Show that R is an equivalence relation.

Show that the equivalence defining R can be written in the form

\[(a - b)(a + b + 1) \equiv 0(\bmod 5).\]

Hence, or otherwise, determine the equivalence classes.

reflexive: \(a(a + 1) \equiv a(a + 1)(\bmod 5)\), therefore aRa     R1

symmetric: \(aRb \Rightarrow a(a + 1) = b(b + 1) + 5N\)     M1

\( \Rightarrow b(b + 1) = a(a + 1) - 5N \Rightarrow bRa\)     A1

transitive:

EITHER

\(aRb{\text{ and }}bRc \Rightarrow a(a + 1) = b(b + 1) + 5M{\text{ and }}b(b + 1) = c(c + 1) + 5N\)     M1

it follows that \(a(a + 1) = c(c + 1) + 5(M + N) \Rightarrow aRc\)     M1A1

OR

\(aRb{\text{ and }}bRc \Rightarrow a(a + 1) \equiv b(b + 1)(\bmod 5){\text{ and}}\)

\(b(b + 1) \equiv c(c + 1)(\bmod 5)\)     M1

\(a(a + 1) - b(b + 1) \equiv 0(\bmod 5);{\text{ }}b(b + 1) - c(c + 1) \equiv 0(\bmod 5)\)     M1

\(a(a + 1) - c(c + 1) \equiv 0\bmod 5 \Rightarrow a(a + 1) \equiv c(c + 1)\bmod 5 \Rightarrow aRc\)     A1

[6 marks]

the equivalence can be written as

\({a^2} + a - {b^2} - b \equiv 0(\bmod 5)\)     M1

\((a - b)(a + b) + a - b \equiv 0(\bmod 5)\)     M1A1

\((a - b)(a + b + 1) \equiv 0(\bmod 5)\)     AG

[3 marks]

the equivalence classes are

{1, 3, 6, 8, 11}

{2, 7, 12}

{4, 5, 9, 10}     A4

Note: Award A3 for 2 correct classes, A2 for 1 correct class.

[4 marks]

Candidates knew the properties of equivalence relations but did not show sufficient working out in the transitive case. Others did not do the modular arithmetic correctly, still others omitted the \(\bmod(5)\) in part or throughout.

Candidates knew the properties of equivalence relations but did not show sufficient working out in the transitive case. Others did not do the modular arithmetic correctly, still others omitted the \(\bmod (5)\) in part or throughout.

Candidates knew the properties of equivalence relations but did not show sufficient working out in the transitive case. Others did not do the modular arithmetic correctly, still others omitted the \(\bmod(5)\) in part or throughout.