{\text{ }}11,{\text{ }} \ldots \}$    A1

$\{ 7,{\text{ }}14,{\text{ }}21,{\text{ }}28,{\text{ }} \ldots \}$    A1

[5 marks]

Most candidates were familiar with the terminology of the required conditions to be satisfied for a relation to be an equivalence relation. The execution of the proofs was variable. It was grating to see such statements as $R$ is symmetric because $aRb = bRa$ or $aRa = {a^n} - {a^n} = 0$, often without mention of $\bmod p$, and such responses were not fully rewarded.

This was not well answered. Few candidates displayed a strategy to find the equivalence classes.