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.